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Case Study Contents
Joey, a recent college graduate, has just started his first job and is trying to be smart in managing his finances for the next several years. Given his wages and some basic knowledge about the state of the economy, Joey would like to calculate how much he can spend each year (consume) on rent, groceries, entertainment, and leisure. Joey first considers a plan for the next three years.
Joey recalls studying in college an economic model called the life cycle model. The life cycle theory of consumption was developed in the 1950s by Franco Modigliani and his student, Richard Brumberg ([1]). The main premise of the theory is that people make consumption decisions with the goal of maintaining a stable level of income over their lifetime. The implication is that individuals save for retirement by building up income during their working years to spend during their retirement.
To solve his own
three-period life cycle problem, Joey needs the following information: period: length of time in each segment. In the example, Joey's period is one year but it could be one day, one month, or any other specified economic period. wages: income received in each period. The model assume that income in each period is known and does not change within the period. utility function: utility refers to the perceived value of a good. A utility function assigns a number to indicate the value associated with a good or service. For example, if Joey consumes 3 slices of pizza, he perceives 5 units of happiness. However, he likely experiences less utility (happiness) with each additional slice. A common utility function is \(log(x)\), in which the value increases rapidly from 0 but levels off as \(x\) increases. interest rate: rate at which interest is paid by borrowers for the use of money that they borrow from a lender. In this model, the interest rate is expressed as a number between 0 and 1. discount factor: measure of how people value time. The discount factor is a measure of how much less something is worth if it is received in the future. In this model, the discount rate is expressed as a number between 0 and 1.
Then, the objective of the
three-period life cycle problem is to determine how much Joey can consume in each period so as to maximize his utility subject to the lifetime budget constraint. The lifetime budget constraint is the key assumption in the life cycle model; the assumption is that consumption over the life cycle depends entirely on the present value of the lifetime income.
We present a mathematical formulation of the three-period life cycle model.
Set P = set of periods = {1, 2, 3} Parameters \(w_p\) = wage income in period \(p\), \(\forall p \in P\) \(r\) = interest rate \(\beta\) = discount factor Decision Variables \(c_p\) = consumption in period \(p\), \(\forall p \in P\) Objective Function Let \(u()\) be the utility function and let \(u(c_p)\) be the utility value associated with consuming \(c_p\). Utility in future periods is discounted by a factor of \(\beta\). Then, the objective function is to maximize the total discounted utility:
maximize \(u(c_1) + \beta u(c_2) + \beta^{2} u(c_3)\)
Constraints The main constraint in the life cycle model is the lifetime budget constraint, which asserts that, over the life cycle, the present value of consumption equals the present value of wage income. From above, \(r\) is the interest rate; therefore, \(R = 1 + r\) is the gross interest rate. If I invest one dollar in this period, then I receive \(R\) dollars in the next period. The expression for the present value of the consumption stream over the life cycle is
\[c_1 + \frac{c_2}{R} + \frac{c_3}{R^{2}}.\]
Similarly, the expression for the present value of the wage income stream over the life cycle is
\[w_1 + \frac{w_2}{R} + \frac{w_3}{R^{2}}.\]
The lifetime budget constraint states that the present value of the two streams must be equal:
\[c_1 + \frac{c_2}{R} + \frac{c_3}{R^{2}} = w_1 + \frac{w_2}{R} + \frac{w_3}{R^{2}}.\]
To avoid numerical difficulties, we add constraints requiring the consumption variables to take a non-negative value:
\(c_1 \geq 0.0001, c_2 \geq 0.0001, c_3 \geq 0.0001\)
Providing an initial starting point is helpful for some solvers; as an example, one possible starting point for this example is
\(c_1 \approx 1, c_2 \approx 1, c_3 \approx 1\)
To solve the three-period life cycle consumption problem, we need to specify a utility function and the values of the parameters. The solution specifies the amount that Joey should consume in each period to maximize his utility. Note that, in the next case study, the Life Cycle Consumption Problem, we generalize the model from three periods to \(n\) periods.
To solve your own three-period life cycle problems, check out the Three-Period Life Cycle Problem demo.
$Title Three-period Life Cycle problem Set p period /1*3/ ; Scalar B discount factor /0.9/; Scalar i interest rate /0.2/ ; Scalar R gross interest rate ; R = 1+i ; $macro u(c) -exp(-c) Parameter w(p) wage income in period p / 1 2 2 3 3 0 / ; Parameter lbnds(p) lower bounds of consumption / 1*3 0.0001 / ; Parameter initvals(p) initial value guess of consumption / 1*3 1 / ; Positive Variables c(p) consumption expenditure in period p , PVc present value of consumption expenditures , PVw present value of wage income ; Variable Z objective ; Equations defPVc definition of PVc , defPVw definition of PVw , budget lifetime budget constraint , obj objective function ; defPVc .. PVc =e= sum(p, c(p) / power(R, p.val - 1)) ; defPVw .. PVw =e= sum(p, w(p) / power(R, p.val - 1)) ; budget .. PVc =e= PVw ; obj .. Z =e= sum(p, u(c(p))*power(B, p.val - 1)) ; model NumericalExample /defPVc, defPVw, budget, obj/ ; c.lo(p) = lbnds(p) ; c.l(p) = initvals(p) ; solve NumericalExample using nlp maximizing Z ; Modigliani, F. and R. Brumberg. 1954. "Utility Analysis and the Consumption Function." In K. Kurihara, ed., Post-Keynesian Economics. New Brunswick: Rutgers University Press. |
Could you explain to me, what is the meaning of "
Effective population size ($N_e$)"?
I would appreciate an example as well.
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Could you explain to me, what is the meaning of "
Effective population size ($N_e$)"?
I would appreciate an example as well.
I'll add an informal answer to complement @Remi.b's excellent answer. In a very simple sense, you can think of the effective population size as the number of reproducing (breeding) individuals in a population. Nature Education has a very good (and free) Scitable article on Genetic Drift and Effective Population Size. The article makes four points, which I've annotated below.
Populations of many species contain individuals that have not yet reached sexual maturity, have passed an age of reproductive capability, or might have a genetic condition that prevents reproduction. All of these individuals count towards the census population size (actual number of individuals present) but not count towards effective population size because they can't reproduce.
All individuals in the population are equally likely to reproduce.
Mating is random.
Not all indivduals (assuming individuals that meet assumption #1 above) are equally likely to reproduce. For example, the process of sexual selection means that some individuals are much more likely to reproduce than other individuals. Sexual selection is not random, and lowers effective population size.
Population sizes for some species can fluctuate over many generations. For example, freshwater mussels have declined rapidly in population size due to habitat loss and other factors. Nearly 72% of the species in the United State and Canada are endangered, threatened or of conservation concern (Williams et al. 1993). The population sizes for the majority of mussels species are but a fraction of what they once were. This is the case for many endangered species. This too lowers effective population size.
Here's a simple mathematical treatment to highlight points 1-3. Consider a population of zebras. Zebras like the Plains Zebra (
Equus quagga) form harems where one male mates with several females. He prevents other males from mating with the females in his harem. Let's make the following assumptions: 1) the population has 250 males and 250 females, all capable of breeding and 2) each male that is able to form a harem gets five females. If so, then only 50 males will actually get to reproduce. The census size is 500 individuals but the effective population size, as shown in Remi.b's answer, is calculate by
$$N_e = \frac{4N_m N_f}{N_m+N_F}$$
where, $N_e$ is the effective population size, $N_m$ is the number of breeding males and $N_f$ is the number of breeding females. Given assumptions 1 and 2 above, $N_m$ is 50 and $N_f$ is 250 (50 males $\times$ 5 females each). Therefore,
$$N_e = \frac{4\times 50 \times 250}{50 + 250} = 166.67$$
Clearly, the effective population size is much smaller than the census population size.
Short answer: The effective population size of a population is the corresponding population size of a idealized (fisherian) population that would function in the same way with respect to genetic drift and inbreeding as the focal population under interest.
Long answer, below.
Definitions: Heterogenity: the probability that two randomly sampled alleles in the population are identical (identical by descent). Census Population Size: The census population size ($N$) is the number of individuals in the population. Fisherian population
A fisherian population is an idealized population of size $N$, where all individuals mate at random, there is no selection, there is no population structure, there is no mutation and generations are non-overlapping. There probably are no perfectly fisherian populations in nature!
Genetic drift causes a decrease in heterogeneity. As the genetic drift is a function of the population size (see this post) this decrease of heterogeneity is a function of the population size. The heterogeneity at time $t$ is:
$$ H_t = H_{0}\left( 1-\frac{1}{2N} \right)^t$$
, where $H_0$ is the heterogeneity at $t=0$. You can verify by yourself that as $t$ increase , $H_t$ decreases. Also, as $N$ decreases, $H_t$ decreases. In other words, the longer you observe, the more heterogeneity gets lost and the smaller is the population, the higher is the loss of heterogeneity.
Effective Population Size
Let
Pop_A be a real population. The effective population size ($Ne$) of
Pop_A is the size of an idealized fisherian population that would have the same loss of heterogeneity than
Pop_A.
You can read the above sentence twice to make sure you understand it. It basically is the definition of the concept of effective population size.
In other words: Let $N$ be the population size of
Pop_A. If
Pop_A is a perfect fisherian population, then its loss of heterozygosity through time is given by:
$$ H_t = H_{0}\left( 1-\frac{1}{2N} \right)^t$$
If
Pop_A is not a perfect fisherian population, then it's decrease in heterozygosity is greater than the one given by the above formula. The decrease in heterozygosity of
Pop_A corresponds to the decrease of heterozygosity of an idealize fisherian population of a different size. This size is called the effective population size $Ne$. So, we can rewrite the loss of heterogosity through time of
Pop_A as:
$$ H_t = H_{0}\left( 1-\frac{1}{2Ne} \right)^t$$
(Note $N$ has been replaced by $Ne$)
, where $Ne$ is the effective population size of
Pop_A.
Usually $Ne < N$, meaning that a real population undergoes more genetic drift than does a fisherian population of the same census size. There are many reasons that may yield a real population to have a higher drift than it would be expected with the same census size but respecting the assumptions of the fisherian population. Below I discuss three reasons:
Examples Unequal sex-ratio
Think as a first example of a population where the sex-ratio is extremely biased to the point of having one male for 99 females. In such population, half of the alleles of a given locus at the next generation will come directly from the father. In such situations, the genetic drift is much higher than what would be expected from a fisherian population of size 100. Therefore $Ne<N=100$. To calculate $Ne$ when the sex-ratio in unequal you can use this formula:
$$Ne=\frac{4N_mN_f}{N_m+N_f}$$
, where $N_m$ and $N_f$ are the census populatino size of males and females respectively. You can verify that when $N_f = N_m$, then $N_f + N_m = N = Ne$.
Variation in population size through time
If your population size varies throughout a given time (a year for example) and that at time $t=3$ (a given period of the year) the census population size is given by $N_i$, then the effective population size is:
$$Ne = \frac{1}{\frac{1}{t}\sum_{i=1}^t\frac{1}{N_i}}$$
This is particularly relevant for insects for example which population size varies a lot between seasons.
Selection
Think now of a population where there is a great variance in fitness (there is selection). In such population, maybe all offspring will be produced by only a tenth of the current population. As a result the effective population is ten times smaller than the census population size.
You can probably find all these informations from wikipedia: http://en.wikipedia.org/wiki/Effective_population_size
Let me know if the concept of the effective population makes more sense to you now. |
First, note that if $p$ and $q$ are any two statements, the implication $p\implies q$ is logically equivalent to $(\lnot q)\implies(\lnot p)$. This equivalence is known as "taking the contrapositive".
Because of this, your statements $(I)$ and $(II)$ are equivalent. That is, the first two properties you mention are equivalent (and they follow from $L \leq_m L'$). The second two properties are also equivalent. Your question then becomes:
$(I)$ If $L \leq_m L'$ and $L \in RE$, can we conclude $L' \not\in RE$ ?
In general, no, we can not conclude that.
For instance, take $L=\{0\}$. This is a decidable language (we can test the input string against $0$, and accept iff they are equal), so it is also a $RE$ language.
Now, note that we can many-one reduce $L$ to absolutely any set $L'$ except the empty set and the full set ($\Sigma^*$). This includes non-$RE$ sets like the complement of the halting problem.
Indeed, to find a reduction, take any word $w\in L'$ and any other word $x\notin L'$. Then, the reduction function behaves as follows: it tests its input, if it is equal to $0$, then it returns $w$, otherwise it return $x$.
By construction, this is a reduction, so $L \leq_m L'$, even if $L\in RE$ and $L'\notin RE$. |
In Computational Intractability, we often come across a need to reduce Vertex Cover (VC) problem to a Subset Sum problem, mostly to prove Subset Sum is NP-Complete. I also see a reduction in the line of -
. However, not able to find out (understand) exactly what is this t. I think providing a concrete example will help many like me. Request any one to provide a concrete graph with |VC| 3 or something manageable, and then show exact value of t i.e. what would be the exact instance of subset sum problem. Finally, explaining/clearly mentioning elements of S would be of great help. A graph G has VC of size k if and only if there is a subset S that sums exactly to t
In Computational Intractability, we often come across a need to reduce Vertex Cover (VC) problem to a Subset Sum problem, mostly to prove Subset Sum is NP-Complete. I also see a reduction in the line of -
In Computational Intractability, we often come across a need to reduce Vertex Cover (VC) problem to a Subset Sum problem...
We do?
... mostly to prove Subset Sum is NP-Complete.
There's no particular reason to go down that route. Karp [1] defined the Knapsack problem as: given $a_1, \dots, a_r, b\in\mathbb{Z}$, is there a set $S\subseteq \{1, \dots, r\}$ such that $\sum_{i\in S}a_i=b$? This is one variant of what is now called Subset Sum. If you prefer to define Subset Sum such that $b$ is always zero, we'll come back to that in a minute.
Karp shows that Subset Sum is
NP-complete by the chain of reductions$$\text{SAT}\leq \text{3SAT} \leq \text{Chromatic Number}\leq\text{Exact Cover}\leq\text{Subset Sum}\,.$$In particular, the reduction from Exact Cover to Subset Sum produces an instance where the $a_i$'s and $b$ are all positive so, if you want to define Subset Sum as "is there a subset whose total is zero?", you can set $a_{r+1}=-b$.
Since Subset Sum and Vertex Cover are both
NP-complete, there is clearly a reduction between them. However, you shouldn't expect that there's a "nice" reduction where a small VC instance naturally transforms into a small Subset Sum instance that makes you say, "Aha, now I understand." And that applies to most pairs of NP-complete problems. In complexity theory courses, we teach the simple, intuitive reductions, usually between problems that are somehow similar, or between SAT or 3SAT and a problem that isn't about Boolean formulas. That can give the impression that there's a natural reduction between any pair of NP-complete problems; in reality, there usually isn't.
[1] Richard M. Karp,
Reducibility among combinatorial problems. In Complexity of Computer Computations, Plenum Press, 1972. (PDF) |
I can't find some of the values I need for this calculation, which I'm trying to do prior to an experiment around superfluidity. Specifically I am trying to get a rough estimate of the first harmonic for second sound in superfluid $^4$He, in a tube 10cm long, at 1.6K. Closed at both ends. Subscript n relates to the normal fluid fraction, s to the superfluid component. The velocity of second sound is given by
$$c = \sqrt{\frac{TS^2\rho_s}{C\rho_n}}$$
Then since $c = f \lambda$ and $\lambda = \frac{L}{2}$,
$$f = \frac{2}{L} \sqrt{\frac{TS^2\rho_s}{C\rho_n}}$$ L, T I know, I can remove $\rho_s$ using $\rho_s v_s + \rho_n v_n = 0$ so that the equation becomes
$$f = \frac{2}{L} \sqrt{-\frac{TS^2v_n}{Cv_s}}$$ But $\frac{-v_n}{v_s}$ will be positive overall since either $v_n$ or $v_s$ will be defined as negative, the superfluid and normal components move in opposite directions I think. So I just need to find the entropy S of Helium 4 at 1.6K and its heat capacity, since I know the velocities of first and second sound at 1.6K, but I can't find those values anywhere. Is there another way to calculate this, or could someone direct me to some tables of constants? I have tried Kaye and Laby, and the internet in general, with no luck.
Most likely my approach is incorrect, although I haven't been able to find an alternative. |
It some sense, it is unfortunate that the first two problems that many students encounter are the free particle and the infinite square well, because both problems are quite subtle from a technical standpoint.
Let's take a look at your questions first:
then they claim that k is a real non-negative number (why? I have no idea)
There's no reason $k$ should be non-negative - DanielSank mentions this thoroughly in his answer. It must, however, be real, and this forces the
energy to be non-negative.
On the other hand, when the text go on to solve infinite potential well, they at first glance do something completely different, despite the problem, within the well, is the very same
The problems are
not the same - the difference in domain radically alters the nature of the problem. In the case of the infinite square well, the Hamiltonian has a discrete spectrum, and therefore a complete set of eigenvalue/eigenstate pairs, from which one can construct any arbitrary function in the Hilbert space.
On the other hand, in the case of the free particle, the Hamiltonian has a continuous spectrum. This means that it does not have
any eigenstates in the Hilbert space, so the above recipe fails spectacularly. It does, on the other hand, yield a set of unphysical quasi-states, which do not live in the Hilbert space, but from which we can construct the physical solutions which do. From there, we can look at the time-evolution of the physical states by examining the time-evolution of the unphysical ones from which it is built.
I will skim over each problem, and give you some context for why one might take slightly different approaches to them. I will be somewhat technical, but hopefully not overly so.
When you write down a problem in quantum mechanics, you need to define a Hamiltonian operator and a Hilbert space of states on which it may act. If the system corresponds to some subset of physical space (say, a particle on a line) then a good guess for the underlying Hilbert space is $L^2(\mathbb R)$, which essentially consists of all functions $f:\mathbb R \rightarrow \mathbb C$ such that
$$\int_{\mathbb R} |f(x)|^2 dx < \infty$$
equipped with the inner product
$$ \langle f,g\rangle = \int_\mathbb R \bar f(x)\cdot g(x) \ dx$$
This is a good choice because we interpret $\int_a^b |f(x)|^2 dx$ as being proportional to the probability that the particle is measured to have a position in the interval $[a,b]$ (with equality holding if $f$ is normalized). We should at the very least demand that a state be normalizable - otherwise this interpretation falls apart.
Once we have decided on the Hilbert space, we need a Hamiltonian operator $\hat H$ which is self-adjoint on our Hilbert space. Notice that $\hat H$ needs to be
self-adjoint, not merely Hermitian. Those words are often taken to be synonyms, but they are not. The distinction has to do with the domains on which unbounded operators such as $\hat X,\hat P,$ and $\hat H$ are allowed to act; I won't actually carry out any such calculations, but I'll mention when operators are Hermitian but not self-adjoint.
Anyway, once we decide on a Hamiltonian, we typically seek to find its eigenstates, from which we can build our general solution. This is not always possible (e.g. in the case of a free particle), but it's a reasonable starting point. Let's take a look at the actual problems in your question.
Free Particle
A free particle is just a particle on a line, so our Hilbert space should be $L^2(\mathbb R)$. The momentum operator $\hat P$ operates on a state like this:
$$\hat P \psi := -i\hbar \psi'$$
and the Hamiltonian operator is simply
$$\hat H\psi := \frac{1}{2m}(\hat P \circ \hat P) \psi = -\frac{\hbar^2}{2m} \psi '' $$
We seek eigenfunctions of the Hamiltonian operator - that is, we seek functions $f\in L^2(\mathbb R)$ such that
$$ -\frac{\hbar^2}{2m} f''= E f $$$$\implies f'' = -\frac{2mE}{\hbar^2} f$$
for some eigenvalue $E$. There are three non-trivial subtleties here:
An arbitrary element $f\in L^2(\mathbb R)$ is not even continuous, much less twice differentiable, and Even if $f$ is twice differentiable, there is no guarantee that $f''\in L^2(\mathbb R)$, and The two linearly independent candidates, namely the complex exponentials $e^{\pm ikx}$, are not even in $L^2(\mathbb R)$ themselves
The first two issues can be swept under the rug by only considering functions $f\in L^2(\mathbb R)$ which are twice differentiable, and whose second derivatives are also in $L^2(\mathbb R)$ (the set of such functions is called the Hilbert-Sobolev space $\mathcal H^2(\mathbb R)$).
The third issue is more problematic. The typical way to handle it is to treat the complex exponentials as "quasi-states" which are unphysical, but still satisfy the above differential equation. If $f=\exp(ikx)$, then
$$f'' = -k^2 f = -\frac{2mE}{\hbar^2}f$$
We then ask whether we could build a state in $L^2(\mathbb R)$ out of such states. We can, and the result is a Fourier transform:
$$ f = \int_{\mathbb R} g(k) e^{-ikx} dk $$
From Parseval's theorem,
$$\int_{\mathbb{R}} |f|^2 \ dx = \int_{\mathbb R} |g|^2 \ dk$$
So as long as $\int_{\mathbb R}|g|^2 \ dk < \infty$ then we are guaranteed that the resulting $f\in L^2(\mathbb R)$, and therefore corresponds to a physical state.
The time dependent state is then
$$ f(x,t) = U(t,0) f(x,0) = e^{-i\hat H t} \int_{\mathbb R} g(k) e^{-ikx} dk$$$$ = \int_{\mathbb R} g(k) e^{-ikx} e^{-i \frac{\hbar^2 k^2}{2m} t} \ dk$$
Notes:
The plane waves $e^{-ikx}$ are not solutions to the time-independent Schrodinger equation, because the TISE is an eigenvalue equation on the Hilbert space $L^2(\mathbb R)$, and the complex exponentials do not belong to that space. Even so, we can build elements of $L^2(\mathbb R)$ from these unphysical quasi-states via superposition, which takes the form of a Fourier transform over momentum space; the resulting physical states evolve just like a superposition of the unphysical ones. The TISE does not have any eigenfunctions or eigenvalues in $L^2(\mathbb R)$, which means there is no such thing as physical states with definite energy for this system.
The last point is crucial and often slightly overlooked. We often claim that self-adjoint operators have complete sets of eigenstates, but this is generally not true if the spectrum of the operator is continuous, as it is here.
As an aside - it's easy to show that $\hat H$ is Hermitian, but somewhat tricky to show that it self-adjoint, which is a stronger requirement. The same is true of $\hat P$ - it is self-adjoint (and therefore Hermitian), but it has a continuous spectrum, and so it has no
physical eigenstates (though as before, we can treat the complex exponentials as quasi-states of definite momentum).
Infinite Square Well
In this problem, the particle is artificially restricted to lie in the interval $I=[0,L]$. The Hilbert space $L^2(\mathbb R)$ is no longer an acceptable choice, so we make the modification that the Hilbert space underlying our system is now $L^2(I)$.
We actually need to make another restriction - we demand that physical states $\psi$ obey the boundary conditions $\psi(0)=\psi(L)=0$. Therefore, our true Hilbert space $h$ (I'm sorry, we're running out of h's) is given by
$$h :=\{\psi \in L^2(I) \big| \psi(0)=\psi(L)=0\}$$
As before, we take the Hamiltonian operator to be
$$ \hat H \psi := -\frac{\hbar^2}{2m} \psi ''$$
which we permit to act on functions which live in the domain $\mathcal D_H$, where
$$\mathcal D_H := \{\psi \in \mathcal H^2(I) \big| \psi(0)=\psi(L)=0\}$$
Unlike the previous case, the TISE actually yields solutions which belong to our Hilbert space, and which take the form
$$\psi_n = c_n \sin\big(\frac{n\pi x}{L}\big)$$
with eigenvalues$$E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}$$
Notes:
As before, the Hamiltonian operator is self-adjoint on this Hilbert space. However, since the spectrum of $\hat H$ is purely discrete, it does have a complete spanning set of eigenstates, as well as a notion of physical states of definite energy. One might imagine that the same is true of $\hat P$. However, it can be shown that $\hat P$ is not self-adjoint on this Hilbert space. Therefore, not only is there no such thing as a state of definite momentum, there isn't even a well-defined notion of what momentum is.
Note that $\hat P$ as defined above actually takes elements of $h$
out of the Hilbert space. This is due to our extremely restrictive boundary condition that $\psi(0)=\psi(L)=0$. If we loosen it to periodic boundary conditions, so that $\psi(0)=\psi(L)$ but they are not necessarily zero, then it turns out that we can create a self-adjoint momentum operator. This corresponds to a particle on a ring, and is an important problem to look at later. |
Problem: I was looking for a function which is $ \epsilon - \delta - continuous $ but is not $ s-continuous $ at some point.
Here are the definitions :
$ s-continuous $ : An internal function $f \subset^* \mathbb{R} \times \mathbb{R}$ is called $ s-continuous $ at a point $x \in \mathbb{R}$,, then its *-extension $^*f$ is $s-continuous$ at the point $x$.
$ \epsilon - \delta-continuous $ : An internal function $f \subset^* \mathbb{R} \times ^*\mathbb{R}$ is called $ \epsilon - \delta-continuous $ at a point $x_0 \in ^* \mathbb{R}$ if for any $\epsilon \in ^* \mathbb{R}^+$, then there exists a $ \delta \in ^* \mathbb{R}^+ $ , such that $ (|x-x_0|< \delta) \implies (|f(x) - f(x_0)| < \epsilon)$.
I am thinking about the function $ f(x) = x^2$ which is a $\delta - \epsilon - continuous $ but not sure about $ s-continuity $. Hoping for your help. |
Positive solutions for Kirchhoff-Schrödinger-Poisson systems with general nonlinearity
1.
School of Mathematics and Statistics, South-Central University for Nationalities, Wuhan, 430074, China
2.
School of Mathematics and Statistics, Hubei Engineering University, Xiaogan, 432000, China
$\left\{ \begin{gathered} - \left( {a + b\int_{{\mathbb{R}^3}} {{{\left| {\nabla u} \right|}^2}{\text{d}}x} } \right)\Delta u + \mu \phi \left( x \right)u =f\left( u \right)\;\;\;&{\text{in}}\;\;{{\mathbb{R}}^3}, \hfill \\ - \Delta \phi =\mu {u^2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;&{\text{in}}\;\;{{\mathbb{R}}^3}, \hfill \\ \end{gathered} \right.$
$a>0,b≥q0 $
$μ>0 $
$f∈ C(\mathbb{R},\mathbb{R}) $
$f $
$μ $ Keywords:Kirchhoff-Schrödinger-Poisson system, Berestycki-Lions type nonlinearity, nonhomogeneous, multiple solutions, asymptotic behavior. Mathematics Subject Classification:Primary:35J50;35A01;Secondary:35B40. Citation:Dengfeng Lü. Positive solutions for Kirchhoff-Schrödinger-Poisson systems with general nonlinearity. Communications on Pure & Applied Analysis, 2018, 17 (2) : 605-626. doi: 10.3934/cpaa.2018033
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Existence and concentration behavior of positive solutionsfor a Kirchhoff equation in $\mathbb{R}^{3} $,
[18]
Y. He and G. Li,
Standing waves for a class of Kirchhoff type problemsin $\mathbb{R}^{3} $ involving critical Sobolev exponents,
[19]
J. Hirata, N. Ikoma and K. Tanaka,
Nonlinear scalar field equations in $R^{N}$: mountain pass and symmetric mountain pass approaches,
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L. Jeanjean,
On the existence of bounded Palais-Smale sequences and application to a Landesman-Lazer-type problem set on $ R^{N}$,
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Multiple solutions for a nonhomogeneous Schrödinger-Maxwell system in $\mathbb{R}^{3} $,
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Y. Li, F. Li and J. Shi,
Existence of a positive solution to Kirchhoff type problems without compactness conditions,
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G. Li and H. Ye,
Existence of positive ground state solutions for the nonlinear Kirchhoff type equations in $\mathbb{R}^{3} $,
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J. Wang, L. Tian, J. Xu and F. Zhang,
Multiplicity and concentration of positive solutions for a Kirchhoff type problem with critical growth,
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J. Zhang, J. Marcos do Ó and M. Squassina, Schrödinger-Poisson systems with a general critical nonlinearity
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G. Zhao, X. Zhu and Y. Li,
Existence of infinitely many solutions to a class of Kirchhoff-Schrödinger-Poisson system,
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Existence and concentration result for the Kirchhoff type equations with general nonlinearities,
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X. He and W. Zou,
Existence and concentration behavior of positive solutionsfor a Kirchhoff equation in $\mathbb{R}^{3} $,
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Y. He and G. Li,
Standing waves for a class of Kirchhoff type problemsin $\mathbb{R}^{3} $ involving critical Sobolev exponents,
[19]
J. Hirata, N. Ikoma and K. Tanaka,
Nonlinear scalar field equations in $R^{N}$: mountain pass and symmetric mountain pass approaches,
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L. Jeanjean,
On the existence of bounded Palais-Smale sequences and application to a Landesman-Lazer-type problem set on $ R^{N}$,
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Y. Jiang, Z. Wang and H.-S. Zhou,
Multiple solutions for a nonhomogeneous Schrödinger-Maxwell system in $\mathbb{R}^{3} $,
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Y. Li, F. Li and J. Shi,
Existence of a positive solution to Kirchhoff type problems without compactness conditions,
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G. Li and H. Ye,
Existence of positive ground state solutions for the nonlinear Kirchhoff type equations in $\mathbb{R}^{3} $,
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J. Wang, L. Tian, J. Xu and F. Zhang,
Multiplicity and concentration of positive solutions for a Kirchhoff type problem with critical growth,
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J. Zhang, J. Marcos do Ó and M. Squassina, Schrödinger-Poisson systems with a general critical nonlinearity
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G. Zhao, X. Zhu and Y. Li,
Existence of infinitely many solutions to a class of Kirchhoff-Schrödinger-Poisson system,
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Sitong Chen, Junping Shi, Xianhua Tang.
Ground state solutions of Nehari-Pohozaev type for the planar Schrödinger-Poisson system with general nonlinearity.
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Existence of ground state solutions for the planar axially symmetric Schrödinger-Poisson system.
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Non-autonomous Schrödinger-Poisson system in $\mathbb{R}^{3}$.
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Ground state solutions for the fractional Schrödinger-Poisson systems involving critical growth in $ \mathbb{R} ^{3} $.
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Suppose G is a Grassmannian variety. What are cohomology groups $H^i(G,T_G)$ $(i\geq 0)$ of the tangent bundle ? thanks.
This answer is in characteristic zero so that I can use Borel-Bott-Weil; I'm not sure if it's still right in finite characteristic. As Serge says, $H^0(G(k,V), T) = \mathrm{End}(V)/\langle \mathrm{Id} \rangle$. All the other cohomology groups are zero.
Proof sketch: Let $Fl(V)$ be the variety of complete flags in $V$; we'll write such a flag as $F_1 \subset F_2 \subset \cdots \subset F_{n-1} \subset V$. Let $L$ be the line bundle $F_1^{\vee}$ over $Fl(V)$ and let $L'$ be the line bundle $V/F_{n-1}$. (I write $E^{\vee}$ for the dual of the vector bundle $E$.)
Let $\pi$ be the projection map $Fl(V) \to G(k,n)$ taking $(F_{\bullet})$ to $F_k$. Let $S$ be the tautological subbundle on $G(k,n)$ and $Q$ the tautological quotient bundle.
Let $[W]$ be a point of $G(k,V)$, with corresponding $k$-dimensional subspace $W$. The fiber $\pi^{-1}([W])$ is $Fl(W) \times Fl(V/W)$. The sections of $L$ and $L'$ on this fiber are naturally $W^{\vee}$ and $V/W$ respectively. The sections of $L \otimes L'$ are $W^{\vee} \otimes (V/W) = \mathrm{Hom}(W,V/W)$. The line bundles $L$ and $L'$ on this fiber have no higher cohomology.
I claim (but have not checked carefully) that the above paragraph works in families, so $\pi_{\ast}(L \otimes L') \cong \mathrm{Hom}(S, Q)$ and $R^i \pi_{\ast}(L \otimes L') =0$. As is well known, $T_G \cong \mathrm{Hom}(S, Q)$.
So the Serre spectral sequence for $Fl(V) \to G(k,V) \to \mathrm{pt}$ collapses and $H^i(G(k,V), T_G) \cong H^i(Fl(V), L \otimes L')$.
By Borel-Bott-Weil (in characteristic zero), $H^0(Fl(V), L \otimes L')$ is the $GL_V$ irrep indexed by $(1,0,0,\ldots,-1)$. This is an $n^2-1$ dimensional vector space, with the explicit description given by Serge. Also by Borel-Bott-Weil, $H^i(Fl(V), L \otimes L')$ vanishes.
$H^1(G,T_G)$=0`. If $G$ is the Grassmannian of $k$-dimensional linear subspaces in a linear space $E$, then $H^0(G,T_G)$ is isomorphic to the quotient $\mathrm{End}(E)$ modulo multiples of identity. Don't know about higher cohomology. |
Assemble a formula using the numbers $2$, $0$, $1$, and $8$ in any order that equals 109. You may use the operations $x + y$, $x - y$, $x \times y$, $x \div y$, $x!$, $\sqrt{x}$, $\sqrt[\leftroot{-2}\uproot{2}x]{y}$ and $x^y$, as long as all operands are either $2$, $0$, $1$, or $8$. Operands may of course also be derived from calculations e.g. $10+(\sqrt{8*2})!$. You may also use brackets to clarify order of operations, and you may concatenate two or more of the four digits you start with (such as $2$ and $8$ to make the number $28$) if you wish. You may only use each of the starting digits once and you must use all four of them. I'm afraid that concatenation of numbers from calculations is not permitted, but answers with concatenations which get $109$ will get plus one from me.
Double, triple, etc. factorials (n-druple-factorials), such as $4!! = 4 \times 2$ are not allowed, but factorials of factorials are fine, such as $(4!)! = 24!$. I will upvote answers with double, triple and n-druple-factorials which get 109, but will not mark them as correct.
Here are some examples to this problem:
Use 2 0 1 and 8 to make 67 Make numbers 93 using the digits 2, 0, 1, 8 Make numbers 1 - 30 using the digits 2, 0, 1, 8
many thanks to the authors of these questions for inspiring this question. |
I was recently playing around with delta-epsilon proofs for limits, which, when true, have the following form:
$$\forall \epsilon > 0, \exists \delta>0 , p \implies q$$
Where $p = (\text{for all $x$, }0 < |x-a| < \delta)$ and $q = (|f(x) - L| < \epsilon)$ (this isn't directly relevant to my question, just showing what these... events? Statements? Propositions? I don't know the word -- what they look like).
I didn't include the actual thing being shown true above about $f(x)$ (I don't know where it'd go), but the more informal statement is "The limit of $f(x)$ as $x$ approaches $a$ is equal to $L$ if for all $\epsilon > 0$ there exists a $\delta>0$ such that for all $x$, $0 < |x-a| < \delta$ implies $|f(x)-L|<\epsilon$."
I wanted to prove a scenario where the limit was knowingly false ahead of time, and it took me some time to notice that this was the new statement to show in order to prove a limit false:
$$\exists \epsilon > 0, \forall \delta>0 , p \not\implies q$$
Now it's something more like, "The limit of $f(x)$ as $x$ approaches $a$ is not equal to $L$ if there exists an $\epsilon > 0$ such that for all $\delta>0$, for all $x$ $0 < |x-a| < \delta$ does not imply $|f(x)-L|<\epsilon$."
My questions:
Is there a formal proof or understanding as to why negation changes for-alls into there-exists and vice-versa?
I'm a little unsure why De Morgan's laws don't seem to apply between the "statements" (apologies for my imprecision, I don't know what any of these things are called). For instance could we not think of it as $(\forall \epsilon > 0) \land (\exists \delta>0) \land (p \implies q)$ as in, all of these things must be true -- the epsilon condition must be true AND the delta condition must be true AND $p$ must imply $q$. Then if we negate it, why doesn't it become something like $(\exists \epsilon > 0) \lor (\forall \delta>0) \lor (p \not\implies q)$ as in, the epsilon condition must be true OR the delta condition must be true OR $p$ must not imply $q$. I know intuitively this isn't the case but I don't know why, since the original statement makes sense to me as a chain of ANDs but the negation... still feels like it should be a chain of ANDs whereas De Morgan's laws normally have you invert them to ORs. |
HP 17bII+ Silver solver
09-14-2018, 09:20 PM
Post: #1
HP 17bII+ Silver solver
I'm very satisfied with my HP 17bII+ Silver, which I find very powerful, but also nice and not looking too complicated (no [f] and [g] functions on keys like the HP 12c or HP 35s that I used to use at work every day, but a really complete calculator except for trigs and complex numbers calculations).
So the 17BII+ is my new everyday calculator. I don't come back to arguments where a good (HP) calculator is a perfect complement or subsitute to Excel.
I chose the 17BII+ after having carefully studied the programs I use in my day to day work:
- price and costs calculations: can be modeled in the solver, with 4 or 5 equations and share vars
- margin calculations: built-in functions
- time value of money: built-in functions
- time functions: built-in functions
For the few moments I need trigs or complex calculations, I always have Free42 or a real 35s / 15c not really far from me.
After having rebuilt my work environment in the (so powerful) solver, I started to study the behavior of the solver, looking at loops (Σ function), conditional branchings (IF function), menu selection (S function), and the Get and Let functions (G(), L()).
I googled a lot and found an interesting pdf file about the Solvers of the 19B, 17B, 17BII and - according to the author, the 17BII+ Silver.
The file is here : http://www.mh-aerotools.de/hp/documents/...ET-LET.pdf
I tried a few equations in the "Using New and Old Values" chapter, pages 4 and following.
There I found lots of differences with my actual 17BII+ Silver, which I would like to share here with you.
For instance, the following equation found page 4:
Code:
In the next example found page 5:
Code:
Then the equation :
Code:
I don't understand the first 2 cases. In the first one, A should be set to -B, not B, if not using the "old" value of A. In the second one, the solver does not use the "old" value, but it does not also solve the equation, as there is no defined solution.
In the last case, I understand that the equation is evaluated twice before finding an answer. So the old value is used there, but not the way I could expect.
I finally found one - and only one - way to use iterations in the solver, with the equation :
Code:
Note that neither A=G(A)+1 or A=1+G(A) or G(A)+2=A works.
I'm not disappointed, as the solver is a really interesting and useful feature of the calculator, but I'm just surprised not having found more working cases of iterations, or a clear understanding of how the solver works.
Comments are welcomed.
Regards,
Thibault
09-14-2018, 10:14 PM (This post was last modified: 09-14-2018 10:25 PM by rprosperi.)
Post: #2
RE: HP 17bII+ Silver solver
The 17BII+ (Silver Edition) is an excellent machine, in fact it has the best keyboard of any machine made today by HP, but the solver does have a bug, and there are also a few other smaller issues making the solver slightly inferior to the 17B/17BII/19B/19BII/27S version.
See these 3 articles for details:
http://www.hpmuseum.org/cgi-sys/cgiwrap/...ead=242551
http://www.hpmuseum.org/forum/thread-657...l#pid58685
http://www.hpmuseum.org/cgi-sys/cgiwrap/...ead=134189
Overall these are not dramatic issues, and once you understand the solver bug, you likely can create equations that can avoid the issue.
I have most HP machines but the 17BII and 17BII+ are the ones I use most often for real work (vs. playing, exploring or following along interesting threads here).
Edit: added 3rd link
--Bob Prosperi
09-15-2018, 12:05 AM
Post: #3
RE: HP 17bII+ Silver solver
(09-14-2018 09:20 PM)pinkman Wrote: I'm not disappointed, as the solver is a really interesting and useful feature of the calculator, but I'm just surprised not having found more working cases of iterations, or a clear understanding of how the solver works.
Thibault, when Kinpo built the 17bii+ calculator years ago (both the gold one and the silver one), they basically goofed the solver implementation. This has been discussed at length in the HPMuseum forum. The solvers on the 17b and 17bii work fine, just as you would expect them to. If you plan on making significant use of the solver and its incredible capabilities, forget the + and get an original 17b or 17bii. You won't be sorry.
Also, get the manual (it's on the Museum DVD) Technical Applications for the HP-27s and HP-19b. It applies to the 17b as well.
The Sigma function also works fine on the 17b and 17bii.
09-15-2018, 04:36 AM
Post: #4
RE: HP 17bII+ Silver solver
Thanks to both of you for the details, links and advice.
I've read the threads carefully, I did not find them by myself first. It's the end of the night now, I'll try to make few testing later.
09-15-2018, 11:55 PM (This post was last modified: 09-15-2018 11:58 PM by rprosperi.)
Post: #5
RE: HP 17bII+ Silver solver
If you want to really explore the capabilities of the awesome Pioneer Solver, and confidently try to push it without worrying about using L() this way or that, I agree with Don, buy a 17BII and use that for the Solver stuff, but continue to use the 17BII+ for every day stuff.
Here's a very nice 17BII for only $25 (shipping included, in US) and you can even find them cheaper if you're willing to wait:
https://www.ebay.com/itm/HP-17Bll-Financ...3252533550
The 17BII is bug-free for solver use, while the 17BII+ has a much better LCD, readable in a wider range of lighting & use conditions.
In case you haven't seen this yet, here's an example of what can be done with the solver:
http://www.hpmuseum.org/forum/thread-2630.html
--Bob Prosperi
09-16-2018, 09:59 PM
Post: #6
RE: HP 17bII+ Silver solver
Well I'll try to find one, even if I'm not in the US.
I also want to continue using my actual 17bII+ at work, as the solver is powerful enough for me (for the moment), and it looks really good.
Don and you Bob have done a lot to help understand what the solver can do, that's pretty good stuff.
Regards,
Thibault
09-16-2018, 10:23 PM
Post: #7
RE: HP 17bII+ Silver solver
Nah, Don and Gerson are the real Pioneer Solver Masters, I just have a good collection of links.
Of all the various tools built-in to the various calculator models I've explored, the Pioneer Solver is easily the one that most exceeds it's initial apparent capability. This awesome tool must have been incredibly well-tested by the QA team. The fact that the sheer size (and audacity!) of Gerson's Trig formulas still return amazingly accurate results says more about the underlying design and code than any comments I could add.
Enjoy exploring it, and when you've mastered it (or at least tamed it a bit), come back here and share some interesting Solver formulas. There are numerous folks here that enjoy entering the formulas and running some test cases. (well, I suppose "enjoy" is not really the right word about entering the formulas - I guess feel good about accomplishing it successfully is more accurate).
When I see some of these long Solver equations, it reminds of TECO commands back in the PDP-11 days.
--Bob Prosperi
09-16-2018, 11:05 PM
Post: #8
RE: HP 17bII+ Silver solver
If you feel like entering long formulas you can solve the 8-queens problem.
):0)
Cheers
Thomas
09-17-2018, 06:30 AM (This post was last modified: 09-18-2018 09:47 AM by Don Shepherd.)
Post: #9
RE: HP 17bII+ Silver solver
How about a nifty, elegant, simple number base conversion Solver equation for the 17b/17bii, courtesy Thomas Klemm:
BC:ANS=N+(FROM-TO)\(\times \Sigma\)(I:0:LOG(N)\(\div\)LOG(TO):1:L(N:IDIV(N:TO))\(\times\)FROM^I)
Note: either FROM or TO must be 10 unless you are doing HEX conversions
09-17-2018, 07:49 AM
Post: #10
RE: HP 17bII+ Silver solver
09-17-2018, 09:17 AM (This post was last modified: 09-17-2018 10:28 AM by Don Shepherd.)
Post: #11
RE: HP 17bII+ Silver solver
(09-17-2018 07:49 AM)Thomas Klemm Wrote:
Wow, I didn't realize that Thomas. Stunning.
Don
09-17-2018, 10:24 AM
Post: #12
RE: HP 17bII+ Silver solver
As has already been written, unfortunately the solver in this calculator is flawed.
Part of the problems comes from the fact that it evaluates the equation twice which leads to incrementing by two instead of one etc. But there seem to be more quirks.
This is too bad, as the solver is a very capable tool (with G() and L()), while still easy and versatile to use (with the menu buttons).
Accomplishing something similar (solve for one variable today, for another tomorrow) with modern tools like Excel is more complicated.
Martin
09-17-2018, 12:48 PM
Post: #13
RE: HP 17bII+ Silver solver
(09-16-2018 10:23 PM)rprosperi Wrote: Nah, Don and Gerson are the real Pioneer Solver Masters, I just have a good collection of links.
An obvious correction is in order here:
Don, Gerson and Thomas Klemm are the real Pioneer Solver Masters...
No disrespect intended by the omission. A review of past posts of significant solver formulas quickly reveals just how often all three of these three guys were the authors.
--Bob Prosperi
09-28-2018, 01:33 PM
Post: #14
RE: HP 17bII+ Silver solver
Be sure I have great respect of everyone mentioned above
Thanks for the advices, my new old 17bii is now in my hands ($29 on French TAS), I'm just waiting for the next rainy Sunday for pushing the solver to it's limits (mmh, to MY limits I guess).
09-28-2018, 07:47 PM (This post was last modified: 09-28-2018 07:48 PM by Jlouis.)
Post: #15
RE: HP 17bII+ Silver solver
Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?
TIA
09-29-2018, 01:37 AM
Post: #16
RE: HP 17bII+ Silver solver
(09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?
Check out the document by Martin Hepperle in this thread which lists the functions available for each calculator. The document is an excellent introduction to the Solver application.
Solver GET-LET
~Mark
Who decides?
09-29-2018, 02:25 AM
Post: #17
RE: HP 17bII+ Silver solver
(09-29-2018 01:37 AM)mfleming Wrote:(09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?
Thanks mfleming, I should have read the beginning of the manual. It is the same solver for the calculators of my question. I prefer to use the 19bII due the dedicated alpha keyboard, but the 27s is a pleasure to use too.
Cheers
09-29-2018, 11:13 AM
Post: #18
RE: HP 17bII+ Silver solver
09-29-2018, 11:53 AM (This post was last modified: 09-29-2018 05:12 PM by Don Shepherd.)
Post: #19
RE: HP 17bII+ Silver solver
(09-28-2018 07:47 PM)Jlouis Wrote: Just one doubt here, Are the solver of 18C and 19B / BII the same of the 17BII and 27S?Here is a 3-page cross-reference I created a few years ago of Solver functions present in the 19bii, 27s, and 17bii.
solver 1.PDF (Size: 1.08 MB / Downloads: 33)
solver 2.PDF (Size: 968.77 KB / Downloads: 21)
solver 3.PDF (Size: 494.75 KB / Downloads: 19)
09-29-2018, 01:36 PM (This post was last modified: 09-29-2018 01:49 PM by Jlouis.)
Post: #20
RE: HP 17bII+ Silver solver
(09-29-2018 11:53 AM)Don Shepherd Wrote:
Really thanks Dom for taking time to make these documents.
This is what makes MoHC a fantastic community.
Cheers
JL
Edited: Looks like the 19BII is a little more complete than the 27s and that the 17BII is the weaker one.
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Consider the Black-Scholes problem $$\frac{\partial A}{\partial t}+\frac{\sigma^2B^2}{2}\frac{\partial^2A}{\partial B^2}+rB\frac{\partial A}{\partial B}-rA=0 \hspace{3mm}\textrm{and}\hspace{3mm} A(B,T)=f(b),$$ with $B>0$ and $t<T$, and where $\sigma>0$, $T>0$, $r\in \mathbb{R}$ are constants.
Now, let $A(B,t)$ be a solution of the above problem and let it be infinitely differentiable in $B>0$ and $t<T$.
Show that $$A_1(B,t)=B\frac{\partial A}{\partial B}(B,t)$$ satisfies the partial differential equation in the above problem.
So I thought the best thing to do would be to look at the two identities:
$$\frac{\partial}{\partial B}B\left(\frac{\partial A}{\partial B}\right)=\frac{\partial A}{\partial B}+S\frac{\partial^2 A}{\partial B^2}$$ and $$\frac{\partial^2}{\partial B^2}B\left(\frac{\partial A}{\partial B}\right)=2\frac{\partial^2 A}{\partial B^2}+S\frac{\partial^3 A}{\partial B^3}.$$
Then taking the derivative of the Black-Scholes problem (with respect to $B$), I obtain: $$\frac{\partial}{\partial t}\frac{\partial A}{\partial B}+\frac{\sigma^2}{2}\left( 2B\frac{\partial^2A}{\partial B^2}+B^2\frac{\partial^3A}{\partial B^3}\right)+r\left(\frac{\partial A}{\partial B}+B\frac{\partial^2 A}{\partial B^2}\right)-r\frac{\partial A}{\partial B}=0.$$
Then I use the two aforementioned identities and multiply by $B$ to get $$\frac{\partial}{\partial t}\left(S\frac{\partial A}{\partial B}\right)+\frac{\sigma^2}{2}S^2\frac{\partial^2}{\partial B^2}\left(S\frac{\partial A}{\partial B}\right)+rB\frac{\partial}{\partial B}\left(S\frac{\partial A}{\partial B}\right)-rB\frac{\partial}{\partial B}=0.$$ After this point I get rather confused as I can't seem to cancel out the terms? Have I made any mistakes with my method? |
In Peskin/Schroeder there is an explicit calculation showing that the retarded Green's function of the real Klein-Gordon field
$$D_R(x-y) ~\equiv~ \theta(x^0 - y^0) \langle 0 | [\phi(x), \phi(y)] |0\rangle\tag{2.55} $$
fulfills the equation
$$(\partial^2 + m^2) D_R(x-y) = -i\delta^4(x-y).\tag{2.56}$$
I can't follow one specific step in the derivation: It seems like they are doing the substitution
$$(\partial_{x^0}\delta(x^0-y^0))\langle 0 | [\phi(x), \phi(y)] |0\rangle = -\delta(x^0-y^0)\partial_{x^0}\langle 0 | [\phi(x), \phi(y)] |0\rangle. \tag{A} $$
But I don't see how this is justified: If we interpret $(\partial_t \delta(t)) f(t)$ as a distribution and act with it on a test function $g(t)$, we get $$\int (\partial_t \delta(t)) f(t)g(t)dt = - \int \delta(t) \partial_t(f(t)g(t))dt = - \partial_t(f(t)g(t))|_{t=0}.\tag{B} $$
If we act instead with $-\delta(t) \partial_t f(t)$ on $g(t)$ we get $$-(g(t)\partial_t f(t))|_{t=0}.\tag{C} $$
Does somebody have an explanation? |
Like some other terms in mathematics (“algebra” comes to mind),
topology is both a discipline and a mathematical object. Moreover like algebra, topology as a subject of study is at heart an artful mathematical branch devoted to generalizing existing structures like the field of real numbers for their most convenient properties. It is also a favorite subject of mine, ever since my first introduction to it. This is due in large part to its exceedingly simple first principles, which make the wealth of expansion they allow all the more impressive.
It is my intent to discuss some of these starting points here, in the first of a short series of posts toward the goal of presenting one of my favorite results arising in topology:
Moore-Smith convergence, a vast extension of the notion of the limit of a sequence. My representation here follows the explanation given by John L. Kelley in his classic text General Topology, which I recommend to all curious readers. What is a topology?
Definition.By topologyis meant any collection \mathscr{T} of sets satisfying two conditions:
\begin{array}{lrcl}\text{(1)}&A,B\in\mathscr{T}&\Rightarrow&A\cap B\in\mathscr{T};\\\text{(2)}&\mathscr{C}\subset\mathscr{T}&\Rightarrow&\bigcup\{C\in\mathscr{C}\}\in\mathscr{T}\end{array}
It is worthwhile to break this definition down. Condition (1) requires that the intersection of any two elements of the collection \mathscr{T} must itself be a member of \mathscr{T}. Condition (2) states that the union of any subcollection of \mathscr{T} must also belong to \mathscr{T}. These are referred to as
closure to finite intersection and closure to arbitrary union, respectively, in some texts.
Notably, the definition speaks only of a collection of sets with no specification beyond the two conditions. Yet, even with these, one can deduce some further characteristic properties.
\begin{array}{ll}\text{(i)}&\emptyset\in\mathscr{T};\\\text{(ii)}&\bigcup\{T\in\mathscr{T}\}\in\mathscr{T}.\end{array}
Corollary.If \mathscr{T} is a topology, then
Since \emptyset\subset S for every set S, and \mathscr{T}\subset\mathscr{T}, it is enough to apply (2) to both of these cases to prove the corollary. In fact, many texts make the definition X\mathrel{:=}\bigcup\{T\in\mathscr{T}\}, and refer to the pair (X,\mathscr{T}) as the
topological space defined by \mathscr{T}.
This way, the space is given its character by way of the scheme that builds \mathscr{T}, rather than the set X. It is an important distinction, for many topologies are possible on a given set. With that, we can look at some examples.
From Trivial to Complicated 1. The Trivial Topology
Based on the corollary just presented, it is enough to gather a given set X and the empty set \emptyset into a collection \{\emptyset,X\} and have created a topology on X. Because X and \emptyset are its only members, the collection is easily closed to arbitrary union and finite intersection of its elements. This is known as the
trivial or indiscrete topology, and it is somewhat uninteresting, as its name suggests, but it is important as an instance of how simple a topology may be. As per the corollary, every topology on X must contain \emptyset and X, and so will feature the trivial topology as a subcollection. 2. The Discrete Topology
For this example, one can start with an arbitrary set, but in order to better illustrate, take the set of the first three primes: \{2,3,5\}. Suppose we consider the collection of all possible subsets of \{2,3,5\}. This is also referred to as the
power set of \{2,3,5\}, and denoted \wp(\{2,3,5\}). Fortunately, the set is small enough to list exhaustively 1. Here they are listed from top-to-bottom in order of increasing inclusion:
\emptyset \{2\} \{3\} \{5\} \{2,3\} \{2,5\} \{3,5\} \{2,3,5\}
Note these are
all possible subsets of \{2,3,5\}. It is clear any union or intersection of the pieces in the table above exists as an entry, and so this meets criteria (1) and (2). This is a special example, known as the discrete topology. Because the discrete topology collects every existing subset, any topology on \{2,3,5\} is a subcollection of this one.
For example, taking the sets\emptyset,\quad\{5\},\quad\{2,3\},\quad\{2,3,5\}
from the collection in the table is enough to produce a topology
2. Remark.Given a topological space (X,\mathscr{T}), the elements of \mathscr{T} are referred to as open sets. This nomenclature is motivated in the next example. 3. \mathbb{R} and Open Intervals
This example will be more constructive than the previous ones. Consider the set of real numbers, \mathbb{R}. Let us define a special collection \mathscr{T} of subsets of real numbers the following way: a set T belongs to \mathscr{T} if, and only if, for every x\in T, there exist real numbers a and b such that x\in(a,b) and (a,b)\subset T. That is, we say T\in\mathscr{T} to mean T contains an open interval around each of its elements.
