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Coset/Examples/Subgroup of Infinite Cyclic Group Examples of Cosets
Let $G = \gen a$ be an infinite cyclic group.
Let $s \in \Z_{>0}$ be a (strictly) positive integer.
Let $H$ be the subgroup of $G$ defined as:
$H := \gen {a^s}$
Then a complete repetition-free list of the cosets of $H$ in $G$ is:
$S = \set {H, aH, a^2 H, \ldots, a^{s - 1} H}$ Proof
First it is demonstrated that $S$ is complete.
Let $x \in G$ be arbitrary.
Then:
$\exists n \in \Z: x = a^n$
By the Division Theorem:
$n = q s + r$
for $q, r \in \Z$ such that $0 \le r \le s - 1$.
Thus:
\(\displaystyle x\) \(=\) \(\displaystyle a^{q s + r}\) \(\displaystyle \) \(=\) \(\displaystyle a^r \paren {a^s}^q\) \(\displaystyle \) \(=\) \(\displaystyle a^r h\) where $h \in H$
Thus:
$x \in a^r H$
and as $0 \le r \le s - 1$ it follows that $a^r H \in S$.
Thus every $x \in G$ belongs to at least one of the cosets on $S$.
Aiming for a contradiction, suppose: $a^i H = a^j H$
where $0 \le i < j \le s - 1$.
Then from Sundry Coset Results:
$\paren {a^i}^{-1} a^j \in H$
that is:
$a^{j - i} \in H$
and so:
$a^{j - i} = a^{k s}$
for some $k \in \Z$.
But $a$ is of infinite order.
So it follows from Subgroup Generated by Infinite Order Element is Infinite that:
$j - i = k s$
But this contradicts our statement that $0 < j - i < s$.
Hence $S$ must be repetition-free.
$\blacksquare$ |
Regularity properties of a cubically convergent scheme for generalized equations
1.
Laboratoire Analyse Optimisation Contrôle, Dept. de Mathématiques, Université des Antilles et de la Guyane, B.P. 250, 97157 Pointe à Pitre, Guadeloupe, France, France
$ v \in f(x_n)+ \nabla f(x_n)(x_{n+1}-x_n) +\frac{1}{2}\nabla^2 f(x_n) (x_{n+1}-x_n)^2 +G(x_{n+1}).$ $\quad$ (*)
We investigate some stability properties of the method (*) and we study the behavior of the sequences that it generates, more precisely, we show that they inherit some regularity properties from the mapping $f+G$.
Mathematics Subject Classification:Primary: 49J53, 49J40, 90C4. Citation:Michel H. Geoffroy, Alain Piétrus. Regularity properties of a cubically convergent scheme for generalized equations. Communications on Pure & Applied Analysis, 2007, 6 (4) : 983-996. doi: 10.3934/cpaa.2007.6.983
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D meson elliptic flow in non-central Pb-Pb collisions at $\sqrt{s_{NN}}$ = 2.76TeV TypeJournal article Peer reviewed publishedVersion Date2013-09 Author Share
MetadataShow full item record Abstract
Azimuthally anisotropic distributions of $D^0, D^+$ and $D^{*+}$ mesons were studied in the central rapidity region (|y| < 0.8) in Pb-Pb collisions at a centre-of-mass energy $\sqrt{s_{NN}}$ = 2.76 TeV per nucleon-nucleon collision, with the ALICE detector at the LHC. The second Fourier coefficient $\nu_2$ (commonly denoted elliptic flow) was measured in the centrality class 30-50% as a function of the D meson transverse momentum $p_T$, in the range 2-16 GeV/c. The measured $\nu_2$ of D mesons is comparable in magnitude to that of light-flavour hadrons. It is positive in the range 2 < $p_T$ < 6 GeV/c with 5.7 $\sigma$ significance, based on the combination of statistical and systematic uncertainties.
CitationAlme J, Erdal HA, Helstrup H, Hetland KF, Kileng B, Altinpinar S, Djuvsland Ø, Fehlker D, Haaland ØS, Huang M, Kanaki K, Langøy R, Larsen DT, Lien JA, Lønne P, Nystrand J, Rehman Au, Røed K, Røhrich D, Skjerdal K, Ullaland K, Wagner B, Yang S, Øvrebekk G, Bätzing P, Dordic O, Eyyubova G, Kværnø H, Lindal S, Løvhøiden G, Milosevic J, Nilsson MS, Pocheptsov TA, Qvigstad H, Richter M, Skaali TB, Tveter TS, Wikne JC, Abelev B, Adam J, Adamová D, Adare AM, Aggarwal MM, Aglieri Rinella G, Agocs AG, Agostinelli A, Aguilar Salazar S, Ahammed Z, Ahmad N, Ahmad Masoodi A, ALICE C. D meson elliptic flow in noncentral Pb-Pb collisions at root(S)(NN)=2.76 TeV. Physical Review Letters. 2013;111(10):102301 PublisherAmerican Physical Society Collections
Copyright 2013 CERN, for the ALICE Collaboration |
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J/Ψ production and nuclear effects in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-02)
Inclusive J/ψ production has been studied with the ALICE detector in p-Pb collisions at the nucleon–nucleon center of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement is performed in the center of mass rapidity ...
Suppression of ψ(2S) production in p-Pb collisions at √sNN=5.02 TeV
(Springer, 2014-12)
The ALICE Collaboration has studied the inclusive production of the charmonium state ψ(2S) in proton-lead (p-Pb) collisions at the nucleon-nucleon centre of mass energy √sNN = 5.02TeV at the CERN LHC. The measurement was ...
Event-by-event mean pT fluctuations in pp and Pb–Pb collisions at the LHC
(Springer, 2014-10)
Event-by-event fluctuations of the mean transverse momentum of charged particles produced in pp collisions at s√ = 0.9, 2.76 and 7 TeV, and Pb–Pb collisions at √sNN = 2.76 TeV are studied as a function of the ...
Centrality, rapidity and transverse momentum dependence of the J/$\psi$ suppression in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76 TeV
(Elsevier, 2014-06)
The inclusive J/$\psi$ nuclear modification factor ($R_{AA}$) in Pb-Pb collisions at $\sqrt{s_{NN}}$=2.76TeV has been measured by ALICE as a function of centrality in the $e^+e^-$ decay channel at mid-rapidity |y| < 0.8 ...
Multiplicity Dependence of Pion, Kaon, Proton and Lambda Production in p-Pb Collisions at $\sqrt{s_{NN}}$ = 5.02 TeV
(Elsevier, 2014-01)
In this Letter, comprehensive results on $\pi^{\pm}, K^{\pm}, K^0_S$, $p(\bar{p})$ and $\Lambda (\bar{\Lambda})$ production at mid-rapidity (0 < $y_{CMS}$ < 0.5) in p-Pb collisions at $\sqrt{s_{NN}}$ = 5.02 TeV, measured ...
Multi-strange baryon production at mid-rapidity in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2014-01)
The production of ${\rm\Xi}^-$ and ${\rm\Omega}^-$ baryons and their anti-particles in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV has been measured using the ALICE detector. The transverse momentum spectra at ...
Measurement of charged jet suppression in Pb-Pb collisions at √sNN = 2.76 TeV
(Springer, 2014-03)
A measurement of the transverse momentum spectra of jets in Pb–Pb collisions at √sNN = 2.76TeV is reported. Jets are reconstructed from charged particles using the anti-kT jet algorithm with jet resolution parameters R ...
Two- and three-pion quantum statistics correlations in Pb-Pb collisions at √sNN = 2.76 TeV at the CERN Large Hadron Collider
(American Physical Society, 2014-02-26)
Correlations induced by quantum statistics are sensitive to the spatiotemporal extent as well as dynamics of particle-emitting sources in heavy-ion collisions. In addition, such correlations can be used to search for the ...
Exclusive J /ψ photoproduction off protons in ultraperipheral p-Pb collisions at √sNN = 5.02TeV
(American Physical Society, 2014-12-05)
We present the first measurement at the LHC of exclusive J/ψ photoproduction off protons, in ultraperipheral proton-lead collisions at √sNN=5.02 TeV. Events are selected with a dimuon pair produced either in the rapidity ...
Measurement of quarkonium production at forward rapidity in pp collisions at √s=7 TeV
(Springer, 2014-08)
The inclusive production cross sections at forward rapidity of J/ψ , ψ(2S) , Υ (1S) and Υ (2S) are measured in pp collisions at s√=7 TeV with the ALICE detector at the LHC. The analysis is based on a data sample corresponding ... |
Alladi Ramakrishnan Hall
Tensor product decomposition for the general linear supergroup GL(m|n)
Thorsten Heidersdorf
Max Planck Institute, Bonn.
Let $\mathfrak{gl}(n)$ denote the Lie algebra of the general linear group $GL(n). Given two finite dimensional irreducible representations $L(\lambda), L(\mu)$ of $\mathfrak{gl}(n)$, its tensor product decomposition $L(\lambda) \otimes L(\mu)$ is given by the Littlewood-Richardson rule. The situation becomes much more complicated when one replaces $\mathfrak{gl}(n)$ by the general linear Lie superalgebra $\mathfrak{gl}(m|n)$. The analogous decomposition $L(\lambda) \otimes L(\mu)$ is largely unknown. Indeed many aspects of the representation theory of $\mathfrak{gl}(m|n)$ are more akin to the study of Lie algebras and their representations in prime characteristic or to the BGG category $\mathcal{O}$. I will give a survey talk about this problem and explain why some approaches don't work and what can be done about it. This will give me the chance to speak about a) the character formula for an irreducible representation $L(\lambda)$, b) Deligne's interpolating category $Rep(GL_t)$ for $t \in \mathbb{C}$ and c) the process of semisimplification of a tensor category. Done |
3 0 1. Homework Statement
Find the x-component of the ##F_b## (parallell with ##\vec{s}## and the ground) and find the resultant total Force.
Then calculate the total work done by both over a distance of 15 m.
2. Homework Equations
[itex] F_{x} = |F| \cdot \cos v [/itex]
[itex] F_{y} = |F| \cdot \sin v [/itex]
[itex] W = \vec{F} \cdot \vec{s} [/itex]
[itex] W = |F| \cdot |s| \cdot \cos{v} [/itex]
3. The Attempt at a Solution
I am a science teacher myself, and a student recently asked me an interesting question I hadn't thought of before.
The math here isn't all that hard, and the x and y components of ##F_b## are
##F_{bx} = 500 \cdot \cos{30 } = 500N \cdot \frac{\sqrt{3}}{2} = 433.01N## (Thanks, voko)
##F_{by} = 500 \cdot \sin{30 } = 500N \cdot \frac{1}{2} = 250 N ##
Now, the question which confuses me is how and why these 2 forces doesn't match up to be equal to 500. From a purely mathematical point of view, I know that they are added via pythagoras and that the math is correct, so that isn't my issue.
Since this is actual forces we are talking about,
how is the sum of the forces he is transferring to the object NOT equal to the force he is dragging on the rope with?
Again, let me stress that I know the math of how to use pythagoras to add them, but from an intuitive point of view, my student was unable to understand this, and I was unable to explain it intuively.
Last edited: |
The total pressure between two points, (\(a\) and \(b\)) can be calculated with integration as
\[ \Delta P_{ab} = \int_a^b \dfrac{dP}{dx} dx \label{phase:eq:TdeltaP} \tag{39} \] and therefore \[ \Delta P_{ab} = \overbrace
Callstack: at (Under_Construction/Purgatory/Book:_Fluid_Mechanics_(Bar-Meir)__-_old_copy/13:_Multi–Phase_Flow/13.7:_Homogeneous_Models/13.7.1:_Pressure_Loss_Components/13.7.1.4:_Total_Pressure_Loss), /content/body/p/span[1], line 1, column 1
^{friction} +
\overbrace
Callstack: at (Under_Construction/Purgatory/Book:_Fluid_Mechanics_(Bar-Meir)__-_old_copy/13:_Multi–Phase_Flow/13.7:_Homogeneous_Models/13.7.1:_Pressure_Loss_Components/13.7.1.4:_Total_Pressure_Loss), /content/body/p/span[2], line 1, column 1
\overbrace
Callstack: at (Under_Construction/Purgatory/Book:_Fluid_Mechanics_(Bar-Meir)__-_old_copy/13:_Multi–Phase_Flow/13.7:_Homogeneous_Models/13.7.1:_Pressure_Loss_Components/13.7.1.4:_Total_Pressure_Loss), /content/body/p/span[3], line 1, column 1
\label{phase:eq:TdeltaPpart} \tag{40}
\]
Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
Suppose I wanted to travel to one of the recently discovered potentially Earth-like planets such as Kepler 186f that is 490 light years away. Assuming I had a powerful rocket and enough fuel, how long would it take me?
Start by considering what is seen by the people watching you from the Earth. Nothing can travel faster than the speed of light, $c$, so the quickest you could get to Kepler 186f would be if you were travelling at $c$ in which case it would take 490 years. In practice it would take longer than this because you have to accelerate from rest when you leave the Earth and decelerate to a halt again when you get to your destination.
So far this isn’t very interesting. What makes the problem interesting is that clocks on fast moving objects run slow due to time dilation. If you could travel near to the speed of light the time that passes for you will be less than 490 years, and in fact can be a lot less, as we’ll see below.
First let’s take the simple case where you travel at some constant velocity $v$, and we won’t worry about how you accelerated to $v$ or how you’re going to slow down again. We’ll call the distance to the star $d$. For the people watching from Earth the time taken is just the distance you travel divided by your velocity:
$$ t = \frac{d}{v} $$
So if the distance is 490 light years and you’re travelling at the speed of light the time taken is just 490 years. But how much time would you measure on your wristwatch? To do the calculation properly you need to use the Lorentz transformations, but in fact the answer turns out to be very simple. The time you measure, $\tau$, is given by:
$$ \tau = \frac{t}{\gamma} $$
where $t$ is the time measured on Earth and $\gamma$ is the Lorentz factor and is given by:
$$ \gamma = \frac{1}{\sqrt{1 - \tfrac{v^2}{c^2}}} $$
Or if you want the whole expression written out in full, the time you measure is:
$$ \tau = \frac{d}{v} \sqrt{1 - \frac{v^2}{c^2}} $$
To give you a feel for this I’ve done the calculation for the 490 light year trip to Kepler 186f and I’ve drawn a graph of the time you measure as a function of your speed:
The blue line is the travel time as measured on Earth, so it goes to 490 years as $v \rightarrow c$. The red line is the time measured on your wristwatch, which goes to zero as $v \rightarrow c$.
But this isn’t very realistic since it ignores acceleration and deceleration. Suppose instead you travel halfway to the star at constant acceleration, then you flip over and travel halfway at constant deceleration. This allows you to start from rest and end at rest, and you also get a nice artificial gravity during the trip. But how can you calculate the time dilation for a trip that involves acceleration?
The details of the calculation are given in Chapter 6 of Gravitation by Misner, Thorne and Wheeler. I won’t reproduce the calculation here because it’s surprisingly boring. You solve a couple of simultaneous equations to get differential equations for the time, $t$, and distance, $x$, and you solve these two differential equations to get:
$$ t = \frac{c}{a} \sinh\left(\frac{a\tau}{c}\right) \tag{1} $$
$$ x = \frac{c^2}{a} \left(\cosh\left(\frac{a\tau}{c} \right) – 1 \right) \tag{2} $$
In these equations $\tau$ is the time measured on your wristwatch, $t$ is the time measured by the observers on Earth and $x$ is the distance travelled as measured by the observers on Earth. The times $t$ and $\tau$ start at zero at the moment you begin accelerating and leave the Earth. Finally $a$ is your constant acceleration. Note that $a$ is the acceleration you measure i.e. it’s the acceleration shown by an accelerometer you hold while you’re sat in the rocket.
To do the calculation, for example for the trip to Kepler 186f, you take the first half of the journey while the rocket is accelerating and set $x$ to this distance. So for Kepler 186f $x = 245$ light years. Then you solve equation (2) to get the elapsed time on the rocket $\tau$, and finally plug this into equation (1) to get the elapsed time on Earth. This is the time for half the trip, so just double it to get the time for the whole trip. I’ve done this for a range of accelerations to get this graph:
Again the blue line is the time measured on Earth and the red line is your time. At an acceleration of only 0.1g the travel time is already down to 76 years (just doable in a single lifetime) and at a more comfortable 1g the travel time is a shade over 12 years.
Since the values aren't that easy to read off the graph here are some representative values:
$$\begin{matrix} a (/g) & \tau (/\text{years}) & t (/\text{years}) \\ 0.01 & 374.9 & 655.9 \\ 0.1 & 76.8 & 509.0 \\ 1 & 12.1 & 491.9 \\ 10 & 1.7 & 490.2 \end{matrix}$$
Footnotes for non-non-nerds
Assuming you have more than a casual interest in Physics (why else would you be reading this!) there is lots more interesting stuff about accelerated motion. For example you might wonder how the spaceship accelerating at 1g can travel 490 light years in 12.1 years if nothing can travel faster than light. The answer is that the spaceship doesn’t travel 490 light years - the Lorentz contraction caused by its high speed means it travels a much shorter distance.
We’ve got the equations for distance and time above, and you can combine them to work out the velocity as a function of spaceship time $\tau$. I won’t do this since it’s just algebra; instead I’ll just quote the result:
$$ v = c \tanh \left( \frac{a\tau}{c} \right) \tag{3} $$
If the spaceship is travelling at velocity $v$ relative to the Earth and destination star then the Earth and star are travelling at velocity $v$ relative to the spaceship, and the crew of the spaceship see distances contracted by the Lorentz factor:
$$ d’ = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}} $$
When the spaceship sets off its distance to the star is 490 light years, but as it accelerates this distance decreases for two reasons. Firstly (obviously) the ship moves towards the star, but secondly Lorentz contraction makes the remaining distance smaller.
To calculate this effect you work out $x(\tau)$ using equation (2) for the first half of the trip. Since the trip is symmetrical you can reflect about the halfway point to get $x(\tau)$ for the second half of the journey. Then the distance left is just (for Kepler 186f) 490 light years - $x$. Calculate the velocity using equation (3) (again for the first half then reflect about the halfway point). Calculate the Lorentz factor from the velocity and multiply to get the contracted distance left. The results for 1g acceleration look like this:
To make the data clearer I’ve plotted the remaining distance for the last half of the trip on an expanded scale to the right. The discontinuity is where the spaceship switches from acceleration to deceleration. The graph shows that the occupants of the ship see the distance they have left to travel shrink rapidly as their speed increases. Conversely, as they start decelerating the Lorentz contraction decreases and the distance left to travel decreases only slowly until they are close to the destination.
protected by Qmechanic♦ May 2 '14 at 19:02
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Given a feasibility problem with both inequality and equality constraints, I'm interested in the sensitivity of the bounds of the region to changes in the constraints. To help with answering the rather general question, I'm interested in particular for linear equality constraints with perturbations for
Linear Non-linear
inequality constraints. As a sample problem, consider
$$ \begin{align} min \qquad 1 \\ \text{subject to} \hspace{1ex} b-\epsilon \leq \hspace{1ex}&a^{T}x \leq b + \epsilon \\ \mathbf{1}^{T}x &= 1 \end{align} $$
How do the limits of each co-ordinate in the feasible region $x_{i}$ vary as a function of $\epsilon$? Is there a common name for this problem? |
Solving linear equations inequalities algebra
To solving linear equations inequalities algebra means to find the value or values of the variable used in it.
Thus;
(i) to solve the inequality 2x + 4 > 6 means to find the variable x.
(ii) to solve the inequality 5 - 6y $\leq$ 3 means to find the value of y and so on.
The following are the rules must be followed for solving a given linear inequality.
Rule 1 :
On transferring the positive term (in addition) from one side of an inequality to its other side, the sign of that transferred number changes to negative.
Examples :
4x + 6 > 7 ⇒ 4x > 7 - 6, 21 $\geq$ 2x + 8 ⇒ 21 - 8 $\geq$ 2x
7x + 4 $\leq$ 18 ⇒ 7x $\leq$ 18 - 4 and so on.
Rule 2:
On On transferring the negative term (in subtraction) from one side of an inequality to its other side, the sign of that transferred number changes to positive .
Examples :
3x - 6 > 5 ⇒ 3x > 5 + 6, 42 $\geq$ 2x - 9 ⇒ 42 + 9 $\geq$ 2x
9x - 1 $\leq$ 20 ⇒ 9x $\leq$ 20 + 1 and so on.
Rule 3 :
If each term of an equation be multiplied or divided by the same positive number, the sign of inequality remains the same.
Case I :
If p is positive and x < y
x < y ⇒ px < py and $\left ( \frac{x}{p} \right )$ < $\left ( \frac{x}{p} \right )$
x > y ⇒ px > py and $\left ( \frac{x}{p} \right )$ > $\left ( \frac{x}{p} \right )$
x $\leq$ y ⇒ px $\leq$ py and $\left ( \frac{x}{p} \right )$ $\leq$ $\left ( \frac{x}{p} \right )$
x $\geq$ y ⇒ px $\geq$ py and $\left ( \frac{x}{p} \right )$ $\geq$ $\left ( \frac{x}{p} \right )$
For example x $\leq$ 5 ⇒ 3x $\leq$ 3 x 5 and x $\geq$ 5 ⇒ 3x $\geq$ 3 x 5
Rule 4:
If each term of an inequality be multiplied or divided by the same negative number, the sign of inequality will flip (reverse).
Case II : If p is negative,
x < y ⇒ -px > -py and $\left ( \frac{-x}{p} \right )$ > $\left ( \frac{-x}{p} \right )$
x > y ⇒ -px < -py and $\left ( \frac{-x}{p} \right )$ > $\left ( \frac{-x}{p} \right )$
x $\leq$ y ⇒ -px $\geq$ -py and $\left ( \frac{-x}{p} \right )$ $\geq$ $\left ( \frac{-x}{p} \right )$
x $\geq$ y ⇒ -px $\leq$ -py and $\left ( \frac{-x}{p} \right )$ $\leq$ $\left ( \frac{-x}{p} \right )$
Rule 5 :
If the sign of the sides of an inequality are positive or negative, then while doing their reciprocals flip(reverse) the sign of inequality.
(i) x > y $\Leftrightarrow \frac{1}{x}$ < $\frac{1}{y}$ (ii) $x \geq y \Leftrightarrow \frac{1}{x} \leq \frac{1}{y}$
(iii) $x \leq y \Leftrightarrow \frac{1}{x} \geq \frac{1}{y}$ and so on.
11th grade math From solving linear equations inequalities algebra to Home |
Q. In a car race on straight road, car A takes a time t less than car B at the finish and passes finishing point with a speed 'v' more than that of car B.
Both the cars start from rest and travel with constant acceleration $a_1$ and $a_2$ respectively.
Then 'v' is equal to :
Solution:
For A & B let time taken by A is $t_0$ from ques. $v_{A} - v_{B} = v =\left(a_{1} -a_{2}\right)t_{0} -a_{2}t$ ....(i) $ x_{B} =x_{A} = \frac{1}{2} a_{1} t_{0}^{2} = \frac{1}{2} a_{2} \left(t_{0} +t\right)^{2} $ $ \Rightarrow \sqrt{a_{1}t_{0}} = \sqrt{a_{2} } \left(t_{0} +t\right) $ $ \Rightarrow \left(\sqrt{a_{2}} - \sqrt{a_{2}}\right)t_{0} = \sqrt{a_{2}t} $ ......(ii) putting $t_0$ in equation $ v = \left(a_{1} -a_{2}\right) \frac{\sqrt{a_{2}t}}{\sqrt{a_{1} } -\sqrt{a_{2}}} -a_{2} t $ $= \left(\sqrt{a_{1}} + \sqrt{a_{2}}\right) \sqrt{a_{2}t} - a_{2} t \Rightarrow v = \sqrt{a_{1}a_{2}t} $ $ \Rightarrow \sqrt{a_{1}a_{2}t} + a_{2}t - a_{2}t $ Questions from JEE Main 2019 Physics Most Viewed Questions 1. An AC ammeter is used to measure current in a circuit. When a given direct current passes through the circuit, the AC ammeter reads 3 A. When another alternating current passes through the circuit, the AC ammeter reads 4 A. Then the reading of this ammeter, if DC and AC flow through the circuit simultaneously, is 8. A cannon of mass I 000 kg, located at the base of an inclined plane fires a shelI of mass 100 kg in a horizontal direction with a velocity 180 $kmh^{-1}$. The angle of inclination of the inclined plane with the horizontal is 45$^{\circ}$. The coefficient of friction between the cannon and the inclined plane is 0.5. The height, in metre, to which the cannon ascends the inclined plane as a result of the recoiI is (g = 10 $ms^{-1})$ Latest Updates Top 10 Medical Entrance Exams in India Top 10 Engineering Entrance Exams in India JIPMER Results Announced NEET UG Counselling Started NEET Result Announced KCET Result Announced KCET College Predictor JEE Main Result Announced JEE Advanced Score Cards Available AP EAMCET Result Announced KEAM Result Announced UPSEE – Online Applications invited from NRI &Kashmiri Migrants MHT CET Result Announced NEET Rank Predictor Questions Tardigrade |
Performing Topology Optimization with the Density Method
Engineers are given significant freedom in their pursuit of lightweight structural components in airplanes and space applications, so it makes sense to use methods that can exploit this freedom, making topology optimization a popular choice in the early design phase. This method often requires regularization and special interpolation functions to get meaningful designs, which can be a nuisance to both new and experienced simulation users. To simplify the solution of topology optimization problems, the COMSOL® software contains a density topology feature.
About the Density Method for Topology Optimization
As the name suggests, topology optimization is a method that has the ability to come up with new and better topologies for an engineering structure given an objective function and set of constraints. The method comes up with these new topologies by introducing a set of design variables that describe the presence, or absence, of material within the design space. These variables are defined either within every element of the mesh or on every node point of the mesh. Changing these design variables thus becomes analogous to changing the topology. This means that holes in the structure can appear, disappear, and merge as well as that boundaries can take on arbitrary shapes. In addition, the control parameters are somewhat automatically defined and tied to the discretization.
As of COMSOL Multiphysics® software version 5.4, the add-on Optimization Module includes a density topology feature to improve the usability of topology optimization. The feature is designed to be used as a density method (Ref. 3), meaning that the control parameters change a material parameter through an interpolation function. Interpolation functions for solid and fluid mechanics are built into the feature and used in example models throughout the Application Library in COMSOL Multiphysics.
A bracket geometry is topology optimized, leaving only 50% of the material, which contributes the most to the stiffness. The printed bracket geometry.
The density method involves the definition of a control variable field, \theta_c, which is bounded between 0 and 1. In solid mechanics, \theta_c=1 corresponds to the material from which the structure is to be built, while \theta_c=0 corresponds to a very soft material. By default, the void Young’s modulus is 0.1% of the solid Young’s modulus. In fluid mechanics, convention dictates that \theta_c=1 corresponds to fluid, while \theta_c=0 is a (slightly) permeable material with an inverse permeability factor, \alpha; i.e., a damping term is added to the Navier-Stokes equation:
The damping term is 0 in fluid domains, while a large value is used in solid domains. These different values give a good approximation of the no-slip boundary condition on the interface between the domains.
An Introduction to the Density Model Feature
The
Density Model feature supports regularization via a Helmholtz equation (Ref. 1). This introduces a minimum length scale using the filter radius, R_\mathrm{min}:
Here, \theta_c is the raw control variable, which is modified by the optimizer, and \theta_f is the filtered variable. The mesh edge size is the default value for the filter radius. While this works well in terms of regularizing the optimization problem, it’s important to set a fixed length (larger than the mesh edge size) to get mesh-independent results.
Top: The equation for the Helmholtz filter can be solved analytically for a 1D Heaviside function. Bottom: This plot is taken from the MBB beam optimization model. It shows the raw control variables to the left and the filtered version to the right.
The Helmholtz filter gives rise to significant grayscale, which does not have a clear physical interpretation. The grayscale can be reduced by applying a smooth step function in what is referred to as
projection in topology optimization. Projection reduces grayscale, but it also makes it more difficult for the optimizer to converge. The density topology feature supports projection based on the hyperbolic tangent function, and the amount of projection can be controlled with the projection steepness, \beta.
Here, \theta_{\beta} is the projection point.
Plot showing the filtered field to the left and the projected field to the right.
Projection makes it possible to avoid grayscale, but grayscale can still appear if the optimization problem favors it. If the same interpolation function is used for the mass and the stiffness, grayscale is optimal in volume-constrained minimum compliance problems. It is thus common to use interpolation functions that cause intermediate values to be associated with little stiffness relative to their cost (compared to the fully solid value). You can think of this as a
penalization of intermediate values for the material volume factor, and the Density Model interface (shown below) supports two such interpolation schemes for solid mechanics: solid isotropic material with penalization (SIMP) and rational approximation of material properties method (RAMP) interpolation. Darcy interpolation is provided for fluid mechanics. The interpolated variable is called the penalized material volume factor, \theta_p, and is used for interpolating the material parameters, e.g., for SIMP interpolation, the p_\textsc{simp} exponent can be increased to reduce the stiffness of intermediate values, so that grayscale becomes less favorable.
\theta_p %26= \theta_\mathrm{min}(1-\theta_\mathrm{min})\theta^{p_\textsc{simp}}\\
E_p %26= E\theta_p
\end{align}
Here, E is the Young’s modulus of the solid material and E_p is the penalized Young’s modulus to be used throughout all optimized domains.
The Density Model feature is available under Topology Optimization in Component > Definitions . The mesh edge length is taken as the default filter radius and it works well, but it has to be replaced with a fixed value in order to produce mesh-independent results.
The penalized Young’s modulus can be defined as a domain variable, or (as in the case of the bracket model) it can be defined directly in the materials.
Topology optimization with the density method involves varying the Young’s modulus spatially. In this case, it is achieved by going to the material properties and multiplying the solid Young’s modulus with the penalized material volume factor, dtopo1.theta_p.
In summary, the density topology feature adds four variables. The filtered material volume factor is defined implicitly using a dependent variable.
Symbol Description Equation \theta_c Control material volume factor 0\leq\theta_c\leq1 \theta_f Filtered material volume factor \theta_f = R_\mathrm{min}^2\mathbf{\nabla}^2\theta_f + \theta_c \theta Material volume factor \theta = \frac{\tanh(\beta(\theta_f-\theta_{\beta}))+\tanh(\beta\theta_{\beta})}{\tanh(\beta(1-\theta_{\beta}))+\tanh(\beta\theta_{\beta})} \theta_p Penalized material volume factor \theta_p = \theta_\mathrm{min}+(1-\theta_\mathrm{min})\theta^{p_\textsc{simp}} or \theta_p = \frac{q_\mathrm{Darcy}(1-\theta)}{q_\mathrm{Darcy}+\theta}
When the filtering is disabled, the filtered variable becomes undefined and the projection instead uses the control material volume factor directly. If the projection is disabled, the material volume factor still exists, but it becomes identical to the projection input.
Applying Continuation to Avoid Local Minima
When the topology is not too complicated, the default values of the density topology feature work well. This is the case for the MBB beam optimization and topology optimized hook models. If the optimal design is more complicated (such as for the bracket example shown at the top of this post), there might be many local minima. To avoid these minima, you can use continuation in the SIMP exponent and the projection slope. This can be achieved by modifying the initial value expression in the density topology feature and adding a
Parametric Sweep feature, as shown below. As a result, the solver ramps over the specified parameters, using the optimum from the previous case as the initial value for the next optimization step. That is, it starts with a small SIMP exponent and projection slope and then continues to higher values. It is possible to apply continuation by combining a parametric sweep with a study reference. See the Bracket — Topology Optimization tutorial model for details. Objectives and Constraints in Topology Optimization
If the geometry is optimized for a single load case (as shown below to the left), the resulting design will be optimal with respect to that load case. This can seem obvious, but often designers make assumptions about symmetries and the design topology. Unless these assumptions are formalized as constraints, they will not be respected. Therefore, the design shown to the right below uses eight load cases (two load groups times four constraint groups).
Left: The bracket geometry is optimized for a single load case, resulting in an asymmetric design with two loosely connected halves. Right: The bracket geometry with eight load cases.
Designers often have several objectives that need to be weighted. To make an informed decision about these objectives, a designer can trace the Pareto optimal front using several optimizations with different weights.
The Pareto optimal front for the bracket geometry can be traced by varying the weight in a parametric sweep. Animation of the topology optimized bracket. (Download the glTF™ file from the Application Gallery in GLB-file format to rotate the geometry yourself.) Exporting and Importing Topology Optimization Results
It is possible to analyze the result of a topology optimized design with respect to stress concentration and buckling without remeshing. However, if you want to be completely sure that the void phase does not play a role, you can eliminate it by exporting and importing the resulting design, as shown below. The details of this procedure are discussed in a previous blog post.
The contour (left) for the topology optimized MBB beam design is exported and imported as an interpolation curve (right). Next Steps
To learn more about the built-in tools and features for solving optimization problems, check out the Optimization Module product page by clicking the button below.
Further Resources Try using the density feature for topology optimization with these example models: Read more about topology optimization on the COMSOL Blog: References B.S. Lazarov and O. Sigmund, “Filters in topology optimization based on Helmholtz‐type differential equations,” International Journal for Numerical Methods in Engineering, vol. 86, no. 6, pp. 765–781, 2011. F. Wang, B.S. Lazarov, and O. Sigmund, “On projection methods, convergence and robust formulations in topology optimization,” Structural and Multidisciplinary Optimization, vol. 43, pp. 767–784, 2011. M.P. Bendsøe, “Optimal shape design as a material distribution problem,” Structural Optimization, vol. 1, pp. 193–202, 1989.
glTF and the glTF logo are trademarks of the Khronos Group Inc. Comments (0) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
A friend asked me about this problem, on how to calculate the distance an object needs to be from the camera to appear a certain height on the screen.
Easy Solution:
The solution is simpler when dealing with an object centered in the in the screen where a right angle can be created from the camera to the object.
Fig. 1: right angle triangle diagram.
$\figLbl{1}{tanSol}$
Solving the triangle in \ref{fig:tanSol}:
\begin{equation}
\tan{\left( \frac{\theta}{2} \right)} = \frac{h}{2d} \label{eq:tanTriangle} \end{equation} |
Let $a \in \Bbb{R}$. Let $\Bbb{Z}$ act on $S^1$ via $(n,z) \mapsto ze^{2 \pi i \cdot an}$. Claim: The is action is not free if and only if $a \Bbb{Q}$. Here's an attempt at the forward direction: If the action is not free, there is some nonzero $n$ and $z \in S^1$ such that $ze^{2 \pi i \cdot an} = 1$. Note $z = e^{2 \pi i \theta}$ for some $\theta \in [0,1)$.
Then the equation becomes $e^{2 \pi i(\theta + an)} = 1$, which holds if and only if $2\pi (\theta + an) = 2 \pi k$ for some $k \in \Bbb{Z}$. Solving for $a$ gives $a = \frac{k-\theta}{n}$...
What if $\theta$ is irrational...what did I do wrong?
'cause I understand that second one but I'm having a hard time explaining it in words
(Re: the first one: a matrix transpose "looks" like the equation $Ax\cdot y=x\cdot A^\top y$. Which implies several things, like how $A^\top x$ is perpendicular to $A^{-1}x^\top$ where $x^\top$ is the vector space perpendicular to $x$.)
DogAteMy: I looked at the link. You're writing garbage with regard to the transpose stuff. Why should a linear map from $\Bbb R^n$ to $\Bbb R^m$ have an inverse in the first place? And for goodness sake don't use $x^\top$ to mean the orthogonal complement when it already means something.
he based much of his success on principles like this I cant believe ive forgotten it
it's basically saying that it's a waste of time to throw a parade for a scholar or win he or she over with compliments and awards etc but this is the biggest source of sense of purpose in the non scholar
yeah there is this thing called the internet and well yes there are better books than others you can study from provided they are not stolen from you by drug dealers you should buy a text book that they base university courses on if you can save for one
I was working from "Problems in Analytic Number Theory" Second Edition, by M.Ram Murty prior to the idiots robbing me and taking that with them which was a fantastic book to self learn from one of the best ive had actually
Yeah I wasn't happy about it either it was more than $200 usd actually well look if you want my honest opinion self study doesn't exist, you are still being taught something by Euclid if you read his works despite him having died a few thousand years ago but he is as much a teacher as you'll get, and if you don't plan on reading the works of others, to maintain some sort of purity in the word self study, well, no you have failed in life and should give up entirely. but that is a very good book
regardless of you attending Princeton university or not
yeah me neither you are the only one I remember talking to on it but I have been well and truly banned from this IP address for that forum now, which, which was as you might have guessed for being too polite and sensitive to delicate religious sensibilities
but no it's not my forum I just remembered it was one of the first I started talking math on, and it was a long road for someone like me being receptive to constructive criticism, especially from a kid a third my age which according to your profile at the time you were
i have a chronological disability that prevents me from accurately recalling exactly when this was, don't worry about it
well yeah it said you were 10, so it was a troubling thought to be getting advice from a ten year old at the time i think i was still holding on to some sort of hopes of a career in non stupidity related fields which was at some point abandoned
@TedShifrin thanks for that in bookmarking all of these under 3500, is there a 101 i should start with and find my way into four digits? what level of expertise is required for all of these is a more clear way of asking
Well, there are various math sources all over the web, including Khan Academy, etc. My particular course was intended for people seriously interested in mathematics (i.e., proofs as well as computations and applications). The students in there were about half first-year students who had taken BC AP calculus in high school and gotten the top score, about half second-year students who'd taken various first-year calculus paths in college.
long time ago tho even the credits have expired not the student debt though so i think they are trying to hint i should go back a start from first year and double said debt but im a terrible student it really wasn't worth while the first time round considering my rate of attendance then and how unlikely that would be different going back now
@BalarkaSen yeah from the number theory i got into in my most recent years it's bizarre how i almost became allergic to calculus i loved it back then and for some reason not quite so when i began focusing on prime numbers
What do you all think of this theorem: The number of ways to write $n$ as a sum of four squares is equal to $8$ times the sum of divisors of $n$ if $n$ is odd and $24$ times sum of odd divisors of $n$ if $n$ is even
A proof of this uses (basically) Fourier analysis
Even though it looks rather innocuous albeit surprising result in pure number theory
@BalarkaSen well because it was what Wikipedia deemed my interests to be categorized as i have simply told myself that is what i am studying, it really starting with me horsing around not even knowing what category of math you call it. actually, ill show you the exact subject you and i discussed on mmf that reminds me you were actually right, i don't know if i would have taken it well at the time tho
yeah looks like i deleted the stack exchange question on it anyway i had found a discrete Fourier transform for $\lfloor \frac{n}{m} \rfloor$ and you attempted to explain to me that is what it was that's all i remember lol @BalarkaSen
oh and when it comes to transcripts involving me on the internet, don't worry, the younger version of you most definitely will be seen in a positive light, and just contemplating all the possibilities of things said by someone as insane as me, agree that pulling up said past conversations isn't productive
absolutely me too but would we have it any other way? i mean i know im like a dog chasing a car as far as any real "purpose" in learning is concerned i think id be terrified if something didnt unfold into a myriad of new things I'm clueless about
@Daminark They key thing if I remember correctly was that if you look at the subgroup $\Gamma$ of $\text{PSL}_2(\Bbb Z)$ generated by (1, 2|0, 1) and (0, -1|1, 0), then any holomorphic function $f : \Bbb H^2 \to \Bbb C$ invariant under $\Gamma$ (in the sense that $f(z + 2) = f(z)$ and $f(-1/z) = z^{2k} f(z)$, $2k$ is called the weight) such that the Fourier expansion of $f$ at infinity and $-1$ having no constant coefficients is called a cusp form (on $\Bbb H^2/\Gamma$).
The $r_4(n)$ thing follows as an immediate corollary of the fact that the only weight $2$ cusp form is identically zero.
I can try to recall more if you're interested.
It's insightful to look at the picture of $\Bbb H^2/\Gamma$... it's like, take the line $\Re[z] = 1$, the semicircle $|z| = 1, z > 0$, and the line $\Re[z] = -1$. This gives a certain region in the upper half plane
Paste those two lines, and paste half of the semicircle (from -1 to i, and then from i to 1) to the other half by folding along i
Yup, that $E_4$ and $E_6$ generates the space of modular forms, that type of things
I think in general if you start thinking about modular forms as eigenfunctions of a Laplacian, the space generated by the Eisenstein series are orthogonal to the space of cusp forms - there's a general story I don't quite know
Cusp forms vanish at the cusp (those are the $-1$ and $\infty$ points in the quotient $\Bbb H^2/\Gamma$ picture I described above, where the hyperbolic metric gets coned off), whereas given any values on the cusps you can make a linear combination of Eisenstein series which takes those specific values on the cusps
So it sort of makes sense
Regarding that particular result, saying it's a weight 2 cusp form is like specifying a strong decay rate of the cusp form towards the cusp. Indeed, one basically argues like the maximum value theorem in complex analysis
@BalarkaSen no you didn't come across as pretentious at all, i can only imagine being so young and having the mind you have would have resulted in many accusing you of such, but really, my experience in life is diverse to say the least, and I've met know it all types that are in everyway detestable, you shouldn't be so hard on your character you are very humble considering your calibre
You probably don't realise how low the bar drops when it comes to integrity of character is concerned, trust me, you wouldn't have come as far as you clearly have if you were a know it all
it was actually the best thing for me to have met a 10 year old at the age of 30 that was well beyond what ill ever realistically become as far as math is concerned someone like you is going to be accused of arrogance simply because you intimidate many ignore the good majority of that mate |
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...
Player $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants.Then $A$ makes a move, then $B$, and so on...The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...
Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
(The cotangent bundle is naturally a symplectic manifold)
Yeah
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
If someone has the time to quickly check my result, I would appreciate. Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$ Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say |
Sylvain JULIEN
I'm a former student in physics (as such, I'm interested in Riemannian geometry and its connections with general relativity) fond of number theory, especially Hilbert's 8th problem, further generalizations of the Riemann Hypothesis and almost everything related to prime numbers and L Functions.
As a huge fan of the concept of symmetry, I plan to study Galois theory and representation theory of automorphism groups of discrete structures.
I'm presentely trying to find a unity in all those fields by considering the semiring $(\mathcal{M},\times,\otimes,s\mapsto 1,\zeta) $ generated by the set of automorphic L-functions belonging to the Selberg class, which I conjecture might be embedded in some Riemannian manifold (at least for a fixed value of the degree of its elements), the automorphism group thereof could help shed a new light on Grand RH, viewing the critical line as a geodesic invariant under the action of this group. I have no idea whether such an approach is realistic or not, but any help would be greatly appreciated.
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This chapter introduces the Sobolev spaces and self-adjoint elliptic operators on those spaces that will be used through the book. It also introduces basic concepts and tools such as Gelfand triples, the Sobolev embedding theorem, the equivalence between energy norm and the Sobolev space norm, the dual norm, the Green's function, and eigenfunctions.
Given an open set with finite perimeter$\Omega \subset {\open R}^n$, we consider the space$LD_\gamma ^{p}(\Omega )$,$1\les p<\infty $, of functions with pth-integrable deformation tensor on Ω and with pth-integrable trace value on the essential boundary of Ω. We establish the continuous embedding$LD_\gamma ^{p}(\Omega )\subset L^{pN/(N-1)}(\Omega )$. The space$LD_\gamma ^{p}(\Omega )$and this embedding arise naturally in studying the motion of rigid bodies in a viscous, incompressible fluid.
This paper considers the Ricci flow coupled with the harmonic map flow between two manifolds. We derive estimates for the fundamental solution of the corresponding conjugate heat equation and we prove an analogue of Perelman's differential Harnack inequality. As an application, we find a connection between the entropy functional and the best constant in the Sobolev embedding theorem in ℝn.
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For a cost function with robust error function $h(e_i)$ $$E(p)=\sum\limits_{i=0}^nh(e_i)$$ it is possible to find an equivalent weighted $L_2$ cost $$E_w(p)=\sum\limits_{i=0}^nw(e_i)\|e_i\|^2,$$ with $$w=\frac{h'(e)}{|e|}$$
In the above, $p$ stands for the parameters of the model we are optimizing. The notation is weird, but I copied it directly from the text I am reading (the error $e_i$ is a function of the parameters and the input).
My question is, how is this derived? Particularly, how is the equality with the weights derived?
The text I am reading is very vague, so I do not know how to go about the derivation. I would appreciate any pointers.
EDIT: My current attempt
Lets say
$$E(x,p,w)=\sum\limits_{i=1}^nw_ie_i(p,x)$$ But we wish to minimize $$E_h(x,p,w)=\sum\limits_{i=1}^nh\circ e_i(p,x)$$ We desire $$\frac{\partial C}{\partial p}=0\iff\frac{\partial C_h}{\partial p}=0$$ Thus $$\frac{\partial }{\partial p}(w_ie_i(x,p))=\frac{\partial }{\partial p}(h\circ e_i(x,p))$$ $$\implies$$ $$w_i\frac{\partial }{\partial p}e_i(x,p)=h'(e_i(x,p))\frac{\partial }{\partial p}e_i(x,p)$$
Thus
$$w_i=h'(e_i(x,p))$$
But this is missing the factor $\frac{1}{|e|}$ the text provides. What am i doing wrong? |
On APN exponents, characterizations of differentially uniform functions by the Walsh transform, and related cyclic-difference-set-like structures 248 Downloads Part of the following topical collections: Abstract
In this paper, we summarize the results obtained recently in three papers on differentially uniform functions in characteristic 2, and presented at the workshop WCC 2017 in Saint-Petersburg, and we give new results on these functions. Firstly, we recall the recent connection between almost perfect nonlinear (APN) power functions and the two notions in additive combinatorics of Sidon sets and sum-free sets; we also recall a characterization of APN exponents which leads to a property of Dickson polynomials in characteristic 2 previously unobserved, which is generalizable to all finite fields. We also give a new characterization of APN exponents in odd dimension by Singer sets. Secondly, after recalling the recent multiple generalization to differentially \(\delta \)-uniform functions of the Chabaud–Vaudenay characterization of APN functions by their Walsh transforms, we generalize the method to all criteria on vectorial functions dealing with the numbers of solutions of equations of the form \(\sum _{i\in I}F(x+u_{i,a})+L_a(x)+u_a=0\), with \(L_a\) linear; we give the examples of injective functions and of o-polynomials; we also deduce a generalization to differentially \(\delta \)-uniform functions of the Nyberg characterization of APN functions by means of the Walsh transforms of their derivatives. Thirdly, we recall the two notions of componentwise APNness (CAPNness) and componentwise Walsh uniformity (CWU). We recall why CAPN functions can exist only if
n is odd and why crooked functions (in particular, quadratic APN functions) are CWU. We also recall that the inverse of one of the Gold permutations is CWU and not the others. Another potential class of CWU functions is that of Kasami functions. We consider the difference sets with Singer parameters equal to the complement of \(\varDelta _F=\{F(x)+F(x+1)+1; x\in \mathbb {F}_{2^n}\}\) where F is a Kasami function. These sets have another potential property, called the cyclic-additive difference set property, which is related to the CWU property in the case of power permutations ( n odd). We study cyclic-additive difference sets among Singer sets. We recall the main properties of Kasami functions and of the related set \(\varDelta _F\) shown by Dillon and Dobbertin and we observe and prove new expressions for \(\varDelta _F\). KeywordsBoolean function Vectorial function Walsh–Hadamard transform APN function Kasami function Cyclic difference set Notes Acknowledgements
The author is grateful to Stjepan Picek for his kind help with computer investigations.
