text
stringlengths 256
16.4k
|
|---|
I'm taking a course on waves and optics using Young and Freedman's
University Physics, but I'm a bit confused about a couple of things. I've also looked at Griffiths' Introduction to Electrodynamics and Taylor's Classical Mechanics, and the three of them seem to say the following: Young and Freedman A wave is (roughly) a disturbance of a system which can propagate from one region of the system to another. Derives the wave function $y(x,t) = A \cos(kx - \omega t)$ for a sinusoidal wave on a string. Uses that to derive the wave equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$, where $v$ is the phase velocity. Uses Newton's second law to derive the wave equation for a general wave on a string, this time on the form $\frac{\partial^2 y}{\partial x^2} = \frac{\mu}{F} \frac{\partial^2 y}{\partial t^2}$. Compares the two and concludes firstly that $v = \sqrt{F/\mu}$ for a wave on a string, and secondly that the wave equation with coefficient $1/v^2$ is valid for any wave on a string.
That last step confuses me and doesn't seem to follow from the above. The wave equation with coefficient $1/v^2$ has only been shown to be valid for sinusoidal waves, so how can they conclude that the coefficient has that form for all waves on a string?
Griffiths
Recognises the intrinsic vagueness of the concept but provides a tentative description (definition?): "A wave is a disturbance of a continuous medium that propagates with a fixed shape at a constant velocity."
Argues that a general wave function has the form $f(z,t) = g(z - vt)$ for any function $g$. (I find this pretty satisfactory.)
Derives the wave equation $\frac{\partial^2 f}{\partial z^2} = \frac{\mu}{T} \frac{\partial^2 f}{\partial t^2}$ for a general wave on a string in a similar manner to Young and Freedman. Rewrites the coefficient as $1/v^2$ with $v = \sqrt{T/\mu}$ but doesn't simply state that $v$ is the phase velocity.
Starts with the wave function $f(z,t) = g(z - vt)$ and derives the wave equation on the form $\frac{\partial^2 f}{\partial z^2} = \frac{1}{v^2} \frac{\partial^2 f}{\partial t^2}$.
Concludes that $v$ is the phase velocity for
anywave.
Mentions (but does not show) that the general solution is $f(z,t) = g(z - vt) + h(z + vt)$ for some function $h$.
This seems more satisfactory. But something still confuses me: Why start with a general wave on a string? Is it simply for pedagogical reasons, or is it not sufficient to start with the wave function $f(z,t) = g(z - vt)$ and derive the wave equation from that?
Taylor
Also starts with a general wave on a string and derives $\frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2}$ with $c = \sqrt{T/\mu}$, basically the same as Griffiths I think. He claims that $c$ "is the speed with which the waves travel" without arguing for it yet. (I find him rather vague here. It's a bit unclear exactly what kind of claim this is, whether he thinks it follows from the derivation or not, whether he wants the reader to take it for granted, or if he's going to return to it or not.)
Shows that the general solution has the form $u(x,t) = f(x - ct) + g(x + ct)$ for any function $f$ and $g$.
Considers the case where $u(x,t) = f(x - ct)$, interprets this solution physically and sees that it describes a disturbance travelling in the positive $x$-direction at speed $c$. I suppose he has
nowshown that $c$ is the phase velocity.
Griffiths and Taylor seem to do things in opposite directions: Griffiths starts with the wave
function and derives the wave equation, and Taylor does the opposite. This leaves me confused as to what a wave is mathematically, and what relation the mathematical description has to the physical phenomenon.
That is, how is a wave defined? Is it a mathematical model which, by construction, fits many physical phenomena (and thus a wave is simply defined as a disturbance that has a wave function or is a solution to the wave equation - and if so, which of the two, function or equation, or are they equivalent?)? Or is it a more or less well-defined class of physical phenomena that just happens to be described by the wave equation (maybe not always)? In that case, what authority does the wave equation have?
Griffiths later derives differential equations for the $\mathbf{E}$ and $\mathbf{B}$ fields in empty space, compares them to the three-dimensional wave equation and concludes that electromagnetic waves exist. If this similarity to the wave equation
shows and not just suggests that EM waves exist, then it seems to follow that the wave equation has some authority in determining what counts as a wave. He uses the word "imply", but I don't know what that means.
|
Pullback attractors and statistical solutions for 2-D Navier-Stokes equations
1.
University of Warsaw, Institute of Applied Mathematics and Mechanics, Banacha 2, 02-097 Warsaw, Poland
Using time-averages and Banach generalized limits we construct a family of probability measures $\{\mu_t\}_{t\in \IR}$ on the pullback attractor $\{A(t)\}_{t\in \R}$ of the dynamical system associated with a two-dimensional nonautonomous Navier-Stokes flow in a bounded domain. The measures satisfy supp$\mu_t \subset A(t)$ for all $t\in \R$ and also the corresponding Liouville equation and energy equation. In the autonomous case, they reduce to some time-average measure $\mu$ with support included in the global attractor and being a stationary statistical solution of the Navier-Stokes flow.
Mathematics Subject Classification:Primary: 35B41, 35D05, 35D06, 76F2. Citation:Grzegorz Łukaszewicz. Pullback attractors and statistical solutions for 2-D Navier-Stokes equations. Discrete & Continuous Dynamical Systems - B, 2008, 9 (3&4, May) : 643-659. doi: 10.3934/dcdsb.2008.9.643
[1] [2] [3] [4]
Rodrigo Samprogna, Tomás Caraballo.
Pullback attractor for a dynamic boundary non-autonomous problem with Infinite Delay.
[5]
T. Caraballo, J. A. Langa, J. Valero.
Structure of the pullback attractor
for a non-autonomous scalar differential inclusion.
[6]
Zeqi Zhu, Caidi Zhao.
Pullback attractor and invariant measures for the three-dimensional regularized MHD equations.
[7]
Yunmei Chen, Jiangli Shi, Murali Rao, Jin-Seop Lee.
Deformable multi-modal image registration by maximizing Rényi's statistical dependence measure.
[8]
Michael Field, Ian Melbourne, Matthew Nicol, Andrei Török.
Statistical properties of compact group extensions of hyperbolic flows and their time one maps.
[9]
Linfang Liu, Xianlong Fu, Yuncheng You.
Pullback attractor in $H^{1}$ for nonautonomous stochastic reaction-diffusion equations on $\mathbb{R}^n$.
[10]
Yangrong Li, Lianbing She, Jinyan Yin.
Longtime robustness and semi-uniform compactness of a pullback attractor via nonautonomous PDE.
[11]
Wen Tan.
The regularity of pullback attractor for a non-autonomous
[12]
Francesco Di Plinio, Gregory S. Duane, Roger Temam.
Time-dependent attractor for the Oscillon equation.
[13]
Tomás Caraballo, Marta Herrera-Cobos, Pedro Marín-Rubio.
Global attractor for a nonlocal
[14]
John A. D. Appleby, John A. Daniels.
Exponential growth in the solution of an affine stochastic differential equation with an average functional and financial market bubbles.
[15]
Zhen Zhang, Jianhua Huang, Xueke Pu.
Pullback attractors of FitzHugh-Nagumo system on the time-varying domains.
[16]
Theodore Tachim Medjo.
Pullback $ \mathbb{V}-$attractor of a three dimensional globally modified two-phase flow model.
[17]
Wenjia Jing, Panagiotis E. Souganidis, Hung V. Tran.
Large time average of reachable sets and Applications to Homogenization of interfaces moving with oscillatory spatio-temporal velocity.
[18]
Vladimir V. Chepyzhov, Mark I. Vishik.
Trajectory attractor for reaction-diffusion system with diffusion
coefficient vanishing in time.
[19]
Fumio Ishizaki.
Analysis of the statistical time-access fairness index of one-bit feedback fair scheduler.
[20]
Ellen Baake, Michael Baake, Majid Salamat.
The general recombination equation
in continuous time and its solution.
2018 Impact Factor: 1.008
Tools Metrics Other articles
by authors
[Back to Top]
|
Coordinate systems are tools that let us use algebraic methods tounderstand geometry. While the
rectangular(also called Cartesian) coordinates that wehave been using are the most common, some problems are easier toanalyze in alternate coordinate systems.
A coordinate system is a scheme that allows us to identify any pointin the plane or in three-dimensional space by a set of numbers. Inrectangular coordinates these numbers are interpreted, roughlyspeaking, as the lengths of the sides of a rectangle. In
polarcoordinates a point in the plane is identified by a pair of numbers $(r,\theta)$.The number $\theta$ measures the angle between the positive$x$-axis and a ray that goes through the point,as shown in figure 12.1.1; the number$r$ measures the distance from the origin to thepoint. Figure 12.1.1 shows the point withrectangular coordinates $\ds (1,\sqrt3)$ and polar coordinates $(2,\pi/3)$, 2 units from the origin and $\pi/3$ radians from thepositive $x$-axis.
Just as we describe curves in the plane using equations involving $x$ and $y$, so can we describe curves using equations involving $r$ and $\theta$. Most common are equations of the form $r=f(\theta)$.
Example 12.1.1 Graph the curve given by $r=2$. All points with $r=2$ are at distance 2 from the origin, so $r=2$ describes the circle of radius 2 with center at the origin.
Example 12.1.2 Graph the curve given by $r=1+\cos\theta$. We first consider $y=1+\cos x$, as in figure 12.1.2. As $\theta$ goes through the values in $[0,2\pi]$, the value of $r$ tracks the value of $y$, forming the "cardioid'' shape of figure 12.1.2. For example, when $\theta=\pi/2$, $r=1+\cos(\pi/2)=1$, so we graph the point at distance 1 from the origin along the positive $y$-axis, which is at an angle of $\pi/2$ from the positive $x$-axis. When $\theta=7\pi/4$, $\ds r=1+\cos(7\pi/4)=1+\sqrt2/2\approx 1.71$, and the corresponding point appears in the fourth quadrant. This illustrates one of the potential benefits of using polar coordinates: the equation for this curve in rectangular coordinates would be quite complicated.
Each point in the plane is associated with exactly one pair of numbers in the rectangular coordinate system; each point is associated with an infinite number of pairs in polar coordinates. In the cardioid example, we considered only the range $0\le \theta\le2\pi$, and already there was a duplicate: $(2,0)$ and $(2,2\pi)$ are the same point. Indeed, every value of $\theta$ outside the interval $[0,2\pi)$ duplicates a point on the curve $r=1+\cos\theta$ when $0\le\theta< 2\pi$. We can even make sense of polar coordinates like $(-2,\pi/4)$: go to the direction $\pi/4$ and then move a distance 2 in the opposite direction; see figure 12.1.3. As usual, a negative angle $\theta$ means an angle measured clockwise from the positive $x$-axis. The point in figure 12.1.3 also has coordinates $(2,5\pi/4)$ and $(2,-3\pi/4)$.
The relationship between rectangular and polar coordinates is quite easy to understand. The point with polar coordinates $(r,\theta)$ has rectangular coordinates $x=r\cos\theta$ and $y=r\sin\theta$; this follows immediately from the definition of the sine and cosine functions. Using figure 12.1.3 as an example, the point shown has rectangular coordinates $\ds x=(-2)\cos(\pi/4)=-\sqrt2\approx 1.4142$ and $\ds y=(-2)\sin(\pi/4)=-\sqrt2$. This makes it very easy to convert equations from rectangular to polar coordinates.
Example 12.1.3 Find the equation of the line $y=3x+2$ in polar coordinates. We merely substitute: $r\sin\theta=3r\cos\theta+2$, or $\ds r= {2\over \sin\theta-3\cos\theta}$.
Example 12.1.4 Find the equation of the circle $\ds (x-1/2)^2+y^2=1/4$ in polar coordinates. Again substituting: $\ds (r\cos\theta-1/2)^2+r^2\sin^2\theta=1/4$. A bit of algebra turns this into $r=\cos(t)$. You should try plotting a few $(r,\theta)$ values to convince yourself that this makes sense.
Example 12.1.5 Graph the polar equation $r=\theta$. Here the distance from the origin exactly matches the angle, so a bit of thought makes it clear that when $\theta\ge0$ we get the spiral of Archimedes in figure 12.1.4. When $\theta< 0$, $r$ is also negative, and so the full graph is the right hand picture in the figure.
Converting polar equations to rectangular equations can be somewhat trickier, and graphing polar equations directly is also not always easy.
Example 12.1.6 Graph $r=2\sin\theta$. Because the sine is periodic, we know that we will get the entire curve for values of $\theta$ in $[0,2\pi)$. As $\theta$ runs from 0 to $\pi/2$, $r$ increases from 0 to 2. Then as $\theta$ continues to $\pi$, $r$ decreases again to 0. When $\theta$ runs from $\pi$ to $2\pi$, $r$ is negative, and it is not hard to see that the first part of the curve is simply traced out again, so in fact we get the whole curve for values of $\theta$ in $[0,\pi)$. Thus, the curve looks something like figure 12.1.5. Now, this suggests that the curve could possibly be a circle, and if it is, it would have to be the circle $\ds x^2+(y-1)^2=1$. Having made this guess, we can easily check it. First we substitute for $x$ and $y$ to get $\ds (r\cos\theta)^2+(r\sin\theta-1)^2=1$; expanding and simplifying does indeed turn this into $r=2\sin\theta$.
Exercises 12.1
Ex 12.1.1Plot these polar coordinate points on one graph:$(2,\pi/3)$, $(-3,\pi/2)$, $(-2,-\pi/4)$, $(1/2,\pi)$, $(1,4\pi/3)$, $(0,3\pi/2)$.
Find an equation in polar coordinates that has the same graph as the given equation in rectangular coordinates.
Ex 12.1.2$\ds y=3x$(answer)
Ex 12.1.3$\ds y=-4$(answer)
Ex 12.1.4$\ds xy^2=1$(answer)
Ex 12.1.5$\ds x^2+y^2=5$(answer)
Ex 12.1.6$\ds y=x^3$(answer)
Ex 12.1.7$\ds y=\sin x$(answer)
Ex 12.1.8$\ds y=5x+2$(answer)
Ex 12.1.9$\ds x=2$(answer)
Ex 12.1.10$\ds y=x^2+1$(answer)
Ex 12.1.11$\ds y=3x^2-2x$(answer)
Ex 12.1.12$\ds y=x^2+y^2$(answer)
Sketch the curve.
Ex 12.1.13$\ds r=\cos\theta$
Ex 12.1.14$\ds r=\sin(\theta+\pi/4)$
Ex 12.1.15$\ds r=-\sec\theta$
Ex 12.1.16$\ds r=\theta/2$, $\theta\ge0$
Ex 12.1.17$\ds r=1+\theta^1/\pi^2$
Ex 12.1.18$\ds r=\cot\theta\csc\theta$
Ex 12.1.19$\ds r={1\over\sin\theta+\cos\theta}$
Ex 12.1.20$\ds r^2=-2\sec\theta\csc\theta$
In the exercises below, find an equation in rectangular coordinates that has the same graph as the given equation in polar coordinates.
Ex 12.1.21$\ds r=\sin(3\theta)$(answer)
Ex 12.1.22$\ds r=\sin^2\theta$(answer)
Ex 12.1.23$\ds r=\sec\theta\csc\theta$(answer)
Ex 12.1.24$\ds r=\tan\theta$(answer)
|
Yes I think so. We may assume that $\mathcal F$ is the $\sigma$-algebra generated by sets of the form $X_n^{-1}(a,\infty)$ as $n$ runs over the positive integers and $a$ runs over the reals as $\mathcal F$ already contains these sets. In so doing, we might be making $\mathcal F$ smaller, but that only makes the problem harder.
Let $\mathcal P_1,\mathcal P_2,\mathcal P_3,\ldots$ be a sequence of refining finite partitions of $\mathbb R$ such that the intersection of an element of $\mathcal P_n$ for each $n$ consists of at most one point. For example, $\mathcal P_n$ could consist of intervals $[n,\infty)$, $(-\infty,-n)$ and half-open dyadic intervals of length $1/2^n$ between $-n$ and $n$.
Then let $\mathcal F_n$ be the finite sub-algebra of $\mathcal F$ generated by the sets $X_i^{-1}A$ for $1\le i\le n$ and $A$ running over $\mathcal P_n$.Write $\mathcal F_n=\{B^n_1,\ldots B^n_{k_n}\}$. Now there is a measurable map $\Phi$ from $\Omega$ to $\Xi=\prod_{n=1}^\infty\{1,\ldots,k_n\}$ sending $\omega$ to sequence of partition elements that it lies in.
Equip $\Xi$ with the lexicographic ordering. We can then check that $\mathbb P$ induces a measure $\mu$ on $\Xi$. There are at most countably many atoms in $\Omega$ with respect to $\mathcal F$. Let the set of atoms be $A$. We can map each of these atoms disjointly to atoms of the same mass (at least) in the limit measure $\tilde {\mathbb P}$ (some justification needed here, but I'm pretty confident).
First we couple the atoms in both $\tilde{\mathbb P}$ and $\Omega$. That is we define $X$ restricted to the atoms. To finish, use quantile coupling to couple what's left. Given an $\omega$, set $t(\omega)=\mu(\{\omega'\in \Omega\setminus A\colon \Phi(\omega')\le \Phi(\omega)\})$ and finally, define for $\omega$ not in an atom, $X(\omega)=\inf\{s\colon \big(\tilde{\mathbb P}(-\infty,s])-\sum\text{(atoms of $\tilde{\mathbb P}\le s$)} \big)\ge t\}$.
|
Trail Trail, or caster, is the horizontal distance from where the steering axis intersects the ground to where the front wheel touches the ground. The measurement is considered positive if the front wheel ground contact point is behind (towards the rear of the bike) the steering axis intersection with the ground. Most bikes have positive trail, though a few, such as the Python Lowracer have negative trail.
Trail is often cited as an important determinant of bicycle handling characteristics [1], and is sometimes listed in bicycle manufacturers' geometry data, although Wilson and Papodopoulos argue that mechanical trail may be a more important and informative variable.
<math>\it{Trail} = \frac{R_w \cos(A_h) - O_f}{\sin(A_h)}</math>
where <math>R_w</math> wheel radius, <math>A_h</math> is the head angle measured clock-wise from the horizontal and <math>O_f</math> is the fork offset or rake. Trail can be increased by increasing the wheel size, decreasing or slackening the head angle, or decreasing the fork rake or offset. Trail decreases as head angle increases (becomes steeper), as fork offset increases, or as wheel diameter decreases.
Motorcyclists tend to speak of trail in relation to rake angle. The larger the rake angle the larger the trail. Note that, on a bicycle, as rake angle increases, head angle decreases.
Trail can vary as the bike leans or steers. In the case of traditional geometry, trail decreases (and wheelbase increases if measuring distance between ground contact points and not hubs) as the bike leans and steers in the direction of the lean.
[2] Trail can also vary as the suspension activates, in response to braking for example. As telescopic forks compress due to load transfer during braking, the trail and the wheelbase both decrease. [3] At least one motorcycle, the MotoCzysz C1, has a fork with adjustable trail, from 89 mm to 101 mm. [4] References[edit]
Share your opinion
|
Ballico, Edoardo ; Bernardi, Alessandra (2011)
Symmetric tensor rank with a tangent vector : a generic uniqueness theorem. [Preprint]
Full text available as:
Abstract
Let $X_{m,d}\subset \mathbb {P}^N$, $N:= \binom{m+d}{m}-1$, be the order $d$ Veronese
embedding of $\mathbb {P}^m$. Let $\tau (X_{m,d})\subset \mathbb {P}^N$, be the tangent developable of $X_{m,d}$. For
each integer $t \ge 2$ let $\tau (X_{m,d},t)\subseteq \mathbb {P}^N$, be the joint of $\tau (X_{m,d})$ and $t-2$ copies
of $X_{m,d}$. Here we prove that if $m\ge 2$, $d\ge 7$ and $t \le 1 + \lfloor \binom{m+d-2}{m}/(m+1)\rfloor$, then for a general
$P\in \tau (X_{m,d},t)$ there are uniquely determined $P_1,\dots ,P_{t-2}\in X_{m,d}$ and a unique tangent vector $\nu$ of $X_{m,d}$ such
that $P$ is in the linear span of $\nu \cup \{P_1,\dots ,P_{t-2}\}$, i.e. a degree $d$ linear form $f$ associated to $P$ may be written as
$$f = L_{t-1}^{d-1}L_t + \sum _{i=1}^{t-2} L_i^d$$
with $L_i$, $1 \le i \le t$, uniquely determined (up to a constant) linear forms on $\mathbb {P}^m$.
Abstract
Let $X_{m,d}\subset \mathbb {P}^N$, $N:= \binom{m+d}{m}-1$, be the order $d$ Veronese
embedding of $\mathbb {P}^m$. Let $\tau (X_{m,d})\subset \mathbb {P}^N$, be the tangent developable of $X_{m,d}$. For
each integer $t \ge 2$ let $\tau (X_{m,d},t)\subseteq \mathbb {P}^N$, be the joint of $\tau (X_{m,d})$ and $t-2$ copies
of $X_{m,d}$. Here we prove that if $m\ge 2$, $d\ge 7$ and $t \le 1 + \lfloor \binom{m+d-2}{m}/(m+1)\rfloor$, then for a general
$P\in \tau (X_{m,d},t)$ there are uniquely determined $P_1,\dots ,P_{t-2}\in X_{m,d}$ and a unique tangent vector $\nu$ of $X_{m,d}$ such
that $P$ is in the linear span of $\nu \cup \{P_1,\dots ,P_{t-2}\}$, i.e. a degree $d$ linear form $f$ associated to $P$ may be written as
$$f = L_{t-1}^{d-1}L_t + \sum _{i=1}^{t-2} L_i^d$$
with $L_i$, $1 \le i \le t$, uniquely determined (up to a constant) linear forms on $\mathbb {P}^m$.
This work may be freely consulted and used, may be reproduced on a permanent basis in a digital format (i.e. saving) and can be printed on paper with own personal equipment (without availing of third -parties services), for strictly and exclusively personal, research or teaching purposes, with express exclusion of any direct or indirect commercial use, unless otherwise expressly agreed between the user and the author or the right holder. It is also allowed, for the same purposes mentioned above, the retransmission via telecommunication network, the distribution or sending in any form of the work, including the personal redirection (e-mail), provided it is always clearly indicated the complete link to the page of the Alma DL Site in which the work is displayed. All other rights are reserved.
Downloads Downloads
Staff only:
|
Let $\omega$ be a third root of unity, then $\mathbb{Z}[\omega]$ is a PID.
We have $m^3 = n^2 + n + 1 = (n-\omega)(n-\omega^2)$.
$\gcd(n-\omega,n-\omega^2) = \gcd(n-\omega,\omega-\omega^2) \mid (1-\omega)$, and $(1-\omega)$ is the ramified prime lying over $3$ in $\mathbb{Z}[\omega]$, so from unique factorization of $m^3$ we get that either $(n-\omega)$ and $(n-\omega^2)$ are both roots of unity times cubes, or one is a root of unity times $(1-\omega)$ times a cube and the other is a root of unity times $3$ times a cube. In the second case, $m$ is a multiple of $3$, but then $n^2 + n + 1 \equiv 0 \mod 9$, which is impossible.
If $(n-\omega)$ and $(n-\omega^2)$ are cubes, say $a^3$ and $\bar{a}^3$, then their difference $\omega^2-\omega$ is $a^3-\bar{a}^3 = (a-\bar{a})(a^2+a\bar{a}+\bar{a}^2)$. Thus $a-\bar{a}$ is either a root of unity or a root of unity times $(1-\omega)$, and it must be the latter since $a-\bar{a}$ is pure imaginary. Thus $\Im a \le \Im (\omega-\omega^2) = \sqrt{3}$. The same argument applied to $\omega a$ shows that $\Im \omega a \le \sqrt{3}$, and similarly for other roots of unity times $a$, so $a$ is in a hexagon around the origin that is contained in a circle of radius $2$ around the origin, i.e. $|a| \le 2$, so $m = |a|^2 \le 4$. which doesn't give us any solutions.
Finally we have the case that one of $(n-\omega), (n-\omega^2)$ is of the form $\omega a^3$. Then we have $\pm(\omega^2-\omega) = \omega a^3 - \omega^2 \bar{a}^3$. Write $a = x+y\omega$. Then $\omega a^3 - \omega^2 \bar{a}^3 = (\omega-\omega^2)(x^3+y^3-3x^2y)$, so we have $x^3+y^3-3x^2y = \pm 1$, which is a Thue equation. One solution is $x = -1, y = 2$, leading to the solution $n = 18, m = 7$.
Edit: Mathematica claims that the only solutions to $x^3+y^3-3x^2y = 1$ are $(x,y) = (-2, -3), (-1, -1), (-1, 2), (0, 1), (1, 0), (3, 1)$. Mathematica's documentation says it computes an explicit bound on the size of a solution to a Thue equation based on the Baker-Wustholz theorem in order to solve it, and in this case it seems like the bound was small enough.
|
Imagine two rolls with the same diameter and mass. The mass of one roll is concentrated to the center of the roll while the mass of the other roll is concentrated to the edge of the roll. If the two are released from the same slope and the same height continuing to a flat platform, the roll whose mass is concentrated to the edge of the roll rolls considerably farther than the other. The two rolls have the same potential energy at the beginning. The roll with a smaller moment of inertia has greater speed at the end while the other roll has greater rotational energy. Does the other roll roll farther due to the air resistance being greater for the faster roll or because the moment of inertia resists changes in rotational speed more? Or maybe because the roll with a greater moment of inertia has more time to "load" energy on the slope? If you made a formula to count the distance rolled, would time play a part in the formula, or only the starting height and the velocity at the end of the slope?
An easy way to see this, assuming that they both roll without slipping, and that no friction is present is with conservation of energy as hinted in the comment above. The no-slip condition is $v=\omega r$. The total energy of one cylinder is
$$E = \frac{1}{2}(I\omega^2+mv^2)$$
Plugging the condition for no slip and solving for v, with $E=mgh$ :
$$v^2 = \sqrt{\frac{2gh}{\frac{I}{mR^2}+1}}$$
For a hollow cylinder, $I = mR^2$, while for a filled one, it is $I=\frac{mR^2}{2}$. As we can see, the ratio of the speeds when all the potential energy is converted to kinetic will be : $$\frac{v_{full}}{v_{hollow}} = \sqrt{4/3} \approx 1.15$$ So the filled cylinder is indeed faster just by energetic considerations.
To answer your question, neglecting friction, the "time" of the interaction does not matter, and the cylinders will roll infinitely far.
I think however that even if we introduce friction, the result should be roughly the same, assuming friction is not speed-dependent (hence not air resistance), since they both have the same mass and same geometry. That is just a hunch, though, but I think in a lab the primary effect is not due to friction, rather to the different moments of intertia.
If we introduce the effect of speed-dependent friction, then the problem becomes non-trivial. I think that a given roll won't always win given different initial conditions. In other words, with air friction, either roll could be the winner in terms of distance, given the right conditions.
Okay let's call the roll with the centralised mass Roll A, and Roll B is the roll with the more radial, outwardly dispersed mass.
When released, the rolls will begin to experience an acceleration (a torque), which is however many radians per second squared.
Net: each roll will receive the same amount of radians, which will be the circumference of the roll divided by the length of the slope. For future reference, we'll call this phi. If the diameter remains constant, then the roll will always be displaced phi radians, regardless of mass.
Roll A will reach the bottom of the slope first. It has a lower moment of inertia and accelerates quicker. In other words, it takes less torque to displace roll A phi radians than it does for roll B. Just like how it take less force to push a bowl 1 meter than it does to push a car 1 meter.
Each roll is experiencing the same torque (brought on by gravity) for every second it's on the slope. Roll A is being accelerated more than roll B because it has a lower moment of inertia (it's easier to spin because it's mass is all tucked in).
So it all comes down to how long it stays on the slope. The longer it's on the slope, the more rotational energy (angular momentum) it acquires. Since roll B stays on longer, it gets more energy than roll A, which translates into a farther travel.
|
It’s always nice, intellectually, when two apparently unrelated areas collide.
I had an experience of this sort recently with an area of mathematics — one very familiar to me — and an ostensibly completely distinct area of science.
On the one hand, contact geometry — a field of pure mathematics, pure geometry.
And on the other hand, the brain and its functioning. More particularly, the visual cortex, and how it processes incoming signals from the eyes.
Now, contact geometry has lots of applications: arguably it goes back to Huygens’ work on optics. It is closely related to thermodynamics. It is the odd-dimensional sibling of symplectic geometry, which is related to classical mechanics and almost every part of physics.
But applications to neurophysiology? Now that’s new.
Well, it’s only new to me. It’s been in the scientific literature for some time. It goes back at least to a paper from 1989:
William C. Hoffman, “The Visual Cortex is a Contact Bundle”, Applied Mathematics and Computation 32:137-167 (1989). (Also available here.)
And the discussion below is largely based on this article:
What’s the connection?
Contact geometry is the study of contact structures. And a contact structure on a 3-dimensional space \(M\) consists of a plane at each point satisfying some conditions. That is, at each point in the space, we have a plane sitting there. But not just any plane at each point. The planes have to vary smoothly from point to point — having such smoothly varying planes forms a (smooth)
plane field. But moreover, the plane field, which we can call \(\xi\), is required to be non-integrable.
There are various ways to explain non-integrability. To “integrate” a 2-plane field is to find a smooth surface \(S\) in space so that, at every point of \(S\), the tangent plane to \(S\) is given by the plane of \(\xi\) there. At every point \(p\) of the 2-dimensional surface \(S\), the tangent plane is a 2-dimensional plane, which we write as \(T_p S\). If we write \(\xi_p\) for the plane of \(\xi\) at the point \(p\), then the integrability condition can be written as \(\xi_p = T_p S\).
Well that’s what integrability means (roughly) — \(\xi\) is integrable if you can always find a surface tangent to \(\xi\) in this way.
But a contact structure is just the opposite: you can
never find a surface tangent to it in this way! The planes of the plane field \(\xi\) somehow twist and turn so much that you can’t every find a surface tangent to it. You can always find a surface tangent to \(\xi\) at a single point, and you might even be able to find a surface which is tangent to \(\xi\) at some of its points, (perhaps even along a curve on \(S\)), but you’ll never be able to find a surface which is tangent to \(\xi\) at all its points.
(If you’re familiar with differential forms, then the plane field \(\xi\) can be described (locally, at least) as the kernel of a 1-form, \(\xi = \ker (\alpha)\), and then the non-integrability condition is that \(\alpha \wedge d\alpha\) is a volume form. If you’re not familiar with differential forms, don’t worry.)
Contact structures can be hard to visualise. Here is a picture of one contact structure on 3-dimensional space:
You’ll note that, if you consider going from left to right in this picture in a straight line, you can actually stay tangent to the contact planes. A curve like this is called a
Legendrian curve. Let’s call the curve/line \(C\). But the planes twist around \(C\) as you travel along \(C\). This is a characteristic property of contact structures (and in fact, with a few extra technicalities, can be made into an equivalent characterisation).
Another example of a contact structure is a
projectivised tangent bundle. Let’s say what this means. (Actually we’ll only consider one such contact structure: on the projectivised tangent bundle of a plane.)
Consider a 2-dimensional plane; let’s call it \(P\). Let’s even be concrete and call it the \(xy\)-plane, complete with coordinates. So all the points on \(P\) can be written as \((x,y)\).
Now, lay \(P\) flat on the ground, in 3-dimensional space. (More precisely, embed it into \(\mathbb{R}^3\).) We would usually denote points in 3-dimensional space by \((x,y,z)\), but I want to suggestively call the third coordinate \(\theta\), because it will denote an
angle. In any case, the points of \(P\) now lie horizontally along \(\theta = 0\); so they lie at the points \((x,y,0)\) in 3-dimensional space.
Now in 3-dimensional space, through every point of \(P\) there is a vertical line. For instance, through the point \((1,2,0)\) of \(P\) is a line, and the points on this line are all the points of the form \((1,2,\theta)\).
And now the “projective” part of the situation comes in. Pick a point on the plane \(P\): let’s say \((1,2,0)\) again. Now consider
lines on \(P\) through this point. There are many such lines; in fact, infinitely many. But we can specify a line by specifying its direction. And that direction can be specified by an angle \(\theta\). We could have various conventions to measure the angle \(\theta\), but let’s do it in the standard way: \(\theta\) is the angle (measured anticlockwise) from the positive \(x\)-direction, round to the line.
Now at each point \(p = (x,y, \theta)\) in 3-dimensional space, we’ll define a plane \(\xi_p\) as follows. The plane \(\xi_p\) contains the vertical line (i.e. in the \(\theta\) direction) through \(p\); and it also contains a horizontal line through \(p\) in the direction given by the angle \(\theta\). The result is as shown below.
Starting from \(p\) (and the plane there), if you move vertically upward you get to other points of the form \(p’ = (x,y,\theta’)\), with the same \(x,y\) coordinates but different \(\theta\) coordinates. The plane at \(p’\) still contains a vertical line, but the horizontal line has rotated from angle \(\theta\) to angle \(\theta’\). Thus, as you move upwards along a vertical curve, the planes spin around the vertical curve — just as shown in the animation.
It’s a contact structure. Indeed, you can even, if you want,
identify the point \((x,y,\theta)\) with the line through \((x,y)\) in the plane \(P\) with direction given by \(\theta\). In this way, the points in 3-dimensional space correspond to the lines in the plane through various points, and this is the thing referred to as the “projectivised tangent bundle”. (Strictly speaking though, a line at angle \(\theta\) and a line at angle \(\theta + \pi\) point in the same direction, so we should identify points \((x,y,\theta) \sim (x,y,\theta+\pi)\).) What does this have to do with the brain?
Well I’m no neurophysiologist, but the claim is that the neurons in the visual cortex can be regarded functionally as exactly this kind of contact structure. This is not to say that the neurons are planes, or spin around quite like the picture above. But it is to say that neurons in some ways,
functionally, behave like this contact structure.
When you look at an image, the photoreceptors in your eye send signals into your brain. These signals are processed, at a low level, in your visual cortex. They are then processed at a higher level, extracting features, objects and eventually reaching the level of consciousness as the unified visual field which is part of ordinary human experience. However, here we are only interested in the lower-level processing, which extracts basic information from the image projected on the retina. This low-level processing extracts features like which areas of the visual field are light and dark, the shapes of light and dark areas, and importantly for us here, the
orientation of any lines or curves that we see.
The particular area of interest in the visual cortex seems to be an area called “V1”. This area of the brain contains many structures. It contains several “horizontal” layers 1-6, each divided into sublayers; the most important is apparently the sublayer 4C. We’ll call this the “cortical layer”, as it’s the one important for our purposes.
Now it turns out that different points on this cortical layer relate to different points on the retina. Each point in your visual field corresponds gets projected to a different point on your retina, which (roughly speaking) connects to a different point in the cortical layer. The map from the retina to the cortical layer is called a
retinotopy. In fact, beautifully, this map from the retina (which is a surface at the back of your eye) to the cortical layer (Which is a surface in your brain) is a map which appears to preserve angles (but not lengths). In other words, the retinotopy is a conformal map.
Even better, the cells of the cortex are organised into structures called
columns and hypercolumns. Along each hypercolumn, the cells detect curves which point in the same orientation. So there are not only cells which are specialised to detect images arriving at particular points on your retina; there are also cells which are specialised to detect a curve at a particular in a particular orientation.
Functionally, then, the visual cortex behaves like a contact structure. The neurons aren’t arranged in a contact structure, but they behave like one. And this means that various processes in low-level visual processing can be understood in terms of contact geometry.
In particular, the “association field” can be understood in terms of contact geometry, as perhaps also can certain hallucinations — including those seen under the influence of psychedelics like LSD.
Well, it’s definitely the most psychedelic application of contact geometry I’ve seen.
Some further references:
Bressloff et al, Geometric visual hallucinations, Euclidean symmetry and the functional architecture of striate cortex, Phil. Trans. R. Soc. Lond. B (2001) 356, 299–300 (Also here.) David J Field, Anthony Hayes, Robert F Hess, Contour Integration by the Human Visual System: Evidence for a Local “Association Field”, Vision Res. Vol. 33, No. 2, 173–193, 1993. (Also here.) Alessandro Sarti, Giovanna Citti, Jean Petitot, The symplectic structure of the primary visual cortex, Biol Cybern (2008) 98:33–48
|
Forgot password? New user? Sign up
Existing user? Log in
The sum of all values of aaa such that the equation(x2−x+a+1)2=4a(5x2−x+1)(x^2-x+a+1)^2=4a(5x^2-x+1)(x2−x+a+1)2=4a(5x2−x+1)has exactly three distinct real solutions, is of the form n+km\dfrac{n+\sqrt{k}}{m}mn+k, where k,m,nk,m,nk,m,n are integers, k≥0,k\geq 0,k≥0, m≥1m\geq 1m≥1 and mmm is the smallest possible. Find k+n+m.k+n+m.k+n+m.
Problem Loading...
Note Loading...
Set Loading...
|
As a part of my research (in array processing - the specifics are not too exciting :-)), I cant resolve one specific integral.
Assume $r\in\left[0 ,1\right]$, $N\in\mathbb{N}$.
The basic form is $$\int_{\cos{(x)}=-1}^{\cos{(x)}=1}{\frac{1-\cos{\left(Nx\right)}}{N^{2}\left(1-\cos{(x)}\right)+r^{2}\left(1-\cos{\left(Nx\right)}\right)-Nr\left(1-\cos{(x)}-\cos{\left(Nx\right)}+\cos{\left(\left(N-1\right)x\right)}\right)}dx}.$$Searching for hints, I came across the very informative "How to expand $\cos{(nx)}$ with $\cos{(x)}$?" discussion which lead me to investigate the expression using
Chebyshev polynomials of the first kind, in light of the fact that $\cos(nu)=T_n(\cos(u)),$ but still no luck.
Anyhow, as can be seen in the wikipedia page (Chebyshev polynomials ),$$T_n(x)= \sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} (x^2-1)^k x^{n-2k},$$and plugging it into the basic integral results in $$\int_{-1}^{1}{\frac{1-T_N(x)}{N^{2}\left(1-x\right)+r^{2}\left(1-T_N(x)\right)-Nr\left(1-x-T_N(x)+T_{N-1}(x)\right)}\frac{-1}{\sqrt{1-x^{2}}}dx}.$$Considering also the
Chebyshev polynomials of the second kind,$$U_{N}\left(x\right)=\frac{\left(x+\sqrt{x^{2}-1}\right)^{N+1}-\left(x-\sqrt{x^{2}-1}\right)^{N+1}}{2\sqrt{x^{2}-1}},$$and the equality $$T_{N+1}(x) = xT_{N}(x)-\left(1-x^{2}\right)U_{N-1}(x)$$results in$$\int_{-1}^{1}{\frac{1-T_N(x)}{N^{2}\left(1-x\right)+r^{2}\left(1-T_N(x)\right)-Nr\left(1-x+\sqrt{1-x^{2}}\left(\left(x+\sqrt{x^{2}-1}\right)^{N}-\left(x-\sqrt{x^{2}-1}\right)^{N}\right)\right)}\frac{-1}{\sqrt{1-x^{2}}}dx},$$ which has (as the headline states) "almost" quadratic form of the denominator.This is where I am stuck and if someone has an idea for solving it I will be thrilled.
Thank you.
|
J. D. Hamkins and R. Yang, “Satisfaction is not absolute,” to appear in the Review of Symbolic Logic, pp. 1-34, 2014.
@ARTICLE{HamkinsYang:SatisfactionIsNotAbsolute, author = {Joel David Hamkins and Ruizhi Yang}, title = {Satisfaction is not absolute}, journal = {to appear in the Review of Symbolic Logic}, year = {2014}, volume = {}, number = {}, pages = {1--34}, month = {}, note = {}, abstract = {}, keywords = {to-appear}, source = {}, eprint = {1312.0670}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://wp.me/p5M0LV-Gf}, doi = {}, }
$\newcommand\N{\mathbb{N}}\newcommand\satisfies{\models}$
Abstract. We prove that the satisfaction relation $\mathcal{N}\satisfies\varphi[\vec a]$ of first-order logic is not absolute between models of set theory having the structure $\mathcal{N}$ and the formulas $\varphi$ all in common. Two models of set theory can have the same natural numbers, for example, and the same standard model of arithmetic $\langle\N,{+},{\cdot},0,1,{\lt}\rangle$, yet disagree on their theories of arithmetic truth; two models of set theory can have the same natural numbers and the same arithmetic truths, yet disagree on their truths-about-truth, at any desired level of the iterated truth-predicate hierarchy; two models of set theory can have the same natural numbers and the same reals, yet disagree on projective truth; two models of set theory can have the same $\langle H_{\omega_2},{\in}\rangle$ or the same rank-initial segment $\langle V_\delta,{\in}\rangle$, yet disagree on which assertions are true in these structures. On the basis of these mathematical results, we argue that a philosophical commitment to the determinateness of the theory of truth for a structure cannot be seen as a consequence solely of the determinateness of the structure in which that truth resides. The determinate nature of arithmetic truth, for example, is not a consequence of the determinate nature of the arithmetic structure $\N=\{ 0,1,2,\ldots\}$ itself, but rather, we argue, is an additional higher-order commitment requiring its own analysis and justification.
Many mathematicians and philosophers regard the natural numbers $0,1,2,\ldots\,$, along with their usual arithmetic structure, as having a privileged mathematical existence, a Platonic realm in which assertions have definite, absolute truth values, independently of our ability to prove or discover them. Although there are some arithmetic assertions that we can neither prove nor refute—such as the consistency of the background theory in which we undertake our proofs—the view is that nevertheless there is a fact of the matter about whether any such arithmetic statement is true or false in the intended interpretation. The definite nature of arithmetic truth is often seen as a consequence of the definiteness of the structure of arithmetic $\langle\N,{+},{\cdot},0,1,{\lt}\rangle$ itself, for if the natural numbers exist in a clear and distinct totality in a way that is unambiguous and absolute, then (on this view) the first-order theory of truth residing in that structure—arithmetic truth—is similarly clear and distinct.
Feferman provides an instance of this perspective when he writes (Feferman 2013, Comments for EFI Workshop, p. 6-7) :
In my view, the conception [of the bare structure of the natural numbers] iscompletely clear , and thenceall arithmetical statements are definite .
It is Feferman’s `thence’ to which we call attention. Martin makes a similar point (Martin, 2012, Completeness or incompleteness of basic mathematical concepts):
What I am suggesting is that the real reason for confidence in first-order completeness is our confidence in the full determinateness of the concept of the natural numbers.
Many mathematicians and philosophers seem to share this perspective. The truth of an arithmetic statement, to be sure, does seem to depend entirely on the structure $\langle\N,{+},{\cdot},0,1,{\lt}\rangle$, with all quantifiers restricted to $\N$ and using only those arithmetic operations and relations, and so if that structure has a definite nature, then it would seem that the truth of the statement should be similarly definite.
Nevertheless, in this article we should like to tease apart these two ontological commitments, arguing that the definiteness of truth for a given mathematical structure, such as the natural numbers, the reals or higher-order structures such as $H_{\omega_2}$ or $V_\delta$, does not follow from the definite nature of the underlying structure in which that truth resides. Rather, we argue that the commitment to a theory of truth for a structure is a higher-order ontological commitment, going strictly beyond the commitment to a definite nature for the underlying structure itself.
We make our argument in part by proving that different models of set theory can have a structure identically in common, even the natural numbers, yet disagree on the theory of truth for that structure.
Theorem. Two models of set theory can have the same structure of arithmetic $$\langle\N,{+},{\cdot},0,1,{\lt}\rangle^{M_1}=\langle\N,{+},{\cdot},0,1,{\lt}\rangle^{M_2},$$yet disagree on the theory of arithmetic truth. Two models of set theory can have the same natural numbers and a computable linear order in common, yet disagree about whether it is a well-order. Two models of set theory that have the same natural numbers and the same reals, yet disagree on projective truth. Two models of set theory can have a transitive rank initial segment in common $$\langle V_\delta,{\in}\rangle^{M_1}=\langle V_\delta,{\in}\rangle^{M_2},$$yet disagree about whether it is a model of ZFC.
The proofs use only elementary classical methods, and might be considered to be a part of the folklore of the subject of models of arithmetic. The paper includes many further examples of the phenomenon, and concludes with a philosophical discussion of the issue of definiteness, concerning the question of whether one may deduce definiteness-of-truth from definiteness-of-objects and definiteness-of-structure.
|
@DavidReed the notion of a "general polynomial" is a bit strange. The general polynomial over a field always has Galois group $S_n$ even if there is not polynomial over the field with Galois group $S_n$
Hey guys. Quick question. What would you call it when the period/amplitude of a cosine/sine function is given by another function? E.g. y=x^2*sin(e^x). I refer to them as variable amplitude and period but upon google search I don't see the correct sort of equation when I enter "variable period cosine"
@LucasHenrique I hate them, i tend to find algebraic proofs are more elegant than ones from analysis. They are tedious. Analysis is the art of showing you can make things as small as you please. The last two characters of every proof are $< \epsilon$
I enjoyed developing the lebesgue integral though. I thought that was cool
But since every singleton except 0 is open, and the union of open sets is open, it follows all intervals of the form $(a,b)$, $(0,c)$, $(d,0)$ are also open. thus we can use these 3 class of intervals as a base which then intersect to give the nonzero singletons?
uh wait a sec...
... I need arbitrary intersection to produce singletons from open intervals...
hmm... 0 does not even have a nbhd, since any set containing 0 is closed
I have no idea how to deal with points having empty nbhd
o wait a sec...
the open set of any topology must contain the whole set itself
so I guess the nbhd of 0 is $\Bbb{R}$
Btw, looking at this picture, I think the alternate name for these class of topologies called British rail topology is quite fitting (with the help of this WfSE to interpret of course mathematica.stackexchange.com/questions/3410/…)
Since as Leaky have noticed, every point is closest to 0 other than itself, therefore to get from A to B, go to 0. The null line is then like a railway line which connects all the points together in the shortest time
So going from a to b directly is no more efficient than go from a to 0 and then 0 to b
hmm...
$d(A \to B \to C) = d(A,B)+d(B,C) = |a|+|b|+|b|+|c|$
$d(A \to 0 \to C) = d(A,0)+d(0,C)=|a|+|c|$
so the distance of travel depends on where the starting point is. If the starting point is 0, then distance only increases linearly for every unit increase in the value of the destination
But if the starting point is nonzero, then the distance increases quadratically
Combining with the animation in the WfSE, it means that in such a space, if one attempt to travel directly to the destination, then say the travelling speed is 3 ms-1, then for every meter forward, the actual distance covered by 3 ms-1 decreases (as illustrated by the shrinking open ball of fixed radius)
only when travelling via the origin, will such qudratic penalty in travelling distance be not apply
More interesting things can be said about slight generalisations of this metric:
Hi, looking a graph isomorphism problem from perspective of eigenspaces of adjacency matrix, it gets geometrical interpretation: question if two sets of points differ only by rotation - e.g. 16 points in 6D, forming a very regular polyhedron ...
To test if two sets of points differ by rotation, I thought to describe them as intersection of ellipsoids, e.g. {x: x^T P x = 1} for P = P_0 + a P_1 ... then generalization of characteristic polynomial would allow to test if our sets differ by rotation ...
1D interpolation: finding a polynomial satisfying $\forall_i\ p(x_i)=y_i$ can be written as a system of linear equations, having well known Vandermonde determinant: $\det=\prod_{i<j} (x_i-x_j)$. Hence, the interpolation problem is well defined as long as the system of equations is determined ($\d...
Any alg geom guys on? I know zilch about alg geom to even start analysing this question
Manwhile I am going to analyse the SR metric later using open balls after the chat proceed a bit
To add to gj255's comment: The Minkowski metric is not a metric in the sense of metric spaces but in the sense of a metric of Semi-Riemannian manifolds. In particular, it can't induce a topology. Instead, the topology on Minkowski space as a manifold must be defined before one introduces the Minkowski metric on said space. — baluApr 13 at 18:24
grr, thought I can get some more intuition in SR by using open balls
tbf there’s actually a third equivalent statement which the author does make an argument about, but they say nothing about substantive about the first two.
The first two statements go like this : Let $a,b,c\in [0,\pi].$ Then the matrix $\begin{pmatrix} 1&\cos a&\cos b \\ \cos a & 1 & \cos c \\ \cos b & \cos c & 1\end{pmatrix}$ is positive semidefinite iff there are three unit vectors with pairwise angles $a,b,c$.
And all it has in the proof is the assertion that the above is clearly true.
I've a mesh specified as an half edge data structure, more specifically I've augmented the data structure in such a way that each vertex also stores a vector tangent to the surface. Essentially this set of vectors for each vertex approximates a vector field, I was wondering if there's some well k...
Consider $a,b$ both irrational and the interval $[a,b]$
Assuming axiom of choice and CH, I can define a $\aleph_1$ enumeration of the irrationals by label them with ordinals from 0 all the way to $\omega_1$
It would seemed we could have a cover $\bigcup_{\alpha < \omega_1} (r_{\alpha},r_{\alpha+1})$. However the rationals are countable, thus we cannot have uncountably many disjoint open intervals, which means this union is not disjoint
This means, we can only have countably many disjoint open intervals such that some irrationals were not in the union, but uncountably many of them will
If I consider an open cover of the rationals in [0,1], the sum of whose length is less than $\epsilon$, and then I now consider [0,1] with every set in that cover excluded, I now have a set with no rationals, and no intervals.One way for an irrational number $\alpha$ to be in this new set is b...
Suppose you take an open interval I of length 1, divide it into countable sub-intervals (I/2, I/4, etc.), and cover each rational with one of the sub-intervals.Since all the rationals are covered, then it seems that sub-intervals (if they don't overlap) are separated by at most a single irrat...
(For ease of construction of enumerations, WLOG, the interval [-1,1] will be used in the proofs) Let $\lambda^*$ be the Lebesgue outer measure We previously proved that $\lambda^*(\{x\})=0$ where $x \in [-1,1]$ by covering it with the open cover $(-a,a)$ for some $a \in [0,1]$ and then noting there are nested open intervals with infimum tends to zero.
We also knew that by using the union $[a,b] = \{a\} \cup (a,b) \cup \{b\}$ for some $a,b \in [-1,1]$ and countable subadditivity, we can prove $\lambda^*([a,b]) = b-a$. Alternately, by using the theorem that $[a,b]$ is compact, we can construct a finite cover consists of overlapping open intervals, then subtract away the overlapping open intervals to avoid double counting, or we can take the interval $(a,b)$ where $a<-1<1<b$ as an open cover and then consider the infimum of this interval such that $[-1,1]$ is still covered. Regardless of which route you take, the result is a finite sum whi…
W also knew that one way to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ is to take the union of all singletons that are rationals. Since there are only countably many of them, by countable subadditivity this give us $\lambda^*(\Bbb{Q}\cap [-1,1]) = 0$. We also knew that one way to compute $\lambda^*(\Bbb{I}\cap [-1,1])$ is to use $\lambda^*(\Bbb{Q}\cap [-1,1])+\lambda^*(\Bbb{I}\cap [-1,1]) = \lambda^*([-1,1])$ and thus deducing $\lambda^*(\Bbb{I}\cap [-1,1]) = 2$
However, what I am interested here is to compute $\lambda^*(\Bbb{Q}\cap [-1,1])$ and $\lambda^*(\Bbb{I}\cap [-1,1])$ directly using open covers of these two sets. This then becomes the focus of the investigation to be written out below:
We first attempt to construct an open cover $C$ for $\Bbb{I}\cap [-1,1]$ in stages:
First denote an enumeration of the rationals as follows:
$\frac{1}{2},-\frac{1}{2},\frac{1}{3},-\frac{1}{3},\frac{2}{3},-\frac{2}{3}, \frac{1}{4},-\frac{1}{4},\frac{3}{4},-\frac{3}{4},\frac{1}{5},-\frac{1}{5}, \frac{2}{5},-\frac{2}{5},\frac{3}{5},-\frac{3}{5},\frac{4}{5},-\frac{4}{5},...$ or in short:
Actually wait, since as the sequence grows, any rationals of the form $\frac{p}{q}$ where $|p-q| > 1$ will be somewhere in between two consecutive terms of the sequence $\{\frac{n+1}{n+2}-\frac{n}{n+1}\}$ and the latter does tends to zero as $n \to \aleph_0$, it follows all intervals will have an infimum of zero
However, any intervals must contain uncountably many irrationals, so (somehow) the infimum of the union of them all are nonzero. Need to figure out how this works...
Let's say that for $N$ clients, Lotta will take $d_N$ days to retire.
For $N+1$ clients, clearly Lotta will have to make sure all the first $N$ clients don't feel mistreated. Therefore, she'll take the $d_N$ days to make sure they are not mistreated. Then she visits client $N+1$. Obviously the client won't feel mistreated anymore. But all the first $N$ clients are mistreated and, therefore, she'll start her algorithm once again and take (by suposition) $d_N$ days to make sure all of them are not mistreated. And therefore we have the recurence $d_{N+1} = 2d_N + 1$
Where $d_1$ = 1.
Yet we have $1 \to 2 \to 1$, that has $3 = d_2 \neq 2^2$ steps.
|
Activity is a measure of the
effective concentration of a species under non-ideal (e.g., concentrated) conditions. This determines the real chemical potential for a real solution rather than an ideal one. Introduction
Activities and concentrations can both be used to calculate equilibrium constants and reaction rates. However, most of the time we use concentration even though activity is also a measure of composition, similar to concentration. It is satisfactory to use concentration for diluted solutions, but when you are dealing with more concentrated solutions, the difference in the observed concentration and the calculated concentration in equilibrium increases. This is the reason that the activity was initially created.
\[ \large a=e^{\frac {\mu -\mu_o}{RT}} \tag{1}\]
where
\(a\)=Activity \(\mu\) is chemical potential (dependent on standard state) which is Gibbs Energy per mole \(\mu_0\) is the standard chemical potential \(R\) is the gas constant \(T\) is the absolute Temperature Non-ideality in Gases (Fugacity)
Fugacity is the effective pressure for a non-ideal gas. The pressures of an ideal gas and a real gas are equivalent when the chemical potential is the same. The equation that relates the non-ideal to the ideal gas pressure is:
\[ f = \phi P \tag{2}\]
with
\(f\) represents fugacity, \(P\) is the pressure for an ideal gas, and \(\phi\) is the fugacity coefficient.
For an ideal gas, the fugacity coefficient is 1.
Non-ideality in Solutions pH
We have become accustomed to using the equation \( pH= -\log[H^+] \), but this equation not accurate at all concentrations. A better expression for pH is \( pH= -\log[a_H^+] \) which accounts for the
activity, a. The only reason that other indicators may correctly seem to measure the acidity, which was equivalent to \(–\log[H^+]\), is because of the use of Beer’s law, which uses concentration rather than activity. 1 References Christopher G. McCarty and Ed Vitz, Journal of Chemical Education, 83(5), 752 (2006) Atkins, Peter; John W. Locke (2002-01). Atkins' Physical Chemistry(7th ed.). Oxford University Press PAC, 1996, 68, 957 Problems What is the purpose of using activity rather than concentration? If the ratio of fugacity to the pressure of the ideal gas is 1 then what is the activity coefficient? If pH does not= -log[H+] then why do the calculations seem to show the correct acidity? If concentration is not accurate does that mean we should start only using activity instead? Solutions Activity is more accurate in more concentrated solutions The coefficient is one because the gas is in a similar state as an ideal gas. The use of Beer’s law which uses also concentration rather than activity results in the seemingly correct results. Activity is only required highly concentrated solutions.
|
I'm going through the phase estimation algorithm, and wanted to sanity-check my calculations by making sure the state I'd calculated was still normalized. It is, assuming the square of the absolute value of the eigenvalue of the arbitrary unitary operator I'm analyzing equals 1. So, does it? Assuming that the eigenvector of the eigenvalue is normalized.