It is good practice to take the time to
prove this collection defines a topology on \mathbb{R}. To do so, it must be shown that \bigcup\{T\in\mathscr{T}\}=\mathbb{R}, and that \mathscr{T} meets conditions (1) and (2).
Proof. To show \bigcup\{T\in\mathscr{T}\}=\mathbb{R}, it must be verified that \bigcup\{T\in\mathscr{T}\}\subset\mathbb{R} and \mathbb{R}\subset\bigcup\{T\in\mathscr{T}\}. The first containment follows by defining every T\in\mathscr{T} as a subset of \mathbb{R} to begin with, so the reverse containment is all that is left. Let x\in\mathbb{R} be given. Then certainly x\in(x-1,x+1), and surely (x-1,x+1)\in\mathscr{T}, as it contains an open interval around all its points by its very design. Thus x\in\bigcup\{T\in\mathscr{T}\}.
On to proving \mathscr{T} satisfies (1) and (2). For (1), let A,B\in\mathscr{T} be given and suppose
3x\in A\cap B. This holds if, and only if, x\in A and x\in B. Since A and B both belong to \mathscr{T}, there exist real numbers a, b, c, and d such that x\in(a,b)\subset A, and x\in(c,d)\subset B. But this means x\in(a,b)\cap(c,d). Fortunately, two intervals of real numbers may only overlap in one way: this means either c<b or a<d. Without loss of generality, suppose it is the former case, that c<b. Then (a,b)\cap(c,d)=(c,b), and it is so that x\in(c,b), an open interval contained in A\cap B (precisely as desired), and it follows A\cap B\in\mathscr{T}.
To show (2) is much easier. Let \{T_\alpha\}_{\alpha\in\mathscr{A}} be a collection
4of sets belonging to \mathscr{T}, and suppose x\in\bigcup_{\alpha\in\mathscr{A}}T_\alpha. Then there exists an index, say \alpha_0\in\mathscr{A}, such that x\in T_{\alpha_0}. Since T_{\alpha_0}\in\mathscr{T}, there exist real numbers a and b such that x\in(a,b)\subset T_{\alpha_0}. But this means x\in(a,b)\subset\bigcup_{\alpha\in\mathscr{A}}T_\alpha. Since x was chosen arbitrarily, it follows \bigcup_{\alpha\in\mathscr{A}}T_\alpha\in\mathscr{T}.
The proof above shows (\mathbb{R},\mathscr{T}) is a topological space; the collection \mathscr{T} is referred to as the
standard topology on \mathbb{R}. The open sets in this space are all the subsets of real numbers contained in open intervals. Fittingly, then, open intervals are open sets in the standard topology. Conclusion
This first post is meant to express the simple starting points of topology as a subject of study. It only takes the two criteria mentioned here to define a topology of sets, and yet an entire realm of theory rests upon them. This is a recurring theme in topology, algebra, and mathematics in general. Building the fully-featured universes that hold the answers for more specific inquiry: the complete ordered field of real numbers
5 \mathbb{R}, the space \mathcal{C}^{\infty}(\mathbb{C}) of infinitely differentiable functions f\mathrel{:}\mathbb{C}\to\mathbb{C}, the class of all real-valued Lebesgue-integrable functions on \mathbb{R}, each of these requires a well-made foundation.
The next post in this series will cover the nature of sequences in topological spaces, particularly those instances where the convenient features afforded by the real numbers are no longer available. With the metric space structure stripped away, how does one define convergence and limit of sequences? What does it mean for elements in topological spaces to be close when distance is otherwise without definition?
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Footnotes A finite set, say with kelements, has a power set possessing exactly 2 elements. k This claim is an example that can be proven exhaustively. An important point here: by supposing x∈ A∩ B, it is automatically assumed A∩ B≠ Ø. To see the empty set belongs to 𝒯 regardless, note that the statement “if x∈ Ø, then x∈ ( a, b)” is always true. Here, 𝒜denotes an arbitrary collection so that the proof is not too specific; similarly αis just an index marker. The only one like it, up to isomorphism! |
Logistic Regression Introduction
In classification problems, the goal is to predict the class membership based on predictors. Often there are two classes and one of the most popular methods for binary classification is logistic regression (Cox 1958, @Freedman:2009).
Suppose now that \(y_i \in \{0,1\}\) is a binary class indicator. The conditional response is modeled as \(y|x \sim \mbox{Bernoulli}(g_{\beta}(x))\), where \(g_{\beta}(x) = \frac{1}{1 + e^{-x^T\beta}}\) is the logistic function, and maximize the log-likelihood function, yielding the optimization problem
\[ \begin{array}{ll} \underset{\beta}{\mbox{maximize}} & \sum_{i=1}^m \{ y_i\log(g_{\beta}(x_i)) + (1-y_i)\log(1 - g_{\beta}(x_i)) \}. \end{array} \]
CVXR provides the
logistic atom as a shortcut for \(f(z) = \log(1 + e^z)\) to express the optimization problem. One may betempted to use
log(1 + exp(X %*% beta)) as in conventional
R syntax. However, this representation of \(f(z)\) violatesthe DCP composition rule, so the
CVXR parser will reject theproblem even though the objective is convex. Users who wish to employa function that is convex, but not DCP compliant should check thedocumentation for a custom atom or consider a different formulation.
Example
The formulation is very similar to OLS, except for the specification of the objective.
In the example below, we demonstrate a
key feature of
CVXR, thatof evaluating various functions of the variables that are solutions tothe optimization problem. For instance, the log-odds, \(X\hat{\beta}\),where \(\hat{\beta}\) is the logistic regression estimate, is simplyspecified as
X %*% beta below, and the
getValue function of theresult will compute its value. (Any other function of the estimatecan be similarly computed.)
n <- 20m <- 1000offset <- 0sigma <- 45DENSITY <- 0.2set.seed(183991)beta_true <- stats::rnorm(n)idxs <- sample(n, size = floor((1-DENSITY)*n), replace = FALSE)beta_true[idxs] <- 0X <- matrix(stats::rnorm(m*n, 0, 5), nrow = m, ncol = n)y <- sign(X %*% beta_true + offset + stats::rnorm(m, 0, sigma))beta <- Variable(n)obj <- -sum(logistic(-X[y <= 0, ] %*% beta)) - sum(logistic(X[y == 1, ] %*% beta))prob <- Problem(Maximize(obj))result <- solve(prob)log_odds <- result$getValue(X %*% beta)beta_res <- result$getValue(beta)y_probs <- 1/(1 + exp(-X %*% beta_res))
We can compare with the standard
stats::glm estimate.
d <- data.frame(y = as.numeric(y > 0), X = X)glm <- stats::glm(formula = y ~ 0 + X, family = "binomial", data = d)est.table <- data.frame("CVXR.est" = beta_res, "GLM.est" = coef(glm))rownames(est.table) <- paste0("$\\beta_{", 1:n, "}$")knitr::kable(est.table, format = "html") %>% kable_styling("striped") %>% column_spec(1:3, background = "#ececec")
CVXR.est GLM.est \(\beta_{1}\) -0.0305494 0.0305494 \(\beta_{2}\) 0.0023528 -0.0023528 \(\beta_{3}\) -0.0110080 0.0110080 \(\beta_{4}\) 0.0163919 -0.0163919 \(\beta_{5}\) 0.0157186 -0.0157186 \(\beta_{6}\) 0.0006251 -0.0006251 \(\beta_{7}\) -0.0157914 0.0157914 \(\beta_{8}\) -0.0092228 0.0092228 \(\beta_{9}\) 0.0173823 -0.0173823 \(\beta_{10}\) 0.0019102 -0.0019102 \(\beta_{11}\) -0.0100746 0.0100746 \(\beta_{12}\) -0.0269883 0.0269883 \(\beta_{13}\) 0.0233625 -0.0233625 \(\beta_{14}\) 0.0009529 -0.0009529 \(\beta_{15}\) -0.0016264 0.0016264 \(\beta_{16}\) 0.0312156 -0.0312156 \(\beta_{17}\) 0.0038949 -0.0038949 \(\beta_{18}\) -0.0121105 0.0121105 \(\beta_{19}\) 0.0246811 -0.0246811 \(\beta_{20}\) -0.0007025 0.0007025
The sign difference is due to the coding of \(y\) as \((-1, 1)\) for
CVXR rather than \((0, 1)\) for
stats::glm.
So, for completeness, if we were to code the \(y\) as \((0, 1)\), the objective will have to be modified as below.
obj <- -sum(X[y <= 0, ] %*% beta) - sum(logistic(-X %*% beta))prob <- Problem(Maximize(obj))result <- solve(prob)beta_log <- result$getValue(beta)est.table <- data.frame("CVXR.est" = beta_log, "GLM.est" = coef(glm))rownames(est.table) <- paste0("$\\beta_{", 1:n, "}$")knitr::kable(est.table, format = "html") %>% kable_styling("striped") %>% column_spec(1:3, background = "#ececec")
CVXR.est GLM.est \(\beta_{1}\) 0.0305494 0.0305494 \(\beta_{2}\) -0.0023528 -0.0023528 \(\beta_{3}\) 0.0110080 0.0110080 \(\beta_{4}\) -0.0163919 -0.0163919 \(\beta_{5}\) -0.0157186 -0.0157186 \(\beta_{6}\) -0.0006251 -0.0006251 \(\beta_{7}\) 0.0157914 0.0157914 \(\beta_{8}\) 0.0092228 0.0092228 \(\beta_{9}\) -0.0173823 -0.0173823 \(\beta_{10}\) -0.0019102 -0.0019102 \(\beta_{11}\) 0.0100746 0.0100746 \(\beta_{12}\) 0.0269883 0.0269883 \(\beta_{13}\) -0.0233625 -0.0233625 \(\beta_{14}\) -0.0009529 -0.0009529 \(\beta_{15}\) 0.0016264 0.0016264 \(\beta_{16}\) -0.0312156 -0.0312156 \(\beta_{17}\) -0.0038949 -0.0038949 \(\beta_{18}\) 0.0121105 0.0121105 \(\beta_{19}\) -0.0246811 -0.0246811 \(\beta_{20}\) 0.0007025 0.0007025
Now, the results match perfectly.
Session Info
sessionInfo()
## R version 3.6.0 (2019-04-26)## Platform: x86_64-apple-darwin18.5.0 (64-bit)## Running under: macOS Mojave 10.14.5## ## Matrix products: default## BLAS/LAPACK: /usr/local/Cellar/openblas/0.3.6_1/lib/libopenblasp-r0.3.6.dylib## ## locale:## [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8## ## attached base packages:## [1] stats graphics grDevices datasets utils methods base ## ## other attached packages:## [1] kableExtra_1.1.0 CVXR_0.99-6 ## ## loaded via a namespace (and not attached):## [1] gmp_0.5-13.5 Rcpp_1.0.1 highr_0.8 ## [4] compiler_3.6.0 pillar_1.4.1 R.methodsS3_1.7.1## [7] R.utils_2.8.0 tools_3.6.0 digest_0.6.19 ## [10] bit_1.1-14 viridisLite_0.3.0 evaluate_0.14 ## [13] tibble_2.1.2 lattice_0.20-38 pkgconfig_2.0.2 ## [16] rlang_0.3.4 Matrix_1.2-17 rstudioapi_0.10 ## [19] yaml_2.2.0 blogdown_0.12.1 xfun_0.7 ## [22] xml2_1.2.0 httr_1.4.0 Rmpfr_0.7-2 ## [25] ECOSolveR_0.5.2 stringr_1.4.0 knitr_1.23 ## [28] hms_0.4.2 webshot_0.5.1 bit64_0.9-7 ## [31] grid_3.6.0 glue_1.3.1 R6_2.4.0 ## [34] rmarkdown_1.13 bookdown_0.11 readr_1.3.1 ## [37] magrittr_1.5 scales_1.0.0 htmltools_0.3.6 ## [40] scs_1.2-3 rvest_0.3.4 colorspace_1.4-1 ## [43] stringi_1.4.3 munsell_0.5.0 crayon_1.3.4 ## [46] R.oo_1.22.0
Source References
Cox, D. R. 1958. “The Regression Analysis of Binary Sequences.”
Journal of the Royal Statistical Society. Series B (Methodological) 20 (2): 215–42.
Freedman, D. A. 2009.
Statistical Models: Theory and Practice. Cambridge University Press. |
I know how to calculate t-statistics and p-values for linear regression, but I'm trying to understand a step in the derivation. I understand where Student's t-distribution comes from, namely I can show that if $Z$ is a random variable drawn from a standard normal distribution $Z \sim \mathcal{N}(0,1)$ and if $\chi$ is drawn from a $\chi^2_k$ distribution, then the new random variable
$$ T = \frac{Z}{\sqrt{\frac{\chi^2}{k}}} $$
will be drawn from a t-distribution with $k$ degrees of freedom.
The part that confuses me is the application of this general formula to the estimates for the coefficients of a simple linear regression. If I parametrize the linear model as $y = \alpha + \beta x + \varepsilon$, with $\varepsilon$ a random variable with zero mean characterizing the errors, then the best estimate for $\beta$ is
$$ \hat{\beta} = \frac{\bar{x y} - \bar{x} \bar{y} }{\bar{x^2} - \bar{x}^2 } \, , $$
where the bar's indicate sample means. The standard error of $\hat{\beta}$ is
$$ SE(\hat{\beta})^2 = \frac{\sigma^2}{n \left(\bar{x^2} - \bar{x}^2 \right) } \,. $$
Here $\sigma = \sqrt{\operatorname{Var}(\varepsilon)}$. The part I am confused about is why
$$ t \equiv \frac{\hat{\beta}}{SE(\hat{\beta})} $$
is taken to be drawn from a t-distribution, assuming the null hypothesis. If I could write $t$ in the form of the above variable $T$, cleanly identifying the $Z$ and the $\chi$ variables, then everything would be clear. |
This is a fairly old question, but I didn't see an answer that reflected my point of view. So here's my two cents. There are three primary reasons (other than the ones that have been belabored already) that I believe learning integration techniques are worth-while:
Learning integration techniques reinforces the importance of duality.
The Fundamental Theorems are beautiful in part because they encapsulate the inverse relationships between differentiation and integration. It would be a shame to not see these inverse relations carried as far as they could be. For example, we might expect that every property that differentiation satisfies, there might be an inverse relationship that integration satisfies. And, indeed, integration by parts can be thought of as the "opposite" of the product rule. And change of variables can be thought of as the "opposite" of the chain rule. This is the most superficial reason to appreciate integration techniques though.
Integration techniques are actually computationally useful for proofs.
More so than the exercises would have a student believe. Some of the best ways to exemplify this is the discovery of recurrence relations. For example, if we defined$$I_n=\int_0^\pi \sin^n x\, dx$$for every natural number $n$, integration by parts would imply$$I_n=\frac{n-1}{n}I(n-2)\,.$$This small observation formed the basis of John Wallis's proof of the identity$$\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdots=\frac{\pi}{2}$$which is quite beautiful in itself, relating $\pi$ to the odd and even positive integers.
A similar technique with the logarithm could should that$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots =\ln 2$$which is also aesthetically interesting.
Leaning integration techniques informs you about what is
possible and hints at what is impossible. These hints of impossibilities lead to some beautiful mathematics.
There are two quotes that are relevant here. Namely,
"Common integration is only the memory of differentiation..." - Augustus de Morgan
"It can be of no practical use to know that $\pi$ is irrational, but if we can know, it surely would be intolerable not to know." - Edward Charles Titschmarsh
Integration is interesting precisely because it is difficult. Without the FTC, calculating integrals with Riemann (or Upper/Lower) sums is tedious and frequently requires a spark ingenuity which changes dramatically between the functions you want to integrate. With the FTC however, many elementary functions become quite easy to integrate. With these functions mastered, it is natural to move on to more complicated functions to probe the limits of the FTC, integration by parts, change of variables, etc. to see where we eventually hit a wall. The techniques you learn is a roadmap through the genius of past mathematicians that you don't have to repeat, but would have difficulty redeveloping on your own.
Despite the swath of functions you're able to integrate with these techniques, it is interesting that there is a wall. Namely, we cannot integrate$$e^{-x^2}\qquad \frac{\sin x}{x}\qquad\frac{1}{\ln x}$$using the techniques that we learn from Calculus. We would not have been able to find these functions without getting our hands dirty through extended and purposeful effort. Few courses take integration to this extreme, much less prove that we cannot integrate these functions in terms of our other elementary functions.
It should also be mentioned that integration by parts also leads to an elementary proof of the irrationality of $\pi$... |
According to page 53 of Modern Computer Arithmetic (pdf), all of the steps in the Schönhage–Strassen Algorithm cost $O(N \cdot \lg(N))$ except for the recursion step which ends up costing $O(N\cdot \lg(N) \cdot \lg(\lg(N)))$.
I don't understand why the same inductive argument used to show the cost of the recursive step doesn't work for $O(N\cdot \lg(N))$.
Assume that, for all $X < N$, the time $F(X)$ is less than $c \cdot X \cdot \lg(X)$ for some $c$. So the recursive step costs $d \cdot \sqrt{N} F(\sqrt{N})$, and we know this is less than $d \cdot \sqrt{N} c \cdot \sqrt{N} \lg(\sqrt{N}) = \frac{c \cdot d}{2} \cdot N \cdot \lg(N)$ by the inductive hypothesis. If we can show that $d < 2$, then we're done because we've satisfied the inductive step. I'm pretty sure recursion overhead is negligible, so $d \approx 1$ and we have $\frac{c}{2} N \cdot \lg(N)$ left to do the rest. Easy: everything else is $O(N \cdot \lg(N))$ so we can pick a $c$ big enough for it to fit in our remaining time.
Basically, without digging into the details that will contradict this somehow, it
looks like things would work out if we assumed the algorithm costs $O(N \cdot \log(N))$. The same thing seems to happen if I expand the recursive invocations then sum it all up... so where is the penalty coming from?
My best guess is that it has to do with the $\lg(\lg(N))$ levels of recursion, since that's how many times you must apply a square root to get to a constant size. But how do we know each recursive pass is not getting cheaper, like in quickselect?
For example, if we group our $N$ initial items into words of size $O(\lg(N))$, meaning we have $O(N/\lg(N))$ items of size $O(\lg(N))$ to multiply when recursing, shouldn't that only take $O(N/\lg(N) \cdot \lg(N) \cdot \lg(\lg(N)) \cdot \lg(\lg(\lg(N)))) = O(N \cdot \lg(\lg(N)) \cdot \lg(\lg(\lg(N))))$ time to do. Not only is that well within the $N \cdot \lg(N)$ limit, it worked even though I used the larger $N\cdot \lg(N)\cdot \lg(\lg(N))$ cost for the recursive steps (for "I probably made a mistake" values of "worked").
My second guess is that there's some blowup at each level that I'm not accounting for. There are constraints on the sizes of things that might work together to slow down how quickly things get smaller, or to multiply how many multiplications have to be done.
Here's the recursive expansion.
Assume we get $N$ bits and split them into $\sqrt{N}$ groups of size $\sqrt{N}$. Everything except the recursion costs $O(N \lg N)$. The recursive multiplications can be done with $3 \cdot \sqrt{N}$ bits. So we get the relationship:
$M(N) = N \cdot \lg(N) + \sqrt{N} \cdot M(3 \cdot \sqrt{N})$
Expanding once:
$M(N) = N \cdot \lg(N) + \sqrt{N} \cdot (3 \cdot \sqrt{N} \cdot lg(3 \cdot \sqrt{N}) + \sqrt{3 \cdot \sqrt{N}} \cdot M(3 \cdot \sqrt{3 \cdot \sqrt{N}})$
Simplifying:
$M(N) = N \cdot \lg(N) + 3 \cdot N \cdot lg(3 \cdot \sqrt{N}) + \cdot \sqrt{3} \cdot N^{2-\frac{1}{2}} \cdot M(3^{2-\frac{1}{2}} \cdot \sqrt{\sqrt{N}})$
See the pattern? Each term will end up in the form $3^{2-2^{-i}} \cdot N \cdot lg(N^{2^{-i}} 3^{2-2^{-i}})$. So the overall sum is:
$\sum_{i=0}^{\lg(\lg(N))} 3^{2-2^{-i}} \cdot N \cdot \lg(N^{2^{-i}} 3^{2-2^{-i}})$
We can upper bound this by increasing the powers of 3 to just 3^2, since that can only increase the value in both cases:
$\sum_{i=0}^{\lg(\lg(N))} 9 \cdot N \cdot \lg(N^{2^{-i}} 9)$
Which is asymptotically the same as:
$\sum_{i=0}^{\lg(\lg(N))} N \cdot \lg(N^{2^{-i}})$
Moving the power out of the logarithm:
$\sum_{i=0}^{\lg(\lg(N))} N \cdot \lg(N) \cdot 2^{-i}$
Moving variables not dependent on $i$ out:
$N \cdot \lg(N) \sum_{i=0}^{\lg(\lg(N))} 2^{-i}$
The series is upper bounded by 2, so we're upper bounded by:
$N \cdot \lg(N)$
Not sure where the $\lg(\lg(N))$ went. All the twiddly factors and offsets (because many recurrence relations "solutions" are broken by those) I throw in seem to get killed off by the $\lg$ creating that exponentially decreasing term, or they end up not multiplied by $N$ and are asymptotically insignificant. |
Prove that $BPP^{BPP} = BPP$.
Obviously $BPP\subseteq BPP^{BPP}$. So we're left with proving $BPP^{BPP} \subseteq BPP$. Let's consider $L\in BPP^{BPP}$. Then, there's a PTM $M$ to decide $L$ for some $BPP(\alpha, \beta)$ which uses an oracle for some language $O\in BPP$. $O$ also has a $PTM\in BPP$, let's denote it $M_o$. So we could create $M'$, a PTM which behaves exactly as $M$, but everytime $M$ uses the oracle, it simulate $M_o$ on the given input for the oracle.
Now, I want to show that $M\in BPP$. By definition, $BPP=BPP(1/3, 1/3)$, so every time $M$ calls $M_o$ it takes a probabilistic risk basing a deciion upon $M_o$ answer.
We've learned in class that $BPP = BPP(1/3, 1,3) = BPP\left(\frac{1}{3^n}, \frac{1}{3^n} \right)$ and I think I shall explain that even though $M_o$ may return false answers, $M_o$ is still in $BPP$.
Can you help me with doing that?
Thanks
EDIT: Basically, I want to give a lower bound for the probability of $M$, answering correctly. Let's assume that $M$ queries the oracle $n^c$ times for some $n\in\mathbb{N}$. I think the lower bound is $\frac{1}{3^{n^c}}$.
My explanation is that we're assuming the worst case. That is, if the oracle is returning a false answer then we are returning a false answer.
Is that right?
EDIT2 I'm not sure it's $100\%$ correct, since I didn't take into account the probabilistic behavior of $M$ itself. |
In the case of
finite dimensional spaces, measurability can be reduced to the "ordinary" notion of measurability as follows:
A function $f : X \to \Bbb{R}^n$ is measurable if and only if for each $i \in \{1, \dots, n\}$, the component function $f_i : X \to \Bbb{R}$ is measurable.
This can be seen as follows: Measurability is preserved by continuous functions (because these are Borel-measurable and compositions of measurable functions are measurable). Hence, if $f$ is measurable, then each $f_i = \pi_i \circ f$ is also measurable, because $\pi_i : \Bbb{R}^n \to \Bbb{R}, x \mapsto x_i$ is continuous.
On the other hand, each of the maps $\iota_i : \Bbb{R} \to \Bbb{R}^n, x \mapsto (0, \dots, 0, x, 0,\dots,0)$ is continuous (where the $x$-entry is in the $i$-th component). Hence, if each $f_i$ is measurable, then
$$f = \sum_{i=1}^n \iota_i \circ f_i$$
is also measurable.
The same arguments remain valid in the case of $\Bbb{C}^n$. In that case, one can even reduce further to measurability of $\mathrm{Re}(f_i)$ and $\mathrm{Im}(f_i)$ for each $i$.
Regarding completeness, one can easily see the norm equivalence
$$\Vert f \Vert_p \asymp \sum_{i=1}^n \Vert f_i \Vert_p,$$
which allows to reduce the completeness to the $1$-dimensional setting.
Regarding the
Bochner measurability, one can show (cf. Serge Lang, Real and Functional Analysis) that $f : X \to E$ (with a Banach space $E$) is $\mu$- Bochner measurable if and only if There is a $\mu$-null-set $N \subset X$ such that $f(X\setminus N)$ is separable and so that $f|_{X \setminus N}$ is (Borel) measurable in the sense described by you. $f$ vanishes outside of a set of $\sigma$-finite measure.
Here, a function $f$ is called $\mu$-Bochner-measurable, if there is a sequence of simple functions $f_n = \sum_{i=1}^{m_n} \alpha_i^{(n)} \chi_{M_i^{(n)}}$ with each $M_i^{(n)}$ of finite measure and with $f_n(x) \to f(x)$ for $\mu$-almost all $x$.
This shows that as long as $E$ is separable (this is in particular the case if $E$ is finite dimensional) and $\mu$ is $\sigma$-finite (and complete), there is no difference between ordinary measurability and Bochner measurability. Problems occur as soon as the target space $E$ is not separable. |
This is something I found in trying to work on Vince Vatter's excellent question. I have no solution, but a much more precise conjecture.
Recall that a rooted planar tree is a rooted tree where, for every vertex, the children of that vertex come equipped with an order. I will think of the vertices of tree as members of an asexually reproducing species, and therefore use language like "sibling", "cousin", "child", "parent", "generation" etc. When I speak of generations, I am counting from the oldest, so the root is the fist generation; the children of the root are the next generation, and so forth. For any vertex, I will think of the ordering of its children as the birth order of the children.
Define a vertex $v$ to be "crucial" if $v$ is the youngest member of its generation, all the other members of that generation are childless, but $v$ has children. For example, in the tree $$\begin{matrix} & & a & & \\ & & \downarrow & & \\ & & b & & \\ & \swarrow & \downarrow & \searrow \\ c & & d & & e \\ & & \downarrow & & \downarrow \\ & & f & & g \\ & & \downarrow & \searrow & \\ & & h & & i \\ & & & & \downarrow \\ & & & & j \\ \end{matrix}$$ the crucial elements are $a$, $b$, $f$ and $i$. So the root is always crucial.
Let $c(p,q,n)$ be the number of planar trees on $n$ vertices with $p$ crucial elements and where the root has $q$ children. For example, of the $5$ planar trees on $4$ elements, there is one tree each with $(p,q)$ equal to $(1,2)$, $(2,1)$, $(3,1)$, $(2,2)$ and $(1,3)$.
The number of planar trees, $\sum_{p,q} c(p,q,n)$, is the Catalan number $C_{n-1}$ (item $e$ on Stanley's famous list). Also, $\sum_{p} c(p,1,n)$ is $C_{n-2}$ -- this counts planar trees where the root only has one child, so just delete the root to get planar trees on $n-1$ vertices.
Planar trees on $n$ vertices are in bijection with sequences $a_1$, $a_2$, \dots, $a_{n-1}$ of integers such that $a_1 =0$ and $0 \leq a_{i+1} \leq a_i+1$ (item $u$ on Stanley's list). The bijection is to consider $j$ to be in generation $a_j$, with the parent of $j$ being the largest $k$ such that $a_k = a_j-1$ and $k \lt j$. Then put a root at the top in generation $-1$.
The lists of integers in Vatter's question are a subset of lists above. Specifically, they are the ones where the root is the only crucial vertex. (Doug Zare's answer was very helpful in making me realize this description.) So our goal is to show that $\sum_q c(1,q,n)$ is Catalan. After a lot of computation, I came to the following conjecture:
The integer $c(p,q,n)$ only depends on $p+q$ and $n$. The array of integers $c(p+q, n)$ is the Catalan triangle.
I'm bad at indexing, so I'll give the values for planar trees on $10$ vertices. In total, there are $4862$ such trees in total. There are $429$ of them with $(p,q) = (1,2)$ and the same number with $(p,q) = (2,1)$. In total, the number of trees with each value of $(p,q)$ is: $$\left( \begin{array}{ccccccccc} 0 & 429 & 429 & 297 & 165 & 75 & 27 & 7 & 1 \\ 429 & 429 & 297 & 165 & 75 & 27 & 7 & 1 & 0 \\ 429 & 297 & 165 & 75 & 27 & 7 & 1 & 0 & 0 \\ 297 & 165 & 75 & 27 & 7 & 1 & 0 & 0 & 0 \\ 165 & 75 & 27 & 7 & 1 & 0 & 0 & 0 & 0 \\ 75 & 27 & 7 & 1 & 0 & 0 & 0 & 0 & 0 \\ 27 & 7 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 7 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{array} \right)$$ Note that the numbers 1, 7, 27, 75, 165, 297, 429, 429 appear as a row of the Catalan triangle. In particular, proving this proves Vatter's conjecture, since the sum $\sum_q c(1,q,n)$ would then be the sum of a row of the Catalan triangle, and hence a Catalan number.
Federico Ardilla found a matroid whose Tutte polynomial $T_n(x,y)$ obeyed $T_n(1,1) = C_n$ and $T_n(x,y) = T_n(y,x)$. I have checked numerically that $T_n(x,y)$ appears to be $\sum c(p,q,n) x^p y^q$. Federico gives an explicit generating function for his polynomial (Theorem 3.6). From this formula, one can check that the coefficient of $x^p y^q$ depends only on $p+q$. I will add that this is a very unusual property for a Tutte polynomial to have, and I would be interested to know any general property of a matroid which implies it.
Federico gives a combinatorial interpretation for the coefficient of $x^p y^q$ in $T_n(x,y)$ in terms of Dyck paths. For him, $q$ is the number of times the Dyck path returns to $0$, which is easily related to the number of children of the root. However, his $p$ is the number of up steps before the first down step. I do not see why this should be the number of crucial vertices.
What's going on here? |
Chord of a circle:
The line segment joining any two points on the circumference of the circle is known as chord of the circle. Diameter is the longest chord of circle which passes through center of the circle. The figure below depicts a circle and its chord.
In the given circle with ‘O’ as center, AB represents the diameter of the circle (longest chord) , ‘OE’ denotes radius of the circle and CD represents a chord of the circle.
Let us consider the chord CD of the circle and two points P and Q anywhere on the circumference of the circle except the chord as shown. If the end points of the chord CD are joined to the point P, then the angle ∠CPD is known as the angle subtended by the chord CD at point P. The angle ∠ CQD is the angle subtended by chord CD at Q. The angle ∠ COD is the angle subtended by chord CD at the center O.
If we try to establish a relationship between different chords and the angle subtended by them on the center, we see that the longer chord subtends a greater angle at the center. Similarly two chords of equal length subtend equal angle at the center. Let us try to prove this statement.
Theorem 1 : Chords which are equal in length subtend equal angles at the center of the circle. Theorem 1
Proof: From fig. 3, In ∆AOB and ∆POQ
S.No. Statement Reason 1. AB=PQ Chords of equal length (Given) 2. OA = OB = OP = OQ Radius of the same circle 3. \(\bigtriangleup AOB = \bigtriangleup POQ\) SSS axiom of Congruence 4. \(\angle AOB = \angle POQ\) From Statement 3
Note: CPCT stands for congruent parts of congruent triangles.
Converse of theorem 1 also holds true, which states that if two angles subtended by two chords at the center are equal then the chords are of equal length. From fig. 3, if∠AOB =∠POQ, then AB=PQ. Let us try to prove this statement.
Theorem 2:If the angles subtended by the chords of a circle are equal in measure then the length of the chords is equal.
Proof:From fig. 3, In ∆AOB and ∆POQ
S.No. Statement Reason 1. \(\angle AOB = \angle POQ\) Equal angle subtended at centre O (Given) 2. OA = OB = OP = OQ Radii of the same circle 3. \(\bigtriangleup AOB \cong \bigtriangleup POQ\) SAS axiom of Congruence 4. AB = PQ From Statement 3 (CPCT) Theorem 3: Equal chords of a circle are equidistant from the center of the circle. Given: Chords AB and CD are equal in length. Construction: Join A and C with centre O and drop perpendiculars from O to the chords AB and CD. Construction: Join A and C with centre O and drop perpendiculars from O to the chords AB and CD.
S.No Statement Reason 1 \(AP = \frac{AB}{2}\), \(CQ = \frac{CD}{2}\) Perpendicular from centre bisect the chord In \(\bigtriangleup OAP \;\; and \;\; \bigtriangleup OCQ\) 2 \(\angle 1 = \angle 2 = 90^{\circ}\) \(OP \perp AB \;\; and \;\; OQ \perp CD\) 3 \(OA = OC\) Radii of the same circle 4 \(OP = OQ\) Given 5 \(\bigtriangleup OPB \cong \bigtriangleup OQD\)< R.H.S. Axiom of Congruency 6 AP = CQ Corresponding parts of congruent triangle 7 AB = CD From statement (1) and (6)
To know more about chord, circles etc. download BYJU’S The Learning App. |
4:53 AM
i'd like to make a plug for a project i'm involved in, which i may have mentioned here before: a double conference. the first double conference is currently underway, on "higher algebra & mathematical physics" -- see here and here . there have been some really neat talks so far, and all videos are being posted to the PI page (scroll down and then click on the "videos" tab).
@JonathanBeardsley right. if all you want is a statement at the level of $\infty$-categories, then as i think has been mentioned before, all you need to do is trace through the image of the full subcategory on the objects [0] and [1] (to see if the morphisms "s" and "t" get swapped or not)
@HarryGindi i don't think the specific functor is very important, really all that matters is that it's a reedy-cofibrant replacement of the standard inclusion $\Delta \to Cat \to sCat$
@HarryGindi it's dugger--spivak, btw. and also, what are the source and target of the enriched coherent nerve?
@HarryGindi yeah that seems unlikely, in light of this discussion
2 hours later…
7:33 AM
@PraphullaKoushik Since we talked about suitable top-level tags for questions about stacks (and gerbes), I'll add that stacks are explicitly listed as one of the topics in the ag.algebraic-geometry tag-info.
However, it seems from @DenisNardin's response and @DavidRoberts' edit/retag, that your questions are probably closer to .
(Sorry for bothering you with the pings, but since this user explicitly mentioned that they were unsure which tags to use and you have interacted with them here in chat or on the main site, I thought you might have a reasonable advice.)
Of course, if this is too big digression from the topic of this room, the discussion about this can continue in MO editors' lounge - tagging is one of the main topics there.
Hey, I have two very basic questions about localizations of $\infty$-categories:
Let $C$ be a presentable $\infty$-category 1. What extra words do I need to put in to make sure that the following is precisely true: "There's a bijective correspondence between idempotent monads on $C$ and reflective localizations of $C$" (in particula I'm worried about what kind of monads am I allowed to use) 2. Suppose that $C$ is moreover pointed. Then (modulo set theoretical issues which I prefer to ignore for this point) any reflective localization of $C$ gives a coreflective localization. You take the…
Let $C$ be a presentable $\infty$-category
1. What extra words do I need to put in to make sure that the following is precisely true: "There's a bijective correspondence between idempotent monads on $C$ and reflective localizations of $C$" (in particula I'm worried about what kind of monads am I allowed to use)
2. Suppose that $C$ is moreover pointed. Then (modulo set theoretical issues which I prefer to ignore for this point) any reflective localization of $C$ gives a coreflective localization. You take the…
3 hours later…
10:27 AM
1 hour later…
11:27 AM
When C is an Eilenberg-Zilber Reedy Cat (this is equivalent to Regular Skeletal in the sense of Cisinski, which is what I am actually using) admitting a terminal object, I can equip Psh(ΘC) with an 'enriched Joyal model structure' and sPsh(C)-Cat has the enriched model structure coming from the injective model structure for simplicial presheaves
In that situation, I have sketched a proof in my private notes that the enriched HC-realization/nerve pair is a Quillen pair
I might be able to show that it's a QEQ, but I'm not that interested in showing it in the full generality.
Anyway, I'm working now on showing that this Quillen adjunction remains a Quillen adjunction when you do Bousfield localization on both sides
It's straightforward but I haven't written it up yet. I am not that interested in developing it in full generality. The case I have in mind is the case where C=Θ=Θ_ω. I also have a definition of the explicit ω-unstraightening and ω-straightening functors in this case, but I can't quite prove anything about them yet because there is some greatly-needed combinatorial work in understanding the lax join/lax slice for cellular sets combinatorially
The missing combinatorial thing is: If X is an ω-quasicat and x is a vertex of X, then the projection map
X_{/_{lax}x} -> X is an ω-cartesian fibration
X_{/_{lax}x} -> X
is an ω-cartesian fibration
Contingent on that result, ω-unstraightening sends projectively fibrant diagrams \mathfrak{C}(S)^op -> Θ-Set to ω-cartesian fibrations
12:41 PM
@JonathanBeardsley by the way, that question you asked, here's the trick: If you take (C[n])^op and reindex the objects by i mapsto n-i, you are done. That is exactly C([n]^op). Hope that helps.
7 hours later…
8:00 PM
8:38 PM
@SaalHardali The natural map X -> π_0(X) is a reflective localization on based spaces (it's the nullification of the map S^1_+ -> S^0).
The kernel of this map is the class of connected spaces, and the coreflective localization is the natural inclusion X_0 -> X of the basepoint component.
The kernel of that coreflective localization is the family of spaces X whose basepoint component is contractible, and so rather than recovering the localization π_0(X) you're recovering the localization X -> X/X_0 that collapses the basepoint component.
2 hours later…
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This is not a complete answer, but it might help.
The map $T: A_i \mapsto A_i \otimes A_i$ sends, by its very definition, the orthonormal family $(A_i)$ to an orthonormal family. It is therefore an isometry for the Hilbert-Schmidt norms.
But there are not that many completely positive maps $M_n \to M_m$ which are also isometries for the Hilbert-Schmidt-norms. Namely such a map is of the form $T(x)= D \pi(x)$, for some (not necessarily unital) $*$-homomorphism $\pi$ and some positive operator $D=T(1)\in M_m$ commuting with the range of $\pi$. This is an if-and-only-if condition provided that $\|D\|_{HS}=\sqrt n$. This statement is probably known. If you want I can expand the proof I have in mind.
This implies that such a map satisfies $Tr\circ T=c Tr$ for some positive $c=Tr(D)/n$, and more generally that for any $p>0$, $\|Tx\|_p = c_p \|x\|_p$ for $c_p=\|D\|_p/n^{1/p}$.
Coming back to your problem, I do not see how to conclude, you can already find a couple of necessary conditions on the $A_i$'s for the map $T(A_i)=A_i \otimes A_i$ to be completely positive.
Matthew asked for a proof of
A linear map $T:M_n\to M_m$ is completely positive and isometric for the Hilbert-Schmidt norm if and only if $T$ is of the form $T(x)= D \pi(x)$, for some (not necessarily unital) $*$-homomorphism $\pi$ and some positive operator $D=T(1)\in M_m$ commuting with the range of $\pi$, and such that $\|D\|_{HS}=\sqrt n$.
I only prove the "only if" direction. Assume that $T$ is cp and isometric for the Hilbert-Schmidt norm. Using the fact that $T$ is cp, by Stinespring's theorem, there is a (finite dimensional) Hilbert space $H$ and a linear map $V:\mathbb C^m \to H\otimes \mathbb C^n$ such that $T$ can be decomposed as $T(x)=V^* 1_H \otimes x V$. I claim that the assumption that $T$ is isometric implies that $VV^*$ is of the form $A \otimes 1_n$ for some positive $A \in B(H)$ (in particular $V V^*$ commutes with $1\otimes x$ for all $x \in M_n$). This will imply that $T(x) T(y) = T(1) T(xy) = T(xy) T(1)$ for all $x,y \in M_n$, and hence putting $\pi(x) = T(x) T(1)^{-1}$ (with the convention $0/0=0$) we get the proposition.
The claim is not complicated to check. By the trace property, $\langle Tx,Ty \rangle = Tr(VV^* (1\otimes x) V V^* (1\otimes y^* ))$. Writing $VV^* = \sum B_{i,j} \otimes e_{i,j}$, taking $x=e_{i,j}$, $y=e_{s,t}$, and using that $T$ preserves the scalar product, one gets $\langle e_{i,j},e_{s,t}\rangle= Tr(B_{s,i}B_{j,t})$. But $VV^*$ being self-adjoint, this becomes $\delta_{i,s}\delta_{j,t}= \langle B_{s,i},B_{t,j}\rangle$. This implies that $B_{s,i}=0$ if $s\neq i$ and that the matrices $B_{i,i}$ are all of Hilbert-Schmidt norm $1$, and that $\langle B_{i,i},B_{j,j}\rangle=1$. Thus (equality in Cauchy-Schwartz inequality), the $B_{i,i}$'s are all equal, to some matrix $U$. This proves the claim. |
To be crystal clear:
Has anyone tried to incorporate the electrons magnetic dipole moment into the atomic orbital theory?
YES. They've tried and they've succeeded. The electron's spin magnetic dipole is a standard part of atomic physics and quantum chemistry. Anybody attempting to claim that this isn't the case is simply describing their own ignorance about atomic physics rather than the subject as it is known.
Also, to be crystal clear:
The effects are perfectly well-known and have been described for the past 80+ years, but still
The effects are weak, and they are secondary to all sorts of other interactions that happen in atoms, including:
The electrostatic interaction between the electrons and the nucleus The kinetic energy of the electrons The electrostatic interaction between the electrons The Pauli exclusion principle The coupling between the (spin-induced) magnetic dipole moment of the electron and the magnetic dipole moment associated with the orbital motion of the electron (a.k.a. spin-orbit coupling) The relativistic effects associated with the kinetic energy of the electrons, particularly in inner shells of large atoms The effects caused by coupling to the QED vacuum (a.k.a. the Lamb shift) The nuclear recoil from one electron's motion affecting the other electrons The interaction between the electrons' (spin-induced) magnetic dipole moment and the magnetic dipole moment of the nucleus The strict constraints imposed by the quantum mechanics of angular momentum on how two different quantized angular momenta can relate to one another, as applied to the relationship between spins and orbital angular momenta as well as to the spins among themselves
$\ $
Essentially
everything in that list takes precedence to spin-spin interaction via the electrons' magnetic dipole moments, particularly if you're looking for effects on the shape of the spectrum instead of just tiny shifts in the energies of the different levels. "The effect exists" and "the effect is well-understood" do not lead to "the effect matters".
And also, to be crystal clear: the existing theory of atomic physics agrees
perfectly with experiment, often predicting the energies of atomic levels to eighteen significant figures (and climbing, as theoretical and experimental precision increase). If you have an alternative theory or you think that the calculations have been done incorrectly, then you need to be sure that your calculations reproduce the existing experimental data (say, this, matching to these calculations) ─ all of it, to its full eighteen significant figures where that is the experimental precision.
A quick note before moving on: full-blown atomic physics is a highly technical subject, and conciseness in technical communications here (as in every technical field) is at a very steep premium. As regards the magnetic interactions of electrons, in particular, this means that
nobody in the technical literature is going to say "intrinsic magnetic dipole moment" where saying "spin" will suffice. That's the case for electrons' intrinsic magnetic dipole moment, which is always proportional to their spin.
This means that, if you want to play in the big leagues, you need to get behind the usage of the term "spin" as synonymous with "intrinsic magnetic dipole moment", as in e.g. "spin-spin coupling" and "spin-spin interaction", the technical terms for the interactions that you repeatedly (and incorrectly) claim are not considered by the literature. If that usage of terminology bothers you, then tough luck.
Now, the final item on the list above is particularly relevant to your second question:
How one imagine the orientation of the magnetic dipoles of the electrons in each shell?
Here the answer from QM is simply: in complicated ways.
Angular momentum in QM is complicated, particularly because its components are incompatible (i.e. do not commute) with each other, which means that the direction in which the spin is pointing is
not something that QM allows its frameworks to talk about. (Yes, there are questions for which QM explicitly dictates that there are no answers available. Deal with it.)
In particular, that means that the relative orientation between any two electron spins in an atom (and therefore the relative orientation between their magnetic dipole moments) is a question that doesn't have an answer within QM. If that bothers you, go somewhere else.
To be more specific, there are two core (fatal) roadblocks in attempting to build an answer to the question of the relative orientation of the spins in an atom.
The first is that the
only well-defined quantity for the entire shell is the total spin angular momentum. This is a consequence of how the addition of angular momenta works in QM, again as a result of the incompatibility of the different components of angular momentum, and it is a completely-well-established framework that is explained in detail in any QM textbook worth its salt. (With an additional spanner in the works coming from the requirement for wavefunction antisymmetry coming from the Pauli exclusion principle, I should add.)
This means that, say, for the six electrons in the $2p$ shell in neon, the
only quantities that have well-defined values are the magnitude of the total spin, $S^2$, and one component of the total spin, normally taken to be $S_z$. Everything else (including all of the components of the spins of all the individual electrons) generally doesn't have a value.
The second roadblock is the fact that the energy scales for spin-spin interactions are so much lower with the key drivers of atomic structure (namely, the kinetic energy of the electrons, and their electrostatic interactions with the nucleus and with each other), which means that the spin and orbital sectors of the quantum state decouple. This means that the total quantum state factors out to the form$$|\Psi\rangle = |\text{spatial dependence}\rangle \otimes |\text{total spin state}\rangle,$$with a global spatial wavefunction governing the relative positions of the electrons in the atom that is completely decoupled from the spin state. Or, in other words, to the extent that each spin has an orientation (i.e. not very $-$ they are all essentially in a superposition of a wide range of orientations), this is independent of where that electron is in relation to the others.
This has important implications to the question of how big of a player the electron spin-spin magnetic interaction is in atomic structure, because the magnetic dipole-dipole interaction, which has the form$$H = K \left[\frac{\mathbf s_1\cdot \mathbf s_2}{r^3}-3\frac{(\mathbf s_1\cdot \mathbf r)(\mathbf s_2\cdot \mathbf r)}{r^5}\right]$$for $K$ a constant and $\mathbf r$ the relative position of the two spins, is sensitively dependent on the relative orientation of the spins with respect to the spins themselves. (Intuitively: two magnets whose orientations that are frozen in space might attract or repel, depending on how they're positioned relative to those orientations.) In atoms, however, the relative orientations are averaged over, as a consequence of the separation between the spatial and spin sectors of the wavefunction, which further weakens the interaction.
Here it is important to emphasize that
if the electron spin-spin interaction were strong enough to overwhelm the electrostatic components of the dynamics, then it would be conceivable that this separation would break down, and indeed it does break down in heavy atoms. This process is known as a change in angular momentum coupling scheme, from $LS$ coupling to $jj$ coupling, where each electron can have its own total angular momentum (with the individual electrons' angular momenta then further coupled into a single total atomic angular momentum), and it leaves clear traces in the spectrum. However, this is basically always driven by spin- orbit coupling, with spin-spin interaction being at best a minor contribution.
In any case, just to provide a specific reference that gives electron spin-spin interactions the full works of the attention of atomic physics, here you have one:
Mutual spin-orbit and spin-spin interactions in atomic structure calculations. M Jones.
J. Phys. B: At. Mol. Phys. 4, 1422 (1971)
There's plenty more where that came from if you know how to search the literature (and if you don't, then to be frank the material is too technical and you should be reading textbooks until you can), though it does seem that these explicit non-relativistic hamiltonians seem to have been dropped in the more recent literature on the theory of precision spectroscopy, due to the requirement of full-blown quantum field theoretic calculations.
(Also, in case you're wondering just how weak: this paper calculates the energy shifts coming from electron spin-spin coupling for a range of two-electron systems. The largest is in helium, for which the coupling energy is of the order of $\sim 7 \:\mathrm{cm}^{-1}$, or about $0.86\:\rm meV$, as compared to typical characteristic energies of $\sim 20\:\rm eV$, some five orders of magnitude higher, for that system.)
In any case, Eq. (4) in Jones 1971 details the interaction hamiltonian for the process that you asked about, together with other similar contributions in Eqs. (2) and (3), and all of these are worked out in their full glorious detail in the ensuing pages. Does it all look like one big, unenlightening mess? well: boo hoo. Of course it does. The interaction is weak, as compared to the real movers-and-shakers of atomic structure, and it is only worth studying if you're working towards precision spectroscopy with the full strength of the formalism. Most of this will be illegible if you haven't spent significant lengths of time with textbooks and doing their exercises, but that's just the nature of technical communication.
And then again, we've been telling you that there is no going around the need of reading textbooks if you want to understand the physics for
years on end, and you've roundly ignored us, so if you don't understand the primary literature, that's why. |
Computational Fluid Dynamics - CFDPython 5 May 2018
This ‘notebook’ consists of my run-through of Prof. Lorena Barba’s “12 steps to Navier-Stokes” course, because I admit I have a problem with fluids. The Navier-Stokes equations that will be solved in this notebook are the following in conservative form: $$ \frac{\partial \rho}{\partial t} + \nabla \cdot \left(\rho V\right) = 0 $$ $$ \frac{\partial (\rho\vec V)}{\partial t} + \nabla \cdot \left(\rho \,\vec V \otimes \vec V + p\hat I \right) = \rho \vec g + \nabla \cdot \hat \tau + \vec f$$ ... Read More
Meshing Techniques - Cambered Airfoil 10 July 2017
Meshing is a crucial process in obtaining accurate results for various simulations across different fields. In computational fluid dynamics, various meshing techniques are used in grid generation for 2D analyses of airfoils. Some nice run-throughs exist on YouTube, but they mostly deal with symmetric airfoils such as the beloved NACA 0012. I’ll attempt generating preliminary meshes over a cambered airfoil in this post and probably write a project once I’ve been able to perform flow analyses with accurate results. ... Read More
Derivation - Geodesic Equation 19 April 2017
Bernard F. Schutz’s A First Course in General Relativity provides a nice introduction to the difficult subject in my opinion. In Chapter 6, he mentions that one should derive the Euler-Lagrange equations to minimise the spacetime interval of a particle’s trajectory, obtaining the geodesic equation: $$ \frac{\mathrm{d}}{\mathrm{d}\lambda}\left(\frac{\mathrm{d}x^{\gamma}}{\mathrm{d}\lambda}\right) + \Gamma^{\gamma}_{\;\alpha\beta}\frac{\mathrm{d}x^{\alpha}}{\mathrm{d}\lambda}\frac{\mathrm{d}x^{\beta}}{\mathrm{d}\lambda} = 0 $$ Note: At first I derived it from variational principles, but the Euler-Lagrange equations provide a faster route via means of a neat trick. ... Read More
Observation - Indian Classical Music 17 January 2017
I’ve been a rock/metal guitarist for the past 9 or 10 years, mostly concentrating on Western music and its theory; I am by no means an expert, but I’m curious about music theory and like to learn its different applications/interpretations in various styles. I had attended a Remember Shakti concert in 2012 which was my first real introduction to Indian classical musical elements. I know that it’s a fusion band, but I learnt the basic rhythmic elements during the concert thanks to my school’s guitar teacher, with whom I went to the concert. ... Read More
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Research Open Access Published: A note on entire functions sharing a finite set with their difference operators Advances in Difference Equations volume 2019, Article number: 114 (2019) Article metrics
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Abstract
In this note, we will show that an entire function is equal to its difference operator if it has a growth property and shares a set, where the set consists of two entire functions of smaller orders. This result
generalizes a result of Li (Comput. Methods Funct. Theory 12:307–328, 2012 and partially answers Liu’s (J. Math. Anal. Appl. 359:384–393, 2009) question. Introduction and main result
A set is called a unique range set (URSE) for a certain class of entire functions if each inverse image of the set uniquely determines a function from the given class.
Let S be a finite set of some entire functions and f an entire function. Then, a set \(E(f, S)\) is defined as
Assume that
g is another entire function. We say that f and g share S CM if \(E(f, S)=E(g, S)\). Thus, a set S is called URSE if \(E(f, S)=E(g, S)\) where f and g are two entire functions; then \(f=g\).
The first example of a URSE was given by Gross and Yang [3], who considered the zero set of equation \(z + e^{z} = 0\). In view of the fact that this set has infinitely many elements, it is natural to ask whether there exists a finite unique range set or not; the question is proposed by Gross in [4]. In 1995, Yi [5] gave a straightforward answer to Gross’ question and found the URSE \(w^{n} +aw^{m} +b=0\), where \(n>2m+4\) and
a, b satisfy a certain condition. Since then, there have been many efforts to study the problem of constructing unique range sets; see e.g. [6, 7, 24]. There is another study direction on the URSE of entire functions, which is to seek a set S such that if \(E(f, S)=E(f', S)\), then \(f=f'\) for an entire function f. Li and Yang [8] deduced that if \(E(f, S)=E(f', S)\) with S consisting of two distinct constants, then f has specific forms. Later, based on the theory of the normal family, Fang and Zalcman [9] answered the question by proving that there exists a finite set S including three elements such that if \(E(f, S)=E(f', S)\), then \(f=f'\).
In recent years, the Nevanlinna characteristic of \(f(z+\omega)\), the value distribution theory for difference polynomials, the Nevanlinna theory of the difference operator and the difference analogue of the lemma on the logarithmic derivative had been established; see e.g. [2, 10, 11, 16,17,18,19,20,21,22,23]. Due to these theories, there has been recent study of whether the derivative \(f'\) can be replaced by the difference operator \(\Delta_{c}f(z)=f(z+c)-f(z)\) in the above question. In 2009, Liu [2] considered the problem and obtained the result as follows.
Theorem A Suppose that a is a nonzero complex number, and f is a transcendental entire function with finite order. If f and \(\Delta_{c}f\) share \(\{a,-a\}\) CM, then \(\Delta_{c}f(z)=f(z)\) for all \(z\in\mathbb{C}\).
Liu also proposed the question as follows [2]: Let
a and b be two small functions of f with period c. When a transcendental entire function f of finite order and its difference operator \(\Delta_{c}f\) share the set \(\{a, b\}\) CM, what can we say about the relationship between f and \(\Delta_{c}f\)?
In 2012, Li [1] considered the problem and proved the following.
Theorem B Suppose that a, b are two distinct entire functions, and f is a nonconstant entire function with \(\rho(f)\neq1\) and \(\lambda(f)<\rho(f)<\infty\) such that \(\rho(a)<\rho (f)\) and \(\rho(b)<\rho(f)\). If f and \(\Delta_{c}f\) share \(\{a,b\}\) CM, then \(f(z)=\Delta_{c}f(z)\) for all \(z\in\mathbb{C}\).