References 1. 2. 3. 4. 5. 6.Carlet C.: Characterizations of the differential uniformity of vectorial functions by the Walsh transform. IEEE Trans. Inf. Theory (see a preliminary version in IACR ePrint Archive 2017/516).Google Scholar 7.Carlet C.: Componentwise APNness, Walsh uniformity of APN functions and cyclic-additive difference sets. IACR ePrint Archive 2017/528.Google Scholar 8.Carlet C., Mesnager S.: Characterizations of o-polynomials by the Walsh transform (2017). arXiv:1709.03765. 9.Carlet C., Picek S.: On the exponents of APN power functions and Sidon sets, sum-free sets and Dickson polynomials. IACR ePrint Archive 2017/1179.Google Scholar 10. 11.Carlet C., Danger J.-L., Desjardins M., Guilley S., Schaub A.: DIBO functions and white box cryptography (2017).Google Scholar 12.Chabaud F., Vaudenay S.: Links between differential and linear cryptanalysis. In: Proceedings of EUROCRYPT’94, Lecture Notes in Computer Science, vol. 950, pp. 356–365 (1995).Google Scholar 13. 14. 15. 16. 17.Nyberg K.: Perfect non-linear S-boxes. In: Proceedings of EUROCRYPT’ 91, Lecture Notes in Computer Science, vol. 547, pp. 378–386 (1992).Google Scholar 18.Nyberg K.: Differentially uniform mappings for cryptography. In: Proceedings of EUROCRYPT’ 93, Lecture Notes in Computer Science, vol. 765, pp. 55–64 (1994).Google Scholar 19.Nyberg K.: S-boxes and round functions with controllable linearity and differential uniformity. In: Proceedings of Fast Software Encryption 1994, Lecture Notes in Computer Science, vol. 1008, pp. 111–130 (1995).Google Scholar 20.Piret G., Roche T., Carlet C.: PICARO—a block cipher allowing efficient higher-order side-channel resistance. In: Proceedings of ACNS 2012, Lecture Notes in Computer Science, vol. 7341, pp. 311–328 (2012).Google Scholar 21. |
I believe that there is a good alternative that benefits from the geometry of the parameter space and completely eliminates the need for constrained optimization. If you explicitly wanted to make use of Lagrangians, I will definitely not be answering the question, but I thought it might be worthwhile to consider the perspective I will describe.In particular, the approach I will be presenting uses Riemannian manifold optimization tools.
Note that the parameter of interest $\mathbf{X}\in \mathbb{R}^{n \times k}$ is simply an element of the Stiefel Manifold $\mathcal{M}\equiv V_k(\mathbb{R}^n)$ (the set of all orthonormal $k$-frames in $\mathbb{R}^n$):$$\begin{align}\mathbf{X} &\in V_k(\mathbb{R}^n) \subset \mathbb{R}^{n\times k}\\V_k(\mathbb{R}^n) = \{\mathbf{X}&\in\mathbb{R}^{n\times k} : \mathbf{X}^\top\mathbf{X}=\mathbf{I}\}.\end{align}$$
With this definition, we can benefit from the Riemannian structure of $V_k(\mathbb{R}^n)$ and optimize the energy with any
unconstrained optimization algorithm, such as the Riemannian-gradient descent:
\begin{align} &\text{ While } \mathbf{X}_{k} \text{ does not sufficiently minimize } f \\ &\text{$\quad$- Pick a gradient related descent direction }\boldsymbol{\eta}_k\in T_{\mathbf{X}_k}\mathcal{M}\\ &\text{$\quad$- Choose a retraction } R_{\mathbf{X}_k}:T_{\mathbf{X}_k}\mathcal{M}\rightarrow \mathcal{M}.\\ &\text{$\quad$- Choose a step length } \tau_k\in \mathbb{R}.\\ &\text{$\quad$- Set } \mathbf{X}_{k+1}\gets R_{\mathbf{X}_k}(\tau_k\boldsymbol{\eta}_k).\\ &\text{$\quad$- } k\gets k+1\end{align}
Here $f$ denotes the function we optimize:$$f = \frac{1}{2}\|\mathbf{X}\mathbf{X}^\top - \mathbf{A} \|_\mathcal{F}^2 = \frac{1}{2}\text{tr}\big((\mathbf{X}\mathbf{X}^\top-\mathbf{A})(\mathbf{X}\mathbf{X}^\top-\mathbf{A})^\top\big)$$whose gradient reads:$$\nabla_{\mathbf{X}}f = (\mathbf{X} \mathbf{X}^\top -\mathbf{A})\mathbf{X}+(\mathbf{X}\mathbf{X}^\top +(-\mathbf{A})^\top )\mathbf{X}.$$
We generally pick the gradient related direction $\boldsymbol{\eta}$ as the projection of the Euclidean gradient onto the tangent space of the manifold (or for a broad class of manifolds including the Stiefel manifold, we can instead use the logarithmic-map):
$$\boldsymbol{\eta}\triangleq \text{grad }_{\mathbf{X}} f = \Pi_{\mathbf{X}} \Big( -\nabla_{\mathbf{X}} f \Big)$$
For now let us assume that $\tau_k$ is fixed,
e.g. $\tau_k=0.1$.The retraction operator $R(\cdot)$ of the Stiefel manifold is analytically defined or in other words the true exponential map is available. More details on that are given in this math-se post and of course in the seminal paper of Edelman et. al:
Edelman, Alan, Tomás A. Arias, and Steven T. Smith. "The geometry of
algorithms with orthogonality constraints." SIAM journal on Matrix
Analysis and Applications 20.2 (1998): 303-353.
All the Riemannian operators for Stiefel manifold are included in the toolboxes such as Manopt or ROPTLIB. Finally, we can update the obtain the new value $\mathbf{X}_{k+1}$ by:\begin{align}\mathbf{X}_{k+1} = R_{\mathbf{X}_k}(\tau_k \boldsymbol{\eta}_k)\end{align}Note that $\mathbf{X}_{k+1}$ will be an orthonormal $k$-frame obeying the aforementioned constrained naturally. Riemennian gradient descent is the simplest choice of Riemannian optimization and there are many others such as Riemannian-LBFGS or Riemannian-Trust Region. Many of those choices vary in how they compute the step size $\tau_k$, usually by a form of line-search.
More recently, Hu et al. considered a very similar problem to the one of this questions where the minimization with orthonormality constraints (again, Stiefel manifold) is made efficient. This method too, uses the Riemannian structure of the problem. |
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Change to browse by: Bookmark(what is this?) Astrophysics > Cosmology and Nongalactic Astrophysics Title: Using the tilted flat-$Λ$CDM and the untilted non-flat $Λ$CDM inflation models to measure cosmological parameters from a compilation of observational data
(Submitted on 31 Dec 2017 (v1), last revised 26 Jul 2019 (this version, v3))
Abstract: We use the physically-consistent tilted spatially-flat and untilted non-flat $\Lambda$CDM inflation models to constrain cosmological parameter values with the Planck 2015 cosmic microwave background (CMB) anisotropy data and recent Type Ia supernovae measurements, baryonic acoustic oscillations (BAO) data, growth rate observations, and Hubble parameter measurements. The most dramatic consequence of including the four non-CMB data sets is the significant strengthening of the evidence for non-flatness in the non-flat $\Lambda$CDM model, from 1.8$\sigma$ for the CMB data alone to 5.1$\sigma$ for the full data combination. The BAO data is the most powerful of the non-CMB data sets in more tightly constraining model parameter values and in favoring a spatially-closed Universe in which spatial curvature contributes about a percent to the current cosmological energy budget. The untilted non-flat $\Lambda$CDM model better fits the large-angle CMB temperature anisotropy angular spectrum and is more consistent with the Dark Energy Survey constraints on the current value of the rms amplitude of mass fluctuations ($\sigma_8$) as a function of the current value of the nonrelativistic matter density parameter ($\Omega_m$) but does not provide as good a fit to the smaller-angle CMB temperature anisotropy data as does the tilted flat-$\Lambda$CDM model. Some measured cosmological parameter values differ significantly between the two models, including the reionization optical depth and the baryonic matter density parameter, both of whose 2$\sigma$ ranges (in the two models) are disjoint or almost so. Submission historyFrom: Chan-Gyung Park [view email] [v1]Sun, 31 Dec 2017 00:36:06 GMT (9709kb,D) [v2]Tue, 17 Jul 2018 17:39:39 GMT (9668kb,D) [v3]Fri, 26 Jul 2019 17:03:54 GMT (9670kb,D) |
April 6th, 2018, 06:48 AM
# 21
Math Team
Joined: Oct 2011
From: Ottawa Ontario, Canada
Posts: 14,597
Thanks: 1039
Well, Skip generously gave you the steps.
Now, for any triangle ABC where angleBAC is given:
k = angleBAC, v = angleBPC : show that k = 2v
Let u = angleABP
v = 180 - (180 - k - u) - (k + 2u)/2
Multiply by 2:
2v = 360 - 2(180 - k - u) - (k + 2u)
simplify:
2v = k
April 6th, 2018, 06:58 PM
# 22
Senior Member
Joined: Aug 2014
From: India
Posts: 476
Thanks: 1
I got the answer, I didn't assume the triangle is isosceles like you.
In $\Delta{BPC}$
$\widehat{BCP}+\widehat{BPC}+\widehat{PBC} = 180^\circ$
$\widehat{BPC}=180^\circ-(\widehat{BCP}+\widehat{PBC})$
$=180^\circ-(\widehat{BCA}+\frac{180-\widehat{BCA}}{2}+\frac{\widehat{ABC}}{2})$
$=180^\circ-\frac{2\widehat{BCA}+180^\circ-\widehat{BCA}+\widehat{ABC}}{2}$
$=180^\circ-\frac{180^\circ+\widehat{BCA}+\widehat{ABC}}{2}$
$=180^\circ-\frac{180^\circ+180^\circ-\widehat{BAC}}{2}$
$=180^\circ-\frac{180^\circ+180^\circ-40^\circ}{2}$
$=20^\circ$
Last edited by Ganesh Ujwal; April 6th, 2018 at 07:03 PM.
April 6th, 2018, 07:48 PM
# 23
Math Team
Joined: Oct 2011
From: Ottawa Ontario, Canada
Posts: 14,597
Thanks: 1039
Quote:
My last post says ANY triangle (not isosceles).
The isosceles portion was to help YOU,
in order for YOU to get familiar with the process...
Are we now finally finished with this charade?
April 7th, 2018, 07:57 AM
# 25
Math Team
Joined: Oct 2011
From: Ottawa Ontario, Canada
Posts: 14,597
Thanks: 1039
Similar to a British Gentleman to a Lady, I bow at the waist
and tip my hat to you, and depart never to be seen again...
Whoops, my elbow accidentally just hit the "ignore button"...
Tags problem, solve
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I am studying a research paper which is concerned with finding paths from a source nodes to a single sink node keeping in mind 2 things.
1.The security of the path.
2.The optimality of the path.
Research paper transmission cost minimization with vulnerability constraint
The vulnerability model of paper says that the link that is shared by greater number of paths is more vulnerable to be attacked by some attacker. For example if there are 5 source nodes and 1 sink node, and say there is a link shared by 3 out of 5 paths and a link shared by only 1 out of 5 paths, then the latter link is more secure.
It basically defines the vulnerability of a link by the value of $No(e)$, where $e$ is the edge and it applies a constraint that the link cannot be used by more than $No(e)$ paths for the link to be secure, other wise it is likely to get attacked.
Here are important points about the research paper.
To guarantee network security, the constraint vulnerability for each link $e$ is set to $No(e)$, which is a positive integer.
$s$ is the single sink node and $P$ is the set of sources.
Definition 1: An $s-P$ cut (Ecut) is defined as the set of links satisfying that if these links are removed, node $s$ and nodes in set $P$ are disconnected, meaning all the sources will be disconnected from the sink node. The weight of $s-P$ node cut is defined as
$\sum{(No(e))}$ $\forall e \in$ cut $s-P$.
The $s-P$ node cut with minimum weight is defined as $C_{min}$. This also serves as the upper bound for the number of sources allowed to transmit, or mathematically it means
$$ |P| \leq C_{min}$$
Up to this point I have understood it completely. The upper bound on $|P|$ can be very easily derived, by analysing the boundary condition. Suppose we allow exactly $C_{min}$ number of sources to transmit and $C_{min} = \sum{No(e)}$ for all the edges in the cut with minimum weight. Now say $No(e)$ paths go through first edge in the cut, No(e) paths go through second edge in the cut and so on. But if anywhere $No(e)+1$ paths go through an edge keeping everything same(or in other words having sources $C_{min} + 1$) then vulnerability constraint is broken as link can at most be used by $No(e)$ paths.
Further the research paper is:Which I am not comfortable with is
Proposed Algorithms
Let us use a strategy to convert undirected to directed graph with the objective of devising an algorithm with the help of network flow theory.
For each link $(u,v)$ in graph, replace it by two directed arcs, $(u,v)$ and $(v,u)$, the cost of both arcs is same as cost of link in undirected graph. Let us denote the new network with $G(N,A)$, $E$ being replaced by $A$.
Now for each directed arc $(u,v) \in A$, we define $f(u,v)$ as the current unit flow from node $u$ to node $v$. We initialize $f(u,v)$ as $No(u,v)$.
Algorithm 1: Computing the number of source nodes using each arc for the data transmission.
I understood that the number of source nodes using an arc for data transmission(or the number of times an arc is being used in various paths) is
$$r(u,v) = |No(u,v) - f(u,v)|$$
But my question is why was the need for arcs, we could have applied the dijkstra on undirected graph and then could have also got this, what is that I am missing.
Further the paper proposes that the cost $r(u,v)$ is added to each link in the undirected graph but does not specify whether both $r(u,v)$ and $r(v,u)$ is added
The second alogirthm, which i could not understand as to why the deletion of links is happening and also minimal explanation in paper of it is given.
Can somebody explain the second algorithm and its link with the first one and try to explain the bigger picture so that I can understand what is going on. |
This question, like your previous one, seems almost cruel. There don't seem to be
particularly elegant ways to proceed, and the solutions aren't particularly pretty. (Who's inflicting these problems on you?)
In these situations, I often just convert to complex exponentials, via $$\cos u = \frac12\left(e^{iu}+e^{-iu}\right) \qquad \sin u = \frac{1}{2i}\left(e^{iu}-e^{-iu}\right)$$and let
Mathematica do all the symbol-crunching.
Here, defining $p := e^{i\theta}$, the equations become$$\begin{align}p^8 + \phantom{2}p^5 ( x - i y)\phantom{-10p^4\;\,} -\phantom{2} p^3 (x+iy) - 1 &= 0 \\3 p^8 + 2 p^5 (x-iy) - 10 p^4 + 2 p^3 (x+iy) + 3 &= 0 \end{align}$$
At this point, "all we need to do" is eliminate $p$. In a polynomial system, this can be accomplished using the method of resultants, conveniently implemented by
Mathematica's
Resultant function (because no one should be applying this method by hand). The method yields this $\theta$-free relation:
$$\begin{align}0 &= x^{10}+5 x^8 y^2+10 x^6 x^4 + 10 x^4 y^6+5 x^2 y^8+y^{10} \\&-705 x^8+12180 x^6 y^2 -24230 x^4 y^4+12180 x^2 y^6-705 y^8 \\&+122560 x^6-112320 x^4 y^2 -112320 x^2 y^4 +122560 y^6 \\&+599040 x^4-1361920 x^2 y^2 +599040 y^4 \\&+327680 x^2+327680 y^2 \\&-1048576 \end{align}$$
It happens that we can find a lot of $(x^2+y^2)$ groupings, and write, say,
$$\begin{align}
0 &= \left(x^2 + y^2\right)^5 - 705 \left( x^2 + y^2\right)^4+122560\left(x^2+y^2\right)^3+599040\left(x^2+y^2\right)^2+327680\left(x^2+y^2\right)\\
&+5000x^2y^2\left(3\left(x^2 + y^2\right)^2 -96 \left(x^2+y^2\right) -512 -10x^2y^2 \right) \\
& -1048576 \\
\end{align} \tag{$\star$}$$
Which is
marginally better, I suppose. There may be additional clever groupings, but it hardly seems worth the trouble to look for them. ( Note: I can't even guarantee that I transcribed everything properly.) Even so, I hope this helps. $\square$
Addendum. The appearance of complex conjugates in $(2)$ suggests that there may be some benefit in expressing things in terms of $z := x+iy$ and $\overline{z}:=x-iy$. It even helps to work towards this from the outset. Writing $(1)$ and $(2)$ for the original equations, we can obtain these conjugate equations:$$\begin{align}(1) + i(2): &\quad \phantom{5}p^8 - 10 p^4\phantom{z} + \phantom{1}4 p^3 z + 5 = 0 \\(1) - i(2): &\quad 5 p^8 + \phantom{1}4 p^5 \overline{z} - 10 p^4\phantom{z} + 1 = 0 \\\end{align}$$The equations are a little nicer-looking, but they don't really make the task of eliminating $p$ any easier. A computer algebra system doesn't care anyway. Be that as it may, applying the method of resultants yields$$\begin{align}0 &= 16777216 - 5242880 z \overline{z} - 4464640 z^2 \overline{z}^2 - 1000960 z^3 \overline{z}^3 + 30 z^4 \overline{z}^4 - 16 z^5 \overline{z}^5 \\&+2500\left(z^4+\overline{z}^4\right)(-1024-192z\overline{z}+z^2\overline{z}^2) \\&+ 3125 \left(z^8 + \overline{z}^8\right) \end{align}$$The powers of $z\overline{z}$ are nice to see, because they give us $x^2+y^2$. Also, we can manipulate the expression to include powers of $z^2+\overline{z}^2=2(x^2-y^2)$; doing so is surprisingly effective, as the left-over $z\overline{z}$ terms factor:$$16 (z \overline{z}-16)^5 -3125 (z^2 + \overline{z}^2)^4 + 10000\left(z^2 + \overline{z}^2\right)^2 \left(256 + 48 z\overline{z} + z^2 \overline{z}^2\right)= 0$$ Abbreviating with $u:= z\overline{z}=x^2+y^2$ and $v:=\frac12(z^2+\overline{z}^2)=x^2-y^2$, this is
$$(u-16)^5 + 2500 v^2\left(u^2+48u+256\right) - 3125 v^4 = 0 \tag{$\star\star$}$$
This might be
about as good as things get. |
June 17th, 2014, 06:10 PM
# 1
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Joined: May 2014
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Thanks: 0
Showing R is an ordered set with cuts
I started reading "Baby Rudin", and I got to Theorem 1.19 in the appendix of chapter 1.
I think I've got this right, but I'm not quite sure, because my logic trick maybe a fake.
To start things, the members of the set R, are subsets of Q called "cuts" (Dedekind cuts, I think?). A cut is any set alpha, a subset of Q, the with the following properties:
1) alpha is not the empty set and alpha is not the set Q (alpha not equal to Q)
2) If p is in alpha, q is in Q, and q<p, then q is in alpha
3) If p is in alpha, then p<r for some r in alpha
The Rudin says 2) implies two different logically equivalent statements, which I checked:
i) If p is in alpha and q is not in alpha, then p < q (very useful)
ii) If r is not in alpha and r<s, then s is in alpha (I haven't used this one)
Onwards to the heart of my long-winded post. The Rudin says you can define "alpha < beta" to mean alpha is a subset of beta.
The Rudin also says you can prove that at least one of the three relations hold, (alpha < beta) or (beta < alpha) or (alpha not equal to beta)
I followed along with Rudin's consise statements that you can show if (alpha not < beta) and (alpha not = beta) (Rudin actually never wrote (alpha not = beta) as that is redundant), then (beta < alpha).
Fair enough. I said this was the heart of the post a few paragraphs back, so this is the left atrium of the heart of the post.
I'm trying to prove that if (alpha not < beta) and (beta not < alpha) then (alpha = beta).
I start with a proof by contradiction. (My reasoning is in italics)
Suppose there is some alpha and beta in R such that [(alpha not < beta) and (beta not < alpha)] and (alpha not = beta).
Since alpha is not < beta, alpha is not a proper subset of beta. So there is some p in alpha and p is not in beta. Likewise, beta is not < alpha, so there is some q in beta and q is not in alpha. Now using statement i) from the top, since p is in alpha and q is not in alpha, p<q. Likewise, since q is in beta, and p is not in beta, q<p. So we have p<q and q<p. A contradiction. So the supposition is wrong and for all alpha and beta in R, if (alpha not < beta) and (beta not < alpha), then (alpha = beta).
This seemed like a trick. Is this proof right? Please help
June 17th, 2014, 08:08 PM
# 2
Senior Member
Joined: Mar 2012
Posts: 294
Thanks: 88
You would be far better off to use the proper notation, as what you are trying to say isn't clear.
So what I think you are trying to show is:
$\alpha \not\subset \beta$ and $\beta \not\subset \alpha \implies \alpha = \beta$
Note that this isn't true for sets in general, so there must be something special about cuts that makes it true.
The first thing to show is that:
if $\alpha \not\subset \beta$ then $\beta \subseteq \alpha\ (\ast\ast)$
So assume that $\alpha \not\subset \beta$, so that we have $q \in \alpha - \beta$ (this is a set difference, not subtraction. It is important to note that we HAVE to assume that $\alpha,\beta$ are non-empty to be sure $q$ exists).
Now let $t \in \beta$. We need to show that $t \in \alpha$. Since $q,t \in \Bbb Q$, we can use the ordering on $\Bbb Q$ to consider 3 cases:
a. $q < t$
b. $q = t$
c. $t < q$.
If (a), then by (2) we have $q \in \beta$, contradicting our choice of $q$. A similar consideration holds for (b).
So (c) must hold, in which case by (2) again, since $q \in \alpha$, we must have $t \in \alpha$.
Now from $(\ast\ast)$, we know that if $\alpha \not\subset \beta$ one (or both) of the following statements holds:
$\alpha = \beta$
$\beta \subset \alpha$.
So if we stipulate that $\alpha \not\subset \beta$ AND $\beta \not\subset \alpha$, the only possibility is $\alpha = \beta$.
Your proof is "probably right", but you need to be clearer about the distinction between $\subset$ and $\subseteq$ (which amounts to the distinction between $<$ and $\leq$ in $\Bbb Q$).
********************
A shorter proof if you know about partial orders: inclusion defines a partial order on the set of rational cuts. The proof of $(\ast\ast)$ above shows that the trichotomy rule holds on the set of these cuts, and is thus a total order.
********************
The reason why your proof may seem like a "fake" is this: what we really mean by the rational cut $\alpha$ is:
$\{q \in \Bbb Q: q < \alpha\}$, where in the set-builder definition $\alpha$ is a real number. Of course, this won't do, we can't use a real number to define itself, it's circular logic.
Fortunately the fine-patterned order structure of the reals is very closely related to the pattern of the rationals. So we USE this, to "fill in the gaps".
Last edited by Deveno; June 17th, 2014 at 08:14 PM.
June 17th, 2014, 08:27 PM
# 4
Senior Member
Joined: Mar 2012
Posts: 294
Thanks: 88
This forum uses a plug-in called MathJax, which is a LaTex rendering system.
To render LaTex, you enclose symbols within dollar signs for in-line rendering, and double dollar signs for display style, on a separate line, centered below.
LaTex has it's own system of commands and syntax, which I can't explain very well in a short post. But I can give you some common expressions:
a \in A
yields:
$a \in A$
a_0 + a_1x + a_2x^2
yields:
$a_0 + a_1x + a_2x^2$
some other common commands:
\leq = $\leq$
\subseteq = $\subseteq$
\emptyset = $\emptyset$
\neq = $\neq$
\frac{a}{b} = $\frac{a}{b}$
\sqrt{x} = $\sqrt{x}$
\to = $\to$
It's very useful, and I urge you to take some time to try to learn it.
June 18th, 2014, 01:40 PM
# 5
Member
Joined: May 2014
From: None of your business
Posts: 31
Thanks: 0
Ok, what I was trying to prove was:
If $\alpha \nsubseteq \beta$ and $\beta \nsubseteq \alpha$ , then $\alpha = \beta$, for any $\alpha , \beta \in R$.
But since you've proven, albeit differently, that if $\alpha \nsubseteq \beta$, then $\beta \subseteq \alpha$.
It follows that $\alpha \subseteq \beta$ and $\beta \subseteq \alpha$, so by definition of set equality, $\alpha = \beta$.
June 18th, 2014, 06:41 PM
# 6
Senior Member
Joined: Mar 2012
Posts: 294
Thanks: 88
The negation of an or statement is an and statement.
Now $A \not\subseteq B$ means:
"$A$ is neither a proper subset nor the same set as $B$", or:
"$A \not\subset B$" AND "$A \neq B$".
I would stay away from compound negations, they get really confusing to work with.
For real numbers, if $\alpha \not\leq \beta$ and $\beta\not\leq\alpha$, we have an impossible situation, so I don't think that's really what you want to say.
******************
A word about Dedekind's construction: we commonly model the real numbers as an (oriented) line, picking "more positive" (greater than) to mean "to the right". So the idea is, choosing any point $x$ on the line (any real number) divides the real numbers into two disjoint sets:
$(-\infty,x)$ and $[x,\infty)$
Since the rationals are DENSE (between any two real numbers there is a rational number), and their ordering is Archimedean (this essentially means $\Bbb Q$ has no "infinitely large" or "infinitesimally small" elements), we can think of $x$ as being:
$\sup((-\infty,x))$
Now again, this involves using $x$ to define $x$, but the density of rationals, means we can define:
$\sup(\{q in \Bbb Q: q \in (-\infty,x)\})$ and this is the same supremum. WE still have the problem, though, that we're using $x$ to define $x$.
So this is the clever thing Dedekind did: he abstracted the order properties of an interval $(-\infty,x)$, and if we use instead the interval $(-\infty,x) \cap \Bbb Q$, now we have a description of a set which uniquely determines a real number which involves only rational numbers.
In essence, then, what Dedekind did is invoke the TOPOLOGY of the real numbers (specifically, the ORDER topology).
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Math Colloquium: Complexifications and Isometries
Wednesday, November 13, 2019
3:00 PM-4:00 PM
Edward Poon, Embry-Riddle Aeronautical University
Abstract: Given a norm $\| \cdot \|$ on a real Banach space $X$, there is a smallest ‘reasonable’ complexification norm $\| \cdot \|_C$ on the complexified space $X_C$, defined by $$\| x+ iy \|_C = \sup \{\|x \cos \theta + y \sin \theta \| : \theta \in [0, 2\pi]\}$$ for $x,y \in X$. Provided $X$ has a certain finiteness condition (possessed by all finite-dimensional spaces) we characterize the isometries for $\| \cdot \|_C$ in terms of the isometries for $\| \cdot \|$. Contact Information
Georgi Medvedev
gsm29@drexel.edu
Location
Korman Center, Room 243, 15 S. 33rd Street, Philadelphia, PA 19014
Audience |
Periodic solutions for implicit evolution inclusions
1.
Department of Mathematics, National Technical University, Zografou Campus, Athens 15780, Greece
2.
Institute of Mathematics, Physics and Mechanics, 1000 Ljubljana, Slovenia
3.
Faculty of Applied Mathematics, AGH University of Science and Technology, 30-059 Kraków, Poland
4.
Institute of Mathematics "Simion Stoilow" of the Romanian Academy, P.O. Box 1-764, 014700 Bucharest, Romania
5.
Faculty of Education and Faculty of Mathematics and Physics, University of Ljubljana, Kardeljeva ploščad 16, 1000 Ljubljana, Slovenia
We consider a nonlinear implicit evolution inclusion driven by a nonlinear, nonmonotone, time-varying set-valued map and defined in the framework of an evolution triple of Hilbert spaces. Using an approximation technique and a surjectivity result for parabolic operators of monotone type, we show the existence of a periodic solution.
Keywords:Evolution triple, compact embedding, pseudo-monotone map, coercive map, implicit inclusion, periodic solution. Mathematics Subject Classification:Primary: 34A20; Secondary: 35A05, 35R70. Citation:Nikolaos S. Papageorgiou, Vicenţiu D. Rădulescu, Dušan D. Repovš. Periodic solutions for implicit evolution inclusions. Evolution Equations & Control Theory, 2019, 8 (3) : 621-631. doi: 10.3934/eect.2019029
References:
[1] [2] [3] [4] [5] [6] [7]
L. Gasinski and N. S. Papageorgiou,
[8] [9]
S. Hu and N. S. Papageorgiou,
[10]
J.-L. Lions,
[11] [12]
N.S. Papageorgiou, F. Papalini and F. Renzacci,
Existence of solutions and periodic solutions for nonlinear evolution inclusions,
[13]
N. S. Papageorgiou and V. D. Rădulescu,
Periodic solutions for time-dependent subdifferential evolution inclusions,
[14]
N. S. Papageorgiou, V. D. Rădulescu and D. D. Repovš,
Sensitivity analysis for optimal control problems governed by nonlinear evolution inclusions,
[15]
R. Showalter,
[16]
show all references
References:
[1] [2] [3] [4] [5] [6] [7]
L. Gasinski and N. S. Papageorgiou,
[8] [9]
S. Hu and N. S. Papageorgiou,
[10]
J.-L. Lions,
[11] [12]
N.S. Papageorgiou, F. Papalini and F. Renzacci,
Existence of solutions and periodic solutions for nonlinear evolution inclusions,
[13]
N. S. Papageorgiou and V. D. Rădulescu,
Periodic solutions for time-dependent subdifferential evolution inclusions,
[14]
N. S. Papageorgiou, V. D. Rădulescu and D. D. Repovš,
Sensitivity analysis for optimal control problems governed by nonlinear evolution inclusions,
[15]
R. Showalter,
[16]
[1] [2]
A. C. Eberhard, J-P. Crouzeix.
Existence of closed graph, maximal, cyclic pseudo-monotone relations and revealed preference theory.
[3] [4]
Alessandra Celletti, Sara Di Ruzza.
Periodic and quasi--periodic orbits of the dissipative standard map.
[5]
Piernicola Bettiol, Hélène Frankowska.
Lipschitz regularity of solution map of control systems with multiple state constraints.
[6] [7]
Wacław Marzantowicz, Piotr Maciej Przygodzki.
Finding periodic points of a map by use of a k-adic expansion.
[8]
Francisco Balibrea, J.L. García Guirao, J.I. Muñoz Casado.
A triangular map on $I^{2}$ whose $\omega$-limit sets are all compact intervals of $\{0\}\times I$.
[9] [10] [11] [12] [13] [14] [15]
Zenonas Navickas, Rasa Smidtaite, Alfonsas Vainoras, Minvydas Ragulskis.
The logistic map of matrices.
[16]
C. R. Chen, S. J. Li.
Semicontinuity of the solution set map to a set-valued weak vector variational inequality.
[17]
Yi Yang, Robert J. Sacker.
Periodic unimodal Allee maps, the semigroup property and the $\lambda$-Ricker map with Allee effect.
[18] [19] [20]
2018 Impact Factor: 1.048
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Under the auspices of the Computational Complexity Foundation (CCF)
The sign-rank of a matrix $A$ with entries in $\{-1, +1\}$ is the least rank of a real matrix $B$ with $A_{ij} \cdot B_{ij} > 0$ for all $i, j$. Razborov and Sherstov (2008) gave the first exponential lower bounds on the sign-rank of a function in AC$^0$, answering an old question of Babai, Frankl, and Simon (1986). Specifically, they exhibited a matrix $A = [F(x, y)]_{x, y}$ for a specific function $F \colon \{-1, 1\}^n \times \{-1, 1\}^n \rightarrow \{-1, 1\}$ in AC$^0$, such that $A$ has sign-rank $\exp(\Omega(n^{1/3}))$.
We prove a generalization of Razborov and Sherstov's result, yielding exponential sign-rank lower bounds for a non-trivial class of functions (that includes the function used by Razborov and Sherstov). As a corollary of our general result, we improve Razborov and Sherstov's lower bound on the sign-rank of AC$^0$ from $\exp(\Omega(n^{1/3}))$ to $\exp(\tilde{\Omega}(n^{2/5}))$. We also describe several applications to communication complexity, learning theory, and circuit complexity.
Minor corrections to the proofs of Claims 2 and 3.1. We thank Lijie Chen for pointing out these issues.
The sign-rank of a matrix $A$ with entries in $\{-1, +1\}$ is the least rank of a real matrix $B$ with $A_{ij} \cdot B_{ij} > 0$ for all $i, j$. Razborov and Sherstov (2008) gave the first exponential lower bounds on the sign-rank of a function in AC$^0$, answering an old question of Babai, Frankl, and Simon (1986). Specifically, they exhibited a matrix $A = [F(x, y)]_{x, y}$ for a specific function $F \colon \{-1, 1\}^n \times \{-1, 1\}^n \rightarrow \{-1, 1\}$ in AC$^0$, such that $A$ has sign-rank $\exp(\Omega(n^{1/3}))$.
We prove a generalization of Razborov and Sherstov's result, yielding exponential sign-rank lower bounds for a non-trivial class of functions (that includes the function used by Razborov and Sherstov). As a corollary of our general result, we improve Razborov and Sherstov's lower bound on the sign-rank of AC$^0$ from $\exp(\Omega(n^{1/3}))$ to $\exp(\tilde{\Omega}(n^{2/5}))$. We also describe several applications to communication complexity, learning theory, and circuit complexity. |
I'm dealing with this problem for a while, and although I have some intuition, I think I'm missing something, please take a look:
We are given a random sample $X_1, X_2,...,X_n$ from Poisson$(\theta)$ distribution. I'm asked to prove that the conditional distribution of: $X_1, X_2,...,X_{n-1}$ given $Y=\sum_{i=1}^n X_i$ is multinomial.
What I have tried so far:
$P(X_1, X_2,...,X_{n-1} \mid Y = y) = \frac{P(X_1, X_2,...,X_{n-1},Y = y)}{P(Y=y)}$
Then, although I know the $n-1$ samples of $X_i$ are independent among each other, and that I could get their joint pdf easily, the problem comes up when I need to related such joint pdf to the pdf of $Y$. Since, I'm not sure whether they are independent, otherwise, the product of the pdfs would suffice.
I know that $Y \sim \text{Poisson}(n\theta)$.
Any clue? Thanks! |
16 0 Homework Statement A copper wire has an internal diameter ##d_1## of ##10## ##cm## and an external diameter ##d_2## of 20 ##cm##. If the wire is ##20m## long what is the resistance of the wire? Homework Equations ##R=\rho\frac{l}{\Sigma}##
What i did is: ##R=\rho\frac{l}{\Sigma}=\rho\frac{l}{d_2^2\pi-d_1^2\pi}##
The problem is that I don't have ##\rho##. Is there a way to find ##R## without knowing it? Many thanks.
The problem is that I don't have ##\rho##. Is there a way to find ##R## without knowing it?
Many thanks. |
Under the auspices of the Computational Complexity Foundation (CCF)
The $\epsilon$-approximate degree $\widetilde{\text{deg}}_\epsilon(f)$ of a Boolean function $f$ is the least degree of a real-valued polynomial that approximates $f$ pointwise to error $\epsilon$. The approximate degree of $f$ is at least $k$ iff there exists a pair of probability distributions, also known as a dual polynomial, that are perfectly $k$-wise indistinguishable, but are distinguishable by $f$ with advantage $1 - \epsilon$. Our contributions are:
We give a simple new construction of a dual polynomial for the AND function, certifying that $\widetilde{\text{deg}}_\epsilon(f) \geq \Omega(\sqrt{n \log 1/\epsilon})$. This construction is the first to extend to the notion of weighted degree, and yields the first explicit certificate that the $1/3$-approximate degree of any read-once DNF is $\Omega(\sqrt{n})$.
We show that any pair of symmetric distributions on $n$-bit strings that are perfectly $k$-wise indistinguishable are also statistically $K$-wise indistinguishable with error at most $K^{3/2} \cdot \exp(-\Omega(k^2/K))$ for all $k \leq K \leq n/64$.
This implies that any symmetric function $f$ is a reconstruction function with constant advantage for a ramp secret sharing scheme that is secure against size-$K$ coalitions with statistical error $K^{3/2} \exp(-\Omega(\widetilde{\text{deg}}_{1/3}(f)^2/K))$ for all values of $K$ up to $n/64$ simultaneously. Previous secret sharing schemes required that $K$ be determined in advance, and only worked for $f=$ AND.
Our analyses draw new connections between approximate degree and concentration phenomena.
As a corollary, we show that for any $d \leq n/64$, any degree $d$ polynomial approximating a symmetric function $f$ to error $1/3$
must have $\ell_1$-norm at least $K^{-3/2} \exp({\Omega(\widetilde{\text{deg}}_{1/3}(f)^2/d)})$, which we also show to be tight for any $d > \widetilde{\text{deg}}_{1/3}(f)$. These upper and lower bounds were also previously only known in the case $f=$ AND. |
Under the auspices of the Computational Complexity Foundation (CCF)
We call a pseudorandom generator $G_n:\{0,1\}^n\to \{0,1\}^m$ {\em
hard} for a propositional proof system $P$ if $P$ can not efficiently prove the (properly encoded) statement $G_n(x_1,\ldots,x_n)\neq b$ for {\em any} string $b\in\{0,1\}^m$. We consider a variety of ``combinatorial'' pseudorandom generators inspired by the Nisan-Wigderson generator on the one hand, and ... more >>>
Assuming the inractability of factoring, we show that the
output of the exponentiation modulo a composite function $f_{N,g}(x)=g^x\bmod N$ (where $N=P\cdot Q$) is pseudorandom, even when its input is restricted to be half the size. This result is equivalent to the simultaneous hardness of the ... more >>>
We study the complexity of building
pseudorandom generators (PRGs) from hard functions.
We show that, starting from a function f : {0,1}^l -> {0,1} that
is mildly hard on average, i.e. every circuit of size 2^Omega(l) fails to compute f on at least a 1/poly(l) fraction of inputs, we can ... more >>>
We study pseudorandom generator (PRG) constructions $G^f : {0,1}^l \to {0,1}^{l+s}$ from one-way functions $f : {0,1}^n \to {0,1}^m$. We consider PRG constructions of the form $G^f(x) = C(f(q_{1}) \ldots f(q_{poly(n)}))$
where $C$ is a polynomial-size constant depth circuit and $C$ and the $q$'s are generated from $x$ arbitrarily. more >>>
We exhibit an explicitly computable `pseudorandom' generator stretching $l$ bits into $m(l) = l^{\Omega(\log l)}$ bits that look random to constant-depth circuits of size $m(l)$ with $\log m(l)$ arbitrary symmetric gates (e.g. PARITY, MAJORITY). This improves on a generator by Luby, Velickovic and Wigderson (ISTCS '93) that achieves the same ... more >>>
We present a new approach to constructing pseudorandom generators that fool low-degree polynomials over finite fields, based on the Gowers norm. Using this approach, we obtain the following main constructions of explicitly computable generators $G : \F^s \to \F^n$ that fool polynomials over a prime field $\F$:
\begin{enumerate} \item a ... more >>>
We prove that the sum of $d$ small-bias generators $L
: \F^s \to \F^n$ fools degree-$d$ polynomials in $n$ variables over a prime field $\F$, for any fixed degree $d$ and field $\F$, including $\F = \F_2 = {0,1}$.
Our result improves on both the work by Bogdanov and
Viola ... more >>>
We show that any distribution on {-1,1}^n that is k-wise independent fools any halfspace h with error \eps for k = O(\log^2(1/\eps)/\eps^2). Up to logarithmic factors, our result matches a lower bound by Benjamini, Gurel-Gurevich, and Peled (2007) showing that k = \Omega(1/(\eps^2 \cdot \log(1/\eps))). Using standard constructions of k-wise ... more >>>
In this paper, we consider coding schemes for computationally bounded channels, which can introduce an arbitrary set of errors as long as (a) the fraction of errors is bounded with high probability by a parameter p and (b) the process which adds the errors can be described by a sufficiently ... more >>>
We show that the promise problem of distinguishing $n$-bit strings of hamming weight $\ge 1/2 + \Omega(1/\log^{d-1} n)$ from strings of weight $\le 1/2 - \Omega(1/\log^{d-1} n)$ can be solved by explicit, randomized (unbounded-fan-in) poly(n)-size depth-$d$ circuits with error $\le 1/3$, but cannot be solved by deterministic poly(n)-size depth-$(d+1)$ circuits, ... more >>>
The {\em hybrid argument}
allows one to relate the {\em distinguishability} of a distribution (from uniform) to the {\em predictability} of individual bits given a prefix. The argument incurs a loss of a factor $k$ equal to the bit-length of the distributions: $\epsilon$-distinguishability implies only $\epsilon/k$-predictability. ... more >>>
Every pseudorandom generator is in particular a one-way function. If we only consider part of the output of the
pseudorandom generator is this still one-way? Here is a general setting formalizing this question. Suppose $G:\{0,1\}^n\rightarrow \{0,1\}^{\ell(n)}$ is a pseudorandom generator with stretch $\ell(n)> n$. Let $M_R\in\{0,1\}^{m(n)\times \ell(n)}$ be a linear ... more >>>
In 1989, Babai, Nisan and Szegedy [BNS92] gave a construction of a pseudorandom generator for logspace, based on lower bounds for multiparty communication complexity. The seed length of their pseudorandom generator was $2^{\Theta(\sqrt n)}\,\,\,$, because the best lower bounds for multiparty communication complexity are relatively weak. Subsequently, pseudorandom generators for ... more >>>
A Boolean function is said to have maximal sensitivity $s$ if $s$ is the largest number of Hamming neighbors of a point which differ from it in function value. We construct a pseudorandom generator with seed-length $2^{O(\sqrt{s})} \cdot \log(n)$ that fools Boolean functions on $n$ variables with maximal sensitivity at ... more >>>
We construct a pseudorandom generator which fools read-$k$ oblivious branching programs and, more generally, any linear length oblivious branching program, assuming that the sequence according to which the bits are read is known in advance. For polynomial width branching programs, the seed lengths in our constructions are $\tilde{O}(n^{1-1/2^{k-1}})$ (for the ... more >>>
We show that a very simple pseudorandom generator fools intersections of $k$ linear threshold functions (LTFs) and arbitrary functions of $k$ LTFs over $n$-dimensional Gaussian space.
The two analyses of our PRG (for intersections versus arbitrary functions of LTFs) are quite different from each other and from previous analyses of ... more >>>
We study the task of seedless randomness extraction from recognizable sources, which are uniform distributions over sets of the form {x : f(x) = v} for functions f in some specified class C. We give two simple methods for constructing seedless extractors for C-recognizable sources.
Our first method shows that ... more >>>
A recent work of Chattopadhyay et al. (CCC 2018) introduced a new framework for the design of pseudorandom generators for Boolean functions. It works under the assumption that the Fourier tails of the Boolean functions are uniformly bounded for all levels by an exponential function. In this work, we design ... more >>> |
Please help me with these questions, I need urgent help FAST!
1) Let line l\(l_1\) be the graph of 5x + 8y = -9. Line L2 is perpendicular to line L1 and passes through the point (10, 10). If line L2 is the graph of the equation y=mx +b, then find m+b .
2) Find all real numbers x such that \(3x - 7 \le 5x +9\). Give your answer as an interval.
3) Find all real numbers s such that \(-2(s - 7) > 4s + 8\) . Give your answer as an interval.
4) Without using a calculator, order the following numbers from least to greatest: \(\begin{align*} A &= \frac{2^{1/2}}{4^{1/6}}\\ B &= \sqrt[12]{128}\vphantom{dfrac{2}{2}}\\ C&= \left( \frac{1}{8^{1/5}} \right)^2\\ D&= \sqrt{\frac{4^{-1}}{2^{-1} \cdot 8^{-1}}}\\ E&= \sqrt[3]{2^{1/2} \cdot 4^{-1/4}}.\vphantom{dfrac{2}{2}}\end{align*}\)
Give your answer as a list of capital letters separated by commas. For example, if you think that E < B < C< D< A , then you would answer E,B,C,D,A.
Sorry for so many questions, help as fast as you can, please! Thanks to anyone who answers! I'll definitely give you extra likes if you answer!
1)\(5x + 8y = -9 <=> y=-\frac{5}{8}x-\frac{9}{8} \)
λ1=-\(\frac{5 }{8}\)
L2 is perpendicular to line L1 so λ1*λ2=-1 <=> -5/8*λ2=-1
λ2=\(\frac{8}{5}\)
so the l2: y-y1=λ2(x-x1) x1=10,y1=10 because passes through the point (10, 10)
so \(y-10 = \frac{8}{5}(x-10)<=> y=\frac{8}{5}x-\frac{80}{5}+10<=> y=\frac{8}{5}x-\frac{30}{5}\)
m=8/5,b=-30/5
m+b=\(\frac{8}{5}-\frac{30}{5}=\frac{-22}{5}\)
2)\(3x-7≤ 5x+9<=> 3x-5x≤9+7 <=> -2x≤16 <=> x≥ 8\)
\(x≥ 8\)
3)\(-2s+14>4s +8<=> 14-8 > 4s+2s <=> 6>6s <=> 1>s <=> s<1\)
\(s<1\)
.
Hope this helps!
1)\(5x + 8y = -9 <=> y=-\frac{5}{8}x-\frac{9}{8} \)
λ1=-\(\frac{5 }{8}\)
L2 is perpendicular to line L1 so λ1*λ2=-1 <=> -5/8*λ2=-1
λ2=\(\frac{8}{5}\)
so the l2: y-y1=λ2(x-x1) x1=10,y1=10 because passes through the point (10, 10)
so \(y-10 = \frac{8}{5}(x-10)<=> y=\frac{8}{5}x-\frac{80}{5}+10<=> y=\frac{8}{5}x-\frac{30}{5}\)
m=8/5,b=-30/5
m+b=\(\frac{8}{5}-\frac{30}{5}=\frac{-22}{5}\)
2)\(3x-7≤ 5x+9<=> 3x-5x≤9+7 <=> -2x≤16 <=> x≥ 8\)
\(x≥ 8\)
3)\(-2s+14>4s +8<=> 14-8 > 4s+2s <=> 6>6s <=> 1>s <=> s<1\)
\(s<1\)
Hope this helps!
If you want to use "-->" or "==>" in LaTex, it is $\Rightarrow$. |
Strategy Library
In this tutorial we build a strategy combining momentum and mean reversion for the foreign exchange markets from Alina F. Serban's research which was based on research in the equity market by Ronald J. Balvers and Yangru Wu. Serban creates a momentum factor using returns of the last 3 months, and a mean reversion factor as a deviation from the mean price. Using these factors we use regression to predict the returns of the coming month. We apply the strategy from Serban's paper and update the mean reversion factor for to improve its significance level.
In theory when trading foreign exchange the expected return accrued in each currency should be the same when adjusted for exchange rates (uncovered interest parity). This suggests the markets should predominately be mean reverting, however in practice we see short term momentum trends and long term mean reversion. This was phenomenon was first noticed by Chiang and Jiang. We tested the theory on EURUSD, GBPUSD, USDCAD and USDJPY and re-balanced monthly. The model significance level and coefficients are close to those in paper, but the returns and Sharpe Ratios obtained are not as good as what the paper claimed. The algorithm achieved a fairly stable annual return of 11%, 0.8 Sharpe Ratio and 11% drawdown.
The strategy is centered on
uncovered interest parity (UIP) theory. UIP states that the change in the exchange rate should incorporate any interest rate differentials between the two currencies. By looking for patterns in the deviation from UIP we can potential generate abnormal returns. Interest Parity Conditions
UIP states that an investor who borrows money in their home country and lends it in another country with a higher interest rate should expect a zero return due to the changes in exchange rate. In other words:\[1+r_t = (1+r^{i}_t)(\frac{F^{i}_t}{S^{i}_t})\]
Where \(r_t\) is the domestic interest rate, \(r^{i}\) is the foreign interest rate, \(S^{i}\) is the spot exchange rate and \(F^{i}\) is the forward rate. We can also replace F forward rate with expected spot rate:\[1+r_t = (1+r^{i}_t)(\frac{E(S^{i}_{t+1})}{S^{i}_t})\]
Taking logs of the above two equations, we obtain:\[r_t - r^{i}_t = \ln F^{i}_t - \ln S^{i}_t\] \[r_t - r^{i}_t =\ln S^{i}_{t+1} - \ln S^{i}_t\]
The deviation from UIP is denoted by y and defined as follows:\[y^{i}_{t+1} =\ln S^{i}_{t+1} -\ln F^{i}_t\]
Model and Parameter Estimation
Fama and French and Summers constructed a simple model for stock price that is the sum of a random walk and a stationary component - they represent the natural log of the stock price with
x. The stationary component represents the temporary swings in stock price (characterized by coefficient \(\delta \)), the parameter \(\mu \) captures the random walk drift component and a coefficient accounts for the momentum effect \(\rho \). Balvers and Wu construct the log of stock prices as:
Using the equation above Serban adapts it to find the abnormal return in the forex market. The \(\delta \) represents the speed of mean reversion and can differ by country, while the \(\rho \) represents the momentum strength and can vary by country and by lag. The parameter \(\mu \) also varies by country. Accounting for these changes:\[y^{i}_t = -(1 - \delta ^{i})(x^{i}_{t-1} - \mu ^{i}) + \sum_{j=1}^{J}\rho ^{i}(x^{i}_{t-1} - x^{i}_{t-j-1}) + \epsilon ^{u}_t\]
Trading Strategy
The trading strategy from the paper allows \(\mu \) to change by country, while let \(\rho \) and \(\delta \) stay fixed. By applying Ordinary Least Squared(OLS) regression, the model estimates the return y for each currency. We construct the portfolio by taking a long position on the currency with the highest expected return and taking a short position on the currency with the lowest expected return. We hold these positions for one month, and repeat the process each month. There are two exceptions to this strategy: if all expected returns are positive, we take a long position only, and vice versa. To limit the number of parameters we need to estimate and find a solution easily we only allow \(\mu \) to change by country. According to the paper if we let ρ stay fixed and J =3, we can obtain the highest return for this strategy. If so the equation can be simplified as:\[y^{i}_t = -(1-\delta )(x^{i}_{t-1} - \mu ^{i}) + \rho (x^{i}_{t-1} - x^{i}_{t-4}) + \epsilon ^{i}_t\]
When applying the above equation, we found that the scale of the mean reversion for each currency are very different, and this difference in scale is large enough to affect the accuracy of our rank. We made an adjustment to standardize the mean-reversion. While calculating \(\mu \) (the mean of the log prices) we also calculated standard deviation σ. In this tutorial we replace \(x-\mu \) with \(\frac{x - \mu}{\sigma } \). This captures the mean reversion factor better than the author's technique.