Good question. The answer turns out to be
. Yes
You don't even need the vector to be normalized. Watch:
Start with the definition of eigenvalues and eigenvectors:
$$ \begin{align} U|\psi\rangle &= \lambda |\psi\rangle\\ \end{align} $$
Conjugate and transpose both sides of the equation:
$$ \begin{align} \langle\psi|U^\dagger &= \langle \psi| \lambda^*. \end{align} $$
Left multiply each side of line 1 by the corresponding side of line 2.
$$ \begin{align} \langle \psi|U^\dagger\cdot U|\psi \rangle &= \langle \psi | \lambda^* \lambda |\psi\rangle \\ \langle \psi |\psi \rangle &= |\lambda |^2 \langle \psi |\psi \rangle \\ c &= |\lambda|^2 c\\ 1 &= |\lambda|^2 \end{align} $$
If $|\psi\rangle $ is normalized, it just means that $c=1$, which makes no difference in this proof because the $c$ was on both sides of the equation and can be divided out.
@user1271772's answer is excellent, and absolutely the right answer. I just wanted to add in some additional perspective, given recent questions regarding Hamiltonians. Many physicists start from the Hamiltonian being the underlying thing that determines evolution, and unitaries are derived as a consequence. They start from the Schrödinger equation, $$ i\frac{d|\psi\rangle}{dt}=H|\psi\rangle. $$ For a time-invariant Hamiltonian, the solution is $$ |\psi(t)\rangle=e^{-iHt}|\psi(0)\rangle, $$ where $e^{-iHt}$ is unitary because $e^{-iHt}e^{iHt}=\mathbb{I}$. Just stating this solution actually skips over the thing I really want to focus on.
We could expand a generic $|\psi\rangle=\sum_ia_i|i\rangle$, so the Schrödinger equation becomes a series of simultaneous differential equations for the $a_i$: $$ i\frac{da_i}{dt}=\sum_jH_{ij}a_j. $$ Now consider what happens to an eigenvector $|\lambda\rangle=\sum_ib_i|i\rangle$ of $H$: $$ \sum_{j}H_{ij}b_i^\star=\lambda b_i^\star. $$ We can take linear combinations of the $a_i$: $$ i\frac{d\sum_ib_i^\star a_i}{dt}=\sum_{ij}b_i^\star H_{ij}a_j=\lambda\sum_jb_j^\star a_j. $$ Hence, we see that the component $x=\sum_jb_j^\star a_j$ simply satisfies $$ i\frac{dx}{dt}=\lambda x, $$ so $x(t)=e^{-i\lambda t}x(0)$. In other words, a state initially created as an eigenvector $|\lambda\rangle$ stays in that state and just acquires a phase over time $e^{-i\lambda t}|\lambda\rangle$. Hence, eigenvectors of $H$ are also eigenvectors of the unitary $U$, with eigenvalues $e^{-i\lambda t}$, and these have modulus 1.
|
Difference between revisions of "State Feedback"
(→Textbook Contents)
(27 intermediate revisions by the same user not shown) Line 1: Line 1:
{{chheader|Linear Systems|State Feedback|Output Feedback}}
{{chheader|Linear Systems|State Feedback|Output Feedback}}
−
This chapter describes how feedback can be used shape the local behavior of a system. The concept of reachability is introduced and used to investigate how to "design" the dynamics of a system through placement of its eigenvalues. In particular, it will be shown that under certain conditions it is possible to assign the system eigenvalues to arbitrary values by appropriate feedback of the system state.
+
This chapter describes how feedback can be used shape the local behavior of a system. The concept of reachability is introduced and used to investigate how to "design" the dynamics of a system through placement of its eigenvalues. In particular, it will be shown that under certain conditions it is possible to assign the system eigenvalues to arbitrary values by appropriate feedback of the system state.
{{chaptertable begin}}
{{chaptertable begin}}
{{chaptertable left}}
{{chaptertable left}}
== Textbook Contents ==
== Textbook Contents ==
−
{{am05pdf|am08-statefbk|
+
{{am05pdf|am08-statefbk||State Feedback}}
* 1. Reachability
* 1. Reachability
* 2. Stabilization by State Feedback
* 2. Stabilization by State Feedback
+
* 3. State Feedback Design Issues
* 3. State Feedback Design Issues
* 4. Integral Action
* 4. Integral Action
Line 20: Line 21:
== Supplemental Information ==
== Supplemental Information ==
* [[#Frequently Asked Questions|Frequently Asked Questions]]
* [[#Frequently Asked Questions|Frequently Asked Questions]]
+
* Wikipedia entries: [http://en.wikipedia.org/wiki/Controllability Controllability (reachability)], [http://en.wikipedia.org/wiki/State_space_%28controls%29#Feedback state feedback], [http://en.wikipedia.org/wiki/Optimal_control#Linear_quadratic_control LQR]
* Wikipedia entries: [http://en.wikipedia.org/wiki/Controllability Controllability (reachability)], [http://en.wikipedia.org/wiki/State_space_%28controls%29#Feedback state feedback], [http://en.wikipedia.org/wiki/Optimal_control#Linear_quadratic_control LQR]
* [[#Additional Information|Additional Information]]
* [[#Additional Information|Additional Information]]
Line 34: Line 36:
\end{aligned}
\end{aligned}
</amsmath></center>
</amsmath></center>
−
is said to be ''reachable'' if we can find an input <
+
is said to be ''reachable'' if we can find an input <>u(t)</> defined on the interval <>[0, T]</> that can steer the system from a given final point <>x(0) = x_0</> to a desired final point <>x(T) = x_f</>.
−
</p>
+
</p>
<li><p>The ''reachability matrix'' for a linear system is given by
<li><p>The ''reachability matrix'' for a linear system is given by
Line 41: Line 43:
W_r = \left[\begin{matrix} B & AB & \cdots & A^{n-1}B \end{matrix}\right].
W_r = \left[\begin{matrix} B & AB & \cdots & A^{n-1}B \end{matrix}\right].
</amsmath></center>
</amsmath></center>
−
A linear system is reachable if and only if the reachability matrix <
+
A linear system is reachable if and only if the reachability matrix <>W_r</> is invertible (assuming a single intput/single output system). Systems that are not reachable have states that are constrained to have a fixed relationship with each other.
−
</p>
+
</p>
<li><p>A linear system of the form
<li><p>A linear system of the form
Line 61: Line 63:
\det(sI-A) = s^n+a_1 s^{n-1} + \cdots + a_{n-1}s + a_n,
\det(sI-A) = s^n+a_1 s^{n-1} + \cdots + a_{n-1}s + a_n,
</amsmath></center>
</amsmath></center>
−
A reachable linear system can be transformed into reachable canonical form through the use of a coordinate transformation <
+
A reachable linear system can be transformed into reachable canonical form through the use of a coordinate transformation <>z = T x</>.
−
</p>
+
</p>
<li><p>A state feedback law has the form
<li><p>A state feedback law has the form
Line 68: Line 70:
u = -K x + k_r r
u = -K x + k_r r
</amsmath></center>
</amsmath></center>
−
where <
+
where <>r</> is the reference value for the output. The closed loop dynamics for the system are given by
<center><amsmath>
<center><amsmath>
\dot x = (A - B K) x + B k_r r.
\dot x = (A - B K) x + B k_r r.
</amsmath></center>
</amsmath></center>
−
The stability of the system is determined by the stability of the matrix <
+
The stability of the system is determined by the stability of the matrix <>A - BK</>. The equilibrium point and steady state output (assuming the systems is stable) are given by
<center><amsmath>
<center><amsmath>
x_e = -(A-BK)^{-1} B k_r r \qquad y_e = C x_e.
x_e = -(A-BK)^{-1} B k_r r \qquad y_e = C x_e.
</amsmath></center>
</amsmath></center>
−
Choosing <
+
Choosing <>k_r</> as
<center><amsmath>
<center><amsmath>
k_r = {-1}/\left(C (A-BK)^{-1} B\right).
k_r = {-1}/\left(C (A-BK)^{-1} B\right).
</amsmath></center>
</amsmath></center>
−
gives <
+
gives <>y_e = r</>.</p>
<li><p>If a system is reachable, then there exists a feedback law of the form
<li><p>If a system is reachable, then there exists a feedback law of the form
Line 87: Line 89:
</amsmath></center>
</amsmath></center>
the gives a closed loop system with an arbitrary characteristic polynomial. Hence the eigenvalues of a reachable linear system can be placed arbitrarily through the use of an appropriate feedback control law.
the gives a closed loop system with an arbitrary characteristic polynomial. Hence the eigenvalues of a reachable linear system can be placed arbitrarily through the use of an appropriate feedback control law.
−
</p>
+
</p>
−
<li><p>''Integral feedback'' can be used to provide zero steady state error instead of careful calibration of the gain <
+
<li><p>''Integral feedback'' can be used to provide zero steady state error instead of careful calibration of the gain <>K_r</>. An integral feedback controller has the form
<center><amsmath>
<center><amsmath>
u = - k_p (x - x_e) - k_i z + k_r r.
u = - k_p (x - x_e) - k_i z + k_r r.
</amsmath></center>
</amsmath></center>
where
where
−
<center><
+
<center><>
\dot z = y - r
\dot z = y - r
−
</
+
</></center>
−
is the integral error. The gains <
+
is the integral error. The gains <>k_p</>, <>k_i</> and <>k_r</> can be found by designing a stabilizing state feedback for the system dynamics augmented by the integrator dynamics.
−
</p>
+
</p>
<li><p>A ''linear quadratic regulator'' minimizes the cost function
<li><p>A ''linear quadratic regulator'' minimizes the cost function
Line 109: Line 111:
u = -Q_u^{-1} B^T P x.
u = -Q_u^{-1} B^T P x.
</amsmath></center>
</amsmath></center>
−
where <
+
where <>P \in R^{n \times n}</> is a positive definite, symmetric
matrix that satisfies the equation
matrix that satisfies the equation
<center><amsmath>
<center><amsmath>
P A + A^T P - P B Q_u^{-1} B^T P + Q_x = 0.
P A + A^T P - P B Q_u^{-1} B^T P + Q_x = 0.
</amsmath></center>
</amsmath></center>
−
This equation is called the''algebraic
+
This equation is called the ''algebraic
Riccati equation'' and can be solved numerically.
Riccati equation'' and can be solved numerically.
−
</p>
+
</p>
</ol>
</ol>
−
== Exercises ==
+ + + +
== Exercises ==
+
<ncl>State Feedback Exercises</ncl>
<ncl>State Feedback Exercises</ncl>
== Frequently Asked Questions ==
== Frequently Asked Questions ==
<ncl>State Feedback FAQ</ncl>
<ncl>State Feedback FAQ</ncl>
+ + + + + + + + + + + + + + + + + + +
== Additional Information ==
== Additional Information ==
+ + + Latest revision as of 04:06, 19 November 2012
Prev: Linear Systems Chapter 6 - State Feedback Next: Output Feedback
This chapter describes how feedback can be used to shape the local behavior of a system. The concept of reachability is introduced and used to investigate how to "design" the dynamics of a system through placement of its eigenvalues. In particular, it will be shown that under certain conditions it is possible to assign the system eigenvalues to arbitrary values by appropriate feedback of the system state.
Textbook Contents Lecture Materials Supplemental Information Chapter Summary
This chapter describes how state feedback can be used to design the (closed loop) dynamics of the system:
A linear system with dynamics
The
reachability matrixfor a linear system is given by
A linear system is reachable if and only if the reachability matrix is invertible (assuming a single intput/single output system). Systems that are not reachable have states that are constrained to have a fixed relationship with each other.
A linear system of the form
is said to be in
reachable canonical form. A system in this form is always reachable and has a characteristic polynomial given by
A state feedback law has the form
gives .
If a system is reachable, then there exists a feedback law of the form
the gives a closed loop system with an arbitrary characteristic polynomial. Hence the eigenvalues of a reachable linear system can be placed arbitrarily through the use of an appropriate feedback control law.
where
A
linear quadratic regulatorminimizes the cost function
The solution to the LQR problem is given by a linear control law of the form
This equation is called the
algebraicRiccati equation and can be solved numerically.
Additional Exercises
The following exercises cover some of the topics introduced in this chapter. Exercises marked with a * appear in the printed text.
Frequently Asked Questions
Errata
MATLAB code
The following MATLAB scripts are available for producing figures that appear in this chapter.
See the software page for more information on how to run these scripts.
Additional Information
More information on optimal control and the linear quadratic regulator can be found in the Optimization-Based Control supplement:
|
The omega one of chess
Infinite chess is chess played on an infinite edgeless chessboard. Since checkmates, when they occur, do so after finitely many moves, this is an open game and therefore subject to the theory of transfinite game values for open games.
Specifically, the
game value (for white) of a position in infinite chess is defined by recursion. The positions with value $0$ are precisely those in which white has already won. If a position $p$ has white to move, then the value of $p$ is $\alpha+1$ if and only if $\alpha$ is minimal such that white may legally move from $p$ to a position with value $\alpha$. If a position $p$ has black to play, where black has a legal move from $p$, and every move by black from $p$ has a value, then the value of $p$ is the supremum of these values.
The term
omega one of chess refers either to the ordinal $\omega_1^{\mathfrak{Ch}}$ or to $\omega_1^{\mathfrak{Ch}_{\!\!\!\!\sim}}$, depending respectively on whether one is considering only finite positions or also positions with infinitely many pieces.[1] The ordinal $\omega_1^{\mathfrak{Ch}}$ is the supremum of the game values of the winning finite positions for white in infinite chess. The ordinal $\omega_1^{\mathfrak{Ch}_{\!\!\!\!\sim}}$ is the supremum of the game values of all the winning positions for white in infinite chess, including positions with infinitely many pieces. Similarly, $\omega_1^{\mathfrak{Ch}_3}$ and $\omega_1^{{\mathfrak{Ch}_{\!\!\!\!\sim}}_3}$ are the analogous ordinals for infinite three-dimensional chess, as described in .
There is an entire natural hierarchy of intermediate concepts between $\omega_1^{\mathfrak{Ch}}$ and $\omega_1^{\mathfrak{Ch}_{\!\!\!\!\sim}}$, corresponding to the various descriptive-set-theoretic complexities of the positions. For example, we may denote by $\omega_1^{\mathfrak{Ch},c}$ the 'computable' omega one of chess, which is the supremum of the game values of the computable positions of infinite chess. Similarly, one may define $\omega_1^{\mathfrak{Ch},\text{hyp}}$ to be the supremum of the values of the hyperarithmetic positions and $\omega_1^{\mathfrak{Ch},\Delta^1_2}$ to be the supremum of the $\Delta^1_2$ definable positions, and so on.
Since there are only countably many finite positions, whose game values form an initial segment of the ordinals, it follows that $\omega_1^{\mathfrak{Ch}}$ is a countable ordinal. The next theorem shows more, that $\omega_1^{\mathfrak{Ch}}$ is bounded by the Church-Kleene ordinal $\omega_1^{ck}$, the supremum of the computable ordinals. Thus, the game value of any finite position in infinite chess with a game value is a computable ordinal.
$\omega_1^{\mathfrak{Ch}}\leq\omega_1^{\mathfrak{Ch},c}\leq\omega_1^{\mathfrak{Ch},\text{hyp}}\leq\omega_1^{ck}$. Thus, the game value of any winning finite position in infinite chess, as well as any winning computable position or winning hyperarithmetic position, is a computable ordinal. Furthermore, if a designated player has a winning strategy from a position $p$, then there is such a strategy with complexity hyperarithmetic in $p$. $\omega_1^{\mathfrak{Ch}_{\!\!\!\!\sim}}\leq\omega_1$. The value of a winning position $p$ is a $p$-computable ordinal, and hence less than $\omega_1^{ck,p}$. Similarly, $\omega_1^{\mathfrak{Ch}_3}\leq\omega_1^{ck}$.
Evans and Hamkins [1] have proved that the omega one of infinite 3D chess is true $\omega_1$, as large as it can be: $\omega_1^{{\mathfrak{Ch}_{\!\!\!\!\sim}}_3}=\omega_1$. Meanwhile, there remains a large gap in the best-known bounds for $\omega_1^{\mathfrak{Ch}}$ and $\omega_1^{\mathfrak{Ch}_{\!\!\!\!\sim}}$, with Evans and Hamkins finding (computable) infinite positions having value $\omega^3$ and somewhat beyond.
References Evans, C D A and Hamkins, Joel David. Transfinite game values in infinite chess.(under review) www arχiv bibtex
|
In the past, I have argued it is so unlikely my vote will matter that voting is not worthwhile, even if I altruistically account for the vast number of people whom elections affect. That is, I've argued that the expected outcome of my vote—the chance it will swing the election, times the impact that would have on each person, times the number of people the election impacts—is next to nothing. The expected causal effect of voting is so small, I claimed, that it would be more altruistic for me to write a nice letter to my grandmother than to vote.
Last week I finally thought through it carefully... and it turns out I was wrong.
I now think the expected net impact of one vote in a typical US presidential election is on the same order of magnitude as the impact of the election on one eligible voter. So if you care about the way your vote will affect the rest of the country and the world (and you think you know what effect it will have!), voting may be a very valuable use of your time.
In this post, I'll explain my old argument against voting, show why it was wrong, and—with minimal amounts of math—ballpark a very rough estimate of the expected marginal impact of a vote. (If you're not interested in the fallacious argument, just skip ahead.)
Disclaimers This whole post works within the framework of a plurality (whoever gets the most votes wins) election between two parties. The electoral college is somewhat more complicated, with aggregation of votes happening at more local levels. For the purpose of the ballpark arguments in this post, I don't think these details matter, but I'm happy to discuss in the comments if anyone disagrees. I'm also ignoring the possibility that the election will be tied after I vote, and I'm ignoring the fact that very close elections are decided by complicated politcal processes I don't understand. Again, I don't think these matter to first order, but I'm happy to discuss in the comments. There are lotsof arguments for (and against) voting, and their omission does not represent a stance on any of them. I am simply focusing on causal impact. None of the reasoning or math in this post is particularly sophisticated. I write it not because it is interesting but because it is important! And because I owe it to everyone to whom I've made the wrong argument. My thinking went like this: If we anonymize all of the $N$ ballots in an election and then consider them one by one, we can think of each as an identical random variable $X_i$. This random variable can take the value $-1$ (Democrat), $0$ (didn't vote), or $1$ (Republican). Since I don't know who voter $i$ is, and since he or she might make a mistake or forget to vote, $X_i$ could equal any of $-1, 0,$ or $1$, so it has some variance $\sigma^2$. Further, there is some bias to the votes—i.e. people on average slightly prefer one candidate to the other—so $X_i$ has some non-zero expected value $b$. My old, fallacious argument against voting
What is the chance that I swing the election? It's the same as the chance that the election was tied before I voted. The probability of a tie is just
$$P(\sum X_i = 0) = P(\sum X_i \leq 0) - P(\sum X_i \leq -1) \\
= P(\sum X_i - E(X_i) \leq -b \cdot N) - P(\sum X_i - E(X_i) \leq -b \cdot N - 1) \\
= P(\frac{1}{\sigma \sqrt{N}} \sum X_i - E(X_i) \leq \frac{-b \sqrt{N}}{\sigma}) - P(\frac{1}{\sigma \sqrt{N}} \sum X_i - E(X_i) \leq \frac{-b \sqrt{N} - 1/\sqrt{N}}{\sigma})$$
Now, applying the central limit theorem—which tells us that the average of many independent, identically distributed, mean-zero random variables converges to a normal distribution distributed "very tightly" around zero—we can say that the probability of a tie is
$$\approx P(Z \leq \frac{-b \sqrt{N}}{\sigma}) - P(Z \leq \frac{-b \sqrt{N} - 1/\sqrt{N}}{\sigma})) \\
\frac{1}{\sigma \sqrt{N}} \cdot \phi(-b \sqrt{N} / \sigma) = \frac{1}{ \sigma \sqrt{2 \pi N}} e^{-\frac{b^2 N}{2 \sigma^2} }
$$
This is absurdly small for large $N$ (many voters). But more importantly, even when we compute my expected impact, multiplying the chance of swinging the election by the number of voters $N$ and the impact $I$ of the election on each voter, it is still basically zero. For example, if we suppose that the election has a whopping $\$$100,000 effect on each voter, that there are only one million voters, and that voters are biased toward the Democrats by only half a percent (so that $b=0.01$), the expected impact of my vote is
$$(number \ of \ voters)(impact \ per \ voter)(probability \ of \ impact) \\
= N \cdot I \cdot \frac{1}{\sigma \sqrt{ 2 \pi N}} e^{-\frac{b^2 N}{2 \sigma^2} }
= \frac{I}{\sqrt{2 \pi} \sigma} \sqrt{N} e^{-\frac{b^2 N}{2 \sigma^2} }
\approx \frac{10^5}{\sqrt{2 \pi} \cdot 1} \sqrt{10^6} e^{-\frac{10^{-4} \cdot 10^6}{2 \cdot 1} } \\
= \$0.0000000000000077
$$
If that didn't make much sense to you, here's the basic intuition: A million votes being cast is like a million coins being flipped. If each of the coins is weighted a bit toward heads, then after enough flips, more than half will be heads. The chance that there are comparably many heads and tails is exceptionally small; in fact, it falls off exponentially as the number of flips increases. Similarly, the chance that when the average voter is a little biased toward the Democrats, all of the random factors--mis-checkings of boxes, people forgettings to go vote after work--favor the Republicans declines exponentially as the number of voters increases. Meanwhile, the impact of election's outcome grows proportionally to the number of voters. And an exponential decline will always dominate a linear growth, when the numbers are big.
Why that argument was wrongIn the argument above, I said "there is some bias to the votes—i.e. people on average slightly prefer one candidate to the other" and interpreted this by giving each vote some non-zero mean $b$. While it's true that the votes will almost certainlyhave some bias one way or the other, it's suspicious of me to treat this bias as a fixed number, because I am uncertain of its value. Rather, I should model $b$ as another random variable. Of course, I can assign near-zero probability to $b=0$, but even an infinitesimal chance that $b=0$ may matter after I multiply it by $N$, the large number of voters whom the election affects.
Another way to understand the fallacy in my argument is to think I misapplied the central limit theorem. If we interpret $b$ not as the actual bias of each vote, but rather as my
best guessof the bias of each vote, then the $X_i$ are not independent—since if the first hundred I observe have a sample mean that is, say, less than $b$, then I will expect that $b$ was an overestimate and that the one-hundred-and-first vote is less than $b$. Since the $X_i$ aren't independent, I can't apply the (standard) central limit theorem. The (actual) expected impact of voting
Rather than separately considering many sources of uncertainty—what the net bias of the population is, how many people will accidentally check the wrong box, who will forget to show up—we can model them all simultaneously, by thinking about my subjective probability distribution on the sum of the votes.
So, the day of a presidential election, what does my subjective distribution of $\sum X_i$ look like? A quick google search suggests that on the day of an election, betting markets typically reflect about 90% odds in favor of one candidate. If I knew better than the betting markets, I could be making a lot of money, so it's reasonable to assume my beliefs are similar to theirs. That means I assign a 10% chance of the predicted-to-lose candidate getting 50% or more of the vote, and—since there's basically no chance that the predicted loser gets more than 60% of the vote—we can say I assign a 10% chance that the predicted loser gets between 50% and 60% of the vote. So how likely is it there is a tie? We'd expect that if the predicted loser wins, it will be by the skin of their teeth. But to be very conservative, let's say it's as likely that the predicted loser gets 50.0% as it is they get 50.1%, as it is they get 50.2%, ..., all the way to 60%. That is, conditional on winning by between 0% and 10%, each number of votes the predicted loser might get is equally likely. Receiving 50% to 60% of the vote corresponds to receiving $\frac{5N}{10}$ to $\frac{6N}{10}$ actual votes, so there are $\frac{N}{10}$ possible numbers of votes that candidate might receive. So if each such number is equally likely, then there is a $10\% / (\frac{N}{10}) = 1/N$ chance that they get exactly 50% of the votes. If there is a $1/N$ chance that one candidate gets exactly half the votes, then there is a $1/N$ chance that I swing the election. So the expected impact of my vote is just $$(number \ of \ voters)(impact \ per \ voter)(probability \ of \ impact) = N \cdot I \cdot 1/N = I$$
So, the day of a presidential election, what does my subjective distribution of $\sum X_i$ look like? A quick google search suggests that on the day of an election, betting markets typically reflect about 90% odds in favor of one candidate. If I knew better than the betting markets, I could be making a lot of money, so it's reasonable to assume my beliefs are similar to theirs. That means I assign a 10% chance of the predicted-to-lose candidate getting 50% or more of the vote, and—since there's basically no chance that the predicted loser gets more than 60% of the vote—we can say I assign a 10% chance that the predicted loser gets between 50% and 60% of the vote.
So how likely is it there is a tie? We'd expect that if the predicted loser wins, it will be by the skin of their teeth. But to be very conservative, let's say it's as likely that the predicted loser gets 50.0% as it is they get 50.1%, as it is they get 50.2%, ..., all the way to 60%. That is, conditional on winning by between 0% and 10%, each number of votes the predicted loser might get is equally likely. Receiving 50% to 60% of the vote corresponds to receiving $\frac{5N}{10}$ to $\frac{6N}{10}$ actual votes, so there are $\frac{N}{10}$ possible numbers of votes that candidate might receive. So if each such number is equally likely, then there is a $10\% / (\frac{N}{10}) = 1/N$ chance that they get exactly 50% of the votes.
If there is a $1/N$ chance that one candidate gets exactly half the votes, then there is a $1/N$ chance that I swing the election. So the expected impact of my vote is just
$$(number \ of \ voters)(impact \ per \ voter)(probability \ of \ impact) = N \cdot I \cdot 1/N = I$$
Wait, really?
How can this be? No US presidential election has ever come within
onevote. Is it really reasonable to think this might happen?
These questions are tempting, but ultimately misguided. We've never seen a tie before, and—since there is only a $1/N \approx 0.0000003\%$ chance of it happening in each presidential election—we shouldn't expect that we ever will. But on the off-chance that there is a tie, each vote will have a marginal impact whose magnitude is as large as the off-chance is small. Since our brains are bad at understanding both tiny probabilities and huge impacts, and since this problem requires us to weigh the two against each other, we shouldn't really expect this to be intuitive.
Loose ends
So far, I've left all estimates in terms of $I$, which I've called the average impact of the election on a voter. By this, I mean the expected difference in outcomes for an average person if your preferred candidate is selected instead of the other one.
It's important to be aware that $I$ may be negative; you might chose a candidate who will actually do a lot of harm to other voters (not to mention the rest of the world!). If you are really humble, you might think that you have no better idea of what is good for people than does anyone else, in which case your $I$ is about zero, and you'll need to find some other reasons to vote.
However, you might also think that you are better informed or better educated than other voters, or that your values are "better" than theirs in some moral sense. In this case, $I$ could be quite large, since different presidents have significantly different priorities. I'll guess that I'd put my own $I$ in the range of thousands or tens of thousands of dollars (though I cringe at the idea of trying to monetize outcomes across such a wide swath of topics, as well as at having to put a number on something about which I'm so uncertain). This is huge, considering that it will take me at most a few hours.
tl;dr
In case you weren't going to already, you should really vote—and you should make an informed decision about whom to support. It is unfathomably unlikely that you will swing the election, but if you do, you will impact an unfathomably large number of people.
Thanks
To Margaret, for having a conversation about voting that finally prompted me to formalize these arguments. To Jake, for some helpful edits and comments.
|
Say I want to apply a zero-phase Butterworth bandpass filter, implemented in MATLAB using
butter and
filtfilt. So, for an order-$4n$ filter with passband $[f_{\mathrm{min}\,},f_{\mathrm{max}}]$, I believe the transfer function $H$ is given by$$ H(2\pi if) \ = \ \frac{1}{1+\left( \frac{f}{f_\mathrm{max}-f_\mathrm{min}} + \frac{f^{-1}}{f_\mathrm{max}^{-1}-f_\mathrm{min}^{-1}} \right)^{\!2n}}\,. $$
It seems that for any given digital signal $x(t)$ (other than the constant zero signal), there is a critical number $N_x$ such that:
for any $n < N_x$ the results are fairly sensible; but for $n \geq N_x$, the results change considerably and are unreliable.
(1) What is the cause of this behaviour? Is it due to numerical imprecisions that grow large when the stability of the filter is poor?
(2) Is there a straightforward way to determine what this critical number $N_x$ is?
Let me illustrate the phenomenon with a couple of examples:
EXAMPLE 1
Consider the signal $x(t)=\sin(2\pi t)$, $t \in \{0,0.01,0.02\ldots,99.99,100\}$. (This signal has 10001 points altogether.)
Suppose I apply the Butterworth filter with passband $[f_{\mathrm{min}\,},f_{\mathrm{max}}]=[0.5,3]$. Of course,
I expect my filtered signal $x_{\mathrm{filt}}(t)$ to be basically the same as $x(t)$, except: there may be edge effects near the start and/or end; for $n=1$ or $2$, the amplitude is likely to be slightly reduced, as the frequency of the sinusoid $x(t)$ is not very close to the centre frequency $\sqrt{1.5}$ of the filter.
For each $n \leq 5$, the results are as expected. E.g., for $n=5$ (corresponding to filter order 20), the result is as follows:
However, when I now take $n=6$ (filter order 24), I suddenly get a drastically different result:
A wavelet transform of the original and filtered signal for $n=6$ looks as follows:
For even larger $n$, the results simply blow up. For $n=7$, the filtered signal is equal to
NaN for about the first two thirds of the duration, and then starts taking values of the order of $\,-10^{305}$, going up to a maximum value of about $4 \times 10^{231}$.
So in this example, $N_x=6$.
EXAMPLE 2
This is exactly the same as Example 1, except with the sampling frequency considerably reduced.
Consider the signal $x(t)=\sin(2\pi t)$, $t \in \{0,0.08,0.16\ldots,99.92,100\}$. (This signal has 1251 points altogether.)
Once again, suppose I apply the Butterworth filter with passband $[f_{\mathrm{min}\,},f_{\mathrm{max}}]=[0.5,3]$. For $n \leq 16$, I get sensible results. Here is the filtered signal with $n=16$:
I think the "early warning sign" here is that the slight inaccuracy of the filtered signal persists throughout the whole duration, not just at the start and end; for $n \leq 15$, the visible inaccuracy is only at the edges.
For $n=17$, I get quite a bad result:
In this case, Fourier and wavelet transforms are as follows:
Once again, for $n=18$, the results blow up:
(Please note that the values of the filtered signal are extremely large throughout, not just for $t<35$. At $t=70$, the value is about $-6.5 \times 10^{11}$.)
So in this example, $N_x=17$.
REMARK: In many (probably most) other examples, there is no intermediary $n$-value between the $n$-values giving consistent results and the $n$-values for which the filtered signal blows up to extreme values. In this regard (but only this regard), the above two examples seem somewhat exceptional.
|
Here's a way to automate the solution of a system that, like the OP's, is a linear system whose coefficient matrix happens to have a constant eigenvector (constant in that it is independent of $t$).If $\dot x = Ax$ and $A$ is an $n \times n$ matrix that has some constant eigenvectors $v_1,v_2,\dots,v_k$,then completing them to a basis $v_1,v_2,\dots,v_k,v_{k+1},\dots,v_n$, we canform the transition matrix$$P = \left( v1 \mid v_2 \mid \cdots \mid v_n \right) \,.$$Then with $x = Pu$, we get the transformed ODE $\dot u = P^{-1}AP\,u$,which hopefully
DSolve can deal with. It should be able to find the first $k$ solutions, but it might fail to find the last $n-k$.
vars = {x1, x2};
ode = {x1'[t] == x2[t],
x2'[t] == 2 Cos[t]/(3 + 2 Sin[t]) x1[t] -
x2[t] (3 - 2 Cos[t] + 2 Sin[t])/(3 + 2 Sin[t])};
rhs = D[Through[vars[t]], t] /. First@Solve[ode, D[Through[vars[t]], t]];
If[AllTrue[rhs, Internal`LinearQ[#, Through[vars[t]]] &], (* check linearity *)
rhsA = CoefficientArrays[rhs, Through[vars[t]]][[2]],
rhsA = Failure["nonlinear", <||>]
]
(* SparseArray[<<4>>] < check: was linear > *)
ev = Cases[Eigenvectors[rhsA], v_ /; FreeQ[v, t]]
If[Length[ev] < 1, Print["Bummer. No constant eigenvectors."]]
(* {{-1, 1}} < check: got a constant eigenvector > *)
pp = Transpose@Join[ev, NullSpace[ev]]; (* transition matrix *)
aa = Inverse[pp].rhsA.pp // Simplify; (* transform coefficient matrix *)
dsol = DSolve[{u'[t], v'[t]} == aa.{u[t], v[t]}, {u, v}, t] (* check DSolve works *)
(* check: DSolve found a solution
{{u -> Function[{t}, E^-t C[2] + C[1] (-3 + 2 Cos[t])],
v -> Function[{t}, C[1] (3 + 2 Sin[t])]}}
*)
pp.{u[t], v[t]} /. First[dsol] // Simplify; (* pp.{u, v} yields {x1, x2} *)
xsol = Thread[vars -> (Function @@ {t, #} & /@ %)]
(*
{x1 -> Function[t, 6 C[1] - E^-t C[2] - 2 C[1] Cos[t] + 2 C[1] Sin[t]],
x2 -> Function[t, E^-t C[2] + 2 C[1] Cos[t] + 2 C[1] Sin[t]]}
*)
ode /. xsol // Simplify (* check solution *)
(* {True, True} *)
|
Suppose the hazard rate is $\lambda$ the default probability density function follow exponential
$f(t) = \lambda e^{-\lambda t}$
and cumulative probability function is
$F(t) = 1 - e^{-\lambda t}$
the probability of default
within 3 years is
$P(t<3) = F(3) = 1-e^{- 3 \lambda }$
and the
conditional that it default in 3rd year given no default in the first 2 years is
$P(t<3|t>2) = \frac{P(t<3)-P(t<2)}{P(t>2)} = \frac{P(t<3)-P(t<2)}{1-P(t<2)} = \frac{e^{-2 \lambda}-e^{-3 \lambda}}{e^{-2 \lambda}} = 1 -e^{-\lambda} \hspace{0.05in} $ (1)
However, if I consider
event A: no default in first 2 years event B: default in year 3
$P(A \cap B) = P(A) * P(B) ={[P(t<3)-P(t<2)]*}{P(t>2)} \hspace{0.65in}$ (2)
Which one is right for the default probability in year 3? (1) or (2), or neither
|
Can I be a pedant and say that if the question states that $\langle \alpha \vert A \vert \alpha \rangle = 0$ for every vector $\lvert \alpha \rangle$, that means that $A$ is everywhere defined, so there are no domain issues?
Gravitational optics is very different from quantum optics, if by the latter you mean the quantum effects of interaction between light and matter. There are three crucial differences I can think of:We can always detect uniform motion with respect to a medium by a positive result to a Michelson...
Hmm, it seems we cannot just superimpose gravitational waves to create standing waves
The above search is inspired by last night dream, which took place in an alternate version of my 3rd year undergrad GR course. The lecturer talks about a weird equation in general relativity that has a huge summation symbol, and then talked about gravitational waves emitting from a body. After that lecture, I then asked the lecturer whether gravitational standing waves are possible, as a imagine the hypothetical scenario of placing a node at the end of the vertical white line
[The Cube] Regarding The Cube, I am thinking about an energy level diagram like this
where the infinitely degenerate level is the lowest energy level when the environment is also taken account of
The idea is that if the possible relaxations between energy levels is restricted so that to relax from an excited state, the bottleneck must be passed, then we have a very high entropy high energy system confined in a compact volume
Therefore, as energy is pumped into the system, the lack of direct relaxation pathways to the ground state plus the huge degeneracy at higher energy levels should result in a lot of possible configurations to give the same high energy, thus effectively create an entropy trap to minimise heat loss to surroundings
@Kaumudi.H there is also an addon that allows Office 2003 to read (but not save) files from later versions of Office, and you probably want this too. The installer for this should also be in \Stuff (but probably isn't if I forgot to include the SP3 installer).
Hi @EmilioPisanty, it's great that you want to help me clear out confusions. I think we have a misunderstanding here. When you say "if you really want to "understand"", I've thought you were mentioning at my questions directly to the close voter, not the question in meta. When you mention about my original post, you think that it's a hopeless mess of confusion? Why? Except being off-topic, it seems clear to understand, doesn't it?
Physics.stackexchange currently uses 2.7.1 with the config TeX-AMS_HTML-full which is affected by a visual glitch on both desktop and mobile version of Safari under latest OS, \vec{x} results in the arrow displayed too far to the right (issue #1737). This has been fixed in 2.7.2. Thanks.
I have never used the app for this site, but if you ask a question on a mobile phone, there is no homework guidance box, as there is on the full site, due to screen size limitations.I think it's a safe asssumption that many students are using their phone to place their homework questions, in wh...
@0ßelö7 I don't really care for the functional analytic technicalities in this case - of course this statement needs some additional assumption to hold rigorously in the infinite-dimensional case, but I'm 99% that that's not what the OP wants to know (and, judging from the comments and other failed attempts, the "simple" version of the statement seems to confuse enough people already :P)
Why were the SI unit prefixes, i.e.\begin{align}\mathrm{giga} && 10^9 \\\mathrm{mega} && 10^6 \\\mathrm{kilo} && 10^3 \\\mathrm{milli} && 10^{-3} \\\mathrm{micro} && 10^{-6} \\\mathrm{nano} && 10^{-9}\end{align}chosen to be a multiple power of 3?Edit: Although this questio...
the major challenge is how to restrict the possible relaxation pathways so that in order to relax back to the ground state, at least one lower rotational level has to be passed, thus creating the bottleneck shown above
If two vectors $\vec{A} =A_x\hat{i} + A_y \hat{j} + A_z \hat{k}$ and$\vec{B} =B_x\hat{i} + B_y \hat{j} + B_z \hat{k}$, have angle $\theta$ between them then the dot product (scalar product) of $\vec{A}$ and $\vec{B}$ is$$\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos \theta$$$$\vec{A}\cdot\...
@ACuriousMind I want to give a talk on my GR work first. That can be hand-wavey. But I also want to present my program for Sobolev spaces and elliptic regularity, which is reasonably original. But the devil is in the details there.
@CooperCape I'm afraid not, you're still just asking us to check whether or not what you wrote there is correct - such questions are not a good fit for the site, since the potentially correct answer "Yes, that's right" is too short to even submit as an answer
|
Another way to look at this involves frequency domain aliasing due to time-domain sampling of a non-bandlimited function. The prototypical continuous-time Hann window function is (Fig. 1):
$$f(x) = \cases{\frac{1}{2}-\frac{1}{2}\cos(\pi x)&if $-1 < x < 1,$\\0&otherwise.}$$
Figure 1. The prototypical continuous-time Hann window function.
It is related to your $w(n)$ by:
$$w(n) = f\left(\frac{2n}{N-1}-1\right).$$
The Fourier transform (with angular frequency $\omega$) of $f(x)$ is:
$$\begin{align}F(\omega) &= \int_{-1}^{1}f(x)e^{-i\omega x}\\&= \int_{-1}^{1}\frac{1}{2}(1-\cos(2\pi x))e^{-i\omega x}\\&= \frac{1}{2}\operatorname{sinc}(\omega+\pi) + \operatorname{sinc}(\omega) + \frac{1}{2}\operatorname{sinc}(\omega-\pi),\end{align}$$
with this convenience definition of the sinc function (often you'd see $\pi \omega$ instead of $\omega$):
$$\operatorname{sinc}(\omega) = \cases{1 &if $\omega = 0,$\\\frac{\sin(\omega)}{\omega}&otherwise.}$$
The Fourier transform consists of shifted and scaled sinc functions, because $f(x)$ is the product of three complex exponential functions (complex sinusoids) and a rectangular function. The frequency domain counterpart of this is convolution of a sum of three shifted and scaled Dirac delta functions by a sinc function. The sinc function (by the above definition) is zero-valued at all integer multiples of $\pi$ except for at $\omega = 0$, and nowhere else. Likewise, $F(\omega)$ is zero-valued at integer multiples of $\pi$ except for at $\omega \in \{-\pi, 0, \pi\}$, and nowhere else (Fig. 2):
Figure 2. Absolute value of the Fourier transform of the prototypical continuous-time Hann window in dB scale. The horizonal axis is shown in multiples of $2\pi.$ All notches are true zeros.
The tails of $F(\omega)$ go on forever, so $f(x)$ is not bandlimited. When $f(x)$ is sampled this will result in frequency domain aliasing. If $f(x)$ was sampled directly with a sampling period of $1$, the aliasing would be around Nyquist frequency $\omega = \pi$ (the first gray gridline), but based on Eq. 1, you are sampling it more densely with a period $\frac{2}{N-1}$ in $x$, which means aliasing around $\omega =$ $ \pi/\frac{2}{N-1} =$ $ (N-1)\frac{\pi}{2}.$ As long as $N$ is an integer, all unaliased zeros will coincide with aliased zeros. They will thus remain zeros. For example $N = 9$ sees aliasing around $\omega=4\pi$ (Fig. 3):
Figure 3. Aliased spectrum of a sampled Hann window $w(n)$ with $N = 9$. The horizontal axis is the angular frequency $\omega$ with respect to $x$ shown in multiples of $2\pi.$
Because $f(x)$ was defined for the complete real line, the sampled window $w(n)$ is zero-padded both ways to eternity. Its aliased spectrum illustrated in Fig. 3 will be sampled by the discrete Fourier transform (DFT) with a uniform grid of $M$ samples inclusive of the first sample at $\omega = 0$ and exclusive of a one-beyond-last sample at the sampling frequency $\omega = (N-1)\pi.$ Considering that zeros still appear only at values of $\omega$ that are integer multiples of $\pi$, if the factor $M / (N-1)$ is an integer, then all the zeros of the spectrum will be included in the frequency sampling, resulting in your second figure (Hanning window 500). Otherwise, DFT might miss (sometimes barely) some of the zeros. Looks like you had $N=51$, because $500/50 = 10$ is an integer.
|
Search
Now showing items 1-10 of 22
Measurement of $W$ boson angular distributions in events with high transverse momentum jets at $\sqrt{s}=$ 8 TeV using the ATLAS detector
(Elsevier, 2017-02)
The $W$ boson angular distribution in events with high transverse momentum jets is measured using data collected by the ATLAS experiment from proton--proton collisions at a centre-of-mass energy $\sqrt{s}=$ 8 TeV at the ...
Search for new resonances decaying to a $W$ or $Z$ boson and a Higgs boson in the $\ell^+ \ell^- b\bar b$, $\ell \nu b\bar b$, and $\nu\bar{\nu} b\bar b$ channels with $pp$ collisions at $\sqrt s = 13$ TeV with the ATLAS detector
(Elsevier, 2017-01)
A search is presented for new resonances decaying to a $W$ or $Z$ boson and a Higgs boson in the $\ell^+ \ell^- b\bar b$, $\ell\nu b\bar b$, and $\nu\bar{\nu} b\bar b$ channels in $pp$ collisions at $\sqrt s = 13$ TeV with ...
Search for new phenomena in events containing a same-flavour opposite-sign dilepton pair, jets, and large missing transverse momentum in $\sqrt{s}=$ 13 $pp$ collisions with the ATLAS detector
(Springer, 2017-03)
Two searches for new phenomena in final states containing a same-flavour opposite-lepton (electron or muon) pair, jets, and large missing transverse momentum are presented. These searches make use of proton--proton collision ...
Search for dark matter in association with a Higgs boson decaying to $b$-quarks in $pp$ collisions at $\sqrt{s} = 13$ TeV with the ATLAS detector
(Elsevier, 2017-02)
A search for dark matter pair production in association with a Higgs boson decaying to a pair of bottom quarks is presented, using 3.2 $fb^{-1}$ of $pp$ collisions at a centre-of-mass energy of 13 TeV collected by the ATLAS ...
Measurement of the prompt $J/\psi$ pair production cross-section in pp collisions at $\sqrt{s} = 8$ TeV with the ATLAS detector
(Springer, 2017-02)
The production of two prompt $J/\psi$ mesons, each with transverse momenta $p_{\mathrm{T}}>8.5$ GeV and rapidity $|y| < 2.1$, is studied using a sample of proton-proton collisions at $\sqrt{s} = 8$ TeV, corresponding to ...
Search for heavy resonances decaying to a $Z$ boson and a photon in $pp$ collisions at $\sqrt{s}=13$ TeV with the ATLAS detector
(Elsevier, 2017-01)
This Letter presents a search for new resonances with mass larger than 250 GeV, decaying to a $Z$ boson and a photon. The dataset consists of an integrated luminosity of 3.2 fb$^{-1}$ of $pp$ collisions collected at $\sqrt{s} ...
Search for dark matter at $\sqrt{s}=13$ TeV in final states containing an energetic photon and large missing transverse momentum with the ATLAS detector
(Springer, 2017-06)
Results of a search for physics beyond the Standard Model in events containing an energetic photon and large missing transverse momentum with the ATLAS detector at the Large Hadron Collider are reported. As the number of ...
Measurement of forward-backward multiplicity correlations in lead-lead, proton-lead, and proton-proton collisions with the ATLAS detector
(American Physical Society, 2017-06)
Two-particle pseudorapidity correlations are measured in $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV Pb+Pb, $\sqrt{s_{\rm{NN}}}$ = 5.02 TeV $p$+Pb, and $\sqrt{s}$ = 13 TeV $pp$ collisions at the LHC, with total integrated luminosities ...
Performance of the ATLAS trigger system in 2015
(Springer, 2017-05)
During 2015 the ATLAS experiment recorded $3.8 \mathrm{fb}^{-1}$ of proton--proton collision data at a centre-of-mass energy of $13 \mathrm{TeV}$. The ATLAS trigger system is a crucial component of the experiment, responsible ...
Measurements of electroweak $Wjj$ production and constraints on anomalous gauge couplings with the ATLAS detector
(Springer, 2017-07)
Measurements of the electroweak production of a $W$ boson in association with two jets at high dijet invariant mass are performed using $\sqrt{s} = 7$ and $8$ TeV proton--proton collision data produced by the Large Hadron ...
|
Search
Now showing items 1-7 of 7
The ALICE Transition Radiation Detector: Construction, operation, and performance
(Elsevier, 2018-02)
The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ...
Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2018-02)
In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ...
Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(American Physical Society, 2018-02)
The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ...
$\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV
(Springer, 2018-03)
An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ...
π0 and η meson production in proton-proton collisions at √s=8 TeV
(Springer, 2018-03-26)
An invariant differential cross section measurement of inclusive π^0 and η meson production at mid-rapidity in pp collisions at √s=8 TeV was carried out by the ALICE experiment at the LHC. The spectra of π^0 and η mesons ...
Longitudinal asymmetry and its effect on pseudorapidity distributions in Pb–Pb collisions at √sNN = 2.76 TeV
(Elsevier, 2018-03-22)
First results on the longitudinal asymmetry and its effect on the pseudorapidity distributions in Pb–Pb collisions at √sNN = 2.76 TeV at the Large Hadron Collider are obtained with the ALICE detector. The longitudinal ...
Production of deuterons, tritons, 3He nuclei, and their antinuclei in pp collisions at √s=0.9, 2.76, and 7 TeV
(American Physical Society, 2018-02)
Invariant differential yields of deuterons and antideuterons in p p collisions at √ s = 0.9, 2.76 and 7 TeV and the yields of tritons, 3 He nuclei, and their antinuclei at √ s = 7 TeV have been measured with the ALICE ...
|
How would you price the following option on underlying $S$ without dividends?
Time to maturity of option $\tau = 12$ months
Option has a strike $K > 0$ and constant barrier $B > 0$.
$t_0$ is the current point in time, while ${t_1, t_2,...t_{12}}$ are observation (potential trigger) dates and $t_{12}$ is the expiration date.
When the underlying is above the barrier at any of the observation dates, the option is called (similar to up and in call) and the difference between underlying at the observation date $S_{t_i}$ and strike $K$ is payed out.
Assume further for our example:
$ 0 < S_{t_0} < B $
$ K = S_{t_0}$ (ATM)
At each observation date, i.e. ${t_1, t_2,...t_{11}}$, excluding the expiration date, the payoff looks as follows:
$$ \begin{align*} if \quad S_{t_i} \geq B: \quad payoff = max(S_{t_i} - K, 0)\\ elif \quad S_{t_i} < B: \qquad \qquad \qquad payoff = 0 \end{align*}$$
The first case in our setting is per definition positive and implies that option is triggered / redeemed
At maturity / expiration if option hasn't been triggered on one of the observation dates, the payoff is independent of the barrier:
$$ \quad payoff = S_{t_i} - K $$
i.e. you get back the payoff of holding the stock over that the whole maturity when $K = S_{t_0}$
In other words, over the whole maturity the option bears the entire downside potential, but each month the option can be triggered if the stock price level is above the barrier $B$ and is then also redeemed at that observation date $t_i$. At maturity you essentially get back the payoff of holding purely the stock if we assume that $ K = S_{t_0}$.
I´d argue that the profile can be replicated dynamically by shorting a 12 month put and being long 1 month calls at the beginning of each month conditional on being knocked out in the prior period.
|
Answer
$$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$ The process is explained step by step below.
Work Step by Step
$$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$ Let's see to the left side first. $$X=\frac{2(\sin x-\sin^3x)}{\cos x}$$ $$X=\frac{2\sin x(1-\sin^2x)}{\cos x}$$ - As to $(1-\sin^2x)$, recall that $\cos^2x=1-\sin^2x$, meaning that $$X=\frac{2\sin x\cos^2x}{\cos x}$$ $$X=2\sin x\cos x$$ - Now recall that $2\sin x\cos x=\sin 2x$, as stated in double-angle identity for sine. Therefore, $$X=\sin2x$$ That means $$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$ The verification is complete. The equation is an identity.
|
Consider an (arithmetic) Ornstein-Uhlenbeck process as a model of the asset price $X_t$:
$$dX_t = \kappa(\mu-S_t)dt + \sigma dW_t$$
where $\mu$ is the mean-reversion level, $\sigma$ is a volatility parameter, $W_t$ is Brownian motion, and $\kappa$ is the reversion speed. An Ornstein-Uhlenbeck process will revert to the mean infinitely often if $\kappa > 0$ and the characteristic time scale for reversion is $\kappa^{-1}$.
The exact solution over an interval $[0,t]$ is
$$X_t = X_0 e^{-\kappa t} + \mu (1 - e^{-\kappa t}) + \sigma\int_{0}^{t} e^{\kappa(s-t)} dW_s,$$
with expected value
$$E(X_t) = X_o e^{-\kappa t} + \mu (1 - e^{-\kappa t}).$$
To address your question, there are a number of metrics that characterize the time frame of reversion -- the most obvious being the expected first passage time. For a mean reverting process the expected time to revert from level $A> \mu$ to level $B<A$ will, of course, be smaller than the expected time to migrate from level $B$ to $A$.
The formulas are complicated, and in my view of limited practical value (as I discuss below).
Here is a reference:
L. M. Ricciardi and S. Sato, First-passage-time density and moments of theOrnstein-Uhlenbeck process, J. Appl. Probab. 25 (1988) 43—57; MR0929503(89b:60189).
In theory, a profitable trading strategy certainly exists for an asset that exhibits a stable mean-reverting behavior. There are many variations such as putting on a trade when the price deviates from the expected reversion level by a fixed amount in a binary fashion, or entering and taking profit in a continuous fashion.
However, in real markets, assets seldom exhibit stationary mean-reversion with predictable reversion levels. Mean reversion may persist for a period of time and then suddenly break down. The assumption of a stable reversion level is, generally, the weak link in the chain -- leading eventually to a large draw-down.
In practice, a more robust trading strategy allows the expected reversion level to adjust gradually as a moving-average. Position size may also be adjusted based on historical volatility. The graph below shows trading profits (without transaction costs) for a simple mean-reversion strategy applied to the EURSEK currency pair. In this case, position size is proportional to the deviation of the exchange rate from a 1-month or 3-month moving average (using daily NY market closing levels). The position size (long or short) is not allowed to go above a cap based on the percentage deviation form the moving average.
The 3-month moving average adjusts more slowly than a 1-month average -- so the position sizes tend to be bigger -- yielding a bigger mean return but with much bigger drawdown when the mean reversion breaks down (late 2008, for example).
|
Little explorations with HP calculators (no Prime)
03-23-2017, 01:23 PM (This post was last modified: 03-23-2017 01:23 PM by pier4r.)
Post: #21
RE: Little explorations with the HP calculators
(03-23-2017 12:19 PM)Joe Horn Wrote: I see no bug here. Variables which are assigned values should never be used where formal variables are required. Managing them is up to the user.
Ok (otherwise a variable cannot just be feed in a function from a program) but then how come that when I set the flags, that let the function return reals, the variable is purged? The behavior could be more consistent.
Nothing bad, just a quirks like this, when not clear to spot, may lead to other solutions (see the advice of John Keit that I followed)
Wikis are great, Contribute :)
03-24-2017, 01:45 PM (This post was last modified: 03-24-2017 03:23 PM by pier4r.)
Post: #22
RE: Little explorations with the HP calculators
Quote:Brilliant.org
Is there a way to solve this without using a wrapping program? (hp 50g)
I'm trying around some functions (e.g: MSLV) with no luck, so I post this while I dig more on the manual, and on search engines focused on "site:hpmuseum.org" or "comp.sys.hp48" (the official hp forums are too chaotic, so I won't search there, although they store great contributions as well).
edit: I don't mind inserting manually new starting values, I mean that there is a function to find at least one solution, then the user can find the others changing the starting values.
Edit. It seems that the numeric solver for one equation can do it, one has to set values to the variables and then press solve, even if one variable has already a value (it was not so obvious from the manual, I thought that variables with given values could not change it). The point is that one variable will change while the others stay constant. In this way one can find all the solutions.
Wikis are great, Contribute :)
03-24-2017, 03:23 PM (This post was last modified: 03-24-2017 03:24 PM by pier4r.)
Post: #23
RE: Little explorations with the HP calculators
Quote:Same site as before.
This can be solved with multiple applications of SOLVEVX (hp 50g) on parts of the equation with proper observations. So it is not that difficult, just I found it nice and I wanted to share.
Wikis are great, Contribute :)
03-24-2017, 10:04 PM (This post was last modified: 03-24-2017 10:05 PM by pier4r.)