Here, the order \(\rho(f)\) is defined by
and the exponent of convergence of zeros \(\lambda(f)\) is defined by
In Theorem B, the condition \(\rho(f)\neq1\) seems not natural. So one may ask whether it can be removed or not. In this paper, we consider the question and show that the theorem still holds without the condition \(\rho(f)\neq1\). More precisely, we give the specific form of
f in a simple way. Main theorem Suppose that a, b are two distinct entire functions, and f is a nonconstant entire function of finite order with \(\lambda(f)<\rho(f)<\infty\) such that \(\rho(a)<\rho(f)\) and \(\rho(b)<\rho(f)\). If f and \(\Delta_{c}f\) share \(\{a,b\}\) CM, then \(f(z)=Ae^{\mu z}\), where A, μ are two nonzero constants satisfying \(e^{\mu c}=2\). Furthermore, \(f(z)=\Delta_{c}f(z)\). Using the almost same method, we generalize the above result from entire functions to meromorphic functions after the above theorem. Due to the use of almost the same method, we omit the detailed proof. Corollary Suppose that a, b are two distinct entire functions, and f is a nonconstant finite order meromorphic function which has finite poles and with \(\lambda(f)<\rho(f)<\infty\) such that \(\rho(a)<\rho(f)\) and \(\rho(b)<\rho(f)\). If f and \(\Delta_{c}f\) share \(\{a,b\}\) CM, then \(f(z)=Ae^{\mu z}\), where A, μ are two nonzero constants satisfying \(e^{\mu c}=2\). Furthermore, \(f(z)=\Delta_{c}f(z)\).
A similar result can be found in [11]. Before we proceed, we suppose that the reader is familiar with Nevanlinna theory, for example, the
first and second main theorems, and the common notations such as \(T(r,f)\), \(m(r,f)\) and \(N(r,f)\). \(S(r,f)\) denotes any quantity which satisfies \(S(r,f)=o (T(r,f) )\) as \(r\to\infty\), except possibly on a set of finite logarithmic measure; see e.g. [12,13,14]. Proof of Main theorem
We will prove our theorem in this section. Before we turn to its proof, we first give the following results, where the first one is Corollary 2.6 of Chiang and Feng in [10], and the second one is Lemma 3.3 of Bergweiler and Langley in [15].
Lemma 2.1 Let f be a meromorphic function of finite order and let \(\omega_{1}\), \(\omega_{2}\) be two arbitrary complex numbers such that \(\omega_{1}\neq\omega_{2}\). Assume that σ is the order of f, then for each \(\epsilon>0\), we have Lemma 2.2 Let g be a function transcendental and meromorphic in the plane of order less than 1. Set \(h>0\). Then there exists an ϵ- set E such that uniformly in ω for \(|\omega|\leq h\). Proof of main theorem
Note that
f and \(\Delta_{c} f\) share \(\{a, b\}\) CM. So we can set
in which
Q is an entire function. Furthermore, it follows from (2.1) and \(\max\{\rho(a),\rho(b)\}<\rho(f)<\infty\) that Q is a polynomial.
By the Hadamard factorization theorem, we suppose that \(f(z)=h(z)e^{P(z)}\), where
h (≢0) is an entire function and P is a polynomial satisfying
Then
We substitute the forms of
f and \(\Delta_{c} f\) into (2.1) to find
Set \(w_{1}=(h(z+c)e^{P(z+c)-P(z)}-h(z))\). Suppose that \(w_{1}\equiv0\). Then,
Note that \(a\not\equiv b\). Without loss of generality, we suppose that \(a\not\equiv0\). Assume that \(z_{0}\) is a zero of \(e^{P}-\frac{a}{h}\), but not a zero of \(w_{1}\).
It follows from (2.3) and the assumption about sharing that \(z_{0}\) is a zero of \(e^{P}-\frac{a}{w_{1}}\) or \(e^{P}-\frac{b}{w_{1}}\). We denote by \(N_{1}(r,e^{P})\) the reduced counting function of those common zeros of \(e^{P}-\frac{a}{h}\) and \(e^{P(z)}-\frac{a}{w_{1}}\). Similarly, we denote by \(N_{2}(r,e^{P})\) the reduced counting function of those common zeros of \(e^{P}-\frac{a}{h}\) and \(e^{P}-\frac{b}{w_{1}}\). Note that h is a small function respect to \(e^{P}\) ; applying the second fundamental theorem to \(e^{P}\) and the first fundamental theorem to \(e^{P}-\frac{a}{h}\) yields
which implies that either \(N_{1}(r,e^{P})\neq S(r,e^{P})\) or \(N_{2}(r,e^{P})\neq S(r,e^{P})\). We consider the following two cases.
Case 1. \(N_{1}(r,e^{P})\neq S(r,e^{P})\).
Let \(a_{0}\) be a common zero of \(e^{P}-\frac{a}{h}\) and \(e^{P}-\frac{a}{w_{1}}\). Then it is clear that \(a_{0}\) is a zero of \(\frac{a}{h}-\frac{a}{w_{1}}\). If \(\frac{a}{h}-\frac{a}{w_{1}}\not\equiv0\), then
a contradiction. Thus
It leads to
Then, by Lemma 2.1, one has, for any \(\epsilon>0\),
On the other hand, \(m(r,e^{P(z+c)-P(z)})=[A+o(1)]r^{\rho(f)-1}\), where
A is a fixed positive constant.
If \(\rho(f)>1\), by \(\rho(f)>\rho(h)\) and the above estimates of \(m(r,e^{P(z+c)-P(z)})\), we can easily get a contradiction. So, \(\rho(f)\leq1\), which means that \(e^{P(z+c)-P(z)}\) is a nonzero constant, say \(c_{0}\). Then (2.6) reduces to
Note that \(1\geq\rho(f)>\rho(h)\). Then by Lemma 2.2, we know that there exists an
ϵ-set E, as \(z\notin E\) and \(|z|\rightarrow\infty\) such that
So, \(c_{0}=2\) and \(h(z)=h(z+c)\), which means that
h is a periodic function. If h is a nonconstant function, then \(\rho(h)\geq1\), a contradiction. Therefore, h is a constant. Note that \(\deg(P)=\rho(f)\leq1\) and f is a nonconstant entire function. Then \(\deg(P)=1\). Thus, we can set \(f=Ae^{\mu z}\), where A, μ are two nonzero constants.
By the assumption of Case 1,
one sees that \(f-a\) and \(\Delta_{c} f-a\) have common zeros, which are not zeros of a. Assume that \(\alpha_{0}\) is a common zero of \(f-a\) and \(\Delta_{c} f-a\), and not a zero of a. Then \(z_{0}\) is a zero of \(f(z+c)-2a(z)\). Furthermore,
which implies that \(e^{\mu c}=2\). Thus, we get \(\Delta_{c} f=f\), which is the desired result.
Case 2. \(N_{2}(r,e^{P})\neq S(r,e^{P})\).
Let \(b_{0}\) be a common zero of \(e^{P}-\frac{a}{h}\) and \(e^{P}-\frac{b}{w_{1}}\). Then it is clear that \(b_{0}\) is a zero of \(\frac{a}{h}-\frac{b}{w_{1}}\). If \(\frac{a}{h}-\frac{b}{w_{1}}\not\equiv0\), then
a contradiction. Thus
If \(b\equiv0\), then \(\frac{a}{h}\equiv0\), a contradiction. Thus, \(b\not\equiv0\).
We assume that \(c_{0}\) is a zero of \(e^{P}-\frac{b}{h}\), but not a zero of \(w_{1}\). It follows from (2.3) that \(c_{0}\) is a zero of \(e^{P}-\frac{a}{w_{1}}\) or \(e^{P}-\frac{b}{w_{1}}\). We denote by \(N_{3}(r,e^{P})\) the reduced counting function of those common zeros of \(e^{P}-\frac{b}{h}\) and \(e^{P}-\frac{a}{w_{1}}\). Similarly, we denote by \(N_{4}(r,e^{P})\) the reduced counting function of those common zeros of \(e^{P}-\frac{b}{h}\) and \(e^{P}-\frac{b}{w_{1}}\). Then
which implies that either \(N_{3}(r,e^{P})\neq S(r,e^{P})\) or \(N_{4}(r,e^{P})\neq S(r,e^{P})\). If \(N_{4}(r,e^{P})\neq S(r,e^{P})\), similar to Case 1, we get the desired result. So, we assume that \(N_{3}(r,e^{P})\neq S(r,e^{P})\) below. Similar to Case 2, we can deduce that
Note that \(a\not\equiv b\). Thus, \(a\equiv-b\). Again by (2.9), one has \(w_{1}=-h\). We can rewrite it as
a contradiction.
Therefore, the proof of the main theorem is finished. □
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170, 33–46 (2003) Acknowledgements
The authors wish to thank the editors and referees for their great useful suggestions and helpful comments.
Funding
This work was supported by Applied Mathematical Academic Discipline Project of Shanghai Dianji University (16JCXK02), and Humanity and Social Science Youth foundation of Ministry of Education(18YJC630120) and NNSF of China Project (11601521), and the Fundamental Research Fund for Central Universities in China Project (15CX05061A, 15CX05063A, 15CX08011A).
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The authors declare that they have no competing interests.
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It would seem that one way of proving this would be to show the existence of non-algebraic numbers. Is there a simpler way to show this?
As Steve D. noted, a finite dimensional vector space over a countable field is necessarily countable: if $v_1,\ldots,v_n$ is a basis, then every vector in $V$ can be written uniquely as $\alpha_1 v_1+\cdots+\alpha_n v_n$ for some scalars $\alpha_1,\ldots,\alpha_n\in F$, so the cardinality of the set of all vectors is exactly $|F|^n$. If $F$ is countable, then this is countable. Since $\mathbb{R}$ is uncountable and $\mathbb{Q}$ is countable, $\mathbb{R}$ cannot be finite dimensional over $\mathbb{Q}$. (Whether it has a basis or not depends on your set theory).
Your further question in the comments, whether a vector space over $\mathbb{Q}$ is finite dimensional
if and only if the set of vectors is countable, has a negative answer. If the vector space is finite dimensional, then it is a countable set; but there are infinite-dimensional vector spaces over $\mathbb{Q}$ that are countable as sets. The simplest example is $\mathbb{Q}[x]$, the vector space of all polynomials with coefficients in $\mathbb{Q}$, which is a countable set, and has dimension $\aleph_0$, with basis $\{1,x,x^2,\ldots,x^n,\ldots\}$. Added: Of course, if $V$ is a vector space over $\mathbb{Q}$, then it has countable dimension (finite or denumerable infinite) if and only if $V$ is countable as a set. So the counting argument in fact shows that not only is $\mathbb{R}$ infinite dimensional over $\mathbb{Q}$, but that (if you are working in an appropriate set theory) it is uncountably-dimensional over $\mathbb{Q}$.
The cardinality argument mentioned by Arturo is probably the simplest. Here is an alternative: an explicit example of an infinite $\, \mathbb Q$-independent set of reals. Consider the set consisting of the logs of all primes $\, p_i.\,$ If $ \, c_1 \log p_1 +\,\cdots\, + c_n\log p_n =\, 0,\ c_i\in\mathbb Q,\,$ multiplying by a common denominator we can assume that all $\ c_i \in \mathbb Z\,$ so, exponentiating, we obtain $\, p_1^{\large c_1}\cdots p_n^{\large c_n}\! = 1\,\Rightarrow\ c_i = 0\,$ for all $\,i,\,$ by the uniqueness of prime factorizations.
No transcendental numbers are needed for this question. Any set of algebraic numbers of unbounded degree spans a vector space of infinite dimension. Explicit examples of linearly independent sets of algebraic numbers are also relatively easy to write down.
The set $\sqrt{2}, \sqrt{\sqrt{2}}, \dots, = \bigcup_{n>0} 2^{2^{-n}} $ is linearly independent over $\mathbb Q$. (Proof: Any expression of the $n$th iterated square root $a_n$ as a linear combination of earlier terms $a_i, i < n$ of the sequence could also be read as a rational polynomial of degree dividing $2^{n-1}$ with $a_n$ as a root and this contradicts the irreducibility of $X^m - 2$, here with $m=2^n$).
The square roots of the prime numbers are linearly independent over $\mathbb Q$. (Proof: this is immediate given the ability to extend the function "number of powers of $p$ dividing $x$" from the rational numbers to algebraic numbers. $\sqrt{p}$ is "divisible by $p^{1/2}$" while any finite linear combination of square roots of other primes is divisible by an integer power of $p$, i.e., is contained in an extension of $\mathbb Q$ unramified at $p$).
Generally any infinite set of algebraic numbers that you can easily write down and is not dependent for trivial reasons usually is independent. This because the only algebraic numbers for which we have a simple notation are fractional powers, and valuation (order of divisibility) arguments work well in this case. Any set of algebraic numbers where, of the ones ramified at any prime $p$, the amount of ramification is different for different elements of the set, will be linearly independent. (Proof: take the most ramified element in a given linear combination, express it in terms of the others, and compare valuations.)
For the sake of completeness, I'm adding a worked-out solution due to F.G. Dorais from his post.
We'll need two propositions from Grillet's
Abstract Algebra, page 335 and 640:
Proposition: $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$ Proof: Let $(q_n)_{n\in\mathbb{N_0}}$ be an enumeration of $\mathbb{Q}$. For $r\in\mathbb{R}$, take $$A_r:=\sum_{q_n<r}\frac{1}{n!}\;\;\;\;\text{ and }\;\;\;\;A:=\{A_r;\,r\in\mathbb{R}\};$$ the series is convergent because $\sum_{q_n<r}\frac{1}{n!}\leq\sum_{n=0}^\infty\frac{1}{n!}=\exp(1)<\infty$ (recall that $\exp(x)=\sum_{n=0}^\infty\frac{x^n}{n!}$ for any $x\in\mathbb{R}$).
To prove $|A|=|\mathbb{R}|$, assume $A_r=A_{s}$ and $r\neq s$. Without loss of generality $r<s$, hence $A_s=\sum_{q_n<s}\frac{1}{n!}=\sum_{q_n<r}\frac{1}{n!}+\sum_{r\leq q_n<s}\frac{1}{n!}=A_r+\sum_{r\leq q_n<s}\frac{1}{n!}$, so $\sum_{r\leq q_n<s}\frac{1}{n!}=0$, which is a contradiction, because each interval $(r,s)$ contains a rational number.
To prove $A$ is $\mathbb{Q}$-independent, assume $\alpha_1A_{r_1}+\cdots+\alpha_kA_{r_k}=0\;(1)$ with $\alpha_i\in\mathbb{Q}$. We can assume $r_1>\cdots>r_k$ (otherwise rearrange the summands) and $\alpha_i\in\mathbb{Z}$ (otherwise multiply by the common denominator). Choose $n$ large enough that $r_1>q_n>r_2\;(2)$; we'll increase $n$ two more times. The equality $n!\cdot(1)$ reads $n!(\alpha_1\sum_{q_m<r_1}\frac{1}{m!}+\cdots+\alpha_k\sum_{q_m<r_k}\frac{1}{m!})=0$. Rearranged (via $(2)$ when $m=n$), it reads
$$-\alpha_1\sum_{\substack{m<n\\q_m<r_1}}\frac{n!}{m!}-\cdots-\alpha_k\sum_{\substack{m<n\\q_m<r_k}}\frac{n!}{m!}-\alpha_1 =\alpha_1\sum_{\substack{m>n\\q_m<r_1}}\frac{n!}{m!}+\cdots+\alpha_k\sum_{\substack{m>n\\q_m<r_k}}\frac{n!}{m!}. \tag*{(3)}$$
The left hand side (LHS) of $(3)$ is an integer for any $n$. If $n$ is large enough that $(|\alpha_1|+\cdots+|\alpha_k|)\sum_{m=n+1}^\infty\frac{n!}{m!}<1$ holds (such $n$ can be found since $\sum_{m=n+1}^\infty\frac{n!}{m!}=\frac{1}{n+1}\sum_{m=n+1}^\infty\frac{1}{(n+2)\cdot\ldots\cdot m}\leq\frac{1}{n+1}\sum_{m=n+1}^\infty\frac{1}{(m-n-1)!}\leq\frac{1}{n+1}\exp(1)\rightarrow 0$ when $n\rightarrow\infty$), then the absolute value of RHS of $(3)$ is $<1$, and yet an integer, hence $\text{RHS}(3)=0$. Thus $(3)$ reads $\alpha_1=-\sum_{i=1}^{k}\sum_{m<n,q_m<r_i}\alpha_i\frac{n!}{m!}=0\;(\mathrm{mod}\,n)$. If moreover $n>|\alpha_1|$, this means that $\alpha_1=0$. Repeat this argument to conclude that also $\alpha_2=\cdots=\alpha_k=0$.
Since $A$ is a $\mathbb{Q}$-independent subset, by proposition 5.3 there exists a basis $B$ of $\mathbb{R}$ that contains $A$. Then $A\subseteq B\subseteq\mathbb{R}$ and $|A|=|\mathbb{R}|$ and Cantor-Bernstein theorem imply $|B|=|\mathbb{R}|$, therefore $[\mathbb{R}:\mathbb{Q}]=\mathrm{dim}_\mathbb{Q}{}\mathbb{R}=|\mathbb{R}|$. $\quad\blacksquare$
Here is a simple proof that a basis $B$ of $[ \mathbb{R}:\mathbb{Q}]$ has cardinality $|\mathbb{R}|$.
Clearly since $B$ is contained in $R$, $|B| \le| \mathbb{R}|$.
But also $\mathbb{R}=span(B)$ and thus $|\mathbb{R}| =|span(B)| \le |\mathbb{Q}^B|=|B|$. The last equality follows because $B$ is not finite (if it was, then $|\mathbb{R}|=|span(B)| \le |\mathbb{Q}^B|=|\mathbb{N}|$, a contradiction).
Hence $ |\mathbb{R}| \le |B| \le| \mathbb{R}|$, so $|\mathbb{R}|=|B|$
Another simple proof:
Take $P=X^n-p$ for a prime $p$.
By Eisenstein's criterion, it is $\mathbb Q[X]$-irreductible. Therefore, the set of algebraic numbers is of infinite dimension over $\mathbb Q$.
Since $\mathbb R$ is bigger, it works for $\mathbb R$ too.
protected by user26857 Nov 25 '15 at 9:30
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Two objects A and B, of masses 5 kg and 20 kg respectively, are connected by a massless string passing over a frictionless pulley at the top of an inclined plane, as shown in the figure. The coefficient of static friction is mu_s = 0.4 between all surfaces (a) To what angle $\theta$ must the plane be inclined for sliding to commence? (b) What is the tension in the rope, and what are the magnitudes of the friction forces at this critical inclination? (c) At an incline angle of 15$^\circ$, what is the tension in the rope? (d) At an incline angle of 35$^\circ$, what is the tension in the rope?
I was able to solve (a) and (b) by drawing a free body diagram as shown:
Newton's Second Law, setting all accelerations to zero, implies the following relationships:
$N_A = m_A g \cos(\theta)$
$T = m_A g \sin(\theta) + f_2$
$N_B = (m_B + m_A) g \cos(\theta)$
$T + f_1 + f_2 = m_B g \sin(\theta)$
The second equation can be substituted into the fourth equation to give \begin{equation} f_1 + 2f_2 = (m_B - m_A) g \sin(\theta) \;\;\;\; (1) \label{eq:1} \end{equation}
Setting the friction forces to their maximum values $f_{1,{\rm max}} = \mu_s N_B = \mu_s (m_B + m_A) g \cos(\theta)$ and $f_{2,{\rm max}} = \mu_s N_A = \mu_s m_A g \cos(\theta)$ allows these equations to be solved for $\theta = 43^\circ$, $f_2 = 14.33$ N, $f_1 = 71.64$ N and $T = 47.76$ N.
I am however a bit confused about parts (c) and (d) dealing with angles below 43$^\circ$.
I have 5 unknowns: tension, two normal forces and two friction forces, but only four constraints from Newton's Second Law. Equivalently, referring to Eq. (1), the net applied force which the static friction must oppose is fixed, so $f_1+2f_2$ is known, but there is no additional constraint to determine how much $f_1$ opposes and how much $f_2$ opposes. There seems to be a degree of freedom in how $f_1$ and $f_2$ are determined, i.e. a free parameter.
My attempt so far at this is to suppose that for very small inclines we might expect friction to hold the blocks stationary and thus the rope would be slack and the tension $T$ is eliminated from the equations. In this case the free body diagram would be drawn differently, because without the rope, the tendency is for block A to slide downwards across block B:
The maximum incline under these conditions is found by balancing forces:
$m_A g \sin(\theta) = f_2 \le \mu_s m_A g \cos(\theta)$
$\Rightarrow \tan(\theta) \le \mu = 0.4$ $\Rightarrow \theta \le 21.8^\circ$
When the incline is increased above $21.8^\circ$, I am confused about what will happen.The blocks would slide down the slope, but the rope will tighten, and suddenly the tendency for the system would be for the heavier block B to accelerate down the slope, and the lighter block A to accelerate
up the slope (because block B pulls on it via the rope), resulting in a free body diagram as in my original figure.I still do not understand how to calculate tension and the two friction forces in this case.
How do I determine the tension $T$ and friction forces $f_1$ and $f_2$ for incline angles between $21.8^\circ$ and $43^\circ$? For these inclines, there does not seem to be enough constraints to determine every quantity, see for example Eq. (1). Is there an additional constraint I haven't thought of, or have I perhaps drawn my free body diagram incorrectly? |
There are several ways to proceed depending on how much machinery you are willing to use.
$\ell_\infty$ is a von Neumann algebra and thus, by Sakai theorem, it has a unique predual. Nevertheless $M([0,1])$ has plenty of nonisomorphic preduals, for instance $M([0,1])$ and $M([0,1]^2)$ are isomorphic while $C([0,1]^2)$ and $C([0,1])$ are not.
In the particular case of $\ell_\infty$ you can see that it has a unique predual directly. Use that $E^\ast = X_1 \oplus_\infty X_2$ implies that $E = E_1 \oplus_1 E_2$ with $E_1$ and $E_2$ the preduals of $X_1$ and $X_2$ respectively. Iterating will give you that $E$ satisfying that $E^\ast = \ell_\infty$ has a nested family $\ell^1(1) \subset \ell_1(2) \subset ... \ell_1(N)$ dense in $E$. Therefore $E \cong \ell_1$.
There other strategies, what I sugested in the comments was using injectivity.
A Banach space $E$ is called injective iff it satisfies a Hanh-Banach theorem if you use as an endpoint space. I.e.: for every $Y \subset X$ and bounded map $\phi: Y \to E$ there is a bounded extension $\tilde{\phi}: X \to E$ with the same norm.
Observation $\ell_\infty$ is trivially injective because we can apply Hanh-Banach in each of the coordinates.
I think, although I do not have a reference at hand, that all injective Banach spaces are $1$-complemented subsets of spaces of the form $C(K)$, where $K$ is totally disconnected.
It is easy to check that a Banach space is injective iff whenever it sits inside a larger space, it is complemented. Therefore it is enough to see that $M([0,1])$ sits as a subset of a larger space in a way that is not complemented. Probably there are examples of this in the literature, but I cannnot came up with one. The closest thing I can think of is:
Use that $M([0,1]) \cong \mathbb{C} \oplus_1 \mathbb{C} \oplus_1 M(0,1)$, to reduce the problem to the open interval. Since $(0,1)$ is homeomorphic to $\mathbb{R}$, $M(0,1)$ is isomorphic to $M(\mathbb{R})$. Take $M(\mathbb{R}) \subset L^1(\mathbb{R})^{\ast \ast}$, the double dual of $L^1$. Assume that there is a projection $P: L^1(\mathbb{R})^{\ast \ast} \to M(\mathbb{R})$ to reach contradiction. The unit ball $B$ of $B(L^1(\mathbb{R})^{\ast \ast}, M(\mathbb{R}))$ is comapct in the point weak-$\ast$ topology given by the predual $M(\mathbb{R}) = C_b(\mathbb{R})/n$, where $n$ is the preannihilator of $M(\mathbb{R})$. The projections are a closed convex subset $B_0 \subset B$. There is an action of $\mathbb{R}$ over $B_0$ given by $t \mapsto \tau_t \circ P \circ \tau_{-t}$ where $\tau_t$ is the translation. By amenability, we have a fixed point $P_0 \in B_0$ which would be an equivariant projection (a projection commuting with the translation action). If the original $P$ would preserve $1_{\mathbb{R}}$, i.e.$$\langle P(\psi),1_{\mathbb{R}} \rangle = \langle \psi, 1_{\mathbb{R}} \rangle$$ we would get a contradiction right away since an invariant mean $m \in L^1(\mathbb{R})^{\ast \ast}$ would give a $\mathbb{R}$-invariant and nonzero finite measure $P(m)$. I do not know whether you can take $P$ preserving $1$ without loss of generality. |
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1 0
Hi folks!
I'm trying to (numerically) find a steady-state solution for [tex]N_b[/tex] and [tex]N_w[/tex] in the following set of coupled DEs using the software package Matlab: [tex] \left{ \begin{array}{l} \frac{\delta N_b}{\delta t} = P_b(N_b) - N_b \cdot \left( \frac{1}{\tau_b} - \frac{1}{\tau_c}D \right)\\ \frac{\delta N_w}{\delta t} = \frac{N_b}{\tau_c} - \frac{N_w}{\tau_w(N_w)} - P_w(N_w) \end{array} \right. [/tex] where [tex]\tau_b[/tex], [tex]\tau_c[/tex] and [tex]D[/tex] are constants. Which way would be the right one to go?
I'm trying to (numerically) find a steady-state solution for [tex]N_b[/tex] and [tex]N_w[/tex] in the following set of coupled DEs using the software package Matlab:
[tex]
\left{
\begin{array}{l}
\frac{\delta N_b}{\delta t} = P_b(N_b) - N_b \cdot \left( \frac{1}{\tau_b} - \frac{1}{\tau_c}D \right)\\
\frac{\delta N_w}{\delta t} = \frac{N_b}{\tau_c} - \frac{N_w}{\tau_w(N_w)} - P_w(N_w)
\end{array}
\right.
[/tex]
where [tex]\tau_b[/tex], [tex]\tau_c[/tex] and [tex]D[/tex] are constants. Which way would be the right one to go?
Last edited: |
I'm reading the AKS primality test paper as it is found here. I'm confused about a statement in Lemma 4.3:
"Note that $(r, n)$ cannot be divisible by all the prime divisors of $r$ since otherwise $r$ will divide $n^{\left \lfloor{log(B)}\right \rfloor}$ by the observation above."
Here $(r, n)$ is short-hand for $gcd(r, n)$ (I think) and the observation referred to is "the largest value of k for any number of the form $m^k ≤ B = \left \lceil{log^5(n)}\right \rceil$, $m ≥ 2$, is $\left \lfloor{log(B)}\right \rfloor$".
How does the observation above imply that $(r, n)$ cannot be divisible by all prime divisors of $r$?
The furthest I can get to understanding this is follows:
Let $s= \prod_{p\in PRIMES, p | r}p$.
Assume $s | (r,n) \Rightarrow s | n $. Let $k$ be the highest exponent in the product-of-primes representation of $r$. Then $r| s^k$. But then $r|s^k|n^k$. But if we've defined $r$ such that $r \nmid n^k$ then we must conclude that our initial assumption, $s|(r,n)$, was wrong. This logic makes sense.
But the paper is claiming this $k$ is at most $\left \lfloor{log(B)}\right \rfloor$. I get if $r \leq B$, then the highest exponent in the product of primes representation of r is less that $\log(r)$, and therefore is less than $\log(B)$. But we cannot assume $r \leq B$ as that is one of the things the lemma is trying to prove.
Am I missing something? |
Bonam, Satish and Kumar, C Hemanth and Vanjari, Siva Rama Krishna and Singh, Shiv Govind (2018)
Gold Passivated Cu-Cu Bonding At 140°C For 3D IC Packaging And Heterogeneous Integration Applications. In: IEEE 20th Electronics Packaging Technology Conference (EPTC), 4-7 December 2018, Singapore.
Full text not available from this repository. (Request a copy
)
Abstract
In the present modern era of electronic industry has motivated for high performance integration by vertically stacked three dimensional integrated circuits (3D ICs). Electronic interconnections at packaging and die levels, Pbfree solder micro bumps are intended to replace conventional Pb-containing solder joints due to increasing awareness of an environmental conservation, and processing at low thermal budgets. The better alternative for solder is copper, due to its high electrical and thermal properties. But the surface oxidation was the major bottleneck. In this work, we have demonstrated low temperature and low-pressure copper to copper interconnect bonding using optimized thin gold passivation layer. Here the passivation layer over the copper surface was optimized to a thickness of 3nm there by helps in preventing Cu surface oxidation and makes lower surface RMS roughness. High-density surface plane orientations that have been studied using XRD helped in faster diffusion through an interface. Majorly in this work, we have discussed the time taken for copper atoms to diffuse over the ultra-thin passivation layer of gold using Fick's second law approximation. These conditions have been used while bonding. Bonded samples were subjected to various reliability studies in order to confirm the efficacy of the proposed Au passivation based bonded structure. Also, we have observed the Interface quality using TEM, and C-SAM (mode C-Scanning acoustic microscopy) imaging resulting in good quality of bonding. The diffusion of copper atomic species movement across the interface is confirmed by EDS analysis. Low and stable specific contact resistance ($\sim$1.43 $\times$ 10-8 $\Omega$ cm 2 ) at robust conditions are confirmed to be effective and front runner for low temperature, low pressure Cu-Cu bonding for 3D IC packaging and heterogeneous integration.
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The first follows from simple matrix multiplication, $x_\mu = \eta_{\mu\nu}x^\nu$.
As for the difference between $\eta^{\mu\nu}$ and $\eta_{\mu\nu}$, you should remember that one of these two objects is defined as the inverse of the other. Hence there is no reason to suspect that in general $\eta^{\mathfrak {ab}} = \eta_{\mathfrak{ab}}$. Certainly this relation is false for ordinary matrices, $A_{\mathfrak i \mathfrak j} \neq A^{-1}_{\mathfrak{ij}}$. Here indices in fraktur style refer to actual components with respect to some coordinates while upright indices just indicate the type of a tensor.
In general the components $\eta^{\mathfrak{ab}}$ are functions on spacetime! However in special relativity where the metric is flat, it is possible to find coordinate systems such that they are all constant and the metric is also diagonal with entries $\pm 1$ - these are the coordinate systems associated with intertial frames. (It is possible to pass from inertial
frames to intertial coordinates if and only if the metric is flat.) In these coordinate systems indeed $\eta^{\mathfrak {ab}} = \eta_{\mathfrak{ab}}$, but this is only true in very special coordinate systems (and false in general relativity except for the special case of, well, special relativity) and in general $\eta^{\mu\nu}$ and $\eta_{\mu\nu}$ must be distinguished -- they are different objects. |
I'm looking for a mathematical model of resistance in valve. Some relation between drop of pressure $\Delta p$ and volumetric flow rate $Q$.
I got this tip from some colleague to use formula for resistance of tube written below:
\begin{equation} R =\frac{8\mu l}{\pi r^4} \end{equation} where $\mu$ is viscosity of water ($\mu = 10^{-3}\,\mathrm{N\,s\,m^{-2}}$), $l$ is length of the tube and $r$ is its radius. So the relation would be $Q = \frac{\Delta p}{R}$. The problem with this approach is, that it would mean linear dependence going from 0 to some value, as showen below.
However, the data I'm getting from the real system doesn't look like this is what's going on. See the graph below. At time $\mathrm{143\,s}$ the valve is completely opened.
It's a level of a water in tank, which has a tube coming from the bottom. There is the valve I want to model. So the drop of pressure is actually pressure at the bottom oh the tank, to be clear. The way I see it, the level is steadily decreasing with the time for all pressures. The speed of level falling does not decrease with the lower pressure in the tank.
This graph below is data from real system. There is something unrelated going on until the time about 400 s (filling the tank). Around that time the ventil opens. The green line is what it would look like, if I use the constant resistance approach. Doesn't look right.
ADDED:
In this full experiment there is one other valve letting the water out from the tank. That one is already modeled well and cause exponential look of the line. I wanted to show here, that the constant resistance approach doesn't fit the data.
I have gathered some data from sensor that measures flow through the tube. Nobody has any idea what units the sensor gives, but it should be in some linear relation to the $\mathrm{m^3s^{-1}}$. Something like $\mathrm{m^3s^{-1}} = 2\cdot10^{-5} \cdot[sensor data]$. The valve is proportional, so the data is given for few setups of valve (opened on 25 % to 100 %)
It seems the relation will be more complicated. So how can I express the relation between pressure and flow rate in valve?
What is physically happening there? I had an idea, that there is some limit to the flow rate through the ventil. Maybe no metter how much pressure there is, the flow cannot go higher. But others told me that was wrong idea and that doesn't happen in reality. Teacher told me it would ruin the valve, if the pressure was too high. I don't have an experience with that. Is that true?
Any idea how to obtain the information needed for successful model of the valve? |
I am currently reading a derivation of Rayleigh-Jeans law for cavity radiation from Eisberg and Resnick
1 . The authors derive the law by considering a cavity with metallic walls. In the book, the authors state
Now, since electromagnetic radiation is a transverse vibration with the electric field vector $\mathbf{E}$ perpendicular to the propagation direction, and since the propagation direction for this component is perpendicular to the wall in question, its electric field vector $\mathbf{E}$ is parallel to the wall. A metallic wall cannot, however, support an electric field parallel to the surface, since charges can always flow in such a way as to neutralize the electric field. Therefore, $\mathbf{E}$ for this component must always be zero at the wall. That is, the standing wave associated with the $x$-component of the radiation must have a node (zero amplitude) at $x = 0$.
But blackbodies need not be made of metallic walls. The walls can be insulating as well which can support a non-zero electric field. Then how does Rayleigh-Jeans law hold for non-metallic blackbodies (
eg. stars)?
Also, the authors consider a sinusoidal wave-function for the electric field standing wave in the cavity.
The electric field for one-dimensional electromagnetic standing waves can be described mathematically by the function $$E\,(x, t) = E_0 \sin {(2 \pi x / \lambda)} \sin {(2 \pi v t)} \tag{1-6}$$ where $\lambda$ is the wavelength of the wave, $v$ is its frequency, and $E_0$ is its maximum amplitude.
Why do we need that the time and space dependence of the electric field to be sinusoidal? Shouldn't any standing wave which has nodes at the same points suffice?
Since any wave can be described as a sum of sinusoidal components using Fourier analysis, why don't we write the electric field as such and the proceed further?
Reference:
Eisberg, R.; Resnick, R. Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles, 2nd ed.; Wiley: Hoboken, NJ, 1985, pp. 6-12. |
I have encountered a problem in my textbook 'The Bayesian Choice' by Christian P. Robert. It goes something like this:
$"$For a particular case of AR(1) model, $(x_t)_{1\leq t\leq T}$. Where $x_t = \rho x_{t-1}+\epsilon_t$, $\epsilon_t$ i.i.d. $\mathcal{N}(0,\sigma^2)$. Given a sequence of observations till time $T-1$, $x_{1:(T-1)} = (x_1,\dots,x_{(T-1)})$.
The predictive distribution of $x_T$ is then $$x_T\mid x_{1:(T-1)}\sim \int\dfrac{1}{\sqrt{2\pi}}\sigma^{-1}\exp(-(x_T-\rho x_{(T-1)})^2/2\sigma^2)\pi(\rho,\sigma\mid x_{1:(T-1)})d\rho d\sigma.$$
Show that, in this scenario, the joint distribution $\pi(\rho,\sigma^2\mid x_{1:(T-1)})$ is explicit under the conjugate prior, $$\rho\sim\mathcal{N}(0,k\sigma^2),\quad \sigma^2\sim\mathcal{IG}(\alpha,\beta)."$$
Now my problem here is how do I show this? Can I show it without calculationg the joint posterior? I'm assuming this is not possible, so how do I calculate the joint posterior here? (Can I just take the product of the two posteriors? And on that note, how do I know when I can and cannot do this?)
From what I've understood, one can only calculate the joint posterior by the product of each posterior if we know that they are not dependent on each other. In this case it seems like this is what is meant to be done. However, I have not been successful in finding more information on wheter or not this is how you would go about this calculation. Nor have I been able to establish that my interpretation of the requirements for deriving the joint posterior is correct. |
For a fixed value of $n$, I'm not sure that a simpler formula exists. However, for $n=\infty$, the probability that there will be more heads than tails permanently is simply equal to $2p-1$. We can show this as follows.
Let $x_k$ denote the probability of success conditional on the current state being $k$ heads. We have the simple recurrence $x_k=p x_{k+1}+(1-p)x_{k-1}$, which we can rewrite as$$x_k=\frac{1}{p}x_{k-1} - \frac{1-p}{p}x_{k-2}.$$Solutions to this recurrence have the form$x_k = a_1\lambda_1^k+a_2\lambda_2^k$, where the $\lambda_i$ are roots of the characteristic polynomial $z^2-\frac{1}{p}z+\frac{1-p}{p}$:\begin{align}\lambda_1=1, ~~~~~~\lambda_2=\frac{1-p}{p},\end{align}We also have the constraints that $x_0=0$ and for $p>1/2$, $\lim_{k\rightarrow \infty}x_k=1$ (see below). It follows that the only possible solution is $$x_k=1-\left(\frac{1-p}{p}\right)^k.$$Since we start with $0$ heads, then in order to survive forever, we must toss a head in the first toss, and then survive from the state of one head. The probability is therefore $$px_1=p\left(1-\frac{1-p}{p}\right)=p-(1-p)=2p-1.$$
Now the answer to the question "how do we know that $\lim_{k\rightarrow \infty}x_k=1$?" There are two ways to go about this. The first is to start from the Borel-Cantelli lemma as in this answer. But there is also a direct proof.
Let $x_{k,n}$ denote the probability that, starting from a state of $k$ heads, we survive at least $n$ more rounds. Note that $x_{k,0}=1$ for $k>0$, and $x_{0,0}=0$. Note also that the $x_{k,n}$ must satisfy $$x_{k,n}=p x_{k+1,n-1}+(1-p)x_{k-1,n-1}.$$Since our values of $x_k$ computed above satisfy this same recurrence, then by monotinicity we may deduce that $$\left(x_{k,n-1}\geq x_k \forall k\right)\implies \left(x_{k,n}\geq x_k \forall k\right),$$and since this inequality holds for $n=0$, then by induction it must hold for all $n$. From this it follows that $\lim_{n\rightarrow\infty}x_{k,n}\geq x_k$ for all $k$, which in turn implies that $\lim_{k\rightarrow \infty}\lim_{n\rightarrow\infty}x_{k,n}=1$. This in turn implies that $\lim_{n\rightarrow\infty}x_{k,n}=x_k$ for all $k$, as deduced above. |
Here is what I think you need to come to your own conclusion. First I will give a very general overview over lift creation, and then I will look at three wings:
An unmodified wing This wing plus a winglet This wing plus the winglet, but this time folded down into the plane of the wing.
For each I will plot the lift and bending moment distribution. I will assume an elliptic circulation, fully knowing that this is not what most aircraft use. But I have to pick a distribution to make all three cases comparable, and the elliptic one makes things easier. The conclusions can be generalized for other distributions.
This will be a lengthy post (you should know me by now), so thanks to all who persevere through all of it.
Lift creation and induced drag
This topic had been covered before, and I mention it again to show a very simple and elegant way to explain induced drag that does not need vortices. I want to dispel the myth that induced drag is caused by air flowing around the wingtip, and winglets somehow magically can suppress this flow.
Consider a wing with elliptic circulation over span (think of circulation as the product of the local lift coefficient $c_l$ and local chord; it is basically the lift per spanwise increment). The wing bends the air through which it flows slightly downwards, and creates an opposite upwards force, namely lift (Newton's second law). I choose an elliptic distribution because then the downwash is constant over span, which makes the following calculations easier.
The sheet of air coming off behind the wing looks trough-shaped and moves downwards, thereby pressing other air below out of the way and allowing air above to flow inwards and to fill up the vacated volume.
That is how the free vortex is created, and air flowing around the wingtips has only a small part in this.
Induced drag is the consequence of the wing bending the airflow downwards. To simplify things, let's assume the wing is just acting on the air with the density $\rho$ flowing with the speed $v$ through a circle with a diameter equal to the span $b$ of the wing. If we just look at this stream tube, the mass flow is $$\frac{dm}{dt} = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v$$
Lift $L$ is then the impulse change which is caused by the wing. With the downward air speed $v_z$ imparted by the wing, lift is: $$L = \frac{b^2}{4}\cdot\pi\cdot\rho\cdot v\cdot v_z = S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho$$
$S$ is the wing area and $c_L$ the overall lift coefficient. If we now solve for the vertical air speed, we get $$v_z = \frac{S\cdot c_L\cdot\frac{v^2}{2}\cdot\rho}{\frac{b^2}{4}\cdot\pi\cdot\rho\cdot v} = \frac{2\cdot c_L\cdot v}{\pi\cdot AR}$$with $AR = \frac{b^2}{S}$ the aspect ratio of the wing. Now we can divide the vertical speed by the air speed to calculate the angle by which the air has been deflected by the wing. Let's call it $\alpha_w$: $$\alpha_w = arctan\left(\frac{v_z}{v}\right) = arctan \left(\frac{2\cdot c_L}{\pi\cdot AR}\right)$$
The deflection happens gradually along the wing chord, so the mean local flow angle along the chord is just $\alpha_w / 2$. Lift acts perpendicularly to this local flow, thus is tilted backwards by $\alpha_w / 2$. In coefficients, lift is $c_L$, and the backwards component is $\alpha_w / 2 \cdot c_L$. Let's call this component $c_{Di}$: $$c_{Di} = arctan \left(\frac{c_L}{\pi\cdot AR}\right)\cdot c_L$$
For small $\alpha_w$s the arcus tangens can be neglected, and we get this familiar looking equation for the backwards-pointing component of the reaction force: $$c_{Di} = \frac{c_L^2}{\pi\cdot AR}$$
If the circulation over span has an elliptic distribution, the local change in circulation times the local amount of circulation is constant, and the induced drag $c_{Di}$ is at its minimum. If this would be different, a higher local $v_z$ causes a quadratic increase in local induced drag, so the whole wing will create its lift less efficiently.
Now we know we can calculate induced drag and we understand why the vortex sheet behind the wing rolls up, producing two counter-rotating vortices, all without looking at the details of the wingtip. What counts is that the wing is of finite span, so the stream tube which is influenced by the wing is of finite diameter as well. Of course, in reality there is no clear boundary between air which is affected by the wing and other air which is not. There is a diffuse transition the more one moves away from the wing.
Comparison of wingtips
First the geometries: Here are three wingtips in top and front views for comparison:
Now let's look at the circulation distribution of the simple wing tip:
Again, I choose the elliptic distribution for simplicity. The corresponding bending moment looks like that:
No surprises so far. Now we add a winglet and make it work as best as possible. This means we have to give it an angle of attack where it carries the circulation from the wing over on the winglet and completes the elliptic tapering of circulation down to 0 at the tip:
The grey dashed line is the circulation of the original wing. I adjusted the circulation such that both wings produce the same lift. $b_{WL}$ is the span at the winglet tip, and for the bending moment plot I have folded the spanwise coordinate down on the y axis:
Now the bending moment starts at the wingtip with a nonzero value. Since the sideways force of the winglet is parallel to the wing spar, this bending moment contribution is constant over span. But there is more: Now also circulation at the old wingtip location is nonzero, and we get a substantial lift increase at the outer wing stations. This effect is what causes the additional lift and gives the better aileron response that winglets make possible. But it also increases the root bending moment, because this additional lift acts with the lever arm of the outer wing.
How can we compare the induced drag of the wing with winglets to the original wing? The circulation gradient is lower, that helps. Also the diameter of that stream tube is larger, but it is hard to say by how much. The sideways force on the winglet is created by pushing the vortex sheet aft of the winglet sideways out, so the trough-shaped area should become wider. Empirical evidence hints at an increase in diameter of 45% of the winglet span (see chapter 6 for a discussion of several papers on the topic).
Just for the heck of it, let's assume that the diameter really increases in line with winglet span. Then let's compare that to the straight wing extension, where the same diameter can be assumed with much more certainty:
Now also the lift on the folded-down winglet acts upwards, so the circulation at the center of the wing can be reduced even further. However, now it adds a linearly increasing part to the bending moment, and the outer wing section creates more lift, as before with the wing with winglet:
Here, the root bending moment is higher than in the winglet case. This is a second advantage of winglets: They allow to increase maximum lift with less bending moment increase than a wing extension. But the wing extension puts all parts towards the creation of lift, and not some to the useless creation of side force. Both the extended and the winglet-wing have the same surface friction and (when we assume the same diameter of the hypothetical stream tube) the same induced drag. But since the winglet creates some side force, the remaining wing needs to fly at a higher lift coefficient. Also, the intersection of wing and winglet might be as well rounded as possible, this is where early separation starts at higher angles of attack. None of this affects the straight wing extension.
Most evidence shows that winglets improve L/D over the original wing, but folding the winglet down will more than double its effectiveness in lowering drag. Even if we assume that the winglet is just as good as an equal span extension, still the span extension comes out ahead in L/D improvement because all its lift contributes to overall lift, whereas the winglet produces a side force instead. If no separation occurs at the wing-winglet intersection, both will create the same induced and profile drag (pressure and friction), because both have the same wetted surface and the same local circulation. Again, this gives winglets the benefit of equally low induced drag, which is not supported by most measurements.
The extended wingtip in the example above has interesting characteristics. It is a sweptback (raked) wingtip, which causes the local lift curve slope to be lower than that of the straight wing. This increases its maximum angle of attack and - assuming that the local area is bigger than what an elliptic wing shape would dictate - makes it possible to keep a nearly elliptic circulation distribution over a wider angle of attack range. The bigger local area is a sensible precaution against the wing tip stalling first, so a raked wing tip will combine benign stall characteristics and very low induced drag.
Compare this to the winglet, which has to be tailored for one polar point: Since changes in wing angle of attack will not change the incidence of the winglet, it cannot adapt as well to different flow conditions as can the extended wing. In sideslip the winglet will mess up the circulation distribution on the wingtip and will act like a deflected spoiler.
Conclusion
Comparing equal winglets and wing extensions gives these basic characteristics:
Both have the same viscous drag at low angle of attack. Both can create more maximum lift, and both lower induced drag. The wing extension can create most lift for the given increase in wetted surface. The wing extension is more than twice as effective in lowering induced drag. The wing extension gives a better circulation distribution at off-design angle of attack. The wing extension produces the highest root bending moment for a given amount of lift.
How much the bending moment increase will drive up structural mass depends on the aspect ratio of the original wing. Low aspect ratio wings will not suffer much, but stretching high aspect ratio wings will drive up spar mass considerably. But please note that the winglet also causes higher root bending moments, and it creates less bending moment than the wing extension because it creates some side force instead of pure, useful lift. |
And I think people said that reading first chapter of Do Carmo mostly fixed the problems in that regard. The only person I asked about the second pset said that his main difficulty was in solving the ODEs
Yeah here there's the double whammy in grad school that every grad student has to take the full year of algebra/analysis/topology, while a number of them already don't care much for some subset, and then they only have to pass rather the class
I know 2 years ago apparently it mostly avoided commutative algebra, half because the professor himself doesn't seem to like it that much and half because he was like yeah the algebraists all place out so I'm assuming everyone here is an analyst and doesn't care about commutative algebra
Then the year after another guy taught and made it mostly commutative algebra + a bit of varieties + Cech cohomology at the end from nowhere and everyone was like uhhh. Then apparently this year was more of an experiment, in part from requests to make things more geometric
It's got 3 "underground" floors (quotation marks because the place is on a very tall hill so the first 3 floors are a good bit above the the street), and then 9 floors above ground. The grad lounge is in the top floor and overlooks the city and lake, it's real nice
The basement floors have the library and all the classrooms (each of them has a lot more area than the higher ones), floor 1 is basically just the entrance, I'm not sure what's on the second floor, 3-8 is all offices, and 9 has the ground lounge mainly
And then there's one weird area called the math bunker that's trickier to access, you have to leave the building from the first floor, head outside (still walking on the roof of the basement floors), go to this other structure, and then get in. Some number of grad student cubicles are there (other grad students get offices in the main building)
It's hard to get a feel for which places are good at undergrad math. Highly ranked places are known for having good researchers but there's no "How well does this place teach?" ranking which is kinda more relevant if you're an undergrad
I think interest might have started the trend, though it is true that grad admissions now is starting to make it closer to an expectation (friends of mine say that for experimental physics, classes and all definitely don't cut it anymore)
In math I don't have a clear picture. It seems there are a lot of Mickey Mouse projects that people seem to not help people much, but more and more people seem to do more serious things and that seems to become a bonus
One of my professors said it to describe a bunch of REUs, basically boils down to problems that some of these give their students which nobody really cares about but which undergrads could work on and get a paper out of
@TedShifrin i think universities have been ostensibly a game of credentialism for a long time, they just used to be gated off to a lot more people than they are now (see: ppl from backgrounds like mine) and now that budgets shrink to nothing (while administrative costs balloon) the problem gets harder and harder for students
In order to show that $x=0$ is asymptotically stable, one needs to show that $$\forall \varepsilon > 0, \; \exists\, T > 0 \; \mathrm{s.t.} \; t > T \implies || x ( t ) - 0 || < \varepsilon.$$The intuitive sketch of the proof is that one has to fit a sublevel set of continuous functions $...
"If $U$ is a domain in $\Bbb C$ and $K$ is a compact subset of $U$, then for all holomorphic functions on $U$, we have $\sup_{z \in K}|f(z)| \leq C_K \|f\|_{L^2(U)}$ with $C_K$ depending only on $K$ and $U$" this took me way longer than it should have
Well, $A$ has these two dictinct eigenvalues meaning that $A$ can be diagonalised to a diagonal matrix with these two values as its diagonal. What will that mean when multiplied to a given vector (x,y) and how will the magnitude of that vector changed?
Alternately, compute the operator norm of $A$ and see if it is larger or smaller than 2, 1/2
Generally, speaking, given. $\alpha=a+b\sqrt{\delta}$, $\beta=c+d\sqrt{\delta}$ we have that multiplication (which I am writing as $\otimes$) is $\alpha\otimes\beta=(a\cdot c+b\cdot d\cdot\delta)+(b\cdot c+a\cdot d)\sqrt{\delta}$
Yep, the reason I am exploring alternative routes of showing associativity is because writing out three elements worth of variables is taking up more than a single line in Latex, and that is really bugging my desire to keep things straight.
hmm... I wonder if you can argue about the rationals forming a ring (hence using commutativity, associativity and distributivitity). You cannot do that for the field you are calculating, but you might be able to take shortcuts by using the multiplication rule and then properties of the ring $\Bbb{Q}$
for example writing $x = ac+bd\delta$ and $y = bc+ad$ we then have $(\alpha \otimes \beta) \otimes \gamma = (xe +yf\delta) + (ye + xf)\sqrt{\delta}$ and then you can argue with the ring property of $\Bbb{Q}$ thus allowing you to deduce $\alpha \otimes (\beta \otimes \gamma)$
I feel like there's a vague consensus that an arithmetic statement is "provable" if and only if ZFC proves it. But I wonder what makes ZFC so great, that it's the standard working theory by which we judge everything.
I'm not sure if I'm making any sense. Let me know if I should either clarify what I mean or shut up. :D
Associativity proofs in general have no shortcuts for arbitrary algebraic systems, that is why non associative algebras are more complicated and need things like Lie algebra machineries and morphisms to make sense of
One aspect, which I will illustrate, of the "push-button" efficacy of Isabelle/HOL is its automation of the classic "diagonalization" argument by Cantor (recall that this states that there is no surjection from the naturals to its power set, or more generally any set to its power set).theorem ...
The axiom of triviality is also used extensively in computer verification languages... take Cantor's Diagnolization theorem. It is obvious.
(but seriously, the best tactic is over powered...)
Extensions is such a powerful idea. I wonder if there exists algebraic structure such that any extensions of it will produce a contradiction. O wait, there a maximal algebraic structures such that given some ordering, it is the largest possible, e.g. surreals are the largest field possible
It says on Wikipedia that any ordered field can be embedded in the Surreal number system. Is this true? How is it done, or if it is unknown (or unknowable) what is the proof that an embedding exists for any ordered field?
Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement?
"Infinity exists" comes to mind as a potential candidate statement.
Well, take ZFC as an example, CH is independent of ZFC, meaning you cannot prove nor disprove CH using anything from ZFC. However, there are many equivalent axioms to CH or derives CH, thus if your set of axioms contain those, then you can decide the truth value of CH in that system
@Rithaniel That is really the crux on those rambles about infinity I made in this chat some weeks ago. I wonder to show that is false by finding a finite sentence and procedure that can produce infinity
but so far failed
Put it in another way, an equivalent formulation of that (possibly open) problem is:
> Does there exists a computable proof verifier P such that the axiom of infinity becomes a theorem without assuming the existence of any infinite object?
If you were to show that you can attain infinity from finite things, you'd have a bombshell on your hands. It's widely accepted that you can't. If fact, I believe there are some proofs floating around that you can't attain infinity from the finite.
My philosophy of infinity however is not good enough as implicitly pointed out when many users who engaged with my rambles always managed to find counterexamples that escape every definition of an infinite object I proposed, which is why you don't see my rambles about infinity in recent days, until I finish reading that philosophy of infinity book
The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.The problem often arises in resource allocation where there are financial constraints and is studied in fields such as combinatorics, computer science...
O great, given a transcendental $s$, computing $\min_P(|P(s)|)$ is a knapsack problem
hmm...
By the fundamental theorem of algebra, every complex polynomial $P$ can be expressed as:
$$P(x) = \prod_{k=0}^n (x - \lambda_k)$$
If the coefficients of $P$ are natural numbers , then all $\lambda_k$ are algebraic
Thus given $s$ transcendental, to minimise $|P(s)|$ will be given as follows:
The first thing I think of with that particular one is to replace the $(1+z^2)$ with $z^2$. Though, this is just at a cursory glance, so it would be worth checking to make sure that such a replacement doesn't have any ugly corner cases.
In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that0<|x−pq|<1qn.{\displaystyle 0<\left|x-{\frac {p}...
Do these still exist if the axiom of infinity is blown up?
Hmmm...