Data Description
The paper used monthly exchange rate data for the Canadian Dollar/USD, German Mark/Euro, UK Pound/USD and Japanese Yen/USD, from 1978 to 2008. Due to data availability, we used Euro/USD instead of German Mark/Euro, and the earliest data available starts from 2004. Each time we launch the strategy we use all of the available historical data prior to the start date to build the OLS model and uses that model for the entire backtest. The paper used 1/3 of their data as the training dataset and the rest of the test set. We directly test our model on backtesting, because QuantConnect makes this easier.
In order to apply the model, we need to first pull history data to build it. The project can be briefly divided into four parts: the historical data request, model training, prediction and execution.
Step 1: Request Historical Data
The first function takes two arguments: symbol and number of daily data points requested. This function requests historical QuoteBars and builds it into a pandas DataFrame. For more information about pandas DataFrame, please refer to the help documentation DataFrame. The calculate_return function takes a DataFrame as an argument to calculate the mean and standard deviation of the log prices, and create new columns for the DataFrame (return, reversal factor and momentum) - it prepares the DataFrame for multiple linear regression.
def get_history(self,symbol, num): data = {} dates = [] history = self.History([symbol], num, Resolution.Daily).loc[symbol]['close'] #request the historical data for a single symbol for time in history.index: t = time.to_pydatetime().date() dates.append(t) dates = pd.to_datetime(dates) df = pd.DataFrame(history) df.reset_index(drop=True) df.index = dates df.columns = ['price'] return df def calculate_return(self,df): #calculate the mean for further use mean = np.mean(df.price) # cauculate the standard deviation sd = np.std(df.price) # pandas method to take the last datapoint of each month. df = df.resample('BM',how = lambda x: x[-1]) # the following three lines are for further experiment purpose # df['j1'] = df.price.shift(1) - df.price.shift(2) # df['j2'] = df.price.shift(2) - df.price.shift(3) # df['j3'] = df.price.shift(3) - df.price.shift(4) # take the return as depend variable df['log_return'] = df.price - df.price.shift(1) # calculate the reversal factor df['reversal'] = (df.price.shift(1) - mean)/sd # calculate the momentum factor df['mom'] = df.price.shift(1) - df.price.shift(4) df = df.dropna() #remove nan value return (df,mean,sd) Step 2: Build Predictive Model
The concat function requests history and joins the results into a single DataFrame. As \(\mu \) varies by country so we assign the mean and standard deviation to the symbol for each currency for future use. The OLS function takes the resulting DataFrame to conduct an OLS regression. We write it into a function because it's easier to change the formula here if we need.
def concat(self): # we requested as many daily tradebars as we can his = self.get_history(self.quoted[0].Value,20*365) # get the clean DataFrame for linear regression his = self.calculate_return(his) # add property to the symbol object for further use. self.quoted[0].mean = his[1] self.quoted[0].sd = his[2] df = his[0] # repeat the above procedure for each symbols, and concat the dataframes for i in range(1,len(self.quoted)): his = self.get_history(self.quoted[i].Value,20*365) his = self.calculate_return(his) self.quoted[i].mean = his[1] self.quoted[i].sd = his[2] df = pd.concat([df,his[0]]) df = df.sort_index() # remove outliers that outside the 99.9% confidence interval df = df[df.apply(lambda x: np.abs(x - x.mean()) / x.std() < 3).all(axis=1)] return df def OLS(self,df): res = sm.ols(formula = 'return ~ reversal + mom',data = df).fit() return res Step 3: Apply Predictive Model
The predict function uses the history for the last 3 months, merges it into a DataFrame and then calculates the updated factors. Using these updated factors (together with the model we built) we calculate the expected return.
def predict(self,symbol): # get current month in string month = str(self.Time).split(' ')[0][5:7] # request the data in the last three months res = self.get_history(symbol.Value,33*3) # pandas method to take the last datapoint of each month res = res.resample('BM',how = lambda x: x[-1]) # remove the data points in the current month res = res[res.index.month != int(month)] # calculate the variables res = self.calculate_input(res,symbol.mean,symbol.sd) res = res.ix[0] # take the coefficient. The first one will not be used for sum-product because it's the intercept params = self.formula.params[1:] # calculate the expected return re = sum([a*b for a,b in zip(res[1:],params)]) + self.formula.params[0] return re def calculate_input(self, df, mean, sd): df['reversal'] = (df.price - mean)/sd df['mom'] = df.price - df.price.shift(3) df = df.dropna() return df
There are a few points of note:
We need historical TradeBars for the last three months. To do this we requested 99 bars and use a pandas DataFrame to extract a data point for the end of each month. We use event schedule to execute the strategy at the first trading day, however, sometimes the first day of the month could be on the 2nd if the 1st falls on a weekend. To fix this we remove the data from the current month, leaving only the last 3 months of data. We start from the second element of res (res[1:]) because res and params are different lengths. This was hard to detect because Python would not throw error when running [a*b for a,b in zip(res,params)] even if the length of the two lists are different. This function also used pandas DataFrame methods extensively. For more information please refer to pandas. Step 4: Initializing the Model
In the Initialize function we prepare the data and conduct a linear regression. The class property 'self.formula' is the result of the OLS regression. We will use this object each time we rebalance the portfolio.
def Initialize(self): self.SetStartDate(2013,6,1) self.SetEndDate(2016,6,1) self.SetCash(10000) self.syls = ['EURUSD','GBPUSD','USDCAD','USDJPY'] self.quoted = [] for i in range(len(self.syls)): self.quoted.append(self.AddForex(self.syls[i],Resolution.Daily,Market.Oanda).Symbol) df = self.concat() self.Log(str(df)) self.formula = self.OLS(df) self.Log(str(self.formula.summary())) self.Log(str(df)) self.Log(str(df.describe())) for i in self.quoted: self.Log(str(i.mean) + ' ' + str(i.sd)) self.Schedule.On(self.DateRules.MonthStart(), self.TimeRules.At(9,31), Action(self.action)) Step 5: Performing Monthly Rebalancing
Every month we rebalance the portfolio using the Schedule Event helper method. The predicted returns are added to the rank array and then sorted by return. The first element in the list is the best return paired with the associated symbol. When all the expected returns in the rank array are positive we only go long the pair with the highest expected return. When all returns are negative, we only go short the pair with the lowest expected return.
def action(self): rank = [] long_short = [] for i in self.quoted: rank.append((i,self.predict(i))) # rank the symbols by their expected return rank.sort(key = lambda x: x[1],reverse = True) # the first element in long_short is the one with the highest expected return, which we are going to long, and the second one is going to be shorted. long_short.append(rank[0]) long_short.append(rank[-1]) self.Liquidate() # the product < 0 means the expected return of the first one is positive and that of the second one is negative--we are going to long and short. if long_short[0][1]*long_short[1][1] < 0: self.SetHoldings(long_short[0][0],1) self.SetHoldings(long_short[1][0],-1) # this means we long only because all of the expected return is positive elif long_short[0][1] > 0 and long_short[1][1] > 0: self.SetHoldings(long_short[0][0],1) # short only else: self.SetHoldings(long_short[1][0],-1)
The following regression output is obtained by backtesting the time period from Jun 2013 to Jun 2016. From the regression statistics, the R-squared value is 1.4% which is smaller than the paper value 3.89%. Our momentum coefficient, ρ, is 0.0633 compared to the paper's 0.042. We obtained 1.0350 mean reversion coefficient (1 + 0.0350), and the paper got 0.9859.
From these results we can say the limited sample size does not impair the feasibility of this model. The t-stats of the coefficients are -4.074 and 1.417 for the reversal factor and momentum factor respectively. The p-value of the reversal factor is very small which means this factor has a very high significance level.
Backtest Sensitivity Results
We performed some rough period sensitivity analysis in different time period(from 2015 to 2018) and summarized the results as the following table:
The paper demonstrates there are inefficiencies in the UIP which can be exploited with a hybrid momentum and mean reversion strategy. Although the sample size of the paper is much larger than ours the parameter and significance level of the two models are very close. The strategies we discussed above keep \(\rho \) fixed and only allow \(\mu \) to change by country. If we let \(\rho \) change by lag the significance level of the model might increase, but this could potentially make modeling more difficult by introducing multicollinearity. To test this we wrote this implementation in the algorithm and commented out the lines. If you are interested in exploring this extension to the model you can change these lines to test your strategy.
Alina F. Serban, Combining mean reversion and momentum trading strategies in foreign exchange markets Online Copy Investopedia, Uncovered Interest Rate Parity. Online Copy Ronald J. Balvers, Yangru Wu, Momentum and mean reversion across national equity markets Online Copy Chiang, T., Jiang, C., 1995. Foreign exchange returns over short and long horizons. International Review of Economics and Finance 4, 267–282. Online Copy Fama, E., 1984. Forward and spot exchange rates. Journal of Monetary Economics Online Copy 14, 319–338. |
Odd Number Theorem Contents Theorem $\displaystyle \sum_{j \mathop = 1}^n \paren {2 j - 1} = n^2$ $S_n = S_{n - 1} + 2 n - 1$
$\blacksquare$
Proof
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
$\displaystyle n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$ Basis for the Induction
$\map P 1$ is true, as this just says $1^2 = 1$.
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis: $\displaystyle k^2 = \sum_{j \mathop = 1}^k \paren {2 j - 1}$ Then we need to show: $\displaystyle \paren {k + 1}^2 = \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}$ Induction Step
This is our induction step:
\(\displaystyle \paren {k + 1}^2\) \(=\) \(\displaystyle k^2 + 2 k + 1\) \(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop = 1}^k \paren {2 j - 1} + 2 k + 1\) Induction Hypothesis \(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop = 1}^k \paren {2 j - 1} + 2 \paren {k + 1} - 1\) \(\displaystyle \) \(=\) \(\displaystyle \sum_{j \mathop = 1}^{k + 1} \paren {2 j - 1}\)
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore: $\displaystyle \forall n \in \N: n^2 = \sum_{j \mathop = 1}^n \paren {2 j - 1}$
$\blacksquare$
Also presented as
This result can also be seen as:
$\displaystyle \sum_{j \mathop = 0}^{n - 1} \paren {2 j + 1} = n^2$
It appears in his
Arithmeticorum Libri Duo, written in $1557$ but not published till $1575$. Sources 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables... (previous) ... (next): $\S 19$: Arithmetic Series: $19.3$ 1971: George E. Andrews: Number Theory... (previous) ... (next): $\text {1-1}$ Principle of Mathematical Induction: Exercise $5$ 1977: Gary Chartrand: Introductory Graph Theory... (previous) ... (next): Appendix $\text{A}.6$: Mathematical Induction 1980: David M. Burton: Elementary Number Theory(revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $1 \ \text {(b)}$ 1982: P.M. Cohn: Algebra Volume 1(2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $10$ 1986: David Wells: Curious and Interesting Numbers... (previous) ... (next): $25$ 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms(3rd ed.) ... (previous) ... (next): $\S 1.2.1$: Mathematical Induction 1997: David Wells: Curious and Interesting Numbers(2nd ed.) ... (previous) ... (next): $25$ |
After working this out I'm confident that I have found out the answer to my question. I shall post it here. Corrections are highly appreciated in the event something is wrong. The main point seems to be:
In Schwarzschild spacetime there are three categories of radial null geodesics: the ingoing, outgoing and the horizon generators.
This has been also a subject of another question of mine which I answered giving the details of this construction.
The ingoing/outgoing radial null geodesics have $r$ as the affine parameter and they have a simple characterization in terms of the Eddington-Finkelstein coordinate systems.
Indeed in the ingoing Eddington-Finkelstein coordinates $(v,r,\theta,\phi)$ the ingoing radial null geodesics are simply the coordinate lines of $r$, parameterized as $(v_0,r,\theta_0,\phi_0)$ by $r$ which is an affine parameter.
In the outgoing Eddington-Finkelstein coordinates $(u,r,\theta,\phi)$ the outgoing radial null geodesics are simply the coordinate lines of $r$ as well, with parameterization $(u_0,r,\theta_0,\phi_0)$ by $r$ which is again an affine parameter.
Regarding these two classes of radial null geodesics, using the Eddington-Finkelstein coordinates is merely a convenient choice of coordinates making them coordinate lines. They can also be written down and studied in Schwarzschild coordinates as I show in the other answer.
Now we come to the horizon generators. These are radial null geodesics with $r$ constant. They are
obviously a separate class since the ingoing/outgoing are parameterized by $r$ which can't be turned into a constant simply by reparemeterization.
It can be shown - and the details are reffered again to the linked question - that this class only has solutions when the constant $r$ is $r = 2M$. These therefore
can't be studied with Schwarzschild coordinates simply because this region is not covered by that chart.
It has two parts, however, each covered by one of the Eddington-Finkelstein coordinates. Working in the ingoing system, it can be seen that such radial null geodesics are parameterized by $$(v,r,\theta,\phi)=(4M\ln \lambda,2M,\theta_0,\phi_0)$$where $\lambda$ is an affine parameter. In the same way, in the outgoing Eddington-Fineklstein system, these radial null geodesics are parameterized by $$(u,r,\theta,\phi)=(-4M\ln\lambda,2M,\theta_0,\phi_0)$$
It can be shown that the surface $r = 2M$ in each of these systems is a null surface and these geodesics are precisely the generators of the surface.
Finally, if one pass to the double-null Eddington-Finkelstein coordinates $(u,v,\theta,\phi)$ a drawback occurs. The surface $r = 2M$ is pushed to infinity of the coordinates. One therefore would like to introduce coordinates pulling it back to finite coordinate value.
A nice and geometrically motivated way is to peform a coordinate transformation satisfying two conditions:
We leave the angular part alone and transform $U = U(u)$ and $V = V(v)$. This ensures that $v$ constant is the same as $V$ constant and $u$ constant is the same as $U$ constant. Thus in such a system, the ingoing/outgoing radial null geodesics
are still coordinate lines. This changes nothing about their affine parameter being $r$.
The coordinates $U,V$ should, when evaluated along the $r = 2M$ generators give the affine parameter. This makes then
well defined and finite at $r = 2M$ by definition.
Inspection of the parameterizations of the horizon generators suggest $U = - e^{-u/4M}$ and $V = e^{v/4M}$. They satisfy the two conditions, the minus sign being a convenience, which doesn't change the fact that $U$ computes an affine parameter along horizon generators.
Finally, it can be seen that along the outgoing/ingoing null geodesics, $U$ and $V$
can't be affine parameters. This is because their coordinate lines, which are the outgoing/ingoing null geodesics, if parameterized by $U$ and $V$ do not satisfy the autoparallel equation in general. This would demand $\Gamma_{UU}^U =0$ and $\Gamma_{VV}^V=0$. But this answer shows these connection coefficients which can be seen to be zero just on the horizon.
So the conclusions are:
The Kruskal Coordinates $U$ and $V$ have as coordinate lines all radial null geodesics. The ingoing ones are $(V,\theta,\phi) = (V_0,\theta_0,\phi_0)$ with $V_0 \neq 0$. These are affinely parameterized by $r$, and when parameterized by $U$
are not affinely parameterized since $\Gamma_{UU}^U\neq 0$ here.
The outgoing ones are $(U,\theta,\phi)=(U_0,\theta_0,\phi_0)$ with $U_0 \neq 0$. These are again affinely parameterized by $r$ and when reparameterized by $V$
are not affinely parameterized since $\Gamma_{VV}^V\neq 0$ here.
The horizon generators are also coordinate lines. The future horizon generators are $(U_0,\theta_0,\phi_0)$ with $U_0 = 0$ and they are affinely parameterized by $V$. The past horizon generators are $(V_0,\theta_0,\phi_0)$ with $V_0 =0$ and are affinely parameterized by $U$. Obviously $r$ can't be a parameter here since it is constant by definition.
So $(U,V)$ and $r$ are affine parameters of radial null geodesics, but each for certain subcategories of said geodesics. |
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Show that there exists a positive integer nnn such that the decimal representations of 3n3^n3n and 7n7^n7n both start with the digits 10.
Note by Finn Hulse 5 years, 5 months ago
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Lol i literally wrote a python program to try to figure this outI got n = 568, 1136, 2098, 2666, 2905, 4196, 4435, ...
Log in to reply
That's awesome!
Check out my new problem!
thanks
Very very nice problem :D
Hint: Rewrite the numbers as (5−2)n(5-2)^n(5−2)n and (5+2)n(5+2)^n(5+2)n and use binomial theorem :D
I don't see how the Binomial Theorem helps. Can you explain your solution?
Look at it again. It is no coincidence that those two numbers have a mean of 555. :D
Expand the first few cases
@Nathan Ramesh – I'm afraid I still don't see it. How does expanding help?
YAY! ABSOLUTELY FANTASTIC! That's exactly what I was looking for @Patrick Corn @Mursalin Habib @Sreejato Bhattacharya! Please write more! I'm really happy that somebody's got it. :D
I believe that this is equivalent to finding an n n n such that the fractional parts of nlog3 n \log 3 nlog3 and nlog7 n \log 7 nlog7 are both smaller than log1.1 \log 1.1 log1.1. (All logs are base 10.)
It seems like this should be no trouble, since log3 \log 3 log3 and log7 \log 7 log7 are both irrational...but I haven't yet written down anything rigorous. Just thought this would help.
Yes, it easily follows from the fact that given any irrational number III and an arbitrarily small positive real r,r,r, there exist integers x,yx,yx,y such that1>x+Iy>1−r.1>x+Iy>1-r.1>x+Iy>1−r.
Or alternatively, one-liner:
hello, n=568n=568n=568 works nicely. Sonnhard
hello,
n=568n=568n=568 works nicely.
Sonnhard
:P
Voted up for the Sonnhard reference :)
@Mursalin Habib – I did not get that. :P
@Finn Hulse – Do you know Dr Sonnhard Graubner?
@Sreejato Bhattacharya – Nope. The link says he doesn't exist. :/
@Finn Hulse – You need to log in to your AoPS account. These two links will give you a fairly good idea of how he posts. ;)
@Sreejato Bhattacharya – But I am logged in. Furthermore, only one of the links you gave works.
I see what you're saying, but there's a really cool approach that doesn't use logarithms at all. :D
Nice question/ thing to prove. I'll attempt it. BTW, do you know the proof?
Yeah dude! This is part of a collection of proof problems I'm collecting from various olympiads. It's gonna be so boss once I've finished. :D
Finn, im eagerly waiting for you cool approach! :)
@Sagnik Saha – I'll leave it to you guys for a while. :D
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84th SSC CGL level Solution Set, 8th on topic Ratio and Proportion
This is the 84th solution set of 10 practice problem exercise for SSC CGL exam and 8th on topic Ratio and Proportion. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set for extracting maximum benefits from this resource.
If you have not taken the test yet, you can refer to
, and then after taking the test come back to this solution. SSC CGL level Question Set 84, Ratio proportion 8 84th solution set—10 problems for SSC CGL exam: topic Ratio and Proportion 8—Answering time 12 mins Problem 1.
If $A:B:C=2:3:4$, then the ratio $\displaystyle\frac{A}{B}:\displaystyle\frac{B}{C}:\displaystyle\frac{C}{A}$ is equal to,
8 : 9 : 12 8 : 9 : 24 8 : 9 : 16 4 : 9 : 16 Solution 1: Problem analysis and solving in mind conceptually by mathematical reasoning
This is a ratio of ratios problem.
In a ratio, a ratio value represents the relative share of the whole for the corresponding ratio variable. For example, in our problem ratio value 2 for A means ratio variable A contributes to 2 portions or shares to the whole amount of $2+3+4=9$ portions. The 3 portions are contributed by B and the 4 by C. This is one fundamental way of looking at a ratio.
Descriptively stated, the ratio means: for each 2 of A, B is 3 and C is 4.
So when in this ratio we divide A by B, B by C and C by A, we can get corresponding ratio values by dividing 2 by 3, 3 by 4 and 4 by 2, the respective shares or portions of A, B and C by shares of B, C and A.
This makes the changed ratio to,
$\displaystyle\frac{A}{B}:\displaystyle\frac{B}{C}:\displaystyle\frac{C}{A}=\displaystyle\frac{2}{3}:\displaystyle\frac{3}{4}:\displaystyle\frac{4}{2}$.
Converting the fractions to integers by multiplying the values by the LCM of denominators 12 we get,
$\displaystyle\frac{A}{B}:\displaystyle\frac{B}{C}:\displaystyle\frac{C}{A}=8:9:24$.
Mentally this conceptual solution can be reached very quickly.
We will show you a more deductive solution to this interesting problem.
Deductive Solution to Problem 1.
$A:B:C=2:3:4$.
From this we get three ratios,
$A:B=2:3$
$B:C=3:4$, and
$C:A=4:2$.
Taking ratio of the first two,
$\displaystyle\frac{A}{B}:\displaystyle\frac{B}{C}=\frac{8}{9}$
And taking ratio of the second and the third,
$\displaystyle\frac{B}{C}:\displaystyle\frac{C}{A}=\frac{6}{16}=\frac{3}{8}$.
Joining the two by multiplying and dividing the second ratio values by 3, we get,
$\displaystyle\frac{A}{B}:\displaystyle\frac{B}{C}:\displaystyle\frac{C}{A}=8:9:24$.
Answer: Option b: 8 : 9 : 24. Key concepts used: Basic ratio concepts -- Portions cioncepts -- Splitting of a ratio -- Joining a ratio -- Ratio of ratios -- Modifying a ratio term and its corresponding value -- Solving in mind. Problem 2.
If $p:q=r:s=t:u=2:3$, then $(mp+nr+ot):(mq+ns+ou)$ is equal to,
1 : 3 3 : 2 1 : 2 2 : 3 Solution 2: Problem analysis and solving in mind by use of actual ratio values based on HCF reintroduced concept : Ratio merging
By the given ratio, $p$, $r$ and $t$ represents $2x$, $2y$ and $2z$ actual values while, $q$, $s$ and $u$ represents, $3x$, $3y$ and $3z$ actual values, where $x$, $y$ and $z$ are the three HCFs reintroduced for the three ratios respectively.
Substituting these values in the target ratio we have,
$(mp+nr+ot):(mq+ns+ou)$
$=(2xm+2yn+2zo):(3xm+3yn+3zo)$
$=2:3$.
Answer: Option d: 2 : 3. Key concepts used: Basic ratio concepts -- HCF reintroduction technique -- Use of actual ratio values in a new ratio -- -- Ratio merging . Solving in mind
This is case of
ratio merging (not joining). When equivalent ratio variables of three (or more) equal ratios are multiplied by same factors for numerator and denominators, the ratio of the sum of their products remains unchanged. Even if similar operations were performed for the numerator and denominator variables instead of all additions, the ratio would have remained unchanged. Though this is an advanced concept on ratios, it becomes believable by using the basic concept to build the advanced concept. Problem 3.
Find the mean proportionals between 2 and 54.
6 and 18 6 and 9 12 and 18 6 and 12 Solution 3: Problem analysis and solving in mind by mean proportional concept and trial
Mean proportionals of two numbers $a$ and $d$ are such that, $a:b=c:d$.
By trial 6 and 18 satisfies this criterion,
$2:6=18:54=2:3$.
Answer: Option a: 6 and 18. Key concepts used: Basic ratio concept -- Mean proportional concept-- Solving in mind. Problem 4.
The reciprocals of squares of the numbers $1\displaystyle\frac{1}{2}$ and $1\displaystyle\frac{1}{3}$ are in the ratio,
81 : 64 4 : 85 64 : 81 8 : 9 Solution 4: Problem analysis and solving in mind by ratio of two fractions concept
Converting the mixed fractions we get, $\displaystyle\frac{3}{2}$ and $\displaystyle\frac{4}{3}$.
Reciprocals of squares of the two are,
$\displaystyle\frac{4}{9}$ and $\displaystyle\frac{9}{16}$.
Ratio of the two will be,
$\displaystyle\frac{\displaystyle\frac{4}{9}}{\displaystyle\frac{9}{16}}=\frac{64}{81}$.
Answer: Option c: 64 : 81. Key concepts used: Basic ratio concept -- Ratio of fractions by division -- . Solving in mind Problem 5.
If 177 is divided into three parts in the ratio, $\displaystyle\frac{1}{2}:\displaystyle\frac{2}{3}:\displaystyle\frac{4}{5}$, then the second part is,
60 75 45 72 Solution 5: Problem analysis and solving in mind using HCF reintroduction technique
Converting the fraction ratio to integer ratio by multiplying by the LCM 30 of the denominators we get,
$15:20:24$.
Total number of portions is then,
$15+20+24=59$, and the value of each portion,
$\displaystyle\frac{177}{59}=3$.
So the value of the second part, that is, 20 portions is 60.
Answer: Option a: 60. Key concept used: Basic ratio concept --- Amount division in a ratio -- Fraction ratio conversion -- Portions concept -- Solving in mind. Problem 6.
If the ratio of two numbers is 1 : 5 and their product is 320, then the difference between squares of these two numbers is,
1536 1024 1435 1256 Solution 6: Problem analysis and solving in mind by HCF reintroduction technique
Assuming $x$ to be the cancelled out HCF, the two numbers are, $x$ and $5x$ respectively and their product,
$5x^2=320$,
Or, $x^2=64$
Difference of squares of the two numbers is,
$25x^2-x^2=24x^2=24\times{64}=1536$.
To speed up we didn't evaluate the numbers.
Answer: Option a: 1536. Key concepts used: Basic ratio concepts -- HCF reintroduction technique -- Speed up technique -- Solving in mind. Problem 7.
If a reduction in number of workers in a factory is in the ratio 15 : 11 and an increment in their wages in the ratio 22 : 25, then the ratio by which the total wage of the workers is decreased is,
3 : 7 3 : 5 5 : 6 6 : 5 Solution 7: Problem analysis and solving in mind by product of ratios concept
Total wage is the product of number of workers and wage per worker. Assuming $x$ and $y$ to be the cancelled out HCFs of the two ratios respectively, we get the ratio of original total wage to the changed total wage as,
$\displaystyle\frac{15xy\times{22}}{11xy\times{25}}=\frac{6}{5}$.
Before change, actual number of workers and wage per worker were $15x$ and $22y$, and after change these became, $11x$ and $25y$. Total wage in both cases was the product of the two. This is why we could take product of the ratios.
Answer: Option Key concepts used: Basic ratio concepts-- Product of ratios -- Total wage as a product of number of workers and wage per worker -- HCF reintroduction technique -- Solving in mind. Problem 8.
Four numbers are in the ratio 1 : 2 : 3 : 4. Their sum is 16. The sum of the first and the fourth number is equal to,
5 80 8 10 Solution 8: Problem analysis and solving in mind portions concept
Total number of portions is 10 of value 16. So each portion value is 1.6 and the sum of the first and fourth numbers is, 5 portions of value, 8.
Answer: Option c: 8. Key concepts used: -- Basic ratio concepts Portions concept -- Solving in mind. Problem 9.
A sum of Rs.730 is divided among A, B and C in such a way that if A gets Rs.3, then B gets Rs.4 and if B gets Rs.3.50 then C gets Rs.3. The share of B exceeds that of C by,
Rs.30 Rs.210 Rs.40 Rs.70 Solution 9: Problem analysis and solving in mind by fraction ratio conversion, ratio joining and portions concept
Assuming $a$, $b$ and $c$ to be the amount of shares of A, B and C,
$a:b=3:4$, and $b:c=3.50:3$.
Converting the second ratio to ratio of integers,
$b:c=7:6$.
We have to join the two ratios to the ratio of $a:b:c$ for which we have to equalize the middle ratio value of $b$ to the LCM 28.
Accordingly converting the first ratio we get,
$a:b=21:28$, and the second ratio to,
$b:c=28:24$.
Now joining the two ratios,
$a:b:c=21:28:24$.
Total number of portions is,
$21+28+24=73$.
As total amount is Rs.730, value of each portion is Rs.10.
Share of B exceeds that of C by 4 portions, that is by Rs.40.
Answer: Option c: Rs.40. Key concepts used: Basic ratio concepts -- Fraction ratio conversion -- Ratio joining -- Portions concept -- Amount division in a ratio -- . Solving in mind Problem 10.
If 378 coins consist of 1 rupee, 50 paise and 25 paise coins whose values are in the ratio of 13 : 11 : 7, the number of 50 paise coins will be,
133 132 136 128 Solution 10: Problem analysis and solving in mind by Ratio value unit conversion
The ratio of values of 1 rupee, 50 paise and 25 paise coins is,
$13:11:7$.
Actual values being product of each with cancelled out HCF, say, $x$, we can get the ratio of their numbers by multiplying each ratio value by the corresponding number of coins making up a rupee (the original unit), that is, by 1, 2 and 4, getting,
$13:22:28$.
Total portions is, $13+22+28=63$ and total portion value, 378. So value of each portion is,
$\displaystyle\frac{378}{63}=6$.
Number of 50 paise coins is then, $22\times{6}=132$.
Effectively we have changed the given ratio value unit of rupee to new ratio value unit of number of coins by multiplying each ratio value with number of coins equivalent to 1 unit of rupee.
Answer: Option b: 132. Key concepts used: Basic ratio concept -- -- Money value ratio to number of coins ratio -- Ratio value unit conversion Solving in mind .
Overall, all the problems could be solved in mind using varieties of concepts and techniques.
For solving these 10 problems, the concepts and techniques used were—basic and rich ratio concepts, HCF reintroduction technique, ratio of fractions by division, division of ratios, fraction ratio conversion, total wage as a product of number of workers and the wage per worker, speed up techniques, product of ratios concept, splitting of ratios, ratio joining, portions concept, ratio merging, mean proportional, amount division in a ratio, money value ratio to number of coins ratio conversion, and ratio value unit conversion. The problems needed an wide array of basic and rich concepts for quick solution.
Just remember, understanding and applying basic and rich concepts should enable you to solve any such problem easily under a minute with no dependence on varieties of formulas.
Resources that should be useful for you or 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests to access all the valuable student resources that we have created specifically for SSC CGL, but section on SSC CGL generally for any hard MCQ test. Concept Tutorials on related topics Efficient solution techniques on related topics SSC CGL level solved question sets on mixture or alligation SSC CGL Tier II level solved question sets on mixture or alligation SSC CGL Tier II level question and solution sets on Ratio and Proportion Other SSC CGL question and solution sets on Ratio and Proportion and Percentage SSC CGL level Solution Set 84, Ratio proportion 8 If you like, you may to get latest content from this place. subscribe |
I'm sorry to bother you with this question, but somehow I fail to find the corresponding passage?
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
The scale-change theorem could be better known as the scaling property or scaling formula, a property of the continuous Fourier transform. It is one avatar of the "invariance" properties ("scale invariance") of some kernel transformations.
It tells you about some proto-version of the Gabor-Weyl-Heisenberg limit: if you compress a signal along the time axis, it will expand or dilate in frequency, and
vice-versa. This is the main duality aspect of the time and frequency variables.
More precisely, if $s(t)\mapsto S(f)$, then $\forall \alpha \neq 0$, and $ S(f)$ is the Fourier transform, then for the compressed signal:
$$ s(\alpha t) \mapsto \frac{1}{|\alpha|} S \left( \frac{f}{|\alpha|}\right) $$
For discrete signals $s[n]$, not all real $\alpha$s dilations make sense in general, because $\alpha n$ may not be a valid integer index.
Discrete equivalents exist for the time reversal ($\alpha = -1$) and integers or inverses of integers ($\alpha = k$ or $\alpha = 1/k$, $k\in \mathbb{Z}$), and fractions in general, somehow corresponding to downsampling and upsampling, addressed in Chapter 4.6, Changing the sampling rate using discrete-time processing, page 167 sq., with a related lecture at Rice.
A recent reference: Generalized Rational Sampling Rate Conversion Polyphase FIR Filter, IEEE Signal Processing Letters, Nov. 2017. For a more historical one, On upsampling, downsampling, and rational sampling rate filter banks, R. Gopinath, 1994. |
1)
---
Why the proof fails
Let $ \{ (pi)_j \}, \{ p_n \} \subseteq \Re; p \in \Re, \forall i,j,n $
And let $ (p1)_j \rightarrow p_1, (p2)_j \rightarrow p_2, (pi)_j \rightarrow p_i $ as $ j \rightarrow \infty $, and $ p_n \rightarrow p $
We show that $ \{ (pi)_j \} $ converges.
Fix $ \epsilon > 0 $, then $ \exists N_1, N_2 \ni \forall n \geq N_1, N_2 $:
Stop here, I'd need to choose more than one $ N_1 $, one for each sequence for what I'm trying to do, and I'm not guaranteed that the $ \sup N_i $ will be finite.
---
From this I'll use the following counterexample:
Let $ p_n = 0 \forall n $, and $ (pi)_j = \{ i, i-1, i-2, \cdots, 1, 0, 0, 0, \cdots \} $
2) Ditto with series.
The partial sums of series are sequences, so the same result should hold.
So, as a counterexample, let $ \Sigma_{j=1}^{\infty} (ai)_j = i + (-1) + (-1) + \cdots $ i times $ \cdots + (-1) + 0 + 0 + \cdots $ And note that the partial sums are like the $ (pi)_j $ of the problem above, and therefore that this counterexample should follow in the same way. Actually it holds for the series... part of the hypothesis was that the $ a_{i,j}\geq 0 $. Just use comparison with the $ a_i $ for convergence of the $ a_{i,i} $ series.
-Coach |
Can someone help me with this question, and maybe add some explanation? I've looked at the other answers on this website about Quadratic questions, but I just don't quite understand it. I've also looked on Wikipedia, but it doesn't help either. Thanks!
In general if $\alpha$ and $\beta$ are the roots of equation $ax^2+bx+c=0$, then,
Sum of the roots, $\alpha+\beta=-\dfrac{\text{coefficient of }x}{\text{coefficient of }x^2}=-\dfrac{b}{a}$, and
Multiplication of roots, $\alpha\beta=\dfrac{\text{constant term}}{\text{coefficient of }x^2}=\dfrac{c}{a}$.
Similarly as $\alpha$ and $\beta$ are the roots of equation $x^2-10x+10=0$
so, $\alpha+\beta=10$ and, $\alpha\beta=10$.
now, \begin{align*} (\alpha-\beta)^2&=(\alpha+\beta)^2-4\alpha\beta\\ &=10^2-4\times10\\ &=100-40\\ &=60 \end{align*}
Hint: $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$
With the usual formula for quadratic equations you get that the solutions are
$$5 \pm \sqrt{15}.$$
Quadratic equations may have $2$,$1$ or none real roots.It depends of the value of discriminant(Δ). If Δ$>0$ it has $2$ roots, if Δ$=0$ it has $1$ and if it is smaller than $0$ it does not have any. The formula for Δ $= b^2-4ac$, where $a$ is the coefficient of $x^2$, $b$ the coefficient of $x$ and $c$ the constant. Then i hope, you know the formula for the roots. Then you let $a$ equal the biggest of these roots and $b$ the other one and you find the value of $(a-b)^2$ |
A trick for triggering expansion is using
\romannumeral:
When due to
\romannumeral (La)TeX does gather together a sequence of digits trailed by a space as the number which it has to convert, expandable tokens get expanded.
When in the end a number is gathered together which is not positive, as the result of the conversion (La)TeX will not deliver any token at all.
Thus one can nicely (ab?)use
\romannumeral for triggering a lot of expansion-work and flipping-arguments-around-work as long as one ensures that in the end
\romannumeral will not find a positive number.
Here is a variant of egreg's answer which does with
\romannumeral and
\exchange instead of
\xdef and
\unexpanded.
\documentclass{article}
\usepackage{environ}
\newcommand\exchange[2]{#2#1}
\NewEnviron{exercise}{%
\expandafter\gdef\expandafter\savedexercises\expandafter{%
\romannumeral0\expandafter\exchange\expandafter{\BODY}{%
\exchange{ }{\expandafter}\savedexercises
\begin{printedexercise}%
}%
\end{printedexercise}%
}%
}
\newcommand{\printexercises}{%
\savedexercises
\gdef\savedexercises{}%
}
\newcommand{\savedexercises}{}
\newtheorem{printedexercise}{Exercise}
\begin{document}
Here we talk about addition and show that $1+1=2$.
\begin{exercise}
Compute $1+2$
\end{exercise}
Here we talk about integrals.
\begin{exercise}
Compute the following integrals:
\begin{itemize}
\item $\displaystyle\int_0^x e^{-t^2}\,dt$
\item $\displaystyle\int_1^x \frac{e^t}{t}\,dt$, for $t>0$.
\end{itemize}
\end{exercise}
Now we can print the exercises.
\printexercises
\end{document}
If you wish to wrap the name of the macro that is to be defined into
\csname..
\endcsname, i.e., if you wish to use
\csname savedexercises\endcsname rather than
\savedexercises, you can take advantage of the fact that (La)TeX does expand expandable tokens while due to
\csname gathering the name of a control sequence token and hereby searching for the matching
\endcsname:
\documentclass{article}
\usepackage{environ}
\newcommand\exchange[2]{#2#1}
\NewEnviron{exercise}{%
\expandafter\gdef\csname savedexercises\expandafter\endcsname\expandafter{%
\romannumeral0\expandafter\exchange\expandafter{\BODY}{%
\exchange{ }{\expandafter\expandafter\expandafter}\csname savedexercises\endcsname
\begin{printedexercise}%
}%
\end{printedexercise}%
}%
}
\newcommand{\printexercises}{%
\csname savedexercises\endcsname
\expandafter\gdef\csname savedexercises\endcsname{}%
}
\expandafter\newcommand\expandafter{\csname savedexercises\endcsname}{}
\newtheorem{printedexercise}{Exercise}
\begin{document}
Here we talk about addition and show that $1+1=2$.
\begin{exercise}
Compute $1+2$
\end{exercise}
Here we talk about integrals.
\begin{exercise}
Compute the following integrals:
\begin{itemize}
\item $\displaystyle\int_0^x e^{-t^2}\,dt$
\item $\displaystyle\int_1^x \frac{e^t}{t}\,dt$, for $t>0$.
\end{itemize}
\end{exercise}
Now we can print the exercises.
\printexercises
\end{document}
Be aware that with the approaches presented by now you cannot use
\printexercises for having exercises occur in arbitrary places. You can have exercises occur only in places of the document which in the source correspond to places behind the exercise-environments.
Perhaps an environment which does read its content under verbatim-catcode-régime for unexpanded-writing it to .aux-file in a way where from the .aux-file it gets read back under verbatim-catcode-régime also for defining a macro where
\scantokens will be applied to, and thus some sort of re-implementation of the
\label-
\ref-mechanism or of the
\tableofcontents-mechanism for verbatimized stuff might make it possible to make exercises printable throughout the entire document.
Implementing such a mechanism might be a nice challenge. But before taking that into consideration at all, exact information is needed on the intended usage and the desired "user-interface", i.e., what additional things you wish to be able to specify in which ways, etc, ... |
51 0
Suppose I want to decompose [tex]A = \left(\begin{array}{cc}4&4\\-3&3\end{array}\right)[/tex].
[tex]A = U \Sigma V^T => A^T A = V \Sigma^2 V^T[/tex] and [tex]A A^T = U \Sigma^2 U^T[/tex]
So 2 independent eigenvectors of [tex]A^T A[/tex] are a basis for the row space of A and 2 independent eigenvectors of [tex]A A^T[/tex] are a basis for the column space of A.
If I pick [tex]v_{1} = \left(\begin{array}{cc}1/\sqrt{2}\\1/\sqrt{2}\end{array}\right)[/tex] [tex]v_{2} = \left(\begin{array}{cc}1/\sqrt{2}\\-1/\sqrt{2}\end{array}\right)[/tex]
and
[tex]u_{1} = \left(\begin{array}{cc}1\\0\end{array}\right)[/tex] [tex]u_{2} = \left(\begin{array}{cc}0\\1\end{array}\right)[/tex]
as these bases, respectively, I get into trouble. Matrix A will not equal the singular value decomposition, there will be a sign error even though these vectors truly are correct eigenvectors.
For those who know him, this is an example that prof. Strang gave during one of his video lectures but he didn't quite see where the error came from. And since he never gave a solution, I hope I can find one here.
I started thinking about this and I noticed that these 2 bases don't have the same orientation. So I tried to find a new basis for one of the 2 spaces, with respect to the eigenvalues: just change one vector into the opposite direction. But the problem now is, this only works half the time. So for the 4 new bases I found (with respect to the eigenvalues) only 2 work out the way it's supposed to.
So if I change [tex]u_{2}[/tex] to [tex]\left(\begin{array}{cc}0\\-1\end{array}\right)[/tex] everything works out. But if I change [tex]u_{1}[/tex] to [tex]\left(\begin{array}{cc}-1\\0\end{array}\right)[/tex] things get into trouble again. Similar for [tex]v_{1}[/tex] and [tex]v_{2}[/tex].
Can anyone tell me what is going on here?
Phew, that was a lot of typing. :tongue2: I hope I didn't make any errors, but my main point is there and that's what counts.
Thanks in advance.
[tex]A = U \Sigma V^T => A^T A = V \Sigma^2 V^T[/tex] and [tex]A A^T = U \Sigma^2 U^T[/tex]
So 2 independent eigenvectors of [tex]A^T A[/tex] are a basis for the row space of A and 2 independent eigenvectors of [tex]A A^T[/tex] are a basis for the column space of A.
If I pick
[tex]v_{1} = \left(\begin{array}{cc}1/\sqrt{2}\\1/\sqrt{2}\end{array}\right)[/tex] [tex]v_{2} = \left(\begin{array}{cc}1/\sqrt{2}\\-1/\sqrt{2}\end{array}\right)[/tex]
and
[tex]u_{1} = \left(\begin{array}{cc}1\\0\end{array}\right)[/tex] [tex]u_{2} = \left(\begin{array}{cc}0\\1\end{array}\right)[/tex]
as these bases, respectively, I get into trouble. Matrix A will not equal the singular value decomposition, there will be a sign error even though these vectors truly are correct eigenvectors.
For those who know him, this is an example that prof. Strang gave during one of his video lectures but he didn't quite see where the error came from. And since he never gave a solution, I hope I can find one here.
I started thinking about this and I noticed that these 2 bases don't have the same orientation. So I tried to find a new basis for one of the 2 spaces, with respect to the eigenvalues: just change one vector into the opposite direction. But the problem now is, this only works half the time. So for the 4 new bases I found (with respect to the eigenvalues) only 2 work out the way it's supposed to.
So if I change [tex]u_{2}[/tex] to [tex]\left(\begin{array}{cc}0\\-1\end{array}\right)[/tex] everything works out. But if I change [tex]u_{1}[/tex] to [tex]\left(\begin{array}{cc}-1\\0\end{array}\right)[/tex] things get into trouble again.
Similar for [tex]v_{1}[/tex] and [tex]v_{2}[/tex].
Can anyone tell me what is going on here?
Phew, that was a lot of typing. :tongue2: I hope I didn't make any errors, but my main point is there and that's what counts.
Thanks in advance. |
The structure of the natural numbers—0, 1, 2, 3, and on to infinity—makes possible a powerful proof technique known as
induction or mathematical induction. The idea behind induction is simple. Let \(P\) be a one-place predicate whose domain of discourse includes the natural numbers. Suppose that we can prove that \(P(0)\) is true. Suppose that we can also prove the statements \(P(0) → P(1), P(1) → P(2), P(2) → P(3)\), and so on. The principal of mathematical induction is the observation that we can then conclude that \(P(n)\) is true for all natural numbers \(n\). This should be clear. Since \(P(0) and P(0) → P(1)\) are true, we can apply the rule of modus ponens to conclude that \(P(1)\) is true. Then, since \(P(1)\) and \(P(1) → P(2)\) are true, we can conclude by modus ponens that \(P(2)\) is true. From \(P(2)\) and \(P (2) → P (3)\), we conclude that \(P (3)\) is true. For any given \(n\) in the set \(N\), we can continue this chain of deduction for \(n\) steps to prove that \(P(n)\) is true.
When applying induction, we don’t actually prove each of the implications \(P (0) → P (1), P (1) → P (2)\), and so on, individually. That would require an infinite amount of work. The whole point of induction is to avoid any infinitely long process. Instead, we prove \(∀k (P (k) → P (k + 1))\) (where the domain of discourse for the predicate \(P\) is \(\mathbb{N}\)). The statement \(∀k (P (k) → P (k + 1))\) summarizes all the infinitely many implications in a single statement. Stated formally, the principle of mathematical induction says that if we can prove the statement \(P(0) ∧ (\forall k (P(k) \to P(k+1))\), then we can deduce that \(∀n P (n)\) (again, with \(N\) as the domain of discourse).
It should be intuitively clear that the principle of induction is valid. It follows from the fact that the list 0, 1, 2, 3, . . . , if extended long enough, will eventually include any given natural number. If we start from \(P (0)\) and take enough steps of the form \(P(k) → P(k + 1)\), we can get \(P(n)\) for any given natural number \(n\). However, whenever we deal with infinity, we are courting the possibility of paradox. We will prove the principle of induction rigorously in the next chapter (see Theorem 2.3), but for now we just state it as a theorem:
Theorem 1.7
Let P be a one-place predicate whose domain of discourse includes the natural numbers. Suppose that \(P (0) ∧ (∀k ∈ N (P (k) → P (k + 1)))\) . Then \(P(n)\) is true for all natural numbers \(n\). (That is, the statement \(∀nP(n)\) is true, where the domain of discourse for \(P\) is the set of natural numbers.)
Mathematical induction can be applied in many situations: you can prove things about strings of characters by doing induction on the length of the string, things about graphs by doing induction on the number of nodes in the graph, things about grammars by doing induction on the number of productions in the grammar, and so on. We’ll be looking at applications of induction for the rest of this chapter, and throughout the remainder of the text. Although proofs by induction can be very different from one another, they all follow just a few basic structures. A proof based on the preceding theorem always has two parts. First, \(P(0)\) is proved. This is called the base case of the induction. Then the statement \(∀k (P (k) → P (k + 1))\) is proved. This statement can be proved by letting \(k\) be an arbitrary element of \(N\) and proving \(P (k) → P (k + 1)\). This in turn can be proved by assuming that \(P (k)\) is true and proving that the truth of \(P (k + 1)\) follows from that assumption. This case is called the inductive case, and P(k) is called the inductive hypothesis or the induction hypothesis. Note that the base case is just as important as the inductive case. By itself, the truth of the statement \(∀k (P (k) → P (k + 1))\) says nothing at all about the truth of any of the individual statements \(P (n)\). The chain of implications \(P (0) → P (1),P(1) → P(2), ..., P(n − 1) → P(n)\) says nothing about \(P(n)\) unless the chain is anchored at the other end by the truth of \(P (0)\). Let’s look at a few examples
Theorem 1.8
The number \(2^{2n} −1\) is divisible by 3 for all natural numbers \(n\).
Proof. Here, P (n) is the statement that \(2^{2n} − 1\) is divisible by 3. Base case: When \(n = 0, 2^{2n} − 1 = 2^{0} − 1 = 1 − 1 = 0\) and 0 is divisible by 3 (since 0 = 3 · 0.) Therefore the statement holds when n = 0. Inductive case: We want to show that if the statement is true for n = k(where k is an arbitrary natural number), then it is true for \(n = k + 1\) also. That is, we must prove the implication \(P (k) → P (k + 1)\). So we assume \(P (k)\), that is, we assume that \(2^{2k}\) is divisible by 3. This means that \(2^{2k} −1 = 3m\) for some integer \(m\). We want to prove \(P(k+1)\), that is, that \(2^{2(k+1)} − 1\) is also divisible by 3:
\(2^{2(k+1)} -1 = 2^{2k +2} -1\) \(= 2^{2k} \cdot 2^{2} -1\) property of exponents \(= 4 \cdot 2^{2k}-1\) \(= 4\cdot 2^{2k} -4+4-1\) \(= 4(2^{2k} -1)+3\) algebra \(= 4(3m) +3\) the inductive hypothesis \(= 3(4m +1)\) algebra
and from the last line we see that \(2^{2k}+1\) is in fact divisible by 3. (The third step—subtracting and adding 4—was done to enable us to use our inductive hypothesis.)