Post: #24
RE: Little explorations with the HP calculators
How do you solve this using a calculator to compute the number?
I solved it translating it in a formula after a bit of thinkering, and I got a number that I may write as "x + 0.343 + O(0.343)" if I'm not mistaken. I used the numeric solver on the hp 50g as helper.
I also needed to prove to myself that the center on the circle is on a particular location to proceed to build the final equation.
Wikis are great, Contribute :)
03-24-2017, 10:54 PM (This post was last modified: 03-24-2017 11:07 PM by Dieter.)
Post: #25
RE: Little explorations with the HP calculators
I think a calculator is the very last thing required here.
(03-24-2017 10:04 PM)pier4r Wrote: I solved it translating it in a formula after a bit of thinkering, and I got a number that I may write as "x + 0.343 + O(0.343)" if I'm not mistaken. I used the numeric solver on the hp 50g as helper.
Numeric solver? The radius can be determined with a simple closed form solution. Take a look at the diagonal through B, O and D which is 6√2 units long.
So 6√2 = 2√2 + r + r√2. Which directly leads to r = 4 / (1 + 1/√2) = 2,343...
Dieter
03-25-2017, 08:17 AM (This post was last modified: 03-25-2017 08:19 AM by pier4r.)
Post: #26
RE: Little explorations with the HP calculators
As always, first comes the mental work to create the formula or the model, but to compute the final number one needs a calculator most of the time.
Quote:Numeric solver? The radius can be determined with a simple closed form solution. Take a look at the diagonal through B, O and D which is 6√2 units long.
I used the numeric solver because instead of grouping r on the left, I just used the formula of one step before - the one without grouping - to find the value. Anyway one cannot just use the diagonal because the picture is well done, one has to prove to himself that O is on the diagonal (nothing difficult, but required), otherwise it may be a step taken for granted.
Wikis are great, Contribute :)
03-27-2017, 12:14 PM
Post: #27
RE: Little explorations with the HP calculators
Quote:Brilliant.org
This one defeated me at the moment. My rusty memory about mathematical relations did not help me. At the end, having the hp50g, I tried to use some visual observations to write down the cartesian coordinates of the points defining the inner square, or observing the length of the sides, so if the side of the inner square is 2r so the sides of the triangle are "s+2r" and "s" from which one can say that "s^2+(s+2r)^2=1" . This plus the knowledge that 4 times the triangles plus the inner square add up to 1 as area. Still, those were not enough for a solution (with or without the hp50g). I end in too ugly/tedious formulae.
Wikis are great, Contribute :)
03-27-2017, 12:54 PM (This post was last modified: 03-27-2017 03:42 PM by Thomas Okken.)
Post: #28
RE: Little explorations with the HP calculators
Consider a half-unit circle jammed into the corner of the first quadrant (so its center is at (0.5, 0.5)). Now consider a radius of that circle sweeping the angles from 0 to pi/2. The tangent on the circle where it meets that radius will intersect the X axis at 1 + tan(phi), and the Y axis at 1 + cot(phi) or 1 + 1 / tan(phi). The triangle formed by the X axis, the Y axis, and this tangent, is like the four triangles in the puzzle, and the challenge is to find phi such that X = Y + 1 (or X = Y - 1). The answer to the puzzle is then obtained by scaling everything down so that the hypotenuse of the triangle OXY becomes 1, and then the diameter of the circle is 1 / sqrt(X^2 + Y^2).
EDIT: No, I screwed up. The intersections at the axes are not at 1 + tan(phi), etc., that relationship is not quite that simple. Back to the drawing board!
Second attempt:
Consider a unit circle jammed into the corner of the first quadrant (so its center is at (1, 1)). Now consider a radius of that circle sweeping the angles from 0 to pi/2. The point P on the circle where that radius intersects it is at (1 + cos(phi), 1 + sin(phi)). The tangent on the circle at that point will have a slope of -1 / tan(phi), and so it will intersect the X axis at Px + Py * tan(phi), or (1 + sin(phi)) * tan(phi) + 1 + cos(phi), and it will intersect the Y axis at (1 + cos(phi)) / tan(phi) + 1 + sin(phi). The triangle formed by the X axis, the Y axis, and this tangent, is like the four triangles in the puzzle, and the challenge is to find phi such that X = Y + 2 (or X = Y - 2). The answer to the puzzle is then obtained by scaling everything down so that the hypotenuse of the triangle OXY becomes 1, and then the radius of the circle is 1 / sqrt(X^2 + Y^2).
Because of symmetry, sweeping the angles from 0 to pi/2 is actually not necessary; you can restrict yourself to 0 through pi/4 and the case that X = Y - 2.
03-27-2017, 02:27 PM (This post was last modified: 03-27-2017 02:28 PM by pier4r.)
Post: #29
RE: Little explorations with the HP calculators
(03-27-2017 12:54 PM)Thomas Okken Wrote: Consider a unit circle jammed into the corner of the first quadrant (so its center is at (1, 1)). Now consider a radius of that circle sweeping the angles from 0 to pi/2. The point P on the circle where that radius intersects it is at (1 + cos(phi), 1 + sin(phi)).
I'm a bit blocked.
http://i.imgur.com/IW4QIeU.jpg
Would be possible to add a quick sketch?
Wikis are great, Contribute :)
03-27-2017, 03:44 PM (This post was last modified: 03-27-2017 03:44 PM by Thomas Okken.)
Post: #30
RE: Little explorations with the HP calculators
(03-27-2017 02:27 PM)pier4r Wrote:(03-27-2017 12:54 PM)Thomas Okken Wrote: Consider a unit circle jammed into the corner of the first quadrant (so its center is at (1, 1)). Now consider a radius of that circle sweeping the angles from 0 to pi/2. The point P on the circle where that radius intersects it is at (1 + cos(phi), 1 + sin(phi)).
OK; I attached a sketch to my previous post.
03-27-2017, 03:55 PM
Post: #31
RE: Little explorations with the HP calculators
Thanks and interesting approach. On brilliant.org there were dubious solutions (that did not prove their assumptions) and just one really cool making use of a known relationship with circles enclosed in right triangles.
Wikis are great, Contribute :)
03-27-2017, 05:29 PM (This post was last modified: 03-27-2017 05:31 PM by pier4r.)
Post: #32
RE: Little explorations with the HP calculators
Quote:Brilliant.org
For this I wrote a quick program, remembering some quality of the mean that after enough iterations it stabilizes. (I should find the right statement though).
Code:
But i'm not sure about the correctness of the approach. Im pretty sure there is a way to compute this with an integral ad then a closed form too.
Anyway this is the result at the moment:
Wikis are great, Contribute :)
03-27-2017, 06:01 PM (This post was last modified: 03-27-2017 07:33 PM by Dieter.)
Post: #33
RE: Little explorations with the HP calculators
(03-27-2017 12:14 PM)pier4r Wrote: ...so if the side of the inner square is 2r so the sides of the triangle are "s+2r" and "s" from which one can say that "s^2+(s+2r)^2=1" . This plus the knowledge that 4 times the triangles plus the inner square add up to 1 as area. Still, those were not enough for a solution
Right, in the end you realize that both formulas are the same. ;-)
The second constraint for s and r could be the formula of a circle inscribed in a triangle. This leads to two equations in two variables s and r. Or with d = 2r you'll end up with something like this:
(d² + d)/2 + (sqrt((d² + d)/2) + d)² = 1
I did not try an analytic solution, but using a numeric solver returns d = 2r = (√3–1)/2 = 0,36603 and s = 1/2 = 0,5.
Edit: finally this seems to be the correct solution. ;-)
Dieter
03-27-2017, 06:14 PM (This post was last modified: 03-27-2017 06:15 PM by pier4r.)
Post: #34
RE: Little explorations with the HP calculators
(03-27-2017 06:01 PM)Dieter Wrote: Right, in the end you realize that both formulas are the same. ;-)
How? One should be from the Pythagoras' theorem, a^2+b^2=c^2 (where I use the two sides of the triangle to get the hypotenuse) the other is the composition of the area of the square, made up from 4 triangles and one inner square. To me they sound as different models for different measurements. Could you explain me why are those the same?
Anyway to me it is great even the numerical solution (actually the one that I search with the hp50g) but I cannot tell you if it is right or not because I did not solve it by myself, other reviewers are needed.
Edit, anyway I remember that a discussed solution mentioned the relationship of a circle inscribed in a triangle, so I guess your direction is right.
Wikis are great, Contribute :)
03-27-2017, 06:30 PM
Post: #35
RE: Little explorations with the HP calculators
(03-27-2017 06:14 PM)pier4r Wrote:(03-27-2017 06:01 PM)Dieter Wrote: Right, in the end you realize that both formulas are the same. ;-)
Just do the math. On the one hand, \( s^2 + (s + 2r)^2 = 1\) from Phythagorus' Theorem as you observed. And your other observation is that
\[ 4 \cdot \underbrace{\frac{1}{2} \cdot s \cdot (s+2r)}_{\text{area of }\Delta}
+ \underbrace{(2r)^2}_{\text{area of } \Box} = 1 \]
Simplify the left hand side:
\[
\begin{align}
4 \cdot \frac{1}{2} \cdot s \cdot (s+2r) + (2r)^2 & =
2s^2+4rs + 4r^2 \\
& = s^2 + s^2 + 4rs + 4r^2 \\
& = s^2 + (s+2r)^2
\end{align} \]
Hence, both formulas are the same.
Graph 3D | QPI | SolveSys
03-27-2017, 06:57 PM (This post was last modified: 03-27-2017 06:57 PM by pier4r.)
Post: #36
RE: Little explorations with the HP calculators
Thanks, I did not worked on the formula, I was more stuck (and somewhat still stuck) on the fact that they should represent different objects/results.
But then again, the square build on the side, it is the square itself. So now I see it. I wanted to see it in terms of "represented objects (1)" not only formulae.
Wikis are great, Contribute :)
03-27-2017, 07:07 PM
Post: #37
RE: Little explorations with the HP calculators
(03-27-2017 06:57 PM)pier4r Wrote:
Are you familiar with the geometric proofs of Phythagorus' Theorem? What I wrote above is just a variation of one of the geometric proofs using areas of regular polygons (triangles, rectangles, squares).
A few geometric proofs: http://www.cut-the-knot.org/pythagoras/
Graph 3D | QPI | SolveSys
03-27-2017, 07:22 PM
Post: #38
RE: Little explorations with the HP calculators
(03-27-2017 07:07 PM)Han Wrote: Are you familiar with the geometric proofs of Phythagorus' Theorem? What I wrote above is just a variation of one of the geometric proofs using areas of regular polygons (triangles, rectangles, squares).
Maybe my choice of words was not the best. I wanted to convey the fact that if I try to model two different events (or objects in this case) and I get the same formula, for me it is not immediate to say "oh, ok, then they are the same object", I have to, how can I say, "see it". So in the case of the problem, I saw it when I realized that the 1^2 is not only equal to the area because also the area is 1^2, it is exactly the area because it models the area of the square itself. (I was visually building 1^2 outside the square, like a duplicate)
Anyway the link you shared is great. I looked briefly and I can say:
- long ago I saw the proof #1
- in school I saw the proof #9
- oh look, the proof #34 would have helped, as someone mentioned
- how many!
Great!
Wikis are great, Contribute :)
03-27-2017, 07:24 PM (This post was last modified: 03-27-2017 07:28 PM by Joe Horn.)
Post: #39
RE: Little explorations with the HP calculators
(03-27-2017 05:29 PM)pier4r Wrote:Quote:Brilliant.org
After running 100 million iterations several times in UBASIC, I'm surprised that each run SEEMS to be converging, but each run ends with a quite different result:
10 randomize
20 T=0:C=0
30 repeat
40 T+=sqr((rnd-rnd)^2+(rnd-rnd)^2):C+=1
50 until C=99999994
60 repeat
70 T+=sqr((rnd-rnd)^2+(rnd-rnd)^2):C+=1
80 print C;T/C
90 until C=99999999
run
99999995 0.5214158234249566646569152059
99999996 0.5214158242970253667174680247
99999997 0.5214158240318481570747604814
99999998 0.5214158247892039896051570164
99999999 0.5214158253601312510245695897
OK
run
99999995 0.5213642776110289008920452545
99999996 0.5213642752079475043717958065
99999997 0.52136427197858201293861314
99999998 0.5213642744828552963477424429
99999999 0.5213642759132547792130043215
OK
run
99999995 0.5213770659191193073147616413
99999996 0.5213770610000764506616015052
99999997 0.5213770617149058467216528505
99999998 0.5213770589414874167694264508
99999999 0.5213770570854305903944611055
OK
So it SEEMS to be zeroing on something close to -LOG(LOG(2)), but I give up.
<0|ɸ|0>
-Joe-
03-27-2017, 07:35 PM (This post was last modified: 03-27-2017 07:37 PM by pier4r.)
Post: #40
RE: Little explorations with the HP calculators
(03-27-2017 07:24 PM)Joe Horn Wrote: 99999999 0.5213770570854305903944611055
Interestingly your number is quite different from mine. Ok that you have a couple of iterations more, but my average was pretty stable for the first 3 digits. I wonder why the discrepancy.
Moreover if you round to the 4th decimal place, you always get 0.5214 if the rounding is adding a "+1" to the last digit in the case the first digit to be excluded is higher than 5.
Wikis are great, Contribute :)
User(s) browsing this thread: 1 Guest(s)
|
This web page says that the microcanonical partition function $$ \Omega(E) = \int \delta(H(x)-E) \,\mathrm{d}x $$ and the canonical partition function $$ Z(\beta) = \int e^{-\beta H(x)}\,\mathrm{d}x $$ are related by the fact that $Z$ is the Laplace transform of $\Omega$. I can see mathematically that this is true, but why are they related this way? Why can we intrepret the integrand in $Z$ as a probability, and what allows us to identify $\beta = \frac{1}{kT}$?
I guess, one could start by considering the microcanonical and canonical ensembles as entirely different concepts. For they represent different statistical ensembles: in the former the energy of the system is fixed while in the second the energy can span all possible values allowed by the energy spectrum of the system with some penalty attributed to high energies.
In this case it is important to notice that the
meaning of the microcanonical measure and, for that matter, of the canonical one is that of probability measures in state space (quantum or classical). Thus, they have to take real values and be integrable/normalizable.
As you noticed, of course, the canonical partition function can be understood as being a Laplace transform of the (microcanonical) density of states. In some respect, it then tells us something about "how many thermostats do we need to capture the physics of the density of state?". This is of course vague-ish because the set of the Boltzmann weights (understood as probabilistic weights) doesn't constitute an orthonormal basis of the space of function.
So I think that for this Laplace transform idea to mean anything, we need to imagine that it is possible, via analytical continuation, to extend $\beta$-values to the complex plane. In such a case then, the inverse Laplace transform may exist and we may be able to understand something.
In particular we can indeed recast the density of state $\Omega(E) \equiv \sum_i \delta(E-E_i)$ where $i$ labels the microstates as follows:
$$\Omega(E) = \mathcal{L}^{-1}\left[ Z(\beta) \right] = \int \: d \beta \:e^{\beta E} Z(\beta)$$
where $\mathcal{L}^{-1}$ denotes the inverse Laplace transform.
The interesting point from the point of view of statistical mechanics is that, for big system sizes i.e. in the thermodynamic limit, we can write that $Z(\beta) \equiv e^{-\beta A(\beta)} \sim e^{-\beta N a(\beta)}$ where $A$ is the Helmholtz free energy and $N$ the number of particles in the system. The latter equality derives from the extensivity of the free energy. Note that $E$ is also extensive in the thermodynamic limit so that $E \sim N e$, we then get:
$$\Omega(E) = \int \: d \beta \:e^{\beta N(e-a(\beta))}$$
If $N$ is very large, one can then perform a saddle point approximation at, say, $\beta = \beta^*$ (which will most likely take a path in the complex plane) and find that:
$$\Omega(E) \sim e^{\beta^* N(e-a(\beta^*))}$$
Assuming the path lies in the complex plane but the saddle is itself on the real line, we then get that each value of $E$, there exists a single thermostat with $\beta$-value $\beta^*$ that is susceptible to accurately model the density of state, and hence all the statistical properties at fixed energy $E$; in the thermodynamic limit that is.
This is what is called the ensemble equivalence between the canonical and the microcanonical ensemble (of course the same can be done when asking "how many fixed energy systems do I need to model my thermostat at inverse temperature $\beta$").
This was the large system limit. Now, it turns out that assuming that microcanonical and canonical partition functions are, if I dare say, Laplace transforms of one another then, one can do things for small systems as well and at any temperature (i.e. even in the quantum regime).
In particular, the density of states is always quite hard to compute while the partition function is more tractable (and yet no easy beast to tame!).
Just to give the idea of what happens, let's consider the case of a 1D harmonic oscillator. In this case we have that the 1-particle partition function reads:
$$z(\beta) = \frac{1}{\sinh(\frac{\beta \hbar \omega}{2})}$$
Now, trying to figure what would be the corresponding density of state we can use the exact relation:
$$\omega(E) = \int d\beta \: \frac{e^{\beta E}}{\sinh(\frac{\beta \hbar \omega}{2})}$$
As I said before, for this integral to converge, it is very likely that the inversion has to be performed within an integrable slab of the complex plane. Moreover, even if we tried to integrate naively, we would encounter a pole at $\beta = 0$. If we now try to avoid the zero pole via an excursion in the complex plane, we would need to integrate over a closed contour, a semi-circle encompassing the top half of the complex plane say. If we do that, we need to account for all the poles that lie on the top half of the imaginary line and cancel the denominator of the integrand.
The corresponding residues give rise to an oscillatory part that is supplementing an average part that is the one we get in the large system size limit discussed above. The origin of this oscillatory behaviour stems from the fact that if we plot the total number of states which have energy below $E$, then, in the quantum case, it takes a "ladder shape" oscillating about a mean that we commonly use in our usual treatment of statistical thermodynamics.
It becomes more interesting beyond 1D and for those who are interested about the 2D harmonic oscillator, you can find a detailed study of the latter here.
At the end of the day, I would say that although the microcanonical and canonical ensembles are not
defined as being related by a Laplace transform, if we allow to extend their definition to the complex plane, then they can be related by a Laplace transform and it enables us to get very important and general results (ensemble equivalence for large systems and oscillatory behaviour of the density of states for small ones) which are by the way well verified numerically and experimentally.
Consider two connected systems $A$ and $B$ in the microcanonical ensemble. Calling $E$ the total (fixed) energy we have $$\Omega(E)=\Omega_A(E_A)\Omega_B(E_B)=\Omega(E_A)\Omega_B(E-E_A).\tag{1}$$ This only states that number of states of the whole system is the product of the number of states of the subsystems but here is really the key, since everything what follows is a direct consequence of $(1)$ and of the maximum entropy principle.
Using the maximum entropy principle with $(1)$, we obtain that at equilibrium between $A$ and $B$ we have $$\frac{1}{\Omega_A(E_A)}\frac{\partial\Omega_A}{\partial E_A}(E_A) =\frac{1}{\Omega_B(E_B)}\frac{\partial\Omega_B}{\partial E_B}(E_B).$$ This number (the relative variation of $\Omega_i$ with respect to the energy $E_i$) is called $\beta$, (up to the constant $k_{\mathrm B}$ which converts into our particular unit system). Consequently, we have at equilibrium $$\frac{\partial}{\partial E}\left(\ln\Omega-\beta E\right)=0,$$ We can therefore define the Legendre transform of $\ln\Omega(E)$ that we call $\ln\mathcal Z(\beta)$. We have $$-\frac{\partial\ln\mathcal Z}{\partial \beta}=E$$ as a consequence of the inverse Legendre transform. This shows that $\mathcal Z(\beta)$ is indeed the canonical partition function.
The Legendre transform actually gives the relation $$ \ln\mathcal Z(\beta)=\ln\Omega(E)-\beta E.$$ Taking the exponential and integrating over $E$ we get the relation $$ \mathcal Z(\beta)=\int \Omega(E)\mathrm e^{-\beta E}\,\mathrm dE.$$ We also get by integrating over $\beta$ the expression of the inverse $\Omega(E)=\int Z(\beta)\mathrm e^{\beta E}\mathrm d\beta$.
Beware that this does not give any information on the way how to perform the integral. The computation of $\Omega$ from $\mathcal Z$ requires actually a complex integration, as the Laplace inverse transform is performed along a contour in the complex plane such that all poles are on the left-hand side.
So the answer to your question could be that this relation is a consequence of the properties of the Legendre transform combined with the fact that $(1)$ implies the additivity of $\ln\Omega$.
I am not sure that it is true in general.
In your equation, $x$ might not be a single continuous variable in real number line. It may consist of a set of variables such as interacting spin system. It can also be discrete variable, and in the most general case, it is just a set of states in some model system. The Laplace transform is not well defined in the later case. Also, if the Hamiltonian is not linear nor quadratic (with canonical transformation), it is questionable whether it is still a Laplace transform.
Just as a quick conceptual contribution to the answers already provided, for any probability distribution defined over $\mathbb{R}_+$, the Laplace transform is the moment/cumulant-generating function, which is equivalent to saying it is the representation of the distribution in terms of its moments. The partition function may look a bit different, since we evaluate $\beta$ not only at zero, but that is precisely what it is.
Working my way backward from the canonical partition function,
$$\begin{align} Z&=\int e^{-\beta E}\,dx\,\,\,\,\textrm{(sum over states)}\\ &=\int e^{-(\beta-\beta_0) E-\beta_0E}\,dx\\ &=\int e^{-\Delta\beta E}\left(e^{-\beta_0E}\right)\,dx\\ &=\int e^{-\Delta\beta E}\rho (E,\beta_0)\,dx,\,\,\,\,\,\,\rho(E,\beta_0)=e^{-\beta_0E}\\ &=\int\left(1-\Delta\beta E + \frac{(\Delta\beta)^2 E^2}{2}+\ldots\right)\rho(E,\beta_0)\,dx\\ &=1-\Delta\beta m_1+ \frac{(\Delta\beta)^2 m_2^2}{2}+\ldots \end{align}$$
where $m_i$ is the $i$'th moment of the distribution $\rho(E,\beta_0)$ which is obtained by taking the appropriate derivative and setting $\Delta\beta=0$, which by the way immediately sets $\beta_0=\beta$.
I have not made any distinction between microcanonical, canoncical, grand-canonical, etc., because all that does is change the functional form of $\rho$ and that prescription is given by thermodynamics. In the microcanonical case, $\rho=\delta(E(x)-\bar{E})$ and does not depend on $\beta$. In the canonical case, $\rho=e^{-\beta E}$. And so on.
So in summary, the partition function is just the representation of the appropriate distribution in terms of its moments (moment-generating function), just as a Fourier series is the representation of a function in terms of its Fourier components.
|
First of all, what's the motivation for the Yang-Mills action and how
should I understand the coupling constants $\theta$ and $g$?
I would say motivation comes from experiments. For instance it is an experimental fact that the electric charge is conserved. The associated current is also conserved, in the sense of$$\partial_\mu J^\mu=0.$$Therefore we can write this current as the curl of a vector potential $A^\mu$. Since the curl of the grad vanish, there is a redundancy$$A^\mu(x)\rightarrow A^\mu(x)-\partial_\mu\Lambda(x),$$giving the same measured current. This redundancy is called gauge symmetry and in the present case it is just $U(1)$. It happens that for some fundamental interactions there are more vector potentials, more charges and we end up with a gauge symmetry based on a non-Abelian group. This is a Yang-Mills theory. For example, the Quantum Chromodynamics, which describes the strong interaction and is based on the $SU(3)$ color symmetry.
The classical fields of the theory must be solutions of equations of motions which are obtained from the action, according to the Hamilton Principle. The quantum fields also require an action in order to quantize it using the path integral method. When constructing the action you have to show how the fields interact. The strength of these interactions are given by the magnitude of the ecoupling constants, which is an experimental input.
How can I get this so called expansion in power series with variable
$\tau$ of the probability amplitude?
Going from a non-interacting quantum field theory to an interacting one is in same sense similar to going from a harmonic oscillator to an anharmonic one. For example, you add a quartic term in the potential, for both cases. Then you solve the equation of motion of the oscillator or the probabilities amplitudes for the fields using perturbation theory. The probabilities in the quantum theory may be obtained from something called generating functional, which involves the exponential the action. Perturbative expansion here means to expand this exponential in powers either of $\hbar$ or of the coupling constants.
What was the motivation to start looking at this duality? A creation
of an everywhere defined (in $\tau$) gauge theory, maybe?
Some context: It was noted since the 70's that some non-Abelian gauge theories admit solutions with stable magnetic charge. Then Montonem and Olive and Goddard, Nuyts and Olive , noticed that they could map the mass spectrum of electric charges of a particular theory to the mass spectrum of magnetic charges of another theory (called dual), as long as they assume a particular map between the couplings of these theories. They conjectured that these theories are electromagnetic dual. This is a non-Abelian generalization of the electromagnetic duality in Maxwell theory. A particular example where this duality is conjectured is between two "copies" of Georgi-Glashow model: $SO(3)\rightarrow SO(2)$ with the Higgs in the adjoint.
$SL(2,\mathbb Z)$ duality: It was shown by Witten that the addition of a topological term ($\theta$-term) to the Georgi-Glashow model gives the spectrum of electric charges$$q_e=e\left(n_e+\frac{\theta}{2\pi}n_m\right),\quad n_e,n_m\in\mathbb Z,$$
This theory also admit dyons, particles with the above electric charge and also a magnetic charge$$q_m=\frac{4\pi}{e}n_m.$$Then if we define$$\tau=\frac{\theta}{2\pi}+\frac{4\pi i}{e^2},$$the charge of the dyon can be written as$$\mathcal Q = q_e+iq_m=e(n_e+\tau n_m).$$
The spectrum of the theory belongs then to a lattice whose vertices give the charges $(n_m,n_e)$. The masses of the dyons (of a theory with coupling $\tau$) are given by$$M(n_e,n_m;\tau)=ve|n_e+n_m|=ve\left|\left(\begin{array}{cc}n_m&n_e\end{array}\right)\left(\begin{array}{c}\sqrt u\tau\\\sqrt u\end{array}\right)\right|,$$where $\sqrt u\equiv ev$ and $v$ is the Higgs vev. In order for a duality happen there must be a mapping between the mass spectrum of two theories, i.e.\begin{equation}\label{sl6}\left|\left(\begin{array}{cc}n_m&n_e\end{array}\right)\left(\begin{array}{c}\sqrt u\tau\\\sqrt u\end{array}\right)\right|=\left|\left(\begin{array}{cc}n'_m&n'_e\end{array}\right)\left(\begin{array}{c}\sqrt{u'}\tau'\\\sqrt{u'}\end{array}\right)\right|.\end{equation}A possible solution is\begin{align*}\left(\begin{array}{c}\sqrt{u'}\tau'\\\sqrt{u'}\end{array}\right)&=e^{i\varphi}\mathcal M\left(\begin{array}{c}\sqrt u\tau\\\sqrt u\end{array}\right),\\\left(\begin{array}{cc}n'_m&n'_e\end{array}\right)&=\left(\begin{array}{cc}n_m&n_e\end{array}\right)\mathcal M^{-1},\end{align*}with $\varphi\in\mathbb R$ and $$\mathcal M=\left(\begin{array}{cc}A&B\\C&D\end{array}\right),\quad \det \mathcal M= 1, \quad A,B,C,D\in\mathbb Z.$$Hence$$\mathcal M\in SL(2,\mathbb Z).$$The group $SL(2,\mathbb Z)$ has two generators$$T=\left(\begin{array}{cc}1&1\\0&1\end{array}\right),\quadS=\left(\begin{array}{cr}0&-1\\1&0\end{array}\right).$$
The duality transformations generated by $T$ and $S$ are called $T$-duality and $S$-duality, respectively. As you can see from the $S$ generator, the $S$-duality invert the coupling constant,$$S:\tau\rightarrow-\frac 1\tau,$$i.e. it is a duality that maps strong coupling to weak coupling. On the other hand $T$ acts as$$T:\tau\rightarrow\tau+1,$$which represents an invariance with respect of $\theta\rightarrow\theta+2\pi$.
It is important to notice that a possible electromagnetic duality is given only by a subgroup of $SL(2,\mathbb Z)$ because it is necessary to consider other quantum number of the particles.
The importance of the $S$-Duality is in the fact that you can use results obtained in a theory with weak coupling (where perturbation theory is valid) in the dual theory which has strong coupling (and perturbation theory is not valid).
|
I am struggling to answer an old general relativity exam question, which is as follows:
"Consider a scalar field $\phi(t,x^i)$ with potential $V(\phi)$ on a general spacetime. Its stress tensor is given as $$T_{\mu\nu} = \nabla_\mu \phi \nabla_\nu \phi - \frac{1}{2} g_{\mu\nu} (\nabla^\alpha \phi \nabla_\alpha \phi)-g_{\mu\nu} V(\phi)\tag{1}$$ Using the equation of motion of this scalar field, $$\nabla^\alpha \nabla_\alpha \phi = \frac{dV(\phi)}{d\phi}\tag{2}$$ show that the stress energy is conserved."
The metric convention used is $(-, +, +, +)$ signature.
To answer this I have been trying to show that this stress-energy tensor is divergence free; $$\nabla_\mu T^{\mu\nu}= 0 \Rightarrow \nabla^\mu T_{\mu\nu} = 0.\tag{3}$$ Doing this I get that $$\nabla^\mu T_{\mu\nu} = (\nabla^\mu \nabla_\mu \phi) \nabla_\nu \phi + \nabla_\mu \phi (\nabla^\mu \nabla_\nu \phi) - \frac{1}{2} \nabla_\nu (\nabla^\alpha \phi \nabla_\alpha \phi) - \nabla_\nu V(\phi)\tag{4}$$ I then use the equation of motion on the first term and the chain rule on the final term (justifying that $V(\phi)$ must be a scalar potential) to cancel them both. Then since $(\nabla^\alpha \phi \nabla_\alpha \phi)$ is a scalar I argue that its covariant derivative is simply the partial derivative. This leaves me with
$$\nabla^\mu T_{\mu\nu} = \nabla_\mu \phi (\nabla^\mu \nabla_\nu \phi) - \frac{1}{2} \partial_\nu (\nabla^\alpha \phi \nabla_\alpha \phi).\tag{5}$$
However I cannot seem to get these terms to cancel. If I use the fact that $\phi$ is a scalar field and therefore its covariant derivative is its partial derivative I get that
$$\nabla^\mu T_{\mu\nu} = \partial_\mu \phi (\partial^\mu \partial_\nu \phi) - \frac{1}{2} \partial_\nu (\partial^\mu \phi \partial_\mu \phi) - \partial_\mu \phi g^{\mu\rho}\Gamma^\sigma_{\rho\nu} \partial_\sigma \phi\tag{6}$$
Which I still cannot cancel.
What am I doing wrong? Should I relax the divergence free condition and simply consider $$\partial^\mu T_{\mu\nu}\tag{7}$$ instead? I have been stuck on this problem for hours and it's really bugging me.
|
A particle moves along the x-axis so that at time t its position is given by $x(t) = t^3-6t^2+9t+11$ during what time intervals is the particle moving to the left? so I know that we need the velocity for that and we can get that after taking the derivative but I don't know what to do after that the velocity would than be $v(t) = 3t^2-12t+9$ how could I find the intervals
Fix $c\in\{0,1,\dots\}$, let $K\geq c$ be an integer, and define $z_K=K^{-\alpha}$ for some $\alpha\in(0,2)$.I believe I have numerically discovered that$$\sum_{n=0}^{K-c}\binom{K}{n}\binom{K}{n+c}z_K^{n+c/2} \sim \sum_{n=0}^K \binom{K}{n}^2 z_K^n \quad \text{ as } K\to\infty$$but cannot ...
So, the whole discussion is about some polynomial $p(A)$, for $A$ an $n\times n$ matrix with entries in $\mathbf{C}$, and eigenvalues $\lambda_1,\ldots, \lambda_k$.
Anyways, part (a) is talking about proving that $p(\lambda_1),\ldots, p(\lambda_k)$ are eigenvalues of $p(A)$. That's basically routine computation. No problem there. The next bit is to compute the dimension of the eigenspaces $E(p(A), p(\lambda_i))$.
Seems like this bit follows from the same argument. An eigenvector for $A$ is an eigenvector for $p(A)$, so the rest seems to follow.
Finally, the last part is to find the characteristic polynomial of $p(A)$. I guess this means in terms of the characteristic polynomial of $A$.
Well, we do know what the eigenvalues are...
The so-called Spectral Mapping Theorem tells us that the eigenvalues of $p(A)$ are exactly the $p(\lambda_i)$.
Usually, by the time you start talking about complex numbers you consider the real numbers as a subset of them, since a and b are real in a + bi. But you could define it that way and call it a "standard form" like ax + by = c for linear equations :-) @Riker
"a + bi where a and b are integers" Complex numbers a + bi where a and b are integers are called Gaussian integers.
I was wondering If it is easier to factor in a non-ufd then it is to factor in a ufd.I can come up with arguments for that , but I also have arguments in the opposite direction.For instance : It should be easier to factor When there are more possibilities ( multiple factorizations in a non-ufd...
Does anyone know if $T: V \to R^n$ is an inner product space isomorphism if $T(v) = (v)_S$, where $S$ is a basis for $V$? My book isn't saying so explicitly, but there was a theorem saying that an inner product isomorphism exists, and another theorem kind of suggesting that it should work.
@TobiasKildetoft Sorry, I meant that they should be equal (accidently sent this before writing my answer. Writing it now)
Isn't there this theorem saying that if $v,w \in V$ ($V$ being an inner product space), then $||v|| = ||(v)_S||$? (where the left norm is defined as the norm in $V$ and the right norm is the euclidean norm) I thought that this would somehow result from isomorphism
@AlessandroCodenotti Actually, such a $f$ in fact needs to be surjective. Take any $y \in Y$; the maximal ideal of $k[Y]$ corresponding to that is $(Y_1 - y_1, \cdots, Y_n - y_n)$. The ideal corresponding to the subvariety $f^{-1}(y) \subset X$ in $k[X]$ is then nothing but $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$. If this is empty, weak Nullstellensatz kicks in to say that there are $g_1, \cdots, g_n \in k[X]$ such that $\sum_i (f^* Y_i - y_i)g_i = 1$.
Well, better to say that $(f^* Y_1 - y_1, \cdots, f^* Y_n - y_n)$ is the trivial ideal I guess. Hmm, I'm stuck again
O(n) acts transitively on S^(n-1) with stabilizer at a point O(n-1)
For any transitive G action on a set X with stabilizer H, G/H $\cong$ X set theoretically. In this case, as the action is a smooth action by a Lie group, you can prove this set-theoretic bijection gives a diffeomorphism
|
I am trying to perform contrast-transfer-function for pictures which were recorded in defocus. To do so, I have to calculate the contrast-transfer function from the Thon rings and fit the function to 1D radial average of the spectrum. Let's say that I have a picture in spatial domain where single pixel is 1.7 Angstrom size, which gives me a Nyquist Limit of resolution 2*1.7 A. So I know, that the resolution limit in my powerspectrum will be 1 / 2*1.7A. How can I connect the radial average from powerspectrum with the resolution? That is the relation of each pixel in the spectrum to resolution? I would be very grateful for some hints! Thanks!
Therefore, your question is about how to read the two dimensional Discrete Fourier Transform and how to interpret its content.
To relate what is mentioned in that answer to your specific question, consider the following CTF, obtained from Wikipedia (Axis added by me):
This is a two dimensional spectrum. If you are familiar with a one dimensional spectrum, it is as if you took that one and rotated it about the centre of the image.
Therefore, low (
spatial) frequencies are towards the centre of the image and high ( spatial) frequencies are towards the edges of the image.
In fact, if we take a cross section of that image (along the direction of the blue line), we see something like this:
And if we shift the origin of the spectrum to its conventional position (for a 1D spectrum), we get:
So, this now describes how well each
spatial frequency goes through the optical system assuming, it is symmetrical.
To relate this to physical frequency, we have to do a bit more work. Now, this is a simple grayscale image, which means that its maximum value is 255. So to normalise the "height" of this curve, we need to simply divide by 255.
The other thing we have to do is to normalise the frequency. This image is 512x512, the DC (or, constant value) is at 0, the Fs is at 512 and the Nyquist frequency is at 256. The formula is a plain simple division of $f = \frac{k}{XDim} \cdot Fs$ where $k$ is the frequency "bin" (of the discrete spectrum) and the fractional part of this we can simply represent in the graph.
Taking all of this into account gives us:
So, now, whatever your Fs turns out to be, you can put it on the diagram and work backwards to find how much each
spatial frequency gets diminished as it passes through this system.
Spatial resolution is measured in "lines" per unit of length. In your case, your pixel size is 1.7Å. The "highest"
frequency ( in any direction) that can theoretically go through a system that is digitised at that level is one black line of width 1.7Å followed immediately by another white line of width 1.7Å and that is one cycle of period 3.4Å.
Therefore, your Nyquist frequency is 1 line per 3.4Å or $\approx 2941176471$ lines per mm.
Hope this helps.
EDIT:
All the plots in this post were created using Octave and the following code:
% Load the image processing packagepkg load image% Load the image that was downloaded from WikipediaI = imread("~/Downloads/CTF.jpg");% The image is 512x512, let's take the cross section of the image to see a slice of the spectrumplot(I(256,:));% Now shift the origin using fftshiftplot(fftshift(I(256,:)));% Add normalised X and Y axis, labels, grid, etcplot((0:511)./512,double(fftshift(I(256,:)))/255.0);xlabel("Normalised Spatial Frequency");ylabel("Normalised Transfer Coefficient");grid on;
|
Set Complement inverts Subsets
Jump to navigation Jump to search
Theorem
Let $S$ and $T$ be sets.
Then:
$S \subseteq T \iff \map \complement T \subseteq \map \complement S$
where:
$S \subseteq \map \complement T \iff T \subseteq \map \complement S$
\(\displaystyle S\) \(\subseteq\) \(\displaystyle T\) \(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle S \cap T\) \(=\) \(\displaystyle S\) Intersection with Subset is Subset \(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \map \complement {S \cap T}\) \(=\) \(\displaystyle \map \complement S\) Complement of Complement \(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \map \complement S \cup \map \complement T\) \(=\) \(\displaystyle \map \complement S\) De Morgan's Laws: Complement of Intersection \(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle \map \complement T\) \(\subseteq\) \(\displaystyle \map \complement S\) Union with Superset is Superset
$\blacksquare$
\(\displaystyle S\) \(\subseteq\) \(\displaystyle T\) \(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle (x \in S\) \(\implies\) \(\displaystyle x \in T)\) Definition of Subset \(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle (x \notin T\) \(\implies\) \(\displaystyle x \notin S)\) Rule of Transposition \(\displaystyle \leadstoandfrom \ \ \) \(\displaystyle (x \in \map \complement T\) \(\implies\) \(\displaystyle x \in \map \complement S)\) Definition of Set Complement \(\displaystyle \map \complement T\) \(\subseteq\) \(\displaystyle \map \complement S\) Definition of Subset
$\blacksquare$
By definition of set complement:
$\map \complement T := \relcomp {\mathbb U} T$
where:
$\mathbb U$ is the universe $\relcomp {\mathbb U} T$ denotes the complement of $T$ relative to $\mathbb U$. Thus the statement can be expressed as: $S \subseteq T \iff \relcomp {\mathbb U} T \subseteq \relcomp {\mathbb U} S$ The result then follows from Relative Complement inverts Subsets.
$\blacksquare$
Sources 1955: John L. Kelley: General Topology... (previous) ... (next): Chapter $0$: Subsets and Complements; Union and Intersection: Theorem $1$ 1960: Paul R. Halmos: Naive Set Theory... (previous) ... (next): $\S 5$: Complements and Powers 1967: George McCarty: Topology: An Introduction with Application to Topological Groups... (previous) ... (next): $\text{I}$: Exercise $\text{B x}$ 1971: Robert H. Kasriel: Undergraduate Topology... (previous) ... (next): $\S 1.6$: Set Identities and Other Set Relations: Exercise $1 \ \text{(b)}$ 1975: Bert Mendelson: Introduction to Topology(3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 3$: Set Operations: Union, Intersection and Complement: Exercise $1 \ \text{(e)}$
|
This is an interesting and not so easy question. Here's my 2 cents:First, you should distinguish between mathematical models for the dynamics of an underlying asset (Black-Scholes, Merton, Heston etc.) and numerical methods designed to calculate financial instruments' prices under given modelling assumptions (lattices, Fourier inversion techniques etc.). ...
Along with Gatheral's book, I'd recommend reading Lorenzo Bergomi's "Stochastic Volatility Modelling". The first 2 chapters are available for download on his website. That being said, let me try to give you the basic picture.Below we assume that the equity forward curve $F(0,t)=\Bbb{E}_0^\Bbb{Q}[S_t]$ is given for all $t$ smaller than some relevant ...
Here's a research note devoted to pricing of CMS by means of a stochastic volatility model. The authors indicate in the Introduction thatan analysis of the coupon structure leads to the conclusion that CMS contracts are particularly sensitive to the asymptotic behavior of implied volatilities for very large strikes. Market CMS rates actually drive the ...
Okay just to wind things down here, I think an important clarification is needed if readers might come and seek to a similar solution.The Geometric Brownian Motion (GBM) is a model of asset prices dynamics which is usually given as follows:$$ dS_t = \mu S_t dt + \sigma S_t dB_t$$where $B_t$ is a standard brownian motion which has several important ...
1. What does it mean by the vol surface is the current view of vol?The local volatility model is calibrated to vanillas prices (and equivalently their implied volatilities), which reflect the market's view of the volatility, in order to use it to use it to price other options that one will hedge with the vanillas.Where a Black-Scholes model (no smile) ...
To recover the Black-Scholes pricing equation, you should first express the standard normal cdf in terms of its characteristic function analogous to the Heston solution:$$N(x) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} Re [\frac{e^{-i\phi x} f(\phi)}{i\phi}] d\phi$$where $f(\phi)$ is the characteristic function of the standard normal distribution:$$...
The way that I understand your question is that you are looking to fit the market prices of European plain vanilla options of a single maturity and then back out the corresponding implied probability density function. There are multiple ways that you could approach your problem.1) Modelling the Market PricesThe market prices of European plain vanilla ...
$X_t$ being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function $f$ of $t$ and $X_t$.Instead one should turn to Itô's lemma, one of the key results of stochastic calculus, which stipulates (assuming $X_t$ is here a continuous, square integrable stochastic process)$$ df(t,X_t) = \frac{...
[Short answer]No closed-form formula in general. You need to resort to numerical methods. Monte Carlo is preferred by most practitioners but you could also use Finite Difference schemes (and sometimes even Fourier inversion techniques depending on the model used and the instruments to be priced).[Long answer]One usually distinguishes between 2 classes ...
Some NotationsIt's easy to get lost so let's introduce some notations and let$$ \sigma : (t, S, K, \tau) \to \sigma(K,\tau; S, t) $$denote the implied volatility smile prevailing at time $t$ when the spot price is $S_t=S$ for an option with strike level $K$ and time to expiry $\tau=T-t$. From here onward, we drop the $t$ argument to keep notations ...
Let $dS_t = \mu_tS_tdt + \sigma_tS_tdW_t$ be the underlying GBM (Geometric Brownian Motion)-like dynamics as in the question.Let $B_t$ a Brownian motion such that $d[B,W]_t = \rho dt$, $\rho\in[-1,1].$CIR (Cox-Ingersoll-Ross) for $\sigma_t^2$ (when combined with GBM-like underlying dynamics, it is the popular Heston SV model)$$d\sigma_t^2 = \kappa(\...
Whenever you use any model to price anything, all you need to do is make sure you model the underlying dynamics that the product you're pricing actually depends on.Any product will be dependent on numerous facets, to varying degrees - this is the same with modelling anything.The modelling that happens in pricing financial derivatives is an integration ...
I'm guessing, and correct me if I'm wrong, you want to create a number of possible paths the stock price could follow with the local volatilty given by GARCH depending on the simulated history, or in pseudocode:N <- numberOfPathsT <- numberOfStepsfor (i in 1:N) {newSeries <- pastPricesfor (t in 1:T) {epsilon <- normrnd(0,1)...
You have to ask yourself what the ultimate purpose of this parameterization is. In their case, they imply the "end-goal is martingale pricing or maximum-likelihood estimation", both of which are ultimately about capturing long-period dynamics rather than intraday or interday behavior.For this reason, the fact that intraday variance may, ahem, vary around ...
There are lots of papers online and here are a few I would suggestmath.umnriskworxG. Dimitroff, J. de KockNowak, SibetzI you have matlab there is an step step example to calibrate SABR model. Since it uses the financial toolbox of matlab for a few functions I dont think you can replicate it in any other language. There must be C++ code available ...
I am not sure if I understood your question correctly but I will try to answer it anyway.If you have a standard normal random vector $z \sim N(\mathbb{0},I_n)$ (where $z,0 \in \mathbb{R}^{n\times1}$ and $I_n \in \mathbb{R}^{n\times n}$ is the identity matrix) and you want to transform it into a multivariate normal $x \sim N(\mu,\Sigma)$ you do it the ...
This effect is coming from the supply and demand in the options markets. Many portfolio managers want (or need) to buy out of the money put options, and many are willing to sell out of the money call options (thereby funding the purchase of put options). Now, when the market goes down, dealers find themselves short vol and they need to buy options to cover ...
I think you did something wrong in translating the input to numerics. As pointed out by dm63 normal vols are quoted in basis points.Using equation A.67a) from the Hagan paper you linked we see (setting $\beta = 0$)$$\sigma_N(K) = \alpha\frac{\xi}{x(\xi)}\left[1+\frac{2-3\rho^2}{24}\nu^2\tau_{exp}\right]$$where $\tau_{exp} = 0.25$ in your example and...
In the context of option pricing, "implied volatility" always refers to the equivalent diffusion coefficient in the geometric Brownian motion (GBM) dynamics that is necessary to match an observed European plain vanilla price for a given strike and maturity.When talking about "model implied volatility smile", what is meant is that:You choose some pricing ...
The following paper is helpful for understanding the point you raise:Hagan et al.: Managing Smile Risk, January 2002, Wilmott 1:84-108The main point is given in the paper:[...] the dynamics of the market smile predicted by local vol modelsis opposite of observed market behavior: when the price of theunderlying decreases, local vol models ...
Well, what you find is that the introduction of stochastic vol changes the delta of your options. So what does this mean? If the new delta reduces the variance of your hedged portfolio versus the pure local vol model , then it means that the introduction of stochastic vol has resulted in a better description of market dynamics versus the pure local vol ...
For pricing, there are a few products whose prices are sensitive to the forward smile and when you compute that with just local vol, it is not realistic.So if you are a seller, you go to the next church and find something that looks kindof reasonable, and that kind of can reconstruct a reasonnable forward smile structure.The game in pricing is to not ...
I am going to supply an answer that is quite similar to SRKX's (which is very very good) because I want to discuss in more detail a few important things. First, you cannot use a stochastic volatility model for the SDE that you've provided as that's GBM with constant diffusion. However, based on what you've said it's obvious you wish to model a discretized ...
You could read it like this:The typical change in equity value is equal to the typical change in asset value, adjusted for the probability of the assets surviving.Note that the formula is not specific to Merton models, it's also true for regular options and their underlyings. It's just that volatility of option prices isn't typically a concern in "...
It's really quite simple. It's just a matter of the fact that we can change measure on the stochastic volatility while not changing the fact that the stock is a martingale. Once we can do this, we have payoffs that have different values under different measures, so the market can't be complete.For clarity, just consider a stock S, a money market account ...
CRR is just a numerical approximation to Black--Scholes. Its main use is in getting American option price. There is no real difference other than slight inaccuracy when using it for Europeans. So no it wouldn't do what you ask.Your questions are philosophical. What is the purpose of the model? if you estimate the volatility from a time series then you can ...
The local vol model has exactly enough freedom to match the individual densities $X_t.$ There is no additional freedom in the local vol model to match even a joint density for a pair of times $(X_t,X_s).$When you ask about the joint density across the continuum of times $t \in [0,T]$ it is pretty easy to show that any local vol model differs from any ...
Quick summary: Your model should still be well specified, as long as:1) You do the analysis on a heavily traded asset, e.g. IBM on NYSE, and2) You use heteroskedasticity-consistent standard errors in your estimation framework, e.g. White's standard errors.I'm going to start the long answer by re-stating the question to make sure I've got it right.Let ...
Because vanilla derivatives with European exercise depend only on total variance , not on it's dynamics in time.If you have a simpler model (like interpolation of these total variances from your volatility surface) you don't have as much of unobservable parameters stochastic volatility models have.Having more parameters (which many times would need to be ...
|
The big issue, I think, is that you need a model of the data to work with. These could be regression models or simply your data fit to a distribution. Your data could theoretically be consistent with a number of models / distributions (such as the normal distribution, proportions of successes arising from a binomial distribution with the appropriate total numbers of trials, or a beta distribution). Since you say that these are fake data just to illustrate the problem, and that you have continuous proportions, I will simply go with the idea that these proportions come from a beta distribution.
In
R, you can model beta distributed data using the betareg package. You should read the vignette (pdf), which is very helpful for understanding the ideas about beta regression and how to use the package to do it. Note in particular that they parameterize the beta distribution differently than the typical parameterization. Instead of using $\alpha$ and $\beta$ (which are, e.g., the
shape1 and
shape2 arguments in the standard ?pbeta, etc., functions), they use the mean proportion, $\mu$, and the precision, $\phi$, which are defined as follows:
\begin{align}\mu &= \frac{\alpha}{(\alpha + \beta)} &\text{and,}& &\phi = \alpha + \beta \\\ &\ &\ & &\ \\ & &\text{thus,}& & \\\ &\ &\ & &\ \\\alpha &= \mu\times\phi &\text{and,}& &\beta = \phi - \alpha\end{align}
We can use these equations to move between the two parameterizations and therefore the two sets of functions. Then we can link the modeling in
betareg to the distribution and quantile functions. The operative transformation will be based on an analogy to how qq-plots can be used to compare two distributions.
First, let's model these data with
betareg and determine if the precision is constant between the groups:
library(betareg)
library(lmtest)
# test of constant precision based on beta distribution:
cp = betareg(x~f, df)
vp = betareg(x~f|f, df)
lrtest(cp, vp)
# Likelihood ratio test
#
# Model 1: x ~ f
# Model 2: x ~ f | f
# #Df LogLik Df Chisq Pr(>Chisq)
# 1 3 9.2136
# 2 4 9.4793 1 0.5314 0.466
For these (fake) data, there is little reason to believe that the precision differs, but for your real data, they might. So I will demonstrate the more complicated version, which you could simplify if you want. At any rate, the next step is to determine the estimated $\alpha$ (
shape1) and $\beta$ (
shape2) parameters for
A and
B by converting the model's coefficients to the distribution parameters (don't forget the link functions!) and using the above formulas to convert between the two parameterizations:
summary(vp)
# ...
# Coefficients (mean model with logit link):
# Estimate Std. Error z value Pr(>|z|)
# (Intercept) -1.3153 0.2165 -6.074 1.25e-09 ***
# fB 1.4260 0.3178 4.487 7.23e-06 ***
#
# Phi coefficients (precision model with log link):
# Estimate Std. Error z value Pr(>|z|)
# (Intercept) 3.0094 0.5695 5.284 1.26e-07 ***
# fB -0.5848 0.7943 -0.736 0.462
# ...
lo.to.prop = function(x){
o = exp(x)
p = o / (o+1)
return(p)
}
alpha.A = lo.to.prop(coef(vp)[1] ) * exp(coef(vp)[3] )
alpha.B = lo.to.prop(coef(vp)[2]+coef(vp)[1]) * exp(coef(vp)[4]+coef(vp)[3])
beta.A = exp( coef(vp)[3] ) - alpha.A
beta.B = exp( coef(vp)[4]+coef(vp)[3]) - alpha.B
Due to the complexity of this, we might do a common-sense check on the final values. (This also suggests another--simpler--way to get these parameters, if you don't need to model them and/or don't need to test if the precisions are the same.)
cbind(c(alpha.A, beta.A), c(alpha.B, beta.B))
# [,1] [,2]
# (Intercept) 4.290073 5.960932
# (phi)_(Intercept) 15.984533 5.336616
# parameterization for beta distribution:
library(fitdistrplus)
fitdist(with(df, x[f=="A"]), "beta")
# Fitting of the distribution ' beta ' by maximum likelihood
# Parameters:
# estimate Std. Error
# shape1 4.290883 2.389871
# shape2 15.987505 9.295580
fitdist(with(df, x[f=="B"]), "beta")
# Fitting of the distribution ' beta ' by maximum likelihood
# Parameters:
# estimate Std. Error
# shape1 5.960404 3.379447
# shape2 5.335963 3.010541
Now, we convert the raw proportions in
B to their corresponding beta probability (that is, how far through their own beta distribution they lie), and convert those probabilities to quantiles of
A's beta distribution:
p.B = pbeta(df$x[df$f=="B"], shape1=alpha.B, shape2=beta.B)
q.B.as.A = qbeta(p.B, shape1=alpha.A, shape2=beta.A)
For a final step, let's take a look at the new values:
cbind(df$x[df$f=="B"], q.B.as.A)
# q.B.as.A
# [1,] 0.50 0.18502881
# [2,] 0.30 0.08784683
# [3,] 0.45 0.15784496
# [4,] 0.60 0.24648412
# [5,] 0.54 0.20838604
# [6,] 0.77 0.38246863
|
Yield Surfaces and Plastic Flow Rules in Geomechanics
In order to ensure safe geotechnical building methods, specific applications require certain foundations and structure reinforcements. Tests are quite expensive to carry out, so simulation can be really useful and even essential. Many numerical models have been developed to give a deep insight into soil behavior. Here, we introduce the most widespread models for soils available in COMSOL Multiphysics and analyze a tunnel excavation example.
Quick Note on Geotechnical Engineering
There seems to be a general trend when it comes to construction. Offshore structures are constructed in deeper and deeper waters; buildings are constructed increasingly close to each other; offshore wind turbines are developed in deep waters far away from the coasts, which are likely to experience extreme loading conditions. Therefore, in recent decades, geotechnical engineers have developed numerical simulations to cope with this construction trend and ensure safe building methods.
“Paris Metro construction 03300288-3”. Licensed under Public domain via Wikimedia Commons. Plasticity and Geomaterials
Materials for which strains or stresses are not released upon unloading are said to behave
plastically. Several materials behave in such a manner, including metals, soils, rocks, and concrete, for example. These give rise to an elastic behavior up to a certain level of stress, the yield stress, at which plastic deformation starts to occur.
The elastic-plastic behavior is path-dependent and the stress depends on the history of deformation. Therefore, the plasticity models are usually written connecting the
rates of stress, rather than stress, and the plastic strain. The most widespread and well-known plasticity model throughout the industry is based on the von Mises yield surface for which plastic flow is not altered by pressure. Therefore, the yield condition and the plastic flow are only based on the deviatoric part of the stress tensor.
However, this model is no longer valid for soil materials since frictional and dilatation effects need to be taken into account. Let’s see how this can be worked out and briefly explain the different soil plasticity models available in the COMSOL Multiphysics® simulation software.