Under a finitist framework where only potential infinity in the form of natural induction exists, define the partial sum:
$$\sum_{k=1}^M \frac{1}{b^{k!}}$$
The resulting partial sums for each M form a monotonically increasing sequence, which converges by ratio test
therefore by induction, there exists some number $L$ that is the limit of the above partial sums. The proof of transcendentally can then be proceeded as usual, thus transcendental numbers can be constructed in a finitist framework
There's this theorem in Spivak's book of Calculus:Theorem 7Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and$$f'...
and neither Rolle nor mean value theorem need the axiom of choice
Thus under finitism, we can construct at least one transcendental number. If we throw away all transcendental functions, it means we can construct a number that cannot be reached from any algebraic procedure
Therefore, the conjecture is that actual infinity has a close relationship to transcendental numbers. Anything else I need to finish that book to comment
typo: neither Rolle nor mean value theorem need the axiom of choice nor an infinite set
> are there palindromes such that the explosion of palindromes is a palindrome nonstop palindrome explosion palindrome prime square palindrome explosion palirome prime explosion explosion palindrome explosion cyclone cyclone cyclone hurricane palindrome explosion palindrome palindrome explosion explosion cyclone clyclonye clycone mathphile palirdlrome explosion rexplosion palirdrome expliarome explosion exploesion |
Where to from here...
Problem: If $U$ and $Y$ are continuous random variables, show that if ${U}$ has uniform $[0,1]$ distribution, then the random variable: $$Y = \tan(\pi U −\frac{\pi}{2})$$ has the Cauchy distribution. Start of a Solution:Firstly, since $U$ is the uniform distribution on $[0,1]$, $u \in [0,1]$. Now, let $F_{Y}(y)$ be the probability distribution function where:$$F_{Y}(y) = \mathbb{P}(\{x \in [0,1] : \tan(\pi u - \pi/2 \leq y \})$$
This is equivalent to: $$F_{Y}(y) = \mathbb{P}(\{ x \in [0,1] : u \leq \frac{1}{\pi}(\arctan( y )) - \frac{1}{2}\}) $$
This is where I need further justification: intuitively, I'd say hey, that condition on the right looks an awful lot like the distribution function I want... but how do I get there from here?
Thanks! |
Here is an extract of the doctoral thesis of C. Lewis under the supervision of D. Joyce (https://people.maths.ox.ac.uk/joyce/theses/LewisDPhil.pdf, 1998):
2.6 Spin Bundles and the Dirac Operator
To consider spin bundles over a $Spin(7)$ manifold $M$, it is usually best to first consider Clifford algebras.
Let $V$ be a finite dimensional vector space with an inner product defined upon it. Let $e_1$, $e_2$, ... , $e_n$ be an orthonormal basis for $V$.
Then the Clifford algebra, $C_n$, of $V$ is defined to be the algebra generated by the elements $e_1$, $e_2$, ... , $e_n$ subject to the relations
$$e_i^2 =-1$$, $$e_ie_j + e_je_i = 0\text{ for }i \neq j$$
Considered as a vector space $C_n$ is of dimension $2n$, spanned by elements of the form
$$e_1^{\delta_1}e_2^{\delta_2}\cdots e_n^{\delta_n}$$
where $\delta_i = 0$ or 1.
Now consider the case $n = 8$. In this case it can be shown that
$$C_8 = \mathbb{R} (16),$$
the algebra of $16\times 16$ matrices with values in $\mathbb{R}$. [Sal, p.171]
Look also here
Thus we may consider $\mathbb{R}$ (16) as a $C_8$ module.
We may define the group $Spin(8)$ as the subset of $C_8$ consisting of all even products $x_1x_2\cdots x_{2r-1}x_{2r}$ of elements of $V$, with each $\|x_i\| = 1$. (Similarly we might have defined $Spin(7)$ as the subset of $C_7$ consisting of all even products $x_1x_2\cdots x_{2r-1}x_{2r}$ of elements of $\mathbb{R}^8$, with each $\|x_i\| = 1$.)
Now let us consider the element $v = e_1e_2\cdots e_8$ of $C_8$. Then $v$ is a involution of $C_8$, and commutes with every element of $Spin(8)$, and hence $\mathbb{R}^{16}$ splits as a $Spin(8)$ module into the eigenspaces of $v$.
Thus $\mathbb{R}^{16}=\Delta_+\oplus\Delta_-$, where $\Delta_+$ is the $+1$ eigenspace of $v$, and $\Delta_-$ is the $-1$ eigenspace of $v$. We call them the positive and negative spin representations of $Spin(8)$.
Now suppose that $M$ is a $Spin(7)$ manifold. Then we have that $M$ is a spin manifold i.e. there exists a spin structure of $M$, a principal $Spin(8)$ bundle $\tilde{E}$ covering the $SO(8)$ bundle of frames for the tangent bundle.
Now since we have a principal $Spin(8)$ bundle, and the two $Spin(8)$ modules (namely $\Delta_+$ and $\Delta_-$), we may form two vector bundles associated to the principal spin bundles by means of the two spin representations.
We call these bundles $S_+$ and $S_-$, the positive and negative spinor bundles, and their sections are known as positive and negative spinors. It is perhaps worth noting at this point that the group $Spin(7)$ is the subgroup of $SO(8)$ preserving a spinor, and hence the manifold $M$ will possess a constant spinor. Thus we will have isomorphisms $S_+ \equiv\Lambda^0\otimes \Lambda^2_7$ and $S_-\equiv \Lambda^1$.
The question is:
Why is it true that $S_+ \equiv\Lambda^0\otimes \Lambda^2_7$ and $S_-\equiv \Lambda^1$?
Any suggestion is welcome. |
Skills to Develop
To learn how to use data transformation if a measurement variable does not fit a normal distribution or has greatly different standard deviations in different groups. Introduction
Many biological variables do not meet the assumptions of parametric statistical tests: they are not normally distributed, the standard deviations are not homogeneous, or both. Using a parametric statistical test (such as an anova or linear regression) on such data may give a misleading result. In some cases, transforming the data will make it fit the assumptions better.
Fig. 4.6.1 Histograms of number of Eastern mudminnows per 75 m section of stream (samples with 0 mudminnows excluded). Untransformed data on left, log-transformed data on right.
To transform data, you perform a mathematical operation on each observation, then use these transformed numbers in your statistical test. For example, as shown in the first graph above, the abundance of the fish species
Umbra pygmaea (Eastern mudminnow) in Maryland streams is non-normally distributed; there are a lot of streams with a small density of mudminnows, and a few streams with lots of them. Applying the log transformation makes the data more normal, as shown in the second graph. Fig. 4.6.2 Eastern mudminnow (Umbra pygmaea).
Here are \(12\) numbers from the mudminnow data set; the first column is the untransformed data, the second column is the square root of the number in the first column, and the third column is the base-\(10\) logarithm of the number in the first column.
Untransformed Square-root
transformed
Log
transformed
38 6.164 1.580 1 1.000 0.000 13 3.606 1.114 2 1.414 0.301 13 3.606 1.114 20 4.472 1.301 50 7.071 1.699 9 3.000 0.954 28 5.292 1.447 6 2.449 0.778 4 2.000 0.602 43 6.557 1.633
You do the statistics on the transformed numbers. For example, the mean of the untransformed data is \(18.9\); the mean of the square-root transformed data is \(3.89\); the mean of the log transformed data is \(1.044\). If you were comparing the fish abundance in different watersheds, and you decided that log transformation was the best, you would do a one-way anova on the logs of fish abundance, and you would test the null hypothesis that the means of the log-transformed abundances were equal.
Back transformation
Even though you've done a statistical test on a transformed variable, such as the log of fish abundance, it is not a good idea to report your means, standard errors, etc. in transformed units. A graph that showed that the mean of the log of fish per \(75m\) of stream was \(1.044\) would not be very informative for someone who can't do fractional exponents in their head. Instead, you should back-transform your results. This involves doing the opposite of the mathematical function you used in the data transformation. For the log transformation, you would back-transform by raising 10 to the power of your number. For example, the log transformed data above has a mean of \(1.044\) and a \(95\%\) confidence interval of \(\pm 0.344\) log-transformed fish. The back-transformed mean would be \(10^{1.044}=11.1\) fish. The upper confidence limit would be \(10^{(1.044+0.344)}=24.4\) fish, and the lower confidence limit would be \(10^{(1.044-0.344)}=5.0\) fish. Note that the confidence interval is not symmetrical; the upper limit is \(13.3\) fish above the mean, while the lower limit is \(6.1\) fish below the mean. Also note that you can't just back-transform the confidence interval and add or subtract that from the back-transformed mean; you can't take \(10^{0.344}\) and add or subtract that.
Choosing the right transformation
Data transformations are an important tool for the proper statistical analysis of biological data. To those with a limited knowledge of statistics, however, they may seem a bit fishy, a form of playing around with your data in order to get the answer you want. It is therefore essential that you be able to defend your use of data transformations.
There are an infinite number of transformations you could use, but it is better to use a transformation that other researchers commonly use in your field, such as the square-root transformation for count data or the log transformation for size data. Even if an obscure transformation that not many people have heard of gives you slightly more normal or more homoscedastic data, it will probably be better to use a more common transformation so people don't get suspicious. Remember that your data don't have to be perfectly normal and homoscedastic; parametric tests aren't extremely sensitive to deviations from their assumptions.
It is also important that you decide which transformation to use before you do the statistical test. Trying different transformations until you find one that gives you a significant result is cheating. If you have a large number of observations, compare the effects of different transformations on the normality and the homoscedasticity of the variable. If you have a small number of observations, you may not be able to see much effect of the transformations on the normality and homoscedasticity; in that case, you should use whatever transformation people in your field routinely use for your variable. For example, if you're studying pollen dispersal distance and other people routinely log-transform it, you should log-transform pollen distance too, even if you only have \(10\) observations and therefore can't really look at normality with a histogram.
Common transformations
There are many transformations that are used occasionally in biology; here are three of the most common:
Log transformation
This consists of taking the log of each observation. You can use either base-\(10\) logs (LOG in a spreadsheet, LOG10 in SAS) or base-\(e\) logs, also known as natural logs (LN in a spreadsheet, LOG in SAS). It makes no difference for a statistical test whether you use base-\(10\) logs or natural logs, because they differ by a constant factor; the base-\(10\) log of a number is just \(2.303…\times \text{the\; natural\; log\; of\; the\; number}\). You should specify which log you're using when you write up the results, as it will affect things like the slope and intercept in a regression. I prefer base-\(10\) logs, because it's possible to look at them and see the magnitude of the original number: \(log(1)=0,\; log(10)=1,\; log(100)=2\), etc.
The back transformation is to raise \(10\) or \(e\) to the power of the number; if the mean of your base-\(10\) log-transformed data is \(1.43\), the back transformed mean is \(10^{1.43}=26.9\) (in a spreadsheet,
"=10^1.43"). If the mean of your base-e log-transformed data is \(3.65\), the back transformed mean is \(e^{3.65}=38.5\) (in a spreadsheet, "=EXP(3.65)". If you have zeros or negative numbers, you can't take the log; you should add a constant to each number to make them positive and non-zero. If you have count data, and some of the counts are zero, the convention is to add \(0.5\) to each number.
Many variables in biology have log-normal distributions, meaning that after log-transformation, the values are normally distributed. This is because if you take a bunch of independent factors and multiply them together, the resulting product is log-normal. For example, let's say you've planted a bunch of maple seeds, then \(10\) years later you see how tall the trees are. The height of an individual tree would be affected by the nitrogen in the soil, the amount of water, amount of sunlight, amount of insect damage, etc. Having more nitrogen might make a tree \(10\%\) larger than one with less nitrogen; the right amount of water might make it \(30\%\) larger than one with too much or too little water; more sunlight might make it \(20\%\) larger; less insect damage might make it \(15\%\) larger, etc. Thus the final size of a tree would be a function of \(\text{nitrogen}\times \text{water}\times \text{sunlight}\times \text{insects}\), and mathematically, this kind of function turns out to be log-normal.
Square-root transformation
This consists of taking the square root of each observation. The back transformation is to square the number. If you have negative numbers, you can't take the square root; you should add a constant to each number to make them all positive.
People often use the square-root transformation when the variable is a count of something, such as bacterial colonies per petri dish, blood cells going through a capillary per minute, mutations per generation, etc.
Arcsine transformation
This consists of taking the arcsine of the square root of a number. (The result is given in radians, not degrees, and can range from \(-\pi /2\; to\; \pi /2\).) The numbers to be arcsine transformed must be in the range \(0\) to \(1\). This is commonly used for proportions, which range from \(0\) to \(1\), such as the proportion of female Eastern mudminnows that are infested by a parasite. Note that this kind of proportion is really a nominal variable, so it is incorrect to treat it as a measurement variable, whether or not you arcsine transform it. For example, it would be incorrect to count the number of mudminnows that are or are not parasitized each of several streams in Maryland, treat the arcsine-transformed proportion of parasitized females in each stream as a measurement variable, then perform a linear regression on these data vs. stream depth. This is because the proportions from streams with a smaller sample size of fish will have a higher standard deviation than proportions from streams with larger samples of fish, information that is disregarded when treating the arcsine-transformed proportions as measurement variables. Instead, you should use a test designed for nominal variables; in this example, you should do logistic regression instead of linear regression. If you insist on using the arcsine transformation, despite what I've just told you, the back-transformation is to square the sine of the number.
How to transform data Spreadsheet
In a blank column, enter the appropriate function for the transformation you've chosen. For example, if you want to transform numbers that start in cell \(A2\), you'd go to cell \(B2\) and enter
=LOG(A2) or =LN(A2) to log transform, =SQRT(A2) to square-root transform, or =ASIN(SQRT(A2)) to arcsine transform. Then copy cell \(B2\) and paste into all the cells in column \(B\) that are next to cells in column \(A\) that contain data. To copy and paste the transformed values into another spreadsheet, remember to use the "Paste Special..." command, then choose to paste "Values." Using the "Paste Special...Values" command makes Excel copy the numerical result of an equation, rather than the equation itself. (If your spreadsheet is Calc, choose "Paste Special" from the Edit menu, uncheck the boxes labeled "Paste All" and "Formulas," and check the box labeled "Numbers.")
To back-transform data, just enter the inverse of the function you used to transform the data. To back-transform log transformed data in cell \(B2\), enter
=10^B2 for base-\(10\) logs or =EXP(B2) for natural logs; for square-root transformed data, enter =B2^2; for arcsine transformed data, enter =(SIN(B2))^2 Web pages
I'm not aware of any web pages that will do data transformations.
SAS
To transform data in SAS, read in the original data, then create a new variable with the appropriate function. This example shows how to create two new variables, square-root transformed and log transformed, of the mudminnow data.
DATA mudminnow;
INPUT location $ banktype $ count; countlog=log10(count); countsqrt=sqrt(count); DATALINES; Gwynn_1 forest 38 Gwynn_2 urban 1 Gwynn_3 urban 13 Jones_1 urban 2 Jones_2 forest 13 LGunpowder_1 forest 20 LGunpowder_2 field 50 LGunpowder_3 forest 9 BGunpowder_1 forest 28 BGunpowder_2 forest 6 BGunpowder_3 forest 4 BGunpowder_4 field 43 ;
The dataset "mudminnow" contains all the original variables ("location", "banktype" and "count") plus the new variables ("countlog" and "countsqrt"). You then run whatever PROC you want and analyze these variables just like you would any others. Of course, this example does two different transformations only as an illustration; in reality, you should decide on one transformation before you analyze your data.
The SAS function for arcsine-transforming X is
ARSIN(SQRT(X)).
You'll probably find it easiest to backtransform using a spreadsheet or calculator, but if you really want to do everything in SAS, the function for taking \(10\) to the \(X\) power is
10**X; the function for taking \(e\) to a power is EXP(X); the function for squaring \(X\) is X**2; and the function for backtransforming an arcsine transformed number is SIN(X)**2. Reference
Picture of a mudminnow from The Virtual Aquarium of Virginia.
Contributor
John H. McDonald (University of Delaware) |
Last week, I wrote about the volume and outer surface area of a spherical cap using different methods, both of which gave the volume as $V = \frac{\pi}{3}R^3 (1-\cos(\alpha))^2(2-\cos(\alpha))$ and the surface area as $A_o = 2\pi R^2 (1-\cos(\alpha))$.
All very nice; however, one of my most beloved heuristics fails in this case.
If you differentiate the area of a circle ($\pi r^2$) with respect to the radius, you get $2\pi r$ - the circumference. If you differentiate the volume of a sphere ($\frac{4}{3}\pi r^3$), you get $4\pi r^2$, the surface area. In a way that can be nicely extended to many shapes, differentiating an $n$-dimensional shape gives you its $n-1$-dimensional boundary.
Only it doesn't work - or even really come close to working - here. $\diff Vr = \pi R^2 (1-\cos(\alpha))^2(2-\cos(\alpha))$, which isn't $2\pi R^2 (1-\cos(\alpha))$. So what's gone wrong?
Well, it's not a nice shape, is what's gone wrong. The rationale behind the heuristic is that if you increase the radius of a circle a tiny amount, you add on a tiny extra sliver around the circumference - and similarly for the sphere's volume and surface area.
With the spherical cap, though, there's a difference: increasing the radius does add a tiny shell around the outside. However, it also scrapes off a tiny circle on the inside of the shape. The area of that circle is $A_i = \pi R^2 \sin^2(\alpha)$, or $\pi R^2 (1 - \cos^2(\alpha))$.
So, increasing the radius should add on the outer surface area, $A_o = 2 \pi R^2(1- \cos(\alpha))$ and remove the inner surface area, $A_i = \pi R^2 (1 - \cos^2(\alpha))$.
$A_o - A_i = \pi R^2 \left[ \left( 2 - 2\cos(\alpha) \right) - (1 - \cos^2(\alpha)) \right] = \pi R^2 \left(1-\cos(\alpha)\right)^2$. This, though, still isn't correct.
Again, don't tell @realityminus3, but this is a post about heuristics rather than rigour. The sum can be fixed if we think about the relative depths of the two bits of the surface area. If you increase the radius by $\delta r$, the curved shell's depth thickens by $\delta r$. However, the circle's depth only increases by $\delta r \cos(\alpha)$ - so the appropriate sum is $A_o - \cos(\alpha) A_i = \pi R^2 \left[ \left(2 - 2\cos(\alpha)\right) - \cos(\alpha)(1- \cos^2(\alpha))\right] = \pi R^2 \left[ 2 - 3\cos(\alpha)+\cos^3(\alpha) \right]$, as required. |
As forest and natural resource managers, we must be aware of how our timber management practices impact the biological communities in which they occur. A silvicultural prescription is going to influence not only the timber we are growing but also the plant and wildlife communities that inhabit these stands. Landowners, both public an(18)}{d private, often require management of non-timber components, such as wildlife, along with meeting the financial objectives achieved through timber management. Resource managers must be cognizant of the effect management practices have on plant and wildlife communities. The primary interface between timber and wildlife is habitat, and habitat is simply an amalgam of environmental factors necessary for species survival (e.g., food or cover). The key component to habitat for most wildlife is vegetation, which provides food and structural cover. Creating prescriptions that combine timber and wildlife management objectives are crucial for sustainable, long-term balance in the system.
So how do we develop a plan that will encompass multiple land use objectives? Knowledge is the key. We need information on the habitat required by the wildlife species of interest and we need to be aware of how timber harvesting and subsequent regeneration will affect the vegetative characteristics of the system. In other words, we need to understand the diversity of organisms present in the community and appreciate the impact our management practices will have on this system.
Diversity of organisms and the measurement of diversity have long interested ecologists and natural resource managers. Diversity is variety and at its simplest level it involves counting or listing species. Biological communities vary in the number of species they contain (richness) and relative abundance of these species (evenness). Species richness, as a measure on its own, does not take into account the number of individuals of each species present. It gives equal weight to those species with few individuals as it does to a species with many individuals. Thus a single yellow birch has as much influence on the richness of an area as 100 sugar maple trees. Evenness is a measure of the relative abundance of the different species making up the richness of an area. Consider the following example.
Example \(\PageIndex{1}\):
Sugar Maple
167
391
Beech
145
24
Yellow Birch
134
31
Both samples have the same richness (3 species) and the same number of individuals (446). However, the first sample has more evenness than the second. The number of individuals is more evenly distributed between the three species. In the second sample, most of the individuals are sugar maples with fewer beech and yellow birch trees. In this example, the first sample would be considered more diverse.
A diversity index is a quantitative measure that reflects the number of different species and how evenly the individuals are distributed among those species. Typically, the value of a diversity index increases when the number of types increases and the evenness increases. For example, communities with a large number of species that are evenly distributed are the most diverse and communities with few species that are dominated by one species are the least diverse. We are going to examine several common measures of species diversity.
Simpson’s Index
Simpson (1949) developed an index of diversity that is computed as:
$$D = \sum^R_{i=1} (\dfrac {n_i(n_i-1)}{N(N-1)})$$
where
ni is the number of individuals in species i, and N is the total number of species in the sample. An equivalent formula is:
$$D = \sum^R_{i=1} p_i^2$$
where \(p_i\) is the proportional abundance for each species and
R is the total number of species in the sample. Simpson’s index is a weighted arithmetic mean of proportional abundance and measures the probability that two individuals randomly selected from a sample will belong to the same species. Since the mean of the proportional abundance of the species increases with decreasing number of species and increasing abundance of the most abundant species, the value of D obtains small values in data sets of high diversity and large values in data sets with low diversity. The value of Simpson’s D ranges from 0 to 1, with 0 representing infinite diversity and 1 representing no diversity, so the larger the value of \(D\), the lower the diversity. For this reason, Simpson’s index is usually expressed as its inverse (1/ D) or its compliment (1- D) which is also known as the Gini-Simpson index. Let’s look at an example.
Example \(\PageIndex{2}\):calculating Simpson’s Index
We want to compute Simpson’s \(D\) for this hypothetical community with three species.
Sugar Maple
35
Beech
19
Yellow Birch
11
First, calculate N.
$$N = 35 + 19 + 11 = 65$$
Then compute the index using the number of individuals for each species:
$$D = \sum^R_{i=1} (\dfrac {n_i(n_i-1)}{N(N-1)}) = (\frac {35(34)}{65(64)} +\frac {19(18)}{65(64)} + \frac {11(10)}{65(64)}) = 0.3947$$
The inverse is found to be:
$$\frac {1}{0.3947} = 2.5336$$
Using the inverse, the value of this index starts with 1 as the lowest possible figure. The higher the value of this inverse index the greater the diversity. If we use the compliment to Simpson’s D, the value is:
$$1-0.3947 = 0.6053$$
This version of the index has values ranging from 0 to 1, but now, the greater the value, the greater the diversity of your sample. This compliment represents the probability that two individuals randomly selected from a sample will belong to different species. It is very important to clearly state which version of Simpson’s D you are using when comparing diversity.
Shannon-Weiner Index
The Shannon-Weiner index (Barnes et al. 1998) was developed from information theory and is based on measuring uncertainty. The degree of uncertainty of predicting the species of a random sample is related to the diversity of a community. If a community has low diversity (dominated by one species), the uncertainty of prediction is low; a randomly sampled species is most likely going to be the dominant species. However, if diversity is high, uncertainty is high. It is computed as:
$$H' = -\sum^R_{i=1} ln(p_i) = ln (\frac {1}{\prod^R_{i=1} p^{p_i}_i})$$
where
pi is the proportion of individuals that belong to species i and R is the number of species in the sample. Since the sum of the pi’s equals unity by definition, the denominator equals the weighted geometric mean of the pi values, with the pi values being used as weights. The term in the parenthesis equals true diversity D and H’=ln( D). When all species in the data set are equally common, all pi values = 1/ R and the Shannon-Weiner index equals ln( R). The more unequal the abundance of species, the larger the weighted geometric mean of the pi values, the smaller the index. If abundance is primarily concentrated into one species, the index will be close to zero.
An equivalent and computationally easier formula is:
$$H' = \frac {N ln \ N -\sum (n_i ln \ n_i)}{N}$$
where
N is the total number of species and ni is the number of individuals in species i. The Shannon-Weiner index is most sensitive to the number of species in a sample, so it is usually considered to be biased toward measuring species richness.
Let’s compute the Shannon-Weiner diversity index for the same hypothetical community in the previous example.
Example \(\PageIndex{3}\):Calculating Shannon-Weiner Index
Species
No. of individuals
Sugar Maple
35
Beech
19
Yellow Birch
11
We know that N = 65. Now let’s compute the index:
$$H' = \dfrac {271.335 - (124.437+55.944+26.377)}{65}=0.993$$ |
Research Open Access Published: Invariant curves for a delay differential equation with a piecewise constant argument Advances in Difference Equations volume 2015, Article number: 19 (2015) Article metrics
1003 Accesses
Abstract
In order to understand the dynamics of a second order delay differential equation with a piecewise constant argument, we investigate invariant curves of the derived planar mapping from the equation. All invariant curves are given in this paper.
Introduction
The study of differential equations with piecewise constant argument (EPCA) initiated in [1, 2]. These equations represent a hybrid of continuous and discrete dynamical systems and combine the properties of both differential and difference equations, hence, they are of importance in control theory and in certain biomedical models [3]. In this paper the second order delay differential equation with a piecewise constant argument
where \(x''(t)\) denotes the second order derivative of \(x(t)\), \([t]\) denotes the greatest integer less than or equal to
t, and \(g:\mathbb {R}\rightarrow \mathbb{R}\) is a continuous or at least piecewise continuous function, is considered. In 1987, Aftabizadeh et al. discussed the oscillatory and periodic properties of the solutions of (1) in [4]. In 1989, Gyori and Ladas investigated linearized oscillations of the solutions of (1) in [5]. Later, Wiener and Cooke considered oscillations of the solutions of systems of two differential equations with piecewise constant arguments in [6].
The invariant curve [7–11] is another interesting problem in the study of dynamics because it can be used to reduce a system to a 1-dimensional one. The problem of invariant curves is actually a part of the research on invariant manifolds. In 1997, Ng and Zhang studied the nonlinear \(C^{1}\) invariant curve of planar mapping \(G:\mathbb {R}^{2}\rightarrow \mathbb{R}^{2}\),
derived from (1) in [12] when
g is nonlinear and gave the conditions that G has linear invariant curves when g is linear. In 2003, Yang et al. investigated nonlinear \(C^{0}\) invariant curves of (2) when g is nonlinear in [13]. So far, nonlinear invariant curves of (2) when g is linear have not been studied. So it is very interesting to look for nonlinear invariant curves of (2) when g is linear. In this paper all the invariant curves of the planar mapping G are given including the linear and nonlinear ones when g is linear. Main results
We discuss invariant curves of the form \(y=f(x)\) for the planar mapping (2). Its invariant curves of the form \(y=f(x)\) satisfy \(f(y)=2y-x-\frac {1}{2}(g(y)+g(x))\), which leads to the iterative functional equation
Considering linear
g and \(g(x)=ax+b\), we compute that
Thus, the invariant curves of planar mapping
G with \(g(x)=ax+b\) can be obtained by solving functional (4). We mainly discuss the generic cases \(a \notin\{-2, 4\}\), but leave the special cases \(a=-2\) and \(a=4\) to the last part of this section. For generic \(a\notin\{-2, 4\}\), (4) with \(b=0\) is of the form discussed in [14, 15]. In order to apply the results of [14], we let
which are the roots of the characteristic polynomial \(P(r):=r^{2} -(2-\frac{a}{2})r +1+\frac{a}{2}\).
(C1)
\(0< r_{1}<1<r_{2}\), if and only if \(-2< a<0\).
(C2)
\(r_{1}=r_{2}=1\), if and only if \(a=0\).
(C3)
\(r_{1}<0<r_{2}\neq1\) and \(r_{1}\ne-r_{2}\), if and only if \(a<-2\).
(C4)
\(r_{1}=r_{2}<0 \), if and only if \(a=16\).
(C5)
\(r_{1}< r_{2}<-1\), if and only if \(a>16\).
Note that the case \(r_{2}>r_{1}>1\) is not listed because the case \(r_{2}>r_{1}>1\) implies \(\frac{(4-a)-(a^{2}-16a)^{\frac{1}{2}}}{4}>1\),
i.e., \(-a>(a^{2}-16a)^{\frac{1}{2}}\), which does not hold, and that the case \(0< r_{1}<r_{2}<1\) is not listed because \(0< r_{1}<r_{2}<1\) implies \(0<\frac{(4-a)+(a^{2}-16a)^{\frac{1}{2}}}{4}<1\), i.e., \(0< a<4\), which contradicts the requirement that \(\Delta=a^{2}-16a\geq0\), and that the case \(0< a<16\) is not listed because in this case (4) with \(b=0\) has no continuous solutions, neither \(r_{1}\) nor \(r_{2}\) is real, by [14]. Since we consider \(a\notin\{-2, 4\}\), none of the case \(r_{1}=0\), the case \(r_{2}=0\), and the case \(r_{1}=-r_{2}\ne0\) is listed. Corresponding to the above list, we have the following results. Theorem 2.1
(i)
If \(-2< a<0\), then a continuous solutions ϕ of (4) with \(b=0\) is either of the piecewise linear form that \(f(x):= r_{i}x\) for \(x>0\), or \(:=0\) for \(x=0\), or \(:=r_{j}x\) for \(x<0\), where \(i,j=1, 2\), or given by where \(x_{n}=\frac{r_{2}^{n}}{r_{2}-r_{1}}(x_{1}-r_{1}x_{0})+ \frac{r_{1}^{n}}{r_{2}-r_{1}}(-x_{1}+r_{2}x_{0})\), \(n \in\mathbb{Z}\), with an arbitrarily chosen \(x_{0}\in(-\infty, +\infty)\) and \(x_{1} \in[r_{1}x_{0}, r_{2}x_{0}]\), and \(f_{n}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{n-1}^{-1}(x)\) for all \(x\in[x_{n}, x_{n +1})\), \(n=1, 2, \ldots \) , \(f_{-n-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac {1}{r_{1}r_{2}}f^{-1}_{-n}(x)\) for all \(x\in[x_{-n}, x_{-n+1})\), \(n=1, 2, \ldots\) , and \(f_{-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac {1}{r_{1}r_{2}}f_{0}(x)\), \(x\in[x_{0}, x_{1})\), with the arbitrarily chosen functions \(f_{0}\) such that \(f_{0}(x_{0})=x_{1}\), \(f_{0}(x_{1})=x_{2}\), and \(r_{1}\leq\frac{f_{0}(x)-f_{0}(y)}{x-y}\leq r_{2}\) for all \(x, y \in [x_{0}, x_{1})\). (ii) If \(a=0\), then (4) with \(b=0\) has a unique continuous solution f and \(f(x)=x+\beta\), where \(\beta\in\mathbb{R}\) is an arbitrary constant. Proof
The proof is a simple application of well-known results in [14]. The result (i) is given by Theorem 2 of [14], where the characteristic roots \(r_{1}\), \(r_{2}\) satisfy \(r_{2}>1>r_{1}>0\) as shown in (C1). We can deduce the result (ii) from Theorem 8 of [14], where \(r_{1}=r_{2}=1\) as shown in (C2). The proof is completed. □
Theorem 2.2
(i)
If \(a<-2\), then (4) with \(b=0\) only has two continuous solutions f and \(f(x)=r_{1}x\) or \(r_{2}x\). (ii) If \(a=16\), (4) with \(b=0\) just has a continuous solution \(f(x)=-3x\). (iii) If \(a>16\), all continuous solutions f of (4) with \(b=0\) are given by where the sequence \(\{x_{n}\}\) is defined by \(x_{n}=\frac {r_{2}^{n}}{r_{2}-r_{1}}(x_{1}-r_{1}x_{0})+ \frac{r_{1}^{n}}{r_{2}-r_{1}}(-x_{1}+r_{2}x_{0})\), \(n\in\mathbb{Z}\), with an arbitrarily chosen \(x_{0}\in(0, +\infty)\) and \(x_{1} \in[r_{1}x_{0}, r_{2}x_{0}]\), and \(f_{2n-1}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{2n-2}^{-1}(x)\), \(x\in [x_{-2n+5},x_{-2n+3})\), \(n=1, 2, \ldots \) , \(f_{2n}(x)=(r_{1}+r_{2})x-r_{1}r_{2}f_{2n-1}^{-1}(x)\), \(x\in [x_{-2n},x_{-2n+2})\), \(n=1, 2, \ldots \) , \(f_{-2n}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac{1}{r_{1}r_{2}} f^{-1}_{-2n+1}(x)\), \(x \in[x_{2n+3}, x_{2n+1})\), \(n=1, 2,\ldots\) , \(f_{-2n-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac {1}{r_{1}r_{2}} f^{-1}_{-2n}(x)\), \(x \in[x_{2n}, x_{2n+2})\), \(n=1, 2,\ldots\) , and \(f_{-1}(x)=(\frac{1}{r_{1}}+\frac{1}{r_{2}})x-\frac{1}{r_{1}r_{2}} f_{0}(x)\), \(x\in[x_{0}, x_{2})\), with an arbitrarily chosen continuous function \(f_{0}\) on \([x_{0},x_{2})\) such that \(f_{0}(x_{0})=x_{1}\), \(f_{0}(x_{2})=x_{3}\), and \(r_{1}\leq\frac {f_{0}(x)-f_{0}(y)}{x-y}\leq r_{2}\), \(\forall x, y \in[x_{0}, x_{2})\). Proof
Firstly, we consider (i). By Theorem 5 in [14], (4) with \(b=0\) only has two continuous solutions
f and \(f(x)=r_{1}x\) or \(r_{2}x\), where the characteristic roots \(r_{1}\), \(r_{2}\) satisfy \(r_{1}<0<r_{2}\neq1\) and \(r_{1}\ne-r_{2}\) as shown in (C3). Next, we consider (ii). By Theorem 6 in [14], (4) with \(b=0\) just has a continuous solution \(f(x)=-3x\), where the characteristic roots \(r_{1}\), \(r_{2}\) satisfy \(r_{1}=r_{2}=-3\) as shown in (C4). Finally, we consider (iii). In order to piecewise construct all solutions of (4) with \(b=0\) we need a partition for the interval \((-\infty, \infty)\). For this purpose we consider a homogeneous linear difference equation
which has the same coefficients as (4) with \(b=0\) correspondingly. Its characteristic equation is
If
f is a solution of (4) with \(b=0\), we easily see that f is invertible. In fact, if \(f(x_{1})=f(x_{2})\), then \(f(f(x_{1}))=f(f(x_{2}))\). Thus, \(x_{1} = x_{2}\) by (4) because \(a\neq-2\), which implies that f is one to one. Next we only need to show that \(f(x)\rightarrow-\infty\) as \(x\rightarrow+\infty\) and \(f(x)\rightarrow+\infty\) as \(x \rightarrow-\infty\) because \(f(x)\rightarrow\pm\infty\) as \(x \rightarrow\pm\infty\), then the left-hand side of (4) with \(b=0\) tends to ±∞ by \(a>16\), but the right-hand side is equal to 0. Otherwise, \(f(x)\) has a finite limit as \(x \rightarrow\infty\), then \(f(f(x))-(2-\frac{a}{2})f(x)\) converges to a finite limit by the continuity of f on the whole of ℝ, but \((1+\frac{a}{2})x\) does not, which contradicts the requirement that \(f(f(x))-(2-\frac{a}{2})f(x)=-(1+\frac{a}{2})x\). Thus, we rewrite (4) in the following equivalent form:
Let \(x_{0}=x\) and \(x_{n+1}=f(x_{n})\) in (11), we have
Furthermore, we can obtain
where \(\Delta f^{n}(x, y)= \frac{f^{n}(x)-f^{n}(y)}{x - y}\) for any \(x\neq y\) and \(n\in\mathbb{Z}\). From (12) we can see that
Since
f is strictly monotonic, \(\Delta f^{n}(x, y)>0\) for even n, which implies \(\Delta f(x, y)-r_{1}\geq0\) and \(-\Delta f(x, y)+r_{2}\geq0\), that is,
Moreover, we can see that \(f(0)=0\) from (13). In what follows, we arbitrarily choose \(x_{0}\in(0, +\infty)\) and \(x_{1}\in[r_{1}x_{0}, r_{2}x_{0}]\) and define a sequence \(\{x_{n}\}\), \(n\in\mathbb{Z}\), by (11). The sequences \(\{x_{2n}\}\), \(\{x_{2n+1}\}\), \(\{x_{-2n}\}\) and \(\{x_{-2n+1}\}\), where \(n=0, 1, 2,\ldots\) , are strictly monotone such that \(x_{2n}\rightarrow+\infty\), \(x_{2n+1}\rightarrow-\infty\), \(x_{-2n} \rightarrow0\), and \(x_{-2n+1} \rightarrow0\) as \(n \rightarrow\infty\). Thus, the sequence \(\{x_{n}\}\), \(n\in\mathbb{Z}\), is a partition of the interval \((-\infty, \infty)\). Next we arbitrarily choose a continuous function defined in the interval \([x_{0}, x_{2})\), satisfying \(f_{0}(x_{0})=x_{1}\), \(f_{0}(x_{2})=x_{3}\), and condition (14). We can recursively define the homeomorphisms \(f_{2n-1}: [x_{-2n+5}, x_{-2n+3})\rightarrow [x_{-2n}, x_{-2n+2})\), \(n=1, 2,\ldots\) , and \(f_{2n}: [x_{-2n}, x_{-2n+2})\rightarrow [x_{-2n+3}, x_{-2n+1})\), \(n=1, 2,\ldots\) , such that
Obviously, \(f_{2n+1}(x_{-2n+3})=x_{-2n}\) and \(f_{2n+1}(x_{-2n+1})=x_{-2n-2}\). Making use of (18), we have \(\frac{1}{r_{2}} \leq\frac{f^{-1}_{2n}(x) - f^{-1}_{2n}(y)}{x-y} \leq\frac{1}{r_{1}}\) for \(x, y \in[x_{-2n}, x_{-2n+2})\). It is easy to deduce that
Furthermore, we again let
By the same argument we can see that
By induction both \(f_{2n-1}\) and \(f_{2n}\) are well defined. Similarly, we can also recursively define the homeomorphisms \(f_{-2n+1}: [x_{2n-2}, x_{2n}) \rightarrow [x_{2n+3},x_{2n+1})\), \(n = 1, 2, \ldots\) , and \(f_{-2n}: [x_{2n+3}, x_{2n+1}) \rightarrow [x_{2n},x_{2n+2})\), \(n = 1, 2, \ldots\) . By the properties of the dual (10) we can obtain
Therefore,
Thus, we can define
f is continuous on ℝ because \(f_{2n}(x_{-2n})=x_{-2n+1}=f_{2n+2}(x_{-2n})\), \(f_{2n+1}(x_{-2n+1})=x_{-2n-2}=f_{2n+3}(x_{-2n+1})\), where \(n=0, 1, 2, \ldots \) , \(f_{1}^{-1}(x_{0}) = f_{-1}(x_{0})\), \(f_{-2n}^{-1}(x_{2n+2})=x_{2n+3}=f_{-2n-2}^{-1}(x_{2n+2})\), and \(f_{-2n+1}^{-1}(x_{2n+3})=x_{2n+2}=f_{-2n-1}^{-1}(x_{2n+3})\), where \(n=1, 2, 3, \ldots\) . we can easily check that f defined in Theorem 2.2 satisfies (4) with \(b=0\) in ℝ. In fact, if \(x\in[x_{-2n},x_{-2n+2})\), \(n=0, 1, 2, \ldots \) , \(f^{2}(x)=f_{2n+1}(f_{2n}(x))=(r_{1}+r_{2})f_{2n}(x)-r_{1}r_{2}x=(r_{1}+r_{2})f(x)-r_{1}r_{2}x\), i.e., \(f^{2}(x)-(r_{1}+r_{2})f(x)-r_{1}r_{2}x=0\). Similarly, we can also check that f satisfies (4) with \(b=0\) for \(x\in[x_{-2n+3},x_{-2n+1})\), \(x\in[x_{2n+3}, x_{2n+1})\), \(x\in[x_{2n}, x_{2n+2})\) and \(x=0\), where \(n=1, 2, 3, \ldots\) . The proof is completed. □ Remark
the same type of equation as the one considered in Theorems 2.1 and 2.2 with vanishing
b, by the replacement \(\tilde{f}(x)=f(x+\xi)-\xi\), where \(\xi=\frac{-b}{(1-r_{1})(1-r_{2})}\), if its characteristic roots \(r_{1}\), \(r_{2}\) are both real but neither of them is equal to 1. In this case solutions can be found from Theorems 2.1 and 2.2. So (4) with \(b\neq0\) can be reduced to (21) except for the case \(a=0\). For the case of \(a=0\) and \(b\neq0\), (4) has no real continuous solutions. In fact, by induction and (4) we can obtain \(f^{n}(x)=nf(x)-(n-1)x-\frac{n(n+1)}{2}b\), \(n\in\mathbb{Z}\). Furthermore, we have \(f^{n+1}(x)-f^{n}(x)=f(x)-x-(n+1)b\), \(n\in\mathbb{Z}\). For an arbitrary \(x\in\mathbb{R}\), \(f^{n+1}(x)-f^{n}(x)\) has the same sign when n takes the values N and − N, where N is a large positive integer, because f is strictly monotonic. But \(f(x)-x-(n+1)b\) has not, which contradicts the requirement \(f^{n+1}(x)-f^{n}(x)=f(x)-x-(n+1)b\), \(n\in\mathbb{Z}\).
In what follows, we consider the case that either \(a=-2\) or \(a=4\), which is not generic.
For \(a=-2\), (4) is of the form \(f^{2}(x)-3f(x)=-b\), from which we get with the replacement \(y=f(x)\): \(f(x)=3x-b\).
For \(a=4\), (4) is of the form
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Cooke, KL, Wiener, J: Retarded differential equations with piecewise constant delays. J. Math. Anal. Appl.
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Shah, SM, Wiener, J: Advanced differential equations with piecewise constant argument deviations. Int. J. Math. Math. Sci.
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Busenberge, S, Cooke, KL: Nonlinearity and Functional Analysis. Academic Press, New York (1982)
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Aftabizadeh, AR, Wiener, J, Xu, J: Oscillatory and periodic properties of delay differential equations with piecewise constant arguments. Proc. Am. Math. Soc.
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Gyori, I, Ladas, G: Linearized oscillations for equations with piecewise constant arguments. Differ. Integral Equ.
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Wiener, J, Cooke, KL: Oscillations in systems of two differential equations with piecewise constant arguments. J. Math. Anal. Appl.
137, 221-239 (1989) 7.
Anosov, DV: Geodesic Flows on Closed Riemannian Manifolds of a Negative Curvature. Trudy Mat. Inst. Im. V.A. Steklova, vol. 90 (1967) (Russian)
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Kuczma, M, Choczewski, B, Ger, R: Iterative Functional Equations. Encyclopedia of Mathematics and Its Applications, vol. 32. Cambridge University Press, Cambridge (1990)
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Matkowski, J: Mean-type mappings and invariant curves. J. Math. Anal. Appl.
384, 431-438 (2011) 10.
Nitecki, Z: Differentiable Dynamics: An Introduction to the Orbit Structure of Diffeomorphisms. MIT Press, London (1971)
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Sternberg, S: On the behaviour of invariant curves near a hyperbolic point of a surface transformation. Am. J. Math.
77, 526-534 (1955) 12.
Ng, CT, Zhang, W: Invariant curves for planar mappings. J. Differ. Equ. Appl.
3, 147-168 (1997) 13.
Yang, D, Zhang, W, Xu, B: Notes on a functional equation related to invariant curves. J. Differ. Equ. Appl.
9, 247-255 (2003) 14.
Matkowski, J, Zhang, W: Method of characteristic for function equations in polynomial form. Acta Math. Sin.
13, 421-432 (1997) 15.
Yang, D, Zhang, W: Characteristic solutions of polynomial-like iterative equations. Aequ. Math.
67, 80-105 (2004) 16.
Zhang, WM, Zhang, W: On continuous solutions of
n-th order polynomial-like iterative equations. Publ. Math. (Debr.) 76, 117-134 (2010) Acknowledgements
The author would like to thank the referees for their valuable comments and suggestions, which have helped to improve the quality of this paper. This work was supported by the general item (L0801) of Zhanjiang Normal University.
Additional information Competing interests
The author declares that he has no competing interests. |
I recently came across this in a textbook (NCERT class 12 , chapter: wave optics , pg:367 , example 10.4(d)) of mine while studying the Young's double slit experiment. It says a condition for the formation of interference pattern is$$\frac{s}{S} < \frac{\lambda}{d}$$Where $s$ is the size of ...
The accepted answer is clearly wrong. The OP's textbook referes to 's' as "size of source" and then gives a relation involving it. But the accepted answer conveniently assumes 's' to be "fringe-width" and proves the relation. One of the unaccepted answers is the correct one. I have flagged the answer for mod attention. This answer wastes time, because I naturally looked at it first ( it being an accepted answer) only to realise it proved something entirely different and trivial.
This question was considered a duplicate because of a previous question titled "Height of Water 'Splashing'". However, the previous question only considers the height of the splash, whereas answers to the later question may consider a lot of different effects on the body of water, such as height ...
I was trying to figure out the cross section $\frac{d\sigma}{d\Omega}$ for spinless $e^{-}\gamma\rightarrow e^{-}$ scattering. First I wrote the terms associated with each component.Vertex:$$ie(P_A+P_B)^{\mu}$$External Boson: $1$Photon: $\epsilon_{\mu}$Multiplying these will give the inv...
As I am now studying on the history of discovery of electricity so I am searching on each scientists on Google but I am not getting a good answers on some scientists.So I want to ask you to provide a good app for studying on the history of scientists?
I am working on correlation in quantum systems.Consider for an arbitrary finite dimensional bipartite system $A$ with elements $A_{1}$ and $A_{2}$ and a bipartite system $B$ with elements $B_{1}$ and $B_{2}$ under the assumption which fulfilled continuity.My question is that would it be possib...
@EmilioPisanty Sup. I finished Part I of Q is for Quantum. I'm a little confused why a black ball turns into a misty of white and minus black, and not into white and black? Is it like a little trick so the second PETE box can cancel out the contrary states? Also I really like that the book avoids words like quantum, superposition, etc.
Is this correct? "The closer you get hovering (as opposed to falling) to a black hole, the further away you see the black hole from you. You would need an impossible rope of an infinite length to reach the event horizon from a hovering ship". From physics.stackexchange.com/questions/480767/…
You can't make a system go to a lower state than its zero point, so you can't do work with ZPE. Similarly, to run a hydroelectric generator you not only need water, you need a height difference so you can make the water run downhill. — PM 2Ring3 hours ago
So in Q is for Quantum there's a box called PETE that has 50% chance of changing the color of a black or white ball. When two PETE boxes are connected, an input white ball will always come out white and the same with a black ball.
@ACuriousMind There is also a NOT box that changes the color of the ball. In the book it's described that each ball has a misty (possible outcomes I suppose). For example a white ball coming into a PETE box will have output misty of WB (it can come out as white or black). But the misty of a black ball is W-B or -WB. (the black ball comes out with a minus). I understand that with the minus the math works out, but what is that minus and why?
@AbhasKumarSinha intriguing/ impressive! would like to hear more! :) am very interested in using physics simulation systems for fluid dynamics vs particle dynamics experiments, alas very few in the world are thinking along the same lines right now, even as the technology improves substantially...
@vzn for physics/simulation, you may use Blender, that is very accurate. If you want to experiment lens and optics, the you may use Mistibushi Renderer, those are made for accurate scientific purposes.
@RyanUnger physics.stackexchange.com/q/27700/50583 is about QFT for mathematicians, which overlaps in the sense that you can't really do string theory without first doing QFT. I think the canonical recommendation is indeed Deligne et al's *Quantum Fields and Strings: A Course For Mathematicians *, but I haven't read it myself
@AbhasKumarSinha when you say you were there, did you work at some kind of Godot facilities/ headquarters? where? dont see something relevant on google yet on "mitsubishi renderer" do you have a link for that?
@ACuriousMind thats exactly how DZA presents it. understand the idea of "not tying it to any particular physical implementation" but that kind of gets stretched thin because the point is that there are "devices from our reality" that match the description and theyre all part of the mystery/ complexity/ inscrutability of QM. actually its QM experts that dont fully grasp the idea because (on deep research) it seems possible classical components exist that fulfill the descriptions...
When I say "the basics of string theory haven't changed", I basically mean the story of string theory up to (but excluding) compactifications, branes and what not. It is the latter that has rapidly evolved, not the former.
@RyanUnger Yes, it's where the actual model building happens. But there's a lot of things to work out independently of that
And that is what I mean by "the basics".
Yes, with mirror symmetry and all that jazz, there's been a lot of things happening in string theory, but I think that's still comparatively "fresh" research where the best you'll find are some survey papers
@RyanUnger trying to think of an adjective for it... nihilistic? :P ps have you seen this? think youll like it, thought of you when found it... Kurzgesagt optimistic nihilismyoutube.com/watch?v=MBRqu0YOH14
The knuckle mnemonic is a mnemonic device for remembering the number of days in the months of the Julian and Gregorian calendars.== Method ===== One handed ===One form of the mnemonic is done by counting on the knuckles of one's hand to remember the numbers of days of the months.Count knuckles as 31 days, depressions between knuckles as 30 (or 28/29) days. Start with the little finger knuckle as January, and count one finger or depression at a time towards the index finger knuckle (July), saying the months while doing so. Then return to the little finger knuckle (now August) and continue for...
@vzn I dont want to go to uni nor college. I prefer to dive into the depths of life early. I'm 16 (2 more years and I graduate). I'm interested in business, physics, neuroscience, philosophy, biology, engineering and other stuff and technologies. I just have constant hunger to widen my view on the world.
@Slereah It's like the brain has a limited capacity on math skills it can store.
@NovaliumCompany btw think either way is acceptable, relate to the feeling of low enthusiasm to submitting to "the higher establishment," but for many, universities are indeed "diving into the depths of life"
I think you should go if you want to learn, but I'd also argue that waiting a couple years could be a sensible option. I know a number of people who went to college because they were told that it was what they should do and ended up wasting a bunch of time/money
It does give you more of a sense of who actually knows what they're talking about and who doesn't though. While there's a lot of information available these days, it isn't all good information and it can be a very difficult thing to judge without some background knowledge
Hello people, does anyone have a suggestion for some good lecture notes on what surface codes are and how are they used for quantum error correction? I just want to have an overview as I might have the possibility of doing a master thesis on the subject. I looked around a bit and it sounds cool but "it sounds cool" doesn't sound like a good enough motivation for devoting 6 months of my life to it |
This is a very deep question.
A proof in terms of numbered formulae and various $\Rightarrow$, resp. $\Leftrightarrow$-signs could be checked by an automated proof checker. On the other hand, a "figure" is just a bitmap, or a pixel heap, and I doubt that an automated proof checker would ever be able to make out what this figure is telling us.
In other words: Figures are viewed at and interpreted by
humans. Sometimes these humans consent in accepting such a figure as proof of some statement, but sometimes they are in error in doing so. When a figure mainly serves to explain a certain concept, say, the derivative of a function $f:\>{\mathbb R}^n\to{\mathbb R}^m$ at some point $p\in{\rm dom}(f)$, then there is not much harm possible, but as soon as there are "cases" involved, say in a geometric proof of $\sin(x+y)=\ldots\ $ for arbitrary angles, the question arises whether the power of $1$ (one) figure is sufficient to prove the general statement. To put it differently: A general statement might involve very different morphologies, only one of which is captured in a single figure.
Concerning your example, it is certainly not sufficient to draw a point $x$ and a circle of radius $\epsilon$ around $x$. But inserting another point $y$ into this circle and drawing a very small circle around $y$ would make the idea of the intended proof clear. Nevertheless, in a course & homework situation it is expected that the idea so obvious in the figure is "verbalized" in a coherent argument. |
I did cointegration test on two identical time series, and the result shows that they are not cointegrated, but intuitively, I think they are.
Can anyone share some thoughts on this? Thanks!
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Let us test that $x$ and $y$ are co-integrated, say that $x_t, y_t \sim I(1)$. In the Engle-Granger we test stationarity of the error term in $$y_t = \alpha + \beta x_t + u_t$$ which we estimate as $$\hat u_t = y_t - \hat \alpha - \hat \beta x_t$$ and find that $\hat \alpha =0$, $\hat \beta = 1$, and $\hat u_t = 0 \; \forall t$.
So now when we Dickey-Fuller test residuals in something like $$\Delta \hat u_t = \gamma_0 + \gamma_1 \hat u_{t-1} + \epsilon_t$$ nothing will be significant and we won't find any co-integration.
I am not precisely schooled in this theory, so I'm not sure if this means these series can't be
referred to as "co-integrated" (clearly they have the same drift) or if this is just a trivial case where the test fails,
Two
integrated series $X_t$ and $Y_t$ are cointegrated if their linear combination (some, not any) $\alpha X_t+\beta Y_t$ is stationary. If you have $P(X_t=Y_t)=1$ for all $t$, then $P(\alpha X_t+\beta Y_t=(\alpha+\beta) X_t)=1$. So according to definition of cointegration $(\alpha+\beta) X_t$ should be stationary, which is identical to $X_t$ being stationary. And here we get the contradiction, since $X_t$ is integrated, hence not stationary.
This was a basic explanation why you received your result. However a lot depends on how the actual statistic is computed. For other statistics or their software implementations you might get that two identical series are cointegrated, but that will not mean that they are. Two identical time series are the degenerate case which no-one checks against, and with degenerate cases you can always get unexpected results.
Your intuition is correct. $X_t$ and $Y_t$ are cointegrated if there exists
some linear combination $\alpha X_t + \beta Y_t$ that is stationary (or more generally, of lower cointegration index --- see for example, Hamilton, pag 571). If $X_t = Y_t$, the above linear combination is zero (hence stationary) whenever $\alpha = -\beta$.
On the other hand, most tests exclude this particular case. The exact reasons depend on the specific test you are using.
Here is an empirical strategy to test for cointegration.
FIRST, check whether both $X_t$ and $Y_t$ contain an unit root.
If they are both non-stationary, and hence $I(1)$, then test for co-integration:
Identical? So $Y_t=X_t$ for all $t$?
Then the difference is zero which is more than just a stationary time-series. They are perfectly collinear.
Now, it depends on the cointegration test you use whether high collinearity will show up as cointegration. If you use Engle-Granger, you do a regression first, which will find $\alpha=0,\beta=1$ and $\epsilon_t=0$. I'm not sure what an ADF test on $0$ will actually do, and this could lead to a numeric instability.