Altogether, we have proved that \(P(0)\) holds and that, for all \(k\), \(P(k) →P(k + 1)\) is true. Therefore, by the principle of induction, \(P(n)\) is true for all \(n\) in \(N\), i.e. \(2^{2n} − 1\) is divisible by 3 for all \(n\) in \(\mathbb{N}\).
The principal of mathematical induction gives a method for proving \(P(n)\) for all \(n\) in the set \(\mathbb{N}\). It should be clear that if \(M\) is any natural number, a similar method can be used to show that \(P(n)\) is true for all natural numbers \(n\) that satisfy \(n ≥ M\). Just start the induction with a base case of \(n=M\) instead of with a base case of \(n=0\). I leave the proof of this extension of the principle of induction as an exercise. We can use the extended principle of induction to prove a result that was first mentioned in Section 1.1.
theorem 1.9
Suppose that a compound proposition contains exactly \(n\) propositional variables, where \(n ≥ 1\). Then there are exactly \(2^{n}\) different ways of assigning truth values to the \(n\) variables.
Proof. Let \(P(n)\) be the statement “There are exactly \(2^{n}\) different ways of assigning truth values to \(n\) propositional variables.” We will use induction to prove the \(P(n)\) is true for all \(n ≥ 1\). Base case: First, we prove the statement \(P(1)\). If there is exactly one variable, then there are exactly two ways of assigning a truth value to that variable. Namely, the variable can be either true or false. Since \(2 = 2^{1}\), \(P(1)\) is true. Inductive case: Suppose that \(P(k)\) is already known to be true. We want to prove that, under this assumption, \(P(k + 1)\) is also true. Suppose that \(p_{1}, p_{2}, . . . , p_{k+1}\) are \(k + 1\) propositional variables. Since we are assuming that \(P(k)\) is true, we know that there are \(2^{k}\) ways of assigning truth values to \(p_{1}, p_{2}, ..., p_{k}\). But each assignment of truth values to \(p_{1}, p_{2}, ..., p_{k}\) can be extended to the complete list \(p_{1}, p_{2}, . . . , p_{k}, p_{k+1}\) in two ways. Namely, \(p_{k+1}\) can be assigned the value true or the value false. It follows that there are \(2·2^{k}\) ways of assigning truth values to \(p_{1}, p_{2}, . . . , p_{k+1}\). Since \(2 · 2_{k} = 2_{k+1}\), this finishes the proof.
The sum of an arbitrary number of terms is written using the symbol \(\sum\). (This symbol is the Greek letter sigma, which is equivalent to the Latin letter \(S\) and stands for “sum.”) Thus, we have
\[\sum_{i =1}^{5} {i^2 = 1^2 +2^2 + 3^2 + 4^2 +5^2} \nonumber\]
\[\sum_{k=3}^{7}{a_k = a_3 + a_4 + a_5 + a_6 + a_7}\nonumber\]
\[\sum_{n=0}^{N}{\frac{1}{n+1} = \frac{1}{0+1} + \frac{1}{1+1} + \frac{1}{2+1} + ... + \frac{1}{N+1}}\nonumber\]
This notation for a sum, using the \(\sum\) operator, is called
summation notation. A similar notation for products uses the symbol \(\Pi\). (This is the Greek letter pi, which is equivalent to the Latin letter P and stands for “product.”) For example,
\[\Pi_{k=2}^{5}{(3k +2) = (3·2 + 2)(3·3+ 2)(3·4 + 2)(3·5 + 2)}\nonumber\]
\[\Pi_{i=1}^{n}{\frac{1}{i} = \frac{1}{1} · \frac{1}{2} ··· \frac{1}{n}·}\nonumber\]
Induction can be used to prove many formulas that use these notations. Here are two examples:
Theorem 1.10
\(\sum_{i=1}^{n}{i = \frac{n(n+1)}{2}}\) for any integer n greater than zero.
Proof. Let \(P(n)\) be the statement \(\sum_{i=1}^{n}{i = \frac{n(n+1)}{2}} \) We use induction to show that \(P(n)\) is true for all \(n \geq 1\). Base case: Consider the case \(n =1\). \(P(1)\) is the statement that \(\sum_{i=1}^{1}{i = \frac{1(1+1)}{2}}\). Since \(\sum_{i=1}^{1}i = 1\) and \(\frac{1(1+1)}{2} = 1\), \(P(n)\) is true. Inductive case: Let \(k > 1\) be arbitrary, and assume that \(P(x)\) is true. We want to show that \(P(k+1)\) is true. \(P(k+1)\) is the statement \(\sum_{i=1}^{k+1}{i = \frac{(k+1)(k+2)}{2}}\). But
\[\sum_{i =1}^{k+1}{i = (\sum_{i=1}^{k}{i}) +(k+1)}\nonumber\]
\[=\frac{k(k+1)}{2} + (k+1)\nonumber\]
\[=\frac{k(k+1)}{2} + \frac{2(k+1)}{2}\nonumber\]
\[=\frac{k(k+1)+2(k+1)}{2}\nonumber\]
\[=\frac{(k+2)(k+1)}{2}\nonumber\]
\[=\frac{(k+1)(k+2)}{2}\nonumber\]
Which is what we wanted to show. This computation completes the induction.
Theorem 1.11
\(\sum_{i=1}^{2}i2^{i-1} = (n-1)·2^{n} +1\) for any natural number \(n > 0\).
Proof. Let P(n) be the statement \(\sum_{i=1}^{2}i2^{i-1} = (n-1)·2^{n} +1\). We use induction to show that \(P(n)\) is true for all \(n > 0\) Base case: Consider the case \(n = 1\). \(P(1)\) is the statement that \(\sum_{i=1}^{1} = (1-1) · 2^{1} +1 \)/ Since each side of this equation is equal to one, this is true. Inductive case: Let \(k > 1\) be arbitrary, and assume that \(P(k)\) is true. We want to show that \(P(k+1)\) is true. \(P(k+1)\) is the statement \(\sum_{i=1}^{k+1} i2^{i-1} = ((k+1)-1)· 2^{k+1} +1\). But, we can compute that
\[\sum_{i=1}^{k+1} i = (\sum_{i=1}^{k} i) + (k+1)\nonumber\]
\[=\frac{k(k+1)}{2} + (k+1)\nonumber\]
\[=\frac{k(k+1)}{2} + \frac{2(k+1)}{2}\nonumber\]
\[=\frac{k(k+1)+2(k+1)}{2}\nonumber\]
\[=\frac{(k+1)(k+1)}{2}\nonumber\]
\[=\frac{(k+1)(k+2)}{2}\nonumber\]
which is what we wanted to show. This computation completes the induction.
Theorem 1.11
\(\sum_{i=1}^{n} i2^{i-1} = (n-1)·2^{n} +1\) for any natural number \(n>0\).
Proof. Let \(P(n)\) be the statement \(\sum_{i=1}^{n}i2^{i-1} = (n-1)·2^{n} +1\). We use induction to show that \(P(n)\) is true for all \(n > 0\). Base case: Consider the case n = 1. P(1) is the statement that \(\sum_{i=1}^{1}i2^{i-1} = (1-1)·2^{1} +1\). Since each side of this equation is equal to one, this is true. Inductive case: Let \(k > 1\) be arbitrary, and assume that \(P(k)\) is true. We want to show that \(P(k+1)\) is true. \(P(k+1)\) is the statement \(\sum_{i=1}^{k+1} i2^{i-1} = ((k+1)-1)·2k+1 +1\). But, we can compute that
\[\sum_{i=1}^{k+1}i2^{i-1} = (\sum_{i=1}^{k}i2^{i-1}) + (k+1)2^{(k+1)-1}\nonumber\]
\[=((k-1)· 2^{k}+1) +(k+1)2^{k}\nonumber\]
\[=((k-1)+(k+1))2^{k}+1\nonumber\]
\[=(k· 2)· 2^{k} +1\nonumber\]
\[=k2^{k+1}+1\nonumber\]
which is what we wanted to show. This completes the induction.
For example, these theorems show that \(\sum_{i=1}^{100} i = 1+2+3+4+···+100 = \frac{100(100+1)}{2} = 5050\) and that \(1·2^{0}+2·2^{1}+3·2^{2}+4·2^{3}+5·2^{4} = (5−1)2^{5}+1 = 129\), as well as infinitely many other such sums.
There is a second form of the principle of mathematical induction which is useful in some cases. To apply the first form of induction, we assume \(P(k)\) for an arbitrary natural number \(k\) and show that \(P(k + 1)\) follows from that assumption. In the second form of induction, the assumption is that \(P(x)\) holds for all \(x\) between 0 and \(k\) inclusive, and we show that \(P(k + 1)\) follows from this. This gives us a lot more to work with when deducing \(P(k + 1)\). We will need this second form of induction in the next two sections. A proof will be given in the next chapter.
Theorem 1.12
Let \(P\) be a one-place predicate whose domain of discourse includes the natural numbers. Suppose that \(P(0)\) is true and that \((P (0) ∧ P (1) ∧ · · · ∧ P (k)) → P (k + 1)\) is true for each natural number \(k ≥ 0\). Then \(P(n)\) is true for every natural number n.
For example, we can use this theorem to prove that every integer greater than one can be written as a product of prime numbers (where a number that is itself prime is considered to be a product of one prime number). The proof illustrates an important point about applications of this theorem:
When proving \(P(k + 1)\), you don’t necessarily have to use the assumptions that \(P(0), P(1), ..., and P(k)\) are true. If \(P(k + 1)\) is proved by any means—possibly including the assumptions—then the statement \((P(0) ∧P(1) ∧ ··· ∧ P(k)) → P(k + 1)\) has been shown to be true. It follows from this observation that several numbers, not just zero, can be “base cases” in the sense that \(P(x + 1)\) can be proved independently of \(P(0)\) through \(P(x)\). In this sense, 0, 1, and every prime number are base cases in the following theorem.
Theorem 1.13
Every natural number greater than one can be written as a product of prime numbers.
Proof. Let \(P(n)\) be the statement “if \(n > 1\), then n can be written as a product of prime numbers.” We will prove that \(P(n)\) is true for all \(n\) by applying the second form of the principle of induction.
Note that \(P(0)\) and \(P(1)\) are both automatically true, since \(n = 0\) and \(n = 1\) do not satisfy the condition that \(n > 1\), and \(P(2)\) is true since 2 is the product of the single prime number 2. Suppose that \(k\) is an arbitrary natural number with \(k > 1\), and suppose that \(P(0), P(1), ..., P(k)\) are already known to be true; we want to show that \(P(k + 1)\) is true. In the case where \(k+1\) is a prime number, then \(k+1\) is a product of one prime number, so \(P(k + 1)\) is true.
Consider the case where \(k + 1\) is not prime. Then, according to the definition of prime number, it is possible to write \(k + 1 = ab\) where \(a\) and \(b\) are numbers in the range from 2 to \(k\) inclusive. Since \(P(0)\) through \(P(k)\) are known to be true, \(a\)and \(b\) can each be written as a product of prime numbers. Since \(k + 1 = ab\), \(k + 1\) can also be written as a product of prime numbers. We have shown that \(P(k+1)\) follows from \(P(0)∧P(1)∧···∧P(k)\), and this completes the induction.
Exercises
1. Use induction to prove that \(n_{3} + 3n_{2} + 2n\) is divisible by 3 for all natural numbers \(n\).
2. Use induction to prove that \(\sum_{i=0}^{n} r^{i} = \frac{1-r^{n+1}}{1-r}\) fora any natural number \(n\) and for any real number \(r\) such that \(r \ne 1\).
3.Use induction to prove that for any natural number n, \(\sum_{i=0}^{n}\frac{1}{2^{i}} = 2 - \frac{1}{2^n}\). In addition to proving this by induction,show that it follows as a corollary of Exercise 2.
4. Use induction to prove that for any natural number n, \(\sum_{i=1}^{n}2^{i} = 2^{n+1} -1\). In addition to proving this by induction, show that it follows as a corollary of Exercise 2.
5.Use induction to prove that for any positive integer n, \(\sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}\).
6.Use induction to prove that for any positive integer n, \(\sum_{i=1}^{n} (2i-1) = n^{2}\).
7.Evaluate the following sums, using results proved in this section and in the previous exercises:
a)\(1+3+5+7+9+11+13+15+17+19\) b)\(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}+\frac{1}{3^6}\) c)\(50+51+52+53+···+99+100\) d)\(1+4+9+16+25+36+49+81+100\) e)\(\frac{1}{2^2} + \frac{1}{2^3} +...+\frac{1}{2^{99}}\)
8.Write each of the sums in the preceding problem using summation notation.
9.Rewrite the proofs of Theorem 1.10 and Theorem 1.11 without using summa- tion notation.
10.Use induction to prove the following generalized distributive laws for propositional logic: For any natural number \(n > 1\) and any propositions \(q, p_{1}, p_{2},..., p_{n},\)
a)\(q∧(p_{1} ∨p_{2} ∨···∨p_{n})=(q∧p_{1})∨(q∧p_{2})∨···∨(q∧p_{n})\)
b)\(q∨(p_{1} ∧p_{2} ∧···∧p_{n})=(q∨p_{1})∧(q∨p_{2})∧···∧(q∨p_{n})\) |
It's a very basic question, but I havn't found the question on google or dsp.stackexchange.com: How does integraion of a signal change the spectrum?
It is the basic property of Fourier Transforms. Given that $g(t)$ and $G(f)$ are Fourier transform pairs i.e.
$g(t) \rightleftharpoons G(f)$ then $\int_{-\infty}^tg(\tau)d\tau \rightleftharpoons \frac{1}{j2\pi f}G(f)$
This assumes that $G(0)=0$. If $G(0)$ is not zeros then the integral of $g(t)$ has a Fourier transform that includes a Dirac delta function. |
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The Marcinkiewicz multipliers are$L^{p}$bounded for$1<p<\infty$on the Heisenberg group$\mathbb{H}^{n}\simeq \mathbb{C}^{n}\times \mathbb{R}$(Müller, Ricci, and Stein). This is surprising in the sense that these multipliers are invariant under a two parameter group of dilations on$\mathbb{C}^{n}\times \mathbb{R}$, while there is no two parameter group of automorphic dilations on$\mathbb{H}^{n}$. The purpose of this paper is to establish a theory of the flag Lipschitz space on the Heisenberg group$\mathbb{H}^{n}\simeq \mathbb{C}^{n}\times \mathbb{R}$that is, in a sense, intermediate between that of the classical Lipschitz space on the Heisenberg group$\mathbb{H}^{n}$and the product Lipschitz space on$\mathbb{C}^{n}\times \mathbb{R}$. We characterize this flag Lipschitz space via the Littlewood–Paley theory and prove that flag singular integral operators, which include the Marcinkiewicz multipliers, are bounded on these flag Lipschitz spaces. |
I came across the following problem:
Show that if $x$ and $y$ are real numbers with $x <y$, then there exists an irrational number $t$ such that $x < t < y$.
We know that $y-x>0$.
By the Archimedean property, there exists a positive integer $n$ such that $n(y-x)>1$ or $1/n < y-x$. There exists an integer $m$ such that $m \leq nx < m+1$ or $\displaystyle \frac{m}{n} \leq x \leq \frac{m+1}{n} < y$.
This is essentially the proof for the denseness of the rationals. Instead of $\large \frac{m+1}{n}$ I need something of the form $\large\frac{\text{irrational}}{n}$. How would I get the numerator? |
At the beginning of this chapter the general idea of a grammar as a set of rewriting or production rules was introduced. For most of the chapter, how- ever, we have restricted our attention to context-free grammars, in which production rules must be of the form \(A \longrightarrow x\) where \(A\) is a non-terminal symbol. In this section, we will consider general grammars, that is, gram- mars in which there is no such restriction on the form of production rules. For a general grammar, a production rule has the form \(u \longrightarrow x,\) where \(u\) is string that can contain both terminal and non-terminal symbols. For convenience, we will assume that u contains at least one non-terminal symbol, although even this restriction could be lifted without changing the class of languages that can be generated by grammars. Note that a context-free grammar is, in fact, an example of a general grammar, since production rules in a general grammar are allowed to be of the form \(A \longrightarrow x .\) They just don’t have to be of this form. I will use the unmodified term grammar to refer to general grammars.2 The definition of grammar is identical to the definition of context-free grammar, except for the form of the production rules:
Definition 4.6.
A
grammar is a 4 -tuple \((V, \Sigma, P, S),\) where:
1. V is a finite set of symbols. The elements of V are the non-terminal symbols of the grammar.
2.\(\Sigma\) is a finite set of symbols such that \(V \cap \Sigma=\emptyset .\) The elements of \(\Sigma\) are the terminal symbols of the grammar. 3. \(P\) is a set of production rules. Each rule is of the form \(u \longrightarrow x\) where \(u\) and \(x\) are strings in \((V \cup \Sigma)^{*}\) and \(u\) contains at least one symbol from V. 4. \(S \in V . S\) is the start symbol of the grammar.
Suppose G is a grammar. Just as in the context-free case, the language generated by \(G\) is denoted by \(L(G)\) and is defined as \(L(G)=\{x \in\) \(\Sigma^{*} | S \Longrightarrow_{G}^{*} x \} .\) That is, a string \(x\) is in \(L(G)\) if and only if \(x\) is a string of terminal symbols and there is a derivation that produces x from the start symbol, S, in one or more steps.
The natural question is whether there are languages that can be generated by general grammars but that cannot be generated by context-free languages. We can answer this question immediately by giving an example of such a language. Let \(L\) be the language \(L=\left\{w \in\{a, b, c\}^{*} | n_{a}(w)=\right.\)\(n_{b}(w)=n_{c}(w) \} .\) We saw at the end of the last section that \(L\) is not context-free. However, L is generated by the following grammar:
\(S \longrightarrow S A B C\)
\(S \longrightarrow \varepsilon\)
\(A B \longrightarrow B A\)
\(B A \longrightarrow A B\)
\(A C \longrightarrow C A\)
\(C A \longrightarrow A C\)
\(B C \longrightarrow C B\)
\(C B \longrightarrow B C\)
\(A \longrightarrow a\)
\(B \longrightarrow b\) \(C \rightarrow c\)
For this grammar, the set of non-terminals is {S, A, B, C} and the set of ter- minal symbols is {a, b, c}. Since both terminals and non-terminal symbols can occur on the left-hand side of a production rule in a general grammar, it is not possible, in general, to determine which symbols are non-terminal and which are terminal just by looking at the list of production rules. However, I will follow the convention that uppercase letters are always non-terminal symbols. With this convention, I can continue to specify a grammar simply by listing production rules.
The first two rules in the above grammar make it possible to produce the strings ε, ABC, ABCABC, ABCABCABC, and so on. Each of these strings contains equal numbers of A’s, B’s, and C’s. The next six rules allow the order of the non-terminal symbols in the string to be changed. They make it possible to arrange the A’s, B’s, and C’s into any arbitrary order. Note that these rules could not occur in a context-free grammar. The last three rules convert the non-terminal symbols A, B, and C into the corresponding terminal symbols a, b, and c. Remember that all the non-terminals must be eliminated in order to produce a string in L(G). Here, for example, is a derivation of the string baabcc using this grammar. In each line, the string that will be replaced on the next line is underlined.
\(S \Longrightarrow \underline{S} A B C\)
\(\Longrightarrow \underline{S} A B C A B C\)
\(\Longrightarrow A B C A B C\)
\(\Longrightarrow B A C A B C\)
\(\Longrightarrow B A A C B C\)
\(\Longrightarrow B A A B C C\)
\(\Longrightarrow b A A B C C\)
\(\Longrightarrow b a A B C C\)
\(\Longrightarrow b a a B C C\)
\(\Longrightarrow b a a b C C\) \(\Longrightarrow b a a b c C\) \(\Longrightarrow b a a b c c\)
We could produce any string in L in a similar way. Of course, this only shows that \(L \subseteq L(G) .\) To show that \(L(G) \subseteq L,\) we can observe that for any string \(w\) such that \(S \Longrightarrow^{*} w, n_{A}(w)+n_{a}(w)=n_{B}(w)+n_{b}(w)=\) \(n_{C}(w)+n_{c}(w) .\) This follows since the rule \(S \Longrightarrow S A B C\) produces strings in which \(n_{A}(w)=n_{B}(w)=n_{C}(w),\) and no other rule changes any of the quantities \(n_{A}(w)+n_{a}(w), n_{B}(w)+n_{b}(w),\) or \(n_{C}(w)+n_{c}(w) .\) After applying these rules to produce a string \(x \in L(G),\) we must have that \(n_{A}(x), n_{B}(x)\) and \(n_{C}(x)\) are zero. The fact that \(n_{a}(x)=n_{c}(x)=n_{c}(x)\) then follows from the fact that \(n_{A}(x)+n_{a}(x)=n_{B}(x)+n_{b}(x)=n_{C}(x)+n_{c}(x) .\) That is, \(x \in L\).
Our first example of a non-context-free language was \(\left\{a^{n} b^{n} c^{n} | n \in \mathbb{N}\right\}\). This language can be generated by a general grammar similar to the previous example. However, it requires some cleverness to force the a’s, b’s, and c’s into the correct order. To do this, instead of allowing A’s, B’s, andC’s to transform themselves spontaneously into a’s, b’s, and c’s, we use additional non-terminal symbols to transform them only after they are in the correct position. Here is a grammar that does this:
\(S \longrightarrow S A B C\)
\(S \longrightarrow X\)
\(B A \longrightarrow A B\)
\(C A \longrightarrow A C\)
\(C B \longrightarrow B C\)
\(X A \longrightarrow a X\)
\(X \rightarrow Y\)
\(Y B \longrightarrow b Y\)
\(Y \longrightarrow Z\)
\(Z C \longrightarrow c Z\)
\(Z \longrightarrow \varepsilon\)
Here, the first two rules produce one of the strings X, XABC, XABCABC,XABCABCABC, and so on. The next three rules allow A’s to move to the left and C’s to move to the right, producing a string of the form XAnBnCn, for some \(n \in \mathbb{N} .\) The rule \(X A \longrightarrow a X\) allows the \(X\) to move through the A’s from left to right, converting A’s to a’s as it goes. After converting theA’s, the X can be transformed into a Y . The Y will then move through theB’s, converting them to b’s. Then, the Y is transformed into a Z, which is responsible for converting C’s to c’s. Finally, an application of the rule \(Z \longrightarrow \varepsilon\) removes the \(Z,\) leaving the string \(a^{n} b^{n} c^{n}\).
Note that if the rule \(X \longrightarrow Y\) is applied before all the \(A^{\prime} s\) have been converted to a’s, then there is no way for the remaining A’s to be converted to a’s or otherwise removed from the string. This means that the derivation has entered a dead end, which can never produce a string that consists of terminal symbols only. The only derivations that can produce strings in the language generated by the grammar are derivations in which the X moves past all the A’s, converting them all to a’s. At this point in the derivation, the string is of the form \(a^{n} X u\) where \(u\) is a string consisting entirely of \(B^{\prime} s\) and \(C\) 's. At this point, the rule \(X \longrightarrow Y\) can be applied, producing the string \(a^{n} Y u .\) Then, if a string of terminal symbols is ever to be produced, the \(Y\) must move past all the \(B^{\prime} s,\) producing the string \(a^{n} b^{n} Y C^{n} .\) You can see that the use of three separate non-terminals, X, Y , and Z, is essential for forcing the symbols in \(a^{n} b^{n} c^{n}\) into the correct order.
For one more example, consider the language \(\left\{a^{n^{2}} | n \in \mathbb{N}\right\} .\) Like the other languages we have considered in this section, this language is not context-free. However, it can be generated by a grammar. Consider the grammar
\(S \longrightarrow D T E\)
\(T \longrightarrow B T A\) \(T \longrightarrow \varepsilon\)
\(B A \longrightarrow A a B\)
\(B a \longrightarrow a B\) \(B E \longrightarrow E\)
\(D A \longrightarrow D\)
\(D a \longrightarrow a D\) \(D E \longrightarrow \varepsilon\)
The first three rules produce all strings of the form \(D B^{n} A^{n} E,\) for \(n \in \mathbb{N}\). Let's consider what happens to the string \(D B^{n} A^{n} E\) as the remaining rules are applied. The next two rules allow a B to move to the right until it reaches the E. Each time the B passes an A, a new a is generated, but a Bwill simply move past an a without generating any other characters. Once the \(B\) reaches the \(E,\) the rule \(B E \longrightarrow E\) makes the \(B\) disappear. Each \(B\) from the string \(D B^{n} A^{n} E\) moves past \(n A^{\prime} s\) and generates \(n\) a's. Since there are \(n B^{\prime} s,\) a total of \(n^{2} a^{\prime} s\) are generated. Now, the only way to get rid of the D at the beginning of the string is for it to move right through all the A’s and a’s until it reaches the E at the end of the string. As it does this, the rule \(D A \longrightarrow D\) eliminates all the \(A^{\prime} s\) from the string, leaving the string \(a^{n^{2}} D E\) . Applying the rule \(D E \longrightarrow \varepsilon\) to this gives \(a^{n^{2}} .\) This string contains no non-terminal symbols and so is in the language generated by the grammar. We see that every string of the form \(a^{n^{2}}\) is generated by the above grammar. Furthermore, only strings of this form can be generated by the grammar.
Given a fixed alphabet Σ, there are only countably many different lan- guages over Σ that can be generated by grammars. Since there are un- countably many different languages over Σ, we know that there are many languages that cannot be generated by grammars. However, it is surpris- ingly difficult to find an actual example of such a language.
As a first guess, you might suspect that just as \(\left\{a^{n} b^{n} | n \in \mathbb{N}\right\}\) is an example of a language that is not regular and \(\left\{a^{n} b^{n} c^{n} | n \in \mathbb{N}\right\}\) is an example of a language that is not context-free, so \(\left\{a^{n} b^{n} c^{n} d^{n} | n \in \mathbb{N}\right\}\) might be an example of a language that cannot be generated by any grammar. However, this is not the case. The same technique that was used to produce a grammar that generates \(\left\{a^{n} b^{n} c^{n} | n \in \mathbb{N}\right\}\) can also be used to produce a grammar for \(\left\{a^{n} b^{n} c^{n} d^{n} | n \in \mathbb{N}\right\} .\) In fact, the technique extends to similar languages based on any number of symbols.
Or you might guess that there is no grammar for the language \(\left\{a^{n} | n\right.\)is a prime number }. Certainly, producing prime numbers doesn’t seem like the kind of thing that we would ordinarily do with a grammar. Nevertheless, there is a grammar that generates this language. We will not actually write down the grammar, but we will eventually have a way to prove that it exists.
The language \(\left\{a^{n^{2}} | n \in \mathbb{N}\right\}\) really doesn't seem all that "grammatical" either, but we produced a grammar for it above. If you think about how this grammar works, you might get the feeling that its operation is more like “computation” than “grammar.” This is our clue. A grammar can be thought of as a kind of program, albeit one that is executed in a non- deterministic fashion. It turns out that general grammars are precisely as powerful as any other general-purpose programming language, such as Java or C++. More exactly, a language can be generated by a grammar if and only if there is a computer program whose output consists of a list contain- ing all the strings and only the strings in that language. Languages that have this property are said to be s
recursively enumerable languages. (This term as used here is not closely related to the idea of a recursive subroutine.) The languages that can be generated by general grammars are precisely the recursively enumerable languages. We will return to this topic in the next chapter.
It turns out that there are many forms of computation that are precisely equivalent in power to grammars and to computer programs, and no one has ever found any form of computation that is more powerful. This is one of the great discoveries of the twentieth century, and we will investigate it further in the next chapter.
Exercises
Find a derivation for the string caabcb, according to the first example grammar in this section. Find a derivation for the string aabbcc, according to the second example grammar in this section. Find a derivation for the stringaaaa, according to the third example grammar in this section.
Consider the third sample grammar from this section, which generates the language \(\left\{a^{n^{2}} | n \in \mathbb{N}\right\}\). Is the non-terminal symbol D necessary in this gram- mar? What if the first rule of the grammar were replaced by \(S \longrightarrow T E\) and the last three rules were replaced by \(A \rightarrow \varepsilon\) and \(E \longrightarrow \varepsilon\) ? Would the resulting grammar still generate the same language? Why or why not?
Find a grammar that generates the language \(L=\left\{w \in\{a, b, c, d\}^{*} | n_{a}(w)=\right.\)\(n_{b}(w)=n_{c}(w)=n_{d}(w) \}\). Let Σ be any alphabet. Argue that the language \(\left\{w \in \Sigma^{*} | \text { all symbols in } \Sigma \text { occur equally often in } w\right\}\) can be generated by a grammar.
For each of the following languages, find a grammar that generates the lan- guage. In each case, explain how your grammar works.
a) \(\left\{a^{n} b^{n} c^{n} d^{n} | n \in \mathbb{N}\right\} \quad\) b) \(\left\{a^{n} b^{m} c^{n m} | n \in \mathbb{N} \text { and } m \in \mathbb{N}\right\}\)
c) \(\left\{w w | w \in\{a, b\}^{*}\right\} \quad\) d) \(\left\{w w w | w \in\{a, b\}^{*}\right\}\)
e) \(\left\{a^{2^{n}} | n \in \mathbb{N}\right\}\) f) \(\left\{w \in\{a, b, c\}^{*} | n_{a}(w)>n_{b}(w)>n_{c}(w)\right\}\) |
Introducing Nonlinear Elastic Materials
Nonlinear elastic materials present nonlinear stress-strain relationships even at infinitesimal strains — as opposed to hyperelastic materials, where stress-strain curves become significantly nonlinear at moderate to large strains. Important materials of this class are Ramberg-Osgood for modeling metals and other ductile materials and nonlinear soils models, such as the Duncan-Chang model.
Power Law
The nonlinear stress-strain behavior in solids was already described 100 years ago by Paul Ludwik in his
Elemente der Technologischen Mechanik. In that treatise, Ludwik described the nonlinear relation between shear stress \tau and shear strain \gamma observed in torsion tests with what is nowadays called Ludwik’s Law:
(1)
For n=1, the stress-strain curve is linear; for n=2, the curve is a parabola; and for n=\infty, the curve represents a perfectly plastic material. Ludwik just described the behavior (
Fließkurve) of what we now call a pseudoplastic material.
In version 5.0 of the COMSOL Multiphysics simulation software, beside Ludwik’s power-law, the Nonlinear Structural Materials Module includes different material models within the family of nonlinear elasticity:
Ramberg-Osgood Power Law Uniaxial Data Bilinear Elastic User Defined
In the Geomechanics Module, we have now included material models intended to represent nonlinear deformations in soils:
Hyperbolic Law Hardin-Drnevich Duncan-Chang Duncan-Selig An Example with Uniaxial Data
The main difference between a nonlinear elastic material and an elastoplastic material (either in metal or soil plasticity) is the reversibility of the deformations. While a nonlinear elastic solid would return to its original shape after a load-unload cycle, an elastoplastic solid would suffer from permanent deformations, and the stress-strain curve would present hysteretic behavior and ratcheting.
Let’s open the Elastoplastic Analysis of a Plate with a Center Hole model, available in the Nonlinear Structural Materials Model Library as
elastoplastic_plate, and modify it to solve for one load-unload cycle. Let’s also add one of the new material models included in version 5.0, the Uniaxial data model, and use the stress_strain_curve already defined in the model.
Here’s a screenshot of what those selections look like:
In our example, the stress_strain_curve represents the bilinear response of the axial stress as a function of axial strain, which can be recovered from Ludwik’s law when n=1.
We can compare the stress distribution after laterally loading the plate to a maximum value. The results are pretty much the same, but the main difference is observed after a full load-unload cycle.
Top: Elastoplastic material. Bottom: Uniaxial data model.
Let’s pick the point where we observed the highest stress and plot the
x-direction stress component versus the corresponding strain. The green curve shows a nonlinear, yet elastic, relation between stress and strain (the stress path goes from a\rightarrow b \rightarrow a \rightarrow c \rightarrow a). The blue curve portraits a hysteresis loop observed in elastoplastic materials with isotropic hardening (the stress path goes from a\rightarrow b \rightarrow d \rightarrow e ).
With the Uniaxial data model, you can also define your own stress-strain curve obtained from experimental data, even if it is not symmetric in both tension and compression.
Further Reading P. Ludwik. Elemente der Technologischen Mechanik “Hypoelasticity“, Chapter 3.3 of Applied Mechanics of Solids Download the Elastoplastic Analysis of a Plate with a Center Hole model Comments (1) CATEGORIES Chemical COMSOL Now Electrical Fluid General Interfacing Mechanical Today in Science |
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The ALICE Transition Radiation Detector: Construction, operation, and performance
(Elsevier, 2018-02)
The Transition Radiation Detector (TRD) was designed and built to enhance the capabilities of the ALICE detector at the Large Hadron Collider (LHC). While aimed at providing electron identification and triggering, the TRD ...
Constraining the magnitude of the Chiral Magnetic Effect with Event Shape Engineering in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(Elsevier, 2018-02)
In ultrarelativistic heavy-ion collisions, the event-by-event variation of the elliptic flow $v_2$ reflects fluctuations in the shape of the initial state of the system. This allows to select events with the same centrality ...
First measurement of jet mass in Pb–Pb and p–Pb collisions at the LHC
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This letter presents the first measurement of jet mass in Pb-Pb and p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV and 5.02 TeV, respectively. Both the jet energy and the jet mass are expected to be sensitive to jet ...
First measurement of $\Xi_{\rm c}^0$ production in pp collisions at $\mathbf{\sqrt{s}}$ = 7 TeV
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The production of the charm-strange baryon $\Xi_{\rm c}^0$ is measured for the first time at the LHC via its semileptonic decay into e$^+\Xi^-\nu_{\rm e}$ in pp collisions at $\sqrt{s}=7$ TeV with the ALICE detector. The ...
D-meson azimuthal anisotropy in mid-central Pb-Pb collisions at $\mathbf{\sqrt{s_{\rm NN}}=5.02}$ TeV
(American Physical Society, 2018-03)
The azimuthal anisotropy coefficient $v_2$ of prompt D$^0$, D$^+$, D$^{*+}$ and D$_s^+$ mesons was measured in mid-central (30-50% centrality class) Pb-Pb collisions at a centre-of-mass energy per nucleon pair $\sqrt{s_{\rm ...
Search for collectivity with azimuthal J/$\psi$-hadron correlations in high multiplicity p-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 and 8.16 TeV
(Elsevier, 2018-05)
We present a measurement of azimuthal correlations between inclusive J/$\psi$ and charged hadrons in p-Pb collisions recorded with the ALICE detector at the CERN LHC. The J/$\psi$ are reconstructed at forward (p-going, ...
Systematic studies of correlations between different order flow harmonics in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV
(American Physical Society, 2018-02)
The correlations between event-by-event fluctuations of anisotropic flow harmonic amplitudes have been measured in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 2.76 TeV with the ALICE detector at the LHC. The results are ...
$\pi^0$ and $\eta$ meson production in proton-proton collisions at $\sqrt{s}=8$ TeV
(Springer, 2018-03)
An invariant differential cross section measurement of inclusive $\pi^{0}$ and $\eta$ meson production at mid-rapidity in pp collisions at $\sqrt{s}=8$ TeV was carried out by the ALICE experiment at the LHC. The spectra ...
J/$\psi$ production as a function of charged-particle pseudorapidity density in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Elsevier, 2018-01)
We report measurements of the inclusive J/$\psi$ yield and average transverse momentum as a function of charged-particle pseudorapidity density ${\rm d}N_{\rm ch}/{\rm d}\eta$ in p-Pb collisions at $\sqrt{s_{\rm NN}}= 5.02$ ...
Energy dependence and fluctuations of anisotropic flow in Pb-Pb collisions at √sNN=5.02 and 2.76 TeV
(Springer Berlin Heidelberg, 2018-07-16)
Measurements of anisotropic flow coefficients with two- and multi-particle cumulants for inclusive charged particles in Pb-Pb collisions at 𝑠NN‾‾‾‾√=5.02 and 2.76 TeV are reported in the pseudorapidity range |η| < 0.8 ... |
Wave energy converters in coastal structures: verschil tussen versies
(→References)
(→Wave energy and wave energy flux)
Regel 18: Regel 18:
== Wave energy and wave energy flux ==
== Wave energy and wave energy flux ==
<p>
<p>
− +
, the time-mean [[waves|wave]] energy density E per unit horizontal area on the water surface (J/m²) is the sum of kinetic and potential energy density per unit horizontal area. The potential energy density is equal to the kinetic energy <ref name="ref1">Mei C.C. (1989) The applied dynamics of ocean surface waves. Advanced series on ocean engineering. World Scientific Publishing Ltd </ref> both contributing half to the time-mean wave energy density E that is proportional to the wave height squared according to linear wave theory <ref name="ref1"/>:
<p>
<p>
<br>
<br>
Regel 25: Regel 25:
(1)
(1)
</div>
</div>
−
<math>E= \frac{1}{8} \rho g
+
<math>E= \frac{1}{8} \rho g ^2</math>
</div>
</div>
Versie van 3 sep 2012 om 09:59 Introduction
Fig 1: Construction of a coastal structure.
Coastal works along European coasts are composed of very diverse structures. Many coastal structures are ageing and facing problems of stability, sustainability and erosion. Moreover climate change and especially sea level rise represent a new danger for them. Coastal dykes in Europe will indeed be exposed to waves with heights that are greater than the dykes were designed to withstand, in particular all the structures built in shallow water where the depth imposes the maximal amplitude because of wave breaking.
This necessary adaptation will be costly but will provide an opportunity to integrate converters of sustainable energy in the new maritime structures along the coasts and in particular in harbours. This initiative will contribute to the reduction of the greenhouse effect. Produced energy can be directly used for the energy consumption in harbour area and will reduce the carbon footprint of harbours by feeding the docked ships with green energy. Nowadays these ships use their motors to produce electricity power on board even if they are docked. Integration of wave energy converters (WEC) in coastal structures will favour the emergence of the new concept of future harbours with zero emissions.
Inhoud Wave energy and wave energy flux
For regular waves, the time-mean wave energy density E per unit horizontal area on the water surface (J/m²) is the sum of kinetic and potential energy density per unit horizontal area. The potential energy density is equal to the kinetic energy
[1] both contributing half to the time-mean wave energy density E that is proportional to the wave height H squared according to linear wave theory [1]:
(1)
[math]E= \frac{1}{8} \rho g H^2[/math]
g is the gravity and [math]H_{m0}^2[/math] the spectral estimate of significant wave height. As the waves propagate, their energy is transported. The energy transport velocity is the group velocity. As a result, the time-mean wave energy flux per unit crest length (W/m) perpendicular to the wave propagation direction, is equal to
[1]:
(2)
[math] P= Ec_{g}[/math]
with [math]c_{g}[/math] the group velocity (m/s). Due to the dispersion relation for water waves under the action of gravity, the group velocity depends on the wavelength λ (m), or equivalently, on the wave period T (s). Further, the dispersion relation is a function of the water depth h (m). As a result, the group velocity behaves differently in the limits of deep and shallow water, and at intermediate depths:
[math](\frac{\lambda}{20} \lt h \lt \frac{\lambda}{2})[/math]
Application for wave energy convertersFor regular waves in deep water:
[math]c_{g} = \frac{gT}{4\pi} [/math] and [math]P_{w1} = \frac{\rho g^2}{32 \pi} H_{m0}^2 T[/math]
The time-mean wave energy flux per unit crest length is used as one of the main criteria to choose a site for wave energy converters. If local data are available ([math]H_{m0}^2 [/math], T) for a sea state through in-situ wave buoys for example, satellite data or numerical modelling, the last equation giving wave energy flux [math]P_{w1}[/math] gives a first estimation. Averaged over a season or a year, it represents the maximal energetic resource that can be theoretically extracted from wave energy. If the directional spectrum of sea state variance F (f,[math]\theta[/math]) is known with f the wave frequency (Hz) and [math]\theta[/math] the wave direction (rad), a more accurate formulation is used:
[math]P_{w2} = \rho g\int\int c_{g}(f,h)F(f,\theta) dfd \theta[/math]
Fig 2: Time-mean wave energy flux along
West European coasts
[2] .
It can be shown easily that equation (4) can be reduced to (3) with the hypothesis of regular waves in deep water. The directional spectrum is deduced from directional wave buoys, SAR images or advanced spectral wind-wave models, known as third-generation models, such as WAM, WAVEWATCH III, TOMAWAC or SWAN. These models solve the spectral action balance equation without any a priori restrictions on the spectrum for the evolution of wave growth.
From TOMAWAC model, the near shore wave atlas ANEMOC along the coasts of Europe and France based on the numerical modelling of wave climate over 25 years has been produced
[3]. Using equation (4), the time-mean wave energy flux along West European coasts is obtained (see Fig. 2). This equation (4) still presents some limits like the definition of the bounds of the integration. Moreover, the objective to get data on the wave energy near coastal structures in shallow or intermediate water requires the use of numerical models that are able to represent the physical processes of wave propagation like the refraction, shoaling, dissipation by bottom friction or by wave breaking, interactions with tides and diffraction by islands.
The wave energy flux is therefore calculated usually for water depth superior to 20 m. This maximal energetic resource calculated in deep water will be limited in the coastal zone:
at low tide by wave breaking; at high tide in storm event when the wave height exceeds the maximal operating conditions; by screen effect due to the presence of capes, spits, reefs, islands,...
Technologies
According to the International Energy Agency (IEA), more than hundred systems of wave energy conversion are in development in the world. Among them, many can be integrated in coastal structures. Evaluations based on objective criteria are necessary in order to sort theses systems and to determine the most promising solutions.
Criteria are in particular:
the converter efficiency : the aim is to estimate the energy produced by the converter. The efficiency gives an estimate of the number of kWh that is produced by the machine but not the cost. the converter survivability : the capacity of the converter to survive in extreme conditions. The survivability gives an estimate of the cost considering that the weaker are the extreme efforts in comparison with the mean effort, the smaller is the cost.
Unfortunately, few data are available in literature. In order to determine the characteristics of the different wave energy technologies, it is necessary to class them first in four main families
[2].
An interesting result is that the maximum average wave power that a point absorber can absorb [math]P_{abs} [/math](W) from the waves does not depend on its dimensions
[4]. It is theoretically possible to absorb a lot of energy with only a small buoy. It can be shown that for a body with a vertical axis of symmetry (but otherwise arbitrary geometry) oscillating in heave the capture (or absorption) width [math]L_{max}[/math](m) is as follows [4]:
[math]L_{max} = \frac{P_{abs}}{P_{w}} = \frac{\lambda}{2\pi}[/math] or [math]1 = \frac{P_{abs}}{P_{w}} \frac{2\pi}{\lambda}[/math]
Fig 4: Upper limit of mean wave power
absorption for a heaving point absorber.
where [math]{P_{w}}[/math] is the wave energy flux per unit crest length (W/m). An optimally damped buoy responds however efficiently to a relatively narrow band of wave periods.
Babarit et Hals propose
[5] to derive that upper limit for the mean annual power in irregular waves at some typical locations where one could be interested in putting some wave energy devices. The mean annual power absorption tends to increase linearly with the wave power resource. Overall, one can say that for a typical site whose resource is between 20-30 kW/m, the upper limit of mean wave power absorption is about 1 MW for a heaving WEC with a capture width between 30-50 m.
In order to complete these theoretical results and to describe the efficiency of the WEC in practical situations, the capture width ratio [math]\eta[/math] is also usually introduced. It is defined as the ratio between the absorbed power and the available wave power resource per meter of wave front times a relevant dimension B [m].
[math]\eta = \frac{P_{abs}}{P_{w}B} [/math]
The choice of the dimension B will depend on the working principle of the WEC. Most of the time, it should be chosen as the width of the device, but in some cases another dimension is more relevant. Estimations of this ratio [math]\eta[/math] are given
[5]: 33 % for OWC, 13 % for overtopping devices, 9-29 % for heaving buoys, 20-41 % for pitching devices. For energy converted to electricity, one must take into account moreover the energy losses in other components of the system.
Civil engineering
Never forget that the energy conversion is only a secondary function for the coastal structure. The primary function of the coastal structure is still protection. It is necessary to verify whether integration of WEC modifies performance criteria of overtopping and stability and to assess the consequences for the construction cost.
Integration of WEC in coastal structures will always be easier for a new structure than for an existing one. In the latter case, it requires some knowledge on the existing coastal structures. Solutions differ according to sea state but also to type of structures (rubble mound breakwater, caisson breakwaters with typically vertical sides). Some types of WEC are more appropriate with some types of coastal structures.
Fig 5: Several OWC (Oscillating water column) configurations (by Wavegen – Voith Hydro).
Environmental impact
Wave absorption if it is significant will change hydrodynamics along the structure. If there is mobile bottom in front of the structure, a sand deposit can occur. Ecosystems can also be altered by change of hydrodynamics and but acoustic noise generated by the machines.
Fig 6: Finistere area and locations of
the six sites (google map).
Study case: Finistere area
Finistere area is an interesting study case because it is located in the far west of Brittany peninsula and receives in consequence the largest wave energy flux along the French coasts (see Fig.2). This area with a very ragged coast gathers moreover many commercial ports, fishing ports, yachting ports. The area produces a weak part of its consumption and is located far from electricity power plants. There are therefore needs for renewable energies that are produced locally. This issue is important in particular in islands. The production of electricity by wave energy will have seasonal variations. Wave energy flux is indeed larger in winter than in summer. The consumption has peaks in winter due to heating of buildings but the consumption in summer is also strong due to the arrival of tourists.
Six sites are selected (see figure 7) for a preliminary study of wave energy flux and capacity of integration of wave energy converters. The wave energy flux is expected to be in the range of 1 – 10 kW/m. The length of each breakwater exceeds 200 meters. The wave power along each structure is therefore estimated between 200 kW and 2 MW. Note that there exist much longer coastal structures like for example Cherbourg (France) with a length of 6 kilometres.
(1) Roscoff (300 meters) (2) Molène (200 meters) (3) Le Conquet (200 meters) (4) Esquibien (300 meters) (5) Saint-Guénolé (200 meters) (6) Lesconil (200 meters) Fig.7: Finistere area, the six coastal structures and their length (google map).
Wave power flux along the structure depends on local parameters: bottom depth that fronts the structure toe, the presence of caps, the direction of waves and the orientation of the coastal structure. See figure 8 for the statistics of wave directions measured by a wave buoy located at the Pierres Noires Lighthouse. These measurements show that structures well-oriented to West waves should be chosen in priority. Peaks of consumption occur often with low temperatures in winter coming with winds from East- North-East directions. Structures well-oriented to East waves could therefore be also interesting even if the mean production is weak.
Fig 8: Wave measurements at the Pierres Noires Lighthouse.
Conclusion
Wave energy converters (WEC) in coastal structures can be considered as a land renewable energy. The expected energy can be compared with the energy of land wind farms but not with offshore wind farms whose number and power are much larger. As a land system, the maintenance will be easy. Except the energy production, the advantages of such systems are :
a “zero emission” port industrial tourism test of WEC for future offshore installations.
Acknowledgement
This work is in progress in the frame of the national project EMACOP funded by the French Ministry of Ecology, Sustainable Development and Energy.