Plasticity of Soils and Rocks
For materials such as soils and rocks, the frictional and dilatational effects cannot be neglected. This whole class of materials is well known to be sensitive to pressure, leading to different tensile and compressive behaviors. The von Mises model presented above is thus not suitable for these types of materials. Instead, yield functions have been worked out to take into account the behavior of frictional materials.
Let’s illustrate the frictional behavior and plastic flow for these materials by considering the block shown here:
The block is loaded as shown by a normal load N and a tangential load Q. Assuming that the block rests on a surface with a coefficient of static friction \mu, according to Coulomb’s law, the maximum force that the block can withstand before sliding is given by F=\mu N. Therefore, the onset of sliding occurs when the following condition is reached:
(1)
The direction of sliding is horizontal. For tangential loads such as f<0, the block will not slide, but as soon as f=0, the block will slide in the direction of the applied load Q. The
Mohr-Coulomb criterion — the first soil plasticity model ever developed — is a generalization of this approach to continuous materials and a multiaxial state of stress. It is defined such that yielding and even rupture occur when a critical condition that combines the shear stress and the mean normal stress is reached on any plane. This condition is stated as below:
(2)
Here, \tau is the shear stress, \sigma is the normal stress, c is the cohesion representing the shear strength under zero normal stress, and \mu=\tan\phi is the coefficient of internal friction coming from the well-known Coulomb model of friction. This equation represents two straight lines in the Mohr plane. A state of stress is safe if all three Mohr’s circles lie between those lines, while it is a critical state (onset of yielding) if one of the three circles is tangent to the lines.
Mohr-Coulomb yield behavior. The Mohr circles are based on the principal stresses \sigma_1, \sigma_2, and \sigma_3. As you can see, one of the circles is tangential to the yield surface, and so the onset of yielding is occurring.
According to the figure above, the stress state is given by \tau=\frac{1}{2}(\sigma_1-\sigma_2)\cos\phi and \sigma=\frac{1}{2}(\sigma_1+\sigma_3)+\frac{1}{2}(\sigma_1-\sigma_3)\sin \phi. The yield criterion and Equation 2 can therefore be re-written in a generalized form as follows:
(3)
It can even be seen as a particular case of a more general family of criteria based on Coulomb friction and written by equations based on invariants of the stress tensor:
(4)
Representation of the Mohr-Coulomb yield function.
The
Mohr-Coulomb criterion defines a hexagonal pyramid in the space of principal stresses, which makes it straightforward for this criterion to be treated analytically. But, the constitutive equations are difficult to handle from a numerical point of view because of the sharp corners (for instance, the normal of this yield surface is undefined at the corners).
In order to avoid the issue associated with the sharp corners, another yield criterion of this family, the
Drucker-Prager yield criterion, has been developed by modifying the von Mises yield criterion to take into account the Coulomb friction, i.e., incorporate a hydrostatic pressure dependency:
(5)
This represents a smooth circular cone in the plane of principal stress, rather than a hexagonal pyramid. If the coefficients \alpha and k are chosen such that they match the coefficients in the
Mohr-Coulomb criterion, as follows:
(6)
the
Drucker-Prager yield surface passes through the inner or outer apexes of the Mohr-Coulomb pyramid, depending on whether the symbol \pm is positive or negative. The plastic flow direction is taken from the so-called “plastic potential”, which can be either the same, associative plasticity, or different, non-associative plasticity, than the onset of yielding (the yield function). Many different non-associative flow rules can be developed.
Using an associative law for the Drucker-Prager model leads the volumetric plastic flow to be nonzero. Therefore, there is a change in volume under compression. However, this is contradictory to the behavior of many soil materials, particularly granular materials. Instead, a non-associative flow rule can be used such that the plastic behavior is isochoric (volume preserving) — a much better reflection of the plastic behavior of granular materials.
Representation of the Drucker-Prager yield function. Non-Associative Law for Soil Plasticity in COMSOL Multiphysics
Next, I will show you how to use a non-associative law for soil plasticity in COMSOL Multiphysics. Non-associative plastic laws can be used regardless of the plasticity model used in the software.
If you’re using the Mohr-Coulomb model, there are basically two different approaches to handling non-associative plasticity. The plastic potential can either be taken from the Drucker-Prager model or be the same as the
Mohr-Coulomb yield function but with a different slope with respect to the hydrostatic axis, i.e., the angle of friction is replaced by the dilatation angle (see screenshot below).
Moreover, when using the Drucker-Prager matched to a
Mohr-Coulomb criterion, it is easy to adapt the dilatation angle to match with the non-associative law that you want to use. For instance, the non-associative law presented above can be worked out by taking the dilatation angle null.
Last but not least, a useful feature called
elliptic cap has been developed to avoid unphysical behavior of the material beyond a certain level of pressure. Indeed, real-life material cannot withstand infinite pressure and still deform elastically. Therefore, to cope with this, we can use the elliptic cap feature available in COMSOL Multiphysics. Soil Plasticity feature settings window.
Let’s try to put into practice everything we’ve learned so far by analyzing the example of a tunnel excavation. This will also be an opportunity to figure out what the effects of the different features we mentioned above are.
Example of a Tunnel Excavation
The simulation of a tunnel excavation process is especially important in predicting the necessary reinforcements that the workers need to use to avoid the collapse of the construction.
The following model aims to simulate the soil behavior during a tunnel excavation. The surface settlement (i.e., the vertical displacement along the free ground surface) and the plastic region are computed and compared between the different soil models used to carry out this simulation. The geometry we’ll use is presented in the figure below. To make our model realistic, infinite elements have been used to enlarge the soil domain, while keeping the computational domain small enough to get the solution in a relatively short time.
The geometry consists of a soil layer that is 100 meters deep and 100 meters wide plus 20 meters of infinite elements. A tunnel 10 meters in diameter is placed 10 meters away from the symmetry axis and 20 meters below the surface.
First of all, we need to add the in-situ stresses in the soil before the excavation of the tunnel. Then, we can compute the elastoplastic behavior once the soil corresponding to the tunnel is removed. The in-situ stresses must be incorporated in this second step. This is fairly straightforward to set up in COMSOL Multiphysics.
We can begin by adding a stationary step where the in-situ stresses will be computed. Then, in a second step but still within the same study, we add a soil plasticity feature. Finally, we compute the solution. In order to get the pre-stresses incorporated into the second step, we should add an Initial Stress and Strain feature under the
Solid Mechanics interface, as shown below. Initial Stress and Strain feature used to incorporate the in-situ stresses from the first step as initial stresses for the second step, during which excavation occurs. The variables solid.sx, solid.sxy, etc. are the x -components of the stress tensor, the xy -components of the stress tensor, etc.
The first plot shows the in-situ stresses computed from the first step. These stresses result from the gravity load.
The von Mises stress in the soil before the excavation of the tunnel.
The second plot shows the stress distribution after excavating the tunnel. In-situ stresses are taken from the first step. Note, as expected, the increase in the von Mises stress around the tunnel as well as the deformation of the tunnel shape.
The von Mises stress in the soil after excavation of the tunnel.
As mentioned previously, while removing the tunnel domain, a plasticity feature is added and the soil experiences a plastic behavior. This is depicted in the figure below of a Drucker-Prager model with associative plastic flow. The plastic region is concentrated around the near surroundings of the tunnel. The analysis of this region is quite important in gaining insight into how the soil is more likely to deform. Therefore, it allows us to handle the necessary reinforcements in order to avoid collapse and get the desired tunnel shape.
Plastic region after excavating the tunnel.
This tunnel excavation simulation has been carried out in four different cases in order to compare the different soil models presented in the previous section as well as understand the influence of the cohesion on the soil’s behavior. The results are shown by taking the surface settlement as the criterion.
Below, we have a 1D plot from which we can observe the following relationship: The lower the cohesion, the greater the deformation. We can also note that the Mohr-Coulomb model tends to, somehow, make the soil stiffer than the Drucker-Prager model. The non-associative law with a null dilatation angle prevents the soil from dilating under compression and so the surface settlement becomes greater.
Surface settlement comparison between different plasticity models and material properties. Further Reading
There are also a couple of other plasticity models for soil, rocks, and concrete available in COMSOL Multiphysics. Please check out the links below to get further information about geotechnical simulations and the Geomechanics Module of COMSOL Multiphysics.
Also, be sure to watch the video on how to build a model of an excavation:
Comments (1) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
|
Let $H \subset G$ be a closed subgroup of a lie group and $G/H$ the homogeneous coset space. There's an exact sequence of adjoint representations of $H$:
$$0 \to \mathfrak{h} \to \mathfrak{g} \to \mathfrak{g/h}\to 0 $$
The canonical principal $H$-bundle $G \to G/H$ gives an exact functor from representations of $H$ to vector bundles over $G/H$. The corresponding sequence of vector bundles is:
$$0 \to G \times_H \mathfrak{h} \to G \times_H \mathfrak{g} \to G \times_H \mathfrak{g/h}\to 0 $$
I'm pretty convinced that $G\times_H \mathfrak{g/h}$ is canonically isomorphic to $T(G/H)$ but a simple proof alludes me so help here would be welcome.
Taking the projective space as an example we have $\mathbb{CP}^n \cong \frac{U(n+1)}{U(n)\times U(1)}$ and the corresponding exact sequence:
$$0 \to U(n+1) \times_{U(n)\times U(1)} \mathfrak{u(n)\times u(1)} \to U(n+1) \times_{U(n)\times U(1)} \mathfrak{u(n+1)} $$
$$\to U(n+1) \times_{U(n)\times U(1)} \mathfrak{\frac{u(n+1)}{u(n)\times u(1)}}\to 0 $$
Where the last vector bundle is the tangent bundle $T\mathbb{CP}^n$.
How does this relate to the euler sequence if at all?
$$0 \to \mathcal{O}_{\mathbb{P}^n} \to \mathcal{O}_{\mathbb{P}^n}(1)^{n+1} \to \mathcal{T}_{\mathbb{P}^n} \to 0$$
|
Answer
$\cos{(-\frac{7\pi}{6})} = -\frac{\sqrt3}{2}$
Work Step by Step
Recall: (1) An angle and its reference angle either have the same trigonometric function values or differ only in signs. (2) $\frac{\pi}{6}$ is a special angle whose cosine value is known to be $\frac{\sqrt3}{2}$. Note that the reference angle of $-\frac{7\pi}{6}$ is $\frac{\pi}{6}$. However, since $-\frac{7\pi}{6}$ is in Quadrant II, then its cosine value is negative. Therefore, $\cos{(-\frac{7\pi}{6})} = -\frac{\sqrt3}{2}$
|
Background of my question is an idea for generating an initial subtour for general symmetric TSPs:
Add to a MST a set of edges with minimal weight sum, that renders the resulting graph free of articulation points remove from that graph all MST-edgesthat appear in at least two shortest paths in the MSTbetween a pair of vertices that is adjacent to the same edge that has been added in the previous step.
The plural for MST was used intentionally to cover modifications due to adding vertex weights.
It is not too hard to come up with an LP formulation of determining the edges of a weight-optimal augmentation of MSTs to biconnected graphs:
for every $(t_u,t_v)$ edge in the MST there must be at least one added edge $e_{ij}\in E\setminus MST$ for which the path $\pi(i,j)$ connecting vertices $i$ and $j$ in the MST contains $(t_u,t_v)$.
Enumerating all such external edges for a given tree-edge $(t_u,t_v)$ is also straight forward:
removing $(t_u,t_v)$ from MST results in two connected components, possibly with an isolated vertex, of which the vertex sets define a complete bipartite graph $K(t_u,t_v) := K(t_v,t_u)$ of which all edges, of course except $(t_u,t_v)$, qualify.
The LP formulation is then
$$\min_{x_{ij}\in [0,1]} \sum_{e_{ij}\in E\setminus MST} {x_{ij}e_{ij}}$$ so that $$\sum_{e_{ij}\in K(t_u,t_v)}{x_{ij}} \ge 1$$
Questions:
has such an augmentation of MSTs already been studied, especially in the context of TSP heuristics of general applicability?
is the solution of the LP formulation integral?
|
I want to solve the problem below
\begin{equation} \begin{aligned} \eta u-\Delta u &=f, &\text{in $\Omega$}\\ \end{aligned} \end{equation}
where $\Omega=(0,1)\times (0,1)$, My boundary are like this $\frac{\partial u}{\partial n} = 0$ if $y=0$ or $1$, and $u=0$ if $x=0$ or $1$. Using ghost nodes can handle the low order discretization of the neumann boundary, but the problem that i'm facing is that I want to include the four corner points in the neumann boundary not in the dirichlet one, and this causes me problem since I cannot use the interior equation on the corners.
I hope my question was clear.
|
Linear Regression with Evenly-Spaced Abscissae
What a boring title. I wish I could come up with something snazzier. One word I learned today is studentization, which is just the normalization of errors in a curve-fitting exercise by the sample standard deviation (e.g. point \( x_i \) is \( 0.3\hat{\sigma} \) from the best-fit linear curve, so \( \frac{x_i - \hat{x}_i}{\hat{\sigma}} = 0.3 \)) —
Studentize me! would have been nice, but I couldn’t work it into the topic for today. Oh well.
I needed a little break from our tour of linear feedback shift registers — and maybe you do too — so I thought I’d share a little insight in curve-fitting I ran across about ten years ago. What we’re going to do today is do a bunch of linear algebra, so that we can handle all the yucky math ahead of time, and then compute something useful and simple without having to touch the linear algebra again.
Let’s look at a linear regression problem. Ugh. Statisticians need to band together and come up with some more exciting terms. All this means is to fit a line to some data.
import numpy as np import matplotlib.pyplot as plt %matplotlib inline np.random.seed(123) t = np.arange(0,1,0.05) n = np.random.randn(*t.shape)*0.02 y = 0.8*t*t + t + n plt.plot(t,y,'.') # Fit a line and a parabola, and plot them t_theo = np.linspace(0, 1, 1000) for deg in [1,2]: p = np.polyfit(t,y,deg) print "Degree %d: p=%s" % (deg,p) plt.plot(t_theo, np.polyval(p,t_theo),label='deg=%d'%deg) plt.legend(loc='best', fontsize=10); Degree 1: p=[ 1.79183292 -0.12683228] Degree 2: p=[ 0.86334813 0.9716522 -0.00380517]
The
numpy.polyfit function is the easy thing to use when fitting any polynomial (linear or not). It uses a least-squares fit, something explored by Carl Friedrich Gauss, and since today is Gauss’s birthday, it seems like a fitting tribute. (Pun intended.)
Here comes the math:
We call
numpy.polyfit on a bunch of data \( (x_i, y_i) \) and we get a vector of coefficients \( p_k \) such that the sum of the squared errors \( E \) is minimized:
$$E = \sum\limits_i (\hat{y}_i - y_i)^2$$
with the estimated value \( \hat{y}_i \) defined as the evaluation of the polynomial at each point:
$$\hat{y}_i = p(x_i) = p_0 + p_1x_i + p_2x_i{}^2 + \ldots + p_nx_i{}^n$$
(Note:
polyfit returns the coefficients from most-significant to least-significant, e.g. it gives you a vector \( \begin{bmatrix}p_n & p_{n-1} & \ldots & p_2 & p_1 & p_0\end{bmatrix} \). Just make sure you’re aware of this order.)
Okay, here I am talking about least-squares, but a few years back I wrote an article on Chebyshev approximation, which is slightly different. Here’s the thing to note:
When you are approximating a functionand trying to fit a polynomial to it, so that you can choose the x-coordinates (abscissae) of the data, and the error is approximation error rather than noise, then Chebyshev approximation is the thing to use. When you are trying to fit a polynomial to noisy data that you’re just stuck with — well, think twice; I would not recommend more than cubic — then least-squares is the best approach. If you have noisy data from an experiment where you have the freedom to choose the x-coordinates — for example, if you are trying to understand the relationship for a sensor between the physical quantity it is sensing, and its output signal — then it’s kind of a gray area whether to use evenly-spaced points or Chebyshev nodes. As the number of samples goes up, the difference between the approaches goes down, and for 10 or more samples, I would just as well choose evenly-spaced data for its neatness.
Basically, Chebyshev approximation exploits some of the smoothness properties of ordinary functions to come up with a result that approximately minimizes the maximum error over an interval. With least-squares approximation, we don’t care about the maximum error so much as the sum of the squares of the error. If you’ve read this and are scratching your head, just forget I mentioned the Chebyshev stuff, and move on.
What does
polyfit do? Well, under the hood it uses least squares to fit a linear sum of basis functions, where the basis functions are \( 1, x, x^2, \ldots, x^n \).
This is the essence of least-squares analysis: given samples \( (x_i, y_i) \), along with some choice of basis functions \( a_0(x), a_1(x), a_2(x), \ldots a_{n-1}(x) \) — watch out, here there are \( n \) terms whereas with \( n \)th degree polynomials there are \( n+1 \) terms — then, through the magic of linear algebra, we can find coefficients \( c_0, c_1, c_2, \ldots c_{n-1} \) such that the best fit function \( \hat{f}(x) = \sum\limits_{j=0}^{n-1} c_ja_j(x) \) minimizes the total squared error \( E = \sum\limits_i (y_i - \hat{f}(x_i))^2 \).
Essentially \( E \) is a “score” that we try to minimize over the entire sample set. The resulting function \( \hat{f}(x) \) usually isn’t going to pass through any of the points, but it stays nearby enough to keep \( E \) as low as possible.
The math to do this is fairly straightforward. You form the following matrix:
$$\mathbf{A} = \begin{bmatrix} a_0(x_0) & a_1(x_0) & a_2(x_0) & \ldots & a_{n-1}(x_0) \cr a_0(x_1) & a_1(x_1) & a_2(x_1) & \ldots & a_{n-1}(x_1) \cr a_0(x_2) & a_1(x_2) & a_2(x_2) & \ldots & a_{n-1}(x_2) \cr \vdots & \vdots & \vdots & & \vdots \cr a_0(x_{m-1}) & a_1(x_{m-1}) & a_2(x_{m-1}) & \ldots & a_{n-1}(x_{m-1}) \cr \end{bmatrix}$$
so that each column represents one basis function evaluated at all the sample points, and each row represents one sample point evaluated using all of the functions, and then we have the matrix equation
$$\mathbf{r} = \mathbf{y} - \mathbf{A}\mathbf{c} $$
where
$$\mathbf{y} = \begin{bmatrix} y_0 \cr y_1 \cr y_2 \cr \vdots \cr y_{m-1} \end{bmatrix}, \qquad \mathbf{c} = \begin{bmatrix} c_0 \cr c_1 \cr c_2 \cr \vdots \cr c_{n-1} \end{bmatrix}$$
It turns out that this has the least-squares solution (which minimizes the vector norm of the residual \( \|\mathbf{r}\|^2 \)) of
$$\mathbf{A}^T\mathbf{A}\mathbf{c} = \mathbf{A}^T\mathbf{y}$$
or
$$\mathbf{c} = (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{y}.$$
Anyway, the
numpy.linalg.lstsq function will do all that for you. You just need to form the \( \mathbf{A} \) matrix yourself. The
numpy.polyfit function will do that part for you, too.
If we wanted to use
numpy.linalg.lstsq instead of
numpy.polyfit for the curve-fitting exercise, we could do it ourselves:
# Fit a line and a parabola, and plot them plt.plot(t,y,'.') t_theo = np.linspace(0, 1, 1000) for deg in [1,2]: # First form the A matrix. # Put basis functions in rows and then transpose. A = np.vstack([t**j for j in xrange(deg+1)]).T # Then call lstsq(), and throw away everything but the coefficients p, _, _, _ = np.linalg.lstsq(A,y) print "Degree %d: p=%s" % (deg,p) y_theo = sum(pj*(t_theo**j) for j,pj in enumerate(p)) plt.plot(t_theo, y_theo, label='deg=%d'%deg) plt.legend(loc='best', fontsize=10); Degree 1: p=[-0.12683228 1.79183292] Degree 2: p=[-0.00380517 0.9716522 0.86334813]
There. Same result, we just have the polynomial coefficients in ascending rather than descending order.
Let’s look at the \( \mathbf{A} \) matrix for the quadratic case, and you can see fairly easily that the columns are just the basis functions \( 1 \), \( x \), and \( x^2 \):
A array([[ 1. , 0. , 0. ], [ 1. , 0.05 , 0.0025], [ 1. , 0.1 , 0.01 ], [ 1. , 0.15 , 0.0225], [ 1. , 0.2 , 0.04 ], [ 1. , 0.25 , 0.0625], [ 1. , 0.3 , 0.09 ], [ 1. , 0.35 , 0.1225], [ 1. , 0.4 , 0.16 ], [ 1. , 0.45 , 0.2025], [ 1. , 0.5 , 0.25 ], [ 1. , 0.55 , 0.3025], [ 1. , 0.6 , 0.36 ], [ 1. , 0.65 , 0.4225], [ 1. , 0.7 , 0.49 ], [ 1. , 0.75 , 0.5625], [ 1. , 0.8 , 0.64 ], [ 1. , 0.85 , 0.7225], [ 1. , 0.9 , 0.81 ], [ 1. , 0.95 , 0.9025]])
I used this technique in LFSR Part XIII with a constant and an exponential function.
Now, here comes the interesting part.
Let’s go back to the least-squares solution:
$$\mathbf{c} = (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{y}$$
which we can write as
$$\mathbf{c} = \mathbf{\Gamma}\mathbf{y}$$
where \( \mathbf{\Gamma} = (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T \) is all that yucky matrix junk that depends only on the basis functions and on the x-coordinate values. The matrix \( \mathbf{\Gamma} \) is
not dependent on the y-coordinate values.
And if we have evenly-spaced x-coordinate values \( x_i = i\Delta x+x_0 \), something special happens. Let’s choose basis functions \( a_0(x) = 1 \) and \( a_1(x_i) = (x_i - x_0)/\Delta x - \bar{i} = i - \bar{i} \) with \( \bar{i} = \frac{m-1}{2} \):
for m in xrange(2, 10): a0 = np.ones(m) a1 = np.arange(m, dtype=np.float64) - (m-1)/2.0 A = np.vstack([a0,a1]).T print "m=%d, A=\n%s" % (m, A) gamma = np.dot( np.linalg.inv(np.dot(A.T, A)), A.T) q = gamma[1,1] - gamma[1,0] print "gamma.T=\n%s" % gamma.T print "q=%f\n-------" % q m=2, A= [[ 1. -0.5] [ 1. 0.5]] gamma.T= [[ 0.5 -1. ] [ 0.5 1. ]] q=2.000000 ------- m=3, A= [[ 1. -1.] [ 1. 0.] [ 1. 1.]] gamma.T= [[ 0.33333333 -0.5 ] [ 0.33333333 0. ] [ 0.33333333 0.5 ]] q=0.500000 ------- m=4, A= [[ 1. -1.5] [ 1. -0.5] [ 1. 0.5] [ 1. 1.5]] gamma.T= [[ 0.25 -0.3 ] [ 0.25 -0.1 ] [ 0.25 0.1 ] [ 0.25 0.3 ]] q=0.200000 ------- m=5, A= [[ 1. -2.] [ 1. -1.] [ 1. 0.] [ 1. 1.] [ 1. 2.]] gamma.T= [[ 0.2 -0.2] [ 0.2 -0.1] [ 0.2 0. ] [ 0.2 0.1] [ 0.2 0.2]] q=0.100000 ------- m=6, A= [[ 1. -2.5] [ 1. -1.5] [ 1. -0.5] [ 1. 0.5] [ 1. 1.5] [ 1. 2.5]] gamma.T= [[ 0.16666667 -0.14285714] [ 0.16666667 -0.08571429] [ 0.16666667 -0.02857143] [ 0.16666667 0.02857143] [ 0.16666667 0.08571429] [ 0.16666667 0.14285714]] q=0.057143 ------- m=7, A= [[ 1. -3.] [ 1. -2.] [ 1. -1.] [ 1. 0.] [ 1. 1.] [ 1. 2.] [ 1. 3.]] gamma.T= [[ 0.14285714 -0.10714286] [ 0.14285714 -0.07142857] [ 0.14285714 -0.03571429] [ 0.14285714 0. ] [ 0.14285714 0.03571429] [ 0.14285714 0.07142857] [ 0.14285714 0.10714286]] q=0.035714 ------- m=8, A= [[ 1. -3.5] [ 1. -2.5] [ 1. -1.5] [ 1. -0.5] [ 1. 0.5] [ 1. 1.5] [ 1. 2.5] [ 1. 3.5]] gamma.T= [[ 0.125 -0.08333333] [ 0.125 -0.05952381] [ 0.125 -0.03571429] [ 0.125 -0.01190476] [ 0.125 0.01190476] [ 0.125 0.03571429] [ 0.125 0.05952381] [ 0.125 0.08333333]] q=0.023810 ------- m=9, A= [[ 1. -4.] [ 1. -3.] [ 1. -2.] [ 1. -1.] [ 1. 0.] [ 1. 1.] [ 1. 2.] [ 1. 3.] [ 1. 4.]] gamma.T= [[ 0.11111111 -0.06666667] [ 0.11111111 -0.05 ] [ 0.11111111 -0.03333333] [ 0.11111111 -0.01666667] [ 0.11111111 0. ] [ 0.11111111 0.01666667] [ 0.11111111 0.03333333] [ 0.11111111 0.05 ] [ 0.11111111 0.06666667]] q=0.016667 -------
It turns out that \( \mathbf{\Gamma} \) has a very special form. The coefficients of its first row \( \mathbf{\Gamma}_0 \) are all identical and are equal to \( \frac{1}{m} \). The coefficients of its second row \( \mathbf{\Gamma}_1 \) are a linear series \( (i-\bar{i})\cdot q \) with \( \bar{i} = \frac{m-1}{2} \) and \( q=\frac{12}{m^3 - m} \).
Time to play that Miles Davis again.
OK, for a linear fit we have
$$\mathbf{c} = \begin{bmatrix}c_0 \cr c_1\end{bmatrix} = \begin{bmatrix}\mathbf{\Gamma}_0 \cr \mathbf{\Gamma}_1\end{bmatrix} \mathbf{y}$$
The value \( c_0 \) is just the mean value of the \( y \) values. That’s easy.
The value \( c_1 \) is a weighted sum of \( y \) values, with weights that form an arithmetic progression.
For \( m=5 \), for instance, \( q=\frac{12}{120} = \frac{1}{10} \) and we get \( c_1 = \frac{1}{10}(-2y_0 -1y_1 +0y_2 + y_3 + 2y_4) \). (Odd values of \( m \) have a zero term in the middle.)
For \( m=6 \), we have \( q=\frac{12}{210} = \frac{2}{35} \) and we get \( c_1 = \frac{2}{35}(-\frac{5}{2}y_0 -\frac{3}{2}y_1 -\frac{1}{2}y_2 + \frac{3}{2}y_3 + \frac{5}{2}y_4) \).
Then the best fit line has the equation \( y = c_0 + c_1(i-\bar{i}) = c_0 + \frac{c_1}{\Delta x}(x_i - x_0) - c_1\bar{i} \).
This is all yucky algebra, and kind of hard to wrap your head around it, so let’s try it on a sample data set to see how it works in practice.
x = np.array([1.1, 1.2, 1.3, 1.4, 1.5, 1.6]) y = np.array([3.0, 4.4, 5.2, 6.8, 8.6, 9.9])
Now here we have \( m=6 \) and \( x_i = 1.1 + 0.1\cdot i = x_0 + i\Delta x \) so \( x_0 = 1.1 \) and \( \Delta x = 0.1 \).
We’re going to calculate \( c_0 \) and \( c_1 \) more explicitly here:
m = 6 q = 12.0/(m**3 - m) c0 = np.mean(y) c1 = q*(-2.5*y[0] - 1.5*y[1] - 0.5*y[2] + 0.5*y[3] + 1.5*y[4] + 2.5*y[5]) x0 = 1.1 dx = 0.1 c0, c1 (6.3166666666666664, 1.391428571428571) line_m = c1/dx ibar = (m-1)/2.0 line_b = c0 - line_m*x0 - c1*ibar line_m, line_b (13.914285714285709, -12.467619047619042) plt.plot(x,y,'.') x_theo = np.arange(1,1.7,0.01) plt.plot(x_theo, line_m*x_theo + line_b);
We can do the same thing in general, not just for a specific data set. What this means is that we can write code to calculate the best-fit line slope and offset
without having to mess around with matrices!
Let’s write a general routine:
def linear_fit_equal_spacing(y, x0, dx): """ Calculates best fit linear coefficients slope, ofs such that e = y - (slope*x + ofs) is minimized in a least-squares sense. The values of x must be equally spaced with x = x0 + i*dx. """ m = len(y) w = 1.0 / m q = 12.0 / (m**3 - m) c0 = 0.0 c1 = 0.0 ibar = (m-1)/2.0 for i,y in enumerate(y): c0 += y*w c1 += y*q*(i-ibar) slope = c1/dx ofs = c0 - slope*x0 - c1*ibar return slope, ofs slope, ofs = linear_fit_equal_spacing(y, 1.1, 0.1) print "slope=%f, ofs=%f" % (slope,ofs) plt.plot(x,y,'.') x_theo = np.arange(1,1.7,0.01) plt.plot(x_theo, slope*x_theo + ofs); slope=13.914286, ofs=-12.467619
Tada!
And that’s my little trick to share for today.
There are more general formulas for estimating linear least squares coefficients, where the x-coordinates are not evenly spaced, and they aren’t much harder to calculate. The common formulas involve sums of the first and second moments of data (\( \sum x, \sum y, \sum x^2, \sum y^2, \sum xy \)).
Linear regression is fitting a linear equation with slope and offset to a series of data points. If we use least-squares fitting, there is a fairly simple set of equations that minimizes the sum of squared errors between the actual data points and the value predicted by the linear equation.
In matrix form, least-squares problems can be expressed as minimizing \( \|\mathbf{r}\|^2 \) where \( \mathbf{r} = \mathbf{y} - \mathbf{A}\mathbf{c} \) for an unknown vector \( \mathbf{c} \). This can be solved as
$$\begin{align}\mathbf{c} &= (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T\mathbf{y} \cr &= \mathbf{\Gamma}\mathbf{y}\end{align}$$
where \( \mathbf{\Gamma} \) is dependent only on the choice of basis functions that make up \( \mathbf{A} \) and the x-coordinates at which they are taken.
If we wish to use a linear equation, and those x-coordinates are evenly spaced, then the form of \( \mathbf{\Gamma} = \begin{bmatrix}\mathbf{\Gamma}_0 \cr \mathbf{\Gamma}_1\end{bmatrix} \) is very simple, and can be computed easily without needing any linear algebra libraries. \( \mathbf{\Gamma}_0 \) is a uniform vector that produces a mean value calculation, and \( \mathbf{\Gamma}_1 \) is an arithmetic progression that weights the two endpoints most strongly.
© 2018 Jason M. Sachs, all rights reserved.
Previous post by Jason Sachs:
Linear Feedback Shift Registers for the Uninitiated, Part XIV: Gold Codes
Next post by Jason Sachs:
Linear Feedback Shift Registers for the Uninitiated, Part XV: Error Detection and Correction
Not boring at all. Thanks for the well written explanations.
To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments.
Registering will allow you to participate to the forums on ALL the related sites and give you access to all pdf downloads.
|
The ordinals of infinite time Turing machines The theory of infinite time Turing machines extends the operation of ordinary Turing machines into transfinite ordinal time. At successor stages of computations, the machines compute as expected, according to the rigid instructions of their finite programs, writing on the tape, moving the head to the left or right and changing to a new state. At limit stages, the information the computation was producing is preserved in a sense: each cell of the tape assumes the limsup of its values going into that limit; the head is reset to the left-most cell and the state is placed in the limit state, a distinguished state like the start state and the halt state.
A real is
writable by such machines if there is a program which on trivial input can write that real on the output tape and then halt. A real is eventually writable if there is a program that on trivial input can write the real on the output tape in such a way that from some point on, the output tape exhibits that real as its final stabilized value, even if the machine does not halt. A real is accidently writable if it appears on one of the tapes during the course of a computation of a program on trivial input. See [1, 2, 3]
Similarly, an ordinal is writable or eventually writable or accidentally writable if it is the order type of a relation coded by such a kind of real.
$\lambda=$ the supremum of the writable ordinals $\zeta=$ the supremum of the eventually writable ordinals $\Sigma=$ the supremum of the accidentally writable ordinals
Welch [4, 5] proved the $\lambda-\zeta-\Sigma$ theorem, asserting that $L_\lambda\prec_{\Sigma_1}L_\zeta\prec_{\Sigma_2}L_\Sigma$, and furthermore $\lambda$ is the least ordinal such that $L_\lambda$ has a $\Sigma_1$-elementary end-extension, and $\zeta$ is least such that $L_\zeta$ has a $\Sigma_2$-elementary end-extension.
References Hamkins, Joel David and Lewis, Andy. Infinite time Turing machines.J Symbolic Logic 65(2):567--604, 2000. www arχiv DOI MR bibtex Hamkins, Joel David. Infinite time Turing machines.Minds and Machines 12(4):521--539, 2002. (special issue devoted to hypercomputation) www arχiv bibtex Hamkins, Joel David. Supertask computation.Classical and new paradigms of computation and their complexity hierarchies23:141--158, Dordrecht, 2004. (Papers of the conference ``Foundations of the Formal Sciences III'' held in Vienna, September 21-24, 2001) www arχiv DOI MR bibtex Welch, Philip. The Lengths of Infinite Time Turing Machine Computations.Bulletin of the London Mathematical Society 32(2):129--136, 2000. bibtex
|
How do I prove that, given $A\in\mathbb C^{n\times n}$, if $A$ is irreducible and $\lambda$ is an eigenvalue of $A$ such that $\lambda\in\partial\left(\displaystyle\bigcup_{i=1}^nK_i\right)$, where $K_i$ is the $i$-th Gerschgorin circle and $\partial(\cdot)$ denotes the boundary of the set $\cdot$, then for all $i=1,\dots,n$, $\lambda\in\partial K_i$?
Taussky's theorem gives us that if $B=(b_{ij})$ is irreducible, diagonally dominant -- that is, $$\tag{1}|b_{ii}|\ge\sum_{j=1,\,j\ne i}^n|b_{ij}|$$ for all $i$ --, and the inequality in (1) is strict for at least one $i$, then $B$ is invertible.
Your result follows from this immediately: Let $A=(a_{ij})$ and $\lambda$ be as in your assumptions, and let $B=(b_{ij})$ be the matrix $A-\lambda I$. Note that the $B$ is still irreducible. Your assumption gives us that, for any $i$, $|b_{ii}|=|\lambda-a_{ii}|\ge\sum_{j=1,\,j\ne i}^n|a_{ij}|=\sum_{j=1,\,j\ne i}^n|b_{ij}|$, that is, (1) holds. However, $B$ is not invertible, since $\lambda$ was an eigenvalue of $A$, and therefore $\lambda-\lambda=0$ is an eigenvalue of $B$.
It follows that the inequality in (1) cannot be strict for any $i$, and therefore we must have equality in (1) for all $i$, which means precisely that $\lambda$ is in the boundary of all Geršgorin circles.
Let me sketch Taussky's result, in case you have not seen the proof. The proof starts along the same lines of the proof of Geršgorin's theorem.
Let $B$ be $n\times n$. The argument assumes that irreducibility of $B$ is equivalent to the
strong connectedness of the graph $G(B)$. Recall that $G(B)$ is the directed graph on $\{1,\dots,n\}$ where (for any $s,l$ with $1\le s,l\le n$) there is an arc from $s$ to $l$, $s\to l$, iff $b_{sl}\ne 0$. The graph is connected iff for any two (not necessarily distinct) vertices $s,l$ there is a (non-empty) directed path $s\to j_1\to\dots\to j_m\to l$. Irreducibility of $B$, on the other hand, means that $B$ is not similar via a permutation to a block upper triangular matrix with more than one block.
We may assume $B$ is $n\times n$ with $n>1$. Suppose $B$ is singular, so there is a vector $x=(x_1,\dots,x_n)^T\ne 0$ with $Bx=0$. By replacing $x$ with a nonzero multiple, if needed, we may assume that $$ 1=\max\{|x_k|\mid 1\le k\le n\}. $$
Let $S=\{j\mid |x_j|=1\}$. Since $Bx=0$, we have that $\sum_j b_{kj}x_j=0x_k=0$ for all $k$, that is, $$ -b_{kk}x_k=\sum_{j=1,\,j\ne k}^nb_{kj}x_j $$ for all $k$. In particular, if $k\in S$, we have that $$ \tag{2}|b_{k,k}|\le\sum_{j=1,\,j\ne k}^n|b_{kj}||x_j|\le\sum_{j=1,\,j\ne k}^n|b_{kj}|. $$ By (1), it follows that we actually have an equality here. Since one of the assumptions is that (1) is strict for at least one $i$, it follows that $S\ne \{1,2,\dots,n\}$.
Let $k\in S$. Since $B$ is irreducible, it cannot be that all $b_{kj}$ with $j\ne k$ are $0$. However, since we have equality in (2), if $j$ is such that $b_{kj}\ne0$, then we must have that $|x_j|=1$, that is, $j\in S$. We have reached a contradiction: Let $i\notin S$. By irreducibility, there is a path $$ k\to j_1\to j_2\to\dots\to j_m\to i $$ in the directed graph $G(B)$ associated to $B$. We just proved that if $s\in S$ and there is an arc $s\to l$ in this graph, then $l\in S$ as well. Since $k\in S$, it follows that $j_1\in S$ and, similarly, $j_2\in S,\dots,j_m\in S$ and, finally, $i\in S$. This is a contradiction.
A nice recent reference for this result is the book
Geršgorin and his circles, by Richard S. Varga. (Coincidentally, I was lecturing on this topic a week ago.)
|
In chapter 9 of the book
Toric varieties by Cox-Little-Schenck several cohomology vanishing theorems for toric varieties are proved or mentioned.
In this question I am interested in references for versions (if existing) of the analogous vanishing theorems for toric Deligne-Mumford stacks (especially the smooth ones, e.g. in the sense of Fantechi-Mann-Nironi). I am happy with assuming we are over $\mathbb{C}$ and the stacks have trivial generic stabilizer.
Edit. To answer the question of Jason Starr in the comments, I think I am mostly interested in a stacky version of Batyrev-Borisov vanishing theorem, which I am reporting below. Nevertheless, I would also be glad to receive answers about the other vanishing theorems that I mentioned or alluded to in my question.
Theorem (Batyrev-Borisov).Let $X=X_{\Sigma}$ be a complete toric variety, and $D=\sum_{\rho\in\Sigma(1)} a_\rho D_\rho$ a nef $\mathbb{Q}$-Cartier divisor on $X$. Then
$$H^p(X,\mathcal{O}(-D))=0\;\;\;\;\forall\; p\neq \dim P_D,$$ $$H^{\dim P_D}(X,\mathcal{O}(-D))=\bigoplus_{m\in \mathrm{relint}(P_D)\cap M}\mathbb{C}\cdot \chi^{-m},$$ where $M$ is the lattice of $\Sigma$ and $P_D$ is a polytope defined by
$$P_D=\{ x \in M \otimes\mathbb{R}\;|\; \langle x,u_{\rho} \rangle \leq -a_{\rho} \; \forall \; \rho\in\Sigma (1)\}$$ $u_{\rho}$ being the minimal generator of the ray $\rho$.
|
I believe that the intersection is not always finitely generated as a normal subgroup. The factor-group $F_g$ of $G$ over the normal subgroup $M=\langle a_1,...,a_g\rangle^N$ is freely generated by the images of $b_1,...,b_g$ which we shall denote by the same letters: $b_1,...,b_g$. Since $M$ is finitely normally generated, it is enough to consider your question for the free group $F_g$. Suppose that $g>2$. Let $N_1=\langle b_1\rangle^N, N_2=\langle b_2\rangle^N$. Then $N_i$ consists of all words in $F$ which become trivial after removing $b_i$, $i=1,2$. The intersection of $N_1$ and $N_2$ consists of all words which become trivial both after removing $b_1$ entrees and after removing $b_2$ entrees. I do not have time to check it now, but I think that $N_1 \cap N_2$ is not finitely generated as a normal subgroup. It is generated as a normal subgroup by all commutators $[b_1, b_2^{\pm u}]$ where $u$ is an arbitrary word in the generators of $F_g$. The factor group $F_g/(N_1\cap N_2)$ is not finitely presented. I will elaborate on that when I have time. Or perhaps somebody can do that for me.
Update. Here is the proof that $N=N_1\cap N_2$ is not finitely generated as a normal subgroup. As I wrote above $N$ is normally generated by all commutators $[b_1, b_2^u]$ where $u$ is an arbitrary word in $b_1,,,,,b_g$. For simplicity assume that $g=3$ (the general case is not significantly different) and denote $b_1, b_2, b_3$ by $a,b,c$. I claim that $N$ is normally generated by commutators $[a, b^{\pm c^n}]$, $n\in \mathbb{Z}$. Clearly each of these commutators is in $N$. We need to show that all relations $[a,b^u]=1$ follow from relations $[a, b^{\pm c^n}]=1$. So suppose that all relations $[a, b^{\pm c^n}]=1$ hold in some group $G=\langle a,b,c\rangle$. In particular, $ab=ba$. Consider a relation $[a, b^u]=1$ for some word $u$. Suppose first that $u$ does not contain the letter $a^{\pm 1}$. Since $b^u=b^{b^ku}$ for every $k$, we can assume that the total degree of $b$ in $u$ is 0. Therefore $u^{-1}$ is a product of conjugates of the form $b^{\pm c^n}$. Note that the relation $[a, b^{\pm u}]=1$ is equivalent to the relation $[a^{u^{-1}},b^{\pm 1}]=1$. Since by our assumption $b^{\pm c^n}$ commutes with $a$, the relation $[a^u,b^{\pm 1}]=1$ follows.
Now suppose that $u$ contains occurrences of $a^{\pm 1}$. Since $b$ and $a$ commute, the relation $[a, b^u]=1$ is equivalent to $[a, b^{a^ku}]=1$ for every $k$, hence we can assume that the total degree of $a$ in $u$ is 0. Hence $u$ is a product of conjugates of the form $a^{\pm v}$ where $v$ is a word in $b,c$. By what we prove in the previous paragraph, $a^{\pm v}$ commutes with $b$ (since $v$ does not have occurrences of $a^{\pm 1}$. Hence $b^u=b$ modulo the relations $[a, b^{c^n}]=1$ and we are done.
We need to show that the factor group $G=F_3/N$ is not finitely presented. By our claim, $G$ has the presentation $\langle a,b,c \mid [a, b^{c^n}]=1, n\in \mathbb{Z}\rangle$. But this is the HNN double (extension with centralizer) of the free group $F_3$ along the infinitely generated subgroup $H=\langle b^{c^n}, n\in \mathbb{Z}\rangle$, so $G$ is indeed not finitely presented.
|
I’m trying to build a deterministic Turing Machine that takes 2 words w1, w2 ∈ {a, b, c, d}* separated in the form of Bw1#w2B and determine whether w2 is a substring of w1. Turing machine should reject eg; b#abcd or aaa#b. In other words, the length of w2 must be less than the length of w1 and w2 must be a substring of w1.
I tried to trace this machine with an example but failed to conclude that the machine is really a right shift machine.
I tried to create a right shift machine.
Which one is wrong and what did I miss?
According to Categorifying CCCs: Computation as a Process, computation or β-reduction process in untyped-lambda calculus is in fact a 2-morphism in category theory.
Can someone please describe me how is it so and elaborate on it?
P.S. I understand that category theory is a mathematical concept but since this specific question is about β-reduction and lambda calculus I have posted it in computer science section.
In information theory, we deal with the quantities $ I(X;Y), H(X),H(Y), H(X|Y), H(Y|X)$ . These are just numbers, but I intuitively think of them as the “measure” of a set of information.
There is at least one special case where this interpretation is exact: suppose there are independent variables $ V_1,…V_n$ , and the variables $ X,Y,Z$ are
tuples of $ V_i$ . Then we can literally think of $ I(X;Y)$ as the measure (entropy) of the intersection of $ X$ and $ Y$ .
But it is not obvious to me whether we can define a more general measure of information, such that mutual information can be interpreted as the measure of the intersection, and analogously for the other information quantities.
Is this possible?
Assume you have a game tree and the features(f1, f2, f3…….fn ) that describe the state of the game at any node. Also assume that you are using depth-limited minimax and always expand up to a fixed depth d. Say you have the following four evaluation functions:
F1=w1∗f1+w2∗f2+w3∗f3+⋯+fn
F2=w1∗f1^2+w2∗f2^2+w3∗f3^2+⋯+fn^2
F3=exp{w1∗f1^2+w2∗f2^2+w3∗f3^2+⋯+fn^2}
F4=w1∗f1∗f2+w2∗f2∗f3+w3∗f3∗f4+⋯+wn∗fn−1∗fn
The weights (w1, w2, w3 …… wn) are same for all evaluation functions. Then select all the statements that are correct:
The optimal sequence of moves for the MAX player would be the different for the function F1 and F2.
The optimal sequence of moves for the MAX player would be the different for the function F2 and F3
The optimal sequence of moves for the MAX player would be the different for the function F1 and F3
The optimal sequence of moves for the MAX player would be necessarily the same for all the functions
None of the above are correct
I need help with this problem that I came across. My understanding on this is that unless weights are changed the evaluation function should be the same. that is, in this case statement 2 and 3 are the only ones correct.
Any ideas ?
What are the category theory structures describing queues and topics? By queues and topics I mean the ones described in systems like Apache Kafka, AcriveMQ or Java Messaging Service.
Where I can find material about that?
what will be kleene star “expansion” of expression $ a^*b^*$ in lexicographic order? I’m confused and I really want to clear my concepts so I can proceed further
What is the usual name of the theory I dubbed “fermionic languages”, for want of a better term?
Provisionally then, say a
fermionic alphabet is a set of symbols, each of which is labeled by a spin: either a (signed) integer, making the symbol a boson, or a half integer, making it a fermion. To make things interesting, add the following conditions: there is at least one fermion in the alphabet, the alphabet is closed under spin inversion: an involutive mapping from each letter to one with a spin of the same magnitude and sign changed. “Scalar” bosons (those with spin 0) are allowed, notrequired, to map to themselves.
Of course, we immediately extend this to finite words, the spin of a word being the sum of the spins of its individual letters: this makes the free monoid on the alphabet an U-semigroup. Then, the sub-languages of the free monoid I called
fermionic languages are those containing >= 1 fermionic word.
The prototypical spin function would map the generators of a group to their orders, assuming they are all finite: in which case, inversion in the group-theoretic sense is composed of the 2 commuting involutions, to wit, spin inversion of the letters and word reversal. I believe anyone dissecting the involutions of a finitely generated group would give a try to the concept, so there must be a name for it, if only in folklore.
However, I emphatically do
NOT require fermion languages to be closed under any of spin inversion, word reversal, nor their composite, which I would call “GT-inversion”. I do not even require the set of spin values of these languages to be closed under change of sign, nor to contain the value 0. All I want is >= 1 fermionic word in the language under study, and immensely valuable properties such as: the empty word is notamong its kin, spin inversion commutes with word reversal.
I’m currently taking the theory of computation at school. Some stuff covered makes sense, but some topics I feel like I’m missing the big picture. Is there an equivalent of theory of computation and/or automaton theory for dummies? I’m a dummy who needs this degree.
There is an information source on the information source alphabet $ A = \{a, b, c\}$ represented by the state transition diagram below:
a) The random variable representing the $ i$ -th output from this information source is represented by $ X_i$ . It is known that the user is now in state $ S_1$ . In this state, let $ H (X_i|s_1)$ denote the entropy when observing the next symbol $ X_i$ , find the value of $ H (X_i|s_1)$ , entropy of this information source, Calculate $ H (X_i|X_{i-1}) $ and $ H (X_i)$ respectively. Assume $ i$ is quite large
How can I find $ H(X_i|s_1)?$ I know that $ $ H(X_i|s_1) = -\sum_{i,s_1} p\left(x_i, s_1\right)\cdot\log_b\!\left(p\left(x_i|s_1\right)\right) = -\sum_{i,j} p\left(x_i, s_1\right)\cdot\log_b\!\left(\frac{p\left(x_i, s_1\right)}{p\left(s_1\right)}\right)$ $ but I don’t know $ p(s_1)$ .
$ $ A=\begin{pmatrix}0.25 & 0.75 & 0\0.5 & 0 & 0.5 \0 & 0.7 & 0.3 \end{pmatrix}.$ $
From matrix I can know that $ p(s_1|s_1)=0.25$ , etc.
But what is the probability of $ s_1$ ? And how can I calculate $ H (X_i|X_{i-1})$ ?
|
I'm not a stats specialist, but I will give it a shot.
First, we can approximate the probability of each event by its empirical probability, i.e. the number of occurrences divided by the total number of trials:
$p(motif_i, condition_j) = \frac{\text{number of occurrences of motif i with condition j}}{ \sum_{i,j} \text{number of occurrences of motif i with condition j}}$
I'll use the shorthands m_1, m_2, c_1, c_2 for motifs and conditions in your table. The approximation gives the following joint distribution $p(m_i,c_j)$:
c_1 c_2
m_1 0.1 0.05
m_2 0.4 0.45
Marginal probabilities can be computed by just summing rows and columns.Have a look at the example there: https://en.wikipedia.org/wiki/Marginal_distributionI.e. here, $p(m_1)=0.15$ and $p(c_1)=0.5$.
Then, the mutual information can be computed from its definition:
$I(motif;condition) = \sum_{i \in [1,2], j \in [1,2]} p(m_i,c_j)\log(\frac{p(m_i,c_j)}{p(m_i)p(c_j)})$
|
ISSN:
1930-5311
eISSN:
1930-532X
All Issues
Journal of Modern Dynamics
January 2010 , Volume 4 , Issue 1
Select all articles
Export/Reference:
Abstract:
Let $M$ be a closed $3$-manifold, and let $X_t$ be a transitive Anosov flow. We construct a diffeomorphism of the form $f(p)=Y_{t(p)}(p)$, where $Y$ is an Anosov flow equivalent to $X$. The diffeomorphism $f$ is structurally stable (satisfies Axiom A and the strong transversality condition); the non-wandering set of $f$ is the union of a transitive attractor and a transitive repeller; and $f$ is also partially hyperbolic (the direction $\RR.Y$ is the central bundle).
Abstract:
We consider the Ricci flow for simply connected nilmanifolds. This translates to a Ricci flow on the space of nilpotent metric Lie algebras. We consider the evolution of the inner product and the evolution of structure constants, as well as the evolution of these quantities modulo rescaling. We set up systems of O.D.E.'s for some of these flows and describe their qualitative properties. We also present some explicit solutions for the evolution of soliton metrics under the Ricci flow.
Abstract:
We consider piecewise cone-hyperbolic systems satisfying a bunching condition, and we obtain a bound on the essential spectral radius of the associated weighted transfer operators acting on anisotropic Sobolev spaces. The bunching condition is always satisfied in dimension two, and our results give a unifying treatment of the work of Demers-Liverani [9] and our previous work [2]. When the complexity is subexponential, our bound implies a spectral gap for the transfer operator corresponding to the physical measures in many cases (for example if $T$ preserves volume, or if the stable dimension is equal to $1$ and the unstable dimension is not zero).
Abstract:
We characterize the volume entropy of a regular building as the topological pressure of the geodesic flow on an apartment. We show that the entropy maximizing measure is not Liouville measure for any regular hyperbolic building. As a consequence, we obtain a strict lower bound on the volume entropy in terms of the branching numbers and the volume of the boundary polyhedrons.
Abstract:
To any self-similar action of a finitely generated group $G$ of automorphisms of a regular rooted tree $T$ can be naturally associated an infinite sequence of finite graphs $\{\Gamma_n\}_{n\geq 1}$, where $\Gamma_n$ is the Schreier graph of the action of $G$ on the $n$-th level of $T$. Moreover, the action of $G$ on $\partial T$ gives rise to orbital Schreier graphs $\Gamma_{\xi}$, $\xi\in \partial T$. Denoting by $\xi_n$ the prefix of length $n$ of the infinite ray $\xi$, the rooted graph $(\Gamma_{\xi},\xi)$ is then the limit of the sequence of finite rooted graphs $\{(\Gamma_n,\xi_n)\}_{n\geq 1}$ in the sense of pointed Gromov-Hausdorff convergence. In this paper, we give a complete classification (up to isomorphism) of the limit graphs $(\Gamma_{\xi},\xi)$ associated with the Basilica group acting on the binary tree, in terms of the infinite binary sequence $\xi$.
Abstract:
N/A.
Readers Authors Librarians Email Alert
Add your name and e-mail address to receive news of forthcoming issues of this journal:
[Back to Top]
|
Let $w$ denote the weight on $A$ so that $1-w$ is the weight on $B$. Recall from the properties of variance that
$\sigma_p^2 = w^2\sigma_A^2 + 2w(1-w)\sigma_A\sigma_B \rho_{AB}+ (1-w)^2\sigma_B^2$
Without loss of generality, assume $\sigma_A \geq \sigma_B$. We wish to show that
$w^2\sigma_A^2 + 2w(1-w)\sigma_A\sigma_B \rho_{AB}+ (1-w)^2\sigma_B^2\leq \sigma_A^2$
Note that
$\sigma_A^2 = \sigma_A^2 (w + (1-w)) ^2 = \sigma_A^2 w^2 + 2w(1-w)\sigma_A^2 + \sigma_A^2(1-w)^2$
Since $\sigma_A \geq \sigma_B$ and $w$, $(1-w)$, and $\sigma_A$ are positive, this means that
$\sigma_A^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B + \sigma_B^2(1-w)^2$
And since the correlation has the property that $-1 \leq \rho_{AB} \leq 1$ and $w$, $(1-w)$, $\sigma_B$ and $\sigma_A$ are all positive, it must be the case that
$\sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B + \sigma_B^2(1-w)^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B\rho_{AB} + \sigma_B^2(1-w)^2$
Therefore
$\sigma_A^2 \geq \sigma_A^2 w^2 + 2w(1-w)\sigma_A\sigma_B\rho_{AB} + \sigma_B^2(1-w)^2$ $\square$
In words, looking at the formula for variance of a convex combination of random variables, the variance is maximized if the correlation between the assets is 1. In this case, the possible portfolio values as a function of $w$ are a straight line segment between $A$ and $B$, which clearly can't have a variance higher than either. Now, if the correlation is less than 1, then any combination of the two will be lower than the straight line case.
Intuitively, the returns to assets $A$ and $B$ will partially cancel each other out any time they are not a fixed multiple of each other. This canceling out behavior reduces the variance of the resulting portfolio. The worst-case scenario is that the two assets are equal to each other, so the portfolio can never have a higher variance than the component asst with the highest variance.
|
I'll assume that you were using a radio tuned to a 1Mhz frequency ($\omega = 6.3\times 10^6$ s$^{-1}$) and that the radio was completely enclosed inside $t=3\,$mm of pure iron.
There are two important effects to consider. (i) How much power is reflected from the iron surface. (ii) How much of the transmitted power makes it through the iron.