I believe that a Johansen test will also lead to a numeric instability.
If this is a serious problem for some automated system, then probably the first thing to do is look for collinearity (take the timeseries covariance matrix and look at the condition number). If they are highly collinear or identical, you should catch it before running it through the cointegration engine. |
Event detail Analysis and PDE Seminar: Asymptotic analysis of Fourier transform on the Heisenberg group when the vertical frequency tends to 0
Seminar | March 6 | 4:10-5 p.m. | 740 Evans Hall
Hajer Bahouri, Université Paris-Est Créteil
In this joint work with Jean-Yves Chemin and Raphael Danchin, we propose a new approach of the Fourier transform on the Heisenberg group. The basic idea is to take advantage of Hermite functions so as to look at Fourier transform of integrable functions as mappings on the set $\tilde {\mathbb H}^d=\mathbb N^d\times \mathbb N^d\times \mathbb R\setminus \{0\}$ endowed with a suitable distance $\hat d $ (whereas with the standard viewpoint the Fourier transform is a one parameter family of bounded operators on $L^2(\mathbb R^d)$). We prove that the Fourier transform of integrable functions is uniformly continuous on $\tilde {\mathbb H}^d$ (for distance $\hat d$), which enables us to extend $\hat f_\mathbb H$ to the completion $\hat {\mathbb H}^d$ of $\tilde {\mathbb H}^d,$ and to get an explicit asymptotic description of the Fourier transform when the `vertical' frequency tends to $0.$ We expect our approach to be relevant for adapting to the Heisenberg framework a number of classical results for the $\mathbb R^n$ case that are based on Fourier analysis. |
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I am completely lost, anyone mind making the new layout understandable? I see text posts, and random questions everywhere, and I can not make sense of it.
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Lost here as well.
Can't see the problems of the week, now it's just a Facebook like mess.
Completely turned off if this is really "new problems every day".
It was awesome having weekly challenges and being able to solve any of the problems, on demand, not having to wait for tomorrow to be fed another problem.
I guess usage data will tell if Brilliant is right with this change or not. Hopefully they'll revert if they're wrong.
Wonder if they rolled this new user interface to a subset of users first?
Log in to reply
I agree completely, I liked looking over all the new questions Sunday night, and then working on them individually throughout the week. It was all orderly, easy to follow, and it's a a big jumble of information now.
To be completely honest, I really don't like it, and I really don't like how often Brilliant changes the layout of the site. In my opinion, it was as it's best (since I've been on) before this last update.
Hey Noah,
Does that help?
to help you guys, in my case my definition of "best problems" are problems whose rating are close to MY problem solving skills. As it is a General Best problem will more likely be either too hard or too easy for me. that was the beauty of the weekly challenges, you were on a path to advancement, now it feels like there are no goals to achieve on the site.
Hi Angel – The topic feeds only show you problems within a range of a few hundred of your current rating. In fact, the range of difficulty you see now should be the same as the weekly problem sets. We're working on doing a better job showing you only problems that are very close to your rating, but it's a work in progress. Similarly, creating the right pacing of problems for each user is a work in progress. There are a lot of updates coming, and hopefully they will make your experience better. Thanks as always for your feedback.
@Suyeon Khim – way to go with the Color Coding on the feed, this helps a great deal!
@Angel Leon – Thanks! Team stayed late to finish it.
@Suyeon Khim – awesome
It does clear things up, I'm just not on-board %100 with the new layout, and that makes a little bit harder to understand it. Thank you for the reply though!
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This paper investigates the behavior of the Min-Sum message passing scheme tosolve systems of linear equations in the Laplacian matrices of graphs and tocompute electric flows. Voltage and flow problems involve the minimization ofquadratic functions and are fundamental primitives that arise in severaldomains. Algorithms that have been proposed are typically centralized andinvolve multiple graph-theoretic constructions or sampling mechanisms that makethem difficult to implement and analyze. On the other hand, message passingroutines are distributed, simple, and easy to implement. In this paper weestablish a framework to analyze Min-Sum to solve voltage and flow problems. Wecharacterize the error committed by the algorithm on general weighted graphs interms of hitting times of random walks defined on the computation trees thatsupport the operations of the algorithms with time. For $d$-regular graphs withequal weights, we show that the convergence of the algorithms is controlled bythe total variation distance between the distributions of non-backtrackingrandom walks defined on the original graph that start from neighboring nodes.The framework that we introduce extends the analysis of Min-Sum to settingswhere the contraction arguments previously considered in the literature (basedon the assumption of walk summability or scaled diagonal dominance) can not beused, possibly in the presence of constraints.
In probability theory and statistics notions of correlation among randomvariables, decay of correlation, and bias-variance trade-off are fundamental.In this work we introduce analogous notions in optimization, and we show theirusefulness in a concrete setting. We propose a general notion of correlationamong variables in optimization procedures that is based on the sensitivity ofoptimal points upon (possibly finite) perturbations. We present a canonicalinstance in network optimization (the min-cost network flow problem) thatexhibits locality, i.e., a setting where the correlation decays as a functionof the graph-theoretical distance in the network. In the case of warm-startreoptimization, we develop a general approach to localize a given optimizationroutine in order to exploit locality. We show that the localization mechanismis responsible for introducing a bias in the original algorithm, and that thebias-variance trade-off that emerges can be exploited to minimize thecomputational complexity required to reach a prescribed level of erroraccuracy. We provide numerical evidence to support our claims.
We apply the Min-Sum message-passing protocol to solve the consensus problemin distributed optimization. We show that while the ordinary Min-Sum algorithmdoes not converge, a modified version of it known as Splitting yieldsconvergence to the problem solution. We prove that a proper choice of thetuning parameters allows Min-Sum Splitting to yield subdiffusive acceleratedconvergence rates, matching the rates obtained by shift-register methods. Theacceleration scheme embodied by Min-Sum Splitting for the consensus problembears similarities with lifted Markov chains techniques and with multi-stepfirst order methods in convex optimization.
We investigate the fundamental principles that drive the development ofscalable algorithms for network optimization. Despite the significant amount ofwork on parallel and decentralized algorithms in the optimization community,the methods that have been proposed typically rely on strict separabilityassumptions for objective function and constraints. Beside sparsity, thesemethods typically do not exploit the strength of the interaction betweenvariables in the system. We propose a notion of correlation in constrainedoptimization that is based on the sensitivity of the optimal solution uponperturbations of the constraints. We develop a general theory of sensitivity ofoptimizers the extends beyond the infinitesimal setting. We present instancesin network optimization where the correlation decays exponentially fast withrespect to the natural distance in the network, and we design algorithms thatcan exploit this decay to yield dimension-free optimization. Our results arethe first of their kind, and open new possibilities in the theory of localalgorithms.
The discovery of particle filtering methods has enabled the use of nonlinearfiltering in a wide array of applications. Unfortunately, the approximationerror of particle filters typically grows exponentially in the dimension of theunderlying model. This phenomenon has rendered particle filters of limited usein complex data assimilation problems. In this paper, we argue that it is oftenpossible, at least in principle, to develop local particle filtering algorithmswhose approximation error is dimension-free. The key to such developments isthe decay of correlations property, which is a spatial counterpart of the muchbetter understood stability property of nonlinear filters. For the simplestpossible algorithm of this type, our results provide under suitable assumptionsan approximation error bound that is uniform both in time and in the modeldimension. More broadly, our results provide a framework for the investigationof filtering problems and algorithms in high dimension.
We investigate the systematic mechanism for designing fast mixing Markovchain Monte Carlo algorithms to sample from discrete point processes under theDobrushin uniqueness condition for Gibbs measures. Discrete point processes aredefined as probability distributions $\mu(S)\propto \exp(\beta f(S))$ over allsubsets $S\in 2^V$ of a finite set $V$ through a bounded set function$f:2^V\rightarrow \mathbb{R}$ and a parameter $\beta>0$. A subclass of discretepoint processes characterized by submodular functions (which includelog-submodular distributions, submodular point processes, and determinantalpoint processes) has recently gained a lot of interest in machine learning andshown to be effective for modeling diversity and coverage. We show that if theset function (not necessarily submodular) displays a natural notion of decay ofcorrelation, then, for $\beta$ small enough, it is possible to design fastmixing Markov chain Monte Carlo methods that yield error bounds on marginalapproximations that do not depend on the size of the set $V$. The sufficientconditions that we derive involve a control on the (discrete) Hessian of setfunctions, a quantity that has not been previously considered in theliterature. We specialize our results for submodular functions, and we discusscanonical examples where the Hessian can be easily controlled.
It has been established under very general conditions that the ergodicproperties of Markov processes are inherited by their conditional distributionsgiven partial information. While the existing theory provides a rather completepicture of classical filtering models, many infinite-dimensional problems areoutside its scope. Far from being a technical issue, the infinite-dimensionalsetting gives rise to surprising phenomena and new questions in filteringtheory. The aim of this paper is to discuss some elementary examples,conjectures, and general theory that arise in this setting, and to highlightconnections with problems in statistical mechanics and ergodic theory. Inparticular, we exhibit a simple example of a uniformly ergodic model in whichergodicity of the filter undergoes a phase transition, and we develop somequalitative understanding as to when such phenomena can and cannot occur. Wealso discuss closely related problems in the setting of conditional Markovrandom fields.
The Dobrushin comparison theorem is a powerful tool to bound the differencebetween the marginals of high-dimensional probability distributions in terms oftheir local specifications. Originally introduced to prove uniqueness and decayof correlations of Gibbs measures, it has been widely used in statisticalmechanics as well as in the analysis of algorithms on random fields andinteracting Markov chains. However, the classical comparison theorem requiresvalidity of the Dobrushin uniqueness criterion, essentially restricting itsapplicability in most models to a small subset of the natural parameter space.In this paper we develop generalized Dobrushin comparison theorems in terms ofinfluences between blocks of sites, in the spirit of Dobrushin-Shlosman andWeitz, that substantially extend the range of applicability of the classicalcomparison theorem. Our proofs are based on the analysis of an associatedfamily of Markov chains. We develop in detail an application of our mainresults to the analysis of sequential Monte Carlo algorithms for filtering inhigh dimension. |
How can one formalize the fact that the law of $X+Z$ where $X \in \mathbb{R}^d$ is any vector-valued random variable and $Z\sim \mathcal{N}(0, \sigma^2 \mathbf{I}_d)$ closely resembles the law of $Z$ if $\sigma^2$ is sufficiently large ? $X$ and $Z$ are supposed independent. Ideally, I would like to prove that some distance/divergence between the law of $X+Z$ and $Z$ approaches zero as $\sigma^2\to \infty$.
If 𝑋 were Gaussian, for example, with a non-zero mean, $𝜇_{𝑋+𝑍}=𝜇_𝑋+𝜇_𝑍≠𝜇_𝑍.$
$\displaystyle \lim_{\sigma\rightarrow \infty} (\mu_{X+Z} - \mu_{Z})=\mu_X \ne 0.$
So the difference between the law of $X+Z$ and the law of $Z$ does not approach zero. |
skills to develop
To learn the distinction between estimation and prediction. To learn the distinction between a confidence interval and a prediction interval. To learn how to implement formulas for computing confidence intervals and prediction intervals.
Consider the following pairs of problems, in the context of Example 10.4.2, the automobile age and value example.
Problem 1 Estimate the average value of all four-year-old automobiles of this make and model. Construct a \(95\%\) confidence interval for the average value of all four-year-old automobiles of this make and model. Problem 2 Shylock intends to buy a four-year-old automobile of this make and model next week. Predict the value of the first such automobile that he encounters. Construct a \(95\%\) confidence interval for the value of the first such automobile that he encounters.
The method of solution and answer to the first question in each pair, (1a) and (2a), are the same. When we set \(x\) equal to \(4\) in the least squares regression equation
\[\hat{y} =−2.05x+32.83\]
that was computed in part (c) of Example 10.4.2, the number returned,
\[\hat{y}=−2.05(4)+32.83=24.63\]
which corresponds to value \(\$24,630\), is an estimate of precisely the number sought in question (1a): the mean \(E(y)\) of all \(y\) values when \(x = 4\). Since nothing is known about the first four-year-old automobile of this make and model that Shylock will encounter, our best guess as to its value is the mean value \(E(y)\) of all such automobiles, the number \(24.63\) or \(\$24,630\), computed in the same way.
The answers to the second part of each question differ. In question (1b) we are trying to estimate a population parameter: the mean of the all the \(y\)-values in the sub-population picked out by the value \(x=4\), that is, the average value of all four-year-old automobiles. In question (2b), however, we are not trying to capture a fixed parameter, but the value of the random variable \(y\) in one trial of an experiment: examine the first four-year-old car Shylock encounters. In the first case we seek to construct a confidence interval in the same sense that we have done before. In the second case the situation is different, and the interval constructed has a different name, prediction interval. In the second case we are trying to “predict” where a the value of a random variable will take its value.
\(100(1−α)\%\) Confidence Interval for the Mean Value of \(y\) at \(x=x_p\)
\[\hat{y}_p ± t_{α∕2} s_ε \sqrt{\dfrac{1}{n}+ \dfrac{(x_p−\overline{x})^2}{SS_{xx}}} \]
where
\(x_p\) is a particular value of \(x\) that lies in the range of \(x\)-values in the sample data set used to construct the least squares regression line; \(\hat{y}_p\) is the numerical value obtained when the least square regression equation is evaluated at \(x=x_p\); and the number of degrees of freedom for \(t_{α∕2}\) is \(df=n−2\).
The assumptions listed in Section 10.3 must hold.
The formula for the prediction interval is identical except for the presence of the number \(1\) underneath the square root sign. This means that the prediction interval is always wider than the confidence interval at the same confidence level and value of \(x\). In practice the presence of the number \(1\) tends to make it much wider.
\(100(1−α)\%\) Prediction Interval for an Individual New Value of of \(y\) at \(x=x_p\)
\[\hat{y}_p ± t_{α∕2} s_ε \sqrt{1+ \dfrac{1}{n}+ \dfrac{(x_p−\overline{x})^2}{SS_{xx}}} \]
where
\(x_p\) is a particular value of \(x\) that lies in the range of \(x\)-values in the data set used to construct the least squares regression line; \(\hat{y}_p\) is the numerical value obtained when the least square regression equation is evaluated at \(x=x_p\); and the number of degrees of freedom for \(t_{α∕2}\) is \(df=n−2\).
The assumptions listed in Section 10.3 must hold.
Example \(\PageIndex{1}\)
Using the sample data of "Example 10.4.2" in Section 10.4 , recorded in Table 10.4.3, construct a \(95\%\) confidence interval for the average value of all three-and-one-half-year-old automobiles of this make and model.
Solution:
Solving this problem is merely a matter of finding the values of \(\hat{y_p},\; \alpha ,\; and\; \; t_{\alpha /2},S_\varepsilon ,\; \bar{x}\; and\; \; SS_{xx}\), and inserting them into the confidence interval formula given just above. Most of these quantities are already known. From Example 10.4.2, \(SS_{xx}=14\; \; and\; \; \bar{x}=4\). From Example 10.5.2, \(S\varepsilon =1.902169814\).
From the statement of the problem \(x_p=3.5\), the value of \(x\) of interest. The value of \(\hat{y_p}\) is the number given by the regression equation, which by Example 10.4.2 is \(\hat{y}=-2.05x+32.83\), when \(x=x_p\), that is, when \(x=3.5\). Thus here \(\hat{y}=-2.05(3.5)+32.83=25.655\).
Lastly, confidence level \(95\%\) means that \(\alpha =1-0.95=0.05\) so \(\alpha /2=0.025\). Since the sample size is \(n=10\), there are \(n-2=8\) degrees of freedom. By Figure 7.1.6, \(t_{0.025}=2.306\). Thus
\[\begin{align*} \hat{y_p}\pm t_{\alpha /2}S_\varepsilon \sqrt{\frac{1}{n}+\frac{(x_p-\bar{x})^2}{SS_{xx}}} &= 25.655\pm (2.306)(1.902169814)\sqrt{\frac{1}{10}+\frac{(3.5-4)^2}{14}}\\ &= 25.655\pm 4.386403591\sqrt{0.1178571429}\\ &= 25.655\pm 1.506 \end{align*}\]
which gives the interval \((24.149,27.161)\).
We are \(95\%\) confident that the average value of all three-and-one-half-year-old vehicles of this make and model is between \(\$24,149\) and \(\$27,161\).
Example \(\PageIndex{2}\)
Using the sample data of Example 10.4.2, recorded in Table 10.4.3, construct a \(95\%\) prediction interval for the predicted value of a randomly selected three-and-one-half-year-old automobile of this make and model.
Solution:
The computations for this example are identical to those of the previous example, except that now there is the extra number \(1\) beneath the square root sign. Since we were careful to record the intermediate results of that computation, we have immediately that the \(95\%\) prediction interval is
\[\begin{align*} \hat{y_p}\pm t_{\alpha /2}S_\varepsilon \sqrt{1+\frac{1}{n}+\frac{(x_p-\bar{x})^2}{SS_{xx}}} &= 25.655\pm 4.386403591\sqrt{1.1178571429}\\ &= 25.655\pm 4 \end{align*}\]
which gives the interval \((21.017,30.293)\).
We are \(95\%\) confident that the value of a randomly selected three-and-one-half-year-old vehicle of this make and model is between \(\$21,017\) and \(\$30,293\).
Note what an enormous difference the presence of the extra number \(1\) under the square root sign made. The prediction interval is about two-and-one-half times wider than the confidence interval at the same level of confidence.
KeyTakaways A confidence interval is used to estimate the mean value of \(y\) in the sub-population determined by the condition that \(x\) have some specific value \(x_p\). The prediction interval is used to predict the value that the random variable \(y\) will take when \(x\) has some specific value \(x_p\). |
Canonization of linear codes over $\mathbb Z$
4 4
1.
Department of Mathematics, University of Bayreuth, 95440 Bayreuth
4 are equivalent if there is a permutation $\pi \in S_n$ of the coordinates and a vector $\varphi \in \{1,3\}^n$ of column multiplications such that $(\varphi; \pi) C = C'$. This generalizes the notion of code equivalence of linear codes over finite fields. n
In a previous paper, the author has described an algorithm to compute the canonical form of a linear code over a finite field. In the present paper, an algorithm is presented to compute the canonical form as well as the automorphism group of a linear code over $\mathbb Z$
. This solves the isomorphism problem for $\mathbb Z$ 4 -linear codes. An efficient implementation of this algorithm is described and some results on the classification of linear codes over $\mathbb Z$ 4 for small parameters are discussed. 4 Keywords:canonization, group action, representative, coding theory, isometry, Automorphism group, $\mathbb Z$ -linear code.. 4 Mathematics Subject Classification:Primary: 05E20; Secondary: 20B25, 94B0. Citation:Thomas Feulner. Canonization of linear codes over $\mathbb Z$ . 4 Advances in Mathematics of Communications, 2011, 5 (2) : 245-266. doi: 10.3934/amc.2011.5.245
References:
[1]
A. Betten, M. Braun, H. Fripertinger, A. Kerber, A. Kohnert and A. Wassermann, "Error-Correcting Linear Codes, Classification by Isometry and Applications,'',
[2] [3]
A. R. Hammons Jr., P. V. Kumar, A. R. Calderbank, N. J. A. Sloane and P. Solé,
[4]
T. Honold and I. Landjev,
[5]
W. C. Huffman and V. Pless, "Fundamentals of Error-Correcting Codes,'',
[6]
M. Kiermaier and J. Zwanzger,
[7]
R. Laue,
[8] [9] [10]
A. A. Nechaev,
[11] [12]
C. C. Sims,
show all references
References:
[1]
A. Betten, M. Braun, H. Fripertinger, A. Kerber, A. Kohnert and A. Wassermann, "Error-Correcting Linear Codes, Classification by Isometry and Applications,'',
[2] [3]
A. R. Hammons Jr., P. V. Kumar, A. R. Calderbank, N. J. A. Sloane and P. Solé,
[4]
T. Honold and I. Landjev,
[5]
W. C. Huffman and V. Pless, "Fundamentals of Error-Correcting Codes,'',
[6]
M. Kiermaier and J. Zwanzger,
[7]
R. Laue,
[8] [9] [10]
A. A. Nechaev,
[11] [12]
C. C. Sims,
[1]
Michael Kiermaier, Johannes Zwanzger.
A $\mathbb Z$
Advances in Mathematics of Communications,
2011, 5
(2)
: 275-286.
doi: 10.3934/amc.2011.5.275 [2]
Martino Borello, Francesca Dalla Volta, Gabriele Nebe.
The automorphism group of a self-dual $[72,36,16]$ code does not contain $\mathcal S_3$, $\mathcal A_4$ or
$D_8$.
[3] [4] [5]
Thomas Feulner.
The automorphism groups of linear codes and canonical representatives of their semilinear isometry classes.
[6] [7] [8] [9]
Helena Rifà-Pous, Josep Rifà, Lorena Ronquillo.
$\mathbb{Z}_2\mathbb{Z}_4$-additive perfect codes in Steganography.
[10]
S. A. Krat.
On pairs of metrics invariant under a cocompact action of a group.
[11]
Tingting Wu, Jian Gao, Yun Gao, Fang-Wei Fu.
$ {{\mathbb{Z}}_{2}}{{\mathbb{Z}}_{2}}{{\mathbb{Z}}_{4}}$-additive cyclic codes.
[12]
Amit Sharma, Maheshanand Bhaintwal.
A class of skew-cyclic codes over $\mathbb{Z}_4+u\mathbb{Z}_4$ with derivation.
[13]
Carlos Matheus, Jean-Christophe Yoccoz.
The action of the affine diffeomorphisms on the relative homology group of certain exceptionally symmetric origamis.
[14]
Xiaojun Huang, Yuan Lian, Changrong Zhu.
A Billingsley-type theorem for the pressure of an action of an amenable group.
[15]
Joaquim Borges, Steven T. Dougherty, Cristina Fernández-Córdoba.
Characterization and constructions of self-dual codes over $\mathbb Z_2\times \mathbb Z_4$.
[16] [17] [18] [19]
François Gay-Balmaz, Cesare Tronci, Cornelia Vizman.
Geometric dynamics on the automorphism group of principal bundles: Geodesic flows, dual pairs and chromomorphism groups.
[20]
2018 Impact Factor: 0.879
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ISSN:
1078-0947
eISSN:
1553-5231
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Discrete & Continuous Dynamical Systems - A
July 2010 , Volume 28 , Issue 3
A special issue
Dedicated to Louis Nirenberg on the Occasion of his 85th Birthday Part II
Select all articles
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Abstract:
"One of the wonders of mathematics is you go somewhere in the world and you meet other mathematicians, and it is like one big family. This large family is a wonderful joy."
Louis Nirenberg, in an interview in the
Notices of the AMS, April 2002.
Louis Nirenberg was born in Hamilton, Ontario on February 28, 1925. He was attracted to physics as a high school student in Montreal while attending the Baron Byng School. He completed a major in Mathematics and Physics at McGill University. Having met Richard Courant, he went to graduate school at NYU and what would become the Courant Institute. There he completed his PhD degree under the direction of James Stoker. He was then invited to join the faculty and has been there ever since. He was one of the founding members of the Courant Institute of Mathematical Sciences and is now an Emeritus Professor.
For more information please click the “Full Text” above.
Abstract:
We prove that global Lipschitz solutions to the linearized Monge-Ampere equation
$ L_$φ$ u$:$=\sum $φ
ij$u_{ij}=0$
must be linear in $2D$. The function φ is assumed to have the Monge-Ampere measure $\det D^2 $φ bounded away from $0$ and $\infty$.
Abstract:
This paper is devoted to the study of solitons whose existence is related to the ratio energy/charge. These solitons are called hylomorphic. In the first part of the paper we prove an abstract theorem on the existence of hylomorphic solitons which can be applied to the main situations considered in literature. In the second part, we apply this theorem to the nonlinear Schrödinger and Klein Gordon equations defined on a lattice.
Abstract:
In this paper we study partial and anisotropic Schauder estimates for linear and nonlinear elliptic equations. We prove that if the inhomogeneous term $f$ is Hölder continuous in the $x_n$-direction, then the mixed derivatives u
xxnare Hölder continuous; if $f$ satisfies an anisotropic Hölder continuity condition, then the second derivatives $D^2 u$ satisfy related anisotropic Hölder continuity estimates. Abstract:
In this paper, we study the local structure and the smoothness of singularities of free boundaries in an optimal partition problem for the Dirichlet eigenvalues. We prove that there is a unique homogeneous blow up(tangent map) at each singular point in the interior of the free boundary. As a consequence we obtain the rectifiability as well as local structures of singularities.
Abstract:
We present some recent results on mean field equations of Liouville type over a closed surface, in presence of Dirac distributions supported at the so called "vortex points". We discuss possible existence and non-existence results as well as uniqueness and multiplicity issues according to the topological and geometrical properties of the surface.
Abstract:
Let (M ,ğ) be an $N$-dimensional smooth (compact or noncompact) Riemannian manifold. We introduce the elliptic
Jacobi-Toda systemon (M ,ğ). We review various recent results on its role in the construction of solutions with multiple interfaces of the Allen-Cahn equation on compact manifolds and entire space, as well as multiple-front traveling waves for its parabolic counterpart. Abstract:
We follow up our work [4] concerning the formation of trapped surfaces. We provide a considerable extension of our result there on pre-scared surfaces to allow for the formation of a surface with multiple pre-scared angular regions which, together, can cover an arbitrarily large portion of the surface. In a forthcoming paper we plan to show that once a significant part of the surface is pre-scared, it can be additionally deformed to produce a bona-fide trapped surface.
Abstract:
We examine the regularity of the extremal solution of the nonlinear eigenvalue problem $\Delta^2 u = \lambda f(u)$ on a general bounded domain $\Omega$ in $ \R^N$, with the Navier boundary condition $ u=\Delta u =0 $ on δΩ. We establish energy estimates which show that for any non-decreasing convex and superlinear nonlinearity $f$ with $f(0)=1$, the extremal solution u
*is smooth provided $N\leq 5$. If in addition $\lim$i$nf_{t \to +\infty}\frac{f (t)f'' (t)}{(f')^2(t)}>0$, then u *is regular for $N\leq 7$, while if $\gamma$:$= \lim$s$up_{t \to +\infty}\frac{f (t)f'' (t)}{(f')^2(t)}<+\infty$, then the same holds for $N < \frac{8}{\gamma}$. It follows that u *is smooth if $f(t) = e^t$ and $ N \le 8$, or if $f(t) = (1+t)^p$ and $N< \frac{8p}{p-1}$. We also show that if $ f(t) = (1-t)^{-p}$, $p>1$ and $p\ne 3$, then u *is smooth for $N \leq \frac{8p}{p+1}$. While these results are major improvements on what is known for general domains, they still fall short of the expected optimal results as recently established on radial domains, e.g., u *is smooth for $ N \le 12$ when $ f(t) = e^t$ [11], and for $ N \le 8$ when $ f(t) = (1-t)^{-2}$ [9] (see also [22]). Abstract:
Limiting profiles of solutions to a 2$\times$2 Lotka-Volterra competition-diffusion-advection system, when the strength of the advection tends to infinity, are determined. The two species, competing in a heterogeneous environment, are identical except for their dispersal strategies: One is just random diffusion while the other is "smarter" - a combination of random diffusion and a directed movement up the environmental gradient. With important progress made, it has been conjectured in [2] and [3] that for large advection the "smarter" species will concentrate near a selected subset of positive local maximum points of the environment function. In this paper, we establish this conjecture in one space dimension, with the peaks located and the limiting profiles determined, under mild hypotheses on the environment function.
Abstract:
We prove that the solution to a parabolic integro-differential equation with a gradient dependence that satisfies a critical power growth becomes immediately Hölder continuous. We also obtain some results in the supercritical case.
Abstract:
We consider the following semilinear elliptic equation on a strip:
$\Delta u-u + u^p=0 \ \mbox{in} \ \R^{N-1} \times (0, L),$
$ u>0, \frac{\partial u}{\partial \nu}=0 \ \mbox{on} \ \partial (\R^{N-1} \times (0, L)) $
where $ 1< p\leq \frac{N+2}{N-2}$. When $ 1 < p <\frac{N+2}{N-2}$, it is shown that there exists a unique L
* >0 such that for L $\leq $L * , the least energy solution is trivial, i.e., doesn't depend on $x_N$, and for L >L * , the least energy solution is nontrivial. When $N \geq 4, p=\frac{N+2}{N-2}$, it is shown that there are two numbers L * < L ** such that the least energy solution is trivial when L $\leq$L *, the least energy solution is nontrivial when L $\in$(L *,L **], and the least energy solution does not exist when L >L **. A connection with Delaunay surfaces in CMC theory is also made. Abstract:
In 1928, motivated by conversations with Keynes, Ramsey formulated an infinite-horizon problem in the calculus of variations. This problem is now classical in economic theory, and its solution lies at the heart of our understanding of economic growth. On the other hand, from the mathematical point of view, it was never solved in a satisfactory manner: In this paper, we give what we believe is the first complete mathematical treatment of the problem, and we show that its solution relies on solving an implicit differential equation. Such equations were first studied by Thom, and we use the geometric method he advocated. We then extend the Ramsey problem to non-constant discount rates, along the lines of Ekeland and Lazrak. In that case, there is time-inconsistency, meaning that optimal growth no longer is a relevant concept for economics, and has to be replaced with equlibrium growth. We briefly define what we mean by equilibrium growth, and proceed to prove that such a path actually exists, The problem, once again, reduces to solving an implicit differential equation, but this time the dimension is higher, and the analysis is more complicated: geometry is not enough, and we have to appeal to the central manifold theorem.
Abstract:
In this paper, we prove interior second derivative estimates of Pogorelov type for a general form of Monge-Ampère equation which includes the optimal transportation equation. The estimate extends that in a previous work with Xu-Jia Wang and assumes only that the matrix function in the equation is regular with respect to the gradient variables, that is it satisfies a weak form of the condition introduced previously by Ma,Trudinger and Wang for regularity of optimal transport mappings. We also indicate briefly an application to optimal transportation.
Abstract:
This is an expository paper dedicated to professor L. Nirenberg for his 85th birthday. First I will discuss my joint works with Z. Zhang and J. Song on the singularity formation of Kähler-Ricci flow. Secondly, I will show a fully nonlinear equation,
scalar V-soliton equation(cf. Section 4, (14)), and some basic results about it. This equation was introduced by G. La Nave and myself in studying the singularity formation of Kähler-Ricci flow. I will also show how this new equation can be applied to studying the singularity formation at finite time. Abstract:
We give a positive lower bound for the principal curvature of the strict convex level sets of harmonic functions in terms of the principal curvature of the domain boundary and the norm of the boundary gradient. We also extend this result to a class of semi-linear elliptic partial differential equations under certain structure condition.
Abstract:
A beautiful and influential subject in the study of the question of smoothness of solutions for the Navier - Stokes equations in three dimensions is the theory of partial regularity. A major paper on this topic is Caffarelli, Kohn & Nirenberg [5](1982) which gives an upper bound on the size of the singular set $S(u)$ of a suitable weak solution $u$. In the present paper we describe a complementary lower bound. More precisely, we study the situation in which a weak solution fails to be continuous in the strong $L^2$ topology at some singular time $t=T$. We identify a closed set in space on which the $L^2$ norm concentrates at this time $T$, and we study microlocal properties of the Fourier transform of the solution in the cotangent bundle T
*(R 3) above this set. Our main result is that $L^2$ concentration can only occur on subsets of T *(R 3) which are sufficiently large. An element of the proof is a new global estimate on weak solutions of the Navier - Stokes equations which have sufficiently smooth initial data. Abstract:
We establish sharp energy estimates for some solutions, such as global minimizers, monotone solutions and saddle-shaped solutions, of the fractional nonlinear equation $(-\Delta)$
1/2$u=f(u)$ in R n. Our energy estimates hold for every nonlinearity $f$ and are sharp since they are optimal for one-dimensional solutions, that is, for solutions depending only on one Euclidian variable.
As a consequence, in dimension $n=3$, we deduce the one-dimensional symmetry of every global minimizer and of every monotone solution. This result is the analog of a conjecture of De Giorgi on one-dimensional symmetry for the classical equation $-\Delta u=f(u)$ in R
n. Abstract:
We imbed an array of thin highly conductive fibers in a surrounding two-dimensional medium with small viscosity. The resulting composite medium is described by a second order elliptic operator in divergence form with discontinuous singular coefficients on an open domain of the plane. We study the asymptotic spectral behavior of the operator when, simultaneously, the viscosity vanishes and the fibers develop fractal geometry. We prove that the spectral measure of the operator converges to the spectral measure of a self-adjoint operator associated with the lower-dimensional fractal limit of the thin fibers. The limit fiber is a compact set that disconnects the initial domain into infinitely many non-empty open components. Our approach is of variational nature and relies on Hilbert space convergence of quadratic energy forms.
Abstract:
In this and the subsequent paper, we are interested in the following nonlinear equation:
$\Delta_g v+\rho(\frac{h^* e^v}{\int_M h^* vd\mu(x)}-1)= 4\pi\sum_{j=1}^N\alpha_j(\delta_{q_i}-1)\quad\text{in }M,$(0.1)
where $(M,g)$ is a Riemann surface with its area $|M|=1$; or
$\Delta v+\rho\frac{h^*e^v}{\int_\Omega h^* e^vdx}=4\pi\sum_{j=1}^N\alpha_j \delta_{q_j}\quad\text{in }\Omega, $ (0.2)
where $\Omega$ is a bounded smooth domain in $ R^2$. Here,
$\rho, \alpha_j$ are positive constants, $\delta_q$ is the Dirac
measure at $q$, and both $h^*$'s are positive smooth functions. In
this paper, we prove a sharp estimate for a sequence of blowing up
solutions $u_k$ to (0.1) or (0.2) with
$\rho_k\rightarrow\rho
*. Among other things, we show that for
equation (0.1),
$\rho_k-\rho_*=\sum_{j=1}^\tau d_j( \Delta \log h^*(p_j)+\rho_*-N^*-2K(p_j)+o(1) )e^{-\frac{\lambda_k}{1+\alpha_j}}, $ (0.3)
and for equation (0.2),
$ \rho_k-\rho_*=\sum_{j=1}^\tau d_j(\Delta \log h^*(p_j)+o(1))e^{-\frac{\lambda_k}{1+\alpha_j}},$ (0.4)
where $\lambda_k\rightarrow+\infty$ and $d_j$ is a constant depending on $p_j$, a blow up point of $u_k$. See section 1 for more precise description. These estimates play an important role when the degree counting formulas are derived. The subsequent paper [19] will complete the proof of computing the degree counting formula.
Abstract:
We study the local behavior of a solution to the Stokes system with singular coefficients in $R^n$ with $n=2,3$. One of our main results is a bound on the vanishing order of a nontrivial solution $u$ satisfying the Stokes system, which is a quantitative version of the strong unique continuation property for $u$. Different from the previous known results, our strong unique continuation result only involves the velocity field $u$. Our proof relies on some delicate Carleman-type estimates. We first use these estimates to derive crucial
optimalthree-ball inequalities for $u$. Taking advantage of the optimality, we then derive an upper bound on the vanishing order of any nontrivial solution $u$ to the Stokes system from those three-ball inequalities. As an application, we derive a minimal decaying rate at infinity of any nontrivial $u$ satisfying the Stokes equation under some a priori assumptions. Abstract:
We prove two mixed versions of the Discrete Nodal Theorem of Davies et. al. [3] for bounded degree graphs, and for three-connected graphs of fixed genus $g$. Using this we can show that for a three-connected graph satisfying a certain volume-growth condition, the multiplicity of the $n$th Laplacian eigenvalue is at most $2[ 6(n-1) + 15(2g-2)]^2$. Our results hold for any Schrödinger operator, not just the Laplacian.
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Change to browse by: Bookmark(what is this?) Astrophysics > High Energy Astrophysical Phenomena Title: NuSTAR Observations of the Accreting Atolls GX 3+1, 4U 1702-429, 4U 0614+091, and 4U 1746-371
(Submitted on 1 Feb 2019)
Abstract: Atoll sources are accreting neutron star (NS) low-mass X-ray binaries. We present a spectral analysis of four persistent atoll sources (GX 3+1, 4U 1702$-$429, 4U 0614+091, and 4U 1746$-$371) observed for $\sim20$ ks each with NuSTAR to determine the extent of the inner accretion disk. These sources range from an apparent luminosity of $0.006-0.11$ of the Eddington limit (assuming the empirical limit of $3.8\times10^{38}$ ergs s$^{-1}$). Broad Fe emission features shaped by Doppler and relativistic effects close to the NS were firmly detected in three of these sources. The position of the disk appears to be close to the innermost stable circular orbit (ISCO) in each case. For GX 3+1, we determine $R_{in}=1.8^{+0.2}_{-0.6}\ R_{\mathrm{ISCO}}$ (90% confidence level) and an inclination of $27^{\circ}-31^{\circ}$. For 4U 1702$-$429, we find a $R_{in}=1.5_{-0.4}^{+1.6}\ R_{\mathrm{ISCO}}$ and inclination of $53^{\circ}-64^{\circ}$. For 4U 0614+091, the disk has a position of $R_{in}=1.3_{-0.2}^{+5.4}\ R_{\mathrm{ISCO}}$ and inclination of $50^{\circ}-62^{\circ}$. If the disk does not extend to the innermost stable circular orbit, we can place conservative limits on the magnetic field strength in these systems in the event that the disk is truncated at the Alfv\'{e}n radius. This provides the limit at the poles of $B\leq6.7\times10^{8}$ G, $3.3\times10^{8}$ G, and $14.5\times10^{8}$ G for GX 3+1, 4U 1702$-$429, and 4U 0614+091, respectively. For 4U 1746$-$371, we argue that the most plausible explanation for the lack of reflection features is a combination of source geometry and strong Comptonization. We place these sources among the larger sample of NSs that have been observed with NuSTAR. Submission historyFrom: Renee Ludlam [view email] [v1]Fri, 1 Feb 2019 19:00:07 GMT (145kb) |
“All characters and events in this show—even those based on real people—are entirely fictional. All celebrity voices are impersonated…..poorly. The following program contains coarse language and due to its content it should not be viewed by anyone.” - South Park disclaimer
South Park follows four fourth grade boys (Stan, Kyle, Cartman and Kenny) and an extensive ensemble cast of recurring characters. This analysis reviews their speech to determine which words and phrases are distinct for each character. Since the series uses a lot of running gags, common phrases should be easy to find.
The programming language R and packages tm, RWeka and stringr were used to scrape South Park episode transcripts from the internet, attribute them to a certain character, break them into ngrams, calculate the log likelihood for each ngram/character pair, and rank them to create a list of most characteristic words/phrases for each character. The results were visualized using ggplot2, wordcloud and RColorBrewer. Full scripts on Github.
Full report Method & Summary Statistics
I used the stringr package to condense each speaker’s text into one large unformatted string. From there, I used the tm package to pre-process the text (to lowercase, remove punctuation, numbers and white space; remove stop words) and form a corpus, which contained more than 30,600 unique words spoken more than 836,000 times. Reducing the sparsity brought that down to about 3,100 unique words. Processing the text reduced it further to our final subset of roughly 1,000 unique words and phrases.
29 characters with the most words were retained, and the remaining 3,958 speakers combined into one “all others” category so as not to lose the text.
speaker words speaker words speaker words cartman 61110 jimmy 3738 narrator 1737 stan 34762 gerald 3285 principal.victoria 1732 kyle 31277 jimbo 3157 jesus 1714 randy 14994 announcer 2900 mayor 1603 butters 13690 wendy 2893 craig 1412 mr..garrison 9436 sheila 2794 reporter 1400 chef 5493 liane 2477 satan 1291 mr..mackey 4829 stephen 2245 linda 1285 sharon 4284 kenny 2112 all.others 152172 Exploring the text
Overall (total words spoken by character / total words spoken), Cartman accounts for the biggest share of words.
It's no surprise that Kenny accounts for a measly 1-2% per season or that Cartman accounts for 15-20% of words (overall, he accounts for 16.4%).
Butters and Randy have become more integral characters over time, leading to a stark increase in their share of words and decrease in Stan and Kyle's share. Chef's share declines over time and ultimately ends completely after his death in season 10. Mr. Garrison trails off after becoming Mrs. Garrison in season 9, returning in season 12's "Eek, a Penis!" episode.
Profanity
Though Kenny doesn't speak as much as the rest of the characters, he swears at a much higher rate.
Individual swear words have been featured in different episodes: shit is common in "You're Getting Old" as well as the record-breaking episode "It Hits the Fan," while hell takes over in successive episodes "Do the Handicapped Go to Hell?" and "Probably." Ass and bitch share the spotlight in "Le Petit Tourette" and bitch shines in "Butters' Bottom Bitch." If you look closely, you can see a marked decreased in ass (red) and increase in fuck (orange). Cartman drove both words' usage, showing either a character development (growing up?) or a shift in preference by the writers.
Identifying Characteristic Words using Log Likelihood
Each corpus was analyzed to determine the most characteristic words for each speaker. Frequent and characteristic words are not the same thing - otherwise words like “I”, “school”, and “you” would rise to the top instead of unique words and phrases like “professor chaos”, “hippies” and “you killed kenny.”
Log likelihood was used to measure the unique-ness of the ngrams by character. Log likelihood compares the occurrence of a word in a particular corpus (the body of a character’s speech) to its occurrence in another corpus (all of the remaining South Park text) to determine if it shows up more or less likely that expected. The returned value represents the likelihood that the corpora are from the same, larger corpus, like a t-test. The higher the score, the more unlikely.
The
chi-square test, or goodness-of-fit test, can be used to compare the occurrence of a word across corpora.
$$\chi^{2} = \sum \frac{(O_i-E_i)^{2}}{E_i}$$
where O = observed frequency and E = expected frequency.
However, flaws have been identified: invalidity at low frequencies (Dunning, 1993) and over-emphasis of common words (Kilgariff, 1996). Dunning was able to show that the log-likelihood statistic was accurate even at low frequencies:
$$2\sum O_i * ln(\frac{O_i}{E_i})$$
Which can be computed from the contingency table below as $$(2 * ((a * log(\frac{a}{E1}) + (b * log(\frac{b}{E2})));$$
$$E1 = (a+c) * \frac{a + b}{N}; E2 = (b+d) * \frac{a + b}{N}$$
Group Corpus.One Corpus.Two Total Word a b a+b Not Word c d c+d Total a+c b+d N=a+b+c+d
Group Cartmans.Text Remaining.Text Total ‘hippies’ 36 5 41 Not ‘hippies’ 28170 144058 172228 Total 28206 144063 172269
Computed:
E1 = 28206 (41/172269) = 6.71 E2 = 144063(41/172269) = 34.28
LL = 2
(36log(36/6.71) + 5*log(5/34.28)) = 101.7
Based on the overall ratio of the word “hippies” in the text, 41/172269 = 0.00023, we would expect to see hippies in Cartman’s speech 6.71 times and in the remaining text 34.28 times. The log likelihood value of 101.7 is significant far beyond even the 0.01% level, so we can reject the null hypothesis that Cartman and the remaining text are one and the same.
Only ngrams that passed a certain threshold were included in the log likelihood test; for unigrams, 50 occurrences, for bigrams, 25, for tri-grams, 15, 4-grams, 10 and 5-grams 5. Each ngram was then compared to all speakers, including those who said it 0 times (using 0.0001 in place of 0 to prevent breaking the log step of the formula). If the number of times the speaker said the word was less than expected, the log likelihood was multiplied by -1 to produce a negative result.
For this analysis, a significance level of 0.001 was chosen to balance the number of significant ngrams with significance. 1.31% of ngrams were found to be significantly characteristic of a given character.
Level Critical.Value p.Value Percent.Sig 5% 3.84 0.05 7.92 1% 6.63 0.01 5.21 0.1% 10.83 0.001 3.66 0.01% 15.13 0.0001 2.83
The results were then filtered to include each word two times or less: once for the speaker most likely to say it (highest LL) and once for the speaker least likely to say it (lowest LL).
Ranking
Finally, the results were ranked using the formula
LL∗ngram.length
Ranking was used to condense the range of log likelihoods (-700 to 1000+). The ranking formula includes ngram length because longer ngrams appear fewer times in the text, leading to lower log likelihoods, but carry more semantic meaning.
References
Dunning, T. (1993) Accurate Methods for the Statistics of Surprise and Coincidence. Computational Linguistics, 19, 1, March 1993, pp. 61-74.
Kilgarriff. A. (1996) Why chi-square doesn’t work, and an improved LOB-Brown comparison. ALLC-ACH Conference, June 1996, Bergen, Norway. |
As promised in the previous article, I plan to review Reference 2,
Active Queue Management with Non-Linear Packet Dropping Function, by D. Augustyn, A. Domanski, and J. Domanska, published in HET-NETs 2010, which discusses a change in the structure of the packet drop probability function using the average queue length in a buffer. I mentioned previously that choosing a linear function of the average queue length can be viewed as a bit of an arbitrary choice, since we’re designing a control mechanism here, and this paper attempts to define a new form of this packet drop probability function.
In summary, the best lesson one can take from this paper is that publication in a journal or conference proceedings does not guarantee that the paper withstands scrutiny. The paper is linked above for the interested reader to peruse himself, and to investigate the claims.
Summary
The paper intended to give a new function to calculate the probability of proactively dropping a packet in a queue in order to prevent a full buffer. It seemed to be presented as an alternative to RED, described in my previous article. The authors define this new function, then set up a simulation in order to examine the effects.
When Orthogonal is Abused
The authors describe using a finite linear combination of orthogonal basis polynomials defined on a finite interval as the underlying mathematical structure.
First, we should discuss what we mean by orthogonal in context of functions.
Orthogonal is most commonly understood in terms of vectors, and when we’re in two dimensions, orthogonal becomes our familiar perpendicular. Orthogonality
Beginning with the familiar notion of perpendicular, we can generalize this to understand orthogonality. The geometric interpretation of two vectors being perpendicular is that the angle between them is 90^{\circ}. Once we leave two and three dimensions (or jump to the space of polynomials, as we’ll do soon), the concept of an angle isn’t as helpful.
Another way to define perpindicular is through an operation known as the
dot product . Suppose we take two 2D vectors, \mathbf{x} and \mathbf{y}. Each vector will have coordinates: \mathbf{x} = (x_{1},x_{2}) and \mathbf{y} = (y_{1}, y_{2}). The dot productis a special type of multiplication defined on vectors, and denoted \cdot: \mathbf{x}\cdot\mathbf{y} = x_{1}y_{1} + x_{2}y_{2}
The dot product can be described in words as the sum of the component-wise multiplication of the coordinates of the two vectors.
Now, we can say that two vectors are perpendicular if their dot product is 0. That is, \mathbf{x} and \mathbf{y} are perpindicular if \mathbf{x}\cdot\mathbf{y} = 0. (This article gives a nice overview of how we move from the algebraic to the geometric definition of perpendicular and orthogonal.)
Remember, perpendicular doesn’t make sense once we get into higher dimensions.
Orthogonal is a more general notion of vectors being perpendicular, and is defined for two vectors (of any length) as their dot product equalling zero. From Dot Product to Inner Product
The dot product is used on vectors, and defines another type of product that is different from the scalar multiplication we know. In fact, we can generalize the notion of a dot product to something called an
, which can be defined on many different spaces than just vectors. We can define operations and products however we like, but for our definition to qualify as an inner product (denoted \langle \cdot, \cdot\rangle), it must meet certain criteria. inner product
For instance, on the set of real valued functions with domain [a,b], we define the inner product of two functions f(x) and g(x) as
\langle f, g\rangle := \int_{a}^{b}f(x)g(x)dx
The concept of orthogonality generalizes to an inner product as well. If the inner product of two functions is zero (as defined above), we say the functions are orthogonal.
Back to the paper
The authors claim to be using a set of orthogonal polynomials to define their drop probability function, and they give the structure of such functions. For a domain [a,b], and for \phi_{j} in the set of polynomials, they define \phi_{j} = (x-a)^{j-1}(b-x). So, for example, \phi_{1} = (b-x), and \phi_{5} = (x-a)^{4}(b-x).
Now, in order to be an orthogonal basis for a space
1, the set of functions that are claimed to form the basis of the set must be pairwise orthogonal. That is, I need to be able to take the inner product of any two functions \phi_{i} and \phi_{j} and get 0. If that isn’t true for even one pair, then the set is not orthogonal.
As it turns out, if we take the inner product of any two functions in the basis set over the domain given, we find that there are
no pairs that are orthogonal. To do this in general, we compute the integral
The integral computation is one of simple polynomial integration, and can be done either by hand or using your favorite software (Mathematica) of choice. What we find here is that this set of functions defined in general this way is never orthogonal, yet the paper claims they are.
Applying to the particular situation of designing a drop probability function, they give the following for average queue length thresholds T_{\min} and T_{\max} p(x,a_{1},a_{2}) = \left\{\begin{array}{lr}0, &x < T_{\min}\\\phi_{0} + a_{1}\phi_{1}(x) + a_{2}\phi_{2}(x),&T_{\min}\leq x \leq T_{\max}\\1,&x > T_{\max}\end{array}\right.
where the basis functions are\begin{aligned}\phi_{0}(x) &= p_{m}\frac{x-T_{\min}}{T_{\max}-T_{\min}}\\\phi_{1}(x) &= (x-T_{\min})(T_{\max}-x)\\\phi_{2}(x) &= (x-T_{\min})^{2}(T_{\max}-x)\end{aligned}
The reader will recognize \phi_{0} as the original drop probability function from the RED algorithm. These functions are absolutely not orthogonal though (as the authors claim), and a simple check as we did above will verify it.
Other issues
Another issue is that these mysterious coefficients a_{1} and a_{2} need to be determined. How, you ask? The authors do not say, other than to note that one can define “a functional” implicit on the unknown p(x,a_{1},a_{2}) that can be minimized to find the optimal values for those coefficients. They write that this mysterious functional
2 can be based on either the average queue length or average waiting time, yet provide no details whatsoever as to the functional they have chosen for this purpose. They provide a figure with a sample function, but give no further details as to how it was obtained.
One other issue I have in their methodology is discussing the order of estimation. For those familiar with all sorts of ways to estimate unknown functions, from Taylor series, to splines, to Fourier series, we know that a function is exactly equal to
an infinite sum of such functions. Any finite sum is an estimation, and the number of terms we use for estimation with a desired accuracy may change with the function being estimated.
For instance, if I want to use Taylor series (a linear combination of polynomials) to estimate a really icky function, how many terms should I include to ensure my accuracy at a point is within 0.001 of the real value? It depends on the function.
3.
The authors simply claim that a second order term is appropriate in this case. The issue I take with that is these queueing management drop probability functions are designed. We’re not estimating a function describing a phenomenon, we are designing a control policy to seek some sort of optimal behavior. Fundamentally, the authors posing this as an estimation problem of some unknown drop probability function is incorrect. This isn’t a law of nature we seek to observe; it’s a control policy we seek to design and implement and optimize. Using language that implies the authors are estimating some “true function” is misleading.
Regarding the simulation itself, since the design was arbitrary, and not based on sound mathematical principles, I cannot give any real comment to the simulations and the results. The authors briefly discuss and cite some papers that explore the behavior of network traffic, and claim to take this into account in their simulations. To those, I cannot comment yet.
Conclusion
Always verify a paper for yourself, and don’t accept anything at face value. Research and technical publications should be completely transparent as to choices and methodologies (and obviously free of mathematical inaccuracies), and derivations and proofs should be present when necessary, even if in an appendix.
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.
Footnotes This discussion on basis functions can get very deep, and the linear algebra can get heavy-ish. Right now I’m really focused on pointing out errors in the paper, so don’t worry too much about the rigorous details of estimating a function by a basis set. I did not make a typo. A functional is a mapping that takes functions as input. Many engineers use a first or second order approximation as a rule of thumb, though it really should still be verified that higher order terms contribute a negligible amount first |
Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.
Dear Uncle Colin,
I need to find an angle! ABC is a triangle with median AD, while angles BAD and CAD are 110º and 20º, respectively. What's angle ACB?
-- Angle Being Evasive, LOL
Hi, ABEL, and thanks for your question! Even if you've used degrees. For heaven's sake, get a grip.
I'd start by drawing a picture, which you can do yourself1, and asking what I can make concrete. If I call the missing angle ACB $\theta$, then I know that CDA is $160-\theta$, CDB is $20+\theta$ and ABC is $50+\theta$. Not all of these will prove useful, but they're good to have.
We also know that the lengths CD and DB are the same, because AD is a median. Let's call that $x$ and the lenght of AB $y$.
I'm going to work with two triangles separately, applying the sine rule to both ACD and ADB.
For ACD, I have $\frac{y}{\sin(\theta)}= \frac{x}{\sin(20)}$. For ADB, I've got $\frac{x}{\sin(110)} = \frac{y}{\sin(50-\theta)}$.
I neither know nor care about $x$ and $y$, and I'm going to eliminate them. In the first triangle, $\frac{y}{x} = \frac{\sin(\theta)}{\sin(20)}$ and in the second, $\frac{y}{x} = \frac{\sin(50-\theta)}{\sin(110)}$.
There's also a nice symmetry: $\sin(110) = \sin(70) = \cos(20)$, which we can use to good effect.
Doing that substitution, we have $\frac{\sin(\theta)}{\sin(20)} = \frac{\sin(50-\theta)}{\cos(20)}$. Rearranging, we get $\frac{\sin(\theta)}{\sin(50-\theta)} = \tan(20)$, which I'll call $T$.
Now, $\sin(\theta) = T \sin(50-\theta)$. Expanding the right hand side gives $\sin(\theta) = T \sin(50)\cos(\theta) - T \cos(50)\sin(\theta)$.
Ah, all of those lovely $\theta$s! You know, we can divide them all by $\cos(\theta)$ and make it even simpler: $\tan(\theta) = T \sin(50) - T \cos(50) \tan(\theta)$
Regroup:
$\tan(\theta) \left[ 1 + T \cos(50) \right] = T\sin(50)$
$\tan(\theta) = \frac{\tan(20)\sin(50)}{1 + \tan(20)\cos(50)}$
Most reasonable people would incur the Mathematical Ninja's displeasure at that point by reaching for a calculator; even a cursory "that's about $\frac{\frac{1}{3}\frac{4}{5}}{1 + \frac{1}{3}\frac{3}{5}} = \frac{\frac{4}{15}}{\frac{6}{5}} = \frac{4}{18}$, about 0.222" would do him (it's actually 0.226, so well done us), and the arctangent of that is in the 12-13 degree region. Or, if you want to do it the easy way, 12.73.