See also Waves Wave transformation Groynes Seawall Seawalls and revetments Coastal defense techniques Wave energy converters Shore protection, coast protection and sea defence methods Overtopping resistant dikes
References Mei C.C. (1989) The applied dynamics of ocean surface waves. Advanced series on ocean engineering. World Scientific Publishing Ltd Mattarolo G., Benoit M., Lafon F. (2009), Wave energy resource off the French coasts: the ANEMOC database applied to the energy yield evaluation of Wave Energy, 10th European Wave and Tidal Energy Conference Series (EWTEC’2009), Uppsala (Sweden) Benoit M. and Lafon F. (2004) : A nearshore wave atlas along the coasts of France based on the numerical modeling of wave climate over 25 years, 29th International Conference on Coastal Engineering (ICCE’2004), Lisbonne (Portugal), 714-726. De O. Falcão A. F. (2010) Wave energy utilization: A review of the technologies. Renewable and Sustainable Energy Reviews, Volume 14, Issue 3, April 2010, Pages 899–918. Babarit A. and Hals J. (2011) On the maximum and actual capture width ratio of wave energy converters – 11th European Wave and Tidal Energy Conference Series (EWTEC’2011) – Southampton (U-K). |
from your posted waveform, i am assuming that this is a unipolar signal. that is
$$ x[n] \ge 0 $$
in audio, it would be the same, except that we would be working on $|x[n]|$ instead.
so first you want a sliding maximum of your signal, where the window length is $L$.
$$ x_1[n] = \max_{0 \le i < L} \Big| x[n-i] \Big| $$
since your input signal $x[n]$ appears to be unipolar, you can leave off the absolute value operation.
then you want to apply a low-pass filter (LPF) to that sliding max. you want the gain of the LPF at DC (0 Hz) to be 1 (or 0 dB gain). a simple first-order LPF is
$$ \begin{align} x_2[n] &= (1-p) \cdot x_1[n] + p \cdot x_2[n-1] \\ \\ &= x_1[n] + p \cdot (x_2[n-1] - x_1[n]) \\ \\ &= x_2[n-1] + (1-p) \cdot (x_1[n] - x_2[n-1]) \\ \end{align} $$
$p$ is the pole value of the LPF and
$$ 0 < 1-p \ll 1 $$
so
$x_2[n]$ will be an envelope for your input signal $x[n]$ and will be delayed by about half of the max window length plus about 4 times the "time constant" of the LPF:
$$d = \frac{L}{2} - \frac{4}{\log(p)} \quad \quad \text{samples}$$.
so, to normalize, you want to invert (compute the reciprocal of) the envelope $x_2[n]$ and multiply your input signal by that inverted envelope. the inverted envelope is
$$ x_3[n] = \frac{A}{x_2[n] + \epsilon} $$
$A$ is the normalized amplitude you want (it can be $A=1$ or $A=$ any other positive number that you like). $\epsilon$ is a tiny number, much smaller than most of your non-zero $x[n]$ that you need to add to the denominator to keep from dividing by zero (which is a
bad thing).
but you should line up the signal and the delayed inverted envelope, so your normalized output is also delayed by $d$ samples (defined above):
$$ y[n] = x_3[n] \cdot x[n-d] $$
the most computationally expensive operation is the sliding max. we were just talking about this sliding max in the
music-dsp mailing list. i will dig up C code for it and post that in a following answer. |
I have been working the afternoon on an intertemporal exercise where I'm blocking on something very basic. Have been looking on previous posts but didn't find a similar question.
We have the utility function of the household : $ U(C_1,C_2,n_2)= \log(c_1) + \beta \log(c_2) - \beta n_2 $
We have an allocation $ \bar y $ for period 1 and none for period 2, the household can only work in period two.
Here is the budget constraint : $ c_1 p_1 + c_2 p_2 = p_1 ( \bar y - x ) + qx + wn_2 + \pi $
And the production function of the firm : $ f(k) = k^\alpha n_2^{1-\alpha} $
What I did until now is replacing $ c_2 $ in the utility function to have it depending on $c_1$,$ n_2$ and $ x $ only. Then I derive it for each variable and I can get the optimal condition for consumption : $ \frac{c_2}{c_1} = \beta \frac{p_1}{p_2} $, the other derivatives gives me : $ w = p_2 c_2 $ and $ q = p_1 $ . By replacing $c_2$ in the budget constraint, I can determine $ c_1$ and then $c_2$, and x is given by the market clearing condition $ \bar y = c_1 + x $.
But I can't find anyway to determine $n_2$, it's usually in a logarithm so we can find it in our First Order Conditions...
Edit : $ x $ is the amount of capital given to the firm by the household, so we sould have $ x=k $ on capital market
$ q $ is the price of capital per unit
$ \pi $ is the profit of the firm , i.e. $ \pi(k,n_2) = p_2 k^\alpha n_2^{1-\alpha} - kq - wn_2 $
Edit 2 : Rereading my question, I realize I should precise : The household can only consume the firm production in period 2, which is why he contributes to the capital and labor of the firm, earning wages and profits. Therefore, we should also have $ f(k,n_2) = c_2 $ I believe
Edit 3 : My profit maximisation gives me : $ \frac{\alpha}{1-\alpha} n_2 w = k q $
Edit 4 after @X recommendation : So I do the lagrangian and derivatives, which gets me :
$ \frac{1}{c_1} = \lambda p_1 $
$ \frac{\beta}{c_2} = \lambda p_2 $
$ \beta = \lambda w $
$ p_1 = q $
After computing, I find : $ c_1 = \frac{p_1 \bar y + w n_2 - \pi}{p_1(1+\beta)} $ and $ x = k = \bar y - c_1 = \frac{\beta p_1 \bar y - w n_2 + \pi}{p_1(1+\beta)} $
But still having trouble regarding my original issue, $ n_2 $ : I did , according to your advices : $ \frac{\alpha}{1-\alpha} n_2 w = k q $
$ \frac{\alpha}{1-\alpha} n_2 \frac{\beta}{\lambda} = \frac{\beta p_1 \bar y - w n_2 + \pi}{p_1(1+\beta)} p_1 $
$ \frac{\alpha}{1-\alpha} n_2 \beta p_1 c_1 = \frac{\beta p_1 \bar y - w n_2 + \pi}{(1+\beta)} $
$ \frac{\alpha}{1-\alpha} n_2 \frac{\beta}{1+\beta}(p_1 \bar y + w n_2 - \pi) = \frac{\beta p_1 \bar y - w n_2 + \pi}{(1+\beta)} $
$ \frac{\alpha}{1-\alpha} n_2 \beta(p_1 \bar y + w n_2 - \pi) = \beta p_1 \bar y - w n_2 + \pi $
$ n_2 \beta(p_1 \bar y + w n_2 - \pi) = \frac{1-\alpha}{\alpha} \frac{\beta p_1 \bar y - w n_2 + \pi}{\beta(p_1 \bar y + w n_2 - \pi)} $
Still pretty stupid as results ... I also used the central planner way and found $ n_2 = 1 - \alpha $ , which is weird.. I'd still like to go at the end of this way if someone has an idea.
Any idea ? Thanks. |
75th SSC CGL level Question Set, 7th on topic fractions, decimals and indices
This is the 75th question set of 10 practice problem exercise for SSC CGL exam and 7th on topic fractions, decimals and indices. A few of the problems may seem large and time-consuming, but can be solved quickly.
Students must complete this question set in prescribed time first and then only refer to the corresponding solution set for extracting maximum benefits from this resource.
You may refer to the related tutorials, question and solution sets listed at the end.
75th question set - 10 problems for SSC CGL exam: 7th on topic Fractions, decimals, indices - time 15 mins Problem 1.
$\displaystyle\frac{13}{48}$ is equal to,
$\displaystyle\frac{1}{3+\displaystyle\frac{1}{1+\displaystyle\frac{1}{16}}}$ $\displaystyle\frac{1}{3+\displaystyle\frac{1}{1+\displaystyle\frac{1}{1+\displaystyle\frac{1}{8}}}}$ $\displaystyle\frac{1}{2+\displaystyle\frac{1}{1+\displaystyle\frac{1}{8}}}$ $\displaystyle\frac{1}{3+\displaystyle\frac{1}{1+\displaystyle\frac{1}{2+\displaystyle\frac{1}{4}}}}$ Problem 2.
When $\left(\displaystyle\frac{1}{2} -\displaystyle\frac{1}{4}+\displaystyle\frac{1}{5}-\displaystyle\frac{1}{6}\right)$ is divided by $\left(\displaystyle\frac{2}{5} -\displaystyle\frac{5}{9}+\displaystyle\frac{3}{5}-\displaystyle\frac{7}{18}\right)$ the result is,
$5\displaystyle\frac{1}{10}$ $3\displaystyle\frac{1}{6}$ $3\displaystyle\frac{3}{10}$ $2\displaystyle\frac{1}{18}$ Problem 3.
On simplification, $3034-(1002\div{20.04})$ is equal to,
$2993$ $2984$ $3029$ $2543$ Problem 4.
Value of $\displaystyle\frac{\displaystyle\frac{5}{3}\times{\displaystyle\frac{7}{51}}\text{ of }\displaystyle\frac{17}{5}-\displaystyle\frac{1}{3}}{\displaystyle\frac{2}{9}\times{\displaystyle\frac{5}{7}}\text{ of }\displaystyle\frac{28}{5}-\displaystyle\frac{2}{3}}$ is,
$\displaystyle\frac{1}{2}$ $4$ $\displaystyle\frac{1}{4}$ $2$ Problem 5.
Value of $\displaystyle\frac{9|3-5|-5|4|\div{10}}{-3(5)-2\times{4}\div{2}}$ is,
$\displaystyle\frac{9}{10}$ $\displaystyle\frac{4}{7}$ $-\displaystyle\frac{8}{17}$ $-\displaystyle\frac{16}{19}$ Problem 6.
The value of
$\left(1+\displaystyle\frac{1}{10+\displaystyle\frac{1}{10}}\right)\left(1+\displaystyle\frac{1}{10+\displaystyle\frac{1}{10}}\right)-$
$\left(1-\displaystyle\frac{1}{10+\displaystyle\frac{1}{10}}\right)\left(1-\displaystyle\frac{1}{10+\displaystyle\frac{1}{10}}\right)$ ÷
$\left[\left(1+\displaystyle\frac{1}{10+\displaystyle\frac{1}{10}}\right)-\left(1-\displaystyle\frac{1}{10+\displaystyle\frac{1}{10}}\right)\right]$ is,
$2$ $\displaystyle\frac{90}{101}$ $\displaystyle\frac{20}{101}$ $\displaystyle\frac{100}{101}$ Problem 7.
The value of $8\displaystyle\frac{1}{2}-\left[3\displaystyle\frac{1}{4}\div{\left\{1\displaystyle\frac{1}{4}-\displaystyle\frac{1}{2}\left(1\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3}-\displaystyle\frac{1}{6}\right)\right\}}\right]$ is
$4\displaystyle\frac{1}{2}$ $\displaystyle\frac{2}{9}$ $4\displaystyle\frac{1}{6}$ $9\displaystyle\frac{1}{2}$ Problem 8.
$\sqrt{\displaystyle\frac{4\displaystyle\frac{1}{7}-2\displaystyle\frac{1}{4}}{3\displaystyle\frac{1}{2}+1\displaystyle\frac{1}{7}}\div{\displaystyle\frac{1}{2+\displaystyle\frac{1}{2+\displaystyle\frac{1}{5-\displaystyle\frac{1}{5}}}}}}$ is equal to,
$1$ $2$ $3$ $4$ Problem 9.
The value of $\displaystyle\frac{1+\displaystyle\frac{1}{2}}{1-\displaystyle\frac{1}{2}}\div{\displaystyle\frac{4}{7}\left(\displaystyle\frac{2}{5}+\displaystyle\frac{3}{10}\right)}\text { of }\displaystyle\frac{\displaystyle\frac{1}{2}+\displaystyle\frac{1}{3}}{\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3}}$ is,
$37\displaystyle\frac{1}{2}$ $18\displaystyle\frac{3}{8}$ $\displaystyle\frac{3}{2}$ $\displaystyle\frac{2}{3}$ Problem 10.
When simplified, the expression $(100)^{\frac{1}{2}}\times{(0.001)^{\frac{1}{3}}}-(0.0016)^{\frac{1}{4}}\times{3^0}+\left(\displaystyle\frac{5}{4}\right)^{-1}$ is equal to,
$1.0$ $1.6$ $0$ $0.8$
You may refer to the corresponding solution set,
, where we have applied special time-saving techniques and methods to quickly solve the problems in mind as far as possible. SSC CGL Solution Set 75 on Fractions decimals and indices 7
The answers to the questions are given below followed by a list of relevant articles of concept tutorials, question and solution sets.
Answers to the questions Problem 1. Answer: Option d: $\displaystyle\frac{1}{3+\displaystyle\frac{1}{1+\displaystyle\frac{1}{2+\displaystyle\frac{1}{4}}}}$. Problem 2. Answer. Option a: $5\displaystyle\frac{1}{10}$. Problem 3. Answer: Option b. 2984. Problem 4. Answer: Option d: $2$. Problem 5. Answer: Option d: $-\displaystyle\frac{16}{19}$. Problem 6. Answer: Option a: $2$. Problem 7. Answer: Option c: $4\displaystyle\frac{1}{6}$. Problem 8. Answer: Option a: 1. Problem 9. Answer. Option a: $37\displaystyle\frac{1}{2}$. Problem 10. Answer: Option b: $1.6$. Note: If you are an SSC CGL aspirant, you may refer to our useful article in which we have included all our student resources on SSC CGL topics as links. 7 steps for sure success in SSC CGL tier 1 and tier 2 tests Watch out for more student resources on SSC CGL in this article.
You may also like to go through the
Fraction, decimals, indices and surds related articles directly by referring to the following articles. Concept tutorials on Fractions, Surds, decimals, indices and related topics Question sets and Solution Sets in Fractions, decimals, indices, surds and related topics SSC CGL level Question set 75 on fractions decimals indices 7 |
Snell's law in its sine-over-sine form is not the best tool to handle situations that involve incidence angles above the critical angle, and the complex angles that you're getting will only lead you astray.
Instead, the way to see things is through an explicit evanescent wave, as in this recent answer of mine. When light refracts, you have an incoming wave of the form$$\mathbf E_\mathrm{in}(\mathbf r,t) = \mathrm{Re}\mathopen{}\left( \mathbf E_{0,\mathrm{in}} e^{i(\mathbf k_\mathrm{in}\cdot \mathbf r- \omega t)}\right)\mathclose{},$$and on the other side you get a transmitted beam of the form$$\mathbf E_\mathrm{tr}(\mathbf r,t) = \mathrm{Re}\mathopen{}\left( \mathbf E_{0,\mathrm{tr}} e^{i(\mathbf k_\mathrm{tr}\cdot \mathbf r- \omega t)}\right)\mathclose{},$$where Snell's law comes in through the matching of the two wavevectors, $\mathbf k_\mathrm{in}$ and $\mathbf k_\mathrm{tr}$. These are subject to two conditions:
The $x$ components need to match, $k_{x,\mathrm{in}} = k_{x,\mathrm{tr}} = k_x$, in order for both waves to match continuously at the boundary. The squares of the wavevectors need to be related to the frequency through $k^2 = n^2\omega^2$, to reflect the change in the phase velocity, and since $\omega$ is the same on both sides, they need to be related to each other through$$k_{z,\mathrm{tr}}^2 +k_x^2 = \frac{n_2^2}{n_1^2}\left(k_x^2 +k_{z,\mathrm{in}}^2 \right).$$
The critical angle comes in when the right-hand side is too small for $k_{z,\mathrm{tr}}^2$ to be positive, i.e. when $n_1$ is so much bigger than $n_2$, and $k_x$ is so big, that the square root in$$k_{z,\mathrm{tr}} = \pm \frac{1}{n_1}\sqrt{ n_2^2k_{z,\mathrm{in}}^2 +(n_2^2-n_1^2)k_x^2 }$$is no longer real.
If this is the case, it's not actually a problem: you just write $k_{z,\mathrm{tr}} = i\kappa_z$, and it gives you a perfectly workable 'transmitted' field, of the form$$\mathbf E_\mathrm{tr}(\mathbf r,t) = \mathrm{Re}\mathopen{}\left( \mathbf E_{0,\mathrm{tr}} e^{-\kappa_{z} z}e^{i(k_x x- \omega t)}\right)\mathclose{},$$where what used to be an oscillatory $e^{ik_z z}$ has turned into the decaying exponential $e^{-\kappa_{z} z}$: this is the evanescent wave, which is the decaying trace of the light that's been unable to make it past the boundary, but which cannot drop discontinuously to zero.
Similarly, if you have a complex index of refraction, it means that your medium absorbs light, and there you have no alternative but to use explicit exponential decays in your description of the beams: if $n$ is complex, then the wavevector still needs to satisfy$$k^2 = n^2\omega^2,$$which means that it too will be complex, and you'll have a decaying field in that medium.
In this situation, I would advise you to stay well away from a complex-angles formalism, and to keep things grounded in the cartesian components of the wavevector - or to be very, very careful with those complex angles if you insist on using them. The reason for this is that the cartesian components are related to the angle and the magnitude via\begin{align}k_{x,\mathrm{tr}} & = k_\mathrm{tr} \sin(\theta_\mathrm{tr}) \\k_{z,\mathrm{tr}} & = k_\mathrm{tr} \cos(\theta_\mathrm{tr}),\end{align}and while it is tempting to make $\theta_\mathrm{tr}$ complex and leave it at that, it's important to note that $k_\mathrm{tr}$ also needs to be complex (because $k^2 = n^2\omega^2$), but that the two imaginary parts need to cancel out when you form $k_{x,\mathrm{tr}} = k_\mathrm{tr} \sin(\theta_\mathrm{tr})$, and that's not easy to visualize at all.
That said, you do have a good point that when the second medium has a nonzero absorbance, it is no longer easy to distinguish between the 'sub-critical' incidence angles and over-critical cases with total internal reflection - and indeed there is no longer any case with full reflection of the incident power, since the transmitted wave always has a nonzero $\mathrm{Re}(k_z)$ and therefore always carries up some amount of power away from the boundary.
The way to analyze this in a way that will bring out the relevant behaviour is to plot the wavevector component $k_z$ in the complex plane as the angle of incidence is shifted from $0$ to $\pi/2$. If $n_2$ is real, then $k_z$ will be real at normal incidence, it will approach the origin as $\theta_\mathrm{in}$ increases, and once it reaches the origin it will advance back up along the imaginary axis.
If $n_2$ has a nonzero imaginary part, though, then $k_z$ will follow a similar path but it will be detached: off with a nonzero imaginary part before reaching the origin, and then up the imaginary axis but with a positive real part. The image below shows this behaviour, tracking $k_z$ for $\theta_\mathrm{in}$ between $0$ and $\pi/2$, and with $n_2$ between $2$ (dark blue) to $2+i$ (light blue).
Mathematica source through
Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.com/f5Zz6.png"]
As you can see, if $\mathrm{Im}(n_2)$ is very small (which it is in real materials), then you can define the critical angle as the 'kink' when $k_z$ is close to the origin. However, as that imaginary part increases, the concept of critical angle simply makes less and less sense: you just have a decaying wave away from the boundary, and the incidence angle has very little effect on how fast that wave decays. It mostly stops being a useful concept to begin with, so it doesn't make a lot of sense to dig deep into the mathematics to look for a precise definition of it in that regime. |
You can evaluate this using the Björck-Pereyra algorithm for solving Vandermonde systems, because you are evaluating $b^\top V^{-1}$ with $b=(2,0,\frac23,0,\frac25,0,\ldots)$, and the algorithm is known to be forward-stable (see
Error analysis of the Björck-Pereyra algorithms for solving Vandermonde systems by Nick Higham, http://www.maths.manchester.ac.uk/~higham/narep/narep108.pdf).
Note: it seems that this analysis relies on the property that $0\leq x_1<x_2<\cdots <x_n$ (which is equivalent to $V$ being totally positive), as well as the elements of $b$ having alternating signs (which ensures there's no catastrophic cancellation in the subtractions below), in which case the errors are independent of the condition number, and it will still work in the more general case, without $0\leq x_1$, but the error bounds will be different. In any case, it takes $O(n^2)$ time, and avoids the problem with evaluating/integrating $L_i$'s, so it might be worth it even then, but I hadn't realized this point when I started writing this answer. You might be able to just map $x\mapsto \frac12(x+1)$ if that works for your problem.
I wrote a small Julia program to check that this actually works, and it gives $O(\epsilon_{\mathrm{mach}})$ relative errors.
module VandermondeInverse
using SpecialMatrices
function main(n=8)
X = Rational{BigInt}[k//(n-1) for k=0:n-1]
# X = convert(Vector{Rational{BigInt}}, linspace(-1, 1, n))
x = convert(Vector{Float64}, X)
A = convert(Matrix{Rational{BigInt}}, Vandermonde(X))
b = [i%2==0 ? 2//(i+1) : 0 for i=0:n-1]
println("Norm: ", norm(A, Inf))
println("Norm of inverse: ", norm(inv(A), Inf))
println("Condition number: ", cond(convert(Matrix{Float64}, A)))
ans = A'\b
println("True answer: ", ans)
B = convert(Matrix{Float64}, A)
c = convert(Vector{Float64}, b)
println("Linear solve: ", norm((B'\c - ans)./ans, Inf))
d = vec(c')
for k=1:n, l=n:-1:k+1
d[l] -= x[k]*d[l-1]
end
for k=n-1:-1:1, l=k:n
if l > k
d[l] /= x[l]-x[l-k]
end
if l < n
d[l] -= d[l+1]/(x[l+1] - x[l-k+1])
end
end
println("Neville elimination: ", norm((d-ans)./ans, Inf))
nothing
end
end
V = VandermondeInverse
Output:
julia> V.main(14)
Norm: 14.0
Norm of inverse: 1.4285962612120493e10
Condition number: 5.2214922998851654e10
True answer: Rational{Int64}[3202439130233//2916000,-688553801328731//52390800,19139253128382829//261954000,-196146528919726853//785862000,6800579086408939//11642400,-43149880138884259//43659000,32567483200938127//26195400,-7339312362348889//6237000,48767438804485271//58212000,-69618881108680969//157172400,44275410625421677//261954000,-2308743351566483//52390800,11057243346333379//1571724000,-209920276397//404250]
Linear solve: 1.5714609387747318e-8
Neville elimination: 1.3238218572356314e-15
If
X isn't positive like in this test, then it seems the relative errors are of the same order as with a regular linear solve.
Why $b^\top V^{-1}$? It's actually a very useful common trick for working with polynomials of all types, but especially the Lagrange interpolating polynomials, converting the problem to a matrix form. The condition that defines $L_i$ is that $L_i(x_j)=\delta_{ij}$. Let $\alpha_{jk}$ be the coefficients of $L_k$, i.e., $$ L_k(x) = \sum_{j,k} \alpha_{j,k}x^j = (1,x,x^2,\ldots,x^n)^\top (\alpha_{0k},\ldots,\alpha_{nk}),$$ and $L$ be the whole matrix of coefficients, arranged by columns:$$ L = \begin{pmatrix} \alpha_{00}& \cdots & \alpha_{0n}\\\vdots &&\vdots\\ \alpha_{n0}&\cdots&\alpha_{nn} \end{pmatrix}. $$Because of the definition of $L_k$ above as a vector product, multiplying $L$ on the left by $(1,x,\ldots,x^n)$ yields$$ (1,x,x^2,\ldots,x^n)L = (L_0(x),L_1(x),\ldots,L_n(x)). $$Using the condition $L_k(x_j)=\delta_{jk}$, this means that$$ \begin{pmatrix} 1&x_0&x_0^2&\cdots&x_0^n\\ \vdots\\ 1&x_n&x_n^2&\cdots&x_n^n \end{pmatrix} L = I, $$so $L = V^{-1}$, where $V$ is the Vandermonde matrix of $x_j$'s.
Finally, since $\int_{-1}^{1}x^k\,\mathrm{d}x = \frac{1+(-1)^k}{k+1}$, we have$$ \int_{-1}^{1}L_k(x)\,\mathrm{d}x = \sum_j \alpha_{jk}\frac{1+(-1)^k}{k+1} = (2,0,\tfrac23,0,\tfrac25,0,\ldots)^\top (\alpha_{0k},\ldots,\alpha_{nk}).$$So the $n+1$ numbers you are looking for, for $k=0\ldots n$, are given by $(2,0,\tfrac23,0,\ldots)^\top L$, where $L=V^{-1}$ is the inverse of the Vandermonde matrix. |
Under the auspices of the Computational Complexity Foundation (CCF)
We survey some of the recent results on the complexity of recognizing
n-dimensional linear arrangements and convex polyhedra by randomized algebraic decision trees. We give also a number of concrete applications of these results. In particular, we derive first nontrivial, in fact quadratic, ... more >>>
We derive new time-space tradeoff lower bounds and algorithms for exactly computing statistics of input data, including frequency moments, element distinctness, and order statistics, that are simple to calculate for sorted data. In particular, we develop a randomized algorithm for the element distinctness problem whose time $T$ and space $S$ ... more >>>
The approximate degree of a Boolean function $f: \{-1, 1\}^n \to \{-1, 1\}$ is the minimum degree of a real polynomial that approximates $f$ to within error $1/3$ in the $\ell_\infty$ norm. In an influential result, Aaronson and Shi (J. ACM 2004) proved tight $\tilde{\Omega}(n^{1/3})$ and $\tilde{\Omega}(n^{2/3})$ lower bounds on ... more >>> |
We have seen that there are context-free languages that are not regular. The natural question arises, are there languages that are not context-free? It’s easy to answer this question in the abstract: For a given alphabet Σ, there are uncountably many languages over Σ, but there are only countably many context-free languages over Σ. It follows that most languages are not context-free. However, this answer is not very satisfying since it doesn’t give us any example of a specific language that is not context-free.
As in the case of regular languages, one way to show that a given lan- guage L is not context-free is to find some property that is shared by all context-free languages and then to show that L does not have that prop- erty. For regular languages, the Pumping Lemma gave us such a property. It turns out that there is a similar Pumping Lemma for context-free lan- guages. The proof of this lemma uses parse trees. In the proof, we will need a way of representing abstract parse trees, without showing all the details of the tree. The picture
represents a parse tree which has the non-terminal symbol A at its root and the string x along the “bottom” of the tree. (That is, x is the string made up of all the symbols at the endpoints of the tree’s branches, read from left to right.) Note that this could be a partial parse tree—something that could be a part of a larger tree. That is, we do not require A to be the start symbol of the grammar and we allow x to contain both terminal and non-terminal symbols. The string x, which is along the bottom of the tree, is referred to as the
yield of the parse tree. Sometimes, we need to show more explicit detail in the tree. For example, the picture
represents a parse tree in which the yield is the string xyz. The string y is the yield of a smaller tree, with root B, which is contained within the larger tree. Note that any of the strings x, y, or z could be the empty string.
We will also need the concept of the
height of a parse tree. The height of a parse tree is the length of the longest path from the root of the tree to the tip of one of its branches.
Like the version for regular languages, the Pumping Lemma for context- free languages shows that any sufficiently long string in a context-free language contains a pattern that can be repeated to produce new strings that are also in the language. However, the pattern in this case is more complicated. For regular languages, the pattern arises because any sufficiently long path through a given DFA must contain a loop. For context-free languages, the pattern arises because in a sufficiently large parse tree, along a path from the root of the tree to the tip of one of its branches, there must be some non-terminal symbol that occurs more than once.
Theorem 4.7
(Pumping Lemma for Context-free Languages). Suppose that L is a context-free language. Then there is an integer K such that any string \(w \in L(G)\) with \(|w| \geq K\) has the property that w can be written in the form w = uxyzv where x and z are not both equal to the empty string; |xyz| < K; and
For any \(n \in \mathbb{N},\) the string \(u x^{n} y z^{n} v\) is in \(L\)
Proof. Let \(G=(V, \Sigma, P, S)\) be a context-free grammar for the languageL. Let N be the number of non-terminal symbols in G, plus 1. That is, \(N=|V|+1\). Consider all possible parse trees for the grammar G with height less than or equal to N. (Include parse trees with any non-terminal symbol as root, not just parse trees with root S.) There are only finitely many such parse trees, and therefore there are only finitely many different strings that are the yields of such parse trees. Let K be an integer which is greater than the length of any such string.
Now suppose that w is any string in L whose length is greater than or equal to K. Then any parse tree for w must have height greater than N. (This follows since \(|w| \geq K\) and the yield of any parse tree of height \(\leq N\) has length less than \(K\). Consider a parse tree for w of minimal size, that is one that contains the smallest possible number of nodes. Since the height of this parse tree is greater than N, there is at least one path from the root of the tree to tip of a branch of the tree that has length greater than N. Consider the longest such path. The symbol at the tip of this path is a terminal symbol, but all the other symbols on the path are non- terminal symbols. There are at least N such non-terminal symbols on the path. Since the number of different non-terminal symbols is |V | and since \(N=|V|+1\), some non-terminal symbol must occur twice on the path. In fact, some non-terminal symbol must occur twice among the bottommostN non-terminal symbols on the path. Call this symbol A. Then we see that the parse tree for w has the form shown here:
The structure of this tree breaks the string w into five substrings, as shown in the above diagram. We then have w = uxyzv. It only remains to show that x, y, and z satisfy the three requirements stated in the theorem.
Let T refer to the entire parse tree, let \(T_{1}\) refer to the parse tree whose root is the upper A in the diagram, and let \(T_{2}\) be the parse tree whose root is the lower A in the diagram. Note that the height of is \(T_{1}\) less than or equal to N. (This follows from two facts: The path shown in T1 from its root to its base has length less than or equal to N, because we chose the two occurrences of A to be among the N bottommost non-terminal symbols along the path in T from its root to its base. We know that there is no longer path from the root of T1 to its base, since we chose the path in Tto be the longest possible path from the root of T to its base.) Since any parse tree with height less than or equal to N has yield of length less than \(K,\) we see that \(|x y z|<K\).
If we remove \(T_{1}\) from T and replace it with a copy of \(T_{2}\), the result is a parse tree with yield uyv, so we see that the string uyv is in the language L. Now, suppose that both x and z are equal to the empty string. In that case, w = uyv, so the tree we have created would be another parse tree for w. But this tree is smaller than T, so this would contradict the fact that T is the smallest parse tree for w. We see that x and z cannot both be the empty string.
If we remove \(T_{2}\) from T and replace it with a copy of \(T_{1}\), the result is a parse tree with yield \(u x^{2} y z^{2} v,\) so we see that \(u x^{2} y z^{2} v \in L .\) The two parse trees that we have created look like this:
Furthermore, we can apply the process of replacing \(T_{2}\) with a copy of \(T_{1}\) to the tree on the right above to create a parse tree with yield \(u x^{3} y z^{3} v\). Continuing in this way, we see that \(u x^{n} y z^{n} v \in L\) for all \(n \in \mathbb{N}\). This completes the proof of the theorem.
Since this theorem guarantees that all context-free languages have a certain property, it can be used to show that specific languages are not context-free. The method is to show that the language in question does not have the property that is guaranteed by the theorem. We give two examples.
Corollary 4.8.
Let L be the language \(\left\{a^{n} b^{n} c^{n} | n \in \mathbb{N}\right\}\). Then L is not a context-free language.
Proof. We give a proof by contradiction. Suppose that L is context-free. Then, by the Pumping Lemma for Context-free Languages, there is an integer \(K\) such that every string \(w \in L\) with \(|w| \geq K\) can be written in the form \(w=u x y z v\) where \(x\) and \(z\) are not both empty, \(|x y z|<K,\) and \(u x^{n} y z^{n} v \in L\) for every \(n \in \mathbb{N}\).
Consider the string \(w=a^{K} b^{K} c^{K},\) which is in \(L,\) and write \(w=u x y z v\), where u, x, y, z, and v satisfy the stated conditions. Since \(|x y z|<K\) we see that if xyz contains an a, then it cannot contain a c. And if it contains a c, then it cannot contain an a. It is also possible that xyz is made up entirely of \(b^{\prime} s .\) In any of these cases, the string \(u x^{2} y z^{2} v\) cannot be in L, since it does not contain equal numbers of a’s, b’s, and c’s. But this contradicts the fact that \(u x^{n} y z^{n} v \in L\) for all \(n \in \mathbb{N} .\) This contradiction shows that the assumption that L is context-free is incorrect.
Corollary 4.9.
Let Σ be any alphabet that contains at least two symbols. Let L be the language over Σ defined by \(L=\left\{s s | s \in \Sigma^{*}\right\}\). Then L is not context-free.
Proof. Suppose, for the sake of contradiction, that L is context-free. Then, by the Pumping Lemma for Context-free Languages, there is an integer \(K\) such that every string \(w \in L\) with \(|w| \geq K\) can be written in the form \(w=u x y z v\) where \(x\) and \(z\) are not both empty, \(|x y z|<K,\) and \(u x^{n} y z^{n} v \in L\) for every \(n \in \mathbb{N}\).
Let \(a\) and \(b\) represent distinct symbols in \(\Sigma .\) Let \(s=a^{K} b a^{K} b\) and let \(w=s s=a^{K} b a^{K} b a^{K} b a^{K} b,\) which is in \(L .\) Write \(w=u x y z v,\) where \(u, x, y\) z, and v satisfy the stated conditions.
Since \(|x y z|<K, x\) and \(z\) can, together, contain no more than one \(b\). If either \(x\) or \(y\) contains a \(b,\) then \(u x^{2} y z^{2} v\) contains exactly five \(b\) 's. But any string in L is of the form rr for some string r and so contains an even number of \(b^{\prime} s .\) The fact that \(u x^{2} y z^{2} z\) contains five \(b^{\prime}\) s contradicts the fact that \(u x^{2} y z^{2} v \in L .\) So, we get a contradiction in the case where \(x\) or \(y\) contains a b.
Now, consider the case where x and y consist entirely of a’s. Again since \(|x y z|<K,\) we must have either that \(x\) and \(y\) are both contained in the same group of \(a^{\prime} s\) in the string \(a^{K} b a^{K} b a^{K} b a^{K} b,\) or that \(x\) is contained in one group of a’s and y is contained in the next. In either case, it is easy to check that the string \(u x^{2} y z^{2} v\) is no longer of the form \(r r\) for any string \(r,\) which contradicts the fact that \(u x^{2} y z^{2} v \in L\).
Since we are led to a contradiction in every case, we see that the as- sumption that L is context-free must be incorrect.
Now that we have some examples of languages that are not context- free, we can settle some other questions about context-free languages. In particular, we can show that the intersection of two context-free languages is not necessarily context-free and that the complement of a context-free language is not necessarily context-free.
Theorem 4.10.
The intersection of two context-free languages is not nec- essarily a context-free language.
Proof. To prove this, it is only necessary to produce an example of two context-free languages \(L\) and \(M\) such that \(L \cap M\) is not a context-free languages. Consider the following languages, defined over the alphabet \(\Sigma=\{a, b, c\}\):
\(L=\left\{a^{n} b^{n} c^{m} | n \in \mathbb{N} \text { and } m \in \mathbb{N}\right\}\)
\(M=\left\{a^{n} b^{m} c^{m} | n \in \mathbb{N} \text { and } m \in \mathbb{N}\right\}\)
Note that strings in L have equal numbers of a’s and b’s while strings in \(M\) have equal numbers of \(b^{\prime} s\) and \(c^{\prime} s .\) It follows that strings in \(L \cap M\) have equal numbers of a’s, b’s, and c’s. That is,
\(L \cap M=\left\{a^{n} b^{n} c^{n} | n \in \mathbb{N}\right\}\)
We know from the above theorem that \(L \cap M\) is not context-free. However, both L and M are context-free. The language L is generated by the context- free grammar
\(S \longrightarrow T C\)
\(C \longrightarrow c C\) \(C \longrightarrow \varepsilon\) \(T \longrightarrow a T b\) \(T \longrightarrow \varepsilon\)
and M is generated by a similar context-free grammar.
Corollary 4.11.
The complement of a context-free language is not necessarily context-free.
Proof. Suppose for the sake of contradiction that the complement of every context-free language is context-free.
Let L and M be two context-free languages over the alphabet Σ. By our assumption, the complements \(\overline{L}\) and \(\overline{M}\) are context-free. By Theorem 4.3, it follows that \(\overline{L} \cup \vec{M}\) is context-free. Applying our assumption once again, we have that \(\overline{\overline{L} \cup \overline{M}}\) is context-free. But \(\overline{\overline{L}} \cup \overline{M}=L \cap M,\) so we have that \(L \cap M\) is context-free.
We have shown, based on our assumption that the complement of any context-free language is context-free, that the intersection of any two context- free languages is context-free. But this contradicts the previous theorem, so we see that the assumption cannot be true. This proves the theorem.
Note that the preceding theorem and corollary say only that \(L \cap M\) is not context-free for some context-free languages \(L\) and \(M\) and that \(\overline{L}\) is not context-free for some context-free language L. There are, of course, many examples of context-free languages \(L\) and \(M\) for which \(L \cap M\) and \(\overline{L}\) are in fact context-free.
Even though the intersection of two context-free languages is not necessarily context-free, it happens that the intersection of a context-free language with a regular language is always context-free. This is not difficult to show, and a proof is outlined in Exercise 4.4.8. I state it here without proof:
Theorem 4.12.
Suppose that L is a context-free language and that M is a regular language. Then \(L \cap M\) is a context-free language.
For example, let L and M be the languages defined by \(L=\{w \in\)\(\{a, b\}^{*} | w=w^{R} \}\) and \(M=\left\{w \in\{a, b\}^{*} | \text { the length of } w \text { is a multiple }\right.\)of 5\(\} .\) since \(L\) is context-free and \(M\) is regular, we know that \(L \cap M\) is context-free. The language \(L \cap M\) consists of every palindrome over the alphabet \(\{a, b\}\) whose length is a multiple of five.
This theorem can also be used to show that certain languages are not context-free. For example, consider the language \(L=\left\{w \in\{a, b, c\}^{*} | n_{a}(w)\right.\)\(=n_{b}(w)=n_{c}(w) \} .\) (Recall that \(n_{x}(w)\) is the number of times that the symbol x occurs in the string w.) We can use a proof by contradiction to show that L is not context-free. Let M be the regular language defined by the regular expression \(a^{*} b^{*} c^{*} .\) It is clear that \(L \cap M=\left\{a^{n} b^{n} c^{n} | n \in \mathbb{N}\right\} .\) If \(L\) were context-free, then, by the previous theorem, \(L \cap M\) would be context-free. However, we know from Theorem 4.8 that \(L \cap M\) is not context-free. So we can conclude that L is not context-free.
Exercises
Show that the following languages are not context-free:
a) \(\left\{a^{n} b^{m} c^{k} | n>m>k\right\}\)
b) \(\left\{w \in\{a, b, c\}^{*} | n_{a}(w)>n_{b}(w)>n_{c}(w)\right\}\) c) \(\left\{w w w | w \in\{a, b\}^{*}\right\}\) d) \(\left\{a^{n} b^{m} c^{k} | n, m \in \mathbb{N} \text { and } k=m * n\right\}\) e) \(\left\{a^{n} b^{m} | m=n^{2}\right\}\)
2. Show that the languages \(\left\{a^{n} | \mathrm{n} \text { is a prime number }\right\}\) and \(\left\{a^{n^{2}} | n \in \mathbb{N}\right\}\) are not context-free. (In fact, it can be shown that a language over the alphabet{a} is context-free if and only if it is regular.)
3. Show that the language \(\left\{w \in\{a, b\}^{*} | n_{a}(w)=n_{b}(w) \text { and } w \text { contains the }\right.\) string baaab as a substring} is context-free.
4. Suppose that M is any finite language and that L is any context-free language. Show that the language L \ M is context-free. (Hint: Any finite language is a regular language.) |
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Cauchy's Inequality Theorem $\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$
where all of $r_i, s_i \in \R$.
For any $\lambda \in \R$, we define $f: \R \to \R$ as the function:
$\displaystyle \map f \lambda = \sum {\paren {r_i + \lambda s_i}^2}$
Now:
$\map f \lambda \ge 0$ Hence:
\(\displaystyle \forall \lambda \in \R: \map f \lambda\) \(\equiv\) \(\displaystyle \sum {\paren {r_i^2 + 2 \lambda r_i s_i + \lambda^2 s_i^2} } \ge 0\) \(\displaystyle \) \(\equiv\) \(\displaystyle \sum {r_i^2} + 2 \lambda \sum {r_i s_i} + \lambda^2 \sum {s_i^2} \ge 0\)
This is a quadratic equation in $\lambda$.
$\displaystyle a \lambda^2 + b \lambda + c = 0: a = \sum {s_i^2}, b = 2 \sum {r_i s_i}, c = \sum {r_i^2}$ The discriminant of this equation (that is $b^2 - 4 a c$) is: $D := \displaystyle 4 \paren {\sum {r_i s_i} }^2 - 4 \sum {r_i^2} \sum {s_i^2}$
But we have:
$\forall \lambda \in \R: \map f \lambda \ge 0$ From this contradiction it follows that: $D \le 0$
which is the same thing as saying:
$\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \paren {\sum {r_i s_i} }^2$
$\blacksquare$
From the Complex Number form of the Cauchy-Schwarz Inequality, we have:
$\displaystyle \sum \left|{w_i}\right|^2 \left|{z_i}\right|^2 \ge \left|{\sum w_i z_i}\right|^2$
where all of $w_i, z_i \in \C$.
As elements of $\R$ are also elements of $\C$, it follows that: $\displaystyle \sum \left|{r_i}\right|^2 \left|{s_i}\right|^2 \ge \left|{\sum r_i s_i}\right|^2$
where all of $r_i, s_i \in \R$.
But from the definition of modulus, it follows that: $\displaystyle \forall r_i \in \R: \left|{r_i}\right|^2 = r_i^2$ Thus: $\displaystyle \sum {r_i^2} \sum {s_i^2} \ge \left({\sum {r_i s_i}}\right)^2$
where all of $r_i, s_i \in \R$.
$\blacksquare$
Also known as Source of Name
This entry was named for Augustin Louis Cauchy.
It is a special case of the Cauchy-Bunyakovsky-Schwarz Inequality.
Sources 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics(5th ed.) ... (previous) ... (next): Entry: Cauchy-Schwarz inequality for sums For a video presentation of the contents of this page, visit the Khan Academy. |
Under the auspices of the Computational Complexity Foundation (CCF)
Suppose $\mathrm{Est}$ is a randomized estimation algorithm that uses $n$ random bits and outputs values in $\mathbb{R}^d$. We show how to execute $\mathrm{Est}$ on $k$ adaptively chosen inputs using only $n + O(k \log(d + 1))$ random bits instead of the trivial $nk$ (at the cost of mild increases in the error and failure probability). Our algorithm combines a variant of the INW pseudorandom generator (STOC '94) with a new scheme for shifting and rounding the outputs of $\mathrm{Est}$. We prove that modifying the outputs of $\mathrm{Est}$ is necessary in this setting, and furthermore, our algorithm's randomness complexity is near-optimal in the case $d \leq O(1)$. As an application, we give a randomness-efficient version of the Goldreich-Levin algorithm; our algorithm finds all Fourier coefficients with absolute value at least $\theta$ of a function $F: \{0, 1\}^n \to \{-1, 1\}$ using $O(n \log n) \cdot \mathrm{poly}(1/\theta)$ queries to $F$ and $O(n)$ random bits (independent of $\theta$), improving previous work by Bshouty et al. (JCSS '04).
Improved presentation
Suppose $\mathrm{Est}$ is a randomized estimation algorithm that uses $n$ random bits and outputs values in $\mathbb{R}^d$. We show how to execute $\mathrm{Est}$ on $k$ adaptively chosen inputs using only $n + O(k \log(d + 1))$ random bits instead of the trivial $nk$ (at the cost of mild increases in the error and failure probability). Our algorithm combines a variant of the INW pseudorandom generator (STOC '94) with a new scheme for shifting and rounding the outputs of $\mathrm{Est}$. We prove that modifying the outputs of $\mathrm{Est}$ is necessary in this setting, and furthermore, our algorithm's randomness complexity is near-optimal in the case $d \leq O(1)$. As an application, we give a randomness-efficient version of the Goldreich-Levin algorithm; our algorithm finds all Fourier coefficients with absolute value at least $\theta$ of a function $F: \{0, 1\}^n \to \{-1, 1\}$ using $O(n \log n) \cdot \mathrm{poly}(1/\theta)$ queries to $F$ and $O(n)$ random bits (independent of $\theta$), improving previous work by Bshouty et al. (JCSS '04).
Improved presentation
Suppose $\mathsf{Est}$ is a randomized estimation algorithm that uses $n$ random bits and outputs values in $\mathbb{R}^d$. We show how to execute $\mathsf{Est}$ on $k$ adaptively chosen inputs using only $n + O(k \log(d + 1))$ random bits instead of the trivial $nk$ (at the cost of mild increases in the error and failure probability). Our algorithm combines a variant of the INW pseudorandom generator (STOC '94) with a new scheme for shifting and rounding the outputs of $\mathsf{Est}$. We prove that modifying the outputs of $\mathsf{Est}$ is necessary in this setting, and furthermore, our algorithm's randomness complexity is near-optimal in the case $d \leq O(1)$. As an application, we give a randomness-efficient version of the Goldreich-Levin algorithm; our algorithm finds all Fourier coefficients with absolute value at least $\theta$ of a function $F: \{0, 1\}^n \to \{-1, 1\}$ using $O(n \log n) \cdot \poly(1/\theta)$ queries to $F$ and $O(n)$ random bits (independent of $\theta$), improving previous work by Bshouty et al. (JCSS '04).
Improved presentation, added appendix C, simplified abstract
We introduce the concept of a randomness steward, a tool for saving random bits when executing a randomized estimation algorithm $\mathrm{Est}$ on many adaptively chosen inputs. For each execution, the steward chooses the randomness of $\mathrm{Est}$ and, crucially, is allowed to modify the output of $\mathrm{Est}$. The notion of a steward is inspired by adaptive data analysis, introduced by Dwork et al. (STOC '15). Suppose $\mathrm{Est}$ outputs values in $\R^d$, has $\ell_{\infty}$ error $\epsilon$, has failure probability $\delta$, uses $n$ coins, and will be executed $k$ times. For any $\gamma > 0$, we construct a computationally efficient steward with $\ell_{\infty}$ error $O(\epsilon d)$, failure probability $k \delta + \gamma$, and randomness complexity $n + O(k \log(d + 1) + (\log k) \log(1/\gamma))$. To achieve these parameters, the steward uses a pseudorandom generator for what we call the block decision tree model, combined with a scheme for shifting and rounding the outputs of $\mathrm{Est}$. The generator is a variant of the INW generator (STOC '94) for space-bounded computation. We also give variant stewards that achieve tradeoffs in parameters.
As an application of our steward, we give a randomness-efficient version of the Goldreich-Levin algorithm; our algorithm finds all Fourier coefficients with absolute value at least $\theta$ of a function $F: \{0, 1\}^n \to \{-1, 1\}$ using $O(n \log n) \cdot \mathrm{poly}(1/\theta)$ queries to $F$ and $O(n)$ random bits (independent of $\theta$), improving previous work by Bshouty et al. (JCSS '04). We also give time- and randomness-efficient algorithms for estimating the acceptance probabilities of many adaptively chosen Boolean circuits and for simulating any algorithm with an oracle for $\mathrm{promise\mbox{-}BPP}$ or $\mathrm{APP}$. We prove that nontrivial pseudorandom generators for this setting do not exist, i.e. modifying the outputs of $\mathrm{Est}$ is necessary. We also prove a randomness complexity lower bound of $n + \Omega(k) - \log(\delta'/\delta)$ for any "one-query steward" with failure probability $\delta'$, which nearly matches our upper bounds in the case $d \leq O(1)$.
Added sections 1.5.3 and 7.1, changed terminology, fixed typos.
We introduce the concept of a randomness steward, a tool for saving random bits when executing a randomized estimation algorithm $\mathrm{Est}$ on many adaptively chosen inputs. For each execution, the chosen input to $\mathrm{Est}$ remains hidden from the steward, but the steward chooses the randomness of $\mathrm{Est}$ and, crucially, is allowed to modify the output of $\mathrm{Est}$. The notion of a steward is inspired by adaptive data analysis, introduced by Dwork et al. Suppose $\mathrm{Est}$ outputs values in $\mathbb{R}^d$, has $\ell_{\infty}$ error $\epsilon$, has failure probability $\delta$, uses $n$ coins, and will be executed $k$ times. For any $\gamma > 0$, we construct a computationally efficient steward with $\ell_{\infty}$ error $O(\epsilon d)$, failure probability $k \delta + \gamma$, and randomness complexity $n + O(k \log(d + 1) + (\log k) \log(1/\gamma))$. To achieve these parameters, the steward uses a pseudorandom generator for what we call the block decision tree model, combined with a scheme for shifting and rounding the outputs of $\mathrm{Est}$. (The generator is a variant of the INW generator for space-bounded computation.) We also give variant stewards that achieve tradeoffs in parameters.