To figure this out we need the properties of iron; a conductivity $\sigma = 10^7$ S/m, a relative permittivity $\epsilon_r \simeq 1$ and a relative permeability $\mu_r \simeq 10^4$ (for 99.9% pure iron).
First we check whether iron works as a good conductor at these frequencies by noting that $\sigma/{\epsilon_r \epsilon_0 \omega} = 1.8\times 10^{11}$; i.e. $\gg 1$ and therefore a good conductor.
The modulus of the impedance of a conductor is given by $\eta_{\rm Fe} = (\mu_r \mu_0 \sigma / \omega)^{1/2} = 0.14$ $\Omega$.
So, now the relevant equations are:Electric field transmission at the air/iron interface (assuming normal incidence)
$$\frac{E_t}{E_i} = \frac{2 \eta_{\rm Fe}}{\eta_0 + \eta_{\rm Fe}} \simeq 2\frac{\eta_{\rm Fe}}{\eta_0}\, ,$$.where $\eta_0 = 377$ $\Omega$.
The EM waves then propogate into the metal but are exponentially attenuated on a scale defined by the "skin depth" $\delta = (2/\mu_r \mu_0 \sigma \omega)^{1/2} = 1.59 \times 10^{-6}\,$m.
Thus, after traversing a thickness $t$, the E-field is attenuated by $\exp(-t/\delta)$.
Finally the wave emerges through the iron/air interface on the other side and we use the transmission formula again but with the labels swapped on the impedance values.
Hence the ratio of the net transmitted electric field to the incident electric field is given approximately by$$ R = 2 \frac{\eta_{\rm Fe}}{\eta_0}\, \exp(-t/\delta)\, 2 \frac{\eta_0}{\eta_0 + \eta_{\rm Fe}} = 4 \frac{\eta_{\rm Fe}}{\eta_0}\, \exp(-t/\delta)\, .$$
For the numbers I've assumed $R \simeq 0$ because the wave traverse $>1000$ skin depths to get through the iron! The transmitted power is $\propto R^2$.
So my conclusion is that enclosing within 3mm of pure iron would certainly block AM radio.
How might this not work? Perhaps the iron you used is very impure and the permeability is orders of magnitudes lower? If $\mu_r =1$ then $\eta_{\rm Fe} = 0.0014\,\Omega$, $\delta = 1.59\times 10^{-4}\,$m. Thus 3mm is still 18 skin depths. The conductivity I assumed is unlikely to be much lower, so I'm a bit confused as to why it wouldn't work.
The demo I use in my lectures is wrapping a mobile phone in aluminium foil. In principle, this is much more marginal because though the frequencies are higher, the foil thickness is much lower, but it certainly works.
|
Projection is Surjection/Family of Sets Theorem
Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets.
Let $\displaystyle \prod_{\alpha \mathop \in I} S_\alpha$ be the Cartesian product of $\family {S_\alpha}_{\alpha \mathop \in I}$.
Let each of $S_\alpha$ be non-empty.
For each $\beta \in I$, let $\displaystyle \pr_\beta: \prod_{\alpha \mathop \in I} S_\alpha \to S_\beta$ be the $\beta$th projection on $\displaystyle S = \prod_{\alpha \mathop \in I} S_\alpha$.
Then $\pr_\beta$ is a surjection. Proof
Consider the $\beta$th projection.
Let $x_\beta \in S_\beta$.
Let $\map x \beta = x_\beta$
Suppose $\gamma \in I: \gamma \ne \beta$.
As $S_\gamma \ne \O$ it is possible to use the axiom of choice to choose $\map x \gamma \in S_\gamma$.
Then:
$\displaystyle x \in \prod_{\alpha \mathop \in I} S_\alpha$
and:
$\map {\pr_\beta} x = \map x \beta$
Hence the result.
$\blacksquare$
|
65 4 Homework Statement In a double slit experiment let d=5.00 D=30.0λ. Estimate the ratio of the intensity of the third order maximum with that of the zero-order maximum. Homework Equations interference diffraction Homework Statement:In a double slit experiment let d=5.00 D=30.0λ. Estimate the ratio of the intensity of the third order maximum with that of the zero-order maximum. Homework Equations:interference diffraction
i guess the goal is this equation
##I_{(\theta)}=I_0 \times(cos^2\beta)\times \left ( \frac {sin\alpha} \alpha \right)^2##
then i do
## D\sin \theta = 3\lambda##
##\sin\theta= \frac {3\lambda} D, \space \theta=5.74^0##
##\beta= \frac {\pi d} \lambda \sin \theta##
##\alpha = \frac {\pi D} \lambda \sin \theta \space##
substituting the data
##\alpha=\frac {\pi 30.0\lambda} \lambda \frac {3\lambda} {30.0\lambda}\space##
next
##\beta= \frac {\pi 5.00} \lambda \frac {3\lambda} {30.0\lambda}## i don't know how to solve this one, and solve the rest of the problem, how do i get rid of ##\space\lambda\space## at denominator?
any help please?
|
J. D. Hamkins and J. Reitz, “The set-theoretic universe $V$ is not necessarily a class-forcing extension of HOD,” ArXiv e-prints, 2017. (manuscript under review)
@ARTICLE{HamkinsReitz:The-set-theoretic-universe-is-not-necessarily-a-forcing-extension-of-HOD, author = {Joel David Hamkins and Jonas Reitz}, title = {The set-theoretic universe {$V$} is not necessarily a class-forcing extension of {HOD}}, journal = {ArXiv e-prints}, year = {2017}, volume = {}, number = {}, pages = {}, month = {September}, note = {manuscript under review}, abstract = {}, keywords = {under-review}, source = {}, doi = {}, eprint = {1709.06062}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://jdh.hamkins.org/the-universe-need-not-be-a-class-forcing-extension-of-hod}, }
Abstract.In light of the celebrated theorem of Vopěnka, proving in ZFC that every set is generic over $\newcommand\HOD{\text{HOD}}\HOD$, it is natural to inquire whether the set-theoretic universe $V$ must be a class-forcing extension of $\HOD$ by some possibly proper-class forcing notion in $\HOD$. We show, negatively, that if ZFC is consistent, then there is a model of ZFC that is not a class-forcing extension of its $\HOD$ for any class forcing notion definable in $\HOD$ and with definable forcing relations there (allowing parameters). Meanwhile, S. Friedman (2012) showed, positively, that if one augments $\HOD$ with a certain ZFC-amenable class $A$, definable in $V$, then the set-theoretic universe $V$ is a class-forcing extension of the expanded structure $\langle\HOD,\in,A\rangle$. Our result shows that this augmentation process can be necessary. The same example shows that $V$ is not necessarily a class-forcing extension of the mantle, and the method provides a counterexample to the intermediate model property, namely, a class-forcing extension $V\subseteq V[G]$ by a certain definable tame forcing and a transitive intermediate inner model $V\subseteq W\subseteq V[G]$ with $W\models\text{ZFC}$, such that $W$ is not a class-forcing extension of $V$ by any class forcing notion with definable forcing relations in $V$. This improves upon a previous example of Friedman (1999) by omitting the need for $0^\sharp$.
In 1972, Vopěnka proved the following celebrated result.
Theorem. (Vopěnka) If $V=L[A]$ where $A$ is a set of ordinals, then $V$ is a forcing extension of the inner model $\HOD$.
The result is now standard, appearing in Jech (Set Theory 2003, p. 249) and elsewhere, and the usual proof establishes a stronger result, stated in ZFC simply as the assertion: every set is generic over $\HOD$. In other words, for every set $a$ there is a forcing notion $\mathbb{B}\in\HOD$ and a $\HOD$-generic filter $G\subseteq\mathbb{B}$ for which $a\in\HOD[G]\subseteq V$. The full set-theoretic universe $V$ is therefore the union of all these various set-forcing generic extensions $\HOD[G]$.
It is natural to wonder whether these various forcing extensions $\HOD[G]$ can be unified or amalgamated to realize $V$ as a single class-forcing extension of $\HOD$ by a possibly proper class forcing notion in $\HOD$. We expect that it must be a very high proportion of set theorists and set-theory graduate students, who upon first learning of Vopěnka’s theorem, immediately ask this question.
Main Question. Must the set-theoretic universe $V$ be a class-forcing extension of $\HOD$?
We intend the question to be asking more specifically whether the universe $V$ arises as a bona-fide class-forcing extension of $\HOD$, in the sense that there is a class forcing notion $\mathbb{P}$, possibly a proper class, which is definable in $\HOD$ and which has definable forcing relation $p\Vdash\varphi(\tau)$ there for any desired first-order formula $\varphi$, such that $V$ arises as a forcing extension $V=\HOD[G]$ for some $\HOD$-generic filter $G\subseteq\mathbb{P}$, not necessarily definable.
In this article, we shall answer the question negatively, by providing a model of ZFC that cannot be realized as such a class-forcing extension of its $\HOD$.
Main Theorem. If ZFC is consistent, then there is a model of ZFC which is not a forcing extension of its $\HOD$ by any class forcing notion definable in that $\HOD$ and having a definable forcing relation there.
Throughout this article, when we say that a class is definable, we mean that it is definable in the first-order language of set theory allowing set parameters.
The main theorem should be placed in contrast to the following result of Sy Friedman.
Theorem. (Friedman 2012) There is a definable class $A$, which is strongly amenable to $\HOD$, such that the set-theoretic universe $V$ is a generic extension of $\langle \HOD,\in,A\rangle$.
This is a postive answer to the main question, if one is willing to augment $\HOD$ with a class $A$ that may not be definable in $\HOD$. Our main theorem shows that in general, this kind of augmentation process is necessary.
It is natural to ask a variant of the main question in the context of set-theoretic geology.
Question. Must the set-theoretic universe $V$ be a class-forcing extension of its mantle?
The mantle is the intersection of all set-forcing grounds, and so the universe is close in a sense to the mantle, perhaps one might hope that it is close enough to be realized as a class-forcing extension of it. Nevertheless, the answer is negative.
Theorem. If ZFC is consistent, then there is a model of ZFC that does not arise as a class-forcing extension of its mantle $M$ by any class forcing notion with definable forcing relations in $M$.
We also use our results to provide some counterexamples to the intermediate-model property for forcing. In the case of set forcing, it is well known that every transitive model $W$ of ZFC set theory that is intermediate $V\subseteq W\subseteq V[G]$ a ground model $V$ and a forcing extension $V[G]$, arises itself as a forcing extension $W=V[G_0]$.
In the case of class forcing, however, this can fail.
Theorem. If ZFC is consistent, then there are models of ZFC set theory $V\subseteq W\subseteq V[G]$, where $V[G]$ is a class-forcing extension of $V$ and $W$ is a transitive inner model of $V[G]$, but $W$ is not a forcing extension of $V$ by any class forcing notion with definable forcing relations in $V$. Theorem. If ZFC + Ord is Mahlo is consistent, then one can form such a counterexample to the class-forcing intermediate model property $V\subseteq W\subseteq V[G]$, where $G\subset\mathbb{B}$ is $V$-generic for an Ord-c.c. tame definable complete class Boolean algebra $\mathbb{B}$, but nevertheless $W$ does not arise by class forcing over $V$ by any definable forcing notion with a definable forcing relation.
More complete details, please go to the paper (click through to the arxiv for a pdf).
J. D. Hamkins and J. Reitz, “The set-theoretic universe $V$ is not necessarily a class-forcing extension of HOD,” ArXiv e-prints, 2017. (manuscript under review)
@ARTICLE{HamkinsReitz:The-set-theoretic-universe-is-not-necessarily-a-forcing-extension-of-HOD, author = {Joel David Hamkins and Jonas Reitz}, title = {The set-theoretic universe {$V$} is not necessarily a class-forcing extension of {HOD}}, journal = {ArXiv e-prints}, year = {2017}, volume = {}, number = {}, pages = {}, month = {September}, note = {manuscript under review}, abstract = {}, keywords = {under-review}, source = {}, doi = {}, eprint = {1709.06062}, archivePrefix = {arXiv}, primaryClass = {math.LO}, url = {http://jdh.hamkins.org/the-universe-need-not-be-a-class-forcing-extension-of-hod}, }
|
What is the fundamental difference between convolutional neural networks and recurrent neural networks? Where are they applied?
Basically, a CNN saves a set of weights and applies them spatially. For example, in a layer, I could have 32 sets of weights (also called feature maps). Each set of weights is a 3x3 block, meaning I have 3x3x32=288 weights for that layer. If you gave me an input image, for each 3x3 map, I slide it across all the pixels in the image, multiplying the regions together. I repeat this for all 32 feature maps, and pass the outputs on. So, I am learning a few weights that I can apply at a lot of locations.
For an RNN, it is a set of weights applied temporally (through time). An input comes in, and is multiplied by the weight. The networks saves an internal state and puts out some sort of output. Then, the next piece of data comes in, and is multiplied by the weight. However, the internal state that was created from the last piece of data also comes in and is multiplied by a different weight. Those are added and the output comes from an activation applied to the sum, times another weight. The internal state is updated, and the process repeats.
CNN's work really well for computer vision. At the low levels, you often want to find things like vertical and horizontal lines. Those kinds of things are going to be all over the images, so it makes sense to have weights that you can apply anywhere in the images.
RNN's are really good for natural language processing. You can imagine that the next word in a sentence will be highly influenced by the ones that came before it, so it makes sense to carry that internal state forward and have a small set of weights that can apply to any input.
However, there are many more applications. In addition, CNN's have performed well on NLP tasks. There are also more advanced versions of RNN's called LSTM's that you could check out.
For an explanation of CNN's, go to the Stanford CS231n course. Especially check out lecture 5. There are full class videos on YouTube.
For an explanation of RNN's, go here.
Recurrent neural networks (RNNs) are artificial neural networks (ANNs) that have one or more recurrent (or cyclic) connections, as opposed to just having feed-forward connections, like a feed-forward neural network (FFNN).
These cyclic connections are used to keep track of temporal relations or dependencies between the elements of a sequence. Hence, RNNs are suited for sequence prediction or related tasks.
In the picture below, you can observe an RNN on the left (that contains only one hidden unit) that is equivalent to the RNN on the right, which is its "unfolded" version. For example, we can observe that $\bf h_1$ (the hidden unit at time step $t=1$) receives both an input $\bf x_1$ and the value of the hidden unit at the previous time step, that is, $\bf h_0$.
The cyclic connections (or the weights of the cyclic edges), like the feed-forward connections, are learned using an optimisation algorithm (like gradient descent) often combined with back-propagation (which is used to compute the gradient of the loss function).
The convolution is an operation that takes two functions, $\bf f$ and $\bf h$, as input and produces a third function, $\bf g = f \circledast h$, where the symbol $\circledast$ denotes the convolution operation. In the context of CNNs, the input function $\bf f$ can e.g. be an image (which can be thought of as a function from 2D coordinates to RGB or grayscale values). The other function $\bf h$ is called the "kernel" (or filter), which can be thought of as (small and square) matrix (which contains the output of the function $\bf h$). $\bf f$ can also be thought of as a (big) matrix (which contains, for each cell, e.g. its grayscale value).
In the context of CNNs, the
convolution operation can be thought of as dot product between the kernel $\bf h$ (a matrix) and several parts of the input (a matrix).
In the picture below, we perform an
element-wise multiplication between the kernel $\bf h$ and part of the input $\bf h$, then we sum the elements of the resulting matrix, and that is the value of the convolution operation for that specific part of the input.
To be more concrete, in the picture above, we are performing the following operation
\begin{align} \sum_{ij} \left( \begin{bmatrix} 1 & 0 & 0\\ 1 & 1 & 0\\ 1 & 1 & 1 \end{bmatrix} \otimes \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix} \right) = \sum_{ij} \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix} = 4 \end{align}
where $\otimes$ is the element-wise multiplication and the summation $\sum_{ij}$ is over all rows $i$ and columns $j$ (of the matrices).
To compute all elements of $\bf g$, we can think of the kernel $\bf h$ as being slided over the matrix $\bf f$.
In general, the kernel function $\bf h$ can be fixed. However, in the context of CNNs, the kernel $\bf h$ represents the learnable parameters of the CNN: in other words, during the training procedure (using e.g. gradient descent and back-propagation), this kernel $\bf h$ (which thus can be thought of as a matrix of weights) changes.
In the context of CNNs, there is often more than one kernel: in other words, it is often the case that a sequence of kernels $\bf h_1, h_2, \dots, h_k$ is applied to $\bf f$ to produce a sequence of convolutions $\bf g_1, g_2, \dots, g_k$. Each kernel $\bf h_i$ is used to "detect different features of the input", so these kernels are different from each other.
A
down-sampling operation is an operation that reduces the input size while attempting to maintain as much information as possible. For example, if the input size is a $2 \times 2$ matrix $\bf f = \begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix}$, a common down-sampling operation is called the max-pooling, which, in the case of $\bf f$, returns $3$ (the maximum element of $\bf f$).
CNNs are particularly suited to deal with high-dimensional inputs (e.g. images), because, compared to FFNNs, they use a smaller number of learnable parameters (which, in the context of CNNs, are the kernels). So, they are often used to e.g. classify images.
What is the fundamental difference between RNNs and CNNs? RNNs have recurrent connections while CNNs do not necessarily have them. The fundamental operation of a CNN is the convolution operation, which is not present in a standard RNN. CNN vs RNN A CNN will learn to recognize patterns across space while RNN is useful for solving temporal data problems. CNNs have become the go-to method for solving any image data challenge while RNN is used for ideal for text and speech analysis. In a very general way, a CNN will learn to recognize components of an image (e.g., lines, curves, etc.) and then learn to combine these components to recognize larger structures (e.g., faces, objects, etc.) while an RNN will similarly learn to recognize patterns across time. So a RNN that is trained to convert speech to text should learn first the low level features like characters, then higher level features like phonemes and then word detection in audio clip. CNN
A convolutional network (ConvNet) is made up of layers. In a convolutional network (ConvNet), there are basically three types of layers:
Convolution layer Pooling layer Fully connected layer
Of these, the convolution layer applies convolution operation on the input 3D tensor. Different filters extract different kinds of features from an image. The below GIF illustrates this point really well:
Here the filter is the green 3x3 matrix while the image is the blue 7x7 matrix.
Many such layers passes through filters in CNN to give an output layer that can again be a NN Fully connected layer or a 3D tensor.
For example, in the above example, the input image passes through convolutional layer, then pooling layer, then convolutional layer, pooling layer, then the 3D tensor is flattened like a Neural Network 1D layer, then passed to a fully connected layer and finally a softmax layer. This makes a CNN.
RNN Recurrent Neural Network(RNN) are a type of Neural Network where the output from previous step are fed as input to the current step.
Here, $x_{t-1}$, , $x_{t}$ and $x_{t+1}$ are the values of inputs data that occur at a specific time steps and are fed into the RNN that goes through the hidden layers namely $h_{t-1}$, , $h_{t}$ and $h_{t+1}$ which further produces output $o_{t-1}$, , $o_{t}$ and $o_{t+1}$ respectively.
On a basic level, an RNN is a neural network whose next state depends on its past state(s), while a CNN is a neural network that does dimensionality reduction (make large data smaller while preserving information) via convolution. See this for more info on convolutions
|
To finish the job initiated by Sivaram Ambikasaran, first you can say that $\mu(x)=\alpha x+ \beta 1/x$
The stochastic integral term with $\beta=0$ case is answered by Sivaram Ambikasaran.
For the $\alpha=0$ case you have using Itô's lemma :
$dLnB_t= \frac{1}{B_t}dB_t-\frac{1}{2.B_t^2}dt$
So integrating this gives you :
$\beta.\int_0^T \frac{dB_s}{B_s}=\beta.(Ln(\frac{B_T}{B_0})+1/2\int_0^T \frac{1}{B_s^2}ds)$
If you start your Brownian Motion at 0, the problem is not well defined. If $B_0\not=0$ then returning to the intial problem (still with $\alpha=0$) you have :
$ \int_0^T \mu(B_s) dB_s - \frac{\int_0^T (\mu(B_s))^2 ds}{2} =
Ln(\frac{B_T}{B_0})+\frac{\beta.(1-\beta)}{2}.\int_0^T \frac{1}{B_s^2}ds$ if I am not mistaken.
For the general case ($\alpha\not= 0$ and $\beta\not= 0$) just adds the solutions ( i.e. respectively for $\alpha=0$ and $\beta=0$) and also add the cross product term from $\int_0^T (\mu(B_s))^2 ds$ (which should be something like $-T.\alpha.\beta$).
Regards
|
In combinatorics there are quite many such disproven conjectures. The most famous of them are:
1) Tait conjecture:
Any 3-vertex connected planar cubic graph is Hamiltonian
The first counterexample found has 46 vertices. The "least" counterexample known has 38 vertices.
2) Tutte conjecture:
Any bipartite cubic graph is Hamiltonian
The first counterexample found has 96 vertices. The "least" counterexample known has 54 vertices.
3) Thom conjecture
If two finite undirected simple graphs have conjugate adjacency matrices over $\mathbb{Z}$, then they are isomorphic.
The least known counterexample pair is formed by two trees with 11 vertices.
4) Borsuk conjecture:
Every bounded subset $E$ of $\mathbb{R}^n$can be partitioned into $n+1$ sets, each of which has a smaller diameter, than $E$
In the first counterexample found $n = 1325$. In the "least" counterexample known $n = 64$.
5) Danzer-Gruenbaum conjecture:
If $A \subset \mathbb{R}^n$ and $\forall u, v, w \in A$ $(u - w, v - w) > 0,$ then $|A| \leq 2n - 1$
This statement is not true for any $n \geq 35$
6) The Boolean Pythagorean Triple Conjecture:
There exists $S \subset \mathbb{N}$, such that neither $S$, nor $\mathbb{N} \setminus S$ contain Pythagorean triples
This conjecture was disproved by M. Heule, O. Kullman and V. Marek. They proved, that there do exist such $S \subset \{n \in \mathbb{N}| n \leq k\}$, such that neither $S$, nor $\{n \in \mathbb{N}| n \leq k\} \setminus S$ contain Pythagorean triples, for all $k \leq 7824$, but not for $k = 7825$
7) Burnside conjecture:
Every finitely generated group with period n is finite
This statement is not true for any odd $n \geq 667$
8) Otto Shmidt conjecture:
If all proper subgroups of a group $G$ are isomorphic to $C_p$, where $p$ is a fixed prime number, then $G$ is finite.
Alexander Olshanskii proved, that there are continuum many non-isomorphic counterexamples to this conjecture for any $p > 10^{75}$.
9) Von Neuman conjecture
Any non-amenable group has a free subgroup of rank 2
The least known finitely presented counterexample has 3 generators and 9 relators
10) Word problem conjecture:
Word problem is solvable for any finitely generated group
The "least" counterexample known has 12 generators.
11) Leinster conjecture:
Any Leinster group has even order
The least counterexample known has order 355433039577.
12) Rotman conjecture:
Automorphism groups of all finite groups not isomorphic to $C_2$ have even order
The first counterexample found has order 78125. The least counterexample has order 2187. It is the automorphism group of a group with order 729.
13) Rose conjecture:
Any nontrivial complete finite group has even order
The least counterexample known has order 788953370457.
14) Hilton conjecture
Automorphism group of a non-abelian group is non-abelian
The least counterexample known has order 64.
15)Hughes conjecture:
Suppose $G$ is a finite group and $p$ is a prime number. Then $[G : \langle\{g \in G| g^p \neq e\}\rangle] \in \{1, p, |G|\}$
The least known counterexample has order 142108547152020037174224853515625.
16) Moreto conjecture:
Let $S$ be a finite simple group and $p$ the largest prime divisor of $|S|$. If $G$ is a finite group with the same number of elements of order $p$ as $S$ and $|G| = |S|$, then $G \cong S$
The first counterexample pair constructed is formed by groups of order 20160 (those groups are $A_8$ and $L_3(4)$)
17) This false statement is not a conjecture, but rather a popular mistake done by many people, who have just started learning group theory:
All elements of the commutant of any finite group are commutators
The least counterexample has order 96.
If the numbers mentioned in this text do not impress you, please, do not feel disappointed: there are complex combinatorial objects "hidden" behind them.
|
Answer
We can find the domain of $sin ~\theta$: domain: $-\infty \lt \theta \lt \infty$ We can find the domain of $cos ~\theta$: domain: $-\infty \lt \theta \lt \infty$ We can find the domain of $tan ~\theta$: domain: $\theta \neq \frac{\pi}{2}+\pi~n$, for any integer $n$ We can find the domain of $csc ~\theta$: domain: $\theta \neq \pi~n$, for any integer $n$ We can find the domain of $sec ~\theta$: domain: $\theta \neq \frac{\pi}{2}+\pi~n$, for any integer $n$ We can find the domain of $cot ~\theta$: domain: $\theta \neq \pi~n$, for any integer $n$
Work Step by Step
We can find the domain of $sin ~\theta$: $sin~\theta = \frac{x}{r}$ Since $r$ is always positive, $sin ~\theta$ is defined for all real numbers. domain: $-\infty \lt \theta \lt \infty$ We can find the domain of $cos ~\theta$: $cos~\theta = \frac{y}{r}$ Since $r$ is always positive, $cos ~\theta$ is defined for all real numbers. domain: $-\infty \lt \theta \lt \infty$ We can find the domain of $tan ~\theta$: $tan~\theta = \frac{sin~\theta}{cos~\theta}$ $tan ~\theta$ is undefined when $cos~\theta = 0$ $tan ~\theta$ is undefined when $\theta = \frac{\pi}{2}+\pi~n$, for any integer $n$ domain: $\theta \neq \frac{\pi}{2}+\pi~n$, for any integer $n$ We can find the domain of $csc ~\theta$: $csc~\theta = \frac{1}{sin~\theta}$ $csc ~\theta$ is undefined when $sin~\theta = 0$ $csc ~\theta$ is undefined when $\theta = \pi~n$, for any integer $n$ domain: $\theta \neq \pi~n$, for any integer $n$ We can find the domain of $sec ~\theta$: $sec~\theta = \frac{1}{cos~\theta}$ $sec ~\theta$ is undefined when $cos~\theta = 0$ $sec ~\theta$ is undefined when $\theta = \frac{\pi}{2}+\pi~n$, for any integer $n$ domain: $\theta \neq \frac{\pi}{2}+\pi~n$, for any integer $n$ We can find the domain of $cot ~\theta$: $cot~\theta = \frac{cos~\theta}{sin~\theta}$ $cot ~\theta$ is undefined when $sin~\theta = 0$ $cot ~\theta$ is undefined when $\theta = \pi~n$, for any integer $n$ domain: $\theta \neq \pi~n$, for any integer $n$
|
Set-theoretic arguments often make use of the fact that a particular property $\varphi$ is
local, in the sense that instances of the property can be verified by checking certain facts in only a bounded part of the set-theoretic universe, such as inside some rank-initial segment $V_\theta$ or inside the collection $H_\kappa$ of all sets of hereditary size less than $\kappa$. It turns out that this concept is exactly equivalent to the property being $\Sigma_2$ expressible in the language of set theory. Theorem. For any assertion $\varphi$ in the language of set theory, the following are equivalent: $\varphi$ is ZFC-provably equivalent to a $\Sigma_2$ assertion. $\varphi$ is ZFC-provably equivalent to an assertion of the form “$\exists \theta\, V_\theta\models\psi$,” where $\psi$ is a statement of any complexity. $\varphi$ is ZFC-provably equivalent to an assertion of the form “$\exists \kappa\, H_\kappa\models\psi$,” where $\psi$ is a statement of any complexity.
Just to clarify, the $\Sigma_2$ assertions in set theory are those of the form $\exists x\,\forall y\,\varphi_0(x,y)$, where $\varphi_0$ has only bounded quantifiers. The set $V_\theta$ refers to the rank-initial segment of the set-theoretic universe, consisting of all sets of von Neumann rank less than $\theta$. The set $H_\kappa$ consists of all sets of hereditary size less than $\kappa$, that is, whose transitive closure has size less than $\kappa$.
Proof. ($3\to 2$) Since $H_\kappa$ is correctly computed inside $V_\theta$ for any $\theta>\kappa$, it follows that to assert that some $H_\kappa$ satisfies $\psi$ is the same as to assert that some $V_\theta$ thinks that there is some cardinal $\kappa$ such that $H_\kappa$ satisfies $\psi$.
($2\to 1$) The statement $\exists \theta\, V_\theta\models\psi$ is equivalent to the assertion $\exists\theta\,\exists x\,(x=V_\theta\wedge x\models\psi)$. The claim that $x\models\psi$ involves only bounded quantifiers, since the quantifiers of $\psi$ become bounded by $x$. The claim that $x=V_\theta$ is $\Pi_1$ in $x$ and $\theta$, since it is equivalent to saying that $x$ is transitive and the ordinals of $x$ are precisely $\theta$ and $x$ thinks every $V_\alpha$ exists, plus a certain minimal set theory (so far this is just $\Delta_0$, since all quantifiers are bounded), plus, finally, the assertion that $x$ contains every subset of each of its elements. So altogether, the assertion that some $V_\theta$ satisfies $\psi$ has complexity $\Sigma_2$ in the language of set theory.
($1\to 3$) This implication is a consequence of the following absoluteness lemma.
Lemma. (Levy) If $\kappa$ is any uncountable cardinal, then $H_\kappa\prec_{\Sigma_1} V$. Proof. Suppose that $x\in H_\kappa$ and $V\models\exists y\,\psi(x,y)$, where $\psi$ has only bounded quantifiers. Fix some such witness $y$, which exists inside some $H_\gamma$ for perhaps much larger $\gamma$. By the Löwenheim-Skolem theorem, there is $X\prec H_\gamma$ with $\text{TC}(\{x\})\subset X$, $y\in X$ and $X$ of size less than $\kappa$. Let $\pi:X\cong M$ be the Mostowski collapse of $X$, so that $M$ is transitive, and since it has size less than $\kappa$, it follows that $M\subset H_\kappa$. Since the transitive closure of $\{x\}$ was contained in $X$, it follows that $\pi(x)=x$. Thus, since $X\models\psi(x,y)$ we conclude that $M\models \psi(x,\pi(y))$ and so hence $\pi(y)$ is a witness to $\psi(x,\cdot)$ inside $H_\kappa$, as desired. QED
Using the lemma, we now prove the remaining part of the theorem. Consider any $\Sigma_2$ assertion $\exists x\,\forall y\, \varphi_0(x,y)$, where $\varphi_0$ has only bounded quantifiers. This assertion is equivalent to $\exists\kappa\, H_\kappa\models\exists x\,\forall y\,\varphi_0(x,y)$, simply because if there is such a $\kappa$ with $H_\kappa$ having such an $x$, then by the lemma this $x$ works for all $y\in V$ since $H_\kappa\prec_{\Sigma_1}V$; and conversely, if there is an $x$ such that $\forall y\, \varphi_0(x,y)$, then this will remain true inside any $H_\kappa$ with $x\in H_\kappa$.
QED
In light of the theorem, it makes sense to refer to the $\Sigma_2$ properties as the
locally verifiable properties, or perhaps as semi-local properties, since positive instances of $\Sigma_2$ assertions can be verified in some sufficiently large $V_\theta$, without need for unbounded search. A truly local property, therefore, would be one such that positive and negative instances can be verified this way, and these would be precisely the $\Delta_2$ properties, whose positive and negative instances are locally verifiable.
Tighter concepts of locality are obtained by insisting that the property is not merely verified in
some $V_\theta$, perhaps very large, but rather is verified in a $V_\theta$ where $\theta$ has a certain closeness to the parameters or instance of the property. For example, a cardinal $\kappa$ is measurable just in case there is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$, and this is verified inside $V_{\kappa+2}$. Thus, the assertion “$\kappa$ is measurable,” has complexity $\Sigma^2_1$ over $V_\kappa$. One may similarly speak of $\Sigma^n_m$ or $\Sigma^\alpha_m$ properties, to refer to properties that can be verified with $\Sigma_m$ assertions in $V_{\kappa+\alpha}$. Alternatively, for any class function $f$ on the ordinals, one may speak of $f$-local properties, meaning a property that can be checked of $x\in V_\theta$ by checking a property inside $V_{f(\theta)}$.
This post was made in response to a question on MathOverflow.
|
Rendezvous with Geostationary Destinations
Geostationary orbits are nontrivial to maintain. They are perturbed by non-round-earth forces represented by the J_{22} term in the spherical harmonics of the gravity field, forming attractors at 75 degrees east (above a point slightly east of the Maldives and south of India) and 105 west (west of the Galapagos, directly south of Denver, Colorado). Evading these attractors can take as much as 1.715 m/s delta V per year.
The North-South Perturbation
The north/south perturbation by the moon and sun are much larger.
For the following analysis, I will approximate both the orbit of the Earth and moon as circles. I also assume that R_S \gg R_M \gg R_G and approximately equal tidal forces at near and far sides of the orbit. A more exact analysis suitable for precision rendezvous would numerically compute the actual multibody elliptical orbits, the spherical harmonic gravity terms for Earth and Moon, gravitational contributions from Jupiter and Venus, and optical perturbations. For now, we are looking for an approximation of maximum \Delta V , but do not forget that the launch, position, and velocity for vehicle in a mature launch loop system will be timed and optically measured to nanoseconds and micrometers, and continuously measured and controlled to this precision throughout the transfer orbit.
R_G
4.216 e4 km
geostationary orbit diameter
\mu_M
4.903 e3 km
moon's standard gravitational parameter
R_M
3.844 e5 km
moon/earth semimajor orbit radius
\mu_S
1.327e11 km
sun standard gravitational parameter
R_S
1.496 e8 km
earth/sun semimajor orbit radius
Lunar North-South Perturbation
The moon's orbit takes it above and below the equatorial plane, which causes out-of-plane tidal forces. The Moon's orbital plane is inclined 5.145 degrees from the ecliptic plane (the plane of the earth's orbit), and precesses over a
nodal period of 18.6 years. The earth's axial tilt is 23°26'21" or 23.439°, so the moon's orbital plane is inclined between 18.294° and 28.584° with respect to the equatorial plane, varying sinusoidally between one and the other over the nodal period.
According to Boden (in Larsen and Wertz) , counteracting these tidal forces on a geostationary object requires a \Delta V budget of as much as 102.67 \cos{ \alpha } \sin{ \alpha } m/s per year ( \equiv ~ 51.335 \sin { 2 \alpha } ), where \alpha is the angle between the equatorial (geostationary orbit) plane and the and the Moon's orbital plane. Boden claims the worst case is 36.93 m/s per year, which implies an \alpha of 23.00°. Actually, the worst case \alpha is 28.584°, so using assuming the first "102.67..." formula and its double-angle "51.3345 ..." equivalent, the worst case annual \Delta V budget is more like 43.13 m/s per year. Let's see if we can derive that 102.67 m/s/y number.
Over the course of a month, the moon moves north of the equatorial plane, through it, then south of the equatorial plane, and back. When the moon is on the equatorial plane, there is no perturbation north or south, though there are tidal forces radially on an object in geostationary orbit. The in-plane tidal acceleration is approximately 2 ~ R_G ~ \mu_M \sin{ \theta } / {R_M}^3 , where \theta is the orbital angle difference between the orbiting object and the moon.
As the moon moves around its orbit, it moves north or south. This puts a northerly component of R_M ~ sin ~ \omega on the moon's position, where \omega is the argument of periapsis, the angle around the moon's orbit from the ascending node relative to the equatorial plane. We can approximate the tidal acceleration as 2 ~ R_G ~ \mu_M \sin{ \theta } \sin{ \omega } \sin{ \alpha } / {R_M}^3 . Perhaps a more accurate analysis would have a sin{ \alpha } \cos{ \alpha }~ \equiv ~ { 1 \over 2 } ~ \sin{ 2 \alpha } factor like Boden. Let's just see if we get in the ballpark.
The thrust needed to counteract the acceleration is the absolute value of the acceleration, and that absolute value can be integrated over time:
Integrated thrust ~ = ~ \int 2 ~ R_G ~ \mu_M ~ ~ sin ~ \theta ~ sin ~ \omega ~ sin ~ \alpha ~ / {R_M}^3 dt
MORE LATER
References
Larsen and Wertz (1999). Space Mission Analysis and Design, Kluwer, 3rd Ed., Page 157 (Chapter Author Daryl G. Boden, PhD, USNA). (?) Soop, E. M. (1994). Handbook of Geostationary Orbits. Springer
|
We have seen how to solve a restricted collection of differential equations, or more accurately, how to attempt to solve them—we may not be able to find the required anti-derivatives. Not surprisingly, non-linear equations can be even more difficult to solve. Yet much is known about solutions to some more general equations.
Suppose $\phi(t,y)$ is a function of two variables. A more general class of first order differential equations has the form $\ds\dot y=\phi(t,y)$. This is not necessarily a linear first order equation, since $\phi$ may depend on $y$ in some complicated way; note however that $\ds\dot y$ appears in a very simple form. Under suitable conditions on the function $\phi$, it can be shown that every such differential equation has a solution, and moreover that for each initial condition the associated initial value problem has exactly one solution. In practical applications this is obviously a very desirable property.
Example 19.4.1 The equation $\ds\dot y=t-y^2$ is a first order non-linear equation, because $y$ appears to the second power. We will not be able to solve this equation.
Example 19.4.2 The equation $\ds\dot y=y^2$ is also non-linear, but it is separable and can be solved by separation of variables.
Not all differential equations that are important in practice can besolved exactly, so techniques have been developed to approximatesolutions. We describe one such technique,
Euler's Method, which is simple though not particularlyuseful compared to some more sophisticated techniques.
Suppose we wish to approximate a solution to the initial value problem $\ds\dot y=\phi(t,y)$, $\ds y(t_0)=y_0$, for $t\ge t_0$. Under reasonable conditions on $\phi$, we know the solution exists, represented by a curve in the $t$-$y$ plane; call this solution $f(t)$. The point $\ds (t_0,y_0)$ is of course on this curve. We also know the slope of the curve at this point, namely $\ds\phi(t_0,y_0)$. If we follow the tangent line for a brief distance, we arrive at a point that should be almost on the graph of $f(t)$, namely $\ds(t_0+\Delta t, y_0+\phi(t_0,y_0)\Delta t)$; call this point $\ds(t_1,y_1)$. Now we pretend, in effect, that this point really is on the graph of $f(t)$, in which case we again know the slope of the curve through $\ds(t_1,y_1)$, namely $\ds\phi(t_1,y_1)$. So we can compute a new point, $\ds(t_2,y_2)=(t_1+\Delta t, y_1+\phi(t_1,y_1)\Delta t)$ that is a little farther along, still close to the graph of $f(t)$ but probably not quite so close as $\ds(t_1,y_1)$. We can continue in this way, doing a sequence of straightforward calculations, until we have an approximation $\ds(t_n,y_n)$ for whatever time $\ds t_n$ we need. At each step we do essentially the same calculation, namely $$(t_{i+1},y_{i+1})=(t_i+\Delta t, y_i+\phi(t_i,y_i)\Delta t).$$ We expect that smaller time steps $\Delta t$ will give better approximations, but of course it will require more work to compute to a specified time. It is possible to compute a guaranteed upper bound on how far off the approximation might be, that is, how far $\ds y_n$ is from $f(t_n)$. Suffice it to say that the bound is not particularly good and that there are other more complicated approximation techniques that do better.
Example 19.4.3 Let us compute an approximation to the solution for $\ds\dot y=t-y^2$, $y(0)=0$, when $t=1$. We will use $\Delta t=0.2$, which is easy to do even by hand, though we should not expect the resulting approximation to be very good. We get $$\eqalign{ (t_1,y_1)&=(0+0.2,0+(0-0^2)0.2) = (0.2,0)\cr (t_2,y_2)&=(0.2+0.2,0+(0.2-0^2)0.2) = (0.4,0.04)\cr (t_3,y_3)&=(0.6,0.04+(0.4-0.04^2)0.2) = (0.6,0.11968)\cr (t_4,y_4)&=(0.8,0.11968+(0.6-0.11968^2)0.2) = (0.8,0.23681533952)\cr (t_5,y_5)&=(1.0,0.23681533952+(0.6-0.23681533952^2)0.2) = (1.0,0.385599038513605)\cr} $$ So $y(1)\approx 0.3856$. As it turns out, this is not accurate to even one decimal place. Figure 19.4.1 shows these points connected by line segments (the lower curve) compared to a solution obtained by a much better approximation technique. Note that the shape is approximately correct even though the end points are quite far apart.
If you need to do Euler's method by hand, it is useful to construct a table to keep track of the work, as shown in figure 19.4.2. Each row holds the computation for a single step: the starting point $(t_i,y_i)$; the stepsize $\Delta t$; the computed slope $\phi(t_i,y_i)$; the change in $y$, $\Delta y=\phi(t_i,y_i)\Delta t$; and the new point, $(t_{i+1},y_{i+1})=(t_i+\Delta t,y_i+\Delta y)$. The starting point in each row is the newly computed point from the end of the previous row.
$(t,y)$ $\Delta t$ $\phi(t,y)$ $\Delta y=\phi(t,y)\Delta t$ $(t+\Delta t,y+\Delta y)$ $(0,0)$ $0.2$ $0$ $0$ $(0.2,0)$ $(0.2,0)$ $0.2$ $0.2$ $0.04$ $(0.4,0.04)$ $(0.4,0.04)$ $0.2$ $0.3984$ $0.07968$ $(0.6,0.11968)$ $(0.6,0.11968)$ $0.2$ $0.58\ldots$ $0.117\ldots$ $(0.8,0.2368\ldots)$ $(0.8,0.236\ldots)$ $0.2$ $0.743\ldots$ $0.148\ldots$ $(1.0,0.385\ldots$
It is easy to write a short function in Sage to do Euler's method.
Euler's method is related to another technique that can help inunderstanding a differential equation in a qualitative way. Euler'smethod is based on the ability to compute the slope of a solutioncurve at any point in the plane, simply by computing $\phi(t,y)$. Ifwe compute $\phi(t,y)$ at many points, say in a grid, and plot a smallline segment with that slope at the point, we can get an idea of howsolution curves must look. Such a plot is called a
slope field. A slope field for $\ds\phi=t-y^2$ is shown in figure 19.4.3; comparethis to figure 19.4.1.With a little practice, one can sketch reasonably accurate solutioncurves based on the slope field, in essence doing Euler's methodvisually.
Even when a differential equation can be solved explicitly, the slope field can help in understanding what the solutions look like with various initial conditions. Recall the logistic equation from exercise 13 in section 19.1, $\ds\dot y = ky(M-y)$: $y$ is a population at time $t$, $M$ is a measure of how large a population the environment can support, and $k$ measures the reproduction rate of the population. Figure 19.4.4 shows a slope field for this equation that is quite informative. It is apparent that if the initial population is smaller than $M$ it rises to $M$ over the long term, while if the initial population is greater than $M$ it decreases to $M$.
Exercises 19.4
In problems 1–4, compute the Euler approximations for the initial value problem for $0\le t\le 1$ and $\Delta t=0.2$. If you have access to Sage, generate the slope field first and attempt to sketch the solution curve. Then use Sage to compute better approximations with smaller values of $\Delta t$.
Ex 19.4.1$\ds\dot y=t/y$, $y(0)=1$(answer)
Ex 19.4.2$\ds\dot y=t+y^3$, $y(0)=1$(answer)
Ex 19.4.3$\ds\dot y=\cos(t+y)$, $y(0)=1$(answer)
Ex 19.4.4$\ds\dot y=t\ln y$, $y(0)=2$(answer)
|
Rendezvous with Geostationary Destinations
Geostationary orbits are nontrivial to maintain. They are perturbed by non-round-earth forces represented by the J_{22} term in the spherical harmonics of the gravity field, forming attractors at 75 degrees east (above a point slightly east of the Maldives and south of India) and 105 west (west of the Galapagos, directly south of Denver, Colorado). Evading these attractors can take as much as 1.715 m/s delta V per year.
The North-South Perturbation
The north/south perturbation by the moon and sun are much larger.
For the following analysis, I will approximate both the orbit of the Earth and moon as circles. I also assume that R_S \gg R_M \gg R_G and approximately equal tidal forces at near and far sides of the orbit. A more exact analysis suitable for precision rendezvous would numerically compute the actual multibody elliptical orbits, the spherical harmonic gravity terms for Earth and Moon, gravitational contributions from Jupiter and Venus, and optical perturbations. For now, we are looking for an approximation of maximum \Delta V , but do not forget that the launch, position, and velocity for vehicle in a mature launch loop system will be timed and optically measured to nanoseconds and micrometers, and continuously measured and controlled to this precision throughout the transfer orbit.
R_G
4.216 e4 km
geostationary orbit diameter
\mu_M
4.903 e3 km
moon's standard gravitational parameter
R_M
3.844 e5 km
moon/earth semimajor orbit radius
\mu_S
1.327e11 km
sun standard gravitational parameter
R_S
1.496 e8 km
earth/sun semimajor orbit radius
Lunar North-South Perturbation
The moon's orbit takes it above and below the equatorial plane, which causes out-of-plane tidal forces. The Moon's orbital plane is inclined 5.145 degrees from the ecliptic plane (the plane of the earth's orbit), and precesses over a
nodal period of 18.6 years. The earth's axial tilt is 23°26'21" or 23.439°, so the moon's orbital plane is inclined between 18.294° and 28.584° with respect to the equatorial plane, varying sinusoidally between one and the other over the nodal period.
According to Boden (in Larsen and Wertz) , counteracting these tidal forces on a geostationary object requires a \Delta V budget of as much as 102.67 \cos{ \alpha } \sin{ \alpha } m/s per year ( \equiv ~ 51.335 \sin { 2 \alpha } ), where \alpha is the angle between the equatorial (geostationary orbit) plane and the and the Moon's orbital plane. Boden claims the worst case is 36.93 m/s per year, which implies an \alpha of 23.00°. Actually, the worst case \alpha is 28.584°, so using assuming the first "102.67..." formula and its double-angle "51.3345 ..." equivalent, the worst case annual \Delta V budget is more like 43.13 m/s per year. Let's see if we can derive that 102.67 m/s/y number.
Over the course of a month, the moon moves north of the equatorial plane, through it, then south of the equatorial plane, and back. When the moon is on the equatorial plane, there is no perturbation north or south, though there are tidal forces radially on an object in geostationary orbit. The in-plane tidal acceleration is approximately 2 ~ R_G ~ \mu_M \sin{ \theta } / {R_M}^3 , where \theta is the orbital angle difference between the orbiting object and the moon.
As the moon moves around its orbit, it moves north or south. This puts a northerly component of R_M ~ sin ~ \omega on the moon's position, where \omega is the argument of periapsis, the angle around the moon's orbit from the ascending node relative to the equatorial plane. We can approximate the tidal acceleration as 2 ~ R_G ~ \mu_M \sin{ \theta } \sin{ \omega } \sin{ \alpha } / {R_M}^3 . Perhaps a more accurate analysis would have a \sin{ \alpha } \cos{ \alpha } ~ \equiv ~ { 1 \over 2 } ~ \sin{ 2 \alpha } factor like Boden. Let's just see if we get in the ballpark.
The thrust needed to counteract the acceleration is the absolute value of the acceleration, and that absolute value can be integrated over time:
Integrated ~ thrust ~ = ~ \int{ \left| 2 ~ R_G ~ \mu_M ~ \sin{ \theta } \sin{ \omega } \sin{ \alpha } / {R_M}^3 \right| dt }
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ = ~ 2 ~ R_G ~ \mu_M ~ / {R_M}^3 ~ \int{ \left| \sin{ \theta } \sin{ \omega } \sin{ \alpha } \right| dt }
\sin{ \alpha } varies slowly, over the 18.6 year nodal period. We are more interested in its maximum, as that is what we must design the thrust our north and south correction thrusters and fuel supply for, assuming frequent resupply. For the monthly \sin{ \omega } and daily \sin { theta } terms, we can assume the maximum of each ( = 1 ) for the maximum correction thruster thrust, and the average of each ( = 2 / \pi ) for fuel supply.
So the maximum thrust per second ( at maximum \alpha } is:
Maximum thrust \approx ~ 2 ~ R_G ~ \mu_M ~ \sin { \alpha_{max} } / {R_M}^3 ~ = ~ 3.48e-6 m/s
2
And the average thrust per year ( at maximum \alpha } is:
Average thrust \approx ~ \left( 8 \over \pi^2 \right) ~ year~ R_G ~ \mu_M ~ \sin { \alpha_{max} } / {R_M}^3 ~ = ~ 44.5 m/s/y
That is 3% above the Boden value (after correction), so we are close. If we use the sin*cos correction from Boden, we come in 10% low, 39.1 m/s/y , so there is still something slightly wrong in my analysis. Use the Boden numbers.
Solar North-South Perturbation
The perturbation from the sun is simpler; the inclination of the equatorial plane to the ecliptic plane ( \gamma = 23.439° ) does not change much over a few decades. By similar reasoning, the maximum solar thrust per second is:
Maximum thrust \approx ~ 2 ~ R_G ~ \mu_S ~ \sin { \gamma } / {R_S}^3 ~ = ~ 1.33e-6 m/s
2
And the average solar thrust per year is:
Average thrust \approx ~ \left( 8 \over \pi^2 \right) ~ year~ R_G ~ \mu_S ~ \sin { \gamma } / {R_S}^3 ~ = ~ 17.0 m/s/y
The latter number is 18% lower than Boden, so there is something more than slightly wrong. Again, use the Boden numbers.
Correction Delta V for an off-equator Launch Loop
For a launch loop exactly on the equator, the destination GEO station will be easy to rendezvous with, requiring at most a few meters per second \Delta V for rendezvous.
Launch loops to GEO may be sited 5 degrees south of the equator to avoid equatorial weather. That means the minimum equatorial inclination is 5°. A geostationary transfer orbit to a geostationary CaptureRail has an apogee velocity of 1335 m/s, and we must have zero north-south velocity for that to work right (unless the rail is spinning diagonally). So, the vehicle's north-south velocity must be corrected by approximately 1335 m/s sin( 5° ) or 116 m/s in the 18500 seconds transfer fractional orbit time, an acceleration of 6.3e-3 meters per second per second. Assuming a 5 ton vehicle, that is a thrust of 31 newtons. If that is made with a laser ablative thruster with an effective exhaust velocity of 5000 meters per second, that is a propellant rate of 6.3 grams per second, or about 120 kilograms of propellant expended, and a power delivery rate of 80 kW . A planar thrust panel will be inefficient, so both the power and the propellant expended will be higher than that.
Off-equator launch loops will require some as-yet-unimagined modification to capture rail, or expend a lot of propellant, to arrive. This is bothersome, and violates the "minimal propellant in orbit" goal. More invention needed.
MORE LATER
References
Larsen and Wertz (1999). Space Mission Analysis and Design, Kluwer, 3rd Ed., Page 157 (Chapter Author Daryl G. Boden, PhD, USNA). (?) Soop, E. M. (1994). Handbook of Geostationary Orbits. Springer
|
Search
Now showing items 1-1 of 1
Anisotropic flow of inclusive and identified particles in Pb–Pb collisions at $\sqrt{{s}_{NN}}=$ 5.02 TeV with ALICE
(Elsevier, 2017-11)
Anisotropic flow measurements constrain the shear $(\eta/s)$ and bulk ($\zeta/s$) viscosity of the quark-gluon plasma created in heavy-ion collisions, as well as give insight into the initial state of such collisions and ...
|
Update. I've improved the argument to use only the consistency of $T$. (2/7/12): I corrected some over-statements previously made about Robinson's Q.
I claim that for every statement $\varphi$, there is a variant wayto express it, $\psi$, which is equivalent to the originalstatement $\varphi$, but which is formally independent of anyparticular desired consistent theory $T$.
In particular, if $\varphi$ is your favorite natural open question,whose truth value is unknown, then there is an equivalentformulation of that question which exhibits formal independence inthe way you had requested. In this sense,
every open question isequivalent to an assertion with the property you have requested. Itake this to reveal certain difficult subtleties with your project.
Theorem. Suppose that $\varphi$ is any sentence and $T$ is any consistent theory containing weak arithmetic. Then there is another sentence $\psi$ such that $\text{PA}+\text{Con}(T)$ proves that $\varphi$ and $\psi$ are equivalent. $T$ does not prove $\psi$. $T$ does not prove $\neg\psi$.
Proof. Let $R$ be the Rosser sentence for $T$, the self-referential assertion that for any proof of $R$ in $T$, there is a smaller proof of $\neg R$. The Gödel-Rosser theorem establishes that if $T$ is consistent, then $T$ proves neither $R$ nor $\neg R$. Formalizing the first part of this argument shows that $\text{PA}+\text{Con}(T)$ proves that $R$ is not provable in $T$ and hence that $R$ is vacuously true. Formalizing the second part of this argument shows that $\text{Con}(T)$ implies $\text{Con}(T+R)$, and hence by the incompleteness theorem applied to $T+R$, we deduce that $T+R$ does not prove $\text{Con}(T)$. Thus, $T+R$ is a strictly intermediate theory between $T$ and $T+\text{Con}(T)$.
Now, let $\psi$ be the assertion $R\to (\text{Con}(T)\wedge \varphi)$. Since $\text{PA}+\text{Con}(T)$ proves $R$, it is easy to see by elementary logic that $\text{PA}+\text{Con}(T)$ proves that $\varphi$ and $\psi$ are equivalent.
The statement $\psi$, however, is not provable in $T$, since if it were, then $T+R$ would prove $\text{Con}(T)$, which it does not by our observations above.
Conversely, $\psi$ is not refutable in $T$, sinceany such refutation would mean that $T$ proves that the hypothesisof $\psi$ is true and the conclusion false; in particular, itwould require $T$ to prove the Rosser sentence $R$, which it does not by the Gödel-Rosser theorem. QED
Note that any instance of non-provability from $T$ will require the consistency of $T$, and so one cannot provide a solution to the problem without assuming the theory is consistent.
The observation of the theorem has arisen in some of the philosophical literature you mayhave in mind, based on what you said in the question. For example, the claim of the theorem is mentioned in Haim Gaifman's newpaper "
On ontology and realism in mathematics," which we read in my course last semesteron the philosophy of set theory; see the discussion on page 24 of Gaifman's paper and specifically footnote 35, where he credits a fixed-point argument to Torkel Franzen, and an independent construction to Harvey Friedman.
My original argument (see edit history) used the sentence $\text{Con}(T)\to(\text{Con}^2(T)\wedge\varphi)$, where $\text{Con}^2(T)$ is the assertion $\text{Con}(T+\text{Con}(T))$, and worked under the assumption that $\text{Con}^2(T)$ is true, relying on the fact that $T+\text{Con}(T)$ is strictly between $T$ and this stronger theory. The current argument uses the essentially similarly idea that $T+R$ is strictly between $T$ and $T+\text{Con}(T)$, thereby reducing the consistency assumption.
|
Let $\phi$ be a scalar field in an interacting theory ($\phi^3$ or $\phi^4$, for example). If $|0\rangle$ is the vacuum of the interacting theory and $P^\mu$ is the four-momentum operator, we have that
$$\langle 0 | \phi(x) | 0 \rangle = \langle 0 | e^{-iPx} \phi(0) e^{iPx} | 0 \rangle = \langle 0 | \phi(0) | 0 \rangle $$
In chapter 5 of his QFT book, Srednicki says that
We would like $ \langle 0 | \phi(0) | 0 \rangle$ to be zero. This is because we would like $a_1^\dagger (\pm \infty)$, when acting on $|0\rangle$, to create a single particle state. We do
notwant $a_1^\dagger (\pm \infty)$ to create a linear combination of a single particle state and the ground state.