Hope that helps!
-- Uncle Colin |
Research Open Access Published: Random periodic solution for a stochastic SIS epidemic model with constant population size Advances in Difference Equations volume 2018, Article number: 64 (2018) Article metrics
792 Accesses
4 Citations
Abstract
In this paper, a stochastic susceptible-infected-susceptible (SIS) epidemic model with periodic coefficients is formulated. Under the assumption that the total population is fixed by
N, an analogue of the threshold \({R_{0}^{T}}\) is identified. If \({R_{0}^{T} > 1}\), the model is proved to admit at least one random periodic solution which is nontrivial and located in \((0,N)\times(0,N)\). Further, the conditions for persistence and extinction of the disease are also established, where a threshold is given in the case that the noise is small. Comparing with the threshold of the autonomous SIS model, it is generalized to its averaged value in one period. The random periodic solution is illuminated by computer simulations. Introduction
Mathematical epidemiology has made a significant progress in better understanding of the disease transmissions. Many epidemic models are described by the ordinary differential equations. In the real world, epidemic dynamics is always affected by the environmental noise, thus investigating the influence of the noises on dynamics of the epidemics is of interest to the researchers. The stochastic population model is always formulated by constructing the discrete time Markov chain based on the deterministic model (see [1] for example). By this way, the common SIR model with stochastic perturbations (see [2–6]) is like
For model (1), Lin and Jiang [2] studied the existence of a positive and global solution, then they showed sufficient conditions for the survival and extinction of the disease. Lahrouz and Omari [3] proved the existence of stationary distribution, which indicates the disease will continue to exist forever. Zhao [4] gave the threshold for this stochastic SIR epidemic model. Recently, by supposing the coefficients are periodic, Liu et al. [5] gave sufficient conditions for the existence of a random periodic solution. For more about model (1) and its extensions, one can refer to [6, 7] and the references therein.
Different from the method mentioned above and from the experimental point of view, the parameters of the model are always estimated by using the regression method with certain confidence intervals, which reveals that the parameters can exhibit random fluctuations to some extent. Then, from the point of view of the parameter randomization, many researchers formulated and studied the stochastic epidemic models with randomized parameters (see [8–12] for example). By supposing that the contact rate is affected by the noise like
Gray et al. [12] established a classical stochastic SIS epidemic model of the form
where \(S(t)\) represents the number of individuals susceptible to the disease at time t, and \(I(t)\) represents the number of infected individuals.
N is a constant input of new members into the population per unit time; β is the transmission coefficient between compartments S and I; μ means the natural death rate; δ is the recovery rate from infectious individuals to the susceptible; \(B ( t )\) is a standard Brownian motion on the complete probability space \(( {\Omega,{\mathcal{F}},{{ ( {{\mathcal {F}_{t}}} )}_{t \ge0}},{P}} )\) with the intensity \({\sigma ^{2}} > 0\). The authors proved that this model has a unique global positive solution and derived the existence of stationary distribution.
Inspired by the work of Capasso and Serio [13], Lin et al. [14] introduced a saturated incidence rate \({\frac{{\beta SI}}{{1 + aI}}}\) into epidemic model (2), where
a is a positive constant, and \(\frac{{\beta I}}{{1 + aI}}\) measures the infection force of the disease with inhibition effect due to the crowding of the infective. This model reads as follows:
It is interesting to see that the authors gave a complete threshold for any size of noise. Moreover, they proved that the model has the ergodic property and derived the expression for its invariant density.
However, in the real word, many infectious diseases of humans, such as measles, mumps, rubella, chickenpox, diphtheria, pertussis, and influenza, fluctuate over time with seasonal variation [15]. This implies that the corresponding mathematical models may have the periodic solutions. Therefore, it is important to investigate the periodic dynamics of epidemic models. For more about the periodic properties of the epidemic model, one can see [16–19] and the references cited therein. At the same time, one can easily find that some diseases mentioned above always do not have significant effects on the total population size. Consequently, in this paper, the total population is assumed to be a positive constant, denoted by
N.
Motivated by the above, we present a stochastic SIS epidemic model with periodic coefficients as follows:
The main concerns of this paper are as follows:
What is the condition for the existence of a random periodic and positive solution of this model?
Under what conditions will the microorganism survive or will be washed out?
Is there a threshold which more or less helps to determine the survival of the microorganism?
For simplicity, we denote \({x^{*}} = \mathop{\sup} _{t \ge0} \{ {x ( t )} \}\) and \({x_{*}} = \mathop{\inf} _{t \ge 0} \{ {x ( t )} \}\) for a function \(x(t)\) defined on \([0,\infty)\). \(R_{0}^{T} = \frac{{\frac{1}{T}\int_{0}^{T} { [ {\beta ( t )N - \frac{{{\sigma^{2}} ( t ){N^{2}}}}{2}} ]\,dt} }}{{\frac{1}{T}\int_{0}^{T} { [ {\mu ( t ) + \delta ( t )} ]\,dt} }}\). One can easily check that \(\Gamma = \{ { ( {S,x} ) \in R_{+} ^{2}:S + x = N} \}\) is the positive invariant set of model (4), which is a crucial property for the proof of a periodic solution. In view of the biological meanings, we assume that the coefficients of model (4) are continuous, positive, bounded, and
T-periodic functions on \([0,\infty)\). Then \({\mu_{*}} > 0\).
The existence of the uniquely positive solution can be proved by following the standard procedure in [12], so we omit it. In the following, we mainly focus on finding the suitable condition for the existence of a random periodic solution, persistence, and extinction of (4). Main contributions of this paper are as follows.
Theorem 1.1 If \({R_{0}^{T} > 1}\) holds, then model (4) has at least one random positive T- periodic solution in Γ. Example 1
Considering model (4), we choose \(N = 1\), \(\mu ( t ) = 0.3 + 0.2\cos t\), \(\beta ( t ) = 0.7 + 0.3\sin t\), \(\delta ( t ) = 0.1\), \(a ( t ) = 0.5 + 0.3\sin2t\), and \(\sigma ( t ) = 0.3 + 0.2\sin t\) with the initial value \(( {S ( 0 ),I ( 0 )} )= ( {0.6,0.4} )\). Clearly, the coefficients are all positive 2
π-periodic functions. Compute that \(R_{0}^{T} = 1.6125 > 1\). Then Theorem 1.1 implies that model (4) has a 2 π-periodic solution which lies in \((0,1)\), see Figure 1. Remark 1
Khasminskii [20] said that a Markov process \(X(t)\) is
T-periodic if and only if its transition probability function is T-periodic and the function \(P_{0}(t, A) =P(X(t)\in A)\) satisfies the equation
The following theorems concern the persistence in mean and extinction of model (4).
Theorem 1.2 Let \(( {S ( t ),I ( t )} )\) be the solution of (4) with the initial value \(( {S ( 0 ),I ( 0 )} )\in R_{+}^{2}\). If \({R_{0}^{T} > 1}\), then the disease will be persistent in mean, i. e., Theorem 1.3 Let \(( {S ( t ),I ( t )} )\) be the solution of model (4) with the initial value \(( {S ( 0 ),I ( 0 )} )\in R_{+}^{2}\). If one of the two assumptions holds (A)
\(\mathop{\sup} _{t \ge0} ( {{\sigma^{2}} ( t )N - \mu ( t )} ) \le0\)
and\(R_{0}^{T} < 1\), (B)
\(\frac{1}{T}\int_{0}^{T} {{ [ {\frac{{{{ [ {N\beta ( s ) - \theta ( {\mu ( s ) + \delta ( s )} )} ]}^{2}}}}{{2{N^{2}}{\sigma^{2}} ( s )}} - ( {1 - \theta} ) ( {\mu ( s ) + \delta ( s )} )} ]}\,ds< 0}\)
holds for any constant\(\theta\in[0,1)\), then the disease I goes extinct exponentially, namely Moreover, \(( {S ( t ),I ( t )} )\) exponentially tends to \(( {N,0} )\) as \(t\rightarrow\infty\). Remark 2
From Theorems 1.2 and 1.3, under the assumption: \(\mathop{\sup} _{t \ge0} ( {{\sigma^{2}} ( t )N - \mu ( t )} ) \le0\), the disease is persistent if \({R_{0}^{T} > 1}\), while it goes extinct if \({R_{0}^{T} < 1}\). So we consider \({R_{0}^{T} }\) as the threshold of the stochastic model (4).
Remark 3 Proofs
First, we introduce some results concerning the periodic Markov process.
Definition 2.1
(see [20])
A stochastic process \(\xi(t) = \xi(t, \omega ), t\in R\), is said to be periodic with period
T if, for every finite sequence of numbers \(t_{1}, t_{2}, \ldots, t_{n}\), the joint distribution of random variables \(\xi(t_{1} + h), \ldots, \xi(t_{n} + h)\) is independent of h, where \(h = kT\ (k = \pm1,\pm2, \ldots)\).
Consider the stochastic differential equation
Lemma 2.2
(see [20])
Suppose that the coefficient of (6) is T- periodic in t and satisfies the condition for some constant \(B>0\) in every cylinder \(I \times\mathcal{D}\); and suppose further that there exists a function \(V(t, x) \in C^{2}\) in \(R^{n}\) which is T- periodic in t and satisfies the following conditions: where the operator \(\mathcal{L}\) is given by \(\mathcal{L} = \frac{\partial}{{\partial t}} + \sum_{l = 1}^{n} {{f_{l}}} ( {t,x} )\frac{\partial}{{\partial{x_{l}}}} + \frac{1}{2}\sum_{i,j = 1}^{n} {{g_{i}}} ( {t,x} ){g_{j}} ( {t,x} )\frac{{{\partial^{2}}}}{{\partial{x_{i}}\,\partial{x_{j}}}}\). Then there exists a solution of (6) which is a T- periodic Markov process. Proof of Theorem 1.1
Set \((S(0),I(0))\in\Gamma\), then \((S(t),I(t))\in\Gamma\) for all \(t>0\). Let \(y = \ln\frac{N}{{N - I}}\), then \(y(t)\in R_{+}\) such that
Since \({R_{0}^{T} > 1}\), we can choose \(p>0\) such that \(\frac{1}{T}\int_{0}^{T} {\lambda ( t )\,dt} >0\). Define a \(C^{2}\)-function \(V:[0,\infty) \times R_{+}^{2}\rightarrow R\) by
where \(\gamma ( t ) = \frac{{p\frac{1}{T}\int_{0}^{T} {\lambda ( t )\,dt} \int_{t}^{t + T} {{e^{p\int_{s}^{t} {\lambda ( u )\,du} }}\,ds} }}{{1 - {e^{ - p\int_{0}^{T} {\lambda ( s )\,ds} }}}}\) is the uniquely positive
T-periodic solution of the equation
Applying the Itô formula to model (7), we get
where
Here,
In view of (8), we obtain
Define a bounded closed set \(\mathcal{D} = [ { \frac{1}{r},r} ]\), where
r is a sufficiently large positive number. Then
On the other hand,
Remark 4 Proof of (B) in Theorem 1.3
By the Itô formula, from (4) we have
After taking integration, we get
where \(M(t) = \int_{0}^{t} {\frac{{\sigma ( s )S ( s )}}{{1 + a ( s )I ( s )}}\,dB ( s )} \) is a local continuous martingale such that \(\mathop{\lim} _{t \to\infty} \frac{{M(t)}}{t} = 0\) due to the strong law of large numbers for martingales. Then we conclude
The proof is complete. □
Concluding remarks
In this paper, a stochastic SIS epidemic model with periodic coefficients is formulated and studied. First, we define a parameter \({R_{0}^{T}}\). Under assumption that the total population is fixed by
N, we show that the model has at least one random periodic solution which is nontrivial and located in \((0,N)\times(0,N)\) if \({R_{0}^{T} > 1}\). These may give better understanding of how the periodic seasonal variation affects the disease. Then, several conditions for persistence in mean and extinction of the disease are also established. In detail,
When \({R_{0}^{T}}<1\), the disease will go extinct with probability 1 under extra mild conditions.
When \({R_{0}^{T}}>1\), the disease will be persistent in mean.
In case that the noise is small, it is clear that \({R_{0}^{T}}\) is the threshold of model (4) which can be used easily to determine whether the disease will survive or not.
Let \(\sigma\equiv0\), we have \(R^{T} = \frac{{\frac{1}{T}\int_{0}^{T} { [ {\beta ( t )N } ]\,dt} }}{{\frac{1}{T}\int _{0}^{T} { [ {\mu ( t ) + \delta ( t )} ]\,dt} }}\), which is the threshold of the corresponding deterministic SIS model. Clearly, \({R_{0}^{T}} < R^{T}\), this means that the disease may go extinct due to the noises, while the deterministic SIS model predicts its survival. (B) in Theorem 1.3 shows that large noise can lead the disease to die out. In general, the noise has negative effects on persistence of the disease.
Finally, we want to address one conjecture on extinction of the disease, i.e.,
When \({R_{0}^{T}}<1\), the disease modeled by (4) will almost surely go extinct without any extra condition.
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Acknowledgements
This work was supported by NSFC (No. 11671260, 11501148) and Shanghai Leading Academic Discipline Project (No. XTKX2015).
Ethics declarations Competing interests
The authors declare that they have no competing interests.
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Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. |
Given a hypergraph $H=(V,E)$ and a set $X\subseteq V$ of vertices, let $int(X)$ be the number of distinct intersections of edges with $X$, i.e. $$int(X)=|\{S\subseteq X, \exists e\in E, e\cap X=S\}|.$$
$X$ is called
shattered if $int(X)=2^{|X|}$, i.e. if $int(X)$ reaches its maximum feasible value.
Question:Is the following claim true?
Claim:If $H$ has a set $X$ with $int(X)\geq 2^{|X|}/2$, then $H$ has a shattered set of size at least $|X|/2$. Remarks: $\bullet$ The two constants "2" are somewhat arbitrary here, I really want to find out if this claim is true for some pair of (not necessarily equal) constants. $\bullet$ The claim is true if $|X|\leq 4$ but my proof cannot be extended to higher values. $\bullet$ My motivation comes from questions related to the Vapnis-Chervonenkis dimension of a hypergraph, i.e. the biggest size of a shattered set. |
There are several possibilities for your confusion - and you bring up several related concepts (Work, potential energy, kinetic energy). When this happens I think it's helpful to isolate one thing which you know must, absolutely be true. The definition of work (for a constant force) is
$$W=\vec{F}\cdot \Delta \vec{x}=F\Delta x \cos\theta.$$
First ask "what is the work due to gravity?". Then $F=mg$, $\Delta x =x_f-x_i=$ 25 m (call this $h$), but the angle between $\vec{F}$ and $\Delta x$ is 180$^\circ$, so $cos(180)=-1$, and $W_g=-mgh=-1000$ J (the work due to gravity, as you've correctly identified).
In my view, this makes the most intuitive sense of all the things you've said. Force of gravity is acting to pull the object downward, so it's doing negative work on the object.
So, what now? We have various options for understanding the rest of your question. Let's try conservation of energy:
$$\Delta P+\Delta K=W$$
Ok, but when you write this equation, you need to know what "the system" is - so you can complete the statement, "I am looking for the change in potential and kinetic energy of
the system". In this case, "the system" is the block, and the work being done on the block is the work by your hand. That's what makes the block go up, that's what adds energy to the system.
So, the work done by your hand is opposite the work done by gravity, and $W_h=mgh$. Now, in order to understand $\Delta K$, we need to know what the initial and final velocities are (which we have not used yet, or even needed). You've said they are both zero - fine, then we understand very well that
$$\Delta P=W_h=mgh$$
Slight aside: How can you tell when the change in potential energy is zero? Potential energy is how much energy an object "could use to do something". When you raise a book a certain distance, the book wants to move downward. Therefore, it has gained potential energy "to do something". This makes sense to me, but the definition is work is a bit more unambiguous so that's why I gave that first. |
Chances are, if you took anything away from that high school or college statistics class you were dragged into, you remember some vague notion about the Central Limit Theorem. It’s likely the most famous theorem in statistics, and the most widely used. Most introductory statistics textbooks state the theorem in broad terms, that as the sample size increases, the sample distribution of the sum of the sample elements will be approximately normally distributed, regardless of the underlying distribution. Many things used in statistical inference as justification in a broad variety of fields, such as the classical z-test, rely on this theorem. Many conclusions in science, economics, public policy, and social studies have been drawn with tests that rely on the Central Limit Theorem as justification. We’re going to dive into this theorem a bit more formally, and discuss some counterexamples to this theorem. Not every sequence of random variables will obey the conditions of theorem, and the assumptions are a bit more strict than are used in practice.
First off, the counterexamples I’m going to discuss come from the book Counterexamples in Probability, written by Jordan M. Stoyanov. This book has many great counterexamples, but is quite heavy for a non mathematician. That’s why I’m going to pull examples and discuss them in detail. Counterexamples are important in mathematics, because they remind us where the limitations of the theory are.[
Fair warning here. This post will get a little math-heavy. Stick with me, and we’ll walk through the valley of the shadow of math together.] The Central Limit Theorem, now in formal attire
First, we’ll state the actual Central Limit Theorem. I’m going to make comments as we go to point some things out, so watch the footnotes. From Stoyanov:
Let S_{n} = \sum_{i=1}^{n}X_{i}
3, \sigma_{i}^{2} be the variance of the random variable X_{i}, A_{n} = \text{E}[S_{n}] = \sum_{i=1}^{n}\text{E}[X_{i}] 4, and s_{n}^{2} = \text{Var}[S_{n}] = \sum_{i=1}^{n}\sigma_{i}^{2} 5.
Then the sequence \{X_{n}\} obeys the Central Limit Theorem if\lim\limits_{n \to \infty}\mathbb{P}\left[\frac{S_{n}-A_{n}}{s_{n}} \leq x\right] = \Phi(x) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{x}e^{-u^{2}/2}du
for all x \in \mathbb{R}
I’d like to talk about this for a bit. What this is saying is that the limit of the probability that the centered (subtract the expected value of the sum from the random variable sum) and scaled (divide by the variance of the sum) sum of random variables is less than some number x converges to the probability that a standard normal random variable is less than that x. The limit here is important. Limits are what happens when we go on forever, and this theorem doesn’t state anything about how fast we get there. That speed is important, because many people just use the sample-size-is-over-30 rule of thumb (which has no mathematical justification, by the way) to assume that after 30 units in your sample, you’ll get close enough. Depending on the underlying distribution, you may need 100, 1000, or billions of terms.
One other note here. This theorem is general enough to handle random variables from the same probability space that have different variances. If we assume the variances are all the same \sigma^{2}, and that the means are all the same \mu, then you get this more “typical” formulation:\lim\limits_{n \to \infty}\mathbb{P}\left[\frac{S_{n}-n\mu}{\sigma \sqrt{n}} \leq x \right] = \Phi(x)
Counterexample
The Central Limit Theorem looked pretty broad. We just needed a sequence of independent random variables on the same probability space. So as long as we have that, we’re good, right? Not exactly. Let’s work through a fairly simple example.
Let’s suppose we have a random variable Y that can take on two possible values: \pm 1, each with probability 1/2. That is, P(Y = 1) = P(Y = -1) = 1/2.
Suppose now that we just make a sequence by repeating Y. That is, we define \{Y_{k}, k \geq 1\}, but each Y_{k} is really just Y.
6. We’re now going to build a new random sequence \{X_{k}, k \geq 1\} from these \{Y_{k}, k \geq 1\}. Let each X_{k} = \frac{\sqrt{15}Y_{k}}{4^{k}} 7. This means that each time we get an X_{k}, we flip the coin to see if Y_{k} = \pm 1, so
with equal probability regarding the sign.
Now,E[Y] = 1 \cdot 1/2 + -1\cdot 1/2 = 0
and\text{Var}[Y] = E[Y^{2}]-E[Y]^{2} = (1\cdot 1/2 + 1\cdot 1/2)-0^{2} = 1
8
Now, since X_{k} is just a constant multiple of Y_{k}, \text{E}[X_{k}] = 0 \text{E}[S_{n}] = \sum_{k=1}^{n}\text{E}[X_{k}] = 0
Now we have to calculate \text{Var}[S_{n}]
9\begin{aligned}\text{Var}[X_{k}] &= \text{E}[X_{k}^{2}]-\text{E}[X_{k}]^{2}\end{aligned}
We already know that \text{E}[X_{k}] = 0, so we just need to find \text{E}[X_{k}]^{2} \begin{aligned}\text{E}[X_{k}]^{2} &=\frac{15\cdot 1^{2}}{(4^{k})^{2}}\cdot P\left(X_{k}=\frac{\sqrt{15}\cdot 1}{4^{k}}\right)+\frac{15\cdot -1^{2}}{(4^{k})^{2}}\cdot P\left(X_{k}=\frac{\sqrt{15}\cdot -1}{4^{k}}\right)\\&=\frac{15\cdot 1^{2}}{(4^{k})^{2}}\cdot\frac{1}{2}+\frac{15\cdot -1^{2}}{(4^{k})^{2}}\cdot\frac{1}{2}\\&=\frac{15}{16^{k}}\end{aligned}
Now we can use the linearity of the variance operator to quickly find \text{Var}[S_{n}]:\begin{aligned}\text{Var}[S_{n}] &= \sum_{k=1}^{n}\text{Var}[X_{k}]\\&= \sum_{k=1}^{n}\frac{15}{16^{k}}\end{aligned}
This sum is the partial sum of a geometric series. So\begin{aligned}\text{Var}[S_{n}] &= \sum_{k=1}^{n}\text{Var}[X_{k}]\\&= \sum_{k=1}^{n}\frac{15}{16^{k}}\\&= \frac{15}{16}\left(\dfrac{1-\left(\frac{1}{16}\right)^{n}}{1-\frac{1}{16}}\right)\\&= 1-\left(\frac{1}{16}\right)^{n}\end{aligned}
Since we are discussing convergence, we really only care about really really large n. So as n gets very large, \frac{1}{16^{n}} is basically 0, so \text{Var}[S_{n}] \approx 1 when n is large. So the \sqrt{15} inserted in the definition of X_{k} seemed to come from left field, but we needed it to ensure the variance of the sum that we wanted.
Now, we’re going to note that P(|S_{n}| \leq 1/2) = 0. Why? Because |X_{1} + X_{2} +\ldots + X_{n}| > 1/2 always.
10
This means that the interval \left[\frac{-1}{2},\frac{1}{2}\right] is off limits for this sum to possibly converge to. Since the variance of the sum is approximately 1, there’s no scaling factor that will “squish” it to fit into this interval. This mean that there are real numbers x such that\mathbb{P}\left[\frac{S_{n}-A_{n}}{s_{n}} \leq x\right] \neq \Phi(x)
ever, meaning it can never converge there. That means this sequence of random variables violates the Central Limit Theorem. No amount of approximation, large sample size, etc can justify its use in this case.
Conclusion
This example may have seemed a bit artificial and constructed, but it serves to show that the Central Limit Theorem is not universal. As in, there are sequences of random variables that cannot be manipulated into fitting the hypotheses at all. This is different than listing the cautions about using the Central Limit Theorem when the sequence does actually obey. In those cases, the distinction is a bit more subtle, because we have to really look at the underlying distribution to see how fast the scaled and centered sum converges to a standard normal distribution. We’ll tackle that issue in another post.
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Footnotes This is really important here. If the sequence of random variables isn’t independent, do not pass go, do not collect $200. You do not fit the CLT. At best you can look for some approximation argument. We’re not going to get into measure theory in this post. Just keep in mind here that the capital Sigma is the sample space of possible outcomes, and P is the probability distributions. No need to get into Borel sigma-algebras just yet. just the sum of the first n terms. This is the expected value (typically called the mean) of the sum of n of these random variables. The expectation operator is linear, so the expectation of a sum is the sum of expectations. I’ll write a quick post on linear operators later. This is the variance of the sum of random variables, given by the sum of variances of the individual ones Like repeating 1s for an entire column in Excel. I can’t believe I just made that analogy. Yea, sometimes counterexamples look oddly specific or contrived. Powerful theorems can be hard to break, which is a good thing. Nonetheless, a theorem is only true forever and always and for every single thing if you can prove that no counterexample, no matter how weird, can be created Basic calculations of mean and variance, especially for discrete random variables, can be found almost anywhere Bear with me. We’re going to have to do a little work here. The first term dominates heavily over the other terms, and in absolute value is very close to 1. The absolute value of the partial sum will never actually hit 1/2 because of this. I realize this is a bit hand-wavy, but I’m already getting pretty deep in the weeds, and I don’t want to detract from the overall message here. This is provable. |
I recently came across this in a textbook (NCERT class 12 , chapter: wave optics , pg:367 , example 10.4(d)) of mine while studying the Young's double slit experiment. It says a condition for the formation of interference pattern is$$\frac{s}{S} < \frac{\lambda}{d}$$Where $s$ is the size of ...
The accepted answer is clearly wrong. The OP's textbook referes to 's' as "size of source" and then gives a relation involving it. But the accepted answer conveniently assumes 's' to be "fringe-width" and proves the relation. One of the unaccepted answers is the correct one. I have flagged the answer for mod attention. This answer wastes time, because I naturally looked at it first ( it being an accepted answer) only to realise it proved something entirely different and trivial.
This question was considered a duplicate because of a previous question titled "Height of Water 'Splashing'". However, the previous question only considers the height of the splash, whereas answers to the later question may consider a lot of different effects on the body of water, such as height ...
I was trying to figure out the cross section $\frac{d\sigma}{d\Omega}$ for spinless $e^{-}\gamma\rightarrow e^{-}$ scattering. First I wrote the terms associated with each component.Vertex:$$ie(P_A+P_B)^{\mu}$$External Boson: $1$Photon: $\epsilon_{\mu}$Multiplying these will give the inv...
As I am now studying on the history of discovery of electricity so I am searching on each scientists on Google but I am not getting a good answers on some scientists.So I want to ask you to provide a good app for studying on the history of scientists?
I am working on correlation in quantum systems.Consider for an arbitrary finite dimensional bipartite system $A$ with elements $A_{1}$ and $A_{2}$ and a bipartite system $B$ with elements $B_{1}$ and $B_{2}$ under the assumption which fulfilled continuity.My question is that would it be possib...
@EmilioPisanty Sup. I finished Part I of Q is for Quantum. I'm a little confused why a black ball turns into a misty of white and minus black, and not into white and black? Is it like a little trick so the second PETE box can cancel out the contrary states? Also I really like that the book avoids words like quantum, superposition, etc.
Is this correct? "The closer you get hovering (as opposed to falling) to a black hole, the further away you see the black hole from you. You would need an impossible rope of an infinite length to reach the event horizon from a hovering ship". From physics.stackexchange.com/questions/480767/…
You can't make a system go to a lower state than its zero point, so you can't do work with ZPE. Similarly, to run a hydroelectric generator you not only need water, you need a height difference so you can make the water run downhill. — PM 2Ring3 hours ago
So in Q is for Quantum there's a box called PETE that has 50% chance of changing the color of a black or white ball. When two PETE boxes are connected, an input white ball will always come out white and the same with a black ball.
@ACuriousMind There is also a NOT box that changes the color of the ball. In the book it's described that each ball has a misty (possible outcomes I suppose). For example a white ball coming into a PETE box will have output misty of WB (it can come out as white or black). But the misty of a black ball is W-B or -WB. (the black ball comes out with a minus). I understand that with the minus the math works out, but what is that minus and why?
@AbhasKumarSinha intriguing/ impressive! would like to hear more! :) am very interested in using physics simulation systems for fluid dynamics vs particle dynamics experiments, alas very few in the world are thinking along the same lines right now, even as the technology improves substantially...
@vzn for physics/simulation, you may use Blender, that is very accurate. If you want to experiment lens and optics, the you may use Mistibushi Renderer, those are made for accurate scientific purposes.
@RyanUnger physics.stackexchange.com/q/27700/50583 is about QFT for mathematicians, which overlaps in the sense that you can't really do string theory without first doing QFT. I think the canonical recommendation is indeed Deligne et al's *Quantum Fields and Strings: A Course For Mathematicians *, but I haven't read it myself
@AbhasKumarSinha when you say you were there, did you work at some kind of Godot facilities/ headquarters? where? dont see something relevant on google yet on "mitsubishi renderer" do you have a link for that?
@ACuriousMind thats exactly how DZA presents it. understand the idea of "not tying it to any particular physical implementation" but that kind of gets stretched thin because the point is that there are "devices from our reality" that match the description and theyre all part of the mystery/ complexity/ inscrutability of QM. actually its QM experts that dont fully grasp the idea because (on deep research) it seems possible classical components exist that fulfill the descriptions...
When I say "the basics of string theory haven't changed", I basically mean the story of string theory up to (but excluding) compactifications, branes and what not. It is the latter that has rapidly evolved, not the former.
@RyanUnger Yes, it's where the actual model building happens. But there's a lot of things to work out independently of that
And that is what I mean by "the basics".
Yes, with mirror symmetry and all that jazz, there's been a lot of things happening in string theory, but I think that's still comparatively "fresh" research where the best you'll find are some survey papers
@RyanUnger trying to think of an adjective for it... nihilistic? :P ps have you seen this? think youll like it, thought of you when found it... Kurzgesagt optimistic nihilismyoutube.com/watch?v=MBRqu0YOH14
The knuckle mnemonic is a mnemonic device for remembering the number of days in the months of the Julian and Gregorian calendars.== Method ===== One handed ===One form of the mnemonic is done by counting on the knuckles of one's hand to remember the numbers of days of the months.Count knuckles as 31 days, depressions between knuckles as 30 (or 28/29) days. Start with the little finger knuckle as January, and count one finger or depression at a time towards the index finger knuckle (July), saying the months while doing so. Then return to the little finger knuckle (now August) and continue for...
@vzn I dont want to go to uni nor college. I prefer to dive into the depths of life early. I'm 16 (2 more years and I graduate). I'm interested in business, physics, neuroscience, philosophy, biology, engineering and other stuff and technologies. I just have constant hunger to widen my view on the world.
@Slereah It's like the brain has a limited capacity on math skills it can store.
@NovaliumCompany btw think either way is acceptable, relate to the feeling of low enthusiasm to submitting to "the higher establishment," but for many, universities are indeed "diving into the depths of life"
I think you should go if you want to learn, but I'd also argue that waiting a couple years could be a sensible option. I know a number of people who went to college because they were told that it was what they should do and ended up wasting a bunch of time/money
It does give you more of a sense of who actually knows what they're talking about and who doesn't though. While there's a lot of information available these days, it isn't all good information and it can be a very difficult thing to judge without some background knowledge
Hello people, does anyone have a suggestion for some good lecture notes on what surface codes are and how are they used for quantum error correction? I just want to have an overview as I might have the possibility of doing a master thesis on the subject. I looked around a bit and it sounds cool but "it sounds cool" doesn't sound like a good enough motivation for devoting 6 months of my life to it |
Introduction Installation User guide API reference Examples Additional resources Project information Optimizing a cognitive model¶
The purpose of this example is to illustrate how Nengo DL can be used to optimize a more complex cognitive model, involving the retrieval of information from highly structured semantic pointers. We will create a network that takes a collection of information as input (encoded using semantic pointers), and train it to retrieve some specific element from that collection.
[1]:
%matplotlib inlinefrom urllib.request import urlretrieveimport zipfileimport matplotlib.pyplot as pltimport nengoimport nengo.spa as spaimport numpy as npimport tensorflow as tfimport nengo_dl
The first thing to do is define a function that produces random examples of structured semantic pointers. Each example consists of a collection of role-filler pairs of the following form:
\(TRACE_0 = \sum_{j=0}^N Role_{0,j} \circledast Filler_{0,j}\)
where terms like \(Role\) refer to simpler semantic pointers (i.e., random vectors), the \(\circledast\) symbol denotes circular convolution, and the summation means vector addition. That is, we define different pieces of information consisting of Roles and Fillers, and then we sum the information together in order to generate the full trace. As an example of how this might look in practice, we could encode information about a dog as
\(DOG = COLOUR \circledast BROWN + LEGS \circledast FOUR + TEXTURE \circledast FURRY + ...\)
The goal of the system is then to retrieve a cued piece of information from the semantic pointer. For example, if we gave the network the trace \(DOG\) and the cue \(COLOUR\) it should output \(BROWN\).
[2]:
def get_data(n_items, pairs_per_item, vec_d, vocab_seed): # the vocabulary object will handle the creation of semantic # pointers for us rng = np.random.RandomState(vocab_seed) vocab = spa.Vocabulary(dimensions=vec_d, rng=rng, max_similarity=1) # initialize arrays of shape (n_inputs, n_steps, vec_d) traces = np.zeros((n_items, 1, vec_d)) cues = np.zeros((n_items, 1, vec_d)) targets = np.zeros((n_items, 1, vec_d)) # iterate through all of the examples to be generated for n in range(n_items): role_names = ["ROLE_%d_%d" % (n, i) for i in range(pairs_per_item)] filler_names = ["FILLER_%d_%d" % (n, i) for i in range(pairs_per_item)] # create key for the 'trace' of bound pairs (i.e. a # structured semantic pointer) trace_key = 'TRACE_' + str(n) trace_ptr = vocab.parse('+'.join( "%s * %s" % (x, y) for x, y in zip(role_names, filler_names))) trace_ptr.normalize() vocab.add(trace_key, trace_ptr) # pick which element will be cued for retrieval cue_idx = rng.randint(pairs_per_item) # fill array elements correspond to this example traces[n, 0, :] = vocab[trace_key].v cues[n, 0, :] = vocab["ROLE_%d_%d" % (n, cue_idx)].v targets[n, 0, :] = vocab["FILLER_%d_%d" % (n, cue_idx)].v return traces, cues, targets, vocab
Next we’ll define a Nengo model that retrieves cued items from structured semantic pointers. So, for a given trace (e.g., \(TRACE_0\)) and cue (e.g., \(Role_{0,0}\)), the correct output would be the corresponding filler (\(Filler_{0,0}\)). The model we’ll build will perform such retrieval by implementing a computation of the form:
\(TRACE_0 \:\: \circledast \sim Role_{0,0} \approx Filler_{0,0}\)
That is, convolving the trace with inverse of the given cue will produce (approximately) the associated filler. More details about the mathematics of how/why this works can be found here.
We can create a model to perform this calculation by using the
nengo.networks.CircularConvolution network that comes with Nengo.
[3]:
seed = 0dims = 32minibatch_size = 50n_pairs = 2with nengo.Network(seed=seed) as net: # use rectified linear neurons to ensure differentiability net.config[nengo.Ensemble].neuron_type = nengo.RectifiedLinear() net.config[nengo.Connection].synapse = None # provide a pointer and a cue as input to the network trace_inp = nengo.Node(np.zeros(dims)) cue_inp = nengo.Node(np.zeros(dims)) # create a convolution network to perform the computation # specified above cconv = nengo.networks.CircularConvolution(5, dims, invert_b=True) # connect the trace and cue inputs to the circular # convolution network nengo.Connection(trace_inp, cconv.input_a) nengo.Connection(cue_inp, cconv.input_b) # probe the output out = nengo.Probe(cconv.output)
In order to assess the retrieval accuracy of the model we need a metric for success. In this case we’ll say that a cue has been successfully retrieved if the output vector is more similar to the correct filler vector than it is to any of the other vectors in the vocabulary.
[4]:
def accuracy(output, vocab, targets, t_step=-1): # provide the probed output data, the vocab, # the target vectors, and the time step at which to evaluate # get output at the given time step output = output[:, t_step, :] # compute similarity between each output and vocab item sims = np.dot(vocab.vectors, output.T) idxs = np.argmax(sims, axis=0) # check that the output is most similar to the target acc = np.mean(np.all(vocab.vectors[idxs] == targets[:, 0], axis=1)) return acc
Now we can run the model on some test data to check the baseline retrieval accuracy. Since we used only a small number of neurons in the circular convolution network, we should expect mediocre results.
[5]:
# generate some test inputstest_traces, test_cues, test_targets, test_vocab = get_data( minibatch_size, n_pairs, dims, vocab_seed=seed)test_inputs = {trace_inp: test_traces, cue_inp: test_cues}# run the simulator for one time step to compute the network outputswith nengo_dl.Simulator( net, minibatch_size=minibatch_size, seed=seed) as sim: sim.step(data=test_inputs)print('Retrieval accuracy: ', accuracy(sim.data[out], test_vocab, test_targets))
|#####################Building network (36%) | ETA: 0:00:00
/home/travis/build/nengo/nengo-dl/nengo_dl/simulator.py:102: UserWarning: No GPU support detected. It is recommended that you install tensorflow-gpu (`pip install tensorflow-gpu`). "No GPU support detected. It is recommended that you "
Build finished in 0:00:01Optimization finished in 0:00:00Construction finished in 0:00:00Retrieval accuracy: 0.04
These results indicate that the model is only rarely performing accurate retrieval, which means that this network is not very capable of manipulating structured semantic pointers in a useful way.
We can visualize the similarity of the output for one of the traces to get a sense of what this accuracy looks like (the similarity to the correct output is shown in red).
[6]:
plt.figure(figsize=(10, 5))bars = plt.bar(np.arange(len(test_vocab.vectors)), np.dot(test_vocab.vectors, sim.data[out][0, 0]))bars[np.where(np.all(test_vocab.vectors == test_targets[0, 0], axis=1))[0][0]].set_color("r")plt.ylim([-1, 1])plt.xlabel("Vocabulary items")plt.ylabel("Similarity");
We can see that the actual output is not particularly similar to this desired output, which illustrates that the model is not performing accurate retrieval.
Now we’ll train the network parameters to improve performance. We won’t directly optimize retrieval accuracy, but will instead minimize the mean squared error between the model’s output vectors and the vectors corresponding to the correct output items for each input cue. We’ll use a large number of training examples that are distinct from our test data, so as to avoid explicitly fitting the model parameters to the test items.
To make this example run a bit quicker we’ll download some pretrained model parameters by default. Set
do_training=True to train the model yourself.
[7]:
sim = nengo_dl.Simulator(net, minibatch_size=minibatch_size, seed=seed)# pick an optimizer and learning rateoptimizer = tf.train.RMSPropOptimizer(2e-3)do_training = Falseif do_training: # create training data and data feeds train_traces, train_cues, train_targets, _ = get_data( n_items=5000, pairs_per_item=n_pairs, vec_d=dims, vocab_seed=seed+1) train_data = {trace_inp: train_traces, cue_inp: train_cues, out: train_targets} # train the model sim.train(train_data, optimizer, n_epochs=100) sim.save_params("./spa_retrieval_params")else: # download pretrained parameters urlretrieve( "https://drive.google.com/uc?export=download&" "id=1jm7EUt7P7IFMsxmoBFXX3me-NsDMw_85", "spa_retrieval_params.zip") with zipfile.ZipFile("spa_retrieval_params.zip") as f: f.extractall() # load parameters sim.load_params('./spa_retrieval_params')
Build finished in 0:00:01Optimization finished in 0:00:00Construction finished in 0:00:00
We can now recompute the network outputs using the trained model on the test data. We can see that the retrieval accuracy is significantly improved. You can modify the dimensionality of the vectors and the number of bound pairs in each trace to explore how these variables influence the upper bound on retrieval accuracy.
[8]:
sim.step(data=test_inputs)print('Retrieval accuracy: ', accuracy(sim.data[out], test_vocab, test_targets))sim.close()
Retrieval accuracy: 0.74
[9]:
plt.figure(figsize=(10, 5))bars = plt.bar(np.arange(len(test_vocab.vectors)), np.dot(test_vocab.vectors, sim.data[out][0, 0]))bars[np.where(np.all(test_vocab.vectors == test_targets[0, 0], axis=1))[0][0]].set_color("r")plt.ylim([-1, 1])plt.xlabel("Vocabulary items")plt.ylabel("Similarity");
Check out this example for a more complicated version of this task/model, in which a structured semantic pointer is built up over time by binding together sequentially presented input items. |
Editor’s note: see this appendix for supporting proofs.
Fields are among the most convenient algebraic structures, preserving much of the arithmetic we know and love from familiar fields like the rationals \mathbb{Q} and the real numbers \mathbb{R}.
Now, it is unnecessary that a set possess infinitely many elements to possibly constitute a field (under the right binary operations). Indeed, Dr. Rachel Traylor recently invited readers to finite field theory by way of GF(4), the field with four elements. In this post, I propose to discuss the simplest of all possible fields, namely \text{GF}(2).
What is \text{GF}(2)?
As Dr. Traylor explains in her post on GF(4), a
field is built via a nonempty set \mathbb{F} and two binary operations, typically denoted + and \cdot when no further specifying is needed. Speaking abstractly, \text{GF}(2) is the case when \mathbb{F} is taken to be any set of two elements, say, \{a,b\}, satisfying the operation tables below 1.
+ a b a a b b b a
\cdot a b a a a b a b
So, what makes \text{GF}(2) simplest?
Briefly put, there is no field with fewer elements than has \text{GF}(2). Why is this so? The operations + and \cdot each require an identity element, and these must be distinct. As a result, any field must contain
at least two elements. And, as it happens, those are enough to define a fully-fledged field.
As it is the most basic of fields, it might be expected that \text{GF}(2) is only trivially interesting, or only appears in coverage of the subject on the theoretical front. (No judgment here; sometimes the simplest examples of structures are illustrative of very little. See the trivial topology or trivial algebra of sets on a set X.) However, \text{GF}(2) is the mathematical representation of many dualities we encounter on a daily basis. We will take a look at some of the prominent cases.
Even & Odd
Let’s denote by {\bf e} and {\bf o} arbitrarily chosen even and odd integers, respectively.
Truly, in this way, we are defining two equivalence classes on the set of integers, \mathbb{Z}, by way of modulo 2 addition.
Reminder, we say an integer m is
even to mean there exists an integer k such that m=2k, and m is odd to mean there exists an integer j such that m=2j+1. Collect these in a set: \{{\bf e},{\bf o}\}, and consider this set under ordinary addition + and multiplication \cdot .
These results are summarized in tables. For instance, in the + table below, we can think of {\bf e}+{\bf o} as(2k)+(2j+1)=2(k+j)+1\equiv{\bf o},
since k+j is also an integer.
+ {\bf e} {\bf o} {\bf e} {\bf e} {\bf o} {\bf o} {\bf o} {\bf e}
\cdot {\bf e} {\bf o} {\bf e} {\bf e} {\bf e} {\bf o} {\bf e} {\bf o}
From these tables, {\bf e} serves as additive identity for the field, and {\bf o} the multiplicative identity.
Binary
Even more readily encountered than even and odd is binary arithmetic. We have a series of posts here on the theory of coding, and all of it rests on \{0,1\} when taken with the operations of +_2 (addition modulo 2) and \cdot .
+_2 0 1 0 0 1 1 1 0
\cdot 0 1 0 0 0 1 0 1
The similarities to the tables for (\{{\bf e},{\bf o}\},+,\cdot) are evident. With binary, 0 is the additive (modulo 2) identity, and 1 is the multiplicative identity. This seems to follow naturally. After all, 0 is an even integer, and 1 is an odd integer, so these elements are easily believed to work alike. With that in mind, I move to an example with a less immediate connection.
Truth & Falsehood
Now I want to consider the set \{{\bf T},{\bf F}\} of truth values
2 true and false, respectively.
It is worthwhile to stop and think on which
operations make sense for this set. Two are needed to construct a field. Just as even and odd integers may be combined by adding and multiplying, mathematical statements may be combined via disjunction (“or,” \vee), conjunction (“and,” \wedge), and implication (“if, then,” \Rightarrow). For this case, I am interested in the special “exclusive or,” also called XOR 3, denoted by \oplus, and conjunction.
\oplus {\bf F} {\bf T} {\bf F} {\bf F} {\bf T} {\bf T} {\bf T} {\bf F}
\wedge {\bf F} {\bf T} {\bf F} {\bf F} {\bf F} {\bf T} {\bf F} {\bf T} Opposite… in Precisely the Same Way
The only thing setting these examples apart is an exchange of symbols. Truly,
a, {\bf e}, 0, and {\bf F},
are interchangeable, just as are
b, {\bf o}, 1, and {\bf T}.
What matters is that these individual elements
behave in exactly the same way with respect to their operations. In the language of algebra, it is said
(\{a,b\},+,\cdot),
(\{{\bf e},{\bf o}\},+,\cdot),
(\{0,1\},+_2,\cdot), and
(\{{\bf F},{\bf T}\},\oplus,\wedge)
are
isomorphic, that is, structurally equivalent.
Definition (Isomorphism of fields). We say two fields(\mathbb{F},+,\times) and(\mathbb{H},\boxplus,\boxtimes) areisomorphic to mean there exists a function\varphi\mathrel{:}\mathbb{F}\to\mathbb{H} such that\varphi is one-to-one, onto, and, for allx,y\in\mathbb{F},
\varphi(x+y)=\varphi(x)\boxplus\varphi(y);\quad\varphi(x\times y)=\varphi(x)\boxtimes\varphi(y).
In the case of \text{GF}(2), proving isomorphism amounts to simply defining the function that swaps the symbols as needed. For example, to show (\{{\bf e},{\bf o}\},+,\cdot) and (\{0,1\},+_2,\cdot) are isomorphic, define
4\varphi\mathrel{:}\{{\bf e},{\bf o}\}\to\{0,1\} by putting \varphi({\bf e})=0 and \varphi({\bf o})=1. Concluding Remarks
Mathematics in many respects is the art
5 of extension and generalization. The goal is frequently to take an existing object, assume an aerial viewpoint to learn its structure and what makes it work. (This is often carried out by seeing how few pieces of the whole are really needed to accomplish those key characteristics.)
With the right perspective, even these sets of opposites can bear likeness. I take comfort in that.
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Footnotes For proof that the triple ({ a, b}, +, ·) with + and · defined as in the tables satisfies all requirements of a field, see appendix. This is perhaps the easiest case proven exhaustively. This is the lingo used when discussing formal mathematical logic. Shortly, a mathematical statement must be either true or false, cannot be both, and cannot be neither. Similar to the linguistic usage, where “Option A XOR Option B” means “either Option A or Option B, but not both, and not neither”. I encourage all readers to prove this function is an isomorphism by definition. The curious may find full proof in the appendix attached. And purest science, by the way. |
Area is the size of a two-dimensional surface. It is defined as the amount of two-dimensional space occupied by an object. Area formulas have many practical applications in building, farming, architecture, science. The area of a shape can be determined by placing the shape over a grid and counting the number of squares that covers the entire space.For example area of square can be calculated using $a^{2}$ where, a is the length of it’s side.
List of Formulas
Figures Area Formula Variables Area of
Rectangle
Area = l $\times$ w l = length
w = width
Area of Square Area = $a^{2}$ a = sides of square Area of a Triangle Area = $\frac{1}{2}$bh b = base
h = height
Area of a Circle Area = $\pi$$r^{2}$ r= radius of circle Area of a Trapezoid Area =$\frac{1}{2}$(a + b)h a =base 1
b = base 2
h = vertical height
Area of Ellipse Area = $\pi$ab a = radius of major axis
b = area of minor axis |
In my beamer presentation, I have three images and a merged pdf that I want to animate using the
animate package. The set up is as follows: the three images are lined up next to each other, and I want to replace the third image with the animation. That is, on the first slide, I want the three images to be next to each other, and on the second slide, I want the left two images to be next to the animation (so the animation is replacing the third image).
Here is a picture:
Here is a MWE to show what I tried to do:
\documentclass[8pt]{beamer}\usetheme{Luebeck}\usefonttheme{serif}\setbeamertemplate{itemize item}{\color{black} $\vcenter{\hbox{\tiny$\bullet$}}$} %style of item\setbeamertemplate{footline}[frame number]{}\setbeamertemplate{navigation symbols}{}\usepackage{xcolor}\usepackage{graphicx}\setlength{\parskip}{0.75em}\usepackage{animate}\usepackage{tikzsymbols}\begin{document}\begin{frame}\begin{figure} \centering \includegraphics<1->[width = 0.33\textwidth]{cont_lasso.pdf} \includegraphics<1->[width = 0.33\textwidth]{cont_ridge.pdf} \includegraphics<1>[width = 0.33\textwidth]{cont_enet.pdf} \animategraphics<2>[autoplay, loop, width = 0.33\textwidth]{30}{enet_ani_merged}{}{}\end{figure}The \textit{shape} of the penalty can give some idea of the type of shrinkage imposed on the model.\begin{itemize} \item Sharp corners $\to$ sparsity! \Laughey[1.5][yellow][pink] \item Round corners $\to$ only shrinkage!\end{itemize}\end{frame}\end{document}
The file names of the three images are
cont_lasso.pdf,
cont_ridge.pdf, and
cont_enet.pdf, and the name of the merged pdf for the animation is
enet_ani_merged.pdf.
EDIT
For clarification, I want to make it such that the animation automatically starts playing when I begin the second slide.
After some experimentation, it seems my issue is because putting the
animategraphics command inside a
figure environment is not allowed, and not because of the overlays. So I now want to know, how am I able to put the two graphics and the animation in the same environment? |
In this lesson, we'll derive an equation which will allow us to calculate the wavefunction (which is to say, the collection of probability amplitudes) associated with any ket vector \(|\psi⟩\). Knowing the wavefunction is very important since we use probability amplitudes to calculate the probability of measuring eigenvalues (i.e. the position or momentum of a quantum system).
Newton's second law describes how the classical state {\(\vec{p_i}, \vec{R_i}\)} of a classical system changes with time based on the initial position and configuration \(\vec{R_i}\), and also the initial momentum \(\vec{p_i}\). We'll see that Schrodinger's equation is the quantum analogue of Newton's second law and describes the time-evolution of a quantum state \(|\psi(t)⟩\) based on the following two initial conditions: the energy and initial state of the system.
In this lesson, we'll give a broad overview and description of single-variable calculus. Single-variable calculus is a big tool kit for finding the slope or area underneath any arbitrary function \(f(x)\) which is smooth and continuous. If the slope of \(f(x)\) is constant, then we don't need calculus to find the slope or area; but when the slope of \(f(x)\) is a variable, then we must use a tool called a derivative to find the slope and another tool called an integral to find the area.
The wavefunction \(\psi(L,t)\) is confined to a circle whenever the eigenvalues L of a particle are only nonzero on the points along a circle. When the wavefunction \(\psi(L,t)\) associated with a particle has non-zero values only on points along a circle of radius \(r\), the eigenvalues \(p\) (of the momentum operator \(\hat{P}\)) are quantized—they come in discrete multiples of \(n\frac{ℏ}{r}\) where \(n=1,2,…\) Since the eigenvalues for angular momentum are \(L=pr=nℏ\), it follows that angular momentum is also quantized.
In this lesson we'll explain why there is structure and "clumpyness" in the universe. In other words, why there is more stuff here than over there. Now, if all of the distribution of matter and energy in the universe was initially completely uniform (meaning homogenous and isotropic), then galaxies, stars, and people would have never formed. Non-uniform density would never arise in such a universe. But due to the time-energy uncertainty principle, the distribution of matter and energy must have had been randomly distributed throughout space at the beginning of the universe. Since the distribution was truly random, there would always be regions of space with slight more matter and energy than other. These slight non-uniformities ("imprinted" by the uncertainty principle) in matter and energy density throughout space near the beginning of the universe is the origin of the slight non-uniformities that we see in the CMBR.
In general, if a quantum system starts out in any arbitrary state, it will evolve with time according to Schrödinger's equation such that the probability \(P(L)\) changes with time. In this lesson, we'll prove that if a quantum system starts out in an energy eigenstate, then the probability \(P(L)\) of measuring any physical quantity will not change with time.
Superconductors are the key to unlocking the future of transportation and electrical transmission. They enable the most efficient approaches to these industrial processes known to present science. A maglev vehicle, to borrow Jeremy Rifkin's wording, will shrink the dimensions of space and time by allowing distant continental and inter-continental regions to be accessed in, well, not much time at all. But superconductors also offer unprecedented efficiency: they eliminate the problem of atoms colliding with other atoms and would allow vehicle to "slide" across enormous distances with virtually no loss of energy and it would allow a loop of current to persist longer than the remaining lifetime of the universe. Much of the damage accumulated in the components of vehicles can, in some way or another, be traced to the friction against the road; maglev transportation circumvents this issue. |
I have been reading Bleichenbacher's 1998 paper on a forged message attack on RSA. The paper assumes access to an Oracle that takes a ciphertext $c$ and will check the decrypted text for valid PKCS #1 padding and returns the validity of the padding. This side-channel can be used for an attack since we can send a forged ciphertext by selecting random integers $s$ and compute:
$c' = cs^{e}\, \mbox{mod}\, n$
If $c'$ is PKCS conforming we know that the two most significant bytes of $ ms = c'^{d} \mbox{mod}\,n$ were equal to 0x00 0x02. If we define
$B=2^{8(k-2)}$
where $k$ is the length of $n$ in bytes, then we know
$2B\leq ms\,\mbox{mod}\,n<3B $
Now we have a range $M_{0}$ in which we know $ms$ to lie:
$M_{0} = {[2B, 3B-1]}$
The attack now proceeds by iteratively generating more valid forged ciphertexts for integers $s_{i}$ and with the gained knowledge reduce the range of $M_{i}$
I am having trouble understanding Step 3 in the paper which deals with narrowing the set of solutions:
$M_{i} \leftarrow \cup_{a, b, r} \lbrace [\mbox{max}(a, [\frac{2B+rn}{s_{i}}]), \mbox{min}(b, [\frac{3B-1+rn}{s_{i}}])]\rbrace$
for all $[a,b] \in M_{i-1}$ and $\frac{as_{i}-3B+1}{n}\leq r< \frac{bs_{i}-2B}{n}$
How would this be implemented in code? It looks to me that we would need to compute each range for every $r$ within the specified bounds e.g.:
for (r=r_min; r<r_max; ++r) max(a, x); min(b, y);
However this seems to be intractable to me and I think I am making an error in trying to convert the math to code. Anyone maybe see where I am going wrong? |
Suppose $(x_i)_{i\in\mathbb{N}}$ a set of strictly positive numbers such that $L=\sum_{i\in\mathbb{N}}x_i$ is finite.
Suppose that $(X_i)_{i\in\mathbb{N}}$ is a set independant (real-valued) random variables, each uniformly distributed in $[-x_i;x_i]$.
I am interested in $S_n=\sum_{i=0}^nX_i$ when $n\rightarrow\infty$.