As applications of our steward, we give time- and randomness-efficient algorithms for estimating the acceptance probabilities of many adaptively chosen Boolean circuits and for simulating any algorithm with an oracle for $\mathrm{promise\mbox{-}BPP}$ or $\mathrm{APP}$. We also give a randomness-efficient version of the Goldreich-Levin algorithm; our algorithm finds all Fourier coefficients with absolute value at least $\theta$ of a function $F: \{0, 1\}^n \to \{-1, 1\}$ using $O(n \log n) \cdot \mathrm{poly}(1/\theta)$ queries to $F$ and $O(n)$ random bits, improving previous work by Bshouty et al. Finally, we prove a randomness complexity lower bound of $n + \Omega(k) - \log(\delta'/\delta)$ for any steward with failure probability $\delta'$, which nearly matches our upper bounds in the case $d \leq O(1)$. |
I'm trying verify that a 2nd order finite difference in space and time approximation of the 1D wave equation is really 2nd order. My Matlab implementation tells me otherwise - I'm not sure of what I've done incorrectly.
For simplicity I've decided to use the smooth solution $u(x,t)=\cos{(ct)}\cos{x}$. The PDE has periodic boundary conditions:
$$ u_{tt} = c^2u_{xx},~~~u(x,0)=\cos{x},~~~u_t(x,0)=0,~~~u(-2\pi,t)=u(2\pi,t). $$ I've taken $c=1$, and the domain to be $[-2\pi,2\pi]$.
The finite difference scheme I'm using is centered second order in space and time: $$ \frac{u(x_j,t_{i+1})-2u(x_j,t_{i})+u(x_j,t_{i-1})}{(\Delta t)^2} = c^2\frac{u(x_{j+1},t_{i})-2u(x_{j},t_{i})+u(x_{j-1},t_{i})}{(\Delta x)^2}. $$
To approximate the initial condition $u_t(x,0)=0$, I tried both the first order forward approximation $$ \frac{u(x_j,t_{j+1})-u(x_j,t_{j})}{\Delta t}=0, $$ and the second order centered approximation $$ \frac{u(x_j,t_{j+1})-u(x_j,t_{j-1})}{2\Delta t}=0. $$ When I fix $\Delta t$ and refine $\Delta x$, both of these approximations to $u_t(x,0)=0$ seem to give an order of approximately 1.5 according to my Matlab code. The largest error looks like it occurs near the boundaries $x=\pm 2\pi$. Can anyone see what I'm doing incorrectly?
a = -2*pi;b = 2*pi;NN = 2.^(3:9);for jj = 1 : length(NN) n = NN(jj); %Number of grid points dx = (b-a)/(n-1); %Spatial mesh spacing x = a : dx : b; %Mesh in x direction c = 1; u = @(x,t) cos(c*t).*cos(x); u0 = u(x,0.0)'; e = ones(n,1); A = spdiags([e -2*e e], -1:1, n, n); %Laplacian matrix A(1,end) = 1; A(end,1) = 1; %Periodic boundary conditions T = 1; %Final simulation time dt = 1e-3; %Temporal mesh spacing nsteps = (floor(T/dt)); %Number if time steps %u_old = u0; %First order approxiation of u_t(x,0)=0. %u_now = u0; %Second order approxiation of u_t(x,0)=0. u_old = u0; u_now = u0 + 0.5*(dt*c/dx)^2*(A* u0); %(approximates u(x,dt)) for ii = 2 : nsteps %First time step was directly above u_new = 2*u_now - u_old + (dt*c/dx)^2*(A* u_now); u_old = u_now; u_now = u_new; end LI(jj) = norm(u_new - u(x,T)',inf); L2(jj) = dx*norm(u_new - u(x,T)',2); if ( jj > 1 ) %Look at order of convergence log ( L2(jj-1)/L2(jj) ) / log ( 2 ) endendfigure(1)semilogy( L2,'-')figure(2)plot(x,u_new,'ro-',x,u(x,T),'--') |
Normally I would just dismiss the formula, but I found it in two different sources (both particle physics sources though). One book talked about the vacuum bubble expansion of the integral:\int \frac{1}{[k^2-m^2][(k-p)^2-m^2]}=\int \frac{1}{[k^2-m^2]^2}-\int \frac{p^2}{[k^2-m^2]^3}...
This is probably a dumb question, but I have a book that claims that if you have a function of the momentum squared, f(p2), that:\frac{d}{dp^2}f=\frac{1}{2d}\frac{\partial }{\partial p_\mu}\frac{\partial }{\partial p^\mu}fwhere the d in the denominator is the number of spacetime...
The Christoffel symbols for your first metric in R3 are the same as the ones for your second metric in S2, provided you ignore all Christoffel symbols that have a radius index. Ben Niehoff provided a general formula, but is it just coincidence that the Christoffel symbols are the same for the...
Consider a vector field on a sphere given by \vec{V}= \hat{e}_\phi . This vector field is not parallel transported on a circle at any latitude except the equator. Yet your equation above seems to say that it's parallel transported along any path. That is, \nabla_ie_j = 0 implies u^i \nabla_ie_j...
The derivative straddles two different tangent spaces, since it is a difference of vector fields at two different points. For example in uniform circular motion, the derivative of the velocity points towards the center of the circle. Only a tangential acceleration can be written as a linear...
How can the derivative of a basis vector at a point be the linear combination of tangent vectors at that point?For example, if you take a sphere, then the derivative of the polar basis vector with respect to the polar coordinate is in the radial direction. How can something in the radial...
When you say that a (nongeodesic) path rotates with respect to a local orthonormal frame, do you mean that you parallel transport the tangent vector 'u' of the path from point A of the old frame to the orthonormal frame of the new point B via adding: -\Gamma^{i}_{jk}u^j dx^k , and compare this...
If you are traveling in a circle on a circle of latitude, and always pointing your arm North, then doesn't your arm always make the same angle with respect to your path (90 degrees)? Isn't that the pictorial definition of parallel transport?The parallel transport equation is DV^i=0 along a...
If you walk at constant latitude with your arm always sticking towards the North pole, is that parallel transport of your arm?The equations don't seem to say it is.The vector field would be \vec{V}=V(\theta)e_\theta . The component of the vector only depends on theta, but at constant...
I prefer the \Lambda 's to have an inverse sign on them, but that's all right.Basically that equation says two reference frames have the same metric, or that the metric tensor is invariant under Lorentz transformation.This implies that two observers take dot products in the same way...
Assume k=0. Then if you measure \rho_0, then don't you have H_0? Or vice versa: if you measure H_0 don't you have \rho_0?That's what seems to be implied by the equation:H^2 = \frac{8 \pi G}{3}\rho, so I'm still not sure how to get a_0
I'm a little confused about the density \rho in the equation:H^2 = \left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3}\rho - \frac{kc^2}{a^2}Measuring \rho at a single instant in time seems easy. But \rho changes with time. The time dependence of \rho is given as...
You've been very helpful and I don't wish to bother you anymore, especially on something you never use :).I'm taking a "math methods in physics course", and I have an end-of-semester presentation in a few days, and I was going to choose to present two topics: relativistic quantum mechanics on...
The duffell.org notes were really good. You sure know how to find good notes!Why SO(3,1)? In the duffell.org notes, the structure group for a vector bundle is GL(4).I guess you can have any group for your structure group, so long as the group maps the fiber onto itself. So for example, the...
I was reading the Carroll notes you mentioned (great notes by the way), and he mentions adding to each point of the manifold an internal vector space. At each point there were already tangent and cotangent spaces, but now he is adding more vector spaces. He calls the collection of all vector...
So if you have a massless particle, say a graviton, which is a 2nd-ranked tensor, or a neutrino which is a spinor, then would the bending be exactly the same as for light which is a vector? I would add scalar particles but I'm unaware of any that are massless.So somehow the mass=0 free...
Can the bending of light because of gravity be derived from the Maxwell equations written in curved space time, i.e.,\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}F^{\mu\nu}})=0In all the examples the bending of light is treated as a massless particle travelling on a light-like geodesic (if I...
Thanks a lot for your responses. They have been very helpful.I have a question about the terminology here. Usually the term extrinsic refers to the embedding space of a manifold, and intrinsic refers to the manifold itself. But in this context, you are using extrinsic to mean the spacetime...
There seems to be a bit of asymmetry between fermions and bosons here. Bosons and fermions both transform under the Lorentz group SO(4). When adding general relativity with GL(4), bosons in their vector representations can transform under this group rather easily. But fermions in their spinor...
I have a question about the nomenclature of the "spin connection".To me "spin" implies quantum mechanics.However, as I understand it, the spin connection is just a part of the covariant derivative of a vector that is written in tetrad (i.e., local frame) components. So isn't the "spin...
My knowledge about this subjected is very limited, but can't you just use the inverse mapping of \phi, and then the pull back of this inverse transformation would be the "natural" form you seek? And doesn't what Flanders say about 0-forms and pull backs apply to k-forms? You always get a natural...
What is the benefit of expressing Maxwell's equation in the language of differential forms? Differential forms seem to be inferior to the language of tensors. Sure you can do fancy things with the exterior derivative and hodge star, but with tensors you can derive those same identities with... |
Ground state solutions for fractional scalar field equations under a general critical nonlinearity
1.
Universidade Federal de Campina Grande, 58429-970, Campina Grande - PB - Brazil
2.
Universidade de Brasilia, Campus Universitário Darcy Ribeiro, 70910-900, Brasília - DF - Brazil
3.
Universidade de São Paulo, Departamento de Matemática - IME, Rua do Matão 1010, 05508-090, São Paulo - SP - Brazil
$ (- \Delta)^{\alpha}u = g(u) \ \ \mbox{in} \ \ \mathbb{R}^{N}, \ \ u \in H^{\alpha}(\mathbb R^N) $
$ (-\Delta)^{\alpha} $
$ \alpha\in (0,1) $
$ N\geq2 $
$ N = 1 $
$ \alpha = 1/2 $
$ g $
$ N\geq2 $
$ N = 1 $ Mathematics Subject Classification:Primary: 35J20; Secondary: 35A15, 35B33. Citation:Claudianor O. Alves, Giovany M. Figueiredo, Gaetano Siciliano. Ground state solutions for fractional scalar field equations under a general critical nonlinearity. Communications on Pure & Applied Analysis, 2019, 18 (5) : 2199-2215. doi: 10.3934/cpaa.2019099
References:
[1]
C. O. Alves, M. Montenegro and M. A. Souto,
Existence of a ground state solution for a nonlinear scalar field equation with critical growth,
[2]
C. O. Alves, J. M do Ó and O. H. Miyagaki,
Concentration phenomena for fractional elliptic equations involving exponential critical growth,
[3] [4]
A unified approach to symmetrization,
[5]
B. Barrios, E. Colorado, R. Servadei and F. Soria,
A critical fractional equation with concave-convex nonlinearities,
[6] [7] [8]
C. Bucur and E. Valdinoci,
[9] [10]
X. Chang and Z-Q Wang,
Ground state of scalar field equations involving a fractional Laplacian with general nonlinearity,
[11] [12]
S. Dipierro, M. Medina and E. Valdinoci, Fractional elliptic problems with critical growth in the whole of $\mathbb{R}^n$, Lecture Notes. Appunti. Edizioni della Normale, Scuola Normale di Pisa (2017), arXiv: 1506.01748. doi: 10.1007/978-88-7642-601-8. Google Scholar
[13]
S. Dipierro, M. Medina, I. Peral and E. Valdinoci,
Bifurcation results for a fractional elliptic equation with critical exponent in $\mathbb R^{n}$,
[14]
J. M. do Ó, O. H. Miyagaki and M. Squassina,
Ground states of nonlocal scalar field equations with Trudinger-Moser critical nonlinearity,
[15] [16] [17]
P. Felmer, A. Quass and J. Tan,
Positive solutions of nonlinear Schrödinger equation with the fractional Laplacian,
[18] [19] [20]
R. Lehrer, L. A. Maia and M. Squassina,
Asymptotically linear fractional Schrödinger equations,
[21] [22] [23]
G. Palatucci and A. Pisante,
Improved Sobolev embeddings, profile decomposition, and concentration-compactness for fractional Sobolev spaces,
[24] [25] [26] [27]
J. J. Zhang, J. M. do Ó and M. Squassina,
Fractional Schrödinger-Poisson systems with a general subcritical or critical nonlinearity,
show all references
References:
[1]
C. O. Alves, M. Montenegro and M. A. Souto,
Existence of a ground state solution for a nonlinear scalar field equation with critical growth,
[2]
C. O. Alves, J. M do Ó and O. H. Miyagaki,
Concentration phenomena for fractional elliptic equations involving exponential critical growth,
[3] [4]
A unified approach to symmetrization,
[5]
B. Barrios, E. Colorado, R. Servadei and F. Soria,
A critical fractional equation with concave-convex nonlinearities,
[6] [7] [8]
C. Bucur and E. Valdinoci,
[9] [10]
X. Chang and Z-Q Wang,
Ground state of scalar field equations involving a fractional Laplacian with general nonlinearity,
[11] [12]
S. Dipierro, M. Medina and E. Valdinoci, Fractional elliptic problems with critical growth in the whole of $\mathbb{R}^n$, Lecture Notes. Appunti. Edizioni della Normale, Scuola Normale di Pisa (2017), arXiv: 1506.01748. doi: 10.1007/978-88-7642-601-8. Google Scholar
[13]
S. Dipierro, M. Medina, I. Peral and E. Valdinoci,
Bifurcation results for a fractional elliptic equation with critical exponent in $\mathbb R^{n}$,
[14]
J. M. do Ó, O. H. Miyagaki and M. Squassina,
Ground states of nonlocal scalar field equations with Trudinger-Moser critical nonlinearity,
[15] [16] [17]
P. Felmer, A. Quass and J. Tan,
Positive solutions of nonlinear Schrödinger equation with the fractional Laplacian,
[18] [19] [20]
R. Lehrer, L. A. Maia and M. Squassina,
Asymptotically linear fractional Schrödinger equations,
[21] [22] [23]
G. Palatucci and A. Pisante,
Improved Sobolev embeddings, profile decomposition, and concentration-compactness for fractional Sobolev spaces,
[24] [25] [26] [27]
J. J. Zhang, J. M. do Ó and M. Squassina,
Fractional Schrödinger-Poisson systems with a general subcritical or critical nonlinearity,
[1]
Mingqi Xiang, Binlin Zhang.
A critical fractional
[2]
Yinbin Deng, Wentao Huang.
Least energy solutions for fractional Kirchhoff type equations involving critical growth.
[3]
Miaomiao Niu, Zhongwei Tang.
Least energy solutions for nonlinear Schrödinger equation involving the fractional Laplacian and critical growth.
[4] [5]
Yinbin Deng, Yi Li, Wei Shuai.
Existence of solutions for a class of p-Laplacian type equation
with critical growth and potential vanishing at infinity.
[6]
Hui Zhang, Jun Wang, Fubao Zhang.
Semiclassical states for fractional Choquard equations with critical growth.
[7]
Hua Jin, Wenbin Liu, Jianjun Zhang.
Multiple solutions of fractional Kirchhoff equations involving a critical nonlinearity.
[8]
Qingfang Wang.
The Nehari manifold for a fractional Laplacian equation involving critical nonlinearities.
[9]
Xudong Shang, Jihui Zhang, Yang Yang.
Positive solutions of nonhomogeneous fractional Laplacian problem with critical exponent.
[10]
Sami Aouaoui.
On some semilinear equation in $R^4$ containing a Laplacian term and involving nonlinearity with exponential growth.
[11]
Patrizia Pucci, Mingqi Xiang, Binlin Zhang.
A diffusion problem of Kirchhoff type involving the nonlocal fractional
[12] [13] [14]
M. Grossi, P. Magrone, M. Matzeu.
Linking type solutions for elliptic equations with indefinite nonlinearities up to the critical growth.
[15] [16]
Maoding Zhen, Jinchun He, Haoyuan Xu, Meihua Yang.
Positive ground state solutions for fractional Laplacian system with one critical exponent and one subcritical exponent.
[17]
Maria-Magdalena Boureanu.
Fourth-order problems with Leray-Lions type operators in variable exponent spaces.
[18] [19]
Hua Jin, Wenbin Liu, Huixing Zhang, Jianjun Zhang.
Ground states of nonlinear fractional Choquard equations with Hardy-Littlewood-Sobolev critical growth.
[20]
Alexander Quaas, Aliang Xia.
Existence and uniqueness of positive solutions for a class of logistic type elliptic equations in $\mathbb{R}^N$ involving fractional Laplacian.
2018 Impact Factor: 0.925
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Analytic results for the linear stability of the equilibrium point in Robe's restricted elliptic three-body problem
1.
Department of Mathematics, Zhejiang University, Hangzhou 310027, Zhejiang, China
2.
School of Mathematics and Statistics, Northeastern University at Qinhuangdao, Qinhuangdao 066004, Hebei, China
We study the Robe's restricted three-body problem. Such a motion was firstly studied by A. G. Robe in [
Keywords:Restricted three-body problem, equilibrium point, linear stability, Maslov-type $ω$-index. Mathematics Subject Classification:Primary:70F07, 70H14;Secondary:34C25. Citation:Qinglong Zhou, Yongchao Zhang. Analytic results for the linear stability of the equilibrium point in Robe's restricted elliptic three-body problem. Discrete & Continuous Dynamical Systems - A, 2017, 37 (3) : 1763-1787. doi: 10.3934/dcds.2017074
References:
[1] [2] [3]
P. P. Hallen and D. N. Rana,
The Existence and stability of equilibrium points in the Robe's restricted three-body problem,
[4]
X. Hu, Y. Long and S. Sun,
Linear stability of elliptic Lagrangian solutions of the classical planar three-body problem via index theory,
[5] [6]
X. Hu and S. Sun,
Index and stability of symmetric periodic orbits in Hamiltonian systems with its application to figure-eight orbit,
[7] [8] [9] [10] [11] [12] [13] [14]
A. K. Shrivastava and D. Garain,
Effect of perturbation on the location of liberation point in the Robe restricted problem of three bodies,
[15]
K. T. Singh, B. S. Kushvah and B. Ishwar,
Stability of triangular equilibrium points in Robe's generalised restricted three body problem,
[16]
Q. Zhou and Y. Long,
Equivalence of linear stabilities of elliptic triangle solutions of the planar charged and classical three-body problems,
[17] [18]
Q. Zhou and Y. Long,
The reduction on the linear stability of elliptic Euler-Moulton solutions of the $n$-body problem to those of $3$-body problems,
show all references
References:
[1] [2] [3]
P. P. Hallen and D. N. Rana,
The Existence and stability of equilibrium points in the Robe's restricted three-body problem,
[4]
X. Hu, Y. Long and S. Sun,
Linear stability of elliptic Lagrangian solutions of the classical planar three-body problem via index theory,
[5] [6]
X. Hu and S. Sun,
Index and stability of symmetric periodic orbits in Hamiltonian systems with its application to figure-eight orbit,
[7] [8] [9] [10] [11] [12] [13] [14]
A. K. Shrivastava and D. Garain,
Effect of perturbation on the location of liberation point in the Robe restricted problem of three bodies,
[15]
K. T. Singh, B. S. Kushvah and B. Ishwar,
Stability of triangular equilibrium points in Robe's generalised restricted three body problem,
[16]
Q. Zhou and Y. Long,
Equivalence of linear stabilities of elliptic triangle solutions of the planar charged and classical three-body problems,
[17] [18]
Q. Zhou and Y. Long,
The reduction on the linear stability of elliptic Euler-Moulton solutions of the $n$-body problem to those of $3$-body problems,
[1] [2]
Hadia H. Selim, Juan L. G. Guirao, Elbaz I. Abouelmagd.
Libration points in the restricted three-body problem: Euler angles, existence and stability.
[3] [4]
Jaime Angulo Pava, César A. Hernández Melo.
On stability properties of the Cubic-Quintic Schródinger equation with $\delta$-point interaction.
[5] [6] [7]
Yeping Li, Jie Liao.
Stability and $ L^{p}$ convergence rates of planar diffusion waves for three-dimensional bipolar Euler-Poisson systems.
[8]
Xiaojun Chang, Tiancheng Ouyang, Duokui Yan.
Linear stability of the criss-cross orbit in the equal-mass three-body problem.
[9]
Niraj Pathak, V. O. Thomas, Elbaz I. Abouelmagd.
The perturbed photogravitational restricted three-body problem: Analysis of resonant periodic orbits.
[10]
Florin Diacu, Shuqiang Zhu.
Almost all 3-body relative equilibria on $ \mathbb S^2 $ and $ \mathbb H^2 $ are inclined.
[11]
Jean-Baptiste Caillau, Bilel Daoud, Joseph Gergaud.
Discrete and differential homotopy in circular restricted three-body control.
[12]
Frederic Gabern, Àngel Jorba, Philippe Robutel.
On the accuracy of restricted three-body models for the Trojan motion.
[13]
Tuan Anh Dao, Michael Reissig.
$ L^1 $ estimates for oscillating integrals and their applications to semi-linear models with $ \sigma $-evolution like structural damping.
[14] [15]
Edcarlos D. Silva, José Carlos de Albuquerque, Uberlandio Severo.
On a class of linearly coupled systems on $ \mathbb{R}^N $ involving asymptotically linear terms.
[16]
Qunyi Bie, Haibo Cui, Qiru Wang, Zheng-An Yao.
Incompressible limit for the compressible flow of liquid crystals in $ L^p$ type critical Besov spaces.
[17]
Nicholas J. Kass, Mohammad A. Rammaha.
Local and global existence of solutions to a strongly damped wave equation of the $ p $-Laplacian type.
[18] [19]
Jaume Llibre, Y. Paulina Martínez, Claudio Vidal.
Phase portraits of linear type centers of polynomial Hamiltonian systems with Hamiltonian function of degree 5 of the form $ H = H_1(x)+H_2(y)$.
[20]
Dajana Conte, Raffaele D'Ambrosio, Beatrice Paternoster.
On the stability of $\vartheta$-methods for stochastic Volterra integral equations.
2018 Impact Factor: 1.143
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Definition) A point $x\in X$ is a limit point of S if every ball $B(x;r)$ contains infinitely many points from $S$.
A point $x\in X$ is called an isolated point of S if $\exists r > 0$ such that $B(x;r)\cap S = \{x\}$.
Problem) $S\subset X^{metric}$. Let $S_1$ be the set of limit points of S. Let $S_2$ be the set of isolated points of S.
Show that $\bar S = S_1\cup S_2$ and $S_1 \cap S_2 = \emptyset$.
Suppose x is not a limit point of S. Then $\exists \epsilon > 0$ such that $B(x;\epsilon)$ contains only finitely points of S. By shrinking $\epsilon$, we can assume that $B(x;\epsilon)$ contains no points of S other than possibly x itself. This means x is an isolated point. So $S_1\cap S_2 = \emptyset$. (Until now, is this right?)
Then I confused about limit point and closure. Originally I thought $\bar S=S_1$. But it is wrong and I lost my way. |
Definition:Krull Dimension of Ring Definition
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
$\map {\operatorname {dim_{Krull} } } R = \sup \set {n \in \N: \exists p_0, \ldots, p_n \in \Spec R: \mathfrak p_0 \subsetneqq \mathfrak p_1 \subsetneqq \cdots \subsetneqq \mathfrak p_n}$
where $\Spec R$ is the prime spectrum of $R$.
Also denoted as
The
Krull dimension can also be denoted $\operatorname{K-dim}$ or simply $\dim$, if there is no confusion. Also see Krull's Theorem which proves the existence of a prime ideal Source of Name
This entry was named for Wolfgang Krull. |
Suppose we are using Simulated Annealing (SA) to minimize a cost function $L:\mathbb{R} \to \mathbb{R}$. Here is my algorithm:
(1). Randomly choose a $x_0 \in \mathbb{R}$. Set $x=x_0$ and $T=T_0$.
for k=1:m
(2). Propose a new point $x_{new}$ by sampling $N(x, \sigma^2)$, where $\sigma$ is a constant.
(3). If $L(x_{new}) < L(x)$,
then set $x=x_{new}$.
else If $u \leq \exp(-\frac{L(x_{new}) - L(x)}{T})$, where u is a sample of $U(0,1)$, then set $x=x_{new}$.
(4). Lower $T$ according to some scheme, such as $T=cT$ for a constant $c \in (0,1)$.
end for for k
Someone (whose credibility is doubted by me) said that in step (2), the propose distribution for a new point $x_{new}$ should be $N(x, T)$ which changes as $T$ changes, instead of $N(x, \sigma^2)$ fixed for all $T$ values. The majority of the sources I have found do not mention much about the proposal distribution, including not mentioning if it depends on the temperature. But I found a similar claim on page 183 of this book Neuro-fuzzy and soft computing by Jang:
In conventional SA, also known as Boltzmann machines, the generating function is a Gaussian probability density function: $$x_{new} - x \sim N(0, T \times I_{n,n})$$ where $T$ is the temperature, and $n$ is the dimension of the space under exploration. It has been proven in ref (Geman and Geman's Stochastic relaxation, Gibbs distribution and the Bayesian Restoration in images, PAMI 1984) that a Boltzmann machine using the aforementioned generating function can find a global optimum of $L(x)$ if the temperature $T$ is reduced not faster than $T_0/\ln(k)$.
My understanding is that both choices for the proposal distribution are correct, since SA uses the Metropolis–Hastings (M-H) algorithm for sampling the target distribution with density $$p(x) = Z_T \exp(-\frac{x}{T}),$$ where $Z_T$ is a normalization factor for $p$ to be a probability density function. Under either of the two choices for the propose distribution of $x_{new}$, i.e. either $N(x, T)$ or $N(x, \sigma^2)$, according to the M-H algorithm (see the Appendix below), the accept probability of proposed state $x_{new}$ given current state $x$ in step (3) should be $$ a_{ij} = \min(1, \exp(-\frac{L(x_{new}) - L(x)}{T})), $$ which is exactly the same as in the step (3) of SA algorithm. So I think the two choices for the propose distribution of $x_{new}$, i.e. $N(x, T)$ and $N(x, \sigma^2)$, would both work. Am I correct?
Does the SA algorithm under both choices for the proposal distribution converge to the optimal solution in some probabilistic sense, possibly under some additional conditions?
If they do in question 2, which one is better, in terms of some standard(s)? For example, which one converges more quickly?
Thanks and regards!
Appendix: Calculating acceptance probability in the M-H algorithm.
Let
$p_{i}$ be the target density at state $i$, $h_{ij}$ be the proposal density for transition to state $j$ given current state $i$, $a_{ij}$ be the accept probability of proposed state $j$ given current state $i$.
By the detailed balance equation, after choosing the proposal density $h$, the accept probabilities $a$ is computed as $$ a_{ij} = \min(1, \frac{p_{j} h_{ji}}{p_{i} h_{ij}}). $$
If $h$ is symmetric, i.e. $h_{ij}=h_{ji}$, then $$ a_{ij} = \min(1, \frac{p_{j}}{p_{i}}). $$ |
Let AC and CE be perpendicular line segments,each of length 18.Suppose B and D are the mid points of AC and CE respectively. If F is the intersection of EB and AD ,then the area of triangle DEF is
My attempt is Area of $\Delta \space DEF=($ Area of $\Delta \space ACD$+ Area of $\Delta \space BCE$ - Area of quadrilateral BCDF)/2
as Triangles ABF and DEF are congruent.
Now Quadrilateral BCDEF can be divided by line segment CF into $\Delta \space BCF \space and \space \Delta CFD$
Now $\angle ADC = tan^{-1}(9/18) \implies tan^{-1}(\frac{1}{2}) $
Similarly we can also get $\angle CBF$ and as $\Delta BCF$ is congruent to $\Delta DCF$ , $\angle BFC \space = \angle DFC$ and as $\angle BCF \space = \angle DCF= 45^{\circ}$ , we can get $\angle BFC \space and \space \angle DFC$
Finally using the sine rule we can get BF,CF,FD,CD using which we can calculate the area of the quadrilateral BCDF as a sum of triangles BFC and DFC
We can easily get the area of triangles ACD and BCE as they are right angled triangles and subtracting it by the new found area of quadrilateral BCDF, we can get our result.
I do not want this process. I want an easier approach to this problem. Please suggest with reasons. |
作者：S. Zeynep Özal Kurşungöz 来源：[J].Annales de la faculté des sciences de Toulouse, 2019, Vol.28 (2), pp.329-356CEDRAM 摘要：Let $\Omega ^0$ be a bounded domain in $\mathbb{C}^n$ and $\mathcal{E}$ be a compact subset of $\Omega ^0$ such that $\Omega := \Omega ^0 \setminus \mathcal{E}$ is connected. This paper deals with the study of the extension properties of the pluricomplex Green function of $\Omega...
作者：Maxime Zavidovique 来源：[J].Annales de la faculté des sciences de Toulouse, 2019, Vol.28 (2), pp.209-224CEDRAM 摘要：We show that viscosity solutions of evolutionary weakly coupled systems of Hamilton–Jacobi equations can be approximated by iterated twisted Lax–Oleinik like operators. We establish convergence to the solution of the iterated scheme and discuss further properties of the appr...
作者：Jean Ruppenthal 来源：[J].Annales de la faculté des sciences de Toulouse, 2019, Vol.28 (2), pp.225-258CEDRAM 摘要：The present paper is a complement to the $L^2$-theory for the $\overline{\partial }$-operator on a Hermitian complex space $X$ of pure dimension $n$ with isolated singularities, presented in [17] and [13]. The general philosophy is to use a resolution of singularities $\pi :...
作者：Ibrahima Hamidine 来源：[J].Annales de la faculté des sciences de Toulouse, 2019, Vol.28 (2), pp.357-396CEDRAM 摘要：We propose an approach of Mellin type for the approximation of integration currents or the effective realization of normalized Green currents associated with a cycle $ \bigwedge _1^m[\mathrm{div}(s_j)] $, where $s_j $ is a meromorphic section of a line bundle $ \mathscr{L}_j \rig...
作者：Jean-François Coulombel 来源：[J].Annales de la faculté des sciences de Toulouse, 2019, Vol.28 (2), pp.259-327CEDRAM 摘要：The aim of this article is to propose a systematic study of transparent boundary conditions for finite difference approximations of evolution equations. We try to keep the discussion at the highest level of generality in order to apply the theory to the broadest class of problems...
作者：Damien Calaque 来源：[J].Annales de la faculté des sciences de Toulouse, 2019, Vol.28 (1), pp.67-90CEDRAM 摘要：We prove that shifted cotangent stacks carry a canonical shifted symplectic structure. We also prove that shifted conormal stacks carry a canonical Lagrangian structure. These results were believed to be true, but no written proof was available in the Artin case.
作者：Nicolas Lerner 来源：[J].Annales de la faculté des sciences de Toulouse, 2019, Vol.28 (1), pp.1-9CEDRAM 摘要：The mathematician Yuri Egorov died on October 6, 2018 in Toulouse. This text outlines two fundamental aspects of his work, the quantification of canonical transformations and the study of subelliptic operators.
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The R help manual cites the Fisher letter to the Australian Journal of Statistics.
If the observations in a $2 \times 2$ table are distinctly out of proportion (and indeed in other cases also) we may wish to set limits to the true product ratio, e.g. the observed table
$$ \begin{array}{cc} 10 & 3 \\ 2 & 15 \end{array}$$
gives a crude ratio of 25. How small could the true ratio be in reasonable consistency with the data? If the expectation in the four classes were
$$ \begin{array}{cc} 10-x & 3+x \\ 2+x & 15-x \end{array}$$
the true ratio would be $(10-x)(15-x)/(3+x)(2+x)$m and $\chi^2$ for the observations would be:
$$\chi^2 = x^2 \left( \frac{1}{10-x} + \frac{1}{3+x} + \frac{1}{2+x} + \frac{1}{15-x} \right)$$
so if $x$ were 3.0, $$\chi^2 = 3^2 (0.59286) = 5.3357$$ with one degree of freedom.
The exact probability of such a small sample of 30 giving 10 or more in the first quadrant is the partial sum of a hypergeometric series, and not easy to calculate for if $\xi$ stand for the theoretical product ratio, the frequencies of 0 to 12 in the quadrant will be proportional to the terms:
$$ 1, \frac{13 \times 12}{1\times 6}\xi, \frac{13\times 12 \times 12 \times 11}{1 \times 2 \times 6 \times 7}\xi^2, \ldots, \frac{13!12!5!}{(13-r)!(12-r)!(5+r)!}\xi^i,\ldots$$
It would not be too difficult, as in the exact test for disproportionality, to calcuate the last three terms for any chosen value of $\xi$, but for the ratio of these to the whole we would require the sum of the entire series or $$F(-13, -12, 6, \xi)$$ which would be best obtained by calculating all the terms and summing them, a process too lengthy to be recommended.
Using Yates' adjustment, however, we can at once find: $$\chi^2_c = (2.5)^2 0.59286 = 3.7054$$.
Further taking $x=3.1$ we have
$$ \chi^2_c = (2.6)^2(0.58717) = 3.9693$$
Interpolating for the tabular entry 3.841 it appears that $x=3.0501$ and the cross product ratio is 2.718.
So that it may be inferred from the data that the true cross-product ratio exceeds 2.718 unless a coincidence of one in forty has occurred, Similar limits can be set in both directions and at all limits of probability. |
Geometric Progression
A sequence of non-zero numbers is called geometric progression. The abbreviated form is G.P. In G.P. the ratio of a term and the term preceding to it is always same or constant. The constant ratio is called common ratio of the G.P.
Definition of Geometric Progression
Let a sequence $a_{1},a_{2},a_{3},a_{4},a_{5},,...,a_{n}$,... is called a G.P if
$\frac{a_{n+1}}{a_{n}}$ = constant for all n $\epsilon $ N.
Example 1 :
The sequence 6,18,54,162,... is a G.P because
$\frac{18}{6} = \frac{54}{18} = \frac{162}{54}$ = 3 which is constant.
Here the sequence of G.P with first term 6 and common ratio 3.
Example 2 :
The sequence $\frac{1}{3},\frac{-1}{2},\frac{3}{4},\frac{-9}{8}$,,... is a G.P. because
$\frac{-1}{2} \div \frac{1}{3} = \frac{3}{4} \div \frac{-1}{2} = \frac{-3}{2}$ which is constant
Here the sequence of G.P with first term $\frac{1}{3}$ and common ratio $\frac{-3}{2}$.
Example 3 :
4,-2,1,$\frac{-1}{2}$,...is a G.P because
(-2) $\div (4) = (1) \div (-2) = \frac{-1}{2}$ which is constant.
Example 4 :
Show that the sequence given by $a_{n} = 3(2^{n})$, for all
n $\epsilon $ N, is a G.P. Also, find its common ratio.
Solution :
We have $a_{n} = 3(2^{n})$
∴ $a_{n+1} = 3(2^{n+1})$
So, $\frac{a_{n+1}}{a_{n}} = \frac{3(2^{n+1})}{3(2^{n})}$
$\frac{a_{n+1}}{a_{n}}$ = 2 which a constant for all n $\epsilon $N.
So the given sequence is a G.P. with common ratio 2.
Geometric series
If $a_{1},a_{2},a_{3},a_{4},...a_{n}$is a G.P the the expression $a_{1}+ a_{2} + a_{3}+....+a_{n}... is called geometric series.
Note that the geometric series is finite or infinite according to as the corresponding G.P. consists of finite or infinite number of terms. Practice questions
I. Check whether the following sequence are in G.P. or not.
(i) -5, 15, -45, 135, ...
(ii) 0.5, 3.5, 24.5, 171.5, ...
(iii) 2, 6, 16, 54, ...
(iv) -11, 22, -44, 88, ...
(iv) 1/2, 1, 2, 4, 8,...
(v) −1, 1, 4, 8, ...
II. The following sequence is in G.P , find the first and the common ratio.
(i) −1, 6, −36, 216, ...
(ii) −3, −15, −75, −375, ...
(iv) 2, $\frac{1}{2}, \frac{1}{8},\frac{1}{128}, \frac{1}{512}$,...
(v) −24, −144, −864,...11th grade math
From Geometric progression to Home |
ISSN:
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August 2017 , Volume 37 , Issue 7
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Abstract:
We consider the semilinear elliptic equation
[ 5]. In this paper, we prove this for higher dimensions depending on the nonlinearity
where
which improves a similar result by Brezis and Vázquez
[ 4]. Abstract:
We combine the classical Gromov-Hausdorff metric [
Gromov- Hausdorff distance between maps of possibly different metric spaces. The latter is then combined with Walters's topological stability [ GH-stable homeomorphism. We prove that there are topologically stable homeomorphism which are not topologically GH-stable. Also that every topological GH-stable circle homeomorphism is topologically stable. Afterwards, we prove that every expansive homeomorphism with the pseudo-orbit tracing property of a compact metric space is topologically GH-stable. This is related to Walters's stability theorem [ Abstract:
Abstract. In this paper, the equivariant degree theory is used to analyze the occurrence of the Hopf bifurcation under effectively verifiable mild conditions. We combine the abstract result with standard interval polynomial techniques based on Kharitonov's theorem to show the existence of a branch of periodic solutions emanating from the equilibrium in the settings relevant to robust control. The results are illustrated with a number of examples.
Abstract:
The goal of this paper is to study the family of singular perturbations of Blaschke products given by
Abstract:
Let
In this paper, instead of using the conventional extension method introduced by Caffarelli and Silvestre, we employ a direct
method of moving planes for the fractional Laplacian to obtain the monotonicity and symmetry of the positive solutions of a semi-linear equation involving the fractional Laplacian. By using the integral definition of the fractional Laplacian, we first introduce various maximum principles which play an important role in the process of moving planes. Then we establish the monotonicity and symmetry of positive solutions of the semi-linear equations involving the fractional Laplacian. Abstract:
In this paper we consider a definition of Morse-Smale evolution process that extends the notion of Morse-Smale dynamical system to the nonautonomous framework. In particular we consider nonautonomous perturbations of autonomous systems. In this case our definition of Morse-Smale evolution process holds for perturbations of Morse-Smale autonomous systems with or without periodic orbits. We establish that small nonautonomous perturbations of autonomous Morse-Smale evolution processes derived from certain nonautonomous differential equations are Morse-Smale evolution processes. We apply our results to examples of scalar parabolic semilinear differential equations generating evolution processes and possessing periodic orbits.
Abstract:
We prove locally in time the existence of the unique smooth solution (including smooth interface) to the multidimensional free boundary problem for the thin film equation with the mobility
n = 2 in the case of partial wetting. We also obtain the Schauder estimates and solvability for the Dirichlet and the Neumann problem for a linear degenerate parabolic equation of fourth order. Abstract:
We study
k-bonacci substitutions through the point of view of thermodynamic formalism. For each substitution we define a renormalization operator associated to it and examine its iterates over potentials in a certain class. We also study the pressure function associated to potentials in this class and prove the existence of a freezing phase transition which is realized by the only ergodic measure on the subshift associated to the substitution. Abstract:
The quadratic nonlinear wave equation on a one-dimensional torus with small initial values located in a single Fourier mode is considered. In this situation, the formation of metastable energy strata has recently been described and their long-time stability has been shown. The topic of the present paper is the correct reproduction of these metastable energy strata by a numerical method. For symplectic trigonometric integrators applied to the equation, it is shown that these energy strata are reproduced even on long time intervals in a qualitatively correct way.
Abstract:
The paper is concerned with a system of two coupled time-independent Gross-Pitaevskii equations in
in mahler's expansion p Abstract:
In this paper, we formulate a conjecture for a measure-preservation criterion of 1-Lipschitz functions defined on the ring
Z of p p-adic integers, in terms of Mahler's expansion. We then provide evidence for this conjecture in the case that p= 3, and verify that it also holds for a wider class of 1-Lipschitz functions that are everywhere differentiable on Z , which we call $\mathcal{ B}$-functions, in the sense of Anashin. p Abstract:
Let
Abstract:
We consider the defocusing energy-critical nonlinear Schrödinger equation with inverse-square potential
Abstract:
We give a fairly complete characterization of the exact components of a large class of uniformly expanding Markov maps of
Abstract:
In this paper, we study the existence of SRB measures for infinite dimensional dynamical systems in a Banach space. We show that if the system has a partially hyperbolic attractor with nontrivial finite dimensional unstable directions, then it has an SRB measure.
Abstract:
This paper studies the local existence of strong solutions to the Cauchy problem of the 2D simplified Ericksen-Leslie system modeling compressible nematic liquid crystal flows, coupled via $ρ$ (the density of the fluid), $u$ (the velocity of the field), and $d$ (the macroscopic/continuum molecular orientations). Notice that the technique used for the corresponding 3D local well-posedness of strong solutions fails treating the 2D case, because the $L^p$-norm ($p>2$) of the velocity $u$ cannot be controlled in terms only of $ρ^{\frac{1}{2}}u$ and $\nabla u$ here. In the present paper, under the framework of weighted approximation estimates introduced in [J. Li, Z. Liang, On classical solutions to the Cauchy problem of the two-dimensional barotropic compressible Navier-Stokes equations with vacuum, J. Math. Pures Appl. (2014) 640-671] for Navier-Stokes equations, we obtain the local existence of strong solutions to the 2D compressible nematic liquid crystal flows.
Abstract:
A type of nonautonomous
n-dimensional state-dependent delay differential equation (SDDE) is studied. The evolution law is supposed to satisfy standard conditions ensuring that it can be imbedded, via the Bebutov hull construction, in a new map which determines a family of SDDEs when it is evaluated along the orbits of a flow on a compact metric space. Additional conditions on the initial equation, inherited by those of the family, ensure the existence and uniqueness of the maximal solution of each initial value problem. The solutions give rise to a skew-product semiflow which may be noncontinuous, but which satisfies strong continuity properties. In addition, the solutions of the variational equation associated to the SDDE determine the Fréchet differential with respect to the initial state of the orbits of the semiflow at the compatibility points. These results are key points to start using topological tools in the analysis of the long-term behavior of the solution of this type of nonautonomous SDDEs. Abstract:
In this paper, we study a class of nonlinear Schrödinger equations involving the fractional Laplacian and the nonlinearity term with critical Sobolev exponent. We assume that the potential of the equations includes a parameter $λ$. Moreover, the potential behaves like a potential well when the parameter λ is large. Using variational methods, combining Nehari methods, we prove that the equation has a least energy solution which, as the parameter λ large, localizes near the bottom of the potential well. Moreover, if the zero set int
Abstract:
We prove new variational properties of the spatial isosceles orbits in the equal-mass three-body problem and analyze their linear stabilities in both the full phase space
θ as the parameter. This set of orbits always lies in a symmetric subspace Γ and we show that their linear stabilities in the full phase space Abstract:
We study parametrised families of piecewise expanding interval mappings
a. This is similar to a result by D.Schnellmann, but with different assumptions. Abstract:
In this paper, we study an anomalous diffusion model of Kirchhoff type driven by a nonlocal integro-differential operator. As a particular case, we are concerned with the following initial-boundary value problem involving the fractional $p$-Laplacian $\left\{ \begin{array}{*{35}{l}} {{\partial }_{t}}u+M([u]_{s, p}^{p}\text{)}(-\Delta)_{p}^{s}u=f(x, t) & \text{in }\Omega \times {{\mathbb{R}}^{+}}, {{\partial }_{t}}u=\partial u/\partial t, \\ u(x, 0)={{u}_{0}}(x) & \text{in }\Omega, \\ u=0\ & \text{in }{{\mathbb{R}}^{N}}\backslash \Omega, \\\end{array}\text{ }\ \ \right.$ where $[u]_{s, p}$ is the Gagliardo $p$-seminorm of $u$, $Ω\subset \mathbb{R}^N$ is a bounded domain with Lipschitz boundary $\partialΩ$, $1 < p < N/s$, with $0 < s < 1$, the main Kirchhoff function $M:\mathbb{R}^{ + }_{0} \to \mathbb{R}^{ + }$ is a continuous and nondecreasing function, $(-Δ)_p^s$ is the fractional $p$-Laplacian, $u_0$ is in $L^2(Ω)$ and $f∈ L^2_{\rm loc}(\mathbb{R}^{ + }_0;L^2(Ω))$. Under some appropriate conditions, the well-posedness of solutions for the problem above is studied by employing the sub-differential approach. Finally, the large-time behavior and extinction of solutions are also investigated.
Abstract:
This paper deals with a characterization of asymptotic stability for a class of dynamical systems in terms of smooth Lyapunov pairs. We point out that well known converse Lyapunov results for differential inclusions cannot be applied to this class of dynamical systems. Following an abstract approach we put an assumption on the trajectories of the dynamical systems which demands for an estimate of the difference between trajectories. Under this assumption, we prove the existence of a $C^∞$-smooth Lyapunov pair. We also show that this assumption is satisfied by differential inclusions defined by Lipschitz continuous set-valued maps taking nonempty, compact and convex values.
Abstract:
The original proof of Dacorogna-Moser theorem on the prescribed Jacobian PDE,
Abstract:
In this article we investigate the nonlinear stability of Hasimoto solitons, in energy space, for a fourth order Schrödinger equation (4NLS) which arises in the context of the vortex filament. The proof relies on a suitable Lyapunov functional, at the
Abstract:
We study nonlinear elliptic equations of strong $p(x)$-Laplacian type to obtain an interior Calderón-Zygmund type estimates by finding a correct regularity assumption on the variable exponent $p(x)$. Our proof is based on the maximal function technique and the appropriate localization method.
Abstract:
We analyze the effect of Robin boundary conditions in a mathematical model for a mitochondria swelling in a living organism. This is a coupled PDE/ODE model for the dependent variables calcium ion contration and three fractions of mitochondria that are distinguished by their state of swelling activity. The model assumes that the boundary is a permeable 'membrane', through which calcium ions can both enter or leave the cell. Under biologically relevant assumptions on the data, we prove the well-posedness of solutions of the model and study the asymptotic behavior of its solutions. We augment the analysis of the model with computer simulations that illustrate the theoretically obtained results.
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For $k \in \mathbb N$, we define $Q(k)$ as follows:
$Q(k): $ let $x, y \in \mathbb N$ and $x$ is a multiple of $3$ and $k = x + y$, then $FUN(x, y)$ terminates and returns $x + y$
I will prove $Q(k)$ using induction
Base Case: let $k = 0 \to iff (x = 0, y = 0)$
By line 1, 2, 5, $FUN(x, y)$ terminates and returns $0 + 0 = x + y$ as wanted
Inductive step: Let $k > 0$. Suppose $Q(j)$ holds whenever $0 \leq j < k$ [I.H]
What to prove: $Q(k)$ holds
Since $k > 0$, it follows that line 3-7 could run since line 1 is not satisfied.
Therefore two cases: $y > 0$ and $y \leq 0$
Case 1: If $y > 0$, then line 3, 4, and 7 will run.
Since $0 \leq k - 1 < k$, this mean IH will apply to $FUN(x+3, y - 1)$
By IH, $FUN(x+3, y - 1)$ terminates and returns $x + 3 + y - 1$
By line 3, 4 and 7. $FUN(x, y)$ terminates and returns $x + 3 + y - 1 - 2 = x + y$ by algebra, and as wanted
Case 2: If $y \leq 0$, then line 5, 6, and 7 will run since line 1, 3 are not satisfied
Since $0 \leq k - 1 < k$, this mean IH will apply to $FUN(x - 3, y )$
By IH, $FUN(x - 3, y )$ terminates and returns $x - 3 + y$
By line 5, 6 and 7. $FUN(x, y)$ terminates and returns $x - 3 + y + 3 = x + y$ by algebra, and as wanted
Therefore $Q(k)$ holds as wanted
This is my attempt above, not sure if I'm correct. I'm pretty confused on how to use IH in this or if my input size is even good. Is it correct? |
Equivalence of Definitions of Conic Section Contents Theorem
Let $\theta$ be half the opening angle of $C$.
Then $K$ is a
conic section, whose nature depends on $\phi$.
A
conic section is a plane curve which can be specified in terms of: a given straight line $D$ known as the directrix a given point $F$ known as a focus a given constant $\epsilon$ known as the eccentricity. $(1): \quad q = \epsilon \, p$
Then $K$ is a
conic section. Equation $(1)$ is known as the focus-directrix property of $K$. Proof
Hence:
$\beta = \angle PDQ$
Thus $PF$ and $PR$ are the same length:
$(1): \quad PR = PF$
From the right triangle $\triangle PQR$:
$PQ = PR \cos \alpha$
From the right triangle $\triangle PQD$:
$PQ = PD \sin \beta$
Thus:
\(\displaystyle PR \cos \alpha\) \(=\) \(\displaystyle PD \sin \beta\) \(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {PR} {PD}\) \(=\) \(\displaystyle \frac {\sin \beta} {\cos \alpha}\) \(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {PF} {PD}\) \(=\) \(\displaystyle \frac {\sin \beta} {\cos \alpha}\) from $(1)$ \(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {PF} {PD}\) \(=\) \(\displaystyle \frac {\cos \gamma} {\cos \alpha}\) where $\gamma = \angle DPQ$
Let us define:
$e = \dfrac {\cos \gamma} {\cos \alpha}$ It is noted that: $(1): \quad$ when $\gamma < \alpha$, that is, the tilting angle is less than half the opening angle of the cone, the conic section is an ellipse $(2): \quad$ when $\gamma = \alpha$, that is, the tilting angle equals half the opening angle of the cone, the conic section is a parabola $(3): \quad$ when $\gamma > \alpha$, that is, the tilting angle is greater than half the opening angle of the cone, the conic section is a hyperbola $(4): \quad$ when $\gamma = 0$, that is, the slicing plane is parallel to the plane of the circle $C$., the conic section is a circle and there is no directrix
The result follows.