Here $a_1^\dagger (\pm \infty)$ is the creation operator $a^\dagger$ for a momentum $\mathbf{k}_1$ taken at time $\pm \infty$, which (according to the book) guarantees that the particle is located away from the origin. In other words, we define $|k\rangle = \lim\limits_{t \to -\infty} a^\dagger(\mathbf{k}, t) |0\rangle$.
It seems to me that Srednicki wants $\langle 0 | k \rangle = 0$, which sounds reasonable. But applying the LSZ formula for the special case of one initial particle and zero final particles, I get $\langle 0 | k \rangle = i (2\pi)^4 m^2 \langle 0 | \phi(0) | 0 \rangle \delta^4(k)$ (the $2\pi$'s might be off). This is nonzero only when $k^\mu=0$, a.k.a. never, since $k$ must be on shell. So why must we ask that the fields's VEV be zero, when it seems that $\langle 0 | k \rangle = 0$ anyway?
|
Consider all non-empty subsets \(S_1,S_2,\ldots,S_{2^n-1}\) of \(\{1,2,3,\ldots,n\}\). Let \(A=(a_{ij})\) be a \((2^n-1)\times(2^n-1)\) matrix such that \[a_{ij}=\begin{cases}1 & \text{if }S_i\cap S_j\ne \emptyset,\\0&\text{otherwise.}\end{cases}\] What is \(\lvert\det A\rvert\)?
(This is the last problem of this semester. Good luck with your final exam!)
GD Star Rating loading...
Compute \(\int_0^1 \frac{x^k-1}{\log x}dx\).
The best solution was submitted by Myeongjae Lee (이명재), 2012학번. Congratulations!
Here is his Solution of Problem 2012-22.
Alternative solutions were submitted by 박민재 (2011학번, +3), 서기원 (수리과학과 2009학번, +2), 김태호 (수리과학과 2011학번, +2), 임현진 (물리학과 2010학번, +2), 조위지 (Stanford Univ. 물리학과 박사과정, +3), 박훈민 (대전과학고 2학년, +3).
GD Star Rating loading...
Prove that for each positive integer \(n\), there exist \(n\) real numbers \(x_1,x_2,\ldots,x_n\) such that \[\sum_{j=1}^n \frac{x_j}{1-4(i-j)^2}=1 \text{ for all }i=1,2,\ldots,n\] and \[\sum_{j=1}^n x_j=\binom{n+1}{2}.\]
GD Star Rating loading...
Let \(n\) be a fixed positive integer and let \(p\in (0,1)\). Let \(D_n\) be the determinant of a random \(n\times n\) 0-1 matrix whose entries are independent identical random variables, each of which is 1 with the probability \(p\) and 0 with the probability \(1-p\). Find the expected value and variance of \(D_n\).
The best solution was submitted by Myeongjae Lee (이명재), 2012학번. Congratulations!
Here is his Solution of Problem 2012-21.
Alternative solutions were submitted by 박민재 (2011학번, +3), 김태호 (수리과학과 2011학번, +3), 임현진 (물리학과 2010학번, +3), 김지홍 (수리과학과 2007학번, +2), 서기원 (수리과학과 2009학번, +2).
GD Star Rating loading...
Compute \(\int_0^1 \frac{x^k-1}{\log x}dx\).
GD Star Rating loading...
Let \(A=(a_{ij})\) be an \(n\times n\) upper triangular matrix such that \[a_{ij}=\binom{n-i+1}{j-i}\] for all \(i\le j\). Find the inverse matrix of \(A\).
The best solution was submitted by Minjae Park (박민재), 2011학번. Congratulations!
Here is his Solution of Problem 2012-20.
Alternative solutions were submitted by 서기원 (수리과학과 2009학번, +3), 이명재 (2012학번, +3), 김태호 (수리과학과 2011학번, +3), 임현진 (물리학과 2010학번, +3), 박훈민 (대전과학고 2학년, +3), 윤성철 (홍익대학교 수학교육학과 2009학번, +3), 어수강 (서울대학교 수리과학부 석사과정, +3).
GD Star Rating loading...
Let \(n\) be a fixed positive integer and let \(p\in (0,1)\). Let \(D_n\) be the determinant of a random \(n\times n\) 0-1 matrix whose entries are independent identical random variables, each of which is 1 with the probability \(p\) and 0 with the probability \(1-p\). Find the expected value and variance of \(D_n\).
GD Star Rating loading...
Let \(A=(a_{ij})\) be an \(n\times n\) upper triangular matrix such that \[a_{ij}=\binom{n-i+1}{j-i}\] for all \(i\le j\). Find the inverse matrix of \(A\).
GD Star Rating loading...
|
Let $\mathcal{G}$ be the set of isomorphism classes of finite groups.
There is an operation $\mathrm{Aut} : \mathcal{G} \rightarrow \mathcal{G}$ which gives the automorphism group of a given group, up to isomorphism. In the most general setting, my question is:
What possible eventual behaviour can arise from iterating the operation $\mathrm{Aut}$?
More precisely, there is a trichotomy of possible eventual behaviours:
Static:a group such as $D_8$ or $S_8$, which is isomorphic to its automorphism group. Periodic:a group which enters a cycle of length $\geq 2$. Divergent:a group which never enters a cycle, and therefore grows without bound.
I can exhibit lots of examples of static groups, and groups which eventually evolve into static groups after many steps, such as:
$$ C_{2879} \mapsto C_{2878} \mapsto C_{1438} \mapsto C_{718} \mapsto C_{358} \mapsto C_{178} \mapsto C_{88} \\ \mapsto C_2 \times C_2 \times C_2 \times C_5 \mapsto PSL(2,7) \times C_4 \mapsto PGL(2, 7) \times C_2 $$
However, I don't have any examples of periodic or divergent groups.
Does there exist a periodic group (duplicate of Periodic Automorphism Towers )? Does there exist a divergent group ( nota duplicate of Does $\mathrm{Aut}(\mathrm{Aut}(...\mathrm{Aut}(G)...))$ stabilize? because the answer does not consider the group merely up to isomorphism)? What is the eventual behaviour of the prime cyclic group $C_{41}$ (the smallest prime cyclic group whose behaviour I have been unable to track)?
I conjecture that $C_{41}$ does diverge, simply because after several iterations it spawns a direct sum involving lots of copies of $C_2$:
$$ C_{41} \mapsto C_{40} \mapsto C_4 \times C_2 \times C_2 \mapsto D_4 \mapsto F_4/Z \times C_2 \\ \mapsto (S_4 \wr C_2) \times C_2 \\ \mapsto (S_4 \wr C_2) \times C_2 \times C_2 \\ \mapsto (S_4 \wr C_2) \times S_4 \times C_2 \times C_2 \\ \mapsto (S_4 \wr C_2) \times S_4 \times S_4 \times C_2 \times C_2 \times C_2 \times C_2 $$
Finally, for the most ambitious question of all:
Does there exist an algorithm which, when given a description of a Turing machine, outputs a finite group which is divergent if and only if the Turing machine does not halt?
|
How to Model Optical Anisotropic Media with COMSOL Multiphysics®
On a bright evening in 1669, Professor Erasmus Bartholinus looked through a piece of an Icelandic calcite crystal he had placed onto a bench. He observed when he covered text on the bench with the stone, it appeared as a double image. The observed optical phenomenon, called birefringence, involves a beam of light that splits into two parallel beams while emerging out of a crystal. Here, we demonstrate a modeling approach for this effect.
Understanding Anisotropic Materials
The beam of light that Erasmus Bartholinus observed traveling straight through the crystal is called an ordinary ray. The other light beam, which bends while traveling through the crystal, is an extraordinary ray. Anisotropic materials, such as the crystal from the stone and bench experiment described above, are found in applications ranging from detecting harmful gases to beam splitting for photonic integrated circuits.
Ordinary and extraordinary rays traveling through an anisotropic crystal.
In a physical context, when an unpolarized electromagnetic beam of light propagates through an anisotropic dielectric material, it polarizes the dielectric domain, leading to a distribution of charges known as
electric dipoles. This phenomenon leads to induced fields within the anisotropic dielectric material, wherein two kinds of waves experience two different refractive indices (ordinary and extraordinary).
The ordinary wave is polarized perpendicular to the principal plane and the extraordinary wave is polarized parallel to the principal plane, where the principal plane is spanned by the optic axis and the two propagation directions in the crystal. Because of this behavior, the waves propagate with different velocities and trajectories.
Introducing Anisotropy Within Silicon Waveguides
In a previous blog post, we discussed silicon and how its derivative, silicon dioxide, is used extensively in photonic integrated chips due to its compatibility with the CMOS fabrication technique. Bulk silicon, which has an isotropic property, is used to develop prototypes for photonic integrated chips. However, due to unique optical properties such as splitting beams and polarization-based optical effects, anisotropy comes into play at a later stage.
Anisotropy in silicon photonics occurs unintentionally due to the annealing process while fabricating the waveguide. The difference in thermal expansion between the core and cladding causes geometry mismatch due to stress optical effects, which results in effects such as mode splitting and pulse broadening. Anisotropy could also be intentionally introduced by varying the porosity of silicon dioxide. This enables researchers to work with a range of effective refractive indices from silicon dioxide (n ~1.44) to air (n ~1), giving them the edge to perform very sensitive optical sensor applications.
Optical Modes of Propagation
To perform qualitative analyses of anisotropic media, researchers investigate how optical energy propagates within planar waveguides (also known as modes of propagation). In planar waveguides, we define modes using E^{x}_{p,q} and E^{y}_{p,q} terminology (Ref. 2), where
x and y depict the direction of polarization and p and q depict the number of maxima in the x– and y-coordinates.
Picture it this way: You are walking on an E^{x}_{2,1} “landscape” (as shown below). The “winds” (polarization) are along the ±x direction, and you encounter two distinct peaks when traveling from the –
x to + x direction. When you move from the – y to + y direction, you observe both of the peaks simultaneously. Mode analysis of the planar waveguide. Top row, left to right: E^{x}_{1,1} and E^{y}_{1,1}. Middle row, left to right: E^{x}_{1,2} and E^{y}_{1,2}. Bottom row, left to right: E^{x}_{2,1} and E^{y}_{2,1}. The arrow plot represents the electric field; contour and surface plot represent out-of-plane power flow (red is high and blue is low magnitude). Analyzing Anisotropic Structures in the COMSOL Multiphysics® Software
Before launching a beam of light through a waveguide using a laser source, it is important to know which optical modes could persist within a specified core/cladding dimension of the waveguide. Performing a mode analysis using a full vectorial finite element tool, such as the COMSOL Multiphysics® software, could be very helpful to qualitatively and quantitatively analyze the optical modes and dispersion curve respectively.
Introducing Diagonal Anisotropy
Performing a modal analysis on any isotropic material requires the definition of a single complex value, while in the case of an anisotropic material, a full tensor relative permittivity approach is required. The electric permittivity essentially relates the electric field with the material property. Here,
tensor refers to a 3-by-3 matrix that has both diagonal (\epsilon xx, \epsilon yy, \epsilon zz) and off-diagonal (\epsilon xy, \epsilon xz, \epsilon yx, \epsilon yz, \epsilon zx, \epsilon zy) terms as shown below.
\epsilon_{xx}&\epsilon _{xy}&\epsilon _{xz}\\
\epsilon _{yx}&\epsilon _{yy}&\epsilon _{yz}\\
\epsilon _{zx}&\epsilon _{zy}&\epsilon _{zz}\\
\end{bmatrix}
However, for all materials, you can find a coordinate system in which you only have nonzero diagonal elements in the permittivity tensor, whereas the off-diagonal elements are all zero. The three coordinate axes in this rotated coordinate system are the principal axes of the material and, correspondingly, the three values for the diagonal elements in the permittivity tensor are called the principal permittivities of the material.
There are basically two kinds of anisotropic crystal: uniaxial and biaxial crystal. With a suitable choice of coordinate system, where only the diagonal elements of the permittivity tensor are nonzero, in terms of optical properties,
uniaxial crystal considers only the diagonal terms, that is \epsilon = \epsilon xx = ( yy n) o 2, \epsilon = ( zz n) e 2, where nand o nare the ordinary and extraordinary refractive indices. However, when \epsilon_{xx}\neq \epsilon_{yy} \neq \epsilon_{zz}, it is known as a e biaxial crystal.
To put this argument into a modeling perspective, we can extend the buried rib waveguide example from this blog post on silicon photonics design. We can perform a modal analysis on the 2D cross section of the waveguide with the square core and cladding length of 4 um and 20 um, respectively (shown below). The operating wavelength for all the cases is considered as 1.55[um].
Schematic of 3D buried rib optical waveguide where the mode analysis was performed at the inlet 2D cross section. The intensity plot and arrow plot representing the mode and polarization of E-fields respectively. Core of the rib waveguide depicting the optic axis (red) along the x -axis and the principal axis (blue).
In the classic case of a uniaxial material, we assume the optic axis (i.e.,
c-axis) is along the principal x-axis (as shown above) and consider the diagonal relative permittivity \epsilon and \epsilon yy terms (which are orthogonal to the zz c-axis) as the square of ordinary refractive index (~1.5199 2~ 2.31). The \epsiloncomponent element that is along the xx c-axis is considered as the square of extraordinary refractive index (~1.4799 2~ 2.19) (as per Ref. 3). In addition, the off-diagonal terms are considered zero (as shown below) and the cladding has an isotropic relative permittivity (~1.4318 2). The optical modes derived are the 6 modes shown above. Note the difference in the refractive indices: “ n– xx n” is known as birefringence, where yy n= \sqrt{\epsilon_{xx}} and xx n= \sqrt{\epsilon_{yy}}. yy
\begin{bmatrix}
2.19 & 0 & 0 \\
0 & 2.31 & 0\\
0 & 0 & 2.31\\
\end{bmatrix}
Relative permittivity tensor with diagonal elements. Dispersion Curves
By evaluating the optical modes, we can visually comprehend the behavior of the optical waveguide. However, the dispersion curves could also be handy for performing quantitative analyses. A dispersion curve represents the variation of the effective refractive index with respect to the length of the waveguide or the operating frequency.
Diagonal Anisotropy
A modal analysis is performed while parametrically sweeping the length of the waveguide from 0.5 um to 4 um to derive the dispersion curve for the anisotropic core, as shown in the figure below. We assume the earlier case stated, with diagonal anisotropy terms of the core (i.e., \epsilon
= 2.19, \epsilon xx = \epsilon yy = 2.31 and all of the diagonal elements are zero). The results are compared with Koshiba et al. (Ref. 3). zz Dispersion curve with transverse anisotropic core. Off-Diagonal Transverse Anisotropy (XY Plane)
When the optic axis (i.e.,
c-axis) lies in XY plane and makes an angle of \theta with the x-axis, the diagonal components \epsilon , \epsilon xx , \epsilon yy and off-diagonal components \epsilon zz and \epsilon xy are nonzero, while the rest of the components are zero. The full relative permittivity tensor could be evaluated by using the rotation matrix [R] as shown below, where the rotation matrix [R] is specifically for rotating the yz c-axis in the XY plane. \epsilon is the square of the extraordinary refractive index (~2.19), because the xx c-axis lies along the principal x-axis, while \epsilon and \epsilon yy are the square of the ordinary refractive index (~2.31). The off-diagonal elements \epsilon zz and \epsilon xy are derived from the multiplication of the matrices as stated below. yz The c -axis lying in the XY plane and making an angle of \theta with the x -axis.
\begin{bmatrix}
cos(\theta) & -sin(\theta) & 0 \\
sin(\theta) & cos(\theta) & 0\\
0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
\epsilon_{xx} & 0 & 0 \\
0 & \epsilon_{yy} & 0 \\
0 & 0 & \epsilon_{zz} \\
\end{bmatrix}
\begin{bmatrix}
cos(\theta) & sin(\theta) & 0 \\
-sin(\theta) & cos(\theta) & 0\\
0 & 0 & 1\\
\end{bmatrix}
\begin{bmatrix}
(\epsilon_{xx}) cos^2(\theta) + (\epsilon_{yy}) sin^2(\theta) & (\epsilon_{xx}) sin(\theta) cos(\theta)-(\epsilon_{yy}) sin(\theta) cos(\theta) & 0 \\
(\epsilon_{xx}) sin(\theta)cos(\theta)-(\epsilon_{yy})sin(\theta)cos(\theta) & (\epsilon_{yy}) cos^2(\theta) + (\epsilon_{xx}) sin^2(\theta) & 0\\
0 & 0 & \epsilon_{zz}\\
\end{bmatrix}
The relative permittivity tensor ε is treated along with a rotation matrix, rotating the c -axis in the XY plane with angle \theta.
Finally, the modal analysis of the waveguide with off-diagonal anisotropic core and isotropic cladding, where the optic axis makes angles of 0, 15, 30, and 45 degrees with respect to the principal
x-axis, as shown below. Here, it could be observed that the direction of the in-plane magnetic field changes according to the change in the angle of the optic axis. The dispersion curve could also be plotted by parametrically sweeping the length of the core and cladding from 0.5 um to 4 um, while considering the angle \theta as 45°. The dispersion curve tends to be similar to the dispersion curve of the diagonal anisotropy, as discussed above. Mode analysis, including off-diagonal terms, for θ = 0° (top-left), θ = 15° (top-right), θ = 30° (bottom-left), and θ = 45° (bottom-right). The figure represents the magnetic field lines within the core for different rotation angles. Off-Diagonal Longitudinal Anisotropy (YZ Plane)
Finally, when considering the longitudinal anisotropy where the optic axis (i.e.,
c-axis) lies in the YZ plane and makes an angle of \phi with the y-axis, the diagonal components \epsilon , \epsilon xx , \epsilon yy and the off-diagonal components \epsilon zz and \epsilon yz are nonzero, while the rest of the components are zero. The relative permittivity tensor could be evaluated by using the rotation matrix [R] as shown below, where the rotation matrix [R] is specifically for rotating the zy c-axis in the YZ plane. \epsilon is the square of the extraordinary refractive index (~2.19), because the yy c-axis lies along the principal y-axis, while \epsilon , \epsilon xx is the square of the ordinary refractive index (~2.31). The off-diagonal elements \epsilon zz and \epsilon yz are derived from the multiplication of the matrices as stated below. zy The c -axis lying in the YZ plane and making an angle of \phi with the x -axis.
\begin{bmatrix}
1 & 0 & 0 \\
0 & cos(\phi) & -sin(\phi)\\
0 & sin(\phi) & cos(\phi) \\
\end{bmatrix}
\begin{bmatrix}
\epsilon_{xx} & 0 & 0 \\
0 & \epsilon_{yy} & 0 \\
0 & 0 & \epsilon_{zz} \\
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 \\
0 & cos(\phi) & sin(\phi)\\
0 & -sin(\phi) & cos(\phi) \\
\end{bmatrix}
\begin{bmatrix}
\epsilon_{xx} & 0 & 0 \\
0 & (\epsilon_{yy}) cos^2(\phi) + (\epsilon_{zz}) sin^2(\phi) & (\epsilon_{yy})sin(\phi)cos(\phi)-(\epsilon_{zz}) sin(\phi)cos(\phi)\\
0 & (\epsilon_{yy})sin(\phi)cos(\phi)-(\epsilon_{zz}) sin(\phi)cos(\phi) & (\epsilon_{zz}) cos^2(\phi) + (\epsilon_{yy}) sin^2(\phi)\\
\end{bmatrix}
The relative permittivity tensor ε is treated along with a rotation matrix, rotating in the YZ plane with angle \phi.
A modal analysis is then performed where the length of the waveguide is parametrically swept from 0.5 um to 4 um to derive the dispersion curve for the longitudinal anisotropic core, as shown in the figure below. In this case, \phi = 45° (i.e., the
c-axis lies in the YZ plane and makes 45° with the y-axis) (Ref. 3). Dispersion curve with longitudinal anisotropic core. Final Thoughts on Modeling Anisotropic Materials
In this blog post, we performed qualitative analyses (modes of propagation) and quantitative analyses (dispersion curves) of the anisotropic optical waveguide using modal analysis in COMSOL Multiphysics. Diagonal anisotropy as well as off-diagonal transverse and longitudinal anisotropy were considered to derive their dispersion relationships. These types of analyses give us more flexibility when carrying out optimization of material and geometric parameters to help us gain an in-depth and intuitive understanding of the physics of anisotropic materials.
A simple tutorial model to help you started would be the Step-Index Fiber, which involves mode analysis over a 2D cross section of the 3D optical fiber.
Next Steps
To try these models, click the button below. This will take you to the Application Gallery, where you can download the related MPH-files as long as you have a COMSOL Access account and valid software license.
Updated List of Blog Posts in the Silicon Photonics Series References E. Hecht, Optics, Pearson. E.A.J. Marcatili, “Dielectric rectangular waveguide and directional coupler for integrated optics”, Bell Syst. Tech. J., vol. 48, pp. 2071–2102, 1969. M. Koshiba, K. Hayata, and M. Suzuki, “Finite-element solution of anisotropic waveguides with arbitrary tensor permittivity,” Journal of Lightwave Technology, vol. 4, no. 2, pp. 121–126, 1986. Comments (4) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science
|
(This post is a continuation of my previous one on Liouville structures , and hence is quite technical. It’s aimed at students of geometry, not the general public. It assumes you know some differential geometry. If you know what a Liouville structure is, read on.)
We’re going to take Liouville structures and move them into 3 dimensions, to obtain
contact structures.
As we’ve seen, a Liouville structure on a surface \(S\) (which necessarily has boundary, or is non-compact) is given by a 1-form \(\beta\) such that \(d\beta\) is non-degenerate. Then \(d\beta\) is a symplectic form on \(S\), and \(\beta\) is dual to a vector field \(X\) via the symplectic form, i.e. \(\iota_X d\beta = \beta\). This structure has the nice property that \(X\) points along \(\ker \beta\), and the flow of \(X\) expands \(\beta\) and \(d\beta\) exponentially. In equations, \(\beta(X) = 0\), \(L_X \beta = \beta\), and \(L_X d\beta = d\beta\).
Let’s now go into the
next dimension and consider a 3-dimensional space (or 3-manifold) \(M = S \times [0,1]\). We can use \(x,y\) as coordinates on \(S\), and \(t\) as a coordinate on \([0,1]\). So this is a thickening of \(S\); you can think of \(S\) as horizontal, and \([0,1]\) as vertical, with the coordinate \(t \in [0,1]\) measuring height. (However \(M\) certainly has boundary; and if \(S\) is non-compact, then the same is true for \(M\).)
On this 3-manifold \(M\), let’s consider a 1-form. The 1-form \(\beta\) is no more interesting here than it is on \(S\), but let’s add to it a form using the third dimension. We’ll just add to \(\beta\) the simplest possible such form, \(dt\).
So define a 1-form \(\alpha\) on \(M\) by \(\alpha = \beta + dt\).
Previously, we considered the example of a Liouville structure on \(S = \mathbb{R}^2\), given by \(\beta = x \; dy\). We will continue this running example. We obtain \(M = \mathbb{R}^2 \times [0,1]\), with the 1-form \(\alpha = x \; dy + dt\).
It’s harder to draw pictures of 1-forms in 3 dimensions, but it is possible! Just as in the 2-dimensional case, we can draw the
kernel of the 1-form. But whereas the kernel of a (nowhere zero) 1-form on a surface is a line field, the kernel of a (nowhere zero) 1-form on a 3-manifold is a plane field.
Now it’s not difficult to see that, in our example, no matter where you are, the \(x\)-direction always lies in \(\ker \alpha\). For \(\alpha\) only contains a \(dy\) and a \(dt\) term; if you feed it a vector in the \(x\)-direction, you get zero. But to get a second linearly independent vector in the kernel, you need to take a more carefully chosen combination of the \(y\) and \(t\) directions. We can check that the combination \(\partial_y – x \partial_t\) lies in the kernel; here \(\partial_y, \partial_t\) denote unit vectors in the \(y\) and \(t\) directions.
\[ \alpha(\partial_y – x \partial_t) = (x \; dy + dt)(\partial_y – x \partial_t) = x \; dy(\partial_y) – dt(x \partial_t) = x-x=0. \] Thus, at each point \((x,y,t)\) in \(M\), \(\ker \alpha\) is spanned by \(\partial_x\) and \(\partial_y – x \partial_t\). We can write \[ \ker \alpha = \langle \partial_x, \; \partial_y – x \partial_t \rangle. \]
Let’s try to visualise \(\ker \alpha\) geometrically. It’s a plane field: there is a plane at each point in \(\mathbb{R}^2 \times [0,1]\). The plane at \((x,y,t)\) is spanned by \(\partial_x\) and \(\partial_y – x \partial_t\). When \(x=0\), the plane is spanned by \(\partial_x\) and \(\partial_y\): it’s horizontal. As \(x\) varies, \(\partial_x\) always points along the plane, and so the planes “spin” around each line in the \(x\)-direction. As \(x\) increases from \(0\), the plane tilts: rather than \(\partial_y\), the plane contains \(\partial_y -x \partial_t\), and this negative \(\partial_t\) component becomes larger as \(x\) increases increases; as \(x\) becomes large, the plane becomes close to vertical. And similarly as \(x\) decreases from \(0\): the plane gains a positive \(\partial_t\) component.
This is in fact quite a standard geometric object, at least if you’re into the field of
contact geometry. This plane field given by the kernel of \(x \; dy + dt\) is also known (possibly after some relabllings of variables, and some sign changes) as the standard contact structure on \(\mathbb{R}^3\).
This plane field has some great properties. Try to find a surface which runs tangent to it! That is, try to find a surface in \(\mathbb{R}^2 \times [0,1]\), which at every point is tangent to this plane field. Even if you try to find a tiny surface, you will not succeed — because this plane field has a property called
non-integrability.
Frobenius’ theorem in differential geometry tells you when a plane field is integrable (i.e. tangent to a surface) or not. (In fact, it says this not just about 2-dimensional plane fields in 3 dimensions, but in general about any-dimensional plane fields in any number of dimensions.) There is a vector-field version, and a differential form version.
The vector-field version of Frobenius’ theorem says to take the Lie bracket of two vector fields spanning the plane field: if the Lie bracket points along the planes of the plane field, it’s integrable; otherwise, it is not. In our example we obtain
\[ [ \partial_x, \; \partial_y – x \partial_t ] = [\partial_x, \; \partial_y] – [\partial_x, \; x \partial_t] = 0 – \partial_x (x \partial_t) + x \partial_t \partial_x \\ = – \partial_t – x \partial_x \partial_t + x \partial_t \partial_x = – \partial_t. \] Here we repeatedly used the fact that the partial derivatives commute, e.g. \(\partial_x \partial_y = \partial_y \partial_x\) or \([\partial_x, \partial_y] = 0\), and the product rule, in the slightly cryptic form of \(\partial_x x = 1 + x \partial_x\). The result is \(– \partial_t\), which is definitely not a linear combination of \(\partial_x\) and \(\partial_y – x \partial_t\).
The differential form version of Frobenius’ theorem says to take the wedge product \(\alpha \wedge d\alpha\), which will be a 3-form: if it is zero, then the plane field is integrable; otherwise, it is not. In our example we obtain
\[ \alpha \wedge d\alpha = (x \; dy + dt) \wedge (dx \wedge dy) \\ = dt \wedge dx \wedge dy \neq 0, \] which is the standard Euclidean volume form and is definitely nowhere zero.
So, if you believe Frobenius’ theorem, or at least one or the other of these two variants, then we’ve shown that the plane field is
nowhere integrable. This is exactly the definition of a contact structure: a plane field which is nowhere integrable. (This is the definition in 3 dimensions, at least.) So our plane field is a contact structure, in fact, it’s essentially the standard contact structure on \(\mathbb{R}^3\). A 1-form whose kernel is a contact structure is called a contact form. So in this example, the 1-form \(\alpha = \beta + dt = x \; dy + dt\) is a contact form.
Another nice feature comes from considering the surface \(S = S \times \{0\}\), sitting inside \(S \times [0,1]\), and how it intersects the contact planes. At every point of \(S\) is the horizontal surface \(S\) (which is a plane), and the plane \(\ker \alpha\). How do these planes intersect? Well, we saw that the planes of \(\ker \alpha\) were spanned by \(\partial_x\) and \(\partial_y – x \partial_t\). The first of these actually points along \(S\), while the second does not — except when \(x = 0\), when the planes are tangent to \(S\). So the contact planes intersect \(S\) along the \(x\) direction.
Now the lines in the \(x\) directions along \(S\) have come up in our previous discussion — they are nothing but the kernel of the Liouville form \(\beta\)! So the planes of \(\ker \alpha\) cut \(S\) along the same line as \(\ker \beta\).
Thus, the lines at each point of \(S\), given by looking at how it intersects the contact plane \(\ker \alpha\), form a
line field on \(S\); and if you integrate it, you get a foliation. We saw last time that \(\ker \beta\) also forms a line field on \(S\), which also integrates to a foliation. What we’ve found here in our example is that the foliations on \(S \times \{0\}\) given by \(\ker \beta\), and by intersection with \(\ker \alpha\), coincide. (The foliations coincide when \(x \neq 0\). When \(x = 0\), both foliations are singular, so they are in fact equal as singular foliations.)
When you have a contact form \(\alpha\) on a 3-manifold, and a surface \(S\) in that 3-manifold, you can always consider, in a similar way, how the contact planes intersect the tangent plane to \(S\), at each point of \(S\). The result is a (singular) line field on \(S\), which integrates to a (singular) foliation: this is called the
characteristic foliation \(\mathcal{F}\) of \(S\). In fancy language, at a point \(p \in S\), the characteristic foliation is given by the intersection of the tangent plane \(T_p S\) to \(S\), with the contact plane \(\ker \alpha_p\) there: \[ \mathcal{F}_p = T_p S \cap \ker \alpha_p. \] In simple language, the characteristic foliation of a surface is the pattern of how it intersects the contact planes.
What we’ve found, in our example, is that the foliation \(\ker \beta\) coincides with the characteristic foliation on \(S \times \{0\}\). And in fact the same occurs not just on \(S \times \{0\}\), but on any slice \(S \times \{t\}\).
Let’s now consider our second example from last time: \(\beta = \frac{1}{2} (x \; dy – y \; dx)\) on the surface \(S = \mathbb{R}^2\). This was the “radial” example where \(X\) and \(\ker \beta\) both pointed out directly out along lines from the origin.
In this example, we obtain on \(M = \mathbb{R^2} \times [0,1]\) the 1-form \(\alpha = \beta + dt = \frac{1}{2} (x \; dy – y \; dx) + dt\). You can check that along the \(t\)-axis \(x=y=0\), \(\ker \alpha\) is a horizontal plane, spanned by \(\partial_x\) and \(\partial_y\). At other points, the kernel is spanned by \(x \partial_x + y \partial_y\), which points horizontally radially outward from the \(t\)-axis, and \(y \partial_x – x \partial_y + \frac{x^2+y^2}{2} \; dt\), which has an angular component \(y \partial x – x \partial y\) (which is in the \(\theta\) direction in cyclindrical coordinates), and a \(dt\) component. As \(x^2 + y^2\) becomes larger, i.e. further from the \(t\)-axis, the \(dt\) term becomes larger and the planes become more vertical. The result is a plane field known as the
standard cylindrically symmetric contact structure, because it’s cylindrically symmetric, and it’s a contact structure.
It’s perhaps not surprising that if you take something on the plane which is radially symmetric, and then just make a 3-d thing out of it, which doesn’t change in the new (\(t\)) dimension, then you get something which is cylindrically symmetric.
Let’s check that we actually have a contact form, by applying the differential form version of Frobenius’ theorem:
\[ \alpha \wedge d\alpha = \left( \frac{1}{2} x \; dy – \frac{1}{2} y \; dx + dt \right) \wedge \left( dx \wedge dy \right) \\ = dt \wedge dx \wedge dy. \] Most of the terms immediately cancel out in the wedge product because of the anti-symmetry: all that’s left is the standard 3-dimensional Euclidean volume form, which is definitely nowhere zero.
You can also check that on \(S \times \{0\}\), or in fact on any slice \(S \times \{t\}\), the characteristic foliation points radially outward from the origin. (Indeed, we just saw that the radial vector field \(x \partial_x + y \partial_y\) lies in \(\ker \alpha\).) This again coincides with the foliation \(\ker \beta\) from the Liouville 1-form.
In fact, this argument works generally, not just on this example. The differential form version of Frobenius’ theorem can always easily be applied to show \(\alpha\) is a contact form. Moreover, it’s not difficult to show that the characteristic foliation coincides with the foliation from the Liouville 1-form.
PROPOSITION. Starting from any Liouville 1-form \(\beta\) on any surface \(S\), the 1-form \(\alpha = \beta + dt\) is a contact form on \(M = S \times [0,1]\).
Moreover, for each \(t \in [0,1]\), the characteristic foliation on \(S \times \{t\}\) coincides with the foliation of \(\ker \beta\).
In other words, at each \((p,t) \in S \times [0,1]\), the intersection of \(\ker \alpha\) with the tangent space to \(S \times \{t\}\) is equal to \(\ker \beta\).
\[ \ker \alpha_{(p,t)} \cap T_{(p,t)} (S \times \{t\}) = \ker \beta_{(p,t)} \] PROOF. Consider \(\alpha \wedge d\alpha\); we’ll show it is nowhere zero. \[ \alpha \wedge d\alpha = (\beta + dt) \wedge d\beta = \beta \wedge d\beta + dt \wedge d\beta \] Now \(\beta \wedge d\beta\) is a 3-form, but \(\beta\) is a 1-form on the surface \(S\) only: it has no \(t\)-component. So \(\beta \wedge d\beta\) is a 3-form… on the 2-dimensional space \(S\)! Consequently it must be zero.
Thus \(\alpha \wedge d\alpha = dt \wedge d\beta\). Now we use the fact that \(\beta\) is a Liouville 1-form: this means that \(d\beta\) is a non-degenerate 2-form on \(S\). When we wedge it with \(dt\) then we must get a non-degenerate 3-form on \(S \times [0,1]\). This means \(\alpha\) is a contact form.
Now consider a point \((p,t) \in S \times [0,1]\), and a tangent vector \(V\) to \(S \times \{t\}\) there. So \(V\) is horizontal: it has no \(t\) component, and \(dt(V) = 0\). So \(\beta(V)= 0\) if and only if \(\alpha(V) = \beta(V) + dt(V) = 0\). Consequently for a vector \(V \in T_{(p,t)} S \times \{t\}\), we have \(V\) lies in \(\ker \alpha\) iff it lies in \(\ker \beta\). In other words, \(\ker \alpha \cap T_{(p,t)} S \times \{t\} = \ker \beta\). QED
Furthermore, this proof works in higher dimensions too. As long as you have a Liouville structure on a manifold \(S\) of any dimension \(2n\) (Liouville structures only exist in even dimension; this was discussed in the prequel), you’ll get a contact form on \(S \times [0,1]\) this way. (The only change is that \(\beta \wedge d\beta\) is a \((2n+1)\)-form on a \((2n)\)-manifold.)
In any case, this means that Liouville geometry leads naturally to contact geometry: we just add another dimension!
Indeed, sometimes this construction is called a
contactisation!
Liouville geometry is another name for symplectic geometry where the symplectic form is exact, i.e. the symplectic form is \(d\beta\). So contactisation upgrades symplectic geometry — the mathematics of classical Hamiltonian mechanics — to an odd-dimensional counterpart.
Another interesting thing to note is that in this “contactisation” construction, the 1-form \(\alpha = \beta + dt\) doesn’t actually depend on \(t\); it’s invariant under translations in the \(t\) direction. We saw in the second example that a radially symmetric Liouville form gave rise to a cylindrically symmetric contact structure.
And indeed, when you take \(\alpha = \beta + dt\), the 1-form \(\beta\) on the surface \(S\) doesn’t depend on \(t\); and \(dt\) is just the same \(dt\) regardless of what \(t\) is!
Such a contact form is sometimes called “vertically invariant”.
In fancy language, it means that the flow of the vertical vector field \(\partial_t\) preserves the contact form. In even fancier language, it means that the Lie derivative of the contact form \(\alpha\) in the direction of \(\partial_t\) is zero:
\[ L_{\partial_t} \alpha = 0. \]
A vector field whose flow preserves a contact structure is often called a
contact vector field. So \(\partial_t\) is a contact vector field.
Our 3-manifold \(M\), the thickened surface \(S \times [0,1]\), can be considered as a family of surfaces \(S_t = S \times {t}\), for \(t \in [0,1]\). And there is a vector field \(\partial_t\), transverse to all of these surfaces.
Emmanuel Giroux discovered in the 1990s that, when you have a surface in a contact 3-manifold with a contact vector field transverse to it, very nice stuff happens. He called such a surface
convex.
So Liouville geometry leads naturally not just to contact geometry, but to
convex surfaces in contact geometry. More on that, another day.
|
Answer
The shortest day in Minneapolis is $~~8.6~hours$ The longest day in Minneapolis is $~~15.4~hours$
Work Step by Step
We can convert $44.88^{\circ}$ to units of radians: $L = (44.88^{\circ})(\frac{\pi~rad}{180^{\circ}}) = 0.7833~rad$ We can convert $-23.44^{\circ}$ to units of radians: $D = (-23.44^{\circ})(\frac{\pi~rad}{180^{\circ}}) = -0.4091~rad$ We can calculate the shortest day in Minneapolis: $cos(0.1309~H) = -tan~D~tan~L$ $cos(0.1309~H) = -tan~(-0.4091)~tan~(0.7833)$ $cos(0.1309~H) = 0.431746$ $0.1309~H = cos^{-1}(0.431746)$ $0.1309~H = 1.12436$ $H = \frac{1.12436}{0.1309}$ $H = 8.6~hours$ The shortest day in Minneapolis is $~~8.6~hours$ We can calculate the longest day in Minneapolis: $cos(0.1309~H) = -tan~D~tan~L$ $cos(0.1309~H) = -tan~(0.4091)~tan~(0.7833)$ $cos(0.1309~H) = -0.431746$ $0.1309~H = cos^{-1}(-0.431746)$ $0.1309~H = 2.017224$ $H = \frac{2.017224}{0.1309}$ $H = 15.4~hours$ The longest day in Minneapolis is $~~15.4~hours$
|
Graduate Colloquium Spring 2019
McHenry Library Room 4130
Refreshments served at 3:30 in room 4161
For further information, please contact Graduate Student John McHugh or call 831-459-2969
Thursday April 11, 2019 Christy Hightower, Science & Engineering Distinguished Librarian, UCSC Scholarly communication is the system through which the written or oral results of research and other scholarly writings are created, evaluated for quality, distributed, located by other researchers, obtained in full text (or other media) for reading or consumption, and preserved for future use. The system includes both formal means of communication, such as publication in peer-reviewed books or journals, and informal channels, such as the communication or discussion of scholarship through email and social media. Libraries and librarians support researchers throughout the entire cycle of scholarly communication. Scholarly Communication Trends, Tools and Tips: Five Things Every Mathematics Graduate Student Should Know
As you can imagine, the topic of scholarly communication is a large one! This talk will focus on five trends, tools or tips that every math graduate student should be aware of. Topics covered will include some tips for locating and obtaining books and articles, the use of citation management software for organizing your notes about the research articles, books, blogs, and websites that you are reading and downloading, as well as copyright basics for authors.
Thursday April 18, 2019
No Seminar
Thursday April 25, 2019 Alejandro Bravo Doddoli, UCSC From Euler-Lagrange Equation to Hamel Equation Euler-Lagrange equations are very well know for being a powerful way to attack variational problems, and specially, in the context of mechanics, to give a covariant definition of Newton equations. Given a manifold $M$ with local coordinates $(q_1. \dots, q_n)$, there is a natural way to give coordinates to the tangent bundle as follow, $(q_1, \dots, q_n, v_1, \dots, v_n)$ and a vector $v \in TM$ at a point $(q_1, \dots, q_n,)$ is given by $ v_1 \frac{\partial }{\partial q_1} + \dots + v_n \frac{\partial }{\partial q_n}$. The Euler-Lagrange approach is taking these coordinates as a local coordinates. However, the set $\{ \frac{partial }{ \partial q_i} \}$ is not the unique base for the tangent space of $M$. we can consider a set of $n$ lineal independent vectors field $\{ X_i \}$ and write the vector $v$ in this base as $v = \sum_{i=1}^n u_i X_i$, where $u_i$ are called quasi-velocities, this approach is more general since the vector fields may not commute, hence they cannot be generated by a local coordinates. Hamel equations are Euler Lagrange equations written in the local coordinates $(q_1, \dots, q_n, u_1, \dots, u_n)$. This new approch can be very helpful for systems with symmetries.We will review some examples such as the rigid body, pendulum and chaplying slight. Thursday May 2, 2019 Alejandro Bravo Doddoli, UCSC Introduction to Geometry of Goursat Distribution and Bifurcation Theory
A Goursat flag is a chain $D_s \subset D_{s-1} \subset \dots \subset D_{1} \subset D_0 = TM$. of subbundles of the tagent bundle $TM$ such that $conrank D_i = i$ and $D_{i-1}$ kis generated by the vector fields in $D_i$ and their Lie brackets. Engel, Goursat, and Cartan studied these flags and establish a normal form them, valid at generic point $M$. After that Kumpera, ruiz and Mormil discovered a Goursat flags with singularities, and the number of these grows exponentially with the conrank $s$. Finally Montgomery classify all the singularities.
The distribution associated with problem of $n$-trailers, is a example of the Goursat's flag with all the singularities. Then we can fiend a critical vector fields, which flow is a periodic orbit that foliated the singularity. Since my master degree I wanted to find consequences in the dynamics, crated by the singularity. I will try to explain how much I have found about the persistent of those periodic orbits. Thursday May 9, 2019
No Seminar
Thursday May 16, 2019 TBA Thursday May 23, 2019 TBA Thursday May 30, 2019 TBA Thursday June 6, 2019
No Seminar due to special Math Tea honoring the 2019 Graduating Mathematics MA and PhD students! Please join us in celebrating: McHenry 3rd Floor Breezeway, 3:00pm - 4:30pm
|
Recently I discovered the differential identity
$$ \frac{d^{k+1}}{dx^{k+1}} (1+x^2)^{k/2} = \frac{(1 \times 3 \times \dots \times k)^2}{(1+x^2)^{(k+2)/2}}$$
valid for any odd natural number $k$; for instance $\frac{d^6}{dx^6} (1+x^2)^{5/2} = \frac{225}{(1+x^2)^{7/2}}$. This identity was surprising at first, since usually the repeated application of the product rule and chain rule leads to far messier expressions than this, but there are now several proofs that adequately explain this identity (collected at this blog post of mine). There is also the more general identity
$$ |\frac{d}{dx}|^{2s-1} (1+x^2)^{s-1} = \frac{2^{2s-1}\Gamma(s)}{\Gamma(1-s)} (1+x^2)^{-s}$$
valid for any complex $s$ (if everything is interpreted distributionally), which is related to the isomorphisms between principal series representations of $PGL_2({\bf R})$.
The purpose of my question here is
not to ask for more proofs of this identity (but you are welcome to visit the above-mentioned blog post to contribute another proof, if you wish). Instead, I am asking as to whether this identity (or something close to it) already appears in the literature - I find it hard to believe that such a simple identity has been missed for centuries, given that it could easily have been discovered and proven by (say) Euler. The closest match that I know of so far are the Rodrigues formulae for the classical orthogonal polynomials, but I was not quite able to place the above identity as a special case of these formulae (the exponents don't quite match up).
|
You should consider an unsupervised learning algorithm such as K-nearest neighbor ('KNN').KNN will measure the distance amongst the observations in your space. You can and probably should consider alternative distance functions (besides euclidean) particularly if you are clustering on features such as returns which have outliers. There are quite a few ...
^GSPC is a price index, not a total return index, so it does not include dividends.SPY is an ETF that holds the underlying stocks. When it receives a dividend it keeps it in a cash account (which of course affects the NAV and market value of SPY shares) until the end of the quarter. At that time (on the 3d friday of Mar Jun Sep or Dec) it will pay out the ...
Basically the Total Return Index assumes reinvestments compared to "regular" indices."A total return index is an index that measures the performance of agroup of components by assuming that all cash distributions arereinvested, in addition to tracking the components' pricemovements.1 While it is common to refer to equity based indices,there ...
Put simply, VIX is a spot index (fair value to a variance swap on SPX of constant maturity) that you cannot own as a security. Market participants create futures for you to trade. Futures trade higher than the VIX -- if you long VIX futures, you lose when the futures contract converges to VIX. You therefore have a negative roll-down. VIX ETF doesn't avoid ...
In my mind, there are two questions here:1) How does DB make money given a zero expense ratio?This is covered by Dirk and Lliane. Basically, DB gets cheap funding and stock loan fees in return for paying marketing / index / hedging costs. The ETF investor gets zero expense ratio in return for taking DB credit risk.2) Why does it look like the etf ...
Are there any other mechanisms at play here which might explain this kind of tracking error?Dirk is right, you often lend the titles internally or not, etc.You can also write calls for your index, this is not orthodox, but it's ETF, there is no orthodoxy there...Edit : With the graph and given the outperforming is seasonnal (around May), I think we can ...
In general, it depends on the particular ETF, and should be checked in the prospectus, but one standard way (e.g. SPY) is to do it on a daily basis based on NAV published after the close. E.g. NAV per share = X$, so the expenses taken out would be X*0.03/(100*252). Again usually ETF's have cash component (with aggregated dividends etc), so there's no problem ...
I would look to run a pre-optimization routine over the whole universe of 200+ ETFs. I would use this pre-optimization to reduce the universe to a cardinality that provides optimal diversification effects. You can do that by first looking at pair-wise correlations and then also run optimizations to reduce portfolio variance by utilizing the covariance ...
Quant Guy's answer is quite informative for your question already.Just to add few other things: instead of figuring out the choice of features by your own brain, you could also use machine learning techniques to help in extracting the 'features' for your specific purpose, e.g. risk modeling or returns forecasting or portfolio construction as mentioned by ...
Probably missing something here but if $X$ has $E(X) = \mu$ and $variance(X) = \sigma^2$ then $2X$ has $E(2X) = 2 \mu, variance(2X) = 4\sigma^2$. Thus the sharp ratio defined as $\frac{\mu}{\sigma}$ stays the same for the 2x leveraged and the regular index.
This is a very good question. It can be argued that risk parity is one example of a smart beta strategy.Yet it is important to understand that both are coming from two different directions: risk parity is basically a form of risk management (in the sense of risk-adjustment) because its basic approach lies in diversification - like the alternative methods ...
On more than a few occasions, I have attempted to extrapolate the current trend towards passive allocation to its logical conclusion: more passive allocation means more inefficiency.I am not aware of any research which directly measures the correlation between market efficiency and active versus passive allocation. In general, the level of market ...
No need to scrape the site. That should always be a last resort. The below will import the .csv file you are asking about and save it to a directory of your choice. If you don't want to specify a directory can eliminate dir and any references to it and the file will go straight to your working directory. I usually save data separately hence that option....
If you are looking for the official SEC filings then EDGAR is your best bet. QQQ is still listed under PowerShares, the old (and better IMHO) name for Invesco.POWERSHARES QQQ TRUST, SERIES 1 CIK#: 0001067839This link should get you what you need;https://www.sec.gov/cgi-bin/browse-edgar?action=getcompany&CIK=0001067839&type=&dateb=&owner=...
Delta one trading desks provide synthetic exposure to their clients.OK, so what does that mean? Delta One desks give their clients exposure to a product (stock index, ETF, or even a single stock) without the client actually buying the underlying product.For example, a customer can take their money and buy the stocks in the SP500 index. Or, they can ...
To answer your questions:1) Yes, the above table is correct2) Your results are correct except..... 1X loss = 9.6%.When you combine both positive and negative changes, it is the MEDIAN value that is of interest. Here are some links:http://www.futuresmag.com/Issues/2010/March-2010/Pages/Trading-with-leveraged-and-iinverse-ETFs.aspxhttp://...
There is a good article in Seeking Alpha but if you did a Google Search you probably found it already. Some ETF's work through swaps with a counterpart, but you will never know who the counter-part is. As you said it depends on the type of ETF, with a UCITS ETF you're not supposed to have a big counter-part risk as you own the underlyings, when it's ...
Vanguard S&P 500 index fund tracks the index and not the total return because it pays dividends out to the owners of the fund... some investors reinvest the dividends, some investors spend their dividends, etc., so, because they cannot control the reinvestment and distribute the dividends, they benchmark against the S&P 500 index and not the total ...
I don't have much experience in the matter, but I've been doing some related literature research recently and I think these links can be helpful:A rather recent study from CMEA (possible a bit biased) report by BlackRockA report by Lyxor (asset manager affialiated to Societe Generale)
There are plenty of sites you can get this information from. etfdb.com and etf.com are two of the bigger ones.See this for an example:http://etfdb.com/etfdb-category/europe-equities/http://etfdb.com/tool/etf-stock-exposure-tool/
The fair price can be calculated by [Net Assets / Shares Outstanding].In reality the ETF should trade at a slight premium to this calculation due to the convenience of having many assets bundled in one, thus reducing your brokerage expenses in the form of transaction fees to construct a similar portfolio.From this link: (https://advisors.vanguard.com/...
Leverage: futures usually require much lower margin than their ETF counterparts. For example /ES (E-mini S&P 500 futures) requires about \$4K overnight maintenance margin per contract (may vary by brokerage) to control 50 times the S&P 500 index (currently valued at about \$108K). This is over 20:1 leverage. Furthermore you do NOT pay interest on ...
First, you might find this recent paper by Israeli, Lee and Sridharan (Review of Accounting Studies, forthcoming) interesting. This is the abstract:We examine whether an increase in ETF ownership is accompanied by a decline in pricing efficiency for the underlying component securities. Our tests show an increase in ETF ownership is associated with: (1) ...
The CBOE VIX index is an aggregated spot value calculated from options. The index itself cannot be traded. Volatility ETPs are usually designed to track an underlying index on VIX Futures that is tradeable.In the case of SVXY, it is stated in their prospectus that it tracks The S&P 500 VIX Short-Term Futures Index. Credit Suisse's XIV follows the exact ...
The futures price goes to the spot price as time to maturity declines, not vice-versa. The difference is referred to as basis. That's not really what roll yield is about though. The roll yield aspect is that as the contracts the ETF holds are expiring, they are close to the spot price. However, the next futures contract's price is higher than the price of ...
Here couple ETFs that may satisfy what you are looking for:http://www.quant-shares.com/etf-list/http://www.etc.db.com/GBR/ENG/Institutional/Downloads/ISIN/Factsheets/GB00B4N0QN94http://guggenheiminvestments.com/products/etf/wmcrhttp://etfdb.com/type/investment-style/high-beta/Those include ETFs with a momentum approach, mean-reversion approach, micro ...
It depends a lot on the structure of the ETF, it could be :* In the "terms and conditions" of the (highly possible) total return swap of the fund* Portfolio insurance* Option combination (or cap & floor)I think it's in the swap details, already saw that a few times.
MSCI has country indices for developed markets going back to 1970 in many cases and a decent history for emerging markets (starting 1988). iShares has pretty liquid ETFs for many of the most popular countries and regions, such as EAFE (EFA), Emerging Markets (EEM), Japan (EWJ), Germany (EWG), Canada (EWC), etc.Other major indices with very long histories ...
|
1
Share Trigonometry Quiz for SSC CGL & Railways RRB
4 years ago .
Here is a Trigonometry Quiz for
SSC CGL Railways RRB. This quiz contains important questions matching the exact pattern and syllabus of upcoming exams. Make sure you attempt today’s Quant Quiz for Upcoming Exams to check your preparation level.
If cosx + cos
2x = 1, the numerical value of (sin 12x + 3sin 10x + 3sin 8x + sin 6x – 1) is :
The value of sin
222° + sin 268° + cot 230° is:
If \(\frac{{\sin \theta }}{x} = \frac{{\cos \theta }}{y}\), then sin θ – cos θ is equal to
If tan α = n tan β and sin α = m sin β, then cos
2 α is
The radian measure of 63° 14’51” is
If sec θ + tan θ = 2 + √5, then the value of sin θ + cos θ is:
The least value of 4 cosec
2α + 9sin 2α is:
If 0° < θ < 90° and 2 sec θ = 3 cosec
2θ, then θ is
If sin
2α = cos 3α , then the value of (cot 6α – cot 2α) is
The angle of depression of a point from the top of a 200 m high tower is 45°. The distance of the point from the tower is
More Quizzes from Testbook.com
Co-founder, Testbook.com
1
|
Prescribed energy connecting orbits for gradient systems
1.
Dipartimento di Ingegneria Industriale e Scienze Matematiche, Università Politecnica delle Marche, Via Brecce Bianche, I-60131 Ancona, Italy
2.
Dipartimento di Ingegneria Civile, Edile e Architettura, Università Politecnica delle Marche, Via Brecce Bianche, I-60131 Ancona, Italy
3.