$S_n$ is clearly in $[-L;L]$, so I guess the Central Limit theorem can't apply here ($\sum_{i\in\mathbb{N}}x_i^2$ is finite), so $S_n$ doesn't converge to $\mathcal{N}(0,?)$ since the probability density function of $S_n$ has a finite support ($[-L;L]$).
Can someone help me to find the distribution of $S_n$ when $n\rightarrow\infty$ (is it something like $\mathcal{N}(0,?)$ restricted to $[-L;L]$ ?)
Thanks |
I sure hope I could get into details but I will be straightforward.
a) Is there any way a process involving a photon can be strong?
I suppose you mean "strong" in a phenomenological description of the processes involved. In a loose way of speaking they could be similar if you restrict yourself to discussing conservation of certain quantities e.g.
total Isospin, parity, etc, however it is the fact that these can be considered somehow "accidental" what makes them different in a general playground. The process you describe is an example of this.
b) Are there types of electromagnetic processes for which G-parity is conserved?
Of course, behind this there are two important facts. The first is:
Non-conservation of a property doesn't mean violation under all circumstances.
while the second is:
G-parity relates a rotation in Isospin space ($R_T$) and charge conjugation ($C$).
Electromagnetic interaction is already $C$-invariant so what you could ask, for G-parity to be conserved in the end, is if the process you're studying is invariant under the Isospin rotation involved. I happened to find your question while looking for something related to a problem in Fayyazuddin's "A modern introduction to particle physics". There you can check that:$$|\pi^{\pm}\rangle \xrightarrow{R_T} |\pi^{\mp}\rangle$$$$|\pi^{0}\rangle \xrightarrow{R_T} |\pi^{0}\rangle$$
since $\eta$ (interaction eigenstate) is a singlet you should have similarly $|\eta\rangle \xrightarrow{R_T} |\eta\rangle$. The photon part may seem trivial but remember that in general a photon may be considered to be a superposition of a $I=0$ and a $I=1$ contribution (do not confuse with EW Isospin). If however you could justify $|\gamma\rangle \xrightarrow{R_T} |\gamma\rangle$ the exercise would be over (this is part of my homework) and G-parity conservation would be a reasonable tool to use. |
Difference between revisions of "Probability Seminar"
(→Friday, April 26, Colloquium, B 239 from 4pm to 5pm, Kavita Ramanan, Brown)
(→May 7, Tuesday, Duncan Dauvergne (Toronto))
Line 114: Line 114:
Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems.
Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems.
−
== May 7, '''Tuesday''', Duncan Dauvergne (Toronto) ==
+
== May 7, '''Tuesday''' , Duncan Dauvergne (Toronto) ==
+ + + Revision as of 14:50, 30 April 2019 Spring 2019 Thursdays in 901 Van Vleck Hall at 2:25 PM, unless otherwise noted. We usually end for questions at 3:15 PM.
If you would like to sign up for the email list to receive seminar announcements then please send an email to join-probsem@lists.wisc.edu
January 31, Oanh Nguyen, Princeton
Title:
Survival and extinction of epidemics on random graphs with general degrees
Abstract: We establish the necessary and sufficient criterion for the contact process on Galton-Watson trees (resp. random graphs) to exhibit the phase of extinction (resp. short survival). We prove that the survival threshold $\lambda_1$ for a Galton-Watson tree is strictly positive if and only if its offspring distribution has an exponential tail, settling a conjecture by Huang and Durrett. On the random graph with degree distribution $D$, we show that if $D$ has an exponential tail, then for small enough $\lambda$ the contact process with the all-infected initial condition survives for polynomial time with high probability, while for large enough $\lambda$ it runs over exponential time with high probability. When $D$ is subexponential, the contact process typically displays long survival for any fixed $\lambda>0$. Joint work with Shankar Bhamidi, Danny Nam, and Allan Sly.
Wednesday, February 6 at 4:00pm in Van Vleck 911 , Li-Cheng Tsai, Columbia University
Title:
When particle systems meet PDEs
Abstract: Interacting particle systems are models that involve many randomly evolving agents (i.e., particles). These systems are widely used in describing real-world phenomena. In this talk we will walk through three facets of interacting particle systems, namely the law of large numbers, random fluctuations, and large deviations. Within each facet, I will explain how Partial Differential Equations (PDEs) play a role in understanding the systems..
Title:
Fluctuations of the KPZ equation in d\geq 2 in a weak disorder regime
Abstract: We will discuss some recent work on the Edwards-Wilkinson limit of the KPZ equation with a small coupling constant in d\geq 2.
February 14, Timo Seppäläinen, UW-Madison
Title:
Geometry of the corner growth model
Abstract: The corner growth model is a last-passage percolation model of random growth on the square lattice. It lies at the nexus of several branches of mathematics: probability, statistical physics, queueing theory, combinatorics, and integrable systems. It has been studied intensely for almost 40 years. This talk reviews properties of the geodesics, Busemann functions and competition interfaces of the corner growth model, and presents some new qualitative and quantitative results. Based on joint projects with Louis Fan (Indiana), Firas Rassoul-Agha and Chris Janjigian (Utah).
February 21, Diane Holcomb, KTH
Title:
On the centered maximum of the Sine beta process Abstract: There has been a great deal or recent work on the asymptotics of the maximum of characteristic polynomials or random matrices. Other recent work studies the analogous result for log-correlated Gaussian fields. Here we will discuss a maximum result for the centered counting function of the Sine beta process. The Sine beta process arises as the local limit in the bulk of a beta-ensemble, and was originally described as the limit of a generalization of the Gaussian Unitary Ensemble by Valko and Virag with an equivalent process identified as a limit of the circular beta ensembles by Killip and Stoiciu. A brief introduction to the Sine process as well as some ideas from the proof of the maximum will be covered. This talk is on joint work with Elliot Paquette.
Title: Quantitative homogenization in a balanced random environment
Abstract: Stochastic homogenization of discrete difference operators is closely related to the convergence of random walk in a random environment (RWRE) to its limiting process. In this talk we discuss non-divergence form difference operators in an i.i.d random environment and the corresponding process—a random walk in a balanced random environment in the integer lattice Z^d. We first quantify the ergodicity of the environment viewed from the point of view of the particle. As consequences, we obtain algebraic rates of convergence for the quenched central limit theorem of the RWRE and for the homogenization of both elliptic and parabolic non-divergence form difference operators. Joint work with J. Peterson (Purdue) and H. V. Tran (UW-Madison).
Wednesday, February 27 at 1:10pm Jon Peterson, Purdue
Title:
Functional Limit Laws for Recurrent Excited Random Walks
Abstract:
Excited random walks (also called cookie random walks) are model for self-interacting random motion where the transition probabilities are dependent on the local time at the current location. While self-interacting random walks are typically very difficult to study, many results for (one-dimensional) excited random walks are remarkably explicit. In particular, one can easily (by hand) calculate a parameter of the model that will determine many features of the random walk: recurrence/transience, non-zero limiting speed, limiting distributions and more. In this talk I will prove functional limit laws for one-dimensional excited random walks that are recurrent. For certain values of the parameters in the model the random walks under diffusive scaling converge to a Brownian motion perturbed at its extremum. This was known previously for the case of excited random walks with boundedly many cookies per site, but we are able to generalize this to excited random walks with periodic cookie stacks. In this more general case, it is much less clear why perturbed Brownian motion should be the correct scaling limit. This is joint work with Elena Kosygina.
March 21, Spring Break, No seminar March 28, Shamgar Gurevitch UW-Madison
Title:
Harmonic Analysis on GLn over finite fields, and Random Walks
Abstract: There are many formulas that express interesting properties of a group G in terms of sums over its characters. For evaluating or estimating these sums, one of the most salient quantities to understand is the
character ratio:
$$ \text{trace}(\rho(g))/\text{dim}(\rho), $$
for an irreducible representation $\rho$ of G and an element g of G. For example, Diaconis and Shahshahani stated a formula of this type for analyzing G-biinvariant random walks on G. It turns out that, for classical groups G over finite fields (which provide most examples of finite simple groups), there is a natural invariant of representations that provides strong information on the character ratio. We call this invariant
rank. This talk will discuss the notion of rank for $GL_n$ over finite fields, and apply the results to random walks. This is joint work with Roger Howe (Yale and Texas AM). April 4, Philip Matchett Wood, UW-Madison
Title:
Outliers in the spectrum for products of independent random matrices
Abstract: For fixed positive integers m, we consider the product of m independent n by n random matrices with iid entries as in the limit as n tends to infinity. Under suitable assumptions on the entries of each matrix, it is known that the limiting empirical distribution of the eigenvalues is described by the m-th power of the circular law. Moreover, this same limiting distribution continues to hold if each iid random matrix is additively perturbed by a bounded rank deterministic error. However, the bounded rank perturbations may create one or more outlier eigenvalues. We describe the asymptotic location of the outlier eigenvalues, which extends a result of Terence Tao for the case of a single iid matrix. Our methods also allow us to consider several other types of perturbations, including multiplicative perturbations. Joint work with Natalie Coston and Sean O'Rourke.
April 11, Eviatar Procaccia, Texas A&M Title: Stabilization of Diffusion Limited Aggregation in a Wedge.
Abstract: We prove a discrete Beurling estimate for the harmonic measure in a wedge in $\mathbf{Z}^2$, and use it to show that Diffusion Limited Aggregation (DLA) in a wedge of angle smaller than $\pi/4$ stabilizes. This allows to consider the infinite DLA and questions about the number of arms, growth and dimension. I will present some conjectures and open problems.
April 18, Andrea Agazzi, Duke
Title:
Large Deviations Theory for Chemical Reaction Networks
Abstract: The microscopic dynamics of well-stirred networks of chemical reactions are modeled as jump Markov processes. At large volume, one may expect in this framework to have a straightforward application of large deviation theory. This is not at all true, for the jump rates of this class of models are typically neither globally Lipschitz, nor bounded away from zero, with both blowup and absorption as quite possible scenarios. In joint work with Amir Dembo and Jean-Pierre Eckmann, we utilize Lyapunov stability theory to bypass this challenges and to characterize a large class of network topologies that satisfy the full Wentzell-Freidlin theory of asymptotic rates of exit from domains of attraction. Under the assumption of positive recurrence these results also allow for the estimation of transitions times between metastable states of this class of processes.
April 25, Kavita Ramanan, Brown
Title:
Beyond Mean-Field Limits: Local Dynamics on Sparse Graphs
Abstract: Many applications can be modeled as a large system of homogeneous interacting particle systems on a graph in which the infinitesimal evolution of each particle depends on its own state and the empirical distribution of the states of neighboring particles. When the graph is a clique, it is well known that the dynamics of a typical particle converges in the limit, as the number of vertices goes to infinity, to a nonlinear Markov process, often referred to as the McKean-Vlasov or mean-field limit. In this talk, we focus on the complementary case of scaling limits of dynamics on certain sequences of sparse graphs, including regular trees and sparse Erdos-Renyi graphs, and obtain a novel characterization of the dynamics of the neighborhood of a typical particle. This is based on various joint works with Ankan Ganguly, Dan Lacker and Ruoyu Wu.
Friday, April 26, Colloquium, Van Vleck 911 from 4pm to 5pm, Kavita Ramanan, Brown
Title:
Tales of Random Projections
Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems. |
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Change to browse by: References & Citations Bookmark(what is this?) Physics > Atomic Physics Title: Atomic H over plane: effective potential and level reconstruction
(Submitted on 27 Jan 2019)
Abstract: The behavior of atomic H in a semi-bounded space $z \geq 0$ with the condition of "not going through" the boundary (the surface $z=0$) for the electronic wavefunction (WF) is considered. It is shown that in a wide range of "not going through" condition parameters the effective atomic potential, treated as a function of the distance $h$ from H to the boundary plane, reveals a well pronounced minimum at certain finite but non-zero $h$, which describes the mode of "soaring" of the atom above the plane. In particular cases of Dirichlet and Neumann conditions the analysis of the soaring effect is based on the exact analytical solutions of the problem in terms of generalized spheroidal Coulomb functions. For $h$ varying between the regions $h \gg a_B$ and $h \ll a_B$ both the deformation of the electronic WF and the atomic state are studied in detail. In particular, for the Dirichlet condition the lowest $1s$ atomic state transforms into $2p$-level with quantum numbers $210$, the first excited ones $2s$ --- into $3p$ with numbers $310$, $2p$ with $m=0$ --- into $4f$ with numbers $430$, etc. At the same time, for Neumann condition the whole picture of the levels transmutation changes drastically. For a more general case of Robin (third type) condition the variational estimates, based on special type trial functions, as well as the direct numerical tools, realized by pertinent modification of the finite element method, are used. By means of the latter it is also shown that in the case of a sufficiently large positive affinity of the atom to the boundary plane a significant reconstruction of the lowest levels takes place, including the change of both the asymptotics and the general dependence on $h$. Submission historyFrom: Andrey Tolokonnikov [view email] [v1]Sun, 27 Jan 2019 18:03:44 GMT (730kb) |
This question already has an answer here:
According to Introduction to algorithms by Cormen et al, $$T(n)=2T(n/2)+n\log n$$ is not case 3 of Master Theorem. Can someone explain me why?
And which case of master theorem is it?
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This question already has an answer here:
According to Introduction to algorithms by Cormen et al, $$T(n)=2T(n/2)+n\log n$$ is not case 3 of Master Theorem. Can someone explain me why?
And which case of master theorem is it?
$\log n$ grows slower than $n^\epsilon$ for any $\epsilon>0$. Thus $n\log n$ grows slower than $n^c$ for any $c>1$. However, the third case of the Master theorem requires the existance of a $c>1$ so that $n\log n$ grows at least as fast as $n^c$ (up to a constant factor).
The function is covered by the second case of the Master theorem as given in Wikipedia.
Case 3 of Master Theorem:
If $f(n)=\Omega(n^{\log_b a+\epsilon})$ for some constant $\epsilon > 0$, and if $f (n/b) ≤ cf (n)$ for some constant $c < 1$ and all sufficiently large $n$, then $T (n) = \Theta(f (n))$.
Here $f(n)=n\log n$ and $n^{\log_ba}=n^{\log_2 2}=n^1=n$. So $n^{log_ba+\epsilon}=n^{1+\varepsilon}$. But $n\log n$ is not $\Omega(n^{1+\epsilon})$.
So it is not under case 3. It is also not under second case: If $f(n)=\Theta(n^{\log_ba})$, then second case can be applied. But here $n^{\log_ba}=n$ and $f(n)=n\log n$ is not $\Theta(n)$.
This recurrence can't be solved by Master method. |
When one wants to know that whether a partial function $f \colon \Sigma^{*} \supsetneq \mathrm{dom}(f) \rightarrow \Sigma^{*}$ is Turing-computable, there are two methods that I think they are both useful.
We can define a total function $\bar{f}$ such that $$ \bar{f}(x) = \left\{ \begin{aligned} &f(x),&x \in \mathrm{dom}(f) \\ &\bot,&x \not\in \mathrm{dom}(f) \\ \end{aligned} \right.$$ Thus, $f$ is Turing-computable if $\bar{f}$ is Turing-computable by a TM using $\bar{\Sigma} = \Sigma \cup \{ \bot \}$. (Some answers use this definition. see [1] and [2])
Let $M$ be a TM, and We say $f$ is Turing-computable by $M$ if $$M(x) = \left\{ \begin{aligned} &f(x),&& x \in \mathrm{dom}(f) \\ &\bot,&&x \not\in \mathrm{dom}(f) \\ \end{aligned} \right.$$ where $M(x) = \bot$ means that $M$ never halts on $x$. (This definition is introduced in my books.)
These two definitions are not equivalent, since $M$ can computes some functions whose domain is Turing-recognizable instead of Turing-decidable in sense of definition 2.
I want know that is the definition 2 more powerful? In another word, if $f$ is Turing-computable in sense of definition 1, can we prove that it is also Turing-computable in sense of definition 2? |
A section is cut out of a circular piece of paper having radius four inches, as shown. Points A and B are then glued together to form a right circular cone. What is the circumference of the base of the resulting cone? Express your answer in terms of \(\pi\). (The \(270^\circ\) sector forms the cone.)
circumference of base of cone = length of major arc AB
length of major arc AB = \(\frac{270}{360}\cdot\) circumference of circle
length of major arc AB = \(\frac{270}{360}\cdot2\pi\cdot4"\)
length of major arc AB = \(\frac{3}{4}\cdot2\pi\cdot4"\)
length of major arc AB = \(6\pi"\)_
circumference of base of cone = length of major arc AB
length of major arc AB = \(\frac{270}{360}\cdot\) circumference of circle
length of major arc AB = \(\frac{270}{360}\cdot2\pi\cdot4"\)
length of major arc AB = \(\frac{3}{4}\cdot2\pi\cdot4"\)
length of major arc AB = \(6\pi"\)_ |
Difference between revisions of "Probability Seminar"
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(→May 7, Tuesday Van Vleck 901, 2:25pm,, Duncan Dauvergne (Toronto))
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Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems.
Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems.
−
== May 7, '''Tuesday''' Van Vleck 901, 2:25pm,, Duncan Dauvergne (Toronto) ==
+
== May 7, '''Tuesday''' Van Vleck 901, 2:25pm,, Duncan Dauvergne (Toronto) ==
− + + + + Revision as of 14:51, 30 April 2019 Spring 2019 Thursdays in 901 Van Vleck Hall at 2:25 PM, unless otherwise noted. We usually end for questions at 3:15 PM.
If you would like to sign up for the email list to receive seminar announcements then please send an email to join-probsem@lists.wisc.edu
January 31, Oanh Nguyen, Princeton
Title:
Survival and extinction of epidemics on random graphs with general degrees
Abstract: We establish the necessary and sufficient criterion for the contact process on Galton-Watson trees (resp. random graphs) to exhibit the phase of extinction (resp. short survival). We prove that the survival threshold $\lambda_1$ for a Galton-Watson tree is strictly positive if and only if its offspring distribution has an exponential tail, settling a conjecture by Huang and Durrett. On the random graph with degree distribution $D$, we show that if $D$ has an exponential tail, then for small enough $\lambda$ the contact process with the all-infected initial condition survives for polynomial time with high probability, while for large enough $\lambda$ it runs over exponential time with high probability. When $D$ is subexponential, the contact process typically displays long survival for any fixed $\lambda>0$. Joint work with Shankar Bhamidi, Danny Nam, and Allan Sly.
Wednesday, February 6 at 4:00pm in Van Vleck 911 , Li-Cheng Tsai, Columbia University
Title:
When particle systems meet PDEs
Abstract: Interacting particle systems are models that involve many randomly evolving agents (i.e., particles). These systems are widely used in describing real-world phenomena. In this talk we will walk through three facets of interacting particle systems, namely the law of large numbers, random fluctuations, and large deviations. Within each facet, I will explain how Partial Differential Equations (PDEs) play a role in understanding the systems..
Title:
Fluctuations of the KPZ equation in d\geq 2 in a weak disorder regime
Abstract: We will discuss some recent work on the Edwards-Wilkinson limit of the KPZ equation with a small coupling constant in d\geq 2.
February 14, Timo Seppäläinen, UW-Madison
Title:
Geometry of the corner growth model
Abstract: The corner growth model is a last-passage percolation model of random growth on the square lattice. It lies at the nexus of several branches of mathematics: probability, statistical physics, queueing theory, combinatorics, and integrable systems. It has been studied intensely for almost 40 years. This talk reviews properties of the geodesics, Busemann functions and competition interfaces of the corner growth model, and presents some new qualitative and quantitative results. Based on joint projects with Louis Fan (Indiana), Firas Rassoul-Agha and Chris Janjigian (Utah).
February 21, Diane Holcomb, KTH
Title:
On the centered maximum of the Sine beta process Abstract: There has been a great deal or recent work on the asymptotics of the maximum of characteristic polynomials or random matrices. Other recent work studies the analogous result for log-correlated Gaussian fields. Here we will discuss a maximum result for the centered counting function of the Sine beta process. The Sine beta process arises as the local limit in the bulk of a beta-ensemble, and was originally described as the limit of a generalization of the Gaussian Unitary Ensemble by Valko and Virag with an equivalent process identified as a limit of the circular beta ensembles by Killip and Stoiciu. A brief introduction to the Sine process as well as some ideas from the proof of the maximum will be covered. This talk is on joint work with Elliot Paquette.
Title: Quantitative homogenization in a balanced random environment
Abstract: Stochastic homogenization of discrete difference operators is closely related to the convergence of random walk in a random environment (RWRE) to its limiting process. In this talk we discuss non-divergence form difference operators in an i.i.d random environment and the corresponding process—a random walk in a balanced random environment in the integer lattice Z^d. We first quantify the ergodicity of the environment viewed from the point of view of the particle. As consequences, we obtain algebraic rates of convergence for the quenched central limit theorem of the RWRE and for the homogenization of both elliptic and parabolic non-divergence form difference operators. Joint work with J. Peterson (Purdue) and H. V. Tran (UW-Madison).
Wednesday, February 27 at 1:10pm Jon Peterson, Purdue
Title:
Functional Limit Laws for Recurrent Excited Random Walks
Abstract:
Excited random walks (also called cookie random walks) are model for self-interacting random motion where the transition probabilities are dependent on the local time at the current location. While self-interacting random walks are typically very difficult to study, many results for (one-dimensional) excited random walks are remarkably explicit. In particular, one can easily (by hand) calculate a parameter of the model that will determine many features of the random walk: recurrence/transience, non-zero limiting speed, limiting distributions and more. In this talk I will prove functional limit laws for one-dimensional excited random walks that are recurrent. For certain values of the parameters in the model the random walks under diffusive scaling converge to a Brownian motion perturbed at its extremum. This was known previously for the case of excited random walks with boundedly many cookies per site, but we are able to generalize this to excited random walks with periodic cookie stacks. In this more general case, it is much less clear why perturbed Brownian motion should be the correct scaling limit. This is joint work with Elena Kosygina.
March 21, Spring Break, No seminar March 28, Shamgar Gurevitch UW-Madison
Title:
Harmonic Analysis on GLn over finite fields, and Random Walks
Abstract: There are many formulas that express interesting properties of a group G in terms of sums over its characters. For evaluating or estimating these sums, one of the most salient quantities to understand is the
character ratio:
$$ \text{trace}(\rho(g))/\text{dim}(\rho), $$
for an irreducible representation $\rho$ of G and an element g of G. For example, Diaconis and Shahshahani stated a formula of this type for analyzing G-biinvariant random walks on G. It turns out that, for classical groups G over finite fields (which provide most examples of finite simple groups), there is a natural invariant of representations that provides strong information on the character ratio. We call this invariant
rank. This talk will discuss the notion of rank for $GL_n$ over finite fields, and apply the results to random walks. This is joint work with Roger Howe (Yale and Texas AM). April 4, Philip Matchett Wood, UW-Madison
Title:
Outliers in the spectrum for products of independent random matrices
Abstract: For fixed positive integers m, we consider the product of m independent n by n random matrices with iid entries as in the limit as n tends to infinity. Under suitable assumptions on the entries of each matrix, it is known that the limiting empirical distribution of the eigenvalues is described by the m-th power of the circular law. Moreover, this same limiting distribution continues to hold if each iid random matrix is additively perturbed by a bounded rank deterministic error. However, the bounded rank perturbations may create one or more outlier eigenvalues. We describe the asymptotic location of the outlier eigenvalues, which extends a result of Terence Tao for the case of a single iid matrix. Our methods also allow us to consider several other types of perturbations, including multiplicative perturbations. Joint work with Natalie Coston and Sean O'Rourke.
April 11, Eviatar Procaccia, Texas A&M Title: Stabilization of Diffusion Limited Aggregation in a Wedge.
Abstract: We prove a discrete Beurling estimate for the harmonic measure in a wedge in $\mathbf{Z}^2$, and use it to show that Diffusion Limited Aggregation (DLA) in a wedge of angle smaller than $\pi/4$ stabilizes. This allows to consider the infinite DLA and questions about the number of arms, growth and dimension. I will present some conjectures and open problems.
April 18, Andrea Agazzi, Duke
Title:
Large Deviations Theory for Chemical Reaction Networks
Abstract: The microscopic dynamics of well-stirred networks of chemical reactions are modeled as jump Markov processes. At large volume, one may expect in this framework to have a straightforward application of large deviation theory. This is not at all true, for the jump rates of this class of models are typically neither globally Lipschitz, nor bounded away from zero, with both blowup and absorption as quite possible scenarios. In joint work with Amir Dembo and Jean-Pierre Eckmann, we utilize Lyapunov stability theory to bypass this challenges and to characterize a large class of network topologies that satisfy the full Wentzell-Freidlin theory of asymptotic rates of exit from domains of attraction. Under the assumption of positive recurrence these results also allow for the estimation of transitions times between metastable states of this class of processes.
April 25, Kavita Ramanan, Brown
Title:
Beyond Mean-Field Limits: Local Dynamics on Sparse Graphs
Abstract: Many applications can be modeled as a large system of homogeneous interacting particle systems on a graph in which the infinitesimal evolution of each particle depends on its own state and the empirical distribution of the states of neighboring particles. When the graph is a clique, it is well known that the dynamics of a typical particle converges in the limit, as the number of vertices goes to infinity, to a nonlinear Markov process, often referred to as the McKean-Vlasov or mean-field limit. In this talk, we focus on the complementary case of scaling limits of dynamics on certain sequences of sparse graphs, including regular trees and sparse Erdos-Renyi graphs, and obtain a novel characterization of the dynamics of the neighborhood of a typical particle. This is based on various joint works with Ankan Ganguly, Dan Lacker and Ruoyu Wu.
Friday, April 26, Colloquium, Van Vleck 911 from 4pm to 5pm, Kavita Ramanan, Brown
Title:
Tales of Random Projections
Abstract: The interplay between geometry and probability in high-dimensional spaces is a subject of active research. Classical theorems in probability theory such as the central limit theorem and Cramer’s theorem can be viewed as providing information about certain scalar projections of high-dimensional product measures. In this talk we will describe the behavior of random projections of more general (possibly non-product) high-dimensional measures, which are of interest in diverse fields, ranging from asymptotic convex geometry to high-dimensional statistics. Although the study of (typical) projections of high-dimensional measures dates back to Borel, only recently has a theory begun to emerge, which in particular identifies the role of certain geometric assumptions that lead to better behaved projections. A particular question of interest is to identify what properties of the high-dimensional measure are captured by its lower-dimensional projections. While fluctuations of these projections have been studied over the past decade, we describe more recent work on the tail behavior of multidimensional projections, and associated conditional limit theorems.
May 7, Tuesday Van Vleck 901, 2:25pm,, Duncan Dauvergne (Toronto) |
Hi, I refer to formula (8) in Chapter 1 of H. Davenport, Multiplicative Number Theory, Third Edition, Springer (2000), which says that for primes $q\equiv 3 \bmod 4$:
$$ L\left(\left(\frac{\cdot}{q}\right),1\right) = \frac{\pi}{q^{1/2}\left(2-\left(\frac{2}{q}\right)\right)}\sum_{0<m<q/2}\left(\frac{m}{q}\right).$$
This formula is due to Dirichlet and implies that for primes $q\equiv 3 \bmod 4$, there are more quadratic residues than nonresidues in $(0,q/2)$. It seems that this approach can be mimicked so that a general formula can be produced which says that for primes $q\equiv 3 \bmod 4$, and any prime $r$, one has
$$ L\left(\left(\frac{\cdot}{q}\right),1\right) = \frac{\pi}{q^{1/2}\left(r-\left(\frac{r}{q}\right)\right)}\sum_{0<m<q/2}\left(\frac{m}{q}\right)\left(r-1-2\left\lfloor\frac{mr}{q}\right\rfloor\right). $$
By plugging in $r=2$, one obtains the first formula. By plugging in $r=3$, one obtains
$$ L\left(\left(\frac{\cdot}{q}\right),1\right) = \frac{2\pi}{q^{1/2}\left(3-\left(\frac{3}{q}\right)\right)} \sum_{0<m<q/3} \left(\frac{m}{q}\right). $$
This implies that there are more quadratic residues than nonresidues in $(0,q/3)$ for primes $q \equiv 3 \bmod 4$. However, by (28) and (29) of "Elementary Trigonometric Sums related to Quadratic Residues" by Laradji, Mignotte and Tzanakis, there are as many quadratic residues as nonresidues in $(0,(q-3)/4]$ and more quadratic residues than nonresidues in $[(q+1)/4,(q-1)/2]$ for primes $q \equiv 3\bmod 8$, with the situation reversed when $q \equiv 7 \bmod 8$. Combined with the above, this means there are more quadratic residues than nonresidues in $[(q+1)/4,q/3)$ when $q\equiv 3 \bmod 8$. This last interval has length about $q/12$.
So I'm wondering what is the smallest $\beta>0$ over which one can prove that there are more quadratic residues than nonresidues in an interval of length $\beta q$? (For a positive density of primes $q$.) And also it'd be nice if it is the same interval, eg $(\delta q, (\delta+\beta) q)$. Thanks. |
This is basically a problem of recursive counting. Start with the uncoupled basis state, i.e. the set of states of the form $\vert \ell m_\ell \rangle \vert s m_s\rangle$. There are clearly $(2\ell+1)(2s+1)$ of these, and the job is to reorganize them.
The key counting result is based on the observation that $\vert \ell m_\ell\rangle\vert sm_s\rangle$ is an eigenstate of $\hat J_z=\hat L_z+\hat S_z$ with eigenvalue $M=m_\ell+m_s$. With this in mind organize your $\vert \ell m_\ell\rangle\vert sm_s\rangle$ states so that those with the same value of $M$ are on the same line. Explicitly, for instance, you would have\begin{align}\begin{array}{rlll}M=\ell+s:&\vert \ell \ell\rangle\vert s s\rangle \\M=\ell+s-1:& \vert \ell,\ell-1\rangle\vert ss\rangle&\vert\ell\ell\rangle \vert s,s-1\rangle\\M=\ell+s-2:&\vert \ell,\ell-2\rangle \vert ss\rangle & \vert\ell,\ell-1\rangle\vert s,s-1\rangle&\vert \ell \ell\rangle \vert s,s-2\rangle\\\vdots\qquad& \qquad\vdots\end{array} \end{align}and replace each state with a $\bullet$ to get$$\begin{array}{rlll}\ell+s:&\bullet \\\ell+s-1:&\bullet & \bullet\\\ell+s-2:&\bullet&\bullet&\bullet\\\vdots\qquad&\vdots\end{array}$$Now, if $M=\ell+s$ is the largest value and it occurs once, the value of $j=\ell+s$ must occur once and also all the states $\vert j=\ell+s,m_j\rangle$ will occur once. There is a linear combination of the two states with $M=\ell+s-1$ that will be the state $\vert j=\ell+s,m_j=\ell+s-1\rangle$, there will be a linear combination of the three states with $M=\ell+s-2$ that will be the $\vert j=\ell+s,m_j=\ell+s-2\rangle$ state etc. Since we are only interested in
enumerating the possible resulting values of $j$, and not interested in the actual states per se, we can eliminate from our table the first column since it contains one state with $m_j=\ell+s$, one with $m_j=\ell+s-1$ etc. Eliminating this column yields the reduced table$$\begin{array}{rll}\ell+s-1: & \bullet\\\ell+s-2:&\bullet&\bullet\\\vdots\qquad&\vdots\end{array}$$Since the value of $m_j=\ell+s-1$ occurs once, the value $j=\ell+s-1$ must occur once, and the states $\vert \ell+s-1,m_j\rangle$ will each occur once. We take out those from the list by deleting the first column to obtain a further reduced table$$\begin{array}{rl}\ell+s-2:&\bullet\\\vdots\qquad &\vdots\end{array}$$ The process so continues until exhaustion. In the examples above we have found $j=\ell+s,\ell+s-1$ and the final reduced table of the example, if not empty, would indicate the value of $j=\ell+s-1$. It is clear this process produces a decreasing sequence of $j$. The last value of $j$ is determined by the width of the original table. It is not hard to convince yourself that the width of the table will stop increasing once we reach $M=\vert \ell-s\vert $, and this is the last value of $j$. Thus by exhaustion you find the possible values of $j$ in the range$$\vert \ell-s\vert\le j\le \ell+s\, .$$
As an example consider $\ell=1$ and $s=2$. The original table then looks like$$ \begin{array}{rlll}\frac{3}{2}:&\vert 11\rangle\vert 1/2,1/2\rangle \\\frac{1}{2}:&\vert 10\rangle\vert 1/2,1/2\rangle & \vert 11\rangle\vert 1/2,-1/2\rangle\\-\frac{1}{2}:&\vert 1,-1\rangle\vert 1/2,1/2\rangle&\vert 10\rangle\vert 1/2,-1/2\rangle\\-\frac{3}{2}:&\vert 1,-1\rangle \vert 1/2,-1/2\rangle\end{array}\qquad \to \qquad \begin{array}{rlll}\frac{3}{2}:&\bullet \\\frac{1}{2}:&\bullet & \bullet \\-\frac{1}{2}:&\bullet &\bullet \\-\frac{3}{2}:&\bullet\end{array}$$It is only $2$ column wide, and the width stop growing at $M=1/2$, indicating the possible $j$ in this case are $3/2$ and $1/2$, and indeed$$\vert 1-1/2\vert \le j\le 1+1/2$$Finally, note that the absolute value is required on the left because one could write state $\vert sm_s\rangle\vert \ell m_\ell\rangle$ without affecting the possible values of $j$. |
Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.
Dear Uncle Colin,
I'm struggling a bit with my C4 vectors. Most of it is fine, except when I have to find a point $P$ on a given line such that $\vec{AP}$ is perpendicular to the line, for some known $A$. How do I figure that out?
-- Any Vector Insight Lavishly Appreciated
Hi, AVILA, and thanks for your message!
There are a couple of approaches I like for this, both of which involve the dot product in slightly different ways.
This is the way I was taught. Suppose you've got the equation of a line as $\mathbf{r} = \colvecthree{x_0}{y_0}{z_0} + \lambda \colvecthree{\Delta_x}{\Delta_y}{\Delta_z}$, and a known point $A$ at $\colvecthree{X}{Y}{Z}$. For any point $P$ on the line, $\vec{AP} = \colvecthree{X-x_0}{Y-y_0}{Z-z_0} - \lambda_P \colvecthree{\Delta_x}{\Delta_y}{\Delta_z}$, for some value of $\lambda_P$.
For that to be perpendicular to the line, the scalar (dot) product between $\vec{AP}$ and the direction vector of the line, $\colvecthree{\Delta_x}{\Delta_y}{\Delta_z}$, must be zero. So, all you need to do is solve: $\left(\colvecthree{X-x_0}{Y-y_0}{Z-z_0} - \lambda_P \colvecthree{\Delta_x}{\Delta_y}{\Delta_z}\right)\cdot \colvecthree{\Delta_x}{\Delta_y}{\Delta_z} = 0$ for $\lambda_P$, the only thing you don't know.
You can try to be clever about rearranging and simplifying, but I wouldn't bother: stick in the numbers and get an answer; then put $\lambda_P$ back into the equation of the line to find point $P$.
A way I like slightly better, as I can see the geometry behind it, is to make a right-angled triangle between a known point $Q$ on the line1, known point $A$ and point $P$. It helps if you normalise the direction vector of the line: divide it by its modulus and call the result $\mathbf{\hat v}$.
You know the length of the hypotenuse: it's simply $|\vec{QA}|$.
You can easily figure out the cosine of the angle $A\hat QP$: call the direction vector of the line $\mathbf{v}$ and it's $\cos(\theta) = \frac{\vec{QA} \cdot \mathbf{\hat v}}{|\vec{QA}|}$ .
Then the distance $|\vec{QP}|$ is then $|\vec{QA}| \cos(\theta)$, using standard right-angled trigonometry.
This
is a place where simplification helps: if you notice the $|\vec{QA}|$ on the bottom of the $\cos(\theta)$ fraction, you can see that the distance you want is $d = \vec{QA} \cdot \mathbf{\hat {v}}$.
Now you just need to go this far along the line from $Q$ - which means you need to add on $d$ lots of $\mathbf {\hat v}$.
Your final answer for point $P$ is $\mathbf{q} + \left(\vec{QA} \cdot \mathbf{\hat{v}}\right) \mathbf{\hat{v}}$, where $\mathbf{q}$ is the vector $\vec{OQ}$.
Hope that helps!
-- Uncle Colin |
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Production of charged pions, kaons and protons at large transverse momenta in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV
(Elsevier, 2014-09)
Transverse momentum spectra of $\pi^{\pm}, K^{\pm}$ and $p(\bar{p})$ up to $p_T$ = 20 GeV/c at mid-rapidity, |y| $\le$ 0.8, in pp and Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76 TeV have been measured using the ALICE detector ... |
Let $(M^{n+1},g)$ be a manifold with a Riemannian metric $g$ and let $\nabla$ be its Riemannian connection. We say an immersion $x : N^n \to M^{n+1}$ is (totally)
umbilicif for all $p \in N$, the second fundamental form $B$ of $x$ at $p$ satisfies $$\langle B(X,Y),\eta \rangle(p)=\lambda(p)\langle X,Y \rangle,$$ for all $X,Y \in \mathcal X(N)$ and $\lambda(p) \in \mathbb R$ and for a given unit field $\eta$ normal to $x(N)$; here we are using $\langle \, , \, \rangle$ to dontate the metric $g$ on $M$ and the metric induced by $x$ on $N$.
e) Calculate the mean curvature and the sectional curvature of the umbilic hypersurfaces of the hyperbolic space.
(For the other parts (a)-(d), please see this link: Problem 8.6 of do Carmo, Riemannian Geometry. Nonetheless, my question is only on part (e), specifically on calculating the mean curvature.)
The author gave the following hint:
Hint:Consider the model of $H^n$ as the upper half-space. Let $S \cap H^n$ be the intersection of $H^n$ with a Euclidean $(n-1)$-sphere $S \subset \mathbb R^n$ of radius $1$ and center in $H^n$. Since $S \cap H^n$ is umbilic, all of the directions are principal, and it is enough to calculate the curvature of the curves of intersection of $S \cap H^n$ with the $x_1x_n$-plane. Use the expression obtained in part (b) of this exercise to establish that the mean curvature of $S \cap H^n$ (in the metric of $H^n$) is equal to $1$ if $S$ is tangent to $\partial H^n$, is equal to $\cos \alpha$ if $S$ makes an angle $\alpha$ with $\partial H^n$, and is equal to the "height" of the Euclidean center of $S$ relative to $\partial H^n$, if $S \subset H^n$. To calculate the secitonal curvature, use the Gauss formula.
The expression from part (b) I believe the author is referring to is $$ \left\langle \overline \nabla_X \left(\frac{\eta}{\sqrt{\mu}} \right),Y \right\rangle_{\overline g} = \frac{-2\lambda \mu+\eta(\mu)}{2\mu\sqrt{\mu}}\langle X,Y \rangle_{\overline g}. $$ I am not sure on how to apply this formula to establishing the mean curvature, depending on how $S$ is related to $\partial H^n$ ($S$ is tangent to $\partial H^n$, $S$ makes an angle $\alpha$ with $\partial H^n$, etc.).
Do Carmo defined earlier (page 129 to be exact) in the textbook the
mean curvature to be $\frac 1n(\lambda_1+\cdots+\lambda_n)$, where $\lambda_i=k_i$ are principal curvatures of $f$.
I am sorry for being unable to show much of what I tried on this question. Nonetheless, any hint or first step will be helpful. Thanks! |
I am just wondering if I know enough prerequisites!
A basic understanding of what's going on gain be gained just using Kepler's laws and Newtonian mechanics.
A simple way of dealing with multiple gravitational sources is to selectively ignore all but one of them. This is the patched conic approximation. Which gravitating body is in play? That depends on whether the spacecraft is inside the gravitational sphere of influence of one of the planets. If that is the case, you ignore the Sun and all the other planets. If the spacecraft is outside all planetary spheres of influence, you ignore all of the planets. With this treatment, the spacecraft is always subject to one and only one gravitating body. The spacecraft's trajectory is a piecewise continuous set of Keplerian segments.
Suppose a spacecraft is on an elliptical orbit about the Sun that brings the spacecraft inside of a planet's sphere of influence. At the point where the planet crosses that sphere, the trick is to switch from looking at the trajectory as an elliptical orbit about the Sun to a hyperbolic orbit about the planet. This is a reference frame change. The spacecraft's position and velocity with respect to the planet are the vector differences between the spacecraft's and planet's heliocentric positions and velocities.
The hyperbolic trajectory that results will soon carry the spacecraft out of the planet's sphere of influence. This trajectory will preserve the magnitude of the spacecraft's planet-centered velocity, but not the direction. Another change of reference frames is performed as the spacecraft exits the sphere of influence, but this time back to heliocentric coordinates. While the planetary encounter doesn't change the magnitude of the planet-centered velocity, it does change the magnitude of the Sun-centered velocity.
(Update) Details
I'll first give a brief overview of the Keplerian orbit of a test mass about a central mass. The mass of the test mass is many, many orders of magnitude smaller than that of the central body. A spacecraft orbiting a planet, for example, qualifies as a test mass (mass ratio is 10
-20 or smaller). Key concepts: $\mu$ - The central body's gravitational parameter, conceptually $\mu = GM$ but generally $\mu$ is known to much greater precision than are $G$ and $M$. $\vec r$ - The position of the test mass relative to the central body. $\vec v$ - The velocity of the test mass relative to the central body. $\vec h = \vec r \times \vec v$ - The specific angular momentum of the test mass. $\nu$ - The true anomaly of the test mass, measured with respect to the periapsis point. $e$ - The eccentricity of the orbit of the test mass about the central body. $r$ - The magnitude of $\vec r$. $v$ - The magnitude of $\vec v$. $a$ - The semi-major axis length of the test mass's orbit about the central body. $r = \frac {a(1-e^2)}{1+e\cos\nu}$ - Kepler's first law. $r_p = a(1-e)$ - Periapsis distance; closest approach of the test mass and central body. $v_\infty = \sqrt{\frac \mu {-a}}$ - Hyperbolic orbit excess velocity; the speed as $r\to\infty$. $\frac {v^2}{\mu} = \frac 2 r - \frac 1 a$ - The vis vivaequation, which provides a mechanism for computing $a$. $\vec e = \frac {\vec v \times \vec h}{\mu} - \frac {\vec r}{r}$ - The test mass's eccentricity vector relative to the central body. $\hat x_o = \frac {\vec e}{e}$ - The x-hat axis of the orbit, which points from the central body toward the periapsis point. $\hat z_o = \frac {\vec h}{h}$ - The z-hat axis of the orbit, which points away from the orbital plane, in the direction of positive angular momentum. $\hat y_o = \hat z_o \times \hat x_o$ - The y-hat axis of the orbit, defined to complete an xyzright-handed coordinate system.
In the limit $r\to \infty$, the test mass will suffer no change in speed but will be subject to a change in velocity given by $\Delta \vec v = 2v_\infty \cos \nu_m \, \hat x_o$, where $\nu_m$ is the maximum true anomaly given by $1+e\cos \nu_m = 0$. Thus $\cos \nu_m = -1/e = -1/(1-r_p/a) = -1/(1+v_\infty^2 r_p/\mu)$. The change in velocity is thus given $\Delta \vec v = -v_\infty/(1+v_\infty^2 r_p / \mu) \, \hat x_0$. Note that this says that too low or too high of a hyperbolic excess velocity both result in a small $\Delta v$. The largest $\Delta v$ for a given periapsis distance results when $v\infty = \sqrt{\mu/r_p}$ (in which case the deflection angle is 60°).
Suppose the spacecraft enters a planet's sphere of influence (a sphere of radius $r_{\text{soi}} = a_p (m_p/m_\odot)^{2/5}$ about the planet) at a position $\vec r_0$ with respect to the planet and with some velocity $\vec v_0$ with respect to the planet. The semi-major axis length, specific angular momentum vector, and eccentricity vector of the spacecraft's hyperbolic orbit about the planet can be calculated given this initial state and the planet's gravitational parameter. Per the patched conic approximation, each of these will be a constant of motion. (Note that the "length" in "semi-major axis length" is a bit of a misnomer; it will be negative in the case of a hyperbolic orbit.) The periapsis distance can then be calculated.
Some of the above calculations simplify with the additional assumption that the initial velocity is very close to the hyperbolic excess velocity. (What's one more simplifying assumption on top of the huge assumption of patched conics?) With one more simplifying assumption, that the time spent inside the planet's sphere of influence is small, the $\Delta v$ from the flyby can be approximated as impulsive.
These key simplifying assumptions give mission planning programs something that can be dealt with. This is a very large and complex search space, and some of the optimization parameters are difficult to express numerically. No matter how good a plan is in terms of low Earth departure velocity, a plan that involves a correction burn on July 4 and a planetary encounter on Christmas Day is not a good plan. No matter how good a plan is in terms of nominally low Earth departure velocity, if the plan is extremely sensitive to errors it is not a good plan.
People are still better than machines with regard to weeding out plans that get in the way of people doing what they are wont to do (e.g., taking the Fourth of July off, along with not working from a day or two before Christmas to a day or two after New Years) and people are still better than machines with regard to weeding out plans that are ultra-sensitive to errors. Mission planners still like their porkchop plots. Unfortunately, porkchop plots are computationally expensive to produce. Multiple porkchop plots that string together (i.e., a planetary gravitational assist) are extremely expensive to produce. Multiple planetary encounters (e.g., Cassini) means stringing together a lot of porkchop plots. Hence all the simplifying assumptions.
All those simplifying assumptions mean that the nominal plan is ultimately flawed. Not badly flawed, but flawed nonetheless. A solver that doesn't make all those simplifying assumptions is needed. Unfortunately, there are no practical, generic closed-form solutions to the N-body problem. The only way around this is numerical propagation. Now we can throw all kinds of kinks at the solver: multiple gravitational bodies, some of which have a non-spherical gravity field, relativistic effects, and so on. This is not something that can be done from the start. It is something that can be done to polish up the solutions from an overly-simplified mission planning perspective. Note: I am not disparaging those mission planning efforts. The mission planning search space is so large that simplifying assumptions are an absolute necessity lest we have to wait until the next millennium (985 years away) for a solution. |
This set of Automobile Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Flywheel”.
1. What is the maximum fluctuation of energy?
a) It is the sum of maximum and minimum energies b) It is the ratio of the maximum and minimum energies c) It is the ratio of the mean resisting torque to the work done per cycle d) It is the difference between the maximum and minimum energies View Answer
Explanation: The maximum fluctuation of energy is the difference between the maximum and minimum energies. The fluctuation of energy can be found out by the turning moment diagram for one complete cycle of operation.
2. What is the coefficient of fluctuation of energy?
a) The ratio of the maximum fluctuation of energy to the minimum fluctuation of energy b) The ratio of the maximum fluctuation of energy to the work done per cycle c) The ratio of the minimum fluctuation of energy to the maximum fluctuation of energy d) The ratio of the work done per cycle to the maximum fluctuation of energy View Answer
Explanation: The coefficient of fluctuation of energy (C
E) is the ratio of the maximum fluctuation of energy to the work done per cycle. C E=\(\frac{\Delta E}{(T_{mean} \theta}\) where ΔE = maximum fluctuation of energy, T mean*θ = work done per cycle, T mean= mean torque, and θ = angle turned in radian in one cycle.
3. Which of the following is true about the flywheel?
a) Flywheel regulates speed over a period of time b) Flywheel takes care of the quantity of fluid c) Flywheel controls \(\frac{dN}{dt}\) d) Flywheel regulates the speed by regulating the quantity of charge of the prime mover View Answer
Explanation: Flywheel controls the rate of fluctuation of the speed during a cycle. Flywheel controls the speed in one cycle. It does not control the supply of fuel to the engine.
4. Given that maximum fluctuation of energy is 2000 N-m/s and coefficient of fluctuation of speed is 0.02. What is the mean kinetic energy of flywheel?
a) 50 kN-m/s b) 40 kN-m/s c) 30 kN-m/s d) 20 kN-m/s View Answer
Explanation: The maximum fluctuation of energy in a flywheel = ΔE, mean kinetic energy of the flywheel = E, and coefficient of fluctuation of speed = C
s. ΔE = 2*E*C s⇒ 2000 = 2*E*0.02 ⇒ E = 50 kN-m/s.
5. What is the coefficient of fluctuation of speed?
a) The ratio of the maximum fluctuation of speed to the mean speed b) The ratio of the maximum fluctuation of energy to the mean speed c) The ratio of the minimum fluctuation of energy to the mean speed d) The ratio of the work done per cycle to the mean speed View Answer
Explanation: The coefficient of fluctuation of speed is the ratio of the maximum fluctuation of speed to the mean speed. C
s= (ω 1– ω 2)/ω =2 (ω 1– ω 2) / (ω 1+ω 2) or C s= 2(N 1-N 2) / (N 1+N 2).
6. Which of the following is the expression of the maximum fluctuation of energy in a flywheel where I = mass moment of inertia of the flywheel, C
s = coefficient of fluctuation of speed, ω = mean angular speed of ω 1 and ω 2, and E = mean kinetic energy of the flywheel? a) I*ω*(ω 1+ω 2) b) I*ω*C s c) 2*E 2*C s d) I*ω 2*C s View Answer
Explanation: The maximum fluctuation of energy in a flywheel = I*ω*(ω
1-ω 2) = I*ω 2*C s= 2*E*C s= I*ω 2*C s. The maximum fluctuation of energy can be calculated by using the turning moment diagram.
7. In the single-cylinder, single acting, four-stroke gas engine, the total fluctuation of speed is not to exceed ±3 percent of the mean speed. What is the coefficient of fluctuation of speed?
a) 0.04 b) 0.03 c) 0.09 d) 0.06 View Answer
Explanation: Since the total fluctuation of speed is not to exceed ±3 percent of the mean speed, therefore ω
1– ω 2= 6% ω and coefficient of fluctuation of speed = C s= (ω 1– ω 2)/ω = 0.06.
8. The steam engine has a flywheel which is having a radius of gyration of 1.1 m and mass of 2200 kg. The starting torque is 1550 Nm and assumed constant. What is the angular acceleration of the flywheel?
a) 0.54 rad/s 2 b) 0.58 rad/s 2 c) 0.62 rad/s 2 d) 0.52 rad/s 2 View Answer
Explanation: I = mass moment of inertia of the flywheel, m = mass of the flywheel, k = radius of gyration, T = stating torque, and α = angular acceleration of the flywheel. I = m*k
2= 2200*1.1 2= 2662 kg-m 2. T = I*α ⇒ 1550 = 2662 * α ⇒ α = 0.58 rad/s 2.
9. The engine has the flywheel of mass 7 tons and the radius of gyration is 1.7 m. The fluctuation of energy is 55 kN-m. The mean speed of the engine is 150 rpm. What is the coefficient of fluctuation of speed?
a) 0.021 b) 0.011 c) 0.031 d) 0.041 View Answer
Explanation: The fluctuation of energy = ΔE, the mass of the flywheel = m, the radius of gyration = k, the mean speed of the flywheel = N. ΔE = \(\frac{π^2}{900}\)*m*k
2*N*(N 1-N 2) = \(\frac{π^2}{900}\)*7000*1.7 2*150*(N 1-N 2) = 33277*(N 1-N 2) ⇒ 55000 = 33277*(N 1-N 2) ⇒ (N 1-N 2)= \(\frac{55000}{33277}\) = 1.652 rpm. Coefficient of fluctuation of speed = C s= \(\frac{N_1-N_2}{N}\) = \(\frac{1.652}{150}\) = 0.011.
10. The flywheel stores the potential energy.
a) True b) False View Answer
Explanation: The flywheel stores the rotational energy. The amount of energy that flywheel can store is proportional to the square of the mass and its rotational velocity. The lighter flywheels are used in a sports car for easy and quick recovery of the RPM.
Sanfoundry Global Education & Learning Series – Automobile Engineering.
To practice all areas of Automobile Engineering,
here is complete set of 1000+ Multiple Choice Questions and Answers. |
Description:
The graph is defined by asymmetric adjacency matrix $A$, with elements $A_{ij} = 1$ if $i$ and $j$ are neighboring nodes and $A_{ij} = 0$ otherwise. The hopping is a local Markov process: the particle which arrives at some moment to node $i$ will hop to a neighboring node $j$ with probability $P_{ij}$, independently of the past history. The elements of the transition matrix are $P_{ij} = 0$ if $A_{ij} = 0$, that is if nodes $i$, $j$ are not linked, and for each $i$ one has $\sum_j P_{ij} = 1$.
The main quantity of interest is the probability, $\pi_i(t)$, of finding the particle at node $i$ at time $t$. One can calculate it recursively, applying the Markov property: $$\pi_i(t + 1) = \sum_j\pi_j(t)P_{ji}. ~~~~~~~~(1)$$
Using spectral properties of the matrix $P_{ij}$ , one can show that πi(t) reaches for $t \rightarrow \infty$ a unique stationary state $\pi^*_i$ obeying the following eigenequation: $$\pi^*_i=\sum_j\pi^*_j P_{ji}. ~~~~~~~~(2)$$
........
Let $\psi_i$ be the normalized eigenvector, $\sum_i \psi^2_i=1$, corresponding to the maximal eigenvalue $\lambda$ of the adjacency matrix $A_{ij}$: $$\sum_j A_{ij}\psi_j = \lambda\psi_i.~~~~~~~~(6)$$
The eigenvalue $\lambda$ is clearly in the range $k_{min} \leq \lambda \leq k_{max}$ ,where $k_{min}$ and $k_{max}$ are the maximal and minimal node degrees of the graph, respectively. The Frobenius-Perron theorem tells us that the eigenvetor has all elements of the same sign, so that one can choose $\psi_i > 0$. Let us use this eigenvector to define the following transition matrix: $$P_{ij}=\frac{A_{ij}}{\lambda}\frac{\psi_j}{\psi_i}.~~~~~~~~(7)$$
By construction, the entries $P_{ij}$ are positive if $i$ and $j$ are neighboring nodes. They are also properly normalized:$\sum_j P_{ij} = 1$. A similar onstrution has been reently proposed in the ontext of optimal information coding [7]. The weight (4) is now independent of intermediate nodes: $$P(\gamma^{(t)}_{i_0~i_t}) = \frac{1}{\lambda^t}\frac{\psi_{i_t}}{\psi_{i_0}},~~~~~~~~(8)$$ and thus all trajectories having length $t$ and given endpoints $i_0$ and $i_t$ are equiprobable. For a closed trajectory, the probability (8) depends only on its length $t$. The stationary distribution of MERW is $$\pi^*_i=\psi^2_i~~~~~~~~(9)$$ which is easy to check by combining Eqs. (7) and (2). It is a normalized probability: $\sum_i\pi^*_i=1$, and the detailed balance condition is fulilled: $\pi^*_i P_{ij}=\pi^*_j P_{ji}$.