$\blacksquare$ |
Graph illustrating consumer (red) and producer (blue) surpluses on a supply and demand chart
In mainstream economics,
economic surplus, also known as total welfare or Marshallian surplus (named after Alfred Marshall), refers to two related quantities. Consumer surplus or consumers' surplus is the monetary gain obtained by consumers because they are able to purchase a product for a price that is less than the highest price that they would be willing to pay. Producer surplus or producers' surplus is the amount that producers benefit by selling at a market price that is higher than the least that they would be willing to sell for.
In Marxian economics, the term
surplus may also refer to surplus value, surplus product and surplus labour. Overview
Economist Paul A. Baran introduced the concept of "
economic surplus" to deal with novel complexities raised by the dominance of monopoly capital. With Paul Sweezy, Baran elaborated the importance of this innovation, its consistency with Marx's labor concept of value, and supplementary relation to Marx's category of surplus value. [1]
On a standard supply and demand diagram, consumer surplus is the area (triangular if the supply and demand curves are linear) above the equilibrium price of the good and below the demand curve. This reflects the fact that consumers would have been willing to buy a single unit of the good at a price higher than the equilibrium price, a second unit at a price below that but still above the equilibrium price, etc., yet they in fact pay just the equilibrium price for each unit they buy.
Likewise, in the supply-demand diagram, producer surplus is the area below the equilibrium price but above the supply curve. This reflects the fact that producers would have been willing to supply the first unit at a price lower than the equilibrium price, the second unit at a price above that but still below the equilibrium price, etc., yet they in fact receive the equilibrium price for all the units they sell.
Consumer surplus
Consumer surplus is the difference between the maximum price a consumer is willing to pay and the actual price they do pay. If a consumer would be willing to pay more than the current asking price, then they are getting more benefit from the purchased product than they initially paid. An example of a good with generally high consumer surplus is drinking water. People would pay very high prices for drinking water, as they need it to survive. The difference in the price that they would pay, if they had to, and the amount that they pay now is their consumer surplus. Note that the utility of the first few liters of drinking water is very high (as it prevents death), so the first few liters would likely have more consumer surplus than subsequent liters.
The maximum amount a consumer would be willing to pay for a given quantity of a good is the sum of the maximum price they would pay for the first unit, the (lower) maximum price they would be willing to pay for the second unit, etc. Typically these prices are decreasing; they are given by the individual demand curve. For a given price the consumer buys the amount for which the consumer surplus is highest, where consumer surplus is the sum, over all units, of the excess of the maximum willingness to pay over the equilibrium (market) price. The consumer's surplus is highest at the largest number of units for which, even for the last unit, the maximum willingness to pay is not below the market price
The aggregate consumers' surplus is the sum of the consumer's surplus for all individual consumers. This can be represented graphically as shown in the above graph of the market demand and supply curves.
Calculation from supply and demand
The consumer surplus (individual or aggregated) is the area under the (individual or aggregated) demand curve and above a horizontal line at the actual price (in the aggregated case: the equilibrium price). If the demand curve is a straight line, the consumer surplus is the area of a triangle:
CS = \frac{1}{2} Q_{\mathit{mkt}} \left( {P_{\mathit{max}} - P_{\mathit{mkt}}} \right)
Where P
mkt is the equilibrium price (where supply equals demand), Q mkt is the total quantity purchased at the equilibrium price and P max is the price at which the quantity purchased would fall to 0 (that is, where the demand curve intercepts the price axis). For more general demand and supply functions, these areas are not triangles but can still be found using integral calculus. Consumer surplus is thus the definite integral of the demand function with respect to price, from the market price to the maximum reservation price (i.e. the price-intercept of the demand function): CS = \int^{P_{\mathit{max}}}_{P_{\mathit{mkt}}} D(P)\, dP, where D(P_{\mathit{max}}) = 0.
This shows that if we see a rise in the equilibrium price and a fall in the equilibrium quantity, then consumer surplus falls.
Distribution of benefits when price falls
When supply of a good expands, the price falls (assuming the demand curve is downward sloping) and consumer surplus increases. This benefits two groups of people: Consumers who were already willing to buy at the initial price benefit from a price reduction; also they may buy more and receive even more consumer surplus, and additional consumers who were unwilling to buy at the initial price but will buy at the new price and also receive some consumer surplus.
Consider an example of linear supply and demand curves. For an initial supply curve S
0, consumer surplus is the triangle above the line formed by price P 0 to the demand line (bounded on the left by the price axis and on the top by the demand line). If supply expands from S 0 to S 1, the consumers' surplus expands to the triangle above P 1 and below the demand line (still bounded by the price axis). The change in consumer's surplus is difference in area between the two triangles, and that is the consumer welfare associated with expansion of supply.
Some people were willing to pay the higher price P
0. When the price is reduced, their benefit is the area in the rectangle formed on the top by P 0, on the bottom by P 1, on the left by the price axis and on the right by line extending vertically upwards from Q 0.
The second set of beneficiaries are consumers who buy more, and new consumers, those who will pay the new lower price (P
1) but not the higher price (P 0). Their additional consumption makes up the difference between Q 1 and Q 0. Their consumer surplus is the triangle bounded on the left by the line extending vertically upwards from Q 0, on the right and top by the demand line, and on the bottom by the line extending horizontally to the right from P 1. Rule of one-half
The
rule of one-half estimates the change in consumer surplus for small changes in supply with a constant demand curve. Note that in the special case where the consumer demand curve is linear, consumer surplus is the area of the triangle bounded by the vertical line Q=0, the horizontal line P = P_{mkt} and the linear demand curve. Hence, the change in consumer surplus is the area of the trapezoid with i) height equal to the change in price and ii) mid-segment length equal to the average of the ex-post and ex-ante equilibrium quantities. Following the figure above, \Delta CS = \frac{1}{2} \left( {Q_1 + Q_0 } \right)\left( {P_1 - P_0} \right)
where:
CS = consumers' surplus Q 0 and Q 1 are, respectively, the quantity demanded before and after a change in supply P 0 and P 1 are, respectively, the prices before and after a change in supply See also References ^ Baran, P.A. & Sweezy, P.M. (2012). "Some Theoretical Implications". Monthly Review. 64 (3). Further reading
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Consider:
$$I=\int_{0}^{1}\frac{\ln(x)}{1+x}dx= \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$$You can show this result by noticing since $0<x<1$ that $$\frac{1}{1+x}= \sum_{n=0}^{\infty}(-x)^n$$ and performing term by term integration. You will need integration by parts to do that.
Now you can also show that:
$$-I=\int_{0}^{1}\int_{0}^{1}\frac{1}{1+xy}dydx$$ by expanding the integrand into a geometric series and doing term by term integration with respect to $y$ and the with respect to $x$.
Now, in the double integral, let $x=u+v,y=-u+v$
Then $\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\begin{vmatrix} 1 && 1\\ -1 && 1 \\ \end{vmatrix}=2$
The change of variables causes the double integral to become:
$$\iint_{T} \frac{2}{1+v^2-u^2}dvdu$$ where $T$ is the square with vertices $(0,0),(\frac{-1}{2},\frac{1}{2}),(0,1),(\frac{1}{2},\frac{1}{2})$
Now, we need to compute two double integrals (over two different subregions of the square) and sum them to get the final answer.
First let $\frac{-1}{2}\leq u \leq 0 , -u \leq v \leq 1+u$.
$$\int_{\frac{-1}{2}}^{0}\int_{-u}^{1+u} \frac{2}{1+v^2-u^2}dvdu=\int_{\frac{-1}{2}}^{0}\frac{2\tan^{-1}\left(\frac{1+u}{\sqrt{1-u^2}}\right)+2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du$$
For the first term on the right hand side, let : $z=2\tan^{-1}\left(\frac{1+u}{\sqrt{1-u^2}}\right),dz=\frac{1}{\sqrt{1-u^2}}du$. For the second term, if you draw a right triangle, notice $2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)=2\sin^{-1}\left(u\right)$ so let $z=\sin^{-1}\left(u\right),dz=\frac{1}{\sqrt{1-u^2}}du$ for the second term.When evaluating these two terms with these substitutions, the sum of these terms should be $\frac{\pi^2}{24}$.
On the other hand, let $0\leq u \leq \frac{1}{2} , u \leq v \leq 1-u$.
$$\int_{0}^{\frac{1}{2}}\int_{u}^{1-u} \frac{2}{1+v^2-u^2}dvdu=\int_{0}^{\frac{1}{2}}\frac{2\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right)-2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du$$
For the first term on the right hand side, let $z=2\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right),dz=\frac{-1}{\sqrt{1-u^2}}$ and for the second term, use $2\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)=2\sin^{-1}\left(u\right)$ and let $z=\sin^{-1}\left(u\right),dz=\frac{1}{\sqrt{1-u^2}}du$ for this.When evaluating these two terms with these substitutions, the sum of these terms should be $\frac{\pi^2}{24}$.
So:
$$\iint_{T} \frac{2}{1+v^2-u^2}dvdu=\frac{2\pi^2}{24}=\frac{\pi^2}{12}$$thus, $$I=\int_{0}^{1}\frac{\ln(x)}{1+x}dx=\frac{-\pi^2}{12}$$ |
Example 6.10
For each following figures discuss and state force direction and the momentum that act on the control volume due to.
Example 6.11
Fig. 6.11 Flow out of un symmetrical tank for example
A similar tank as shown in Figure is built with a exit located in uneven distance from the the right and the left and is filled with liquid. The exit is located on the left hand side at the front. What are the direction of the forces that keep the control volume in the same location? Hints, consider the unsteady effects. Look at the directions which the unsteady state momentum in the tank change its value.
Example 6.12
A large tank has opening with area, A. In front and against the opening there a block with mass of 50[\(kg\)]. The friction factor between the block and surface is 0.5. Assume that resistance between the air and the water jet is negligible. Calculated the minimum height of the liquid in the tank in order to start to have the block moving?
Solution 6.12
The solution of this kind problem first requires to know at what accuracy this solution is needed. For great accuracy, the effect minor loss or the loss in the tank opening have taken into account. First assuming that a minimum accuracy therefore the information was given on the tank that it large. First, the velocity to move the block can be obtained from the analysis of the block free body diagram (the impinging jet diagram).
Fig. 6.12 Jet impinging jet surface perpendicular and with the surface.
The control volume is attached to the block. It is assumed that the two streams in the vertical cancel each other. The jet stream has only one component in the horizontal component. Hence,
\[ \label{forceBlock:horz} F = \rho\,A\,{U_{exit}}^2 \tag{86} \] The minimum force the push the block is \[ \label{forceBlock:block} \rho\,A\,{U_{exit}}^2 = m\,g\,\mu \Longrightarrow U_{exit} = \sqrt{ \dfrac{m\,g\,\mu}{ \rho\,A} } \tag{87} \] And the velocity as a function of the height is \(U=\sqrt{\rho\,g\,h}\) and thus \[ \label{forceBlock:sol} h = \dfrac{m\,\mu}{ \rho^2\,A} \tag{88} \] It is interesting to point out that the gravity is relevant. That is the gravity has no effect on the velocity (height) required to move the block. However, if the gravity was in the opposite direction, no matter what the height will be the block will not move (neglecting other minor effects). So, the gravity has effect and the effect is the direction, that is the same height will be required on the moon as the earth. For very tall blocks, the forces that acts on the block in the vertical direction is can be obtained from the analysis of the control volume shown in Figure 6.12. The jet impinged on the surface results in out flow stream going to all the directions in the block surface. Yet, the gravity acts on all these "streams'' and eventually the liquid flows downwards. In fact because the gravity the jet impinging in downwards sled direction. At the extreme case, all liquid flows downwards. The balance on the stream downwards (for steady state) is \[ \label{forceBlock:downGov} \rho\,\overline{U_{out}}^{\,2} \cong \rho\,V_{liquid}\,g + m\,g \tag{89} \] Where \(V_{liquid}\) is the liquid volume in the control volume (attached to the block). The pressure is canceled because the flow is exposed to air. In cases were \(\rho\,V_{liquid}\,g > \rho\,\overline{U_{out}}^{\,2}\) the required height is larger. In the opposite cases the height is smaller. Contributors
Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license. |
I am dealing with a mechanics problem on a unit sphere where all the equations are written in terms of Cartesian components and wish to make a conversion to a system $n, t, s$ where $n$ is the unit normal to the surface, $t$ is tangential to the surface and $s$ is also tangential to the surface and orthogonal to $n$ and $t$.
In theory, this should be easy for a sphere because the normal vector is the same as the radial vector, then you can take the cross product of the normal vector and the unit vector in the $z$ direction to get the tangent vector.
$\textbf{t} = \textbf{n} \: \times \: \hat{\textbf{k}}. $
The second tangential vector is then obtained via
$ \textbf{s} = \textbf{n} \: \times \: \textbf{t}. $
This seems fine, but then say, I have a radial vector with Cartesian components $[r_x \: r_y \: r_z]^T$, but now I want it to have the equivalent tangential-normal components $[r_n \: r_t \: r_s]^T$, so what would the formula be to take the $x$, $y$ and $z$ components and convert them, as there will no longer be a formula to convert as there would be with Cartesian to spherical polar coordinates, for example.
Edit: Added example of conversion.
$\begin{bmatrix} r_{x} \\ r_{y} \\ r_{z} \end{bmatrix} = \begin{bmatrix} n_x & t_x & s_x \\ n_y & t_y & s_y \\ n_z & t_z & s_z \end{bmatrix} \begin{bmatrix} r_{n} \\ r_{t} \\ r_{s} \end{bmatrix}=\begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} r_{n} \\ r_{t} \\ r_{s} \end{bmatrix}$
Edit: Should the matrix giving the conversion actually be:
$\begin{bmatrix} r_{x} \\ r_{y} \\ r_{z} \end{bmatrix} = \begin{bmatrix} \sin \theta \cos \phi & \sin \phi & \cos \theta \cos \phi \\ \sin \theta \sin \phi & - \cos \phi & \cos \theta \sin \phi\\ \cos \phi & 0 &- \sin \theta \end{bmatrix} \begin{bmatrix} r_{n} \\ r_{t} \\ r_{s} \end{bmatrix} $ |
The simple regression linear model looks like $y = X\beta + \epsilon$ and among constraints we have independent white noise $\epsilon_i \sim N(0, \sigma^2)$. Fitting with least mean squares gives the know normal equations in the form:
$\hat{\beta} = (X^TX)^{-1}X^Ty$
To optimize with least mean square one does not use any statistical reasoning. However if you ask questions about the statistical properties of your sample statistic $\hat{\beta}$ you start too need those assumptions.
For example if you ask about statistical properties of your sample statistics this is something else. For example if we ask if the $\hat{\beta}$ is biased you might go with:
$E\hat{\beta} = E[(X^TX)^{-1}X^Ty] = E[(X^TX)^{-1}X^T(X\beta+\epsilon)] = E\beta + E[(X^TX)^{-1}X^T\epsilon]$
If we consider X as fixed constant we have that $E\hat{\beta} = E\beta + (X^TX)^{-1}X^TE[\epsilon] = \beta$ only if $\epsilon$ have independent observations.
If we go for variance of $\hat{\beta}$ we see that has a fixed component $\beta$ and a random noise $(X^TX)^{-1}X^T\epsilon$. So to find the variance of our estimator we look for
$Var(\hat{\beta}) = Var[\beta + (X^TX)^{-1}X^T\epsilon]$
Here the $\beta$ in variance have no effect, we can remove it. See properties of variance here: basic properties of variance on wikipedia. We also use the assumption that $X$ is given so it behaves like a constant term which factors out squared. We have then
$Var(\hat{\beta}) = ((X^TX)^{-1}X^T)^2 Var(\epsilon) = (X^TX)^{-1}Var(\epsilon)$
Now the answer to your question is, if we assume that variance is the same for all noise (additionally to independence) then we can say that
$Var(\beta) = \sigma^2 (X^TX)^{-1}$
And by estimating $\sigma$ you have all the good stuff about coefficients like standard errors, t values, p values, confidence intervals, hypothesis tests and so on. If the variance is not assumed to by heteroskedastic then you have to control it somehow, but things gets more complicated.
Even if the $X$ is not assumed as fixed variable, if is considered random variable a similar assumption on independence like $E[X\epsilon] = 0$ is enough together with heteroskedasticity to have the same results.
Hope that helps. |
Mixture models of only two distributions, $D_1$ and $D_2$, can be considered to have the density $p D_1+(1-p) D_2$, where $0<p<1$. Now $p\neq0,1$ because it would no longer be a mixture. In general for any mixture distribution, software is available to find the best mixture model, e.g., FindDistribution in Mathematica, or mixtools in R as suggested by @omidi. However, in the case given by the OP, there is no overlap between distributions, as a salary of zero is not a salary. There is no particular need to add one to the income to take logarithms, as there is no need to take logarithms. Instead, all that need be done is to assign a Dirac $\delta$ for the $D_1$, zero salaries, i.e., $p \delta(x=0) +(1-p)D_2$, and then find $p$ and $D_2$. Finding $p$ is trivial, as $p=\frac{N_{no}}{N_{no}+N_{yes}}$, where $N_{yes}$ and $N_{no}$ are the number of subjects with and without income, respectively. Finding the best distribution for those with incomes can use software, or prior literature to search for models. However, even then, the distribution of those with income, $D_2$ here, can itself a mixture distribution. Moreover, use of an empirical distribution may be enough to answer the some of the questions the OP needs to answer, and although it's nice to have a theoretical distribution, it would not be absolutely required.
That is, the final distribution could be $p \delta(0) +(1-p)D_{Emp}(x)$, where $p=\frac{N_{no}}{N_{no}+N_{yes}}$. To be clear, that formulation is identical to what the data is, as the empirical distribution is $\frac{\delta(x_i)}{N_{yes}}$.
Other questions: "...set the mean for one of them [
Sic, Gaussian distributions] to be known as equal to 0." That is what the Dirac $\delta(0)$ is; although the first distribution used to create it was historically Cauchy, it makes little difference which limit one uses to create the $\delta$, because its standard deviation is zero. Regarding using a Gaussian for the second distribution, not a good idea as the right tail of income is notoriously heavy, e.g., see Pareto distribution. If a theoretical distribution is desired in R for the OP's data for those having an income, see this.
Finally, I do not fully understand "Another problem might be that I am always predicting the income using a regression and building a ranking from that, rather than running an ordinal regression. What is the best way handle this situation - if the target variable (income) that the ranking is based on is itself available for training data?"
If one desires to predict income using regression, transforming variables may be desirable, but no transformation of $p \delta(0) +(1-p)D_{Emp}(x)$ will have "nice" residuals. That is, the residuals will not be Gaussian or homoscedastic, so the usual default regression techniques will not yield accurate answers. Maybe the way to treat this is to use classifiers to find the probability of finding a job with a yes/no Y-axis variable, and if yes use
glm to find what that salary is for a best theoretical distribution of non-zero salaries. That is, one can then use classifiers to determine predictors of non-zero salary, and when finished, one then has two cases, predictors of getting a job, and predictors of salary if one secures a position. |
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Kaon femtoscopy in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV
(Elsevier, 2017-12-21)
We present the results of three-dimensional femtoscopic analyses for charged and neutral kaons recorded by ALICE in Pb-Pb collisions at $\sqrt{s_{\rm{NN}}}$ = 2.76 TeV. Femtoscopy is used to measure the space-time ...
Anomalous evolution of the near-side jet peak shape in Pb-Pb collisions at $\sqrt{s_{\mathrm{NN}}}$ = 2.76 TeV
(American Physical Society, 2017-09-08)
The measurement of two-particle angular correlations is a powerful tool to study jet quenching in a $p_{\mathrm{T}}$ region inaccessible by direct jet identification. In these measurements pseudorapidity ($\Delta\eta$) and ...
Online data compression in the ALICE O$^2$ facility
(IOP, 2017)
The ALICE Collaboration and the ALICE O2 project have carried out detailed studies for a new online computing facility planned to be deployed for Run 3 of the Large Hadron Collider (LHC) at CERN. Some of the main aspects ...
Evolution of the longitudinal and azimuthal structure of the near-side peak in Pb–Pb collisions at $\sqrt{s_{\rm NN}}=2.76$ TeV
(American Physical Society, 2017-09-08)
In two-particle angular correlation measurements, jets give rise to a near-side peak, formed by particles associated to a higher $p_{\mathrm{T}}$ trigger particle. Measurements of these correlations as a function of ...
J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV
(American Physical Society, 2017-12-15)
We report a precise measurement of the J/$\psi$ elliptic flow in Pb-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV with the ALICE detector at the LHC. The J/$\psi$ mesons are reconstructed at mid-rapidity ($|y| < 0.9$) ...
Enhanced production of multi-strange hadrons in high-multiplicity proton-proton collisions
(Nature Publishing Group, 2017)
At sufficiently high temperature and energy density, nuclear matter undergoes a transition to a phase in which quarks and gluons are not confined: the quark–gluon plasma (QGP)1. Such an exotic state of strongly interacting ...
K$^{*}(892)^{0}$ and $\phi(1020)$ meson production at high transverse momentum in pp and Pb-Pb collisions at $\sqrt{s_\mathrm{NN}}$ = 2.76 TeV
(American Physical Society, 2017-06)
The production of K$^{*}(892)^{0}$ and $\phi(1020)$ mesons in proton-proton (pp) and lead-lead (Pb-Pb) collisions at $\sqrt{s_\mathrm{NN}} =$ 2.76 TeV has been analyzed using a high luminosity data sample accumulated in ...
Production of $\Sigma(1385)^{\pm}$ and $\Xi(1530)^{0}$ in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV
(Springer, 2017-06)
The transverse momentum distributions of the strange and double-strange hyperon resonances ($\Sigma(1385)^{\pm}$, $\Xi(1530)^{0}$) produced in p-Pb collisions at $\sqrt{s_{\rm NN}} = 5.02$ TeV were measured in the rapidity ...
Charged–particle multiplicities in proton–proton collisions at $\sqrt{s}=$ 0.9 to 8 TeV, with ALICE at the LHC
(Springer, 2017-01)
The ALICE Collaboration has carried out a detailed study of pseudorapidity densities and multiplicity distributions of primary charged particles produced in proton-proton collisions, at $\sqrt{s} =$ 0.9, 2.36, 2.76, 7 and ...
Energy dependence of forward-rapidity J/$\psi$ and $\psi(2S)$ production in pp collisions at the LHC
(Springer, 2017-06)
We present ALICE results on transverse momentum ($p_{\rm T}$) and rapidity ($y$) differential production cross sections, mean transverse momentum and mean transverse momentum square of inclusive J/$\psi$ and $\psi(2S)$ at ... |
Under the auspices of the Computational Complexity Foundation (CCF)
We study the complexity of arithmetic in finite fields of characteristic two, $\F_{2^n}$.
We concentrate on the following two problems:
Iterated Multiplication: Given $\alpha_1, \alpha_2,..., \alpha_t \in \F_{2^n}$, compute $\alpha_1 \cdot \alpha_2 \cdots \alpha_t \in \F_{2^n}$.
Exponentiation: Given $\alpha \in \F_{2^n}$ and a $t$-bit integer $k$, compute $\alpha^k \in \F_{2^n}$.... more >>>
The problem of finding a nontrivial factor of a polynomial $f(x)$ over a finite field $\mathbb{F}_q$ has many known efficient, but randomized, algorithms. The deterministic complexity of this problem is a famous open question even assuming the generalized Riemann hypothesis (GRH). In this work we improve the state of the ... more >>>
Fix a prime $p$. Given a positive integer $k$, a vector of positive integers ${\bf \Delta} = (\Delta_1, \Delta_2, \dots, \Delta_k)$ and a function $\Gamma: \mathbb{F}_p^k \to \F_p$, we say that a function $P: \mathbb{F}_p^n \to \mathbb{F}_p$ is $(k,{\bf \Delta},\Gamma)$-structured if there exist polynomials $P_1, P_2, \dots, P_k:\mathbb{F}_p^n \to \mathbb{F}_p$ ... more >>>
Constructing $r$-th nonresidue over a finite field is a fundamental computational problem. A related problem is to construct an irreducible polynomial of degree $r^e$ (where $r$ is a prime) over a given finite field $\F_q$ of characteristic $p$ (equivalently, constructing the bigger field $\F_{q^{r^e}}$). Both these problems have famous randomized ... more >>> |
Unless you transform a two-term surd expression under a square root, you can't solve the surd problem
We have given a brief introduction to surds in our article
and Fractions and decimals part 1, . You may like to go through these before going ahead further in this session. How to solve surds 1, Rationalization
In this session we will learn about
three additional surd techniques to solve surd problems, Unless we free up the surd expression under square root we can't deal in any way with the double square root situation—use of Double square root surds: How to solve surd problems, where a two-term surd expression appears under a second square root. . conversion to square of sum surd expression . Surd term factoring: How to solve surd problems by taking a factor out of a surd term on two sides of an equation—use of Surd coefficient comparison: How to solve a surd problem by comparing and equalizing coefficients of surd and non-surd terms Coefficient equalization of similar variables.
Through solving of chosen SSC CGL level problems we will show how to apply these three different techniques to solve apparently difficult surd problems effectively.
Double square root surd expression and Conversion to square of sum surd expression technique
Three examples of double square root surd expressions are,
$\sqrt{10+2\sqrt{21}}$
$\sqrt{3+2\sqrt{2}}$
$\sqrt{9+4\sqrt{5}}$
In many surd simplification problems such double squre root surds appear. Unless we free up the surd expression from the surrounding square root there is no way we can proceed further.
The obvious and simple way to do this is,
To express the two term surd expression under square root as a square of another two-term surd expression.
Sometimes it is easy, other times transforming to a square of sum a bit tricky.
Let us convert the three examples.
Example 1.
$\sqrt{10+2\sqrt{21}}$
$=\sqrt{7 + 2\times{\sqrt{7}}\times{\sqrt{3}} + 3}$
$=\sqrt{(\sqrt{7}+\sqrt{3})^2}$
$=\sqrt{7}+\sqrt{3}$.
The surd under square root is now free of enclosing square root but in changed form.
We have used the algebraic expression,
$a^2 + 2ab+b^2=(a+b)^2$
and attempted to determine the values of $a$ and $b$ in the surd expression so that the two term expression under square root can be converted to a square of two term surd expression.
Coefficient 2 identifies the middle term: The main help is provided by the middle term of the form of $2ab$. Isolating the coefficient 2, the rest should be broken up into two factors, both or one of which should be a surd, so that after squaring each of these and summing up, result becomes the second numeric term of the original two term surd expression.
In this case, middle term without coefficient 2 is $\sqrt{21}$. It consists of a unique pair of two surd factors, $\sqrt{7}$ and $\sqrt{3}$. After squaring each and taking their sum, we get the result as 10 as expected.
This is is easy to transform.
Example 2
The second example is,
$\sqrt{3+2\sqrt{2}}$
$=\sqrt{2 +2\times{\sqrt{2}}\times{1}+1}$
$=\sqrt{(\sqrt{2}+1)^2}$
This is also easy.
Let us take the third example,
Example 3.
The third example is,
$\sqrt{9+4\sqrt{5}}$.
This example is a little different. The middle term coefficient after excluding 2 is $2\sqrt{5}$. So one term must be 2 and the second $\sqrt{5}$, squaring each and adding we get 9, satisfying the given expression.
So,
$\sqrt{9+4\sqrt{5}}$.
$=\sqrt{(\sqrt{5}+2)^2}$.
We follow the convention of writing the larger term of the surd expression first.
So with a little bit of practice you should be able to deal with such double square root surd expressions.
Notice that the method in all three cases depends on isolating the coefficient 2 of the middle term. What do we do if such a coefficient of 2 is not present in the two term surd expression under square root!
Example 4 of middle term of a double square root surd expression without factor 2
Let us take a very common example of a two term surd expression $(2 + \sqrt{3})$, it is a frequently occurring surd expression. Unfortunately sometimes it is put under a square root,
$\sqrt{2+\sqrt{3}}$.
How should we eliminate the enclosing square root in this case?
The method follows from two step deductive reasoning, If the surd under root is to be freed of square root, it must be expressed as $a^2 +2ab +b^2$. If 2 is absent in any term, we must supply 2 artificially by multiplying and dividing the surd expression under square root itself by 2, and then on start analyzing the expression.
Let us apply this new method to the fourth double square root surd expression.
Our expression was,
$\sqrt{2+\sqrt{3}}$.
Multiplying and dividing the surd expression under square root by 2 we get,
$\displaystyle\frac{1}{\sqrt{2}}\sqrt{4+2\sqrt{3}}$
$=\displaystyle\frac{1}{\sqrt{2}}\sqrt{(\sqrt{3}+1)^2}$
$=\displaystyle\frac{1}{\sqrt{2}}(\sqrt{3}+1)$.
Only this type of expression is a little tricky, but with practice, dealing with any form should become easy.
Before taking up application of these methods in solving actual problems let us show in what form of surd expressions, Surd factoring technique can be applied.
Surd term factoring technique Examples
The following two examples of surd expressions might appear normal expressions with no other action on them possible.
$2 + \sqrt{6}$
$3 + \sqrt{15}$.
If you look closely you will find the surd term under the square root consists of two prime factors. If one of these two factors also matches with the second numeric term, there is a strong case of surd term factoring.
Let us see how.
The first example,
$2+\sqrt{6}=\sqrt{2}(\sqrt{3}+\sqrt{2})$.
We have taken out the factor $\sqrt{2}$ not only out of the surd term, but also out of the numeric term. Often you will find that the transformed surd expression either appears both in numerator and denominator or its complementary surd expression, in this case, $(\sqrt{3}-\sqrt{2})$ appears as another factor, so that the expression immediately gets simplified to a great extent.
Similarly in the second example,
$3+\sqrt{15}=\sqrt{3}(\sqrt{5}+\sqrt{3})$.
In both these cases we have taken the common factor out of both the two terms in the surd expression. Sometimes we take a factor out of the single surd term. Example,
$5+\sqrt{24}=5+2\sqrt{6}=(\sqrt{3}+\sqrt{2})^2$.
By taking out factor 4 from 24 under square root, the expression is so transformed that it could be expressed as a square of sum expression.
With this background, let us now solve a few SSC CGL level surd problems where double square root surd expressions appear as well as we need to use surd term factoring technique.
Solving double square root surd problems and use Surd term factoring when needed Problem example 1.
The value of $\sqrt{5+2\sqrt{6}} - \displaystyle\frac{1}{\sqrt{5+2\sqrt{6}}}$ is,
$\sqrt{5}-1$ $1+\sqrt{5}$ $2\sqrt{2}$ $\sqrt{2}$ Solution 1. Problem example 1.
Let us first simplify the subtractive sum of inverses,
$E=\sqrt{5+2\sqrt{6}} - \displaystyle\frac{1}{\sqrt{5+2\sqrt{6}}}$
$=\displaystyle\frac{4+2\sqrt{6}}{\sqrt{5+2\sqrt{6}}}$
$=\displaystyle\frac{4+2\sqrt{6}}{\sqrt{(\sqrt{3}+\sqrt{2})^2}}$
$=\displaystyle\frac{2(2+\sqrt{6})}{\sqrt{3}+\sqrt{2}}$.
Taking factor $\sqrt{2}$ out of the two terms in the numerator,
$E=\displaystyle\frac{2\sqrt{2}(\sqrt{3}+\sqrt{2})}{\sqrt{3}+\sqrt{2}}$
$=2\sqrt{2}$.
Answer: Option c: $2\sqrt{2}$.
We have not only taken the surd expression in the denominator out of its enclosing square roots, but also identified the pattern of convenient surd factoring case on the numerator by taking the factor $\sqrt{2}$ both out of $2$ and $\sqrt{6}$.
This type of factorization is not readily visible and a closer analytical look is required to identify the possibility. This technique is called which many times makes simplification easy. Surd term factoring In this example, need to rationalize the denominator didn't arise at all. Problem example 2.
The value of $\sqrt{\displaystyle\frac{(\sqrt{12}-\sqrt{8})(\sqrt{3}+\sqrt{2})}{5+\sqrt{24}}}$ is,
$\sqrt{6}-\sqrt{2}$ $2-\sqrt{6}$ $\sqrt{6}-2$ $\sqrt{6}+\sqrt{2}$ Solution 2. Problem example 2.
First we need to simplify the three factor expression and only then think of double square root technique if the situation demands.
Simplifying first factor in the numerator by Surd term factoring,
$\sqrt{12}-\sqrt{8}=2(\sqrt{3}-\sqrt{2})$.
Multiplying this with the second factor using $(a+b)(a-b)=a^2-b^2$, we get in the numerator, the result of just $2$.
Now we need to give attention to the denominator expression. Taking 4 out of square root term 24,
$5+\sqrt{24}=5+2\sqrt{6}=(\sqrt{3}+\sqrt{2})^2$.
So the given expression is transformed to,
$E=\displaystyle\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}$.
Now we need to rationalize the denominator,
$E=\sqrt{2}(\sqrt{3}-\sqrt{2})=\sqrt{6}-2$.
Answer. Option c: $\sqrt{6}-2$.
To solve this problem, we needed to use all three surd problem solving techniques we have learned till now,
Surd rationalization Conversion to square of sum surd expression, and Surd term factoring.
Though the third technique seems to be lightweight in comparison to the other two, sometimes ability to see the factoring possibility becomes crucial. That was the case in the first problem as well as in the second problem.
Let us now solve a third SSC CGL level surd problem.
Problem example 3.
The value of $\displaystyle\frac{1}{\sqrt{12-\sqrt{140}}}-\displaystyle\frac{1}{\sqrt{8-\sqrt{60}}}-\displaystyle\frac{2}{\sqrt{10+\sqrt{84}}}$ is
3 0 1 2 Solution 3: Problem analysis and solving execution
Whenever we meet a surd expression under a square root, we know the surd expression under the square root must be converted to a square of sum. This problem has three such expressions. In each we take out first the factors from the surd term.
Taking these one by one,
$12 - \sqrt{140}=12-2\sqrt{35}=(\sqrt{7}-\sqrt{5})^2$,
$8-\sqrt{60}=8-2\sqrt{15}=(\sqrt{5}-\sqrt{3})^2$, and
$10 +\sqrt{84}=10+2\sqrt{21}=(\sqrt{7}+\sqrt{3})^2$.
Surd sums in the brackets are then the three denominators (after taking out of the square roots). We rationalize to form the simplified expression,
$\displaystyle\frac{\sqrt{7}+\sqrt{5}}{2}-\displaystyle\frac{\sqrt{5}+\sqrt{3}}{2}-\displaystyle\frac{\sqrt{7}-\sqrt{3}}{2}$
$=0$, all terms cancel out.
Answer: Option b: 0.
To solve this third problem also we applied all the three techniques we have learned,
Surd term factoring Conversion to square of sum surd expression, and Surd rationalization.
We will now end this session with application of the fourth surd solving technique, namely,
between two sides of an equation. Surd coefficient comparison and equalization of similar variables Coefficient equalization of similar variables Problem example 4.
If $\displaystyle\frac{4+3\sqrt{3}}{\sqrt{7+4\sqrt{3}}}=A + \sqrt{B}$ then $B-A$ is,
$-13$ $3\sqrt{3}-7$ $13$ $\sqrt{13}$ Solution 4: Problem analysis and solving execution
First we convert the denominator surd expression under square root to a square of sum,
$7 + 4\sqrt{3}=(2+\sqrt{3})^2$.
Rationalizing the transformed denominator $2+\sqrt{3}$, we get the given expession as,
$(4+3\sqrt{3})(2-\sqrt{3})=A + \sqrt{B}$,
Or, $-1+2\sqrt{3}=A+\sqrt{B}$.
As $\sqrt{B}$ is the surd term it must be equal to the surd term on the LHS (irrational and rational can't be added, though we represent an addition we can't arrive at any result by carrying out the addition).
This is what we call
that don't mix together. This is a Coefficient equalization of similar variables . fundamental algebraic principle
Thus,
$A=-1$, and
$2\sqrt{3}=\sqrt{B}$,
Or, $B=12$, and
$B-A=13$.
Answer: Option c: 13.
To solve this last problem also we needed three methods,
First, conversion to square of sum surd expression, Second, surd rationalization, and Third, Coefficient equalization of similar variables.
By applying these three methods along with Surd factoring whenever needed, most of the surd problems can be solved.
In the next tutorial session on how to solve surds, we will take up the more involved problems of surd expression comparison ranking.
Concept tutorials on Fractions, Surds, decimals and related topics How to solve surds part 2, Double square root surds and surd term factoring |
Submitted by Lanowen on Sun, 05/24/2015 - 02:15
A friend asked me about this problem, on how to calculate the distance an object needs to be from the camera to appear a certain height on the screen.
The solution is simpler when dealing with an object centered in the in the screen where a right angle can be created from the camera to the object.
Fig. 1: right angle triangle diagram. $\figLbl{1}{tanSol}$
Solving the triangle in \ref{fig:tanSol}:
\begin{equation}$\vdots$
\tan{\left( \frac{\theta}{2} \right)} = \frac{h}{2d} \label{eq:tanTriangle} \end{equation} |
I have a parallelogram $ABCD$ with side length $AB=58.7 cm$ and $BC=65.8 cm$ and $\angle DAB=120$ degrees.
I need to find the possible value(s) of $\angle CAB$.
Attempt.This seems like a standard question, but I can't seem to figure out whether the angle is ambiguous or not, and why cosine rule would only give one solution but sine rule would give two solutions.
Using cointerior angles we can deduce $\angle ABC=60$ degrees.
$$AC=\sqrt{AB^2+BC^2-2AB\times BC\times \cos(60)}\approx62.55$$
Now to find $\angle CAB$ I have the option of using sine rule or cosine rule.
$$\frac{\sin(\angle CAB)}{65.8}=\frac{\sin(60)}{62.6}$$ $$\angle CAB=65^{\circ}39' \text{ or } 114^{\circ}22'$$
But if I find $\angle CAB$ with cosine rule I get $$\cos(\angle CAB)=\frac{58.7^2+AC^2-65.8^2}{2\times58.7\times AC}$$ $$\angle CAB=65^{\circ}38'$$
Are both solutions valid? Or is only the acute angle valid? How would you tell whether this is an ambiguous case or not?
Thank you! |
Choose the most appropriate word from the options given below to complete the following sentence.
The principal presented the chief guest with a ________, as token of appreciation.
Choose the appropriate word/phrase, out of the four options given below, to complete the following sentence:
Frogs__________.
Choose the word most similar in meaning to the given word:
Educe
Operators $\square,\Diamond\;\mathrm{and}\;\rightarrow$ are defined by $\mathrm a\square\mathrm b\;=\;\frac{\mathrm a-\mathrm b}{\mathrm a+\mathrm b};\;\mathrm a\Diamond\mathrm b\;=\frac{\mathrm a+\mathrm b}{\mathrm a-\mathrm b};\;\mathrm a\rightarrow\mathrm b=\mathrm{ab}$
Find the value of $(66\square6\;)\rightarrow(66\;\Diamond6)$
If logx=(5/7)=-1/3,then the value of x is
The following question presents a sentence, part of which is underlined. Beneath the sentence you find four ways of phrasing the underlined part. Following the requirements of the standard written English, select the answer that produces the most effective sentence.
Tuberculosis, together with its effects, ranks one of the leading causes of death in India.
Read the following paragraph and choose the correct statement.
Climate change has reduced human security and threatened human well being. An ignored reality of human progress is that human security largely depends upon environmental security. But on the contrary, human progress seems contradictory to environment security. To keep up both at the required level is a challenge to be addressed by one and all. One of the ways to curb the climate change may be suitable scientific innovations, while the other may be the Gandhian perspective on small scale progress with focus on sustainability.
Find the missing Value:
A cube of side 3 units is formed using a set of smaller cubes of side 1 unit; Find the proportion of the number of faces of the smaller cubes visible to those which are NOT visible.
Humpty Dumpty sits on a every day while having lunch. The wall sometimes breaks. A person sitting on the wall falls if the wall breaks.
Which one of the statements below is logically valid and can be inferred from the above sentences?
Consider a system of linear equations :
x – 2y +3z = –1
x – 3y + 4z = 1 and
–2x +4y – 3z = k.
The value of k for which the system has infinitely many solutions is _______.
A function f (x) = 1 – x2 + x3 is defined in the closed interval [–1,1]. The value of x, in the open interval (–1,1) for which the mean value theorem is satisfied, is
Suppose $ A $ and $ B $ are two independent events with probabilities $ P(A)\neq0 $ and $ P(B)\neq0 $. Let A and B be their complements Which of the following statement is FALSE?
Let $ z=x+iy $ be a complex variable. Consider that contour integration is performed along the unit circle in anticlockwise direction. Which one of the following statements is NOT TRUE?
The value of $ p $ such that the vector 123 is an eigenvector of the matrix $ \begin{bmatrix}4&1&2\\p&2&1\\14&-4&10\end{bmatrix} $ is_______.
In the circuit shown, at resonance, the amplitude of the sinusoidal voltage (in Volts) across the capacitor is ______ .
In the network shown in the figure, all resistors are identical with R = 300 Ω. The resistance Rab (in Ω) of the network is __________.
In the given circuit, the values of $ V_1 $ and $ V_2 $ respectively are
A region of negative differential resistance is observed in the current voltage characteristics of a silicon PN junction if
A silicon sample is uniformly doped with donor type impurities with a concentration of 1016/cm3. The electron and hole mobilities in the sample are 1200 cm2/V-s and 400 cm2/V-s respectively. Assume complete ionization of impurities. The charge of an electron is 1.6 X 10-19 C. The resistivity of the sample (in Ω-cm) is _______.
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It is our sincere effort to help you. |
Submitted by Lanowen on Sun, 05/24/2015 - 02:15
A friend asked me about this problem, on how to calculate the distance an object needs to be from the camera to appear a certain height on the screen.
The solution is simpler when dealing with an object centered in the in the screen where a right angle can be created from the camera to the object.
Fig. 1: right angle triangle diagram. $\figLbl{1}{tanSol}$
Solving the triangle in \ref{fig:tanSol}:
\begin{equation}$\vdots$
\tan{\left( \frac{\theta}{2} \right)} = \frac{h}{2d} \label{eq:tanTriangle} \end{equation} |
I am looking for the name and a reference for a $\Delta_2^P$-complete problem that looks like the following
Input:A collection of CNF formulas $\phi_i(x_1^i, x_2^i,\dots, x_m^i, z_1, z_2, \dots, z_{i-1})$ for $1 \leq i \leq n$ where the $x_j^i$ are free variables and the $z_i$ variables are bound, and the value of $z_i$ is
true if $\phi_i$ is satisfiable and
false if $\phi_i$ is unsatisfiable.
Output: Whether $\phi_n$ is satisfiable.
I looked at https://cs.stackexchange.com/questions/14251/which-problems-are-hard-for-pnp and at https://mathoverflow.net/questions/2218/characterize-pnp-a-k-a-delta-2p but couldn't find what I'm looking for. I believe I came across a paper mentioning a problem defined similarly to the one above a couple months ago, but I don't remember which paper was listing it. I am not 100% sure about my recollection of the problem definition and this is one of the reasons behind this reference request. |
Suppose $(X^m, g)$ is a closed Riemannian manifold of dimension $m$ with the following properties,
There is a constant $\kappa$ such that $\kappa r^m \leq Vol(B(x, r)) \leq \kappa^{-1} r^m$ for every $r \in (0,1)$.
For every $C^1$ function $f$, we have $(\int_X |f|^{\frac{2m}{m-2}})^{\frac{m-2}{m}} \leq C_S (\int_X |f|^2 + |\nabla f|^2)$.
For every $C^1$ function $f$, we have $\int_X |f|^2 \leq C_P (\int_X |\nabla f|^2 + (\int_X f)^2)$.
Can we say that there is a positive constant $I$ depending only on $m, \kappa, C_S, C_P$ such that $\frac{Area(\partial \Omega)^m}{Vol(\Omega)^{m-1}} \geq I$ for every domain $\Omega$ satisfying $Vol(\Omega) \leq \frac{1}{2} Vol(X)$? |
23rd SSC CGL Tier II level Solution Set, 2nd on topic Ratio and Proportion
This is the 23rd solution set of 10 practice problem exercise for SSC CGL Tier II exam and 2nd on topic Ratio and Proportion. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set for extracting maximum benefits from this resource.
If you have not taken the test yet, you can refer to
, and then after taking the test come back to this solution. SSC CGL Tier II level Question Set 23, Ratio proportion 2 23rd solution set—10 problems for SSC CGL Tier II exam: topic Ratio and Proportion 2—Answering time 12 mins Problem 1.
Ratio of number of boys to the number of girls in a school of 432 students is 5 : 4. When some new boys and girls are admitted, the number of boys increased by 12 and the ratio of boys to girls changed to 7 : 6. The number of new girls admitted is,
12 14 20 24 Solution 1: Problem analysis and solving in mind by multiple of ratio value technique
By introducing the cancelled out HCF in original ratio of boys to girls 5 : 4 as $x$, the total number of students originally was,
$5x+4x=9x=432$,
Or, $x=48$.
The original number of boys and girls were then respectively,
$5x=240$, and
$4x=192$.
After the new admission, with additional 12 new boys number boys changed to, $240+12=252$.
Assuming the number of new girls to be $a$, the new ratio is expressed as,
$\displaystyle\frac{252}{192+a}=\frac{7}{6}$.
Applying multiple of ratio value technique we find that LHS numerator ratio value, $252=7\times{36}$, where 7 is the RHS numerator ratio value. The cancelled out HCF in this new ratio must then be 36.
The LHS denominator will then be,
$6\times{36}=216$.
And number of new girls,
$216-192=24$.
Answer: Option d: 24. Key concepts used: -- Basic ratio concepts Rich ratio concepts -- HCF reintroduction technique -- Multiple of ratio value technique -- Basic factors multiples concept -- Solving in mind.
With conceptual clarity, the problem can easily be solved wholly in mind.
Problem 2.
If by increasing the price of a ticket in the ratio 8 : 11 the number of tickets sold falls in the ratio 23 : 21 then what is the increase in revenue if revenue before increase in price of ticket were Rs.36800?
Rs.7850 Rs.12850 Rs.9400 Rs.21250 Solution 2: Problem analysis and solving in mind by product of ratios concept
By introducing the cancelled out HCF in the first and second ratio as $x$ and $y$, the price of ticket and number of tickets sold before price rise were,
$8x$ and $23y$.
And the revenue before the price rise was,
$8\times{23}xy=36800$.
This is a product of corresponding ratio terms and the product of two HCF's $xy$ can be treated as a single variable. We name it as product of ratios concept.
Simplifying,
$xy=200$.
In the same way price and number of tickets sold after the price rise were respectively,
$11x$ and $21y$.
So the revenue after the price rise became,
$11\times{21}xy=231\times{200}=46200$.
The revenue increase is then,
$46200-36800=9400$.
Answer: Option c: Rs.9400. Key concepts used: Basic ratio concepts -- HCF reintroduction technique -- New variable revenue as product of corresponding terms of two ratios so that product of two introduced HCFs $xy$ is used as a single variable -- -- Product of ratios . Solving in mind Problem 3.
Rs.2420 were divided among A, B and C so that A : B = 5 : 4 and B : C = 9 : 10. Then C gets,
Rs.900 Rs.680 Rs.950 Rs.800 Solution 3: Problem analysis and solving in mind by ratio joining
Share of each would be proportional to the corresponding ratio value in the joint three-party ratio. So we need to join the two ratios to get the three-party joint ratio of shares of A, B and C.
To join the two given ratios, the ratio value corresponding to the middle variable B is to equalized in the two ratios to the LCM of the two values 4 and 9, that is, 36.
The first ratio A : B = 5 : 4 is thus converted to A : B = 45 : 36 by multiplying the numerator and denominator by 9.