CEREMADE (CNRS UMR n° 7534), Université Paris-Dauphine, PSL Research University, Place de Lattre de Tassigny, 75775 Paris Cedex 16, France
We are concerned with conservative systems $ \ddot q = \nabla V(q) $, $ q\in{\mathbb R}^{N} $ for a general class of potentials $ V\in C^1({\mathbb R}^N) $. Assuming that a given sublevel set $ \{V\leq c\} $ splits in the disjoint union of two closed subsets $ \mathcal{V}^{c}_{-} $ and $ \mathcal{V}^{c}_{+} $, for some $ c\in{\mathbb R} $, we establish the existence of bounded solutions $ q_{c} $ to the above system with energy equal to $ -c $ whose trajectories connect $ \mathcal{V}^{c}_{-} $ and $ \mathcal{V}^{c}_{+} $. The solutions are obtained through an energy constrained variational method, whenever mild coerciveness properties are present in the problem. The
connecting orbits are classified into brake, heteroclinic or homoclinic type, depending on the behavior of $ \nabla V $ on $ \partial \mathcal{V}^{c}_{\pm} $. Next, we illustrate applications of the existence result to double-well potentials $ V $, and for potentials associated to systems of duffing type and of multiple-pendulum type. In each of the above cases we prove some convergence results of the family of solutions $ (q_{c}) $. Keywords:Conservative systems, energy constraints, variational methods, brake orbits, homoclinic orbits, heteroclinic orbits, convergence of solutions. Mathematics Subject Classification:Primary: 34C25, 34C37, 49J40, 49J45. Citation:Francesca Alessio, Piero Montecchiari, Andres Zuniga. Prescribed energy connecting orbits for gradient systems. Discrete & Continuous Dynamical Systems - A, 2019, 39 (8) : 4895-4928. doi: 10.3934/dcds.2019200
References:
[1] [2]
F. Alessio, M. L. Bertotti and P. Montecchiari,
Multibump solutions to possibly degenerate equilibria for almost periodic Lagrangian systems,
[3]
F. Alessio and P. Montecchiari,
Entire solutions in $\mathbb{R}^{2}$ for a class of Allen-Cahn equations,
[4] [5] [6]
_____, An energy constrained method for the existence of layered type solutions of NLS equations,
[7] [8] [9] [10]
A. Ambrosetti and M.L. Bertotti, Homoclinics for second order conservative systems, in
[11]
P. Antonopoulos and P. Smyrnelis,
On minimizers of the Hamiltonian system $u'' = \nabla W(u)$ and on the existence of heteroclinic, homoclinic and periodic orbits,
[12]
V. Benci,
Closed geodesics for the Jacobi metric and periodic solutions of prescribed energy of natural Hamiltonian systems,
[13]
V. Benci and F. Giannoni, A new proof of the existence of a brake orbit, in
[14]
M. L. Bertotti and P. Montecchiari,
Connecting orbits for some classes of almost periodic Lagrangian systems,
[15] [16] [17] [18]
G. Fusco, G. F. Gronchi and M. Novaga,
On the existence of connecting orbits for critical values of the energy,
[19] [20] [21] [22] [23] [24]
C. Li,
The study of minimal period estimates for brake orbits of autonomous subquadratic Hamiltonian systems,
[25] [26] [27] [28] [29] [30] [31]
P. Sternberg and A. Zuniga,
On the heteroclinic connection problem for multi-well potentials with several global minima,
[32] [33]
A. Zuniga,
show all references
References:
[1] [2]
F. Alessio, M. L. Bertotti and P. Montecchiari,
Multibump solutions to possibly degenerate equilibria for almost periodic Lagrangian systems,
[3]
F. Alessio and P. Montecchiari,
Entire solutions in $\mathbb{R}^{2}$ for a class of Allen-Cahn equations,
[4] [5] [6]
_____, An energy constrained method for the existence of layered type solutions of NLS equations,
[7] [8] [9] [10]
A. Ambrosetti and M.L. Bertotti, Homoclinics for second order conservative systems, in
[11]
P. Antonopoulos and P. Smyrnelis,
On minimizers of the Hamiltonian system $u'' = \nabla W(u)$ and on the existence of heteroclinic, homoclinic and periodic orbits,
[12]
V. Benci,
Closed geodesics for the Jacobi metric and periodic solutions of prescribed energy of natural Hamiltonian systems,
[13]
V. Benci and F. Giannoni, A new proof of the existence of a brake orbit, in
[14]
M. L. Bertotti and P. Montecchiari,
Connecting orbits for some classes of almost periodic Lagrangian systems,
[15] [16] [17] [18]
G. Fusco, G. F. Gronchi and M. Novaga,
On the existence of connecting orbits for critical values of the energy,
[19] [20] [21] [22] [23] [24]
C. Li,
The study of minimal period estimates for brake orbits of autonomous subquadratic Hamiltonian systems,
[25] [26] [27] [28] [29] [30] [31]
P. Sternberg and A. Zuniga,
On the heteroclinic connection problem for multi-well potentials with several global minima,
[32] [33]
A. Zuniga,
[1]
Yuika Kajihara, Misturu Shibayama.
Variational proof of the existence of brake orbits in the planar 2-center problem.
[2]
Fei Liu, Jaume Llibre, Xiang Zhang.
Heteroclinic orbits for a class of Hamiltonian systems on
Riemannian manifolds.
[3]
E. Canalias, Josep J. Masdemont.
Homoclinic and heteroclinic transfer trajectories between planar Lyapunov orbits in the sun-earth and earth-moon systems.
[4]
Daniel Wilczak.
Abundance of heteroclinic and homoclinic orbits for the hyperchaotic Rössler system.
[5]
B. Buffoni, F. Giannoni.
Brake periodic orbits of prescribed Hamiltonian for indefinite Lagrangian systems.
[6] [7]
Ying Lv, Yan-Fang Xue, Chun-Lei Tang.
Homoclinic orbits for a class of asymptotically quadratic Hamiltonian systems.
[8] [9] [10] [11]
Chen-Chang Peng, Kuan-Ju Chen.
Existence of transversal homoclinic orbits in higher
dimensional discrete dynamical systems.
[12]
Juntao Sun, Jifeng Chu, Zhaosheng Feng.
Homoclinic orbits for first order periodic
Hamiltonian systems with spectrum point zero.
[13]
Boris Buffoni, Laurent Landry.
Multiplicity of homoclinic orbits in quasi-linear autonomous
Lagrangian systems.
[14] [15]
Jun Wang, Junxiang Xu, Fubao Zhang.
Homoclinic orbits for a class of Hamiltonian systems
with superquadratic or asymptotically quadratic potentials.
[16]
Tiantian Wu, Xiao-Song Yang.
A new class of 3-dimensional piecewise affine systems with homoclinic orbits.
[17] [18]
Flaviano Battelli, Ken Palmer.
Transversal periodic-to-periodic homoclinic orbits in singularly
perturbed systems.
[19]
Duanzhi Zhang.
Minimal period problems for brake orbits of nonlinear
autonomous reversible semipositive Hamiltonian systems.
[20]
2018 Impact Factor: 1.143
Tools Metrics Other articles
by authors
[Back to Top]
|
Data Types for Control & DSP
There's a lot of information out there on what data types to use for digital signal processing, but there's also a lot of confusion, so the topic bears repeating.
I recently posted an entry on PID control. In that article I glossed over the data types used by showing "double" in all of my example code. Numerically, this should work for most control problems, but it can be an extravagant use of processor resources. There ought to be a better way to determine what precision you need out of your arithmetic, and what sorts of arithmetic you can use on your processor.
This blog post seeks to answer two questions: "what data types must I use to get the job done?" and "what data types are fast enough on my processor to get the job done?" If you're lucky, there is some overlap in the answers to those two questions and you can go out and design a controller that works. If you're not that lucky, at least you know that you either need to seek out a new, faster processor, or perhaps a new, more forgiving job.
All of these issues are discussed in more depth in my book, Applied Control Theory for Embedded Systems.
For the purposes of this discussion the world of data representation is divided into three: floating point types, integer types, and fixed-point types. You should know what floating point and integer representation are. Fixed-point representation is just a generalization of integer representation, with the weights of the bits (possibly) scaled such that the least-significant bit is a fraction. The notion of non-integer fixed-point arithmetic is discussed in this Wikipedia article: the short story is that if I talk about a number in Q0.31 notation I mean a fixed-point number in the range $-1 < x < 1$.
Each of these three types has advantages and disadvantages. Floating point arithmetic is conceptually easy, but on all but desktop-class processors it is slower than integer or fixed-point arithmetic, and it has some subtle "gotchas" that can bite you if you're not careful. Integer arithmetic uses familiar data types, but in general the scaling for signal processing (and, hence, control systems) is all wrong. Non-integer fixed-point types are not directly supported in common languages, and can be hard for a beginner to wrap their heads around, but are close to optimal for a wide range of problems.
For all of these data types, you have to worry about quantization, for floating point numbers you have to worry about varying quantization effects, and for integer and other fixed-point data types you need to worry about overflow.
For the purposes of this post, I'll make an example PID controller. I'll use $u$ to mean the measurement of the controlled variable, $ut$ to be the target value of the controlled variable, and $y$ to mean the controller output. For all variables, $x_n$ means the value of $x$ at sample time $n$. The variables $k_i, k_p, k_d, k_{dp}$ are the integrator gain, the proportional gain, the derivative gain, and the derivative band-limiting factor, respectively. The math for this controller is
$$xi_n = xi_{n-1} + k_i \left ( ut_n - u_n \right )$$
$$xd_n = xd_n + k_{dp} \left ( \left ( ut_n - u_n \right ) - xd_{n-1} \right )$$
$$y_n = xi_n + k_p u_n + k_d k_{dp} \left ( \left ( ut_n - u_n \right ) - xd_{n-1} \right )$$
The first problem that crops up with this algorithm is the integrator gain. For most control systems, the integrator gain is much less than 1. This means that the factor $k_i \left ( ut_n - u_n \right )$ is, in general, small. Moreover, as you increase the sampling rate, you need to adjust the integrator gain downward. It is up to you to insure that for the smallest possible value of $u_n$, the factor $k_i \left ( ut_n - u_n \right )$ fits in the data type that you have chosen for the integrator state, $xi$.
As a concrete example, consider a system that uses a 16-bit ADC to measure the plant's output variable. Further, let's assume that we scale this output variable to a range $0 \le u_n < 1$. If $n_{ADC}$ is the ADC reading that ranges from 0 to 65535, then we calculate $u_n = \frac{n_{ADC}}{65536}$. Now, further assume that the integrator gain is a not-unreasonable $k_i = 0.0002$, and that the integrator state can fall in the range $-1 < xi < +1$.
With this example, the smallest increment of the ADC can be $\frac{1}{65536}$. This, in turn, means that the smallest increment of the factor $k_i \left ( ut_n - u_n \right )$ can be $\frac{0.0002}{65536}$, or about $3 \cdot 10^{-9}$.
If you store $xi$ in a 32-bit IEEE floating-point variable, then the mantissa has an effective length of 25 bits. When $xi$ has an absolute value greater than $\frac{1}{2}$, the smallest increment that can be added into $xi$ is $2^{-26}$, or about $15 \cdot 10^{-9}$. That's about five times larger than the smallest increment that may occur.
What does all this razz-matazz with numbers mean? It means that in this circumstance, the integrator term of your PID controller is missing out on potentially important information in the feedback. This, in turn, could result in your system getting "stuck" at the wrong value until the error grows to objectionable amounts. In a real-world system, this would mean that you might see a small amount of random drift around your desired target point, or a small oscillation around your desired target point.
To make sure this doesn't happen, you should make sure that the smallest increment that will register in your integrator state is as small or smaller than the smallest increment that can be presented to it. Better yet, make sure that the smallest increment that will register on your integrator state is smaller than about $\frac{1}{8}$ of the smallest increment that will be presented to it.
Determine the smallest increment that your integrator state will absorb. For an integer, this increment is 1. For a signed fractional number that ranges from -1 to 1, with $n$ bits this increment is $2^{-(n-1)}$, or $2^{-31}$ for a 32-bit number. For a 32-bit IEEE floating point number ("float" in most C compilers) that ranges from -1 to 1, this increment can be as high as $2^{-25}$. The increment isn't a constant -- this is one of the lovely things about using floating point. For a 64-bit IEEE floating point number ("double" in most C compilers) that ranges from -1 to 1, this increment can be as high as $2^{-54}$. Again, the increment isn't a constant. Determine the smallest increment that you will present to your integrator. This will be the smallest increment of your sensor (usually an ADC, but you know your system), multiplied by any pre-scaling factors you may apply, then multiplied by the integrator gain. Check which number is bigger, and by how much -- if the smallest increment you'll ever present to the integrator state is eight times bigger than the smallest increment it can register, then you're probably OK.
Astute readers will notice that there's a problem with the controller equation that I show if you're using integers -- when the smallest increment that an integrator can register is $\pm 1$, then you need to swap things around. In this case, you should refrain from scaling the ADC output: let $u_n = n_{ADC}$. Then, move the integrator gain:
$$xi_n = xi_{n-1} + \left ( ut_n - u_n \right )$$
$$xd_n = xd_n + k_{dp} \left ( \left ( ut_n - u_n \right ) - xd_{n-1} \right )$$
$$y_n = k_i xi_n + k_p u_n + k_d k_{dp} \left ( \left ( ut_n - u_n \right ) - xd_{n-1} \right )$$
Now, your integrator state will always register the smallest change in the ADC. You will have to scale all of your variables for the convenience of the mathematics rather than your own convenience, but it'll work.
With fixed-point numbers, quantization is fixed -- either the smallest increment you're presenting to the integrator state is too small to be registered, or it's not. Life isn't so easy with floating point. With floating points, if the value of a state (such as $xi$) is small then the smallest increment that you can add in to it is also small. But as the value of the state grows the smallest increment you can add in also grows -- so if you're dealing with floating point numbers you need to do your calculations based off of the maximum value that the state can take (or the maximum value that you allow the state to take).
Floating point numbers only have problems with getting too big when they grow past the point where small changes in the system inputs can affect them properly. Fixed point numbers, however, can have much more dramatic problems. The problem is called overflow.
Consider the C code snippet:
int a = 30000; printf("The number is %d\n", a + 2768);
Can you say what the output will be? You can't, really.
If you try this on a normal PC, the output will be "32768". However, if you can find a system that uses 16-bit integers and 2's compliment notation (and that has a working printf), the output will most likely be "-32768". The reason for this is that 32768 does not fit into a 2's compliment signed 16-bit integer, and because C tends to be pretty bone-headed about handling this situation. The phenomenon that we've just seen is called overflow.
If you are designing a digital control system (or any digital signal processing system) you need to either design your data paths so that overflow simply cannot happen, or you need to make sure that overflow is handled gracefully.
Designing data paths so that overflow cannot happen is beyond the scope of this paper. If you understand the relevant signal processing theory, you can start from the control system as designed and the maximum possible ranges of all the inputs, and you can compute the largest value for each of the states and intermediate values in the system. If those largest values are all smaller than anything that will overflow, then you don't have to worry.
Designing data paths that handle overflow gracefully is conceptually more simple: at each step in the computation that might result in an increased value (whether it's a state variable or an intermediate value), you test for overflow, and you deal with it gracefully. I have found that in C and C++, the way to do this is to test for overflow and if it happens, let the result take on the greatest value allowed by the data type.
This overflow handling is detailed in my book, but I can give an example using integer math. In this case I'm defining that "overflow" is anything that results in a value greater than INT_MAX/2.
int add(int a, int b) { // Assume that a and b are both smaller than INT_MAX/2 int x = a + b; if (x > INT_MAX / 2) { x = INT_MAX / 2; } else if (x < -(INT_MAX / 2)) { x = -(INT_MAX / 2); } return x; }
There are more sophisticated ways of handling this, especially if you're willing to do some assembly-language programming, but the above code shows the idea.
So far we've dealt with the "what does my data type need to do?" side of the question. The other side of the question is "what can my processor do?" I would like to be able to give you some firm guidelines on how to find this out before the fact -- but I can't. I can give you some rules of thumb to narrow your choices down, but in the end analysis you'll need to write a sample controller, with your intended data types, and then benchmark its performance to figure out how much of the available processor resources it consumes.
The rules of thumb that I can give you are:
Doing the work in fixed-point math is almost always faster than floating-point math done in software. If you have floating-point hardware, it may or may not be faster than fixed-point math (and if it's slower, it'll be a lot closer). Be careful when a processor claims to have floating-point hardware. 32-bit floating point hardware is much more common than 64-bit, and you often have to do some strenuous digging to figure out that you're only going to get 32-bit. 16-bit fixed point will work for a few systems. 32-bit floating point gives more precision than 16-bit fixed point, but less than 32-bit fixed point. 64-bit floating point will probably give you more precision than you'll need -- if this isn't the case please hire me, I want to work on your project! It always takes more clock cycles than you think -- benchmark.
It's a good idea to do your benchmarking early in a project -- you'll often find yourself either needing to adopt fixed-point math, or needing to buy an entirely different processor. If you are a manager, drive your team to choose a processor candidate early, then buy an evaluation board and drive them to benchmark their candidate software. If you are a line engineer, then do everything you can to get your hands on an appropriate evaluation board and give it a whirl. In all cases, try to build time into the project to absorb a false start on the processor selection.
In a perfect world -- at least for the lazy programmer -- you'll always be able to use a processor that can keep up with the system sampling rate while using 64-bit floating point math. While this can happen (sometimes even when you're using a slow 8-bit processor!), you'll often be stuck with using fixed point math of some type, to make your speed numbers.
Previous post by Tim Wescott:
PID Without a PhD
Next post by Tim Wescott:
Stability or insanity
Say, I'm not understanding your PID equations exactly, so please enlighten.
The way I see it, the Proportional constant "Kp" in the last equation should be multiplied by the error ( "Utn - Un" ) and not simply the measured process variable "Un". Also, the derivative state equation seem odd to me, especially in its use of "Xdn" on both sides of the equation. Shouldn't the one on the right side of the derivative state equation be "Xd(n-1)" instead of Xdn ?
Anyway, great read---thanks!
-Myles
Enjoy!
I don't think you really get this until you write some code and something horrible happens. I saw a picture once of a log smashed into the operator's console of a saw mill. The tree was larger than 65.535 inches -- you can probably fill in the blanks, or at least guess at the bug.
Jerry
To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments.
Registering will allow you to participate to the forums on ALL the related sites and give you access to all pdf downloads.
|
LaTeX supports many worldwide languages by means of some special packages. In this article is explained how to import and use those packages to create documents in
Portuguese.
Contents
Portuguese language has some accentuated words. For this reason the preamble of your file must be modified accordingly to support these characters and some other features.
\documentclass{article} %encoding %-------------------------------------- \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} %-------------------------------------- %Portuguese-specific commands %-------------------------------------- \usepackage[portuguese]{babel} %-------------------------------------- %Hyphenation rules %-------------------------------------- \usepackage{hyphenat} \hyphenation{mate-mática recu-perar} %-------------------------------------- \begin{document} \tableofcontents \vspace{2cm} %Add a 2cm space \begin{abstract} Este é um breve resumo do conteúdo do documento escrito em Português. \end{abstract} \section{Seção introdutória} Esta é a primeira seção, podemos acrescentar alguns elementos adicionais e tudo será escrito corretamente. Além disso, se uma palavra é um caminho muito longo e tem de ser truncado, babel irá tentar truncar corretamente, dependendo do idioma. \section{Segunda seção} Esta seção é para ver o que acontece com comandos de texto que definem \[ \lim x = \theta + 152383.52 \] \end{document}
There are two packages in this document related to the encoding and the special characters. These packages will be explained in the next sections.
If your are looking for instructions on how to use more than one language in a single document, for instance English and Portuguese, see the International language support article.
Modern computer systems allow you to input letters of national alphabets directly from the keyboard. In order to handle a variety of input encodings used for different groups of languages and/or on different computer platforms LaTeX employs the
inputenc package to set up input encoding. In this case the package properly displays characters in the Portuguese alphabet. To use this package add the next line to the preamble of your document: \usepackage[utf8]{inputenc} The recommended input encoding is utf-8. You can use other encodings depending on your operating system.
To format LaTeX documents properly you should also choose a font encoding which supports specific characters for Portuguese language, this is accomplished by the
package: fontenc \usepackage[T1]{fontenc} Even though the default encoding works well in Portuguese, using this specific encoding will avoid glitches with some specific characters, e.g., some accented characters might not be directly copyable from the generated PDF and instead are constructed using the base character and an overlayed shifted accent symbol, resulting in two separate symbols if you copy it. The default LaTeX encoding is
OT1.
To extended the default LaTeX capabilities, for proper hyphenation and translating the names of the document elements, import the
babel package for the Portuguese language. \usepackage[portuguese]{babel} As you may see in the example at the introduction, instead of "abstract" and "Contents" the Portuguese words "Resumo" and "Conteúdo" are used.
If you need to use the Brazilian Portuguese localization use
brazilian instead of
portuguese as parameter when importing
babel.
Sometimes for formatting reasons some words have to be broken up in syllables separated by a
- (
hyphen) to continue the word in a new line. For example, matemática could become mate-mática. The package babel, whose usage was described in the previous section, usually does a good job breaking up the words correctly, but if this is not the case you can use a couple of commands in your preamble.
\usepackage{hyphenat} \hyphenation{mate-mática recu-perar}
The first command will import the package
hyphenat and the second line is a list of space-separated words with defined hyphenation rules. On the other side, if you want a word not to be broken automatically, use the
{\nobreak word} command within your document or include it in an
\mbox{word}.
For more information see
|
Post-Newtonian expansions in general relativity are used for finding an approximate solution of the Einstein field equations for the metric tensor. The approximations are expanded in small parameters which express orders of deviations from Newton's law of universal gravitation. This allows approximations to Einstein's equations to be made in the case of weak fields. Higher order terms can be added to increase accuracy, but for strong fields sometimes it is preferable to solve the complete equations numerically. This method is a common mark of Effective Field Theories. In the limit, when the small parameters are equal to 0, the post-Newtonian expansion reduces to Newton's law of gravity. Expansion in 1/ c 2
The
post-Newtonian approximations are expansions in a small parameter, which is the ratio of the velocity of matter, forming the gravitational field, to the speed of light, which in this case is better called the speed of gravity. [1]
In the limit, when the fundamental speed of gravity becomes infinite, the post-Newtonian expansion reduces to Newton's law of gravity.
[2] Expansion in h
Another approach is to expand the equations of general relativity in a power series in the deviation of the metric from its value in the absence of gravity
h_{\alpha \beta} = g_{\alpha \beta} - \eta_{\alpha \beta} \,.
To this end, one must choose a coordinate system in which the eigenvalues of h_{\alpha \beta} \eta^{\beta \gamma} \, all have absolute values less than 1.
For example, if one goes one step beyond linearized gravity to get the expansion to the second order in
h: g^{\mu \nu} \approx \eta^{\mu \nu} - \eta^{\mu \alpha} h_{\alpha \beta} \eta^{\beta \nu} + \eta^{\mu \alpha} h_{\alpha \beta} \eta^{\beta \gamma} h_{\gamma \delta} \eta^{\delta \nu} \,. \sqrt{- g} \approx 1 + \tfrac12 h_{\alpha \beta} \eta^{\beta \alpha} + \tfrac18 h_{\alpha \beta} \eta^{\beta \alpha} h_{\gamma \delta} \eta^{\delta \gamma} - \tfrac14 h_{\alpha \beta} \eta^{\beta \gamma} h_{\gamma \delta} \eta^{\delta \alpha} \,. Hybrid expansion
Sometimes, as with the Parameterized post-Newtonian formalism, a hybrid approach is used in which both the reciprocal of the speed of gravity and masses are assumed to be small.
Uses
The first use of a PN expansion (to first order) was made by Einstein in calculating the Perihelion precession of Mercury's orbit; today it is recognized as a first simple case of the most common use of the PN expansion - solving the General Relativistic Two-Body Problem, which includes the emission of Gravitational waves.
See also References ^ ^ Kopeikin, S., Efroimsky, M., Kaplan, G. (2011). Relativistic Celestial Mechanics of the Solar System. External links "On the Motion of Particles in General Relativity Theory" by A.Einstein and L.Infeld "Gravitational Radiation from Post-Newtonian Sources and Inspiralling Compact Binaries" by Luc Blanchet
This article was sourced from Creative Commons Attribution-ShareAlike License; additional terms may apply. World Heritage Encyclopedia content is assembled from numerous content providers, Open Access Publishing, and in compliance with The Fair Access to Science and Technology Research Act (FASTR), Wikimedia Foundation, Inc., Public Library of Science, The Encyclopedia of Life, Open Book Publishers (OBP), PubMed, U.S. National Library of Medicine, National Center for Biotechnology Information, U.S. National Library of Medicine, National Institutes of Health (NIH), U.S. Department of Health & Human Services, and USA.gov, which sources content from all federal, state, local, tribal, and territorial government publication portals (.gov, .mil, .edu). Funding for USA.gov and content contributors is made possible from the U.S. Congress, E-Government Act of 2002.
Crowd sourced content that is contributed to World Heritage Encyclopedia is peer reviewed and edited by our editorial staff to ensure quality scholarly research articles.
By using this site, you agree to the Terms of Use and Privacy Policy. World Heritage Encyclopedia™ is a registered trademark of the World Public Library Association, a non-profit organization.
|
Prologue: The big $O$ notation is a classic example of the power and ambiguity of some notations as part of language loved by human mind. No matter how much confusion it have caused, it remains the choice of notation to convey the ideas that we can easily identify and agree to efficiently.
I totally understand what big $O$ notation means. My issue is when we say $T(n)=O(f(n))$ , where $T(n)$ is running time of an algorithm on input of size $n$.
Sorry, but you do not have an issue if you understand the meaning of big $O$ notation.
I understand semantics of it. But $T(n)$ and $O(f(n))$ are two different things. $T(n)$ is an exact number, But $O(f(n))$ is not a function that spits out a number, so technically we can't say $T(n)$
$O(f(n))$, if one asks you what's the equals of $O(f(n))$, what would be your answer? There is no answer. value
What is important is the semantics. What is important is (how) people can agree easily on (one of) its precise interpretations that will describe asymptotic behavior or time or space complexity we are interested in. The default precise interpretation/definition of $T(n)=O(f(n))$ is, as translated from Wikipedia,
$T$ is a real or complex valued function and $f$ is a real valued function, both defined on some unbounded subset of the real positive numbers, such that $f(n)$ is strictly positive for all large enough values of $n$. For for all sufficiently large values of $n$, the absolute value of $T(n)$ is at most a positive constant multiple of $f(n)$. That is, there exists a positive real number $M$ and a real number $n_0$ such that
${\text{ for all }n\geq n_{0}, |T(n)|\leq \;Mf(n){\text{ for all }}n\geq n_{0}.}$
Please note this interpretation is considered
the definition. All other interpretations and understandings, which may help you greatly in various ways, are secondary and corollary. Everyone (well, at least every answerer here) agrees to this interpretation/definition/semantics. As long as you can apply this interpretation, you are probably good most of time. Relax and be comfortable. You do not want to think too much, just as you do not think too much about some of the irregularity of English or French or most of natural languages. Just use the notation by that definition.
$T(n)$ is an exact number, But $O(f(n))$ is not a function that spits out a number, so technically we can't say $T(n)$
$O(f(n))$, if one asks you what's the equals of $O(f(n))$, what would be your answer? There is no answer. value
Indeed, there could be no answer, since the question is ill-posed. $T(n)$ does not mean an exact number. It is meant to stand for a function whose name is $T$ and whose formal parameter is $n$ (which is sort of bounded to the $n$ in $f(n)$). It is just as correct and even more so if we write $T=O(f)$. If $T$ is the function that maps $n$ to $n^2$ and $f$ is the function that maps $n$ to $n^3$, it is also conventional to write $f(n)=O(n^3)$ or $n^2=O(n^3)$. Please also note that the definition does not say $O$ is a function or not. It does not say the left hand side is supposed to be equal to the right hand side at all! You are right to suspect that equal sign does not mean equality in its ordinary sense, where you can switch both sides of the equality and it should be backed by an equivalent relation. (Another even more famous example of abuse of the equal sign is the usage of equal sign to mean assignment in most programming languages, instead of more cumbersome
:= as in some languages.)
If we are only concerned about that one equality (I am starting to abuse language as well. It
is not an equality; however, it is an equality since there is an equal sign in the notation or it could be construed as some kind of equality), $T(n)=O(f(n))$, this answer is done.
However, the question actually goes on. What does it mean by, for example, $f(n)=3n+O(\log n)$? This equality is not covered by the definition above. We would like to introduce another convention,
the placeholder convention. Here is the full statement of placeholder convention as stated in Wikipedia.
In more complicated usage, $O(\cdots)$ can appear in different places in an equation, even several times on each side. For example, the following are true for $n\to \infty$.
$(n+1)^{2}=n^{2}+O(n)$
$(n+O(n^{1/2}))(n+O(\log n))^{2}=n^{3}+O(n^{5/2})$
$n^{O(1)}=O(e^{n})$
The meaning of such statements is as follows: for any functions which satisfy each $O(\cdots)$ on the left side, there are some functions satisfying each $O(\cdots)$ on the right side, such that substituting all these functions into the equation makes the two sides equal. For example, the third equation above means: "For any function $f(n) = O(1)$, there is some function $g(n) = O(e^n)$ such that $n^{f(n)} = g(n)$."
You may want to check here for another example of placeholder convention in action.
You might have noticed by now that I have not used the set-theoretic explanation of the big $O$-notation. All I have done is just to show even without that set-theoretic explanation such as "$O(f(n))$ is a set of functions", we can still understand big $O$-notation fully and perfectly. If you find that set-theoretic explanation useful, please go ahead anyway.
You can check the section in "asymptotic notation" of CLRS for a more detailed analysis and usage pattern for the family of notations for asymptotic behavior, such as big $\Theta$, $\Omega$, small $o$, small $\omega$, multivariable usage and more. The Wikipedia entry is also a pretty good reference.
Lastly, there is some inherent ambiguity/controversy with big $O$ notation with multiple variables,1 and 2. You might want to think twice when you are using those.
|
Snell's Law, also known as the Law of Refraction, is an equation that relates the angle of the incident light and the angle of the transmitted light at the interface of two different mediums. Snell's Law can be applied to all materials, in all phases of matter. Most people are familiar with Snell's Law because of the apparent shortening of their legs that is observed when standing in water. Another commonly recognized example of refraction in a material is diamonds. The many facets of the cut diamond combined with a high index of refraction give diamonds the brilliance that they are known for. Snell's Law is especially important for optical devices, such as fiber optics. Snell's Law states that the ratio of the sine of the angles of incidence and transmission is equal to the ratio of the refractive index of the materials at the interface.
Speed of Light and Index of Refraction
Refraction occurs because the speed of the light changes when it passes into a new medium. The speed of light in a medium is given by the following equation:
\[c=nv\]
where n is the refractive index of the material and c is the speed of light in a vacuum. The refractive index can also be determined from the permittivity and permeability of the material. Therefore it is possible to know the optical properties of the material from the electrical properties of the material. Using these properties, the index of refraction is given by the following equation:
\[n=\dfrac{c}{v}=\sqrt{\dfrac{\varepsilon \mu }{\varepsilon_{0} \mu_{0}}}=\sqrt{\varepsilon_{r} \mu_{r}}\]
where ε
r and μ r are the permativity and permeability of the material respectively. Table 1 lists the index of refraction for some common materials.
Vacuum 1.000 Air 1.000293 Water 1.333 Ice 1.309 Plexiglass 1.49 Soda-Lime Glass 1.46 Diamond 2.42 Snell's Law
Snell's law relates the sines of the angles of incidence and transmission to the index of refraction for each material:
\[\dfrac{\sin \theta_{1}}{\sin \theta_{2}}=\dfrac{n_{2}}{n_{1}}\]
It should be noted that the angles are measured from the normal line at the interface (Figure 1).
Figure 1: Refraction at the interface between two materials (taken with permission from Wikipedia)
Additionally, because the index of refraction is related to the speed of light in the material, the following equation is also true:
\[\dfrac{\sin \theta_{1}}{sin \theta_{2}}=\dfrac{v_{1}}{v_{2}}\]
As light crosses the boundary between two different materials, the light will be refracted either at a greater angle or a smaller angle depending on the relative refractive indecies of each material. If the index of refraction for the second material is greater than the first material, then the light will be refracted to a smaller angle.
Theory: Light as a Wave
Snell's Law arises from the wave nature of light. Light can be described in one of several ways: a ray, a particle (photon), and as a wave. The interaction of light with a material can be best described for light when light is thought of as a wave.
The key to understanding the behavior of light at an interface is that Energy must be conserved: the energy of light going in the first material must be the same as the energy of the light in the second material. The energy of light can be described in the following way:
\[E=hf=\dfrac{hv}{\lambda}\]
where f is the frequency of the wave, v is the speed of the wave, h is Plank's constant, and c is the speed of light in a vacuum.
Because energy must be conserved at the boundary, the frequency of the light must remain the same for all materials. This is because the number of waves arriving at the interface per unit time must be the same as the number of waves leaving the interface per unit time; the boundary interface cannot create or destroy waves. This concept is shown in Figure 2.
Figure 2: The number of wavefronts arriving at the interface must be the same at the number leaving the interface. (taken with permission from Wikipedia)
However, the speed that a wave propagates at is dependent on the medium the light is traveling in. This velocity change is due to the interaction between light and the resonance of the electrons in the material which are either tightly or loosely held to the nuclei in the material. It is this change in velocity that causes the refraction of light in a new medium. The velocity of light in a medium is given by the following equation: \[v=\dfrac{1}{\sqrt{\varepsilon \mu }}\]where ε is the the permativity and μ is the permeability of the particular material.
Critical Angle and Total Internal Reflection
An important extension of Snell's law is the concept of Total Internal Reflection and the critical angle. For any combination of mediums, there is an angle for which the refracted light will be perpendicular to the normal (Figure 3). The critical angle is given by the equation:
\[\theta_{c}=\sin^{-1} \left [ \dfrac{n_{1}}{n_{2}} \right ]\]
Figure 3: Refraction, Critical Angle, and Total Internal Reflection at an interface (taken with permission from Wikipedia)
If the incident angle is greater than the critical angle, the light will be completely
reflected back into the original medium. This is shown in the last image of Figure 3. Total Internal Reflection can also occur if there is a significant difference in the refractive index between the two materials. It is this type of total internal reflection that gives rise to fiber optics. Applications and Importance
A very important application of Snell's Law is Fiber Optics. Fiber Optics are used for many applications, from telecommunications to data transmission in high speed servers. Because the fibers are not laid out in straight lines, the light must be guided down the length of the fiber. This is achieved with one of three types of fiber: step-index fiber, graded-index fiber (multimode), or single-mode fiber. These fiber types are shown in Figure 4.
Figure 4: Schematics and comparison of Fiber Optic Cable types.
In step-index fibers, there is a very large decrease in the index of refraction at the interface between the glass fiber and its cladding. This causes the light to move in a zigzag pattern down the length of the fiber, and because of this there is a considerable distortion in the light pulse. It is because of this distortion that step-index fibers are only used for applications that have shorter fiber lengths.
For graded-mode fibers, the index of refraction of the core decreases mas the radius of the fiber gets larger. This is achieved by introducing additives to the fiber glass, such as B
2O 3 and GeO 2. By having the index of refraction change gradually, the light is gradually bent back towards the center of the fiber. Therefore, the distortion of the light is much smaller than it is for the step-index fiber. This means that a longer fiber can be used if the fiber has graded-index core. Graded-index core fibers are generally used for data transfer, such as in local area networks.
The final type of fiber is the single-mode fiber. Single-mode fibers are generally used for long distance networks where there is very little need for a curve in the fiber. In single-mode fibers, the light travels straight down the axis of the fiber and does not experience refraction. This is shown in Figure 4.
Questions Calculation of Transmittance Angle Using values given in Table 1, calculate the angle of transmittance if the incident angle of the light is 26 o. The light is moving from air into glass. Is Snell's Law only for solid materials? Which direction will light bend (towards or away from the normal) if the index of refraction for the second material in the interface is higher than the first? Why? Answers
θ=17.5
No
The transmitted angle will be smaller than the incident angle. So the light will bend towards the normal. This is because the a higher index of refraction in the second material causes a decrease in the speed of the light in the second material.
Additional Links Polarization of Waves (Chemwiki) Refractive Index (Wikipedia) Critical Angle (Wikipedia) Snell's Law (Wikipedia) Graded-Index Fiber (Wikipedia) Single-Mode Fiber (Wikipedia) Optical Fiber (Wikipedia) Fiber Optics (Chemwiki) Glass (Chemwiki) Material Science and Engineering: An Introduction (PDF) Permeability (Wikipedia) Permitivity (Wikipedia) Total Internal Reflection (Wikipedia) Reflection Conditions (Chemwiki) References Flens, Frank. Director of Engineering, Finisar. Personal interview. Hummel, Rolf E. Electronic Properties of Materials.New York: Springer, 2011. Print. Young, Hugh D. , Roger A. Freedman. Sears and Zemansky's University Physics: with Modern Physics. San Fransisco: Pearson Addison Wesley, 2008. Print. Hecht, E., and A. Zajac. Optics. 1979. Print. Shackelford, James F. Introduction to Materials Science for Engineers. Upper Saddle River, NJ: Pearson/Prentice Hall, 2005. Print. Hecht, Eugene. Optics. Reading, MA: Addison-Weley, 2002. Print. Callister, William D., David G. Rethwisch. Materials Science and Engineering: An Introduction. December 2009. Print. Contributors Hannah Flens (UC Davis, Material Science and Engineering)
|
Recap of Lecture 22
Last lecture address two unique aspects of electrons: spin and indistinguishability and how they couple into describing multi-electron wavefunctions. The spin results in an angular momentum that follows the same properties of orbital angular moment including commutators and uncertainty effect. The Slater determinant wavefunction was introduced as a way to consistently address both properties.
The Unsolvable System
The generic multielectron atom including terms for the additional electrons with a general charge \(Z\); e.g.
\[V_{nuclear-electron}(r_1) = -\dfrac {Z}{\left\vert\mathbf{r} - \mathbf{R}\right\vert} \label{8.3.1}\]
in atomic units with \(\left\vert\mathbf{r} - \mathbf{R}\right\vert\) is the distance between the electron and the nucleus, The Hamiltonian must also have terms for electron-electron repulsion (also in atomic units)
\[V_{electron-electron}(r_{12}) = \dfrac {1}{\left\vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} \label{8.3.2}\]
with \(\left\vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert\) is the distance between electron 1 and electron 2. So the proper multi-electron Hamiltonian can be constructed
\[\hat {H} (r_1, r_2, ... r_n) = -\dfrac {\hbar ^2}{2m_e} \sum _i \nabla ^2_i + \sum _i V_{nuclear-electron} (r_i) + \sum _{i \ne j} V_{electron-electron} (r_{ij}) \label{8.3.3}\]
Given what we have learned from the previous quantum mechanical systems we’ve studied, we predict that exact solutions to the multi-electron Schrödinger equation would consist of a family of multi-electron wavefunctions, each with an associated energy eigenvalue. These wavefunctions and energies would describe the ground and excited states of the multi-electron atom, just as the hydrogen wavefunctions and their associated energies describe the ground and excited states of the hydrogen atom. We would predict quantum numbers to be involved, as well.
The fact that electrons interact through their electron-electron repulsion (final term in Equation \(\ref{8.3.3}\)) means that an exact wavefunction for a multi-electron system would be a single function that depends
simultaneously upon the coordinates of all the electrons; i.e., a multi-electron wavefunction:
\[|\Psi (r_1, r_2, \cdots r_i) \rangle \label{8.3.4}\]
Unfortunately, the electron-electron repulsion terms make it impossible to find an exact solution to the Schrödinger equation for many-electron atoms.
The Hartree Approximation
The method for finding best possible one-electron wavefunctions that was published by Hartree in 1948 and improved two years later by Fock. For the Schrödinger equation to be solvable, the variables must
be separable. The variables are the coordinates of the electrons. To separate the variables in a way that retains information about electron-electron interactions, the electron-electron term (Equation \(\ref{8.3.1}\)) must be approximated so it depends only on the coordinates of one electron. Such an approximate Hamiltonian can account for the interaction of the electrons in an average way. The exact one-electron eigenfunctions of this approximate Hamiltonian then can be found by solving the Schrödinger equation. These functions are the best possible one-electron functions.
The Hartree approximation starts by invoking an initial
ansatz that the multi-electron wavefucntion in Equation \(\ref{8.3.4}\) can be expanded as a product of single-electron wavefunctions
\[\Psi(\mathbf{r}_1,\mathbf{r}_2, \ldots, \mathbf{r}_N) \approx \psi_{1}(\mathbf{r}_1)\psi_{2}(\mathbf{r}_2) \ldots \psi_{N}(\mathbf{r}_N) \label{2.3}\]
from which it follows that the electrons are
independent, and interact only via the mean-field Coulomb potential. This yields one-electron Schrödinger equations of the form
\[-\dfrac{\hbar^{2}}{2m} \nabla^{2}\psi_{i}(\mathbf{r}) + V(\mathbf{r})\psi_{i}(\mathbf{r}) = \epsilon_{i}\psi_{i}(\mathbf{r}) \label{2.4}\]
or
\[H_e(r) \psi_{i}(\mathbf{r}) = \epsilon_{i}\psi_{i}(\mathbf{r}) \label{2.4B}\]
where \(V(r)\) is the potential in which the electron moves; this includes both the nuclear-electron interaction
\[V_{nucleus}(\mathbf{r}) = -Ze^{2}\sum_{R} \dfrac{1}{\left\vert\mathbf{r} - \mathbf{R}\right\vert} \label{2.5}\]
and the mean field arising from the \(N-1\) other electrons. We smear the other electrons out into a smooth negative charge density \(\rho(\mathbf{r}')\) leading to a potential of the form
\[V_{electron}(\mathbf{r}) = -e\int d\mathbf{r}^{\prime} \rho(\mathbf{r}^{\prime}) \dfrac{1}{\left\vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} \label{2.6}\]
where
\[\rho(\mathbf{r}) = \sum_{i}^{\text{occupied}}\vert\psi(\mathbf{r})\vert^{2}.\]
The sum over runs over all occupied states.
The wavefunctions that from this approach with the Hamiltonian \(H_e(r)\) involve possess three kinds of energies:
Kinetic Energy
The Kinetic energy of the electron has an average value is computed by taking the expectation value of the kinetic energy operator \[-\dfrac{\hbar^2}{2m} \nabla^2\] with respect to any particular solution \(\phi_j(r)\) to the Schrödinger equation: \[KE = \langle\phi_j| -\dfrac{\hbar^2}{2m} \nabla^2 |\phi_j\rangle \]
Nuclear-Electron Coulombic Attraction Energy
Coulombic attraction energy with the nucleus of charge \(Z\): \[\langle\phi_j| \dfrac{-Z e^2}{\left\vert\mathbf{r} - \mathbf{R}\right\vert} |\phi_j\rangle\]
Electron-Electron Coulombic Repulsion Energy
Coulomb repulsion energies with all of the \(N-1\) other electrons, which are assumed to occupy other atomic orbitals denoted \(\phi_K\), with this energy computed as
\[\sum_{J\neq K} \langle\phi_j(r) \phi_K(r’) |\dfrac{e^2}{|r-r’|} | \phi_j(r) \phi_K(r’)\rangle.\label{8.3.8}\]
The Dirac notation \(\langle\phi_j(r) \phi_K(r’) |\dfrac{e^2}{|r-r’|} | \phi_j(r) \phi_K(r’)\rangle\) is used to represent the six-dimensional Coulomb integral
\[J_{J,K} = \int |\phi_j(r)|^2 |\phi_K(r’)|^2 \dfrac{e^2}{r-r'} dr dr’ \label{8.3.9}\]
that describes the Coulomb repulsion between the charge density \(|\phi_j(r)|^2\) for the electron in \(\phi_j\) and the charge density \(|\phi_K(r’)|^2\) for the electron in \(\phi_K\). Of course, the sum over \(K\) must be limited to exclude \(K=J\) to avoid counting a “self-interaction” of the electron in orbital \(\phi_j\) with itself.
The total energy \(\epsilon_J\) of the orbital \(\phi_j\), is the sum of the above three contributions:
\[\epsilon_J = \langle\phi_j| - \dfrac{\hbar^2}{2m} \nabla^2 |\phi_j\rangle + \langle\phi_j| \dfrac{-Z e^2}{\left\vert\mathbf{r} - \mathbf{R}\right\vert} |\phi_j\rangle + \sum_{J\neq K} \langle\phi_j(r) \phi_K(r’) |\dfrac{e^2}{|r-r’|} | \phi_j(r) \phi_K(r’)\rangle.\label{8.3.10}\]
This treatment of the electrons and their orbitals is referred to as the Hartree-level of theory.
When screened hydrogenic atomic orbitals are used to approximate the \(\phi_j\) and \(\phi_K\) orbitals, the resultant \(\epsilon_J\) values do
not produce accurate predictions. For example, the negative of \(\epsilon_J\) should approximate the ionization energy for removal of an electron from the orbitals \(\phi_j\). Such ionization potentials (IP s) can be measured, and the measured values do not agree well with the theoretical values when a crude screening approximation is made for the atomic orbitals. The Self-Consistant Field (SCF) Approach to the Variational Method
An important unsolved problem in quantum mechanics is how to deal with indistinguishable, interacting particles - in particular electrons which determine the behavior of almost every object in nature. The basic problem is that if particles interact, that interaction must be in the Hamiltonian. So until we know where the particles are, we cannot write down the Hamiltonian, but until we know the Hamiltonian, we can’t tell where the particles are.
The idea is to solve the Schrödinger equation for an electron moving in the potential of the nucleus and all the other electrons. We start with a guess for the trial electron charge density, solve N/2 one-particle Schrödinger equations (initially identical) to obtain N electron wavefunctions. Then we construct the potential for each wavefunction from that of the nucleus and that of all the other electrons, symmetrize it, and solve the N/2 Schrödinger equations again. This method is ideal for a computer, because it is easily written as an algorithm Figure \(\PageIndex{1}\).
Although we are concerned here with atoms, the same methodology is used for molecules or even solids (with appropriate potential symmetries and boundary conditions). This is a variational method, so wherever we refer to wavefunctions, we assume that they are expanded in some appropriate basis set.
Hartree-Fock method
Fock improved on Hartree’s method by using proper "antisymmetrized wavefunctions" (called the Hartree-Fock method) instead of simple one-electron wavefunctions. The generalized Slater determinant for a multi-electrom atom with \(N\) electrons is then
\[ \psi(\mathbf{r}_1, \mathbf{r}_2, \ldots, \mathbf{r}_N)=\dfrac{1}{\sqrt{N!}} \left| \begin{matrix} \chi_1(\mathbf{r}_1) & \chi_2(\mathbf{r}_1) & \cdots & \chi_N(\mathbf{r}_1) \\ \chi_1(\mathbf{r}_2) & \chi_2(\mathbf{r}_2) & \cdots & \chi_N(\mathbf{r}_2) \\ \vdots & \vdots & \ddots & \vdots \\ \varphi_1(\mathbf{r}_N) & \chi_2(\mathbf{r}_N) & \cdots & \chi_N(\mathbf{r}_N) \end{matrix} \right| \label{5.6.96}\]
If we go through the same process as above with this more proper wavefunction we get similar, but not identical single-particle
Hartree-Fock equations:
\[-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi_{i}(\mathbf{r}) + V_{nucleus}(\mathbf{r})\psi_{i}(\mathbf{r}) + V_{electron}(\mathbf{r})\psi_{i}(\mathbf{r}) - \sum_{j} \int d\mathbf{r}^{\prime} \dfrac{\psi^{\star}_{j}(\mathbf{r}') \psi^{\star}_{i}(\mathbf{r}') \psi_{j}(\mathbf{r}) } {\left \vert \mathbf{r} - \mathbf{r}^{\prime} \right\vert} = \epsilon_{i}\psi_{i}(\mathbf{r}). \label{2.9}\]
can be recast as series of Schrödinger-like equations:
\[ \hat {F} | \varphi _i \rangle = \epsilon _i| \varphi _i \rangle \label {8.7.2}\]
where
\(\hat {F}\) is called the Fock operatorand \( \{| \varphi_i \rangle \}\) are the Hatree-Fock orbitals with corresponding energies \(\epsilon_i\).
The Fock operator is a one-electron operator and solving a Hartree-Fock equation gives the energy and Hartree-Fock orbital for one electron.
The nature of the Fock operator reveals how the Hartree-Fock (HF) or Self-Consistent Field (SCF) Method accounts for the electron-electron interaction in atoms and molecules while preserving the idea of
independent atomic orbitals. The wavefunction written as a Slater determinant of spin-orbitals is necessary to derive the form of the Fock operator, which is
\[\hat {F} = \hat {H} ^0 + \sum _{j=1}^N ( 2 \hat {J} _j - \hat {K} _j ) = -\dfrac {\hbar ^2}{2m} \nabla ^2 - \dfrac {Ze^2}{4 \pi \epsilon _0 r} + \sum _{j=1}^N (2\hat {J}_j - \hat {K} _j ) \label {8.7.3}\]
\(\hat{J}\) is the Coulomb operator,defining the electron-electron repulsion energy due to each of the two electrons in the jth orbital. \(\hat {K}\) is the exchange operator, defining the electron exchange energy due to the antisymmetry of the total n-electron wave function. This (so called) "exchange energy" operator, K, is simply an artifact of the Slater determinant.
The Hartree-Fock equations \(h_e \phi_i = \epsilon_i \phi_i\) imply that the orbital energies \(\epsilon_i\) can be written as:
\[\epsilon_i = \langle \phi_i | h_e | \phi_i \rangle = \langle \phi_i | T + V | \phi_i \rangle + \sum_{j({\rm occupied})} \langle \phi_i | J_j - K_j | \phi_i \rangle \label{8.7.6}\]
\[= \langle \phi_i | T + V | \phi_i \rangle + \sum_{j({\rm occupied})} [ J_{i,j} - K_{i,j} ],\label{8.7.7}\]
where \(T + V\) represents the kinetic (\(T\)) and nuclear attraction (\(V\)) energies, respectively. Thus, \(\epsilon_i\) is the average value of the kinetic energy plus Coulombic attraction to the nuclei for an electron in \(\phi_i\) plus the sum over all of the spin-orbitals occupied in \(\psi\) of Coulomb minus Exchange interactions of these electrons with the electron in \(\phi_i\).
Hartree vs. Hartree-Fock
The many-body wave function gives the N-particle distribution function, i.e. \(|Φ(r_1, ..., r_N )|^2\) is the probability density that particle 1 is at \(r_1\), ..., and particle \(N\) is at \(r_N\).
However, when trying to work out the interaction between electrons, what we want to know is the probability of finding an electron at \(r\), given the positions of all the other electrons \(\{r_i\}\). This implies that the electron behaves quantum mechanically when we evaluate its wavefunction, but as a classical point particle when it contributes to the potential seen by the other electrons.
Hartree-Fock is not that bad... but not super great either
Atom Hartree-Fock Energy Experiment \(He\) \(-5.72\) \(-5.80\) \(Li\) \(-14.86\) \(-14.96\) \(Ne\) \(-257.10\) \(-257.88\) \(Ar\) \(-1053.64\) \(-1055.20\) Koopman's Theorem
Koopmans' theorem states that the first ionization energy is equal to the negative of the orbital energy of the highest occupied molecular orbital. Hence, the ionization energy required to generated a cation and detached electron is represented by the removal of an electron from an orbital without changing the wavefunctions of the other electrons. this is called the "frozen orbital approximation."
Within this model, the energy difference between the daughter and the parent can be written as follows (\(\phi_k\) represents the \(k^{th}\) spin-orbital that is added or removed):
Hence for electron detachment (ionization):
\[E_{N-1} - E_N = - \epsilon_k \]
For electron attachement (electron affinity): It is sometimes claimed that Koopmans' theorem also allows the calculation of electron affinities as the energy of the lowest unoccupied molecular orbitals (LUMO) of the respective systems. However, Koopmans' original paper makes no claim with regard to the significance of eigenvalues other than that corresponding to the HOMO. Nevertheless, it is straightforward to generalize the original statement of Koopmans' to calculate the electron affinity in this sense.
\[E_N - E_{N+1} = - \epsilon_{k+1} .\]
Koopman's Theorem in action
Koopman's Theorem argues that you do not need to perform two independent quantum calculations to determine the ionization energy and electron affinity as done in the ab initio homeworks, but only one is needed and the energies of the spin-orbitals are needed.
Contributors Wikipedia
Adapted from "Quantum States of Atoms and Molecules" by David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski
|
Let $F$ be a $p$-adic local field, and $G$ a connected reductive group over $F$. What is the meaning of a "highly ramified character" of $G(F)$? I have seen this terminology in many places in representation theory, e.g. "twisting an irreducible admissible representation of $G(F)$ by a highly ramified character," but I have never seen a formal definition.
If $G = \operatorname{GL}_1$, then a character of $G(F)$ is a continuous homomorphism $F^{\ast} \rightarrow S^1$. It is called unramified if it is trivial on $\mathcal O_F^{\ast}$, and highly ramified if it is only trivial on $1 + \mathfrak p_F^n$ for $n$ large.
If $G = \operatorname{GL}_n$, then the characters of $G(F)$ are often taken to be compositions $$G(F) \xrightarrow{\textrm{det}} F^{\ast} \xrightarrow{\chi} S^1$$
for $\chi$ a character of $F^{\ast}$, and they are called ramified if $\chi$ is ramified.
For a general reductive group $G$ over $F$, I would imagine the discussion of ramified characters to be done in the same way, i.e. by compositions of characters of $F^{\ast}$ with $F$-rational characters of $G$. But there may not exist nontrivial $F$-rational characters of $G$, i.e. the split component of $G$ may be trivial. This occurs for example when $G$ is the unitary group $U(n,n)$ for a quadratic extension $E/F$.
|
Hedging American Swaption
Hello, I priced an American swaption using Black model with swap rates diffusion to find the european (call) price at t.
$$ C_t = (\delta \sum_{j=n+1}^{M+1} Z_t^{T_j})[R(t,T_n,T_m) - \hat{R}]^{+} $$
$$ dR(t,T_n,T_M) = \sigma R(t,T_n,T_M) dW_t $$
Then I found the early exercice boundary via MC Simulation, with this method.
The choice of this method rely on the need to have an explicit criterion for the optimal exercise, and because I had to add a depreciaton factor on the the principal amount (contract linked to a loan).
As you may know, it is not very interesting to compute greeks with finite difference of a Mc Price.
Now I want to hedge this american swaption, so I am trying to calculate the $\Delta$ of the contract.
I know that Malliavin calculus can give good results in this domain, but I can't find any paper for the implementation of the method for an american swaption.
I found this paper for american option, but I am not sure to be able to adapt it.
Do you think this approach could be generalised to American Swaption ? What would be your approach to hedging an American Swaption ?
|
There is far more structure in entanglement than your simplistic definition seems to be implying. Yes, you could ask a question
Is there any entanglement in this pure state $|\psi\rangle$ of $n$ qubits?
Which can be answered by whether or not the entire state can be expressed as$$|\psi\rangle=|\phi_1\rangle|\phi_2\rangle\ldots|\phi_n\rangle$$but as you are rightly realising, you can ask far more structured questions of
is qubit $i$ entangled?
or even
is qubit $i$ entangled with qubit $j$?
or
What sort of entanglement does qubit $i$ share with different qubits, and how much?
The first is answered in much the same way as your original strategy: if it can be written as $|\phi\rangle_i\otimes|\psi\rangle_{1,2,\ldots,n\setminus i}$, then qubit $i$ is not entangled with anything. In your example, qubit 2 is not entangled, but qubits 1 and 3 are.
The second is far more subtle, and perhaps requires a more precise definition of entanglement. Should we say that qubits 1 and 2 in the state $(|000\rangle+|111\rangle)/\sqrt{2}$ are entangled? If you look at just those two qubits, they are in a maximally mixed state, i.e. not entangled, but they are definitely part or an entangled state that cannot be localised to any pair of qubits. Hence the question about the type of entanglement.
Ultimately, this really comes down to a question of why are you interested in entanglement. Usually, it's to do something with the entangled state, so you have a measure of a given state's usefulness (and that is often some form of entanglement measure), and you can ask questions with respect to that particular measure.
|
Answer
$\frac{1}{2\pi}\sqrt{\frac{2a}{m}}$
Work Step by Step
We know that the force is given by $F=-\frac{dU}{dx}=-2ax$. Thus, we use Hooke's Law and the equation for frequency to find: $-kx=-2ax \\ k=2a$ Thus, it follows: $f= \frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{2a}{m}}$
|
We can interpret the
sdr problem as a problem aboutgraphs. Given sets $A_1,A_2,\ldots,A_n$, with $\bigcup_{i=1}^n A_i=\{x_1,x_2,\ldots,x_m\}$, we define a graph with$n+m$ vertices as follows: The vertices are labeled$\{A_1,A_2,\ldots,A_n,x_1,x_2,\ldots x_m\}$, and the edges are$\{\{A_i,x_j\}\mid x_j\in A_i\}$.