I am confused about how to obtain (9). Does any one can give me the detail intermediate steps to obtain (9)? Thanks in advance.
I tried:
$$\pi^*_i=\sum_j\pi^*_jP_{ji}$$ $$j\rightarrow i, P_{ji}=\frac{A_{ji}}{\lambda}\frac{\psi_i}{\psi_j}$$ $$i\rightarrow j, P_{ij}=\frac{A_{ij}}{\lambda}\frac{\psi_j}{\psi_i}$$ $$\pi^*_i=\sum_j\pi^*_jP_{ji}$$ $$=\sum_j(\sum_i\pi^*_i P_{ij})P_{ji}$$ $$=\sum_j\sum_i\pi^*_i P_{ij}P_{ji}$$ $$=\sum_j\sum_i\pi^*_i \frac{A_{ij}}{\lambda}\frac{\psi_j}{\psi_i}\frac{A_{ji}}{\lambda}\frac{\psi_i}{\psi_j}$$ $$????~~~~=\psi^2_i~~~~????$$
I also read ref Maximal_entropy_random_walk. I still confused. |
I saw a solved example in a book (Concepts of Physics by H.C. Verma, volume 2), where there is a body near surface of the earth, the problem is to calculate the increase in mass of the body when it is lifted 1 meter on the surface of earth. The book assumes that the increase in potential energy goes to the mass of the body, and applies Einstein's formula. I am attaching a link to the screenshot of the book. Check question number 7. (Somehow the uploader on this site says file type not supported)
However, the potential energy can also be thought of as integral over all space of the function $\frac { -g_1 \cdot g_2}{4\pi G}$, where $g_1$ and $g_2$ are the gravitational field of the earth and the body, respectively, and the integral can be shown to be equal to the interaction in potential energy (there is one similar problem with electrostatic potential energy in Introduction to Electrodynamics by Griffiths).
Why do we assume that the increased energy is stored as increase in mass of the body, instead of being stored in the field?
Also, why do we assume that only the mass of the body increases, not that of earth? |
Given an integer sequence $\{ a_1, a_2, \ldots, a_N \}$ that has length $N$ and a fixed integer $M\leq N$, the problem is to find a subset $A =\{i_1, \dots, i_M\} \subseteq [N]$ with $1 \leq i_1 \lt i_1 \lt \dots \lt i_M \leq N$ such that
$\qquad \displaystyle \sum_{j=1}^M j \cdot a_{i_j}$
is maximized.
For instance, if the given sequence is $-50; 100; -20; 40; 30$ and $M = 2$, the best weighted sum arises when we choose positions 2 and 4.
So that we get a value $1 \cdot 100 + 2 \cdot 40 = 180$.
On the other hand, if the given sequence is $10; 50; 20$ and $M$ is again 2, the best option is to choose positions 1 and 2 that we get a value $1 \cdot 10 + 2 \cdot 50 = 110$.
To me it looks similar to the maximum subarray problem, but I can think of many examples in which the maximum subarray is not the best solution.
Is this problem an instance of a well studied problem? What is the best algorithm to solve it?
This question was inspired by this StackOverflow question. |
In the present paper, we treat random matrix products on the general lineargroup $\textrm{GL}(V)$, where $V$ is a vector space defined on any local field,when the top Lyapunov exponent is simple, without irreducibility assumption. Inparticular, we show the existence and uniqueness of the stationary measure$\nu$ on $\textrm{P}(V)$ that is relative to the top Lyapunov exponent and wedescribe the projective subspace generated by its support. We observe that thedynamics takes place in a open set of $\textrm{P}(V)$ which has the structureof a skew product space. Then, we relate this support to the limit set of thesemi-group $T_{\mu}$ of $\textrm{GL}(V)$ generated by the random walk.Moreover, we show that $\nu$ has H\"older regularity and give some limittheorems concerning the behavior of the random walk and the probability ofhitting a hyperplane. These results generalize known ones when $T_{\mu}$ actsstrongly irreducibly and proximally (i-p to abbreviate) on $V$. In particular,when applied to the affine group in the so-called contracting case or moregenerally when the Zariski closure of $T_{\mu}$ is not necessarily reductive,the H\"older regularity of the stationary measure together with the descriptionof the limit set are new. We mention that we don't use results from the i-psetting; rather we see it as a particular case.
Let $G$ be a real linear semisimple algebraic group without compact factorsand $\Gamma$ a Zariski dense subgroup of $G$. In this paper, we use aprobabilistic counting in order to study the asymptotic properties of $\Gamma$acting on the Furstenberg boundary of $G$. First, we show that the $K$components of the elements of $\Gamma$ in the KAK decomposition of $G$ becomeasymptotically independent. This result is an analog of a result of Gorodnik-Ohin the context of the Archimedean counting. Then, we give a new proof of aresult of Guivarc'h concerning the positivity of the Hausdorff dimension of theunique stationary probability measure on the Furstenberg Boundary of $G$.Finally, we show how these results can be combined to give a probabilisticproof of the Tit's alternative; namely that two independent random walks on$\Gamma$ will eventually generate a free subgroup. This result answered aquestion of Guivarc'h and was published earlier by the author. Since we'reworking with the field of real numbers, we give here a more direct proof and amore general statement.
We study the transience of algebraic varieties in linear groups. Inparticular, we show that a "non elementary" random walk in SL_2(R) escapesexponentially fast from every proper algebraic subvariety. We also treat thecase where the random walk is on the real points of a semi-simple splitalgebraic group and show such a result for a wide family of random walks. As anapplication, we prove that generic subgroups (in some sense) of linear groupsare Zariski dense.
We show that on an arbitrary finitely generated non virtually solvable lineargroup, any two independent random walks will eventually generate a freesubgroup. In fact, this will hold for an exponential number of independentrandom walks. |
A few years ago, I came up with this proof of Perron's theorem for a class presentation: http://www.math.cornell.edu/~web6720/Perron-Frobenius_Hannah%20Cairns.pdf
I've written an outline of it below so that you don't have to read a link.
It's close in spirit to Wielandt's proof using $$\rho := \sup_{\substack{x \ge 0\\|x| = 1}} \min_j {|{\sum x_i A_{ij}}| \over x_j},$$ but I think it's simpler. (In particular, you don't have to divide by anything or take the sup min of anything.) The exception is the part where you prove that the spectral radius has only one eigenvector, which is exactly the same as Wieland's proof.
I believe it's correct, but it hasn't passed through any kind of verification process aside from being presented in class. And I've had a couple people write me about it, and, for God knows what reason, it comes up on Google in the first couple pages if you search for "Perron-Frobenius."
So I'd appreciate it if you would look at this and see if you see anything wrong with it. And if you don't, I'd like to know if it's original, because if so then I get to feel proud of myself.
Here is the proof:
Let $A > 0$ be a positive $n \times n$ matrix with eigenvalues $\lambda_1, \ldots, \lambda_n$, counted with multiplicity. Let $\rho = \max |\lambda_i|$ be the spectral radius. We want to prove that $\rho$ is a simple eigenvalue of $A$ with a positive eigenvector, and that every other eigenvalue is strictly smaller in absolute value.
Let $\lambda$ be an eigenvalue with $|\lambda| = \rho$, and finally let $\psi$ an eigenvector for $\lambda$. Consider $$\Psi := |\psi| = (|\psi_1|, \ldots, |\psi_n|).$$ Then $A \Psi = A |\psi| \ge |A \psi| = |\lambda \psi| = \rho |\psi| = \rho \Psi$, where "$x \ge y$" means that each coordinate of $x$ is greater than or equal to each coordinate of $y$.
Suppose $A \Psi \ne \rho \Psi$. Then by positivity we have $A^2 \Psi > \rho A \Psi$, which means that by continuity there is some $\varepsilon > 0$ with $A^2 \Psi \ge (\rho + \varepsilon) A \Psi$. Therefore \begin{align*}A^{n+1} \Psi &\ge (\rho + \varepsilon) A^n \Psi \\&\cdots\\&\ge (\rho + \varepsilon)^n A \Psi \ge 0\end{align*} and taking norms we get $\Vert A^{n+1} \Psi \Vert_1 \ge (\rho + \varepsilon)^n \Vert A \Psi \Vert_1$, so the operator 1-norm of $A^n$ is at least $(\rho + \varepsilon)^n$, which is a contradiction with Gelfand's formula $\lim \Vert A^n \Vert^{1/n} = \rho$.
Therefore $A \Psi = \rho \Psi$ and $\rho$ is an eigenvalue with positive eigenvector $\rho \Psi = A \Psi > 0$.
Suppose there is an eigenvalue $\lambda$ with $|\lambda| = \rho$. Let $\psi$ be an eigenvector for $\lambda$. We have seen above that $A \Psi = \rho \Psi = |A \psi|$ or $\sum_j A_{ij} |\psi_j| = |\sum_{ij} A_{ij} \psi_j|$. Fix an index $i$. Then $A_{ij} > 0$ for each row $j$, so $\sum_{ij} A_{ij} \psi_j$ is a weighted sum of $\psi_j$ where all the weights are positive, and its absolute value is the weighted sum of $|\psi_j|$ with the same weights. Those two things can only be equal if all the summands $\psi_j$ all have the same complex argument, so $\psi = e^{i\theta} \psi'$ where $\psi' \ge 0$, and $\lambda \psi' = A \psi' > 0$, so $\lambda > 0$. Therefore $\lambda = \rho$.
Now we know that every eigenvalue with $|\lambda| = \rho$ is $\rho$, and it has one positive eigenvector (and possibly more), but we don't know how many times $\rho$ appears in the list of eigenvalues. That is, we don't know whether it's simple or not.
We can prove that $\rho$ has only one eigenvector by the same argument in Wielandt's proof. We know $\Psi$ is a positive eigenvector. Suppose that there is another linearly independent eigenvector $\psi$. We can pick $\psi$ to be real (because $\mathop{\rm Re} \psi$ and $\mathop{\rm Im} \psi$ are eigenvectors or zero and at least one is linearly independent of $\Psi$). Choose $c$ so $\Psi + c \psi$ is nonnegative and has one zero entry. Then $\rho (\Psi + c \psi) = A(\Psi + c \psi) > 0$ by positivity, but it has one zero entry, which is a contradiction. So there's no other linearly independent eigenvector.
Now that we know there's only one eigenvector, we can prove that $\rho$ is a simple eigenvalue. By the previous reasoning, there is a positive left eigenvector $\Pi$ of $\rho$, so $\Pi A = \rho A$. Then $\Pi > 0$ and $\Psi > 0$, so $\Pi \Psi \ne 0$. Then $\Pi^0 := \{x: \Pi x = 0\}$ is an $(n-1)$-dimensional subspace of $\mathbb R^n$ and $\Psi \notin \Pi^0$, so we can decompose $\mathbb R^n$ into the direct sum $$\mathbb R^n = \mathop{\text{span}}\{\Psi\} \oplus \Pi^0.$$
Both of these spaces are invariant under $A$, because $A \Psi = \rho \Psi$ and $\Pi A x = \rho \Pi x = 0$. Let $x_2, \ldots, x_n$ be a basis of $\Pi^0$. Let $$X = \begin{bmatrix}\Psi&x_2&x_3&\cdots&x_n\end{bmatrix}.$$ Then the invariance means that $$X^{-1}AX = \begin{bmatrix}\rho&0\\0&Y\end{bmatrix}$$ where the top right $0$ says $\Pi^0$ is invariant under $A$ and the lower left $0$ says $\mathop{\text{span}}\{\Psi\}$ is invariant under $A$. Here $Y$ is some unknown $(n-1) \times (n-1)$ matrix.
$A$ is similar to the above block matrix, so the eigenvalues of $A$ are $\rho$ followed by the eigenvalues of $Y$. If $\rho$ is not a simple eigenvalue, then it must be an eigenvalue of $Y$.
Suppose $\rho$ is an eigenvalue of $Y$. Let $\psi'$ be an eigenvector with $Y \psi' = \rho \psi'$. Then $A X {0 \choose \psi'} = \rho X {0 \choose \psi'}$ and $X{0 \choose \psi'}$ is linearly independent of $\Psi = X {1 \choose \mathbf{0}}$. We've already proved that $A$ has only one eigenvector for $\rho$, so that is impossible. Therefore, $\rho$ is not an eigenvalue of $Y$, so $\rho$ is a simple eigenvalue of $A$. That's the last thing we had to prove.
Extending to $A \ge 0$ with $A^n > 0$ works as usual.
Thanks! |
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Research Open Access Published: Existence and multiplicity of solutions for a second-order impulsive differential equation via variational methods Advances in Difference Equations volume 2017, Article number: 46 (2017) Article metrics
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Abstract
In this paper, we study the existence and multiplicity of solutions for an impulsive differential equation via some critical point theory and the variational method. We extend and improve some recent results and reduce conditions.
Introduction
As an important research field of study, the impulsive differential equation has been attracting the attention of several mathematicians. In the early years, the main way to resolve this kind of problems is based on the fixed point theory, the theorem of topological degree, the upper and lower solutions method coupled with the monotone iterative technique, and so on; see for example [1–3]. Recently, many authors have tried to use the variational method and some specific critical point theorems, such as mountain pass lemma, fountain theorem, linking theorem, symmetric mountain pass lemma, and so on, to study the existence (see [4–9]) and multiplicity (see [10–15]) of solutions for some impulsive differential equations.
In [5], authors have shown the variational structure of an impulsive differential equation and proved the existence of a solution by using the mountain pass lemma.
In [10], the authors studied the following equation:
where \(J=[0,1]\), \(0=t_{0}< t_{1}< t_{2}<\cdots<t_{p}<t_{p+1}= 1\), \(f\in C[J\times R,R]\), \(p\in C[J,R^{+}]\), \(q, r\in C[0,T]\).
Then \(Zhang\) in [11] proved the existence of two solutions and the existence of infinitely many solutions of problem given by
In this paper, we study the existence and multiplicity of solutions for the following nonlinear impulsive problem:
where \(J=[0,T]\), \(0=t_{0}< t_{1}< t_{2}<\cdots<t_{p}<t_{p+1}= T\), \(r\in C[0,T]\), \(I_{i}\in C[R,R]\),
λ is a parameter, \(f\in C[J\times R,R] \), with \(F(t,u)=\int_{0}^{u}{f(t,\xi)}\,d\xi \).
We will prove that equation (1) has at least two classical solutions and infinitely many classical solutions under different conditions. Our main results extend the existing result in [5, 10, 11]. We prove the same impulsive problem in [5] cannot only have two solutions but also have infinitely many classical solutions. Compared with [10], we do not require the impulsive functions \(I_{j}\) and
F to satisfy the sublinear growth condition and the superlinear growth condition about \(u^{\mu}\), which such that the problem more general. Different from [11] in which F is a negative function, in this paper, our results relax the restriction of F with a wider range of applications. Preliminaries
Let \(R(t)=\int_{0}^{t}{r(s)}\,ds\), \(M=\max_{t\in [0,T]}e^{-R(t)}\), \(m=\min_{t\in [0,T]}e^{-R(t)}\). Multiplying the first equation of (1) by \(e^{-R(t)}\), we obtain
Multiply the first equation of (2) by \(v \in H_{0}^{1}(0,T)\), integrate it on the interval \([0,T]\). It follows from the boundary conditions \(u(0) = 0\), \(u(T) =0\) that
Now, define the function \(\varphi:H_{0}^{1}(0,T)\rightarrow R \)
where
and for all \(v\in H_{0}^{1}(0,T)\), we have
Definition 1
Let \(\lambda_{k} ( k=1,2,\ldots ) \) be the eigenvalue of the following Dirichlet problem, where \(\lambda_{1}=\frac{\pi^{2}}{T^{2}}\) is the first eigenvalue:
We assume \(X_{k}\) is the feature space corresponding with \(\lambda_{k}\), then \(H_{0}^{1}(0,T)=\overline{\bigoplus_{i\in N}X_{i}}\).
Lemma 1 There exists a constant \(C>0\), such that \(\Vert u\Vert _{\infty}\leq C\Vert u\Vert \), where \(\Vert u\Vert _{\infty}= \max_{t\in[0,T]}\vert u(t)\vert \). Lemma 2
[5]
If \(\lambda>-\frac{m\lambda_{1}}{M}\), then there exist \(0< a_{1}< a_{2}\), such that Remark 1
If \(\vert \lambda \vert <\frac{m\lambda_{1}}{M}\), then \(0< a_{1}< a_{2}<1\). In fact, if \(\lambda>0\), then let \(a_{1}=\frac{1}{2}\), and by the Poincaré inequality we have
Then letting \(a_{2}=\frac{1}{2}+\frac{\lambda M}{2m\lambda_{1}}\), we can get the result. Similarly, if \(-\frac{m\lambda_{1}}{M}<\lambda<0\), then let \(a_{2}=\frac{1}{2}\), and by the Poincaré inequality we have
We can also let \(a_{1}=\frac{1}{2}+\frac{\lambda M}{2m\lambda_{1}}\) to get the result.
Lemma 3
[16]
For the function \(F:M\subseteq E\rightarrow R\) with \(M\neq \emptyset\), \(\min_{u\in M}F(u)=\alpha\) has a solution in the case that the following hold:
\((F_{1})\)
E is a real reflexive Banach space,
\((F_{2})\)
M is bounded and weak sequentially closed,
\((F_{3})\)
F is weak sequentially lower semi- continuous on M, i. e., by definition, for each sequence \(\{u_{n}\}\) in M such that \(u_{n}\rightharpoonup u\) as \(n\rightarrow \infty\), we have \(F(u)\leq \liminf_{n\rightarrow \infty}F(u_{n})\). Lemma 4
[6]
Let E be a real Banach space with \(E=V\bigoplus W\), where V is finite- dimensional. Suppose \(\phi\in C'(E,R)\) satisfies \(P.S\). condition, and:
\((\phi_{1})\)
there are constants \(\rho, \tau> 0\) such that \(\phi |_{\partial B_{\rho}\cap W}\geq \tau\), and
\((\phi_{2})\)
there is \(e\in \partial B_{1}\cap W\) and \(R> \rho\) such that if \(Q=(\overline{B_{R}}\cap V)\oplus\{re|0< r< R\}\), then \(\phi |_{\partial Q}\leq 0\). Then ϕ possesses a critical value \(c\geq \tau \) which can be characterized as \(c=\inf_{h\in \Gamma}\max_{u\in Q}\phi(h(u))\), where \(\Gamma=\{h\in C(\overline{Q},E) ,h|_{\partial Q}=id\}\). Lemma 5
[17]
Let E be an infinite- dimensional real Banach space and \(\varphi\in C^{1}(E,R)\) be even, satisfying the \(P.S\). condition and \(u(0)=0\). If \(E=V\oplus W\), where V is finite- dimensional, and φ satisfies the following conditions:
\((\varphi_{1})\)
there exist constants \(\rho, \tau >0\), such that \(\varphi |_{\partial B_{\rho}\cap W}\geq \tau\),
\((\varphi_{2})\)
for each finite- dimensional subspace \(V_{1}\subset E\), there is \(R=R(V_{1})\) such that, for all \(u\in \{u\in V_{1}, \Vert u\Vert \geq R\}\), we have \(\varphi(u)\leq0\). Then φ has an unbounded sequence of critical values. Main results
In this paper, we assume:
\((H_{1})\) There exists \(\mu> 2\) such that \(\mu F(t,u)\leq uf(t,u)\) and \(I_{i}(u)u\leq\mu\int_{0}^{u}{I_{i}(s)}\,ds<0\).
\((H_{2})\) There exists \(\rho_{0}>0\), \(\delta_{i}>0\) such that \(\int_{0}^{u}{I_{i}(s)}\,ds\geq-\delta_{i}\vert u\vert ^{\mu}\) for all \(\Vert u\Vert \leq \rho_{0}\).
\((H_{3})\) For all \(u\in X_{1}\cup X_{2}\), we have \(\int_{0}^{T}{F(t,u(t))}\,dt\geq\frac{a_{2}\Vert u\Vert ^{2}}{m}\).
The main results are the following theorems.
Theorem 1 Assume \((H_{1})\)-\((H_{3})\) hold and \(\vert \lambda \vert < \frac{m\lambda_{1}}{M}\), then the impulsive problem (1) has at least two weak solutions. Theorem 2 Suppose that \((H_{1})\)-\((H_{2})\) are satisfied and \(\lambda\geq - \frac{m\lambda_{1}}{M}\). If \(f(t,u)\) and \(I_{i}(u)\) are odd about u, then the problem (1) has infinitely many weak solutions.
Next, we give the main lemma used in this paper.
Lemma 6
[12]
Suppose that \((H_{1})\) holds, then for all \(t\in [0,T]\) we can obtain Remark 2
For the convenience of the reader, we denote \(M_{1}=\sup\{F(t,u),t\in [0,T],\vert u\vert =1\}\), \(m_{1}=\inf\{F(t,u),t\in [0,T],\vert u\vert =1\} \).
Lemma 7 Suppose that \((H_{1})\) and \((H_{2})\) hold, then φ satisfies the \(P.S\). condition. Proof
Let \(\{u_{n}\}\in H_{0}^{1}(0,T)\) be such a sequence that \(\{\varphi(u_{n})\}\) is bounded and \(\lim_{n\rightarrow \infty}\varphi'(u_{n})=0\), we will show that \(u_{n}\) has a convergent subsequence. In view of the given condition, there exists \(C_{1}>0\) such that
for all
n. By the definition of φ and (3) we obtain
Combining \((H_{1})\), we have
Hence, \(u_{n}\rightarrow u\) in \(C[0,T]\). Furthermore
Combining with (6), we know \(\Vert u_{n}-u\Vert ^{2}\rightarrow 0\). So,
φ satisfies the \(P.S\). condition. □ Proof of Theorem 1
Because \(H_{0}^{1}(0,T)\) is Hilbert space, \(\overline{B_{\rho}}\) is bounded and weak sequentially closed for all \(\rho >0\). We will show
φ is weak sequentially lower semi-continuous on \(\overline{B_{\rho}}\). In fact suppose \(u_{n}\rightharpoonup u\) in \(H_{0}^{1}(0,T)\), then \(\Vert u\Vert \leq\liminf_{n\rightarrow\infty} \Vert u_{n}\Vert \), and \(u_{n}\rightarrow u\) in \(C[0,T]\), so
By the definition of
φ, we can obtain \(\varphi(u)\leq\liminf_{n\rightarrow \infty}\varphi(u_{n})\). From Lemma 3, there exists \(u_{0}\) such that \(\varphi(u_{0})=\min\{\varphi(u),u\in \overline{B_{\rho}}\}\). Noting that \(\varphi(0)=0\), so \(\varphi(u_{0})\leq 0\). Combining with Lemma 2, Lemma 6, and \((H_{2})\), there exists a small ρ with \(\rho<\rho_{0}\) such that when \(u(t)\in \partial B_{\rho}\) we have
\(\varphi(u_{0})\leq 0\) implies that
φ possesses a critical point \(u_{0}\in B_{\rho}\).
Let \(V=X_{1}\oplus X_{2}, W=\overline{\bigoplus_{i=3}^{\infty}X_{i}}\), then there exists \(\rho, \tau>0\), such that
Hence
φ satisfies the condition \((\phi_{1})\).
Taking \(e\in W\) such that \(\Vert e\Vert =1\). By Lemma 6, we claim that there exist \(C_{2}, M_{2}>0\), such that
By \((H_{1})\), for all \(r>0\), \(u\in V\), we have
Combining with \(a_{2}<1\), there exists \(R>0\), such that the following conclusions hold:
For any \(u(t)\in V\), \(\Vert u\Vert \leq R\), we can obtain
Let \(\Omega=(\overline{B_{R}}\cap V)\oplus\{re|0< r< R\}\), \(\partial\Omega=\Omega_{1}\cup\Omega_{2}\cup\Omega_{3}\), where
So \(\varphi |_{\partial}Q\leq0\). From Lemma 4 we know that
φ has a critical point c with \(c\geq\tau>0\). In other words, there exists \(u_{1}\) such that \(\varphi(u_{1})= c\). Hence \(u_{0}\) and \(u_{1}\) are classical solutions of the impulsive problem (1). □ Proof of Theorem 2
Since \(f(t,u)\) and \(I_{i}\) are odd about
u, \(\varphi(u)\) is even and \(\varphi(0)=0\). Lemma 7 shows that φ satisfies the \(P.S\). condition. In the same way as in Theorem 1, we can verify φ satisfies the condition \((\varphi_{1})\) in Lemma 5. Finally we prove that φ also satisfies the condition \((\varphi_{2})\). According to Lemma 6 and \((H_{1})\), for every \(V_{1}\), \(u\in V_{1}\), there exists \(C_{3}, M_{3}>0\) such that Example
Take \(x(t)\in C[0,T]\), \(t_{1}\in (0,T)\) and \(k>0\). Consider the following impulsive problem:
Conclusion
A second-order impulsive differential equation is considered in this paper. Some necessary conditions for the existence and multiplicity of solutions are presented by critical point theories and variational methods. We also proposed a numerical example to show the advantage.
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Acknowledgements
The authors are very grateful to the anonymous referee for his/her valuable comments and suggestions.
Additional information Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors have made equal contributions. All authors read and approved the final manuscript. |
We take a lot of things for granted: electricity, gas at the pump, and mediocre coffee at the office. Many concepts in basic algebra are also taken for granted, such as cancellation of terms, and commutativity. This post will revisit some basic algebra (think solving for x), but with some of those things we took for granted removed.
(1) Associativity of the operation: For a,b,c \in G, a\ast(b\ast c) = (a\ast b) \ast c (2) Existence of an identity element in the group: There is some e\in G such that for any a \in G, e\ast a = a\ast e = a (3) Existence of inverses in the group: For every a \in G there is a corresponding a^{-1} \in G such that a\ast a^{-1} = a^{-1}\ast a = e
There are a couple natural questions we can discuss first:
Q1: Can a group have more than one identity element?
Axiom (2) just said we had to have one. It never said we had to have
only one. 1 Let’s see what happens if we have some group \langle G, \ast\rangle that has two identity elements, e_{1} and e_{2}. Well, since both are identity elements, then e_{1}\ast a = a for all elements a in G, and e_{2}\ast a = a for all elements a in G.
So let’s stick e_{2} in for a with e_{1}. That means e_{1}\ast e_{2} = e_{2}. But we also know that if we use e_{2} as the identity element, e_{1} as our a, and the fact from Axiom (2) that the identity element commutes with every other group element, then e_{1} \ast e_{2} = e_{1} as well. That means that e_{1} \ast e_{2} = e_{2} and e_{1} \ast e_{2}=e_{1}. The only possible way this can happen (because \ast is an operation) is for e_{1} = e_{2}
What this shows is that there can only be one identity element. We started by assuming there were two different ones, and it resulted in our finding out the two different ones were equal, and thus not different at all.
Q2: Can an element have more than one inverse?
We can use the same trick we just did with Q1 (just on Axiom (3)) to show that every element in the group has exactly one and only one inverse.
Those are a couple seemingly simple questions, but they weren’t specifically answered in the definition of a group, so we had to investigate and answer them ourselves before we could move on. Next, we want to develop a couple rules that come from the group axioms that we can use in solving equations on generic groups.
Cancellation Laws and Handling Inverses
The cancellation laws allow us to cancel out common factors on either side of an equal sign. Stated formally from Pinter
2 If G is a group, and a,b,c are elements of G, then Theorem:
(i) ab = ac \Rightarrow b=c
and
(ii) ba = ca \Rightarrow b=c
The proof isn’t hard, and not really the focus here. What’s important to note is that there are
two different cancellations. One where I cancel a when it is multiplied on the left of b and c, and one where a is multiplied on the right. Multiplying on the left and on the right is not the same thing unless we have commutativity. (Read about commutativity here). Unless we have commutativity as a property of our group (and often we do not, such as in matrix multiplication), ba = ac does not imply that b=c.
I’ve taken away that thing you have all just assumed would be there, and likely relied on in your high school algebra classes.
3 Now when we solve for x, as we will do shortly, we’re going to have to be a little more clever. First, we’ll also need a tool to handle inverses, a more general form of “subtract from both sides” that you used to use. : If G is a group, and a,b are elements of G, then we have the following: Theorem
(i) (ab)^{-1} = b^{-1}a^{-1}
and
(ii)(a^{-1})^{-1} = a
Again, the proof isn’t actually what’s important here, although it’s a fun thing to do. What’s important for this discussion is the understanding of the theorem. The first point tells us how to find the inverse element of a product of two elements. It’s a product of the two individual elements’ inverses, but in reverse order. The common analogy here is the “socks and shoes principle”. First you put on socks, then the shoes. (a, then “multiply”
4 b). To get the socks and shoes off, you have to remove the shoes first (inverse of b) then the socks (inverse of a).
The second one states formally what we would expect to be true. The inverse of an inverse returns the original element.
Let’s do some algebra!
Let’s revisit some of that high school algebra. This time we aren’t doing it on unknown numbers, but rather an entirely general group with an entirely general operation. The cool part? As long as we’re careful about the rules we have and don’t, solving for x is just as easy as it was in high school.
For these next few exercises, all we’re going to get is that a,b,c, and x are in a group G. Our goal? Solve for x.
axb = c
OK, the x isn’t multiplication. It’s the variable we want to solve for. The only rules we get are the two theorems above. We still need to isolate x, so we need to get rid of a and b. Normally, you’d tell me to either subtract or divide by a and b. Here, we use the inverse operation. I’m going to do this almost painfully slow so you get an understanding, not just a mechanism.
First, we’re going to get rid of b. To do that, we multiply
on the right by b^{-1} on both sides:
Now, we know by now how inverses work, so we know that bb^{-1} = e, where e is the identity element of the group G.
5. So, now we have that
But anything multiplied
6 by the identity element is that thing. So
One down, one to go. We need to get rid of a. Notice that it’s being multiplied on the left of x. We can’t just shove an a^{-1} in between a and x, so we have to multiply a^{-1} on the far left of both sides of the equation.\begin{aligned}axbb^{-1} &= cb^{-1}\\axe &= cb^{-1}\\ax&=cb^{-1}\\a^{-1}ax &= a^{-1}cb^{-1}\end{aligned}
You’ve recognized by now that a^{-1}a = e, so\begin{aligned}axbb^{-1} &= cb^{-1}\\axe &= cb^{-1}\\ax&=cb^{-1}\\a^{-1}ax &= a^{-1}cb^{-1}\\ ex &= a^{-1}cb^{-1}\\x &=a^{-1}cb^{-1}\end{aligned}
Done! We did it! To isolate x, we had to reverse what was done to it, and do it to both sides. We multiplied on the left by a and on the right by b, so we had to multiply on the left by a^{-1} and on the right by b^{-1}. The most important thing to note here is that we didn’t have commutativity, meaning that we couldn’t just switch the order of the variables around. We couldn’t guarantee that axb \neq abx \neq xab, and thus the result we got (x = a^{-1}cb^{-1}) is strict. You can’t change the order around and have that equation still be true.
Let’s try one more. Once again, we aren’t explicitly told our generic group has commutativity, so we can’t assume we have it.
x^{2}b = xa^{-1}c
Yikes! An x^{2}. Fear not. Let’s write that a bit differently. x^{2} = xx, soxxb = xa^{-1}c
Holy cancellation law, Batman! No need to get into any scary generalized quadratics here. Recall that we get to cancel the same thing on both sides if they are either on the far left or the far right on both sides. x is multiplied on the left on both sides, and by our awesome cancellation theorem, is redundant. Off it goes, leaving us with\begin{aligned}xxb &= xa^{-1}c\\xb&=a^{-1}c\end{aligned}
From here, the problem is even easier than the first one we did. Multiply on the right by b^{-1} and we’re home free:\begin{aligned}xxb &= xa^{-1}c\\xb&=a^{-1}c\\xbb^{-1} &= a^{-1}cb^{-1}\\xe &= a^{-1}cb^{-1}\\x &= a^{-1}cb^{-1}\end{aligned}
I didn’t really do much to the first problem; I just scared you by multiplying on the left by x.
Conclusion
What we just did in this post was build on our knowledge of groups to solve equations within those groups. The way to go about this is really no different than you used to in high school; it’s just more general and a little more abstract. We didn’t even know what the elements of G were, or the operation for that matter. We didn’t know if the group had a finite number of elements or an infinite number. We didn’t even have commutativity. But we could still solve equations, and in a way that was almost like very elementary algebra. This is the power of algebra and abstraction. The goal is to take what we already knew, reduce it to its skeleton (the group structure), and see what we can still do with it.
Specific groups and operations can get messy. What if we have a molecule we can rotate, translate, or reflect about an axis? What if we do some combination of those actions to a molecule? What if we want to figure out if two molecules are actually the same or chiral
7? Instead of trying to figure out how to play with it, we can abstract the notion of these actions as elements of a group, and use group theory and algebra to answer that question in a very similar manner to the basic equations we just solved here.
Abstraction and generalization make our lives easier. We can rise above a messy, complicated space, and reduce the problem to one that looks as simple as high school algebra.
This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License. Footnotes This is why language is so important in mathematics. Read everything carefully, and read only what is written. Otherwise you can end up assuming you have things or properties you don’t actually have The operation is still there, but we don’t always have to write it. ab is the same as a*b in our case. Just mathematicians bein’ lazy. It happens to the best of us. I’m laughing maniacally right now. I’m going to use multiply as the generic word for our operation. I don’t necessarily mean standard multiplication. This is like dividing by 2 to get rid of multiplying by 2. It doesn’t actually get rid of the 2; it leaves a 1 (the multiplicative identity element) behind, and multiplying by 1 doesn’t change anything again, multiply is just the word we are using for the general operation. Don’t get bogged down that this is regular number multiplication. This is the notion of handedness. Your left and right hand are chiral: mirror images of the other, but no rotations or reflections will transform your left hand into your right hand. A more formal way to state it is that the object and its mirror image are not superimposable. |
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Change to browse by: References & Citations Bookmark(what is this?) Mathematics > Group Theory Title: The influence of cut vertices and eigenvalues on character graphs of solvable groups
(Submitted on 19 Sep 2019)
Abstract: Given a finite group $G$, the character graph, denoted by $\Delta(G)$, for its irreducible character degrees is a graph with vertex set $\rho(G)$ which is the set of prime numbers that divide the irreducible character degrees of $G$, and with $\{p,q\}$ being an edge if there exist a non-linear $\chi\in {\rm Irr}(G)$ whose degree is divisible by $pq$. In this paper, we discuss the influences of cut vertices and eigenvalues of $\Delta(G)$ on the group structure of $G$. Recently, Lewis and Meng proved the character graph of each solvable group has at most one cut vertex. Now, we determine the structure of character graphs of solvable groups with a cut vertex and diameter $3$. Furthermore, we study solvable groups whose character graphs have at most two distinct eigenvalues. Moreover, we investigate the solvable groups whose character graphs are regular with three distinct eigenvalues. In addition, we give some lower bounds for the number of edges of $\Delta(G)$. Submission historyFrom: Ali Iranmanesh [view email] [v1]Thu, 19 Sep 2019 21:11:26 GMT (12kb) |
Television signal splitter consisting of a high-pass filter (left) and a low-pass filter (right). The antenna is connected to the screw terminals to the left of center.
Electronic filters are analog circuits which perform signal processing functions, specifically to remove unwanted frequency components from the signal, to enhance wanted ones, or both. Electronic filters can be:
The most common types of electronic filters are linear filters, regardless of other aspects of their design. See the article on linear filters for details on their design and analysis.
Contents History 1 Classification by technology 2 Passive filters 2.1 Single element types 2.1.1 L filter 2.1.2 T and π filters 2.1.3 Multiple element types 2.1.4 Active filters 2.2 Other filter technologies 2.3 Digital filters 2.3.1 Quartz filters and piezoelectrics 2.3.2 SAW filters 2.3.3 BAW filters 2.3.4 Garnet filters 2.3.5 Atomic filters 2.3.6 The transfer function 3 Classification by topology 4 Classification by design methodology 5 Direct circuit analysis 5.1 Image impedance analysis 5.2 Network synthesis 5.3 See also 6 Notes, references and external links 7 History
The oldest forms of electronic filters are passive analog linear filters, constructed using only resistors and capacitors or resistors and inductors. These are known as RC and RL single-pole filters respectively. More complex multipole LC filters have also existed for many years, and their operation is well understood.
Hybrid filters are also possible, typically involving a combination of analog amplifiers with mechanical resonators or delay lines. Other devices such as CCD delay lines have also been used as discrete-time filters. With the availability of digital signal processing, active digital filters have become common.
Classification by technology Passive filters
Passive implementations of linear filters are based on combinations of resistors (R), inductors (L) and capacitors (C). These types are collectively known as
passive filters, because they do not depend upon an external power supply and/or they do not contain active components such as transistors.
Inductors block high-frequency signals and conduct low-frequency signals, while capacitors do the reverse. A filter in which the signal passes through an inductor, or in which a capacitor provides a path to ground, presents less attenuation to low-frequency signals than high-frequency signals and is therefore a
low-pass filter. If the signal passes through a capacitor, or has a path to ground through an inductor, then the filter presents less attenuation to high-frequency signals than low-frequency signals and therefore is a high-pass filter. Resistors on their own have no frequency-selective properties, but are added to inductors and capacitors to determine the time-constants of the circuit, and therefore the frequencies to which it responds.
The inductors and capacitors are the reactive elements of the filter. The number of elements determines the order of the filter. In this context, an LC tuned circuit being used in a band-pass or band-stop filter is considered a single element even though it consists of two components.
At high frequencies (above about 100 megahertz), sometimes the inductors consist of single loops or strips of sheet metal, and the capacitors consist of adjacent strips of metal. These inductive or capacitive pieces of metal are called stubs.
Single element types
A low-pass electronic filter realised by an RC circuit
The simplest passive filters, RC and RL filters, include only one reactive element, except hybrid LC filter which is characterized by inductance and capacitance integrated in one element.
[1] L filter
An L filter consists of two reactive elements, one in series and one in parallel.
T and π filters
Low-pass π filter
High-pass T filter
Three-element filters can have a 'T' or 'π' topology and in either geometries, a low-pass, high-pass, band-pass, or band-stop characteristic is possible. The components can be chosen symmetric or not, depending on the required frequency characteristics. The high-pass T filter in the illustration, has a very low impedance at high frequencies, and a very high impedance at low frequencies. That means that it can be inserted in a transmission line, resulting in the high frequencies being passed and low frequencies being reflected. Likewise, for the illustrated low-pass π filter, the circuit can be connected to a transmission line, transmitting low frequencies and reflecting high frequencies. Using m-derived filter sections with correct termination impedances, the input impedance can be reasonably constant in the pass band.
[2] Multiple element types
Multiple element filters are usually constructed as a ladder network. These can be seen as a continuation of the L,T and π designs of filters. More elements are needed when it is desired to improve some parameter of the filter such as stop-band rejection or slope of transition from pass-band to stop-band.
Active filters
Active filters are implemented using a combination of passive and active (amplifying) components, and require an outside power source. Operational amplifiers are frequently used in active filter designs. These can have high Q factor, and can achieve resonance without the use of inductors. However, their upper frequency limit is limited by the bandwidth of the amplifiers.
Other filter technologies Digital filters
A general finite impulse response filter with
n stages, each with an independent delay, d i and amplification gain, a i.
Digital signal processing allows the inexpensive construction of a wide variety of filters. The signal is sampled and an analog-to-digital converter turns the signal into a stream of numbers. A computer program running on a CPU or a specialized DSP (or less often running on a hardware implementation of the algorithm) calculates an output number stream. This output can be converted to a signal by passing it through a digital-to-analog converter. There are problems with noise introduced by the conversions, but these can be controlled and limited for many useful filters. Due to the sampling involved, the input signal must be of limited frequency content or aliasing will occur.
Quartz filters and piezoelectrics
Crystal filter with a center frequency of 45 MHz and a bandwidth B
3dB of 12 kHz.
In the late 1930s, engineers realized that small mechanical systems made of rigid materials such as quartz would acoustically resonate at radio frequencies, i.e. from audible frequencies (sound) up to several hundred megahertz. Some early resonators were made of steel, but quartz quickly became favored. The biggest advantage of quartz is that it is piezoelectric. This means that quartz resonators can directly convert their own mechanical motion into electrical signals. Quartz also has a very low coefficient of thermal expansion which means that quartz resonators can produce stable frequencies over a wide temperature range. Quartz crystal filters have much higher quality factors than LCR filters. When higher stabilities are required, the crystals and their driving circuits may be mounted in a "crystal oven" to control the temperature. For very narrow band filters, sometimes several crystals are operated in series.
Engineers realized that a large number of crystals could be collapsed into a single component, by mounting comb-shaped evaporations of metal on a quartz crystal. In this scheme, a "tapped delay line" reinforces the desired frequencies as the sound waves flow across the surface of the quartz crystal. The tapped delay line has become a general scheme of making high-
Q filters in many different ways. SAW filters
SAW (surface acoustic wave) filters are electromechanical devices commonly used in radio frequency applications. Electrical signals are converted to a mechanical wave in a device constructed of a piezoelectric crystal or ceramic; this wave is delayed as it propagates across the device, before being converted back to an electrical signal by further electrodes. The delayed outputs are recombined to produce a direct analog implementation of a finite impulse response filter. This hybrid filtering technique is also found in an analog sampled filter. SAW filters are limited to frequencies up to 3 GHz. The filters were developed by Professor Ted Paige and others.
[3] BAW filters
BAW (bulk acoustic wave) filters are electromechanical devices. BAW filters can implement ladder or lattice filters. BAW filters typically operate at frequencies from around 2 to around 16 GHz, and may be smaller or thinner than equivalent SAW filters. Two main variants of BAW filters are making their way into devices: thin-film bulk acoustic resonator or FBAR and solid mounted bulk acoustic resonators.
Garnet filters
Another method of filtering, at microwave frequencies from 800 MHz to about 5 GHz, is to use a synthetic single crystal yttrium iron garnet sphere made of a chemical combination of yttrium and iron (YIGF, or yttrium iron garnet filter). The garnet sits on a strip of metal driven by a transistor, and a small loop antenna touches the top of the sphere. An electromagnet changes the frequency that the garnet will pass. The advantage of this method is that the garnet can be tuned over a very wide frequency by varying the strength of the magnetic field.
Atomic filters
For even higher frequencies and greater precision, the vibrations of atoms must be used. Atomic clocks use caesium masers as ultra-high
Q filters to stabilize their primary oscillators. Another method, used at high, fixed frequencies with very weak radio signals, is to use a ruby maser tapped delay line. The transfer function see also Filter (signal processing) for further analysis
The transfer function \ H(s) of a filter is the ratio of the output signal \ Y(s) to that of the input signal \ X(s) as a function of the complex frequency \ s:
\ H(s)=\frac{Y(s)}{X(s)}
with \ s = \sigma + j \omega.
The transfer function of all linear time-invariant filters, when constructed of discrete components, will be the ratio of two polynomials in \ s, i.e. a rational function of \ s. The order of the transfer function will be the highest power of \ s encountered in either the numerator or the denominator.
Classification by topology
Electronic filters can be classified by the technology used to implement them. Filters using passive filter and active filter technology can be further classified by the particular electronic filter topology used to implement them.
Any given filter transfer function may be implemented in any electronic filter topology.
Some common circuit topologies are:
Classification by design methodology
Historically, linear analog filter design has evolved through three major approaches. The oldest designs are simple circuits where the main design criterion was the Q factor of the circuit. This reflected the radio receiver application of filtering as Q was a measure of the frequency selectivity of a tuning circuit. From the 1920s filters began to be designed from the image point of view, mostly being driven by the requirements of telecommunications. After World War II the dominant methodology was network synthesis. The higher mathematics used originally required extensive tables of polynomial coefficient values to be published but modern computer resources have made that unnecessary.
[4] Direct circuit analysis
Low order filters can be designed by directly applying basic circuit laws such as Kirchhoff's laws to obtain the transfer function. This kind of analysis is usually only carried out for simple filters of 1st or 2nd order.
RL filter frequency response
Image impedance analysis
This approach analyses the filter sections from the point of view of the filter being in an infinite chain of identical sections. It has the advantages of simplicity of approach and the ability to easily extend to higher orders. It has the disadvantage that accuracy of predicted responses relies on filter terminations in the image impedance, which is usually not the case.
[5]
Constant k filter response with 5 elements
Zobel network (constant R) filter, 5 sections
m-derived filter response, m=0.5, 2 elements
m-derived filter response, m=0.5, 5 elements
Network synthesis
The network synthesis approach starts with a required transfer function and then expresses that as a polynomial equation of the input impedance of the filter. The actual element values of the filter are obtained by continued-fraction or partial-fraction expansions of this polynomial. Unlike the image method, there is no need for impedance matching networks at the terminations as the effects of the terminating resistors are included in the analysis from the start.
[5]
Here is an image comparing Butterworth, Chebyshev, and elliptic filters. The filters in this illustration are all fifth-order low-pass filters. The particular implementation – analog or digital, passive or active – makes no difference; their output would be the same.
As is clear from the image, elliptic filters are sharper than all the others, but they show ripples on the whole bandwidth.
See also Notes, references and external links ^ Dzhankhotov V., Hybrid LC filter for power electronic drives: Theory and Implementation, 2009 ^ The American Radio Relay League, Inc.: "The ARRL Handbook, 1968" page 50 ^ Ash, Eric A; E. Peter Raynes (December 2009). "Edward George Sydney Paige. 18 July 1930 — 20 February 2004" (PDF). Biographical Memoirs of Fellows of the Royal Society 55: 185–200. ^ Bray, J, Innovation and the Communications Revolution, Institute of Electrical Engineers ^ a b Matthaei, Young, Jones Microwave Filters, Impedance-Matching Networks, and Coupling Structures McGraw-Hill 1964
Zverev, Anatol, I (1969). Handbook of Filter Synthesis. John Wiley & Sons. Catalog of passive filter types and component values. The Bible for practical electronic filter design. Williams, Arthur B & Taylor, Fred J (1995). Electronic Filter Design Handbook. McGraw-Hill. National Semiconductor AN-779 application note describing analog filter theory Fundamentals of Electrical Engineering and Electronics – Detailed explanation of all types of filters BAW filters (in French; PDF) Some Interesting Filter Design Configurations & Transformations Analog Filters for Data Conversion
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I am trying to catch up with my statistics. I want to understand what are multivariate Gaussian or multivariate normal distribution. I understand what a univariate Gaussian is any variable $X$ such that $X\sim N(\mu,\sigma^2),\mu\in\mathbb R, \sigma^2>0$ (except for complex numbers, isn't something squared always greater than $0$ ?), that is to say, if $x_i\in X\sim N(0,\sigma^2)$ There is lots of chance that it pertains to something that looks like this :
And as far as
if $x_i\in X\sim N(\mu,\sigma^2)$ with $\mu =\begin{bmatrix}0\\0\end{bmatrix}$ and $\sigma =\begin{bmatrix}1 & 3/5\\3/5 & 2\end{bmatrix}$ (I don't understand the covariance matrix, especially the bottom right corner) It would have be part of something like this :
But then I don't understand its definition given on ... which I found back on Wikipedia :
A random vector $X = (X_1, …, X_k)'$ is said to have the multivariate normal distribution if it satisfies the following equivalent conditions.
Every linear combination of its components $Y = a_1X_1 + … + a_kX_k$ is normally distributed i.e. $a^TX=\sum_{i=1}^na_iX_i$ is Gaussian $\forall a\in\mathbb R^n$ (what $a$ has to do there ?)
Would it mean, if we had three random variables, that combining two of them would be a Gaussian, it would look like the picture above. Yet, it isn't a Gaussian, it's a bi dimensional Gaussian ! |
Experiment 1
You are unhappy with the logging company you hired to thin a stand of red pine. You carefully laid out the skid trails leaving bumper trees to avoid excess damage to the remaining trees. In the contract, it is stated that the logging company would pay a penalty (3 times the stumpage rate) for trees damaged beyond the agreed amount of five or more damaged trees per acre. You want to estimate the number of damaged trees per acre to see if they exceeded this amount. You take 27 samples, from which you compute the sample mean, and then construct a 95% confidence interval about the mean number of damaged trees per acre.
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Enter these data in the first column of the Minitab worksheet and label it “Trees.” Now calculate the sample mean and sample standard deviation.
Stat> Basic Statistics> Display Descriptive Statistics. Select the column with your data in the variable box.
a) sample mean: ____________________________
sample standard deviation: ___________________
Examine the normal probability plot for this data set. Remember, for a sample size less than n = 30, we must verify the assumption of normality if we do not know that the random variable is normally distributed. Go to
GRAPH → PROBABILITY PLOT. Enter the column with your data in the “ Graph variables” box and click OK.
b) Would you say that this distribution is normal?
c) Calculate the 95% confidence interval by hand using \(x \pm t_{\alpha/2}(\frac {s}{\squr {n}})\) and the t-table.
95% CI for the mean number of damaged trees:____________________________________
Now find the 95% confidence interval for the mean using Minitab.
Go to
STAT> Basic Statistics> 1-sample t…Enter data in “Samples in columns.” You do not have to enter the standard deviation but select OPTIONS and set the confidence level (make sure it is for 95%) and select “ Alternative:not equal.”
d) 95% CI for the mean number of damaged trees: __________________________________
e) Do you have enough statistical evidence to state that the logging company has exceeded the damage limit? Why?
Experiment 2
The amount of sewage and industrial pollution dumped into a body of water affects the health of the water by reducing the amount of dissolved oxygen available for aquatic life. If the population mean dissolved oxygen drops below five parts per million (ppm), a level some scientists think is marginal for supplying enough dissolved oxygen for fish, some remedial action will be attempted. Given the expense of remediation, a decision to take action will be made only if there is sufficient evidence to support the claim that the mean dissolved oxygen has DECREASED below 5 ppm. Below are weekly readings from the same location in a river over a two-month time period.
5.2, 4.9, 5.1, 4.2, 4.7, 4.5, 5.0, 5.2, 4.8, 4.6, 4.8
The population standard deviation is unknown and we have a small sample (n≤30). You must verify the assumption of normality. Go to
GRAPH→ PROBABILITY PLOT. Examine the normal probability plot. Does the distribution look normal?
Use DESCRIPTIVE STATISTICS (
Basic Statistics>Display Descriptive Statistics) to get the mean and sample standard deviation.
Now test the claim that the mean dissolved oxygen is less than 5ppm using α = 0.05
a) First, state the null and alternative hypotheses
H0:__________________________________
H1: __________________________________
b) Compute the test statistic by hand \(t = \frac {\bar x - \mu}{s/\sqrt {n}}\)
c) Find the critical value from the t-table: ________________________________________
d) Do you reject the null hypothesis or fail to reject the null hypothesis?
Now use Minitab to do the hypothesis test. Go to
STAT > BASIC STAT > 1-SAMPLE t. Check PERFORM HYPOTHESIS TEST and enter the hypothesized mean (5.00). Click OPTIONS and enter the confidence level (1-α) and select alternative hypothesis (H1). Click OK. Check to see that the null and alternative hypotheses shown in the session window are correct.
e) What is the p-value for this test?
f) Do you reject or fail to reject the null hypothesis?
g) State your conclusion:
Experiment 3
A forester believes that tent caterpillars are doing a significant amount of damage to the growth of the hardwood tree species in his stand. He has growth data from 21 plots before the infestation. Since then, he has re-measured those same plots and wants to know if there has been a significant reduction in the annual diameter growth.
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You need to compute the differences between the
before values and the after values. To create a new variable (diff), type “diff” in the header of the column you want to use. Select CALC>CALCULATOR. In the “Expressions” box, type in the equation “ Before-After.” In the box “ Store results in variable” type “diff.” Click OK.
You now have a new data set of the
differences with which you will complete your analyses. Compute basic descriptive statistics to get the sample mean \(\bar d\) and sample standard deviation \(s_d\) of the differences. Use these statistics to test the claim that there has been a reduction in annual diameter growth. You can answer this question by using either a hypothesis test or confidence interval.
a) H0:____________________________________
H1: ____________________________________
\(t= \frac {\bar d -\mu_d}{s_d/\sqrt {n}}\) or \(\bar d \pm t_{\alpha/2} \frac {s_d}{\sqrt{n}}\)
Do you reject or fail to reject the null hypothesis?
Now let Minitab do the work for you. Select
STAT> Basic Statistics> Paired t… Select SAMPLES IN COLUMNS. Enter the before as the First sample and after data as the Second sample. Select OPTIONS to set the confidence level and alternative hypothesis. Make sure the Test mean is set to 0.0. Click OK.
b) Write the test statistic and p-value
c) Write a complete conclusion that answers the question.
Experiment 4
Alternative energy is an important topic these days and a researcher is studying a solar electric system. Each day at the same time he collected voltage readings from a meter connected to the system and the data are given below. Is there a significant difference in the mean voltage readings for the different types of days? First do an F-test to test for equal variances and then test the means using the appropriate 2-sample t-test based on the results from the F-test. Please state a complete conclusion for this problem. α = 0.05.
Sunny – 13.5, 15.8, 13.2, 13.9, 13.8, 14.0, 15.2, 12.1, 12.9, 14.9
Cloudy – 12.7, 12.5, 12.6, 12.7, 13.0, 13.0, 12.1, 12.2, 12.9, 12.7
F-Test
a) Write the null and alternative hypotheses to test the claim that the variances are not equal.
H0:____________________________________ H1: ____________________________________
Select
STAT>BASIC STAT>2 Variances. In the Data box select “Samples in different columns” and enter Sunny in the First box and Cloudy in the Second box. Click OPTIONS and in Hypothesized Ratio box select Variance1/Variance2. Make sure the Alternative is set at “ Not equal.” Click OK. Look at the p-value for the F-test at the bottom of the output.
b) Do you reject for fail to reject the null hypothesis?
c) Can you assume equal variances?
Now conduct a 2-sample t-test (you should have rejected the null hypothesis in the F-test and assumed unequal variances).
STAT>BASIC STAT>2-Sample t…Select the button for “ Samples in different columns” and put Sunny in the First box and Cloudy in the Second box. Click OPTIONS and set the confidence level and select the correct alternative hypothesis. Set the Test difference at 0.0. Click OK.
d) What is the p-value for this test?
e) Do you reject or fail to reject the null hypothesis? State your conclusion. |
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