Example 4.4.1 Let $A_1=\{a,b,c,d\}$, $A_2=\{a,c,e,g\}$, $A_3=\{b,d\}$, and $A_4=\{a,f,g\}$. The corresponding graph is shown in figure 4.4.1.
Before exploring this idea, we introduce a few basic concepts aboutgraphs. If two vertices in a graph are connected by an edge, we saythe vertices are
adjacent. If a vertex $v$ isan endpoint of edge $e$, we say they are incident. The set of vertices adjacent to $v$ iscalled the neighborhood of $v$, denoted$N(v)$. This is sometimes called the open neighborhood of $v$ todistinguish it from the closed neighborhoodof $v$, $N[v]=N(v)\cup\{v\}$.The degree of a vertex $v$ is the number of edges incident with $v$; it is denoted $\d(v)$.
Some simple types of graph come up often: A
path is a graph $P_n$ on vertices $v_1,v_2,\ldots,v_n$,with edges $\{v_i,v_{i+1}\}$ for $1\le i\le n-1$, and no other edges.A cycle is a graph $C_n$ on vertices$v_1,v_2,\ldots,v_n$ with edges $\{v_i,v_{1+(i\bmod n)}\}$ for $1\le i\le n$,and no other edges; this is a path in which the first and lastvertices have been joined by an edge. (Generally, we require that acycle have at least three vertices. If it has two, then the two arejoined by two distinct edges; when a graph has more than one edge withthe same endpoints it is called a multigraph.If a cycle has one vertex, there is an edge, called a loop, in which a single vertex serves as bothendpoints.) The length of a path or cycle isthe number of edges in the graph. For example, $P_1$ has length 0,$C_1$ has length 1. A complete graph $K_n$ is agraph on $v_1,v_2,\ldots,v_n$ in which every two distinct vertices are joinedby an edge. See figure 4.4.2 for examples.
The graph in figure 4.4.1 is a
bipartite graph.
Definition 4.4.2 A graph $G$ is bipartite if its vertices can be partitioned into two parts, say $\{v_1,v_2,\ldots,v_n\}$ and $\{w_1,w_2,\ldots,w_m\}$ so that all edges join some $v_i$ to some $w_j$; no two vertices $v_i$ and $v_j$ are adjacent, nor are any vertices $w_i$ and $w_j$.
|
5.1: Let \(G\) : \(\{0,1\}^{\lambda} \rightarrow \{0,1\}^{\lambda + \ell}\). Define \(im(G) = \{y \in \{0,1\}^{\lambda + \ell} \mid \exists s : G(s) = y\}\), the set of possible outputs of \(G\). For simplicity, assume that \(G\) is injective (i.e., 1-to-1).
(a). What is \(|im(G)|\), as a function of \(\lambda\) and \(\ell\)?
(b). A string \(y\) is chosen randomly from \(\{0,1\}^{\lambda + \ell}\). What is the probability that \(y \in im(G)\), expressed as a function of \(\lambda\) and \(\ell\)?
5.2: Let \(G : \{0,1\}^{\lambda} \rightarrow \{0,1\}^{\lambda + \ell}\) be an injective PRG. Consider the following distinguisher:
\[\mathscr{A}\\x:=\text{QUERY()}\\\text{for\space all}\space s^{\prime}\in\{0,1\}^{\lambda}:\\\text{if}\space G(s^{\prime})=x\space\text{then\space return 1}\\text{return}\space 0\nonumber\]
(a). What is the advantage of \(?\) in distinguishing \(\mathscr{L}^G_{\text{prg-real}}\) and \(\mathscr{L}^G_{\text{prg-rand}}\)? Is it negligible?
(b). Does this contradict the fact that \(G\) is a PRG? Why or why not?
(c). What happens to the advantage if \(G\) is not injective?
5.3: Let \(G : \{0,1\}^{\lambda} \rightarrow \{0,1\}^{\lambda + \ell}\) be an injective PRG, and consider the following distinguisher:
\[\mathscr{A}\\x:=\text{QUERY()}\\s^{\prime}\space\leftarrow\{0,1\}^{\lambda}\\\text{return}\space G(s^{\prime})\stackrel{?}{=}x\nonumber\]
What is the advantage of ? in distinguishing \(\mathscr{L}^G_{\text{prg-real}}\) from \(\mathscr{L}^G_{\text{prg-rand}}\)?
Hint: When computing \(Pr[? \diamondsuit \mathscr{L}^G_{\text{prg-rand}}\) outputs 1], separate the probabilities based on whether \(x \in im(G)\) or not. (\(im(G)\) is defined in a previous problem)
5.4: For any PRG \(G : \{0,1\}^{\lambda} \rightarrow \{0,1\}^{\lambda + \ell}\) there will be many strings in \(\{0,1\}^{\lambda + \ell}\) that are impossible to get as output of \(G\). Let S be any such set of impossible \(G\)-outputs, and consider the following adversary that has \(S\) hard-coded:
\[\mathscr{A}\\x:=\text{QUERY()}\\\text{return}\space x\stackrel{?}{\in}S\nonumber\]
What is the advantage of ? in distinguishing \(\mathscr{L}^G_{\text{prg-real}}\) from \(\mathscr{L}^G_{\text{prg-rand}}\)? Why does an adversary like this one not automatically break every PRG?
5.5: Show that the scheme from Section 5.2 does not have perfect one-time secrecy, by showing that there must exist two messages \(m_1\) and \(m_2\) whose ciphertext distributions differ.
Hint: There must exist strings \(s_1, s_2 \in \{0,1\}^{2\lambda}\) where \(s_1 \in im(G)\), and \(s_2 \notin im(G)\). Use these two strings to find two messages \(m_1\) and \(m_2\) whose ciphertext distributions assign different probabilities to \(s_1\) and \(s_2\). Note that it is legitimate for an attacker to “know” \(s_1\) and \(s_2\), as these are properties of \(G\) alone, and independent of the random choices made when executing the scheme.
5.6: (a). Let \(?\) be any function. Show that the following function \(G\) is
not a secure PRF (even if ? is a secure PRG!). Describe a successful distinguisher and explicitly compute its advantage:
\[\underline{G(s):}\\\text{return}\space\parallel f(s)\nonumber\]
(b).Let \(G : {0,1}^{\lambda} \rightarrow \{0,1\}^{\lambda + \ell}\) be a candidate PRG. Suppose there is a polynomial-time algorithm \(V\) with the property that it inverts \(G\) with non-negligible probability. That is,
\[\underset{s\leftarrow\{0,1\}^{\lambda}}{\operatorname{Pr}}[V(G(s))=s]\text { is non-negligible. }\nonumber\]
Show that if an algorithm \(V\) exists with this property, then \(G\) is not a secure PRG. In other words, construct a distinguisher contradicting the PRG-security of \(G\) and show that it achieves non-negligible distinguishing advantage.
Note: Don’t assume anything about the output of \(V\) other than the property shown above. In particular, \(V\) might very frequently output the “wrong” thing.
5.7: In the “PRG feedback” construction \(H\) in Section 5.4, there are two calls to \(G\). The security proof applies the PRG security rule to both of them, starting with the first. Describe what happens when you try to apply the PRG security of \(G\) to these two calls in the opposite order. Does the proof still work, or must it be in the order that was presented?
5.8: Let \(\ell^{\prime} > \ell > 0\). Extend the “PRG feedback” construction to transform any PRG of stretch \(\ell\) into a PRG of stretch \(\ell^{\prime}\). Formally define the new PRG and prove its security using the security of the underlying PRG.
5.9: Let \(G : \{0,1\}^{\lambda} \rightarrow \{0,1\}^{3\lambda}\) be a length-tripling PRG. For each function below, state whether it is also a secure PRG. If the function is a secure PRG, give a proof. If not, then describe a successful distinguisher and explicitly compute its advantage.
\[\text { (a) } \left| \begin{array}{l}{\underline{H(s)}\\ {x} :=G(s)} \\ {y :=G(s)} \\ {y :=\text { first } \lambda \text { bits of } x} \\ {z :=\text { last } \lambda \text { bits of } x} \\ {\text { return } G(y)\|G(z)}\end{array}\right.\nonumber\]
\[\text { (b) } \left| \begin{array}{c}{\underline{H(s):}\\{x} :=G(s)} \\ {y :=\text { first } 2 \lambda \text { bits of } x} \\ {\text { return } y}\end{array}\right.\nonumber\]
\[\text { (c) } \left| \begin{array}{l}{\underline{H(s):}\\{x} :=G(s)} \\ {y :=G(s)} \\ {\text { return } x\|y}\end{array}\right.\nonumber\]
\[\text { (d) } \left| \begin{array}{c}{\underline{H(s):}\\{x} :=G(s)} \\ {y :=G\left(0^{\lambda}\right)} \\ {\text { return } x\|y}\end{array}\right.\nonumber\]
\[\text { (e) } \left| \begin{array}{c}{\underline{H(s):}\\{x} :=G(s)} \\ {y :=G\left(0^{\lambda}\right)} \\ {\text { return } x \oplus y}\end{array}\right.\nonumber\]
\[\text{(f)}\left|\begin{array}{l}{/ / H :\{0,1\}^{2 \lambda} \rightarrow\{0,1\}^{3 \lambda}} \\ {\underline{H(s) :}\\{x :=G\left(s_{\text { left }}\right)}} \\ {y :=G\left(s_{\text { right }}\right)} \\ {\text { return } x \oplus y}\end{array}\right.\nonumber\]
\[\text{(g)}\left|\begin{array}{l}{/ / H :\{0,1\}^{2 \lambda} \rightarrow\{0,1\}^{6 \lambda}} \\ {\underline{H(s):}\\{x :=G\left(s_{\text { left }}\right)}} \\ {y :=G\left(s_{\text { right }}\right)} \\ {\text { return } x\|y}\end{array}\right.\nonumber\]
5.10: Let \(G : \{0,1\}^{\lambda} \rightarrow \{0,1\}^{3\lambda}\) be a length-tripling PRG. Prove that each of the following functions is also a secure PRG:
\[\text{(a)}\left|\begin{array}{l}{/ / H :\{0,1\}^{2 \lambda} \rightarrow\{0,1\}^{3 \lambda}} \\ {\underline{H(s)}\\{\text { return } G\left(s_{\text { left }}\right)}}\end{array}\right.\nonumber\]
Note that H includes half of its input directly in the output. How do you reconcile this fact with the conclusion of Exercise 5.6(b)?
\[\text{(b)}\left|\begin{array}{l}{/ / H :\{0,1\}^{2 \lambda} \rightarrow\{0,1\}^{3 \lambda}} \\ {\underline{H(s)}\\{\text { return } G\left(s_{\text { left }}\right)}}\end{array}\right.\nonumber\]
5.11: A frequently asked question in cryptography forums is whether it’s possible to determine which PRG implementation was used by looking at output samples.
Let \(G_1\) and \(G_2\) be two PRGs with matching input/output lengths. Define two libraries \(\mathscr{L}^{G_1}_{\text{which-prg}}\) and \(\mathscr{L}^{G_2}_{\text{which-prg}}\) as follows:
Prove that if \(G_1\) and \(G_2\) are both secure PRGs, then \(\mathscr{L}^{G_1}_{\text{which-prg}} ≋ \mathscr{L}^{G_2}_{\text{which-prg}}\) — that is, it is infeasible to distinguish which PRG was used simply by receiving output samples.
5.12: Prove that if PRGs exist, thenP \(\not=\) NP
Hint: \(\{y | \exists s : G(s) = y\} \in \text{NP}\) Note: This implies that \(\mathscr{L}^{G}_{\text{prg-real}} \not\equiv \equiv \mathscr{L}^G_{\text{prg-rand}} \text{for all} G\). That is, there can be no PRG that is secure against computationally unbounded distinguishers.
|
I'm trying to study the thermodynamics of an N particles system in a fixed volume. In particular, I'm using Metropolis algorithm for simulating the dynamics of such system. My interest is to calculate the average of the potential among the particles, this can be done by evaluating the potential at each step and summing the potential of all steps, the error of such average can be simply evaluated using: $$ \sigma(V) = \sqrt{\frac{\langle V^2 \rangle-\langle V\rangle^2}{M}}$$ where $M$ is the number of steps. But this method allows me to calculate the average potential only at one temperature for simulation. But if we rewrite the average definition we can calculate the average of the potential at any temperature with only one simulation, using: $$\langle V\rangle_{T'} \simeq \frac{\sum_i V(x_i)e^{-(\beta'-\beta)V(x_i)}}{\sum_i e^{-(\beta'-\beta)V(x_i)}} $$ where $\beta' = 1/T'$ and $\beta = 1/T$ with T the temperature of the simulation. My problem is: What's the error of the average evaluated with the last equation? Furthermore, I was wondering if there is any difference if, in a single step of the Metropolis algorithm, I move only one particle and then check if the move is accepted or it's better if I randomly move every particle and then I check if the new state is accepted.
I am afraid that your estimate $\sigma(V)$ underestimates the true error. Implicitly, you made use of the central limit theorem. But you omitted one necessary condition for this theorem to hold: the random variables must be uncorrelated which is absolutely not the case for the measurements of $V$ at each Monte Carlo step. Apart when the system undergoes a phase transition, one expects exponential autocorrelations between the measurements $${1\over M-t}\sum_{i=1}^{M-t} V(x_i)V(x_i+t)-\langle V\rangle^2\sim e^{-t/\tau}$$where $\tau$ is the autocorrelation time, i.e. the number of Monte Carlo steps that are necessary to obtain two statistically uncorrelated configurations of the system. Taking into account these autocorrelations, the error reads $$\sqrt{{\langle V^2\rangle-\langle V\rangle^2\over M}(1+2\tau)}$$Another possibility is to first estimate $\tau$ and then measure $V$, not every Monte Carlo steps, but every $\tau$ steps. Then, the error is given by $\sigma(V)$. The original reference on the subject is H. Muller-Krumbhaar et K. Binder (1973), J. Stat. Phys.
8, 1. but you can find a discussion in any book on Monte Carlo simulations (for e.g. Monte Carlo Methods in Statistical Physics by M.E.J. Newman and G.T. Barkema, Clarendon Press).
The histogram reweighting technique and the error estimation is discussed inA.M. Ferrenberg, D.P. Landau et R.H. Swendsen (1995), Phys. Rev. E
51, 5092. A recent reference on the subject is Error estimation and reduction with cross correlations Martin Weigel and Wolfhard JankePhys. Rev. E 81, 066701 (2010)
|
In thermodynamics one assumes that the energy (or other observables) of the system only depend on a few parameters like the temperature, the particle number and also the volume:
$dE = \left(\frac{\partial E}{\partial T}\right) dT + \left(\frac{\partial E}{\partial N}\right) dN + \left(\frac{\partial E}{\partial V}\right) dV := SdT + \mu dN - pdV$.
Why is the volume enough to describe the whole geometry and details of the container? Why isn't there another term e.g. for the surface? Or some geometry dependence?
On the microscopic level the energy levels depend on the geometry (For example a spherical and a rectangular box have different energy levels.). Also like in the case of a gas trapped in a harmonic oscillator potential $V(x) = k\cdot x^2$ there is no well-defined "volume" of the gas. However if one would increase $k$ one would also increase the energy of the system. So in this case the energy depends on $k$ and not on the volume (Probably there is some way to calculate the characteristic gas "volume" out of $k$, but this would be only a rough estimate.).
Is there any argumentation (from a fundamental point of view) why the energy of a huge system only depends on the volume and one can neglect everything else?
EDIT: As I understood thermodynamics you basically start from the microscopic point of view. There you have a Hamiltonian. Then one imposes a density matrix (micro canonical, canonical, grand canonical). This defines the whole dynamics of the system. Then you take the limit $N \to \infty$ (this allows replacing the sum over all states by the integral). This can be exactly done in the case of a ideal gas in a rectangular container. There you find out that the energy of the system only depends on the temperature $T$, the particle number $N$ and the volume $V$. Calculating the total differential of the energy $E = \left(\frac{\partial E}{\partial T}\right) dT + \left(\frac{\partial E}{\partial N}\right) dN + \left(\frac{\partial E}{\partial V}\right) dV := SdT + \mu dN - pdV$ one can find expressions for $S,\mu$ and $p$ which are defined by this equation.
For all other systems where one cannot solve it exactly one imposes that after taking $N \to \infty$ the energy only depends on $T, N$ and $V$. For me it's completely clear that the energy depends on $N$ (since the Hamiltonian is linear with the particle number) and on $T$ (because it is part of the density matrix).
Apart from that it is also quite clear that there must be a term that represents the change in the potential (i.e. work). According to the first law this work is only given by the change in volume. That is the point I don't understand: why should the change of potential energy only depend on the volume for $N \to \infty$?
Two examples for that:
There are potentials (like the harmonic oscillator, but many more) where there is no fixed border and so there is no sensible definition of volume. In the case of a harmonic oscillator I can also solve it explicitly and find that the change of the potential energy depends on $\omega_x\omega_y\omega_z$ which one can relate to a "typical volume" $V$ of the harmonic oscillator. Using this volume I can the bring the energy dependence in the same form as in the first law. However for any other potential I would need to guess a "typical volume" $V$ (since I can't solve the equations). Even if I can find a equation that looks like the first law and has a term $\sim dV$ the pressure would depend on my "typical volume" (for example if I choose my "typical volume" to be twice as big then the pressure has only half of its value). This sounds pretty strange to me because if I have a gas trapped in a harmonic potential I can measure the pressure and get some well defined value.
My second idea was that maybe one only considers system where the potential is either 0 or infinite. There one has a fixed geometry. But if I don't have a rectangular geometry the energy levels differ. So there is a clear dependence on the geometry. So why in the limit $N \to \infty$ only a term depending on the volume survives? This means that if I change the geometry of my container completely but keep the volume constant nothing changes. I don't see any fundamental reason why this should be the case.
|
There is a following theorem:
$H$ is a commutative Hopf algebra over a field $k$. Then there exists a bijective correspondence between $$\{ \textrm{Hopf subalgebras }K\subset H \} \quad \leftrightarrow\quad\{\textrm{ normal Hopf ideals } I\subset H\}$$
Where Hopf subalgebra is a subalgebra $K$ satisfying $\Delta K\subset K\otimes K$, and a normal Hopf ideal is a Hopf ideal $I$ which satisfies $ad(I)\subset I\otimes A$ where $ad$ is defined by $\mathrm{ad}:A\to A\otimes A$, $a\mapsto a_2\otimes a_1Sa_3$. The correspondence is defined by $$K\mapsto HK^+,\quad I\mapsto H^I:= k\square_{H/I}H=\{h\in H|\Delta h\equiv h\otimes 1\pmod{H\otimes I}\}$$.
This is proven in M.Takeuchi, A CORRESPONDENCE BETWEEN HOPF IDEALS AND SUB-HOPF ALGEBRAS (1972).
The following is my personal impression on the proof. The easy part is to show that $K\mapsto HK^+\mapsto H^{HK^+}$ is $K$, which depends on rather complicated proof of the faithful flatness of $H$ over $K$. For the other way around, the paper uses a variant of argument Chevalley's theorem on algebraic groups, which is not truely "algebraic".
I found some papers which simplifies the proof that $H$ is faithfully flat over $K$. Some of those are: Schneider, Principal Homogeneous Spaces for Arbitrary Hopf Algebras(1990) Masuoka and Wigner, Faithful Flatness of Hopf Algebras (1992)
However, I couldn't find any simplification of the proof of $H(H^I)^+=I$.
I would like to know about more recent result related to this theorem. Is there any simplification on the part that I mentioned? Is there any generalization for, like quantum groups?
I come to this question while I was learning basic theory of algebraic groups or group schemes. Most references construct a quotient by realizing it as an orbit space of certain action on a projective plane. Then I thought there could be a way of constructing it in terms of coordinate rings. I heard that there is a chapter in SGA which deals with a construction of quotient, but I don't have access to french literatures yet... I would appreciate if someone instead explain their approach a bit, or give some references written in english.
Thank you.
|
Let
C be an $(\infty,1)$-category, incarnated as a complete Segal space, hence in particular a bisimplicial set. Is there a model structure on the slice category of bisimplicial sets over C which presents the $(\infty,1)$-presheaf category of C? Ideally, such a model structure would be Quillen equivalent to the contravariant model structure over a quasicategory incarnation of C, and to the projective model structure for simplicial presheaves on a simplicial-category incarnation of C.
Let
Yes. Let $W$ be a complete Segal space, thought of as a simplicial "space" $(W_q)$. The fibrant objects of your model category will be the fibrations $f:X\to W$ such that for each simplicial operator $\delta:[q]\to [p]$ with $\delta(q)=p$, the evident map from $X_p$ to the pullback of $$X_q \xrightarrow{f} W_q \xleftarrow{\delta} W_p$$is a weak equivalence of spaces. (
Edit: in fact, it suffices to require the evident map to the pullback to be a weak equivalence only for $\delta:[0]\to[p]$ with $\delta(0)=p$.)
I worked out some of this years ago, but never finished it; somebody should do this (or perhaps someone has already?). Lurie has done pretty much exactly the same thing in the context of quasi-categories, in HTT.
|
Surjection if Composite is Surjection Theorem Then $g$ is a surjection. Proof
Let $g \circ f$ be surjective.
Fix $z \in S_3$.
Now find an $x \in S_1: \map {g \circ f} x = z$.
The surjectivity of $g \circ f$ guarantees this can be done.
Now find an $y \in S_2: f \paren x = y$.
It follows that:
\(\displaystyle \map g y\) \(=\) \(\displaystyle \map g {\map f x}\) \(\displaystyle \) \(=\) \(\displaystyle \map {g \circ f} x\) Definition of Composition of Mappings \(\displaystyle \) \(=\) \(\displaystyle z\) Choice of $x$
$\blacksquare$
Also see Sources 1965: J.A. Green: Sets and Groups... (previous) ... (next): $\S 3.6$. Products of bijective mappings. Permutations 1965: Seth Warner: Modern Algebra... (previous) ... (next): $\S 5$: Theorem $5.3: \ 2^\circ$ 1967: George McCarty: Topology: An Introduction with Application to Topological Groups... (previous) ... (next): $\text{I}$: Exercise $\text{H}$ 1968: Ian D. Macdonald: The Theory of Groups... (previous) ... (next): Appendix: Elementary set and number theory 1975: T.S. Blyth: Set Theory and Abstract Algebra... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $14 \ \text{(b)}$ 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra... (previous) ... (next): Chapter $4$: Mappings: Exercise $13$ 1982: P.M. Cohn: Algebra Volume 1(2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $2$ 2000: James R. Munkres: Topology(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.4 \ \text{(e)}$
|
Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a^4}{b}+\frac{b^4}{c}+\frac{c^4}{a}\geq a^3+b^3+c^3+25(a-b)(b-c)(c-a)$$
I tried the uvw's technique and BW and more but without some success.
For example, BW does not help here:
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Hence, $$abc\left(\sum\limits_{cyc}\frac{a^4}{b}-\sum\limits_{cyc}a^3-25\prod\limits_{cyc}(a-b)\right)=4(u^2-uv+v^2)a^4+$$ $$+(6u^3+19u^2v-21uv^2+6v^3)a^3+(4u^4+21u^3v-19uv^3+4v^4)a^2+$$ $$+(u^5-u^4v+25u^3v^2-25u^2v^3+4uv^4+v^5)a+uv^5,$$ which is nothing.
|
Given an internal symmetry group, we gauge it by promoting the exterior derivative to its covariant version:
$$ D = d+A, $$
where $A=A^a T_a$ is a Lie algebra valued one-form known as the
connection (or gauge field) and $T$ the algebra generators.
For GR, we would like to do the same thing with the Poincaré group. But the Poincaré group isn't simple, but rather splits into translations $P$ and Lorentz transformations $J$. I would thus expect two species of connections:
$$ D = d + B^a P_a + A^{ab} J_{ab}. $$
But the covariant derivative of GR as usually found in textbooks is:
$$ \nabla_\mu = \partial_\mu +\frac{1}{2}(\omega^{\alpha \beta})_\mu J_{\alpha \beta}, $$
where $\omega$ is the spin connection. It is defined for any object that has a defined transformation under $J$, i.e., under Lorentz transformations, like spinors or tensors. But it makes no mention of the translation generator $P$. What happened? Shouldn't I have this extra gauge field?
|
OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that.
The simplest somewhat-plausible way to handle #1 is to suppose that the value to you of your material assets is roughly proportional to the
logarithm of their total value or, equivalently, that the value to you of your "next" dollar is roughly inversely proportional to your current net wealth. (There's some evidence that it actually grows a bit slower than that, but it'll do).
The way I'll handle #2 is to suppose that the probability that you pay up when I win \$$X$ in the game is roughly proportional to $\frac a{a+X}$ where $a$ is a fairly large number; so when $X$ is small you almost certainly pay and when $X$ is very large you almost certainly don't. (We should expect $a$ to be of the same order of magnitude as your net wealth in dollars; it's the size of payoff at which I think it's equally likely that you will or won't pay up.)
Then, if my net wealth before playing the game is \$$w$ and your pay-or-don't parameter is \$$a$, the "correct" price \$$p$ is such that
$$\frac12\log\frac{w-p}w+\sum_{k=0}^{\infty}\frac1{2^{k+2}}\frac{a}{a+2^k}\log\frac{w-p+2^k}w=0.$$
This isn't the sort of thing we should expect to be able to solve analytically; so far as I know there is no nice closed form for that sum. But we can do it numerically. Here are some specific results. (I cannot guarantee that I haven't messed up the calculations, though the numbers seem plausible enough to me.)
Suppose your $a$ parameter is (\$)1000000, so that your probability of paying up has reduced to 1/2 by the time you owe me a megabuck. And suppose my own net wealth happens also to be one megabuck. Then I should be willing to pay about \$4.87. Suppose my net wealth remains at \$1M but now $a$ is \$$10^9$; even once you owe me a billion dollars you might well pay up. Then the amount I'm willing to pay increases to about ... \$5.46. Suppose you remain a billionaire but I am much poorer, having only \$1000 to my name. Then the amount I'm willing to pay goes down to \$2.98. (It goes down because the poorer I am, the faster the value-to-me of a dollar decreases as I get more.) Suppose we're both poor, so I'm at \$1000 and your $a$ parameter is the same. Then I am willing to pay about \$2.39. Suppose I am a billionaire and you have infinitely much money. Then I am willing to pay about \$12.80.
So I'm fairly comfortable answering the original question as follows:
I am willing to pay somewhere between about \$2 and about \$13, depending on how wealthy we both are and how much I trust you to pay up even if doing so is painful for you.
|
Intro
Hey there reader! It has been quite a while since I wrote a blog post.. but I have had a ton of things on my mind I wanted to write about! I am stoked to be able to write about some of them now!
Since leaving my job in defense to go to graduate school — yes, I am obviously into being a poor student — I have been busy with classes and research and have been introduced to some pretty cool new ideas. One awesome algorithm I was introduced to was related to the estimation of the Range (also called Column Space) of some matrix operator using a randomized algorithm.
To be more rigorous, we can define a matrix operator as $M: \mathbb{R}^n \rightarrow \mathbb{R}^m$ for $m \leq n$, meaning it maps an $n$-dimensional vector into an $m$-dimensional vector. The goal of the algorithm that will be introduced is to essentially find an approximate basis for the range of some matrix $M$ that spans $M$’s column space. Additionally, we would like to ideally estimate this basis with minimal computation. Time to see where we can go with this!
Baseline Approach
So given some matrix $M \in \mathbb{R}^{m \times n}$, it actually is not very tough to find a basis for the range of $M$. The easiest way one could do this would be to do a QR decomposition, i.e. $M = QR$, where the columns of $Q$ would represent orthogonal vectors that span the range of $M$. The sad thing about using this approach directly is the computational complexity is $O(m^2 n)$. Can we do better? The answer: we can.
Randomized Approach
So let us assume we know the approximate rank, $k$, of the matrix $M$ where $k \leq m \leq n$. Let us then assume we can generate some random set of input samples $\Omega \in \mathbb{R}^{n \times k}$ where each column of $\Omega$ is a random vector. If we assume we can compute each element of $\Omega$ in constant complexity, a randomized algorithm can proceed in the following steps with each associated computational complexity:
\begin{align}
&\text{Construct random matrix $\Omega \in \mathbb{R}^{n \times k}$ } &\rightarrow &O(nk) \tag{1}\\ &\text{Get measurements from $A$ by doing $Y = A\Omega$ } &\rightarrow &O(mnk) \tag{2}\\ &\text{Perform QR of $Y \ni Y = QR$ } &\rightarrow &O(mk^2) \tag{3}\\ &\text{Return range estimate, $Q$} &\rightarrow &O(mk) \tag{4} \end{align}
The described algorithm has a dominating complexity of $O(mnk)$, potentially a huge speed up if $k \ll m$ and at least an improvement over the baseline approach which has an $O(m^2 n)$ complexity. The whole idea of this method is to use random inputs from $\Omega$ to extract information from the range of $M$ and then use a QR decomposition to actually use that information to estimate the basis that spans $M$’s range. If one wants to find the most dominant $k$ basis vectors more reliably, you can also throw in a power iteration styled loop into the algorithm. This can modify the algorithm into the following:
\begin{align}
&\text{Construct random matrix $\Omega \in \mathbb{R}^{n \times k}$ } &\rightarrow &O(nk) \tag{1}\\ &\text{Get measurements from $A$ by doing $Y = A\Omega$ } &\rightarrow &O(mnk) \tag{2}\\ &\text{For some small constant $N$ iterations, do: } \tag*{}\\ &\quad\text{Perform QR of $Y \ni Y = QR$ } &\rightarrow &O(mk^2)\tag{i}\\ &\quad\text{Compute $U = A^T Q$ } &\rightarrow &O(mnk)\tag{ii}\\ &\quad\text{Perform QR of $U \ni U = QR$ } &\rightarrow &O(mk^2)\tag{iii}\\ &\quad\text{Compute $Y = AQ$ } &\rightarrow &O(mnk)\tag{iv}\\ &\text{Perform QR of $Y \ni Y = QR$ } &\rightarrow &O(mk^2) \tag{3}\\ &\text{Return range estimate, $Q$} &\rightarrow &O(mk) \tag{4} \end{align}
Note that, again, the overall computational complexity ends up being $O(mnk)$ but in this case the basis found that spans the range of $M$ will be much more precise thanks to the power iteration. For those that like to have some code to look at, the following function can be used to perform the above computation.
import numpy as np def randproj(A, k, random_seed = 17, num_power_method = 5): # Author: Christian Howard # This function is designed to take some input matrix A # and approximate it by the low-rank form A = Q*(Q^T*A) = Q*B. # This form is achieved using randomized algorithms. # # Inputs: # A: Matrix to be approximated by low-rank form # k: The target rank the algorithm will strive for. # random_seed: The random seed used in code so things are repeatable. # num_power_method: Number of power method iterations # set the random seed np.random.seed(seed=random_seed) # get dimensions of A (r, c) = A.shape # get the random input and measurements from column space omega = np.random.randn(c, k) Y = np.matmul(A, omega) # form estimate for Q using power method for i in range(1, num_power_method): Q1, R1 = np.linalg.qr(Y) Q2, R2 = np.linalg.qr(np.matmul(A.T, Q1[:, :k])) Y = np.matmul(A, Q2[:, :k]) Q3, R3 = np.linalg.qr(Y) # get final k orthogonal vector estimates from column space Q = Q3[:, :k] # return the two matrices return Q So what?
Sweet, we found a better way to go about finding the range of some matrix $M$! So what? Why is this even valuable to know?
Well, one huge opportunity for this algorithm is when one wishes to find a low rank approximation for the matrix $M$. The development of the above algorithm, based on the power iteration, actually depended on an implicit assumption that we were approximating $M$ with the form $M \approx Q Q^T M$.
Since the above algorithm finds $Q$, we can in turn approximate $M$ with a low rank factorization of $M \approx Q B$ where $B = Q^T M$. That in and of itself can be very useful to shrink down datasets represented by $M$, given $k \ll m \leq n$.
Another use for this is when we want to perform Principle Component Analysis (PCA) on some matrix $M$. It turns out that the columns of $Q$ are actually the set of the $k$ dominant features vectors we can use to drop a dataset represented by $M$ to $k$ dimensions and $B$ is the dataset in the lower dimensional form. In practice, one would then use $B$ as the independent variable data that a model would be build from.
The badass thing about using the above algorithm for PCA is you get the same orthogonal features you would performing PCA via an SVD decomposition, but you avoid the computational cost of an SVD decomposition!
Just as an example, a dataset of $10,000$ images of handwritten digits from the MNIST dataset were used to test out dimensionality reduction using the above algorithm. Note that each image is made up of $784$ pixels, meaning each point in the dataset is $784$ dimensions. Now to give one an idea of how the dataset looks, the first figure below shows what a random set of digits might look like.
The second figure shows the $36$ features extracted using the randomized range finder algorithm, effectively making the dataset reducible to having $36$ dimensional data points instead of the original $784$ dimensional points!
The above example using this Randomized Range Finder is pretty cool and it certainly can be used in plenty more applications! For example, I personally have used the above algorithm as a stepping stone to building an efficient randomized SVD code.
Additionally, one can modify the above algorithm to be adaptive, meaning we do not need to specify some rank $k$, and can instead allow the algorithm to find the optimal value for $k$ given some tolerance. This can be very useful in the above examples so we can trade off accuracy with run-time/data compression.
Conclusion
As we saw in this post, there are methods we can use to estimate the range of some matrix. Using some pretty simple randomized approaches, we can make this estimation much more efficient at the expense of some minor approximation error. As shown, the resulting Randomized Range Finder can find use in many things, ranging from data compression to feature extraction and more. Basically, this algorithm is pretty cool!
|
Hi, Can someone provide me some self reading material for Condensed matter theory? I've done QFT previously for which I could happily read Peskin supplemented with David Tong. Can you please suggest some references along those lines? Thanks
@skullpatrol The second one was in my MSc and covered considerably less than my first and (I felt) didn't do it in any particularly great way, so distinctly average. The third was pretty decent - I liked the way he did things and was essentially a more mathematically detailed version of the first :)
2. A weird particle or state that is made of a superposition of a torus region with clockwise momentum and anticlockwise momentum, resulting in one that has no momentum along the major circumference of the torus but still nonzero momentum in directions that are not pointing along the torus
Same thought as you, however I think the major challenge of such simulator is the computational cost. GR calculations with its highly nonlinear nature, might be more costy than a computation of a protein.
However I can see some ways approaching it. Recall how Slereah was building some kind of spaceitme database, that could be the first step. Next, one might be looking for machine learning techniques to help on the simulation by using the classifications of spacetimes as machines are known to perform very well on sign problems as a recent paper has shown
Since GR equations are ultimately a system of 10 nonlinear PDEs, it might be possible the solution strategy has some relation with the class of spacetime that is under consideration, thus that might help heavily reduce the parameters need to consider to simulate them
I just mean this: The EFE is a tensor equation relating a set of symmetric 4 × 4 tensors. Each tensor has 10 independent components. The four Bianchi identities reduce the number of independent equations from 10 to 6, leaving the metric with four gauge fixing degrees of freedom, which correspond to the freedom to choose a coordinate system.
@ooolb Even if that is really possible (I always can talk about things in a non joking perspective), the issue is that 1) Unlike other people, I cannot incubate my dreams for a certain topic due to Mechanism 1 (consicous desires have reduced probability of appearing in dreams), and 2) For 6 years, my dream still yet to show any sign of revisiting the exact same idea, and there are no known instance of either sequel dreams nor recurrence dreams
@0celo7 I felt this aspect can be helped by machine learning. You can train a neural network with some PDEs of a known class with some known constraints, and let it figure out the best solution for some new PDE after say training it on 1000 different PDEs
Actually that makes me wonder, are the space of all coordinate choices more than all possible moves of Go?
enumaris: From what I understood from the dream, the warp drive showed here may be some variation of the alcuberrie metric with a global topology that has 4 holes in it whereas the original alcuberrie drive, if I recall, don't have holes
orbit stabilizer: h bar is my home chat, because this is the first SE chat I joined. Maths chat is the 2nd one I joined, followed by periodic table, biosphere, factory floor and many others
Btw, since gravity is nonlinear, do we expect if we have a region where spacetime is frame dragged in the clockwise direction being superimposed on a spacetime that is frame dragged in the anticlockwise direction will result in a spacetime with no frame drag? (one possible physical scenario that I can envision such can occur may be when two massive rotating objects with opposite angular velocity are on the course of merging)
Well. I'm a begginer in the study of General Relativity ok? My knowledge about the subject is based on books like Schutz, Hartle,Carroll and introductory papers. About quantum mechanics I have a poor knowledge yet.
So, what I meant about "Gravitational Double slit experiment" is: There's and gravitational analogue of the Double slit experiment, for gravitational waves?
@JackClerk the double slits experiment is just interference of two coherent sources, where we get the two sources from a single light beam using the two slits. But gravitational waves interact so weakly with matter that it's hard to see how we could screen a gravitational wave to get two coherent GW sources.
But if we could figure out a way to do it then yes GWs would interfere just like light wave.
Thank you @Secret and @JohnRennie . But for conclude the discussion, I want to put a "silly picture" here: Imagine a huge double slit plate in space close to a strong source of gravitational waves. Then like water waves, and light, we will see the pattern?
So, if the source (like a Black Hole binary) are sufficent away, then in the regions of destructive interference, space-time would have a flat geometry and then with we put a spherical object in this region the metric will become schwarzschild-like.
if**
Pardon, I just spend some naive-phylosophy time here with these discussions**
The situation was even more dire for Calculus and I managed!
This is a neat strategy I have found-revision becomes more bearable when I have The h Bar open on the side.
In all honesty, I actually prefer exam season! At all other times-as I have observed in this semester, at least-there is nothing exciting to do. This system of tortuous panic, followed by a reward is obviously very satisfying.
My opinion is that I need you kaumudi to decrease the probabilty of h bar having software system infrastructure conversations, which confuse me like hell and is why I take refugee in the maths chat a few weeks ago
(Not that I have questions to ask or anything; like I said, it is a little relieving to be with friends while I am panicked. I think it is possible to gauge how much of a social recluse I am from this, because I spend some of my free time hanging out with you lot, even though I am literally inside a hostel teeming with hundreds of my peers)
that's true. though back in high school ,regardless of code, our teacher taught us to always indent your code to allow easy reading and troubleshooting. We are also taught the 4 spacebar indentation convention
@JohnRennie I wish I can just tab because I am also lazy, but sometimes tab insert 4 spaces while other times it inserts 5-6 spaces, thus screwing up a block of if then conditions in my code, which is why I had no choice
I currently automate almost everything from job submission to data extraction, and later on, with the help of the machine learning group in my uni, we might be able to automate a GUI library search thingy
I can do all tasks related to my work without leaving the text editor (of course, such text editor is emacs). The only inconvenience is that some websites don't render in a optimal way (but most of the work-related ones do)
Hi to all. Does anyone know where I could write matlab code online(for free)? Apparently another one of my institutions great inspirations is to have a matlab-oriented computational physics course without having matlab on the universities pcs. Thanks.
@Kaumudi.H Hacky way: 1st thing is that $\psi\left(x, y, z, t\right) = \psi\left(x, y, t\right)$, so no propagation in $z$-direction. Now, in '$1$ unit' of time, it travels $\frac{\sqrt{3}}{2}$ units in the $y$-direction and $\frac{1}{2}$ units in the $x$-direction. Use this to form a triangle and you'll get the answer with simple trig :)
@Kaumudi.H Ah, it was okayish. It was mostly memory based. Each small question was of 10-15 marks. No idea what they expect me to write for questions like "Describe acoustic and optic phonons" for 15 marks!! I only wrote two small paragraphs...meh. I don't like this subject much :P (physical electronics). Hope to do better in the upcoming tests so that there isn't a huge effect on the gpa.
@Blue Ok, thanks. I found a way by connecting to the servers of the university( the program isn't installed on the pcs on the computer room, but if I connect to the server of the university- which means running remotely another environment, i found an older version of matlab). But thanks again.
@user685252 No; I am saying that it has no bearing on how good you actually are at the subject - it has no bearing on how good you are at applying knowledge; it doesn't test problem solving skills; it doesn't take into account that, if I'm sitting in the office having forgotten the difference between different types of matrix decomposition or something, I can just search the internet (or a textbook), so it doesn't say how good someone is at research in that subject;
it doesn't test how good you are at deriving anything - someone can write down a definition without any understanding, while someone who can derive it, but has forgotten it probably won't have time in an exam situation. In short, testing memory is not the same as testing understanding
If you really want to test someone's understanding, give them a few problems in that area that they've never seen before and give them a reasonable amount of time to do it, with access to textbooks etc.
|
First off, great question.
Ok I have crunched the numbers. I made a 5 degree of freedom model of the rolling disk/wheel which includes, 2 variables $x$, $y$ for the location on the incline plane (with positive $x$ downwards and $y$ across the plane), $\psi$ for the orientation of the wheel ($\psi=0$ when wheel is facing downwards), $\phi$ for the tilt of the wheel from the incline, and finally $\theta$ the spin of the wheel (with $\dot{\theta}=\omega$).
Included is the incline plane angle $\gamma$ and gravity $g$, and the rolling radius of the wheel $r$. The no-slip condition is described as $\omega r = \dot{x} \cos \psi + \dot{y} \sin \psi $ (the velocity along the $\psi$ direction).
The equations simplify when I use $\dot{x} = \omega r \cos\psi + v_{slip} \sin \psi$ and $\dot{y} = \omega r \sin \psi - v_{slip} \cos \psi $, to get the motion of the orientation and tilt as
$$ \begin{aligned} \ddot{\psi} & = \frac{2 \dot{\phi} ( \dot{\psi} \sin \phi - \omega)}{\cos \phi} + \frac{\sin\phi}{\cos^2 \phi} \left( \frac{2 \dot{\psi} v_{slip}}{r} - \frac{2 g \cos \psi \sin \gamma}{r} \right) \\\ddot{\phi} & = \frac{ \cos\phi \left( \left( \frac{1}{2} I_{disk} \dot{\psi}^2 - m r^2 \dot{\phi}^2 \right) \sin\phi + I_{disk} \omega \dot{\phi} \right)}{\frac{1}{2} I_{disk} + m r^2 \sin^2 \phi} + \frac{g\, m\, r \sin\phi \cos \gamma}{\frac{1}{2} I_{disk} + m r^2 \sin^2 \phi}\end{aligned}$$
where $I_{disk}$ is the mass moment of inertia about the spin axis, and $\frac{1}{2} I_{disk}$ the mass moment of inertia about any other axis of the wheel.
So is the above stable? There are two scenarios that I can come up with. Stable when $\dot{\psi}=0$ and $\dot{\phi}=0$. This can only be true if $\psi=0$ or the orientation is always "downhill", but with a small disturbance it will throw it off. So not stable on this condition. Stable when $\dot{\psi}$ and $\dot{\phi}$ oscillate about zero with appropriate corrective twists $\ddot{\psi}$ and $\ddot{\phi}$ developing. I can only examine this case when $\phi \approx 0$ and $\phi \approx 0$. This has the solution of $$ \dot{\psi} =B \sin(2 \omega t) -A \cos(2 \omega t) \\ \dot{\phi} = B \cos(2 \omega t)+A \sin(2 \omega t) $$
Any coefficients of $A$, $B$ work when $\phi = 0$ and $\phi = 0$, but beyond that no constants would work. Nevertheless it shows that under some conditions the orientation and tilt couple in an
stable oscillation, also known as a wobble to anyone who was rolled a coin on a table.
I think this is a far as anyone can take it without employing numerical methods, and I think this would make for a great thesis for someone.
PS. I used a
MATLAB script with the symbolic toolbox to find this solution, with some custom code I have developed for 3d kinematics and the equations of motion for rigid bodies.
|
We now come to the first of three important theorems that extend theFundamental Theorem of Calculus to higher dimensions. (The FundamentalTheorem of Line Integrals has already done this in one way, but inthat case we were still dealing with an essentially one-dimensionalintegral.) They all share with the Fundamental Theorem the followingrather vague description:
To compute a certain sort of integral over aregion, we may do a computation on the boundary of the regionthat involves one fewer integrations.
Note that this does indeed describe the Fundamental Theorem of Calculus and the Fundamental Theorem of Line Integrals: to compute a single integral over an interval, we do a computation on the boundary (the endpoints) that involves one fewer integrations, namely, no integrations at all.
Theorem 18.4.1 (Green's Theorem) If the vector field ${\bf F}=\langle P,Q\rangle$ and the region $D$ are sufficiently nice, and if $C$ is the boundary of $D$ ($C$ is a closed curve), then $$\dint{D} {\partial Q\over\partial x} -{\partial P\over\partial y} \,dA = \int_C P\,dx +Q\,dy ,$$ provided the integration on the right is done counter-clockwise around $C$.
To indicate that an integral $\ds\int_C$ is being done over a closedcurve in the counter-clockwise direction, we usually write$\ds\oint_C$. We also use the notation $\partial D$ to mean theboundary of $D$
oriented in thecounterclockwise direction. With this notation,$\ds\oint_C=\int_{\partial D}$.
We already know one case, not particularly interesting, in which this theorem is true: If $\bf F$ is conservative, we know that the integral $\ds\oint_C {\bf F}\cdot d{\bf r}=0$, because any integral of a conservative vector field around a closed curve is zero. We also know in this case that $\partial P/\partial y=\partial Q/\partial x$, so the double integral in the theorem is simply the integral of the zero function, namely, 0. So in the case that $\bf F$ is conservative, the theorem says simply that $0=0$.
Example 18.4.2 We illustrate the theorem by computing both sides of $$\int_{\partial D} x^4\,dx + xy\,dy=\dint{D} y-0\,dA,$$ where $D$ is the triangular region with corners $(0,0)$, $(1,0)$, $(0,1)$.
Starting with the double integral: $$\dint{D} y-0\,dA=\int_0^1\int_0^{1-x} y\,dy\,dx= \int_0^1 {(1-x)^2\over2}\,dx=\left.-{(1-x)^3\over6}\right|_0^1={1\over6}.$$
There is no single formula to describe the boundary of $D$, so to compute the left side directly we need to compute three separate integrals corresponding to the three sides of the triangle, and each of these integrals we break into two integrals, the "$dx$'' part and the "$dy$'' part. The three sides are described by $y=0$, $y=1-x$, and $x=0$. The integrals are then $$\eqalign{ \int_{\partial D}\!\!\! x^4\,dx + xy\,dy&= \int_0^1 x^4\,dx+\int_0^0 0\,dy+\int_1^0 x^4\,dx+\int_0^1 (1-y)y\,dy+ \int_0^0 0\,dx+\int_1^0 0\,dy\cr &={1\over5}+0-{1\over5}+{1\over6}+0+0={1\over6}.\cr} $$
Alternately, we could describe the three sides in vector form as $\langle t,0\rangle$, $\langle 1-t,t\rangle$, and $\langle 0,1-t\rangle$. Note that in each case, as $t$ ranges from 0 to 1, we follow the corresponding side in the correct direction. Now $$\eqalign{ \int_{\partial D} x^4\,dx + xy\,dy&= \int_0^1 t^4 + t\cdot 0\,dt + \int_0^1 -(1-t)^4 + (1-t)t\,dt +\int_0^1 0 + 0\,dt\cr &=\int_0^1 t^4\,dt + \int_0^1 -(1-t)^4 + (1-t)t\,dt ={1\over6}.\cr }$$
In this case, none of the integrations are difficult, but the second approach is somewhat tedious because of the necessity to set up three different integrals. In different circumstances, either of the integrals, the single or the double, might be easier to compute. Sometimes it is worthwhile to turn a single integral into the corresponding double integral, sometimes exactly the opposite approach is best.
Here is a clever use of Green's Theorem: We know that areas can be computed using double integrals, namely, $$\dint{D} 1\,dA$$ computes the area of region $D$. If we can find $P$ and $Q$ so that $\partial Q/\partial x-\partial P/\partial y=1$, then the area is also $$\int_{\partial D} P\,dx+Q\,dy.$$ It is quite easy to do this: $P=0,Q=x$ works, as do $P=-y, Q=0$ and $P=-y/2,Q=x/2$.
Example 18.4.3 An ellipse centered at the origin, with its two principal axes aligned with the $x$ and $y$ axes, is given by $${x^2\over a^2}+{y^2\over b^2}=1.$$ We find the area of the interior of the ellipse via Green's theorem. To do this we need a vector equation for the boundary; one such equation is $\langle a\cos t,b\sin t\rangle$, as $t$ ranges from 0 to $2\pi$. We can easily verify this by substitution: $${x^2\over a^2}+{y^2\over b^2}= {a^2\cos^2 t\over a^2}+{b^2\sin^2t\over b^2}= \cos^2t+\sin^2t=1.$$ Let's consider the three possibilities for $P$ and $Q$ above: Using 0 and $x$ gives $$\oint_C 0\,dx+x\,dy=\int_0^{2\pi} a\cos(t)b\cos(t)\,dt= \int_0^{2\pi} ab\cos^2(t)\,dt.$$ Using $-y$ and 0 gives $$\oint_C -y\,dx+0\,dy=\int_0^{2\pi} -b\sin(t)(-a\sin(t))\,dt= \int_0^{2\pi} ab\sin^2(t)\,dt.$$ Finally, using $-y/2$ and $x/2$ gives $$\eqalign{ \oint_C -{y\over2}\,dx+{x\over2}\,dy&= \int_0^{2\pi} -{b\sin(t)\over2}(-a\sin(t))\,dt +{a\cos(t)\over2}(b\cos(t))\,dt\cr &=\int_0^{2\pi} {ab\sin^2t\over2}+{ab\cos^2t\over2}\,dt= \int_0^{2\pi} {ab\over2}\,dt=\pi ab.\cr}$$ The first two integrals are not particularly difficult, but the third is very easy, though the choice of $P$ and $Q$ seems more complicated.
Proof of Green's Theorem. We cannot here prove Green's Theorem in general, but we can do a special case. We seek to prove that $$\oint_C P\,dx +Q\,dy = \dint{D} {\partial Q\over\partial x} -{\partial P\over\partial y} \,dA.$$ It is sufficient to show that $$\oint_C P\,dx=\dint{D}-{\partial P\over\partial y} \,dA\qquad\hbox{and} \qquad\oint_C Q\,dy=\dint{D} {\partial Q\over\partial x}\,dA,$$ which we can do if we can compute the double integral in both possible ways, that is, using $dA=dy\,dx$ and $dA=dx\,dy$.
For the first equation, we start with $$\dint{D}{\partial P\over\partial y}\,dA= \int_a^b\int_{g_1(x)}^{g_2(x)} {\partial P\over \partial y}\,dy\,dx= \int_a^b P(x,g_2(x))-P(x,g_1(x))\,dx.$$ Here we have simply used the ordinary Fundamental Theorem of Calculus, since for the inner integral we are integrating a derivative with respect to $y$: an antiderivative of $\partial P/\partial y$ with respect to $y$ is simply $P(x,y)$, and then we substitute $g_1$ and $g_2$ for $y$ and subtract.
Now we need to manipulate $\oint_C P\,dx$. The boundary of region $D$ consists of 4 parts, given by the equations $y=g_1(x)$, $x=b$, $y=g_2(x)$, and $x=a$. On the portions $x=b$ and $x=a$, $dx=0\,dt$, so the corresponding integrals are zero. For the other two portions, we use the parametric forms $x=t$, $y=g_1(t)$, $a\le t\le b$, and $x=t$, $y=g_2(t)$, letting $t$ range from $b$ to $a$, since we are integrating counter-clockwise around the boundary. The resulting integrals give us $$\eqalign{ \oint_C P\,dx = \int_a^b P(t,g_1(t))\,dt+\int_b^a P(t,g_2(t))\,dt &=\int_a^b P(t,g_1(t))\,dt-\int_a^b P(t,g_2(t))\,dt\cr &=\int_a^b P(t,g_1(t))-P(t,g_2(t))\,dt\cr }$$ which is the result of the double integral times $-1$, as desired.
The equation involving $Q$ is essentially the same, and left as an exercise.
Exercises 18.4
Ex 18.4.1Compute $\ds\int_{\partial D} 2y\,dx + 3x\,dy$, where $D$ is described by $0\le x\le1$, $0\le y\le 1$.(answer)
Ex 18.4.2Compute $\ds\int_{\partial D} xy\,dx + xy\,dy$, where $D$ is described by $0\le x\le1$, $0\le y\le 1$.(answer)
Ex 18.4.3Compute $\ds\int_{\partial D} e^{2x+3y}\,dx + e^{xy}\,dy$, where $D$ is described by $-2\le x\le 2$, $-1\le y\le 1$.(answer)
Ex 18.4.4Compute $\ds\int_{\partial D} y\cos x\,dx + y\sin x\,dy$, where $D$ is described by $0\le x\le \pi/2$, $1\le y\le 2$.(answer)
Ex 18.4.5Compute $\ds\int_{\partial D} x^2y\,dx + xy^2\,dy$, where $D$ is described by $0\le x\le 1$, $0\le y\le x$.(answer)
Ex 18.4.6Compute $\ds\int_{\partial D} x\sqrt{y}\,dx + \sqrt{x+y}\,dy$, where $D$ is described by $1\le x\le 2$, $2x\le y\le 4$.(answer)
Ex 18.4.7Compute $\ds\int_{\partial D} (x/y)\,dx + (2+3x)\,dy$, where $D$ is described by $1\le x\le 2$, $1\le y\le x^2$.(answer)
Ex 18.4.8Compute $\ds\int_{\partial D} \sin y\,dx + \sin x\,dy$, where $D$ is described by $0\le x\le \pi/2$, $x\le y\le \pi/2$.(answer)
Ex 18.4.9Compute $\ds\int_{\partial D} x\ln y\,dx$,where $D$ is described by $1\le x\le 2$, $\ds e^x\le y\le e^{x^2}$.(answer)
Ex 18.4.10Compute $\ds\int_{\partial D} \sqrt{1+x^2}\,dy$, where $D$ is described by $-1\le x\le 1$, $x^2\le y\le 1$.(answer)
Ex 18.4.11Compute $\ds\int_{\partial D} x^2y\,dx - xy^2\,dy$, where $D$ is described by $x^2+y^2\le 1$.(answer)
Ex 18.4.12Compute $\ds\int_{\partial D} y^3\,dx + 2x^3\,dy$, where $D$ is described by $x^2+y^2\le 4$.(answer)
Ex 18.4.13Evaluate $\ds\oint_C (y-\sin(x))\,dx + \cos(x) \, dy$,where $C$ is the boundary of the triangle with vertices $(0,0)$,$(1,0)$, and $(1,2)$ oriented counter-clockwise.(answer)
Ex 18.4.14Finish our proof of Green's Theorem by showing that$\ds\oint_C Q\,dy=\dint{D} {\partial Q\over\partial x}\,dA$.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.