Similarly the second ratio B : C = 9 : 10 is converted to B : C = 36 : 40, by multiplying numerator and denominator by 4.
The joined ratio is then, A : B : C = 45 : 36 : 40.
Value of total number of 121 portions = Rs.2420.
Value of each portion is then,
$\displaystyle\frac{2420}{121}=20$.
And share of C,
$20\times{40}=800$.
Answer: Option d: Rs.800. Key concepts used: Basic ratio concept -- Fund share in proportion to corresponding ratio value -- Ratio joining -- Portions concept -- Share of each of three parties can be determined only if the three share variables are expressed in a single ratio which gives the total fund being shared also -- Dividing in ratios -- Solving in mind. Problem 4.
In a regiment the ratio between the number of officers to soldiers was 3 : 31 before a battle. In the battle 6 officers and 22 soldiers were killed and the ratio became 1 : 13. The number of officers before the battle was,
28 21 38 31 Solution 4: Problem analysis and solving in mind
If $x$ be the cancelled out HCF of the ratio of officers and soldiers before the battle, the ratio after the battle can be expressed as,
$\displaystyle\frac{3x-6}{31x-22}=\frac{1}{13}$.
Cross-multiplying,
$31x-22=39x-78$,
Or, $8x=56$
Or, $x=7$.
So the number of officers before the battle was,
$3x=21$.
Answer: Option b: 21. Key concepts used: Basic ratio concept -- Change in ratio because of change in ratio variable values -- Context awareness -- . Solving in mind Problem 5.
If the three numbers in the ratio 3 : 2 : 5 be such that the sum of their squares is equal to 1862, then which is the middle number?
13 14 15 16 Solution 5: Problem analysis and solving in mind using HCF reintroduction technique
Introducing cancelled out HCF as $x$, the sum of squares of the three number can be expressed as,
$x^2(9+4+25)=38x^2=1862$
Or, $x^2=49$.
Or, $x=7$.
The middle number is then,
$2x=14$.
Answer: Option b: 14. Key concept used: Basic ratio concept --- HCF reintroduction technique -- Solving in mind. Problem 6.
In a college union there are 48 students. The ratio of number of boys to number of girls is 5 : 3. The number of girls to be added to the union so that the ratio of boys to girls becomes 6 : 5 is,
7 6 17 12 Solution 6: Problem analysis and solving in mind
48 students total up to $(5+3)=8$ portions, so each portion value is 6, the number of boys is 30 and the number of girls is 18.
If $y$ be the number of girls to be added to make the ratio of boys to girls to 6 : 5,
$\displaystyle\frac{30}{18+y}=\frac{6}{5}$.
As the LHS denominator needs to have the value of 25 to make the ratio 6 : 5,
$y=25-18=7$.
Answer: Option a: 7. Key concepts used: Basic ratio concepts -- Portions concept -- HCF reintroduction technique -- Factors multiples concept -- Change in ratio variable values -- Solving in mind. Problem 7.
A and B start an enterprise together with A as active partner. A invests Rs.4000 and Rs.2000 more after 8 months. B invests Rs.5000 and withdraws Rs.2000 after 9 months. Being the active partner, A takes Rs.100 as allowance from the profit. What is the share of B if the profit for the year is Rs.6700?
Rs.3250 Rs.3350 Rs.2800 Rs.2700 Solution 7: Problem analysis and solving in mind by use of investment-month concept
Each individual profit share will be proportional to the figure of total investment-month. This will be the sum over a year, of the product of the amount of capital invested during a period and the period duration in months.
Total investment-month for A will be,
$4000\times{12}+2000\times{4}=56000$,
And for B it will be,
$5000\times{12}-2000\times{3}=54000$.
The ratio in which profit will be shared between A and B is then,
$\text{Profit}_A:\text{Profit}_B=56000 : 54000=28:27$, a total of 55 portions.
As A takes special allowance of Rs.100 per month for the whole year from the profit, the profit to be shared reduces to,
$6700-1200=5500$.
As this total profit of Rs.5500 is equivalent to 55 portions to be shared between A and B in the ratio of 28 : 27, each portion value will be Rs.100 and B will get Rs.2700.
Answer: Option Key concepts used: Basic ratio concepts -- Investment-month concept in capital investment -- Profit sharing -- Portions concept -- Solving in mind. Problem 8.
Annual incomes of Abhi and Riju are in the ratio 3 : 2, while the ratio of their expenditures is 5 : 3. If at the end of the year each saves Rs.1000, the annual income of Abhi is,
Rs.9000 Rs.8000 Rs.6000 Rs.7000 Solution 8: Problem analysis and solving in mind using Income expenditure ratios concept, HCF reintroduction and linear equation solution
In the two given ratios, if the cancelled out HCF were $x$ and $y$ respectively then for income and savings of Abhi, we have the relation,
$3x-5y=1000$, and for Riju,
$2x-3y=1000$.
Subtracting, $x=2y$,
Or, $y=\frac{1}{2}x$.
Substituting this value of $y$ in the first equation,
$3x-\frac{5}{2}x=1000$,
Or, $x=2000$.
So annual income of Abhi is,
$3x=6000$.
Answer: Option c: 6000. Key concepts used: -- Basic ratio concepts Rich ratio concept -- Income expenditure ratios concept -- HCF reintroduction technique -- Linear equations -- Solving in mind. Problem 9.
The ratio of sum of salaries of A and B to the difference of their salaries is 11 : 1 and the ratio of sum of the salaries of B and C to the difference of their salaries is also 11 : 1. If A's salary is the highest and C's the lowest, then what is B's salary, given total of their salaries to be Rs.182000?
Rs.50000 Rs.86400 Rs.60000 Rs.72000 Solution 9: Problem analysis and solving in mind by componendo dividendo and ratio joining
If the salaries of A, B and C were $A$, $B$ and $C$, by the first ratio statement,
$(A+B) : (A-B) = 11 : 1$.
Expressing it to a fraction,
$\displaystyle\frac{A+B}{A-B}=\frac{11}{1}$
Applying three-step componendo dividendo,
$\displaystyle\frac{A}{B}=\frac{12}{10}=\frac{6}{5}$.
Similarly by the second ratio statement,
$\displaystyle\frac{B+C}{B-C}=\frac{11}{1}$,
Again applying componendo dividendo,
$\displaystyle\frac{B}{C}=\frac{6}{5}$.
To join the two ratios, the ratio values corresponding to the middle variable B are to be equalized to the LCM of the two values, 6 and 5, that is 30.
Accordingly the first ratio is transformed to,
$\displaystyle\frac{A}{B}=\frac{36}{30}$.
And the second ratio to,
$\displaystyle\frac{B}{C}=\frac{30}{25}$.
Joining the two ratios,
$A : B : C=36 : 30 : 25$ with total number of portions as,
$36x+30x+25x=91x=182000$,
So portion value is,
$x=2000$.
And B's salary is,
$30x=60000$.
Answer: Option c: Rs.60000. Key concepts used: Basic ratio concepts -- Rich ratio concepts -- Componendo dividendo -- HCF reintroduction technique -- Portions concept -- Ratio joining -- . Solving in mind
Being aware of the three step method of componendo dividendo, each of the two operations is executed directly without elaboration.
Problem 10.
What is the fourth proportional to 189, 273 and 153?
117 187 299 221 Solution 10: Problem analysis and solving in mind by fourth proportional concept
The proportional concept is another way of identifying the variables in an equal ratio of four variables,
$A:B=C:D$.
Variable $D$ is the fourth proportional in this ratio of four variablT%he values of the first three variables as given are,
$A=189$,
$B=273$, and
$C=153$.
Thus,
$\displaystyle\frac{189}{273}=\frac{153}{D}$,
So,
$D=\displaystyle\frac{153}{189}\times{273}$
$=\displaystyle\frac{153}{63}\times{91}$
$=17\times{13}$
$=221$.
Answer: Option d: 221. Key concepts used: Basic ratio concept -- -- Fourth proportional concept Solving in mind .
Overall, all of the eight problems could be solved in mind using varieties of concepts and techniques.
For solving these 10 problems, the concepts and techniques used were—basic and rich ratio concepts, HCF reintroduction technique, Multiple of ratio value technique, factors multiples concept, product of ratios concept, dividing in ratios, ratio joining, portions concept, change in ratio variable values, Investment-month concept, profit sharing, context awareness, componendo dividendo and fourth proportional concept. The problems needed an wide array of basic and rich concepts for quick solution.
Just remember, understanding and applying basic and rich concepts should enable you to solve any such problem easily under a minute with no dependence on varieties of formulas.
Resources that should be useful for you or 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests to access all the valuable student resources that we have created specifically for SSC CGL, but section on SSC CGL generally for any hard MCQ test. Concept Tutorials on related topics Efficient solution techniques on related topics SSC CGL level solved question sets on mixture or alligation SSC CGL Tier II level solved question sets on mixture or alligation SSC CGL Tier II level question and solution sets on Ratio and Proportion SSC CGL Tier II level Solution Set 23 Ratio proportion 2 SSC CGL question and solution sets on Ratio and Proportion and Percentage If you like, you may to get latest content from this place. subscribe |
(35S) Statistical Distributions Functions
10-25-2015, 12:19 AM (This post was last modified: 06-15-2017 01:33 PM by Gene.)
Post: #1
(35S) Statistical Distributions Functions
I need programs of Statistical Distributions: t Student, F Snedecor and Chi Square. The Hp 67 Statistical Pakage was excellent, but I am not able to translate them to HP 35s. Does somebody could help me in this task?
10-25-2015, 01:29 PM
Post: #2
RE: HP 35s Statistical Distributions Functions
The HP 35s has two features that are missing in the HP-67:
Thus I wouldn't recommend to translate these HP-67 programs directly to the HP 35s. I just had a quick look at Chi-Square Distribution for the HP-67 and it uses a series approximation to evaluate the cumulative distribution. With the HP 35s you can use numeric integration instead.
Just have a look at the program Normal and Inverse–Normal Distributions in the user's guide (pp 16-11). You can replace the subroutine F that calculates the integrand for the normal distribution. It calculates the "upper tail" area but that is easy to change. Since integration and solver can't be mixed a Newton-iteration is used to calculate the inverse. The calculation of the derivative is just evaluating the probability density function due to the fundamental theorem of calculus.
Let us know if you struggle to translate the formulas of the probability density functions to programs. Using an algebraic equation might be easier but will be probably slower.
HTH
Thomas
10-26-2015, 03:03 PM
Post: #3
RE: HP 35s Statistical Distributions Functions
(10-25-2015 12:19 AM)PedroLeiva Wrote: I need programs of Statistical Distributions: t Student, F Snedecor and Chi Square. The Hp 67 Statistical Pakage was excellent, but I am not able to translate them to HP 35s. Does somebody could help me in this task?
Might I suggest...
NCS35TB Stat Pac
For the HP-35s Calculator
... by Namir Shamas
at NCSSTATPAC
BEST!
SlideRule
10-26-2015, 04:41 PM (This post was last modified: 03-28-2016 02:11 PM by PedroLeiva.)
Post: #4
RE: HP 35s Statistical Distributions Functions
Thank you very much for your recommendation, but Namir programs are not what I need. In Appendix A, there are detailed calculations based on estimates of the inverse probability of each 4 statistical functions. I also need to calculate probability according to a value of each function.
HP 67 Statistical Distributions programs (Normal, Chi Square, "F" and "t" Student) are perfect for this purpose.
10-26-2015, 08:30 PM (This post was last modified: 10-26-2015 08:34 PM by Dieter.)
Post: #5
RE: HP 35s Statistical Distributions Functions
(10-26-2015 04:41 PM)PedroLeiva Wrote: Thank you very much for your recommendation, but Namir programs are not what I need. In Appendix A, there are detailed calculations based on estimates of the inverse probability of each 4 statistical functions. I also need to calculate probability according to a value of each function.
OK – what exactly do you need? The CDFs for Normal, Student, Chi² and Fisher's F distribution? Also their quantile (inverse) functions? What accuracy is required?
BTW, the 34s does all this. I provided some code for these four quantile functions which are evaluated with mostly 30+ digit accuracy. This is possible due to CDFs that are (mostly) equally accurate. These CDFs in turn are calculated using incomplete Beta and Gamma functions, coded by Pauli.
(10-26-2015 04:41 PM)PedroLeiva Wrote: HP 67 Statistical Distributions programs (Normal, Chi Square, "F" and "t" Student) a perfect for this purpose.
Could you provide a pointer to this package? How accurate are the results? Do these programs also calculate the inverse (quantile) functions?
If you really appreciate these HP67 programs so much it should be easy to translate them for the 35s.
Dieter
10-26-2015, 08:39 PM (This post was last modified: 10-28-2015 12:09 AM by Thomas Klemm.)
Post: #6
RE: HP 35s Statistical Distributions Functions
Here's a program for the Student's t-distribution:
Code:
T001 LBL T ; Student's t
You have to initialize the program with the degrees of freedom:
XEQ T ENTER
K?
7
R/S
To calculate the PDF use the program P:
XEQ P ENTER
X?
1.6
R/S
0.110665
PDF[StudentTDistribution[7],1.6]
To calculate the CDF use the program C:
XEQ C ENTER
X?
1.5
R/S
0.911351
CDF[StudentTDistribution[7],1.5]
For the next program set the accuracy to a reasonable value:
1E-6
STO E
To calculate the Inverse CDF use the program I:
XEQ I ENTER
X?
0.99
R/S
2.997951
HTH
Thomas
10-26-2015, 09:25 PM (This post was last modified: 10-26-2015 09:27 PM by Dieter.)
Post: #7
RE: HP 35s Statistical Distributions Functions
(10-26-2015 08:39 PM)Thomas Klemm Wrote: Here's a program for the Student's t-distribution:
Hint: the Normal, Student, Chi² and Fisher quantiles will converge faster if the Halley method (instead of simple Newton) is used. This is easy here since the second derivative simply is the PDF times a factor. Another tweak that can provide better convergence (especially in the Chi² and F case) is solving ln(CDF/p)=0 instead of CDF–p=0.
Both methods are used in the WP34s quantile functions.
Dieter
10-26-2015, 09:28 PM
Post: #8
RE: HP 35s Statistical Distributions Functions
For the Chi-squared distribution you just have to calculate a different PDF:
The factor \(\frac{1}{2^{\frac{k}{2}}\Gamma\left(\frac{k}{2}\right)}\) has to be calculated only once as long as the degrees of freedom k doesn't change. This can be done similar to program T.
The value of the distribution has to be calculated in program D. The other programs don't change.
For the F-distribution you have to calculate:
with parameters \(d_1\) and \(d_2\).
To calculate the beta function we can use:
Together with the program for the Student's t-distribution it should be possible to create the missing programs.
Cheers
Thomas
10-26-2015, 10:29 PM
Post: #9
RE: HP 35s Statistical Distributions Functions
[attachment=2697][attachment=2697][attachment=2697][attachment=2697]
(10-26-2015 08:30 PM)Dieter Wrote:(10-26-2015 04:41 PM)PedroLeiva Wrote: Thank you very much for your recommendation, but Namir programs are not what I need. In Appendix A, there are detailed calculations based on estimates of the inverse probability of each 4 statistical functions. I also need to calculate probability according to a value of each function.
Attached are three HP 67 programs, with descriptions and formulas, some examples or test cases, and the program itself (keystrokes sequence). I need exactly what they allow to calculate, and with the same precision.
I am not a specialist in mathematics; I simply use a lot of statistic and want to become independent of the function tables.
10-26-2015, 11:04 PM
Post: #10
RE: HP 35s Statistical Distributions Functions
(10-26-2015 10:29 PM)PedroLeiva Wrote: Attached are three HP 67 programs, with descriptions and formulas, some examples or test cases, and the program itself (keystrokes sequence). I need exactly what they allow to calculate, and with the same precision.
So you only need the PDFs and CDFs – that's simple and straightforwand on the 35s. Actually the implementation is easier as the 35s features a Gamma function.
(10-26-2015 10:29 PM)PedroLeiva Wrote: I am not a specialist in mathematics; I simply use a lot of statistic and want to become independent of the function tables.
Get a 34s. ;-)
Dieter
10-27-2015, 11:59 AM
Post: #11
RE: HP 35s Statistical Distributions Functions
(10-26-2015 09:28 PM)Thomas Klemm Wrote: For the Chi-squared distribution you just have to calculate a different PDF:
Thomas, with time and patience I will try to program my HP 35s with the suggestions that you just gave me. Thank you
10-27-2015, 02:29 PM
Post: #12
RE: HP 35s Statistical Distributions Functions
(10-26-2015 08:39 PM)Thomas Klemm Wrote: Here's a program for the Student's t-distribution:
Thomas, two questions. What is the sequence of keys to enter STEP C006. To run the label I, the question is A ?, rather X?, right?
10-27-2015, 04:48 PM
Post: #13
RE: HP 35s Statistical Distributions Functions
Code:
C006 ∫FN d T ; integrate pdf
That's ∫ followed by T.
[ ↰ ] [EQN] [ 9 ]
Quote:To run the label I, the question is A ?, rather X?, right?
Input is area (that's why A) and the result is \(x\) so that \(\int_{-\infty}^x pdf(t)\,dt = A\).
You may rename it to something different of course. Just don't use variables that are already used in the program. Thus X might be confusing.
Cheers
Thomas
10-27-2015, 05:22 PM
Post: #14
RE: HP 35s Statistical Distributions Functions
(10-27-2015 04:48 PM)Thomas Klemm Wrote:Something is not working well. When I run the routine I (XEQ I ENTER), with a 1E-6 E value in register E, the program do not stop integration!
10-27-2015, 07:32 PM
Post: #15
RE: HP 35s Statistical Distributions Functions
(10-27-2015 05:22 PM)PedroLeiva Wrote: Something is not working well. When I run the routine I (XEQ I ENTER), with a 1E-6 E value in register E, the program do not stop integration!
The 1E–6 error threshold in E is irrelevant for the integration routine. The 35s "Integrate" function stops if the result is correct to display precision. You probably have set your 35s to ALL display mode, or maybe FIX 9 or SCI 9, so the 35s calculates the integral until is thinks it has all (!) 12 digits (resp. 9) correct. This will take some time. Solution: simply set something like FIX 4 or FIX 6 (you may add this step to the program right before the Integrate command), so you get 4 resp. 6 correct decimals.
BTW, at the moment I am looking at an 35s implementation of the HP67 routines.
Dieter
10-27-2015, 10:25 PM (This post was last modified: 10-27-2015 10:28 PM by Dieter.)
Post: #16
RE: HP 35s Statistical Distributions Functions
(10-27-2015 07:32 PM)Dieter Wrote: BTW, at the moment I am looking at an 35s implementation of the HP67 routines.
OK, here is a first experimental version of the Chi² algorithm used in the HP67 Stat Pac. PDF and CDF are calculated simultaneously, and both are returned in Y and X. Here is the code:
Code:
C001 LBL C
Usage (results as shown in FIX 9 mode):
Code:
[XEQ] C [ENTER] J=?
As usual, comments, suggestions and error reports are welcome. ;-)
Since the main computation routine (C013 ff.) returns both the PDF and the CDF, adding a function that determines the inverse (quantile) is rather simple. I tried an approach using a second-order (Halley) method and it worked well for me. This required about 30 additional lines of code.
Dieter
10-27-2015, 10:29 PM
Post: #17
RE: HP 35s Statistical Distributions Functions
Let's try to pin that down. Dieter already gave a hint: the accuracy in register E and the display precision should match.
But since the program I uses program C I first want to know if you get the same result for this example:
XEQ T ENTER
K?
7
R/S
XEQ C ENTER
X?
1.5
R/S
0.911351
This is just to make sure that program C works fine.
If that's okay please make sure to set FIX 3 and E = 1E-3 before running the next example.
XEQ I ENTER
X?
0.99
R/S
2.997
It should flicker about four times INTEGRATING and then display the result.
If the integration doesn't stop then please add an R/S command between lines I007 and I008 and report both the value displayed and the contents of register X (use VIEW X). Just use R/S to run the next round of the loop for a couple of times.
Hopefully this helps to find the problem.
Cheers
Thomas
PS: Thanks for posting the HP-67 programs. It's always interesting to learn something new. For instance I didn't know about the formula used to calculate the CDF of the Student's t-distribution. But then I must admit that I'm not familiar with statistics. Still I'm curious how they came up with that formula.
10-27-2015, 11:41 PM (This post was last modified: 10-27-2015 11:45 PM by PedroLeiva.)
Post: #18
RE: HP 35s Statistical Distributions Functions
(10-27-2015 10:29 PM)Thomas Klemm Wrote:For routing T the value after R/S is 0.385
For routing C the value is the same as yours, 0.911351
The program did not stop integration again. After R/S adition the figures are:
1° time: x=0.878, y=3.782 E-2; variable X= 1.273
2° time: x=0.5, y=3.850 E-7; variable X= 0.001
# times: the same as 2° time
10-27-2015, 11:48 PM
Post: #19
RE: HP 35s Statistical Distributions Functions
(10-27-2015 07:32 PM)Dieter Wrote:Thank you for your interest Dieter. I am ussing FIX 3, but the problem persist(10-27-2015 05:22 PM)PedroLeiva Wrote: Something is not working well. When I run the routine I (XEQ I ENTER), with a 1E-6 E value in register E, the program do not stop integration!
10-28-2015, 12:11 AM
Post: #20
RE: HP 35s Statistical Distributions Functions
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Consider the 5 x 5 matrix
A=1234551234451233451223451
It is given that A has only one real eigenvalue. Then the real eigenvalue of A is
The rank of the matrix M=51010102366 is
Consider the following statement about the linear dependence of the real valued function $ y_1=1,y_2=x$ and $ y_3=x^2$, over the field of real of real numbers.
I. $ y_1,\;y_2 $ and $y_3$ are linearly independent on -1≤x≤0
II. $ y_1,\;y_2 $ and $y_3$ are linearly dependent on $ 0\leq x\leq1 $
III. $ y_1,\;y_2 $ and $y_3$ are linearly independent on 0≤x≤1
IV. $ y_1,\;y_2 $ and $y_3$ and are linearly dependent on -1≤x≤0
Which one among the following is correct?
Three fair cubical dice are thrown simultaneously. The probably that all three dice have the same number of dots on the faces showing us is (up to third decimal place) ____________.
Consider the following statements for continuous-time linear time invariant (LTI) system.
I. There is no bounded input bounded output (BIBO) stable system with a pole in the right half of the complex plane.
II. There is no casual and BIBO stable system with a pole in the right half of the complex plane.
Consider a single input single output discrete-time system with $ x\left[n\right] $ as input and $ y\left[n\right] $ as output, where the two are related as
$ y\left[n\right]=\left\{\begin{array}{l}\;\;\;\;\;\;\;\;\;\;\;\;n\left|x\left[n\right]\right|,\;\;\;\;\;\;\mathrm{for}\;0\leq\mathrm n\leq10\\x\left[n\right]-x\left[n-1\right],\;\;\;\;\;\;\;\mathrm{otherwise}.\end{array}\right. $
Which one of the following statement is true about the system?
In the circuit shown, the positive angular frequency ω (in radians per second) at which the magnitude of the phase difference between the voltages V1 and V2 equals π4 radians, is_________.
A periodic signal x(t) has a trigonometric Fourier aeries expansion
$ x\left(t\right)=a_0+\sum\limits_{n=1}^\infty\left(a_n\;\cos\omega_0t\;+\;b_n\sin\;n\omega_0t\right) $
If $ x\left(t\right)=-x\left(-t\right)=\left(t-\pi/\omega_0\right) $, we can conclude that
A bar of Gallium Arsenide (GaAs) is doped with Silicon such that the Silicon atoms occupy Gallium and Arsenic sites in the GaAs crystal. Which one of the following statements is true?
An n+ -n Silicon device is fabricated with uniform and nono-degenerate donor doping concentration of ND1 =1 × 1018 cm-3 and ND2 =1 × 1015 cm-3 corresponding to the n+ and n regions respectively. At the operational temperature T, assume complete impurity ionization, kT/q=25mV, and intrinsic carrier concentration to be ni = 1 × 1010 cm-3. What is the magnitude of the built-in potential of this device?
For a narrow base PNP BJT, the excess minority carrier concentration ($\triangle n_E$ for emitter,$\triangle p_B$ for base,$\triangle n_C$ for collector) normalized to equilibrium minority carrier concentration ($n_{E0}$ for emitter,$n_{B0}$ for base,$n_{C0}$ for collector) in the quasi-neutral emitter, base and collector regions are shown below. Which one of the following biasing modes is the transistor operating in?
For the operational amplifier circuit shown, the output saturation voltages are ±15V. The upper and lower threshold voltages for the circuit are, respectively,
A good transconductance amplifier should have
A Miller effect in the context of a Common Emitter amplifier explains
In the latch circuit shown, the NAND gates have non-zero, but unequal proportion delays. The present input condition is: P=Q=’0’. If the input condition is changed simultaneously to P=Q=’1’, the outputs X and Y are
The clock frequency of an 8085 microprocessor is 5 MHz. If the time required to execute an instruction is 1.4 µs, then the number of T-states needed for executing the instruction is
Consider the D-Latch shows in the figure, which is transparent when its clock input CK is high and has zero propagation delay. In the figure, the clock signal CLK1 has a 50% duty cycle and CLK2 is a one-fifth period delayed version of CLK1. The duty at the output at the latch in percentage is__________
The open loop transfer function G(s)=(s+1)sp(s+2)(s+3) where p is an integer, is connected in unity feedback configuration as shown in the figure.
Given that the steady state error is zero for unit and 6 for unit ramp input, the value of the parameter p is_________
Consider a stable system with transfer function G(s)=sp+b1sp-1+···+bpsq+a1sq-1+···+aq where b1, …, bp and a1, … , aq are real valued constant. The slop of the Bode log magnitude curve of G(s) converges to -60 dB /decade as ω→∞. A possible pair of values for p and q is
Which of the following can be the pole-zero configuration of a phase-lag controller (lag compensator)?
This is not the official website of GATE.
It is our sincere effort to help you. |
UGC conducts Jointly with CSIR, the Net exam twice a year for PhD and Lecturership aspirants. For science subjects such as Life Science, Chemical Science etc, the question paper is divided into three parts - Part A, B and C. Part A is on Maths related topics and contains 20 questions any 15 of which have to be answered. The other two parts are on the specific subject chosen by the student.
The questions in Maths are tuned towards judging the problem solving capability of the student using the basic knowledge in maths and not the procedural competence in maths.
The third set of answer and structured solution of 15 Net level questions follow.
This is a set of 15 questions for practicing UGC/CSIR Net exam: ANS Set 3
Answer all 15 questions. Each correct answer will add 2 marks to your score and each wrong answer will deduct 0.5 mark from your score. Total maximum score 30 marks. Time: Proportionate (35 mins). Q1. Rainfall is measured in mm. One day there was a rainfall of 4mm at Kolkata. To collect the rainwater a hemispherical glass bowl of radius 1cm was kept in an open area. How much percent of the bowl was filled up by rainwater? 40 50 60 75 Solution: 4mm rainfall means if the rainfall is collected in a flat bottom container with straight rising sides, the depth of water collected will be 4mm. In other words, rainfall per unit horizontal area is of 4mm column height.
In our case the collecting surface area is $A={\pi}10^2=100{\pi}$ sq mm. The volume of water will be, $V_{1}=A\times{4mm}=400{\pi}$ cu mm. The volume of the hemipherical bowl is $V_{2}=\frac{2}{3}{\pi}10^3$ cu mm. So, $V_{1} : V_{2} = \frac{400{\pi}\times{3}}{2\times{\pi}\times{1000}}=3 : 5=60{\%}$.
Answer: Option c: $60{\%}$. Key concept used: Rainfall concept -- visualizing of the rainfall being collected in a cylindrical vessel of radius of mouth 1cm -- ratio of the volume of cylindrical column of water 4mm deep to the volume of the hemisphere. Q2. Two friends decided to start a race on a bright sunny day. Friend A with usual speed of 8m/sec ran faster than his friend B, whose usual speed was 6m/sec and so it was agreed that the race would be run with a handicap of 40m to A. How much distance would B cover before A overtook him? 120m 160m 80m 100m Solution: When the speedier man starts the race 40m behind the slower man, he will catch up when he covers the intial gap running at a relative speed of 2m/sec which he will do in 20sec. In this time B will cover a distance of 120m. Answer: Option a: 120m. Key concept used: Concept of race -- make the slower man standstill and correspondingly reduce the speed of speedier man to their relative speed -- calculate how much time he takes to cover the initial gap at this relative speed. This is the time taken to catch up. In reality, running at their original speeds, the slower man will cover 120m distance in this 20secs time.
This is the easiest way to deal with races. Races may be between various objects moving at different speeds. It may even be between the hour hand and the minute hand of a clock.
Q3. One day Minti started down in a lift from the 10th floor of a building at the same time as her friend Pinki started up from the ground floor in another lift. They had decided to meet in between rather than wait longer. If Minti's lift covered 3 floors by the time Pinki's covered 2 floors and if they knew it, on which floor will they meet wasting minimum time? 3 4 5 2 Solution: The two lifts will cover a total of 10 floors at a relative speed of 5 floors per unit time embodied in the ratio. It would take two unit time to cover 10 floors and meet. In this time Pinki's lift will be on 4th floor and Minti's lift will come down 6 floors to meet her. Answer: Option b: 4th floor. Key concept used: Relative speed of two moving objects approaching each other is the summation of their individual speeds -- At this speed they will cover the distance between them when they meet -- From ratios of two quantities, their sum can obtained as the sum of the two ratio numbers in terms of ratio unit, actual quantities are not required to be known. Q4. At the end of a year in a business, profit decreased by 15% with cost price remaining the same. If the previous year profit was 20% what is the percent decrease in sale price this year? 12.5% 15% 10% 20% Solution: Profit is always expressed as a percent of cost price or investment. Assuming the cost price as 100 with last year profit as 20%, last year sale price was 120. Profit decreasing by 15%, this year sale price decreased to 105 a decrease of 15 from 120, which is a fall of one eighth or 12.5%. Answer: Option a: 12.5% Key concept used: Clear idea of profit expressed always as a percent of cost price or investment -- but when decrease of sale price was wanted it was on the basis of the original sale price as the basis, not the cost price obviously.
In percentage sums, three things are important. The base of percentage expressed is to be carefully identified and secondly, the percentage is always equivalent to a number uaually a decimal being divided by 100. For example, 20% profit means, this profit is on the base of cost or investment price, that is, 20% of the cost price or one fifth of the cost price.
Thirdly, In absence of actual figures of the quantities, the base quantity of the first percentage expressed can always be taken as 100. All other values will be in terms of this base value.
Q5. $\left(\frac{1}{5}\right)^{3x} = 0.008.$ The value of $(0.25)^x$ is, 22.5 0.25 0.0625 0.5 Solution: $\left(\frac{1}{5}\right)^{3x} = 0.008 = \frac{8}{1000} = \frac{2^3}{10^3}=\left(\frac{2}{10}\right)^3=\left(\frac{1}{5}\right)^3$.
So $x=1$, and $(0.25)^x=0.25$.
Answer: Option b: 0.25. Key concept used: Concept of indices -- RHS analysis to make base $\frac{1}{5}$ equal on both sides so that we get a value of $x$.
Use of
base equalization technique in most of the indices sums solve the problem. Q6. Five boys are sitting in a row. Pintu is to the left of Khokon. Kalua is to the right of Apu. Sunny is in the middle of Kalua and Pintu. Who is to the extreme right of the row? Khokon Apu Pintu Kalua Solution: Examining the last statement we find it carries maximum information by forming a trio of Kalua -- Sunny -- Pintu where Kalua may be on the right or left Sunny. In first statement Pintu appears to the left of Khokon. In this case, Khokon should be on the right of the trio, that is on the right of Pintu.
As Kalua is on the right of Apu
from the second statement, Apu has to be on the left of the trio. Thus, Khokon is left to be the rightmost. Notice, Kalua may be on the right or left of the trio, but both will satisfy the conditions.
The third statement forms the trio -- the first statement puts Khokon on the right of the trio and the second statement puts Apu on the left of the trio.
Alternatively, from first two statements, Khokon and Kalua are the initial candidates to be the rightmost. As Khokon has to be on the right of Pintu from the first statement, the whole of the trio including Kalua has to be on the left of Khokon from the third statement. So Khokon is the rightmost. The relative position of Khokon and Pintu is transferred to the relative position of Khokon and the trio and hence to Khokon and Kalua. This is a better reasoning as it focuses on the end requirement all along.
This approach is another example of applying
End State Analysis Approach. Answer: Option a: Khokon. Key concept used: Reasoning with respect to relative position in a row -- end requirement targetted analysis. Q7. A can do a work in 3 days while B can do the same work in in 4 days. If they work together for a total wage of Rs. 2800 how much does A get? Rs. 1200 Rs. 1900 Rs. 1300 Rs. 1600 Solution: In work wage problem wage earned is in proportion of amount of work done in a day. As A does work faster than B he should get more wage. In 1 day A does $\frac{1}{3}rd$ of the total work and B does $\frac{1}{4}th$ of the total work. Their daily wage will be in proportion of their amount of work in a day, that is, in ratio $\frac{1}{3} : \frac{1}{4} = 4 : 3$, it will be the inverse of number of days in which they do the total work. So total wage unit is 7 which is Rs. 2800. Thus 1 wage unit is Rs. 400 and A will get 4 units that is Rs. 1600. Answer: Option d: Rs. 1600. Key concept used: Wage work concept: the more one works the more wage he will get -- wage of a person is fixed in rupees per day -- wage will be in direct proportion of work amount they do per day or in inverse proportion of number of days they take to do same amount of work. Q8. Remainder of $7^{57}$ divided by 10 is, 7 9 3 1 Solution: Remainder after division by 10 is the unit's digit. Powers of 7 follow a 4 digit cycle of 7, 9, 3, 1. As 57 modulo 4 is 1 (remainder after dividing by 4), the answer is 7. Answer: Option a: 7. Key concept used: Place value system -- remainder concept -- unit's digit cycle of powers. Q9. In $\triangle{PQR}$, S and T are two points on the sides PR and PQ respectively. $\angle{PQR}=\angle{PST}$. If PT = 5 cm, PS = 3 cm and TQ= 3 cm, the length of line SR is,
$\frac{31}{3}$ cm 5 Cm 3 cm $\frac{41}{3}$ cm Solution: In triangles PQR and PST, the three angles are same. $\angle{PQR}=\angle{PST}$, $\angle{QPR}=\angle{TPS}$, and so the third pair of angles $\angle{PRQ}=\angle{PTS}$. Thus these two are similar triangles. Thus ratios of corresponding pair of sides are same which leads to $\frac{PR}{PT}=\frac{PQ}{PS}$, or, $PR=\frac{8}{3}\times{5}=\frac{40}{3}$.
So, $SR = PR - PS = \frac{31}{3}$.
Answer: Option a: $\frac{31}{3}$. Key concept used: Two triangles are similar if three pairs of angles are equal - similar triangle property : ratio of all three pairs of corresponding sides of the two triangles are equal -- rule of identifying corresponding sides in two similar triangles ABC and PQR, where angles A, B and C are equal to angles P, Q and R respectively : take the side opposite to angle A in triangle ABC as numerator of first ratio and the side opposite to its equal angle P in triangle PQR as denominator. This forms the first ratio. Similarly the second and third ratios are formed. All these three ratios will be equal to each other. Q10. Growth of a type of fungus was monitored at regular intervals of time and is shown in the graph below. Around which time is the rate of growth zero?
Near day 1 On day 3 Between days 3 and 5 Between days 5 and 7 Solution: The fungus mass is plotted on y-axis and the slope of the curve at any point gives its rate of growth. The slope at any point is the angle the tangent at the point makes with the x-axis. The more vertical the tangent is more is the rate of growth. Between days 3 and 5 the slope is zero and so growth also is zero. Answer: Option c: Between days 3 and 5. Key concept used: Growth is change per unit time and thus in a growth curve rate of growth is given by the slope of the tangent at the point under consideration -- in any inverted bell shaped curve growth is zero at its maximum value, that is, at the top point where the tangent of horizontal. Q11. There are seven sisters in a house in a village where there is no electricity or any gadget. The activities of the seven sisters in one day are, Sister-1: Reading Novel Sister-2: Cooking Sister-3: Playing Chess Sister-4: Playing Sudoku Sister-5: Washing clothes Sister-6: Gardening
What is Sister-7 doing?
Sleeping Taking bath Out marketing Playing chess Solution: She could have been doing umpteen number of things, but as there is no gadget or electricity and as chess usually can't be played alone, she should be playing chess with her other sister. Answer: Option d: Playing chess. Key concept used: Analyzing given facts for any clue rather than leaving the problem altogether as possibilities are uncertain. Q12. How many squares are there in the following figure?
13 10 4 5 Solution: The four smaller squares and the larger square, in total 5. As the other sides are not completed, the squares are not formed. Answer: Option c: 5. Key concept used: Observation -- square detection. Q13. In a bicycle wheel if there were 10 more spokes, the angle between them would have reduced by six degrees. Can you find out the number of spokes in the wheel? 10 20 30 40 Solution: If $n$ be the number of spokes now, then $\frac{360}{n+10}=\frac{360}{n} - 6=\frac{360 - 6n}{n}$,
Or, $360n = 360n + 3600 - 6n^2 -60n$, Or, $6n^2 + 60n = 3600$, or, $n(n+10)=600.$
n = 20 satisfies this condition.
Answer: Option b: 20. Key concept used: Formation of algebraic relationship -- simplification -- trial and error.
Alternatively, $n(n+10)=600$, or, $n^2 +10n - 600=0$, or, $(n + 30)(n - 20) = 0$.
As $n$ can't be negative -30, it must be 20.
Q14. How many numbers are there from 20 to 40 that have no factor among the numbers from 2 to 10? 3 4 5 6 Solution: It is basically equivalent to the question, "how many prime numbers are there between 20 to 40?." the number is 4: 23, 29, 31 and 37. Answer: Option b: 4. Key concept used: Factor -- prime number - divisibility. Q15. If $2^{4x} + 2^{5x} = 1280$, then x is, 4 5 3 2 Solution: There are two ways to solve the problem. You can test the value of expression with choice values of x. Actually x will be 2 and if you have a good estimate of powers of 2, you may get this answer quickly.
But the more elegant way is to simplify the expression. $2^{4x} + 2^{5x} = 2^{4x}(1 + 2^{x})=1280=2^{8}\times{5}$. Thus, both of the equalities $2^{4x}=2^{8}$ and $(1 + 2^{x})=5$ must be true, both of which give x=2.
Answer: Option d: 2. Key concept used: Algebraic simplification -- factorization and factor equality. |
So we are again talking about $\ce{BrSF5}$ but this post extends to all generally octahedral complexes/ molecules and even quadratic planar geometries. The main requirement is, that the molecule has a $C_{\mathrm{4v}}$ point group. (It may be extended to all non-linear $C_{n\mathrm{v}}\ (n>1)$ point groups.) First of all reconsider the molecular shape again.
Now admittedly the degree of pyramidalization is more than minimal, with less than $1^\circ$ deviation from the plane. However, this has a slight effect on the equatorial $\angle(\ce{F-S-F'})$ bond angle, purely to symmetric reasons.
Basic trigonometric considerations are enough to understand this phenomena. Consider the following scheme, where the peak of the pyramid is represented by $\mathbf{\ce{Y}}=\ce{S}$ and the base is span by $\mathbf{\ce{X_{eq}}}=\ce{F_{eq}}$, $\mathbf{\ce{X'_{eq}}}=\ce{F'_{eq}}$, $\mathbf{\ce{X''_{eq}}}=\ce{F''_{eq}}$ and for completeness $\mathbf{\ce{X'''_{eq}}}=\ce{F'''_{eq}}$.
So we have some basic quantities, that we can address, but in the end we can derive the connection of the two angles without caring about the other quantities, only knowing the point group.
$a$ is the $\mathbf{d}(\ce{F_{eq}\cdots{}F'_{eq}})$ distance $d$ is the $\mathbf{d}(\ce{F_{eq}\cdots{}F''_{eq}})$ distance $s$ is the $\mathbf{d}(\ce{S\cdots{}F_{eq}})$ distance $h$ is the height of the pyramid and it refers to the out of plane shift of $\ce{S}$ compared to the $\ce{F_{eq}\cdots{}F'_{eq}\cdots{}F''_{eq}\cdots{}F'''_{eq}}$ plane $h'$ is another virtual quantity, the angle bisector of $\angle(\ce{F_{eq}-S-F'_{eq}})$ $\alpha$ refers to the bonding angle $\angle(\ce{F_{eq}-S-F''_{eq}})$ $\beta$ refers to the bonding angle $\angle(\ce{F_{eq}-S-F'_{eq}})$
From Pythagoras's Theorem we can easily derive the following connection (square in the scheme):\begin{align} && d^2 &= a^2 + a^2\\ \therefore && d &= \sqrt{2}\cdot a\tag{1}\end{align}
So now we are going to use some trigonometric functions for the angles (look at the triangles):\begin{align} && \sin\left(\frac{\alpha}{2}\right) &= \frac{\frac12d}{s}\\ \therefore && s &= \frac{\frac12d}{\sin\left(\frac{\alpha}{2}\right)}\tag{2}\end{align}
\begin{align} && \sin\left(\frac{\beta}{2}\right) &= \frac{\frac12a}{s}\\ \therefore && s &= \frac{\frac12a}{\sin\left(\frac{\beta}{2}\right)}\tag{3}\end{align}
Let's derive the connection between the angles by equalizing $(2)$ and $(3)$:\begin{align} && \frac{\frac12d}{\sin\left(\frac{\alpha}{2}\right)} &= \frac{\frac12a}{\sin\left(\frac{\beta}{2}\right)}\\ \therefore && \sin\left(\frac{\beta}{2}\right) &= \frac{a}{d}\sin\left(\frac{\alpha}{2}\right)\\ \text{with (1)}\implies && \sin\left(\frac{\beta}{2}\right) &= \frac{1}{\sqrt{2}}\sin\left(\frac{\alpha}{2}\right)\\ \equiv && \sqrt{2} &= \frac{\sin\left(\frac{\alpha}{2}\right)}{\sin\left(\frac{\beta}{2}\right)}\tag{4}\end{align}
From this you can see, that the degree of pyramidalization is codependent on the equatorial bond angles. When the pyramidalization increases, $\alpha\to0$, then the equatorial bond angle has to decrease, $\beta\to0$, i.e. a constant ratio.
In the special case of $\alpha = 180^\circ$ this simplifies to $\beta=90^\circ$. In all other cases you can derive the degree of pyramidalization from the equatorial bond angles and vice versa.
Transformation of $(4)$ yields a function that displays the equatorial bond angle in dependence of the pyramidalization, with $\alpha\in~]0^\circ;180^\circ]$ or $\alpha\in~]0;\pi]$.$$\beta = 2\cdot\arcsin\left[\frac{1}{\sqrt{2}}\cdot\sin\left(\frac{\alpha}{2}\right)\right]$$
Addendum
The out of plane height can also be derived from Pythagoras's theorem (look at right triangle):\begin{aligned} && h^2 &= s^2 + \left(\frac{d}{2}\right)^2\\ \therefore && h &= \sqrt{s^2 + \left(\frac{d}{2}\right)^2}\\ \text{with (1)}&& h &= \sqrt{s^2 + \frac{1}{2}a^2}\\\end{aligned}
This is usually a value that is given for domed complexes, so it is nice to know how to find it. |
First of all, you need to understand what $dS$ is.
$dS$ is the form $dV$ contracted by the vector field $\nu$ on $\partial \Omega$. In other words, to find $dS(v_2,v_3,...,v_{n})$ for tangent vectors $v_2,v_3,...,v_{n}$ to $\partial \Omega$, you just evaluate $dV(\nu, v_2,v_3,...,v_{n})$. The motivation for this definition is that the area of a parallelogram $P$ is the same as the volume of the parallelepiped with base $P$ and height $1$.
So now note that,
$ \frac{\partial f}{\partial x_1} dV = d(f dx_2 \wedge dx_3 \wedge ...\wedge dx_n)$
Letting $ \widehat{dx_1} = dx_2 \wedge dx_3 \wedge ...\wedge dx_n$
Let $p \in \partial \Omega$ and $v_2,v_3,...,v_n \in T_p( \partial \Omega)$ then
$\begin{align*}f(p) \widehat{dx_1}(v_2,v_3,...,v_n) &= f(p) dV(e_1,v_2,...,v_n)\\&=f(p) dV(\textrm{proj}_{\nu(p)}(e_1),v_2,...,v_n)\\&=f(p) dV(\nu_1(p)\nu(p),v_2,...,v_n)\\&=f(p)\nu_1(p) dV(\nu(p),v_2,...,v_n)\\&=f(p)\nu_1(p) dS(v_2,...,v_n)\end{align*}$
so as forms on $\partial S$, we have $f\widehat{dx_1} = f\nu_1dS$. The big step is the second line in the equality above, which is justified by writing $e_1$ as a linear combination of $\nu$ and $v_i$'s. All of the $v_i$ terms die because of the alternating property of forms.
Now we conclude that
$\begin{align*}\int_\Omega \frac{\partial f}{\partial x_1} dV &= \int_\Omega d(f \widehat{dx_i})\\&=\int_{\partial \Omega} f \widehat{dx_i} \text{by Stokes' theorem}\\ &=\int_{\partial \Omega} f \nu_1 dS\end{align*}$
If any of this does not make sense, please ask for follow up in the comments. Notation in differential geometry is horrible, and everyone makes it up for themselves I think. So I understand if some part of this is hard to parse. I am also no differential geometer, but I do a lot of calculus in high dimensional spaces (I work in several complex variables). This kind of thing used to frustrate me a lot. |
65 4 Homework Statement Two identical audio speakers, connected to the same amplifier, produce monochromatic sound waves with a frequency that can be varied between 300 and 600 Hz. The speed of the sound is 340 m/s. You find that, where you are standing, you hear minimum intensity sound a) Explain why you hear minimum-intensity sound b) If one of the speakers is moved 39.8 cm toward you, the sound you hear has maximum intensity. What is the frequency of the sound? c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity? Homework Equations interference Homework Statement:Two identical audio speakers, connected to the same amplifier, produce monochromatic sound waves with a frequency that can be varied between 300 and 600 Hz. The speed of the sound is 340 m/s. You find that, where you are standing, you hear minimum intensity sound
a) Explain why you hear minimum-intensity sound
b) If one of the speakers is moved 39.8 cm toward you, the sound you hear has maximum intensity. What is the frequency of the sound?
c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?
Homework Equations:interference
I have no idea on how to proceed
I started with
## frequency=\frac {speed\space of\space sound} \lambda \space = \frac {340 \frac m s} \lambda ##
then
##d \space sin\alpha \space = \space \frac \lambda 2\space ##
but now i'm stuck
Any help please? |
Previously we've discussed the permutation test and the sign test. We've used the sign test to look at the median (a measure of location) survival time with a censored data point: the original observations were transformed into "successes" and "failures" and throws out a
lot of information.
For location inference, a basic one-sample procedure is the
Wilcoxon Signed-Rank test. It's used to test whether a sample comes from a population with a specified mean or median.
Wilcoxon Signed-Rank Test
Suppose $n$ observations $x_1, x_2, \cdots, x_n$, are a sample from a
symmetric continuous distribution with unknown median, $\theta$. We want to test:
\[\begin{aligned} &H_0: \theta = \theta_0 \\ \text{vs. } &H_1: \theta \neq \theta_0 \end{aligned}\]
The assumption of symmetry implies that under $H_0$:
\[\begin{aligned} d_i = x_i - \theta_0 && i = 1, 2, \cdots, n \end{aligned}\]
Each $d_i = x_i - \theta_0$ is equally likely to be positive or negative (i.e. each $x_i$ is equally likely to be above or below $\theta_0$ under $H_0$ - logic for sign test). The magnitude $|d_i|$ of any size is equally likely to be positive or negative.
A symmetric population distribution has a mean that coincides with the median. In those circumstances the test may also be formulated in terms of means. The test procedure is simple:
Calculate the discrepancies of each observation from $\theta_0$, the median hypothesized under $H_0$. Order these magnitudes (i.e. in absolute value) from smallest ($\text{rank} = 1$) to largest ($\text{rank} = n$). Assign a $+$ sign to ranks corresponding to $d_i > 0$, and a $-$ sign to ranks corresponding to $d_i < 0$.
From here, we can define a few different test statistics. We denote by $S_+$the the sum of positive ranks, and by $S_-$ the sum of ranks associated with negative deviations. These two are equivalent to each other because the total sum of ranks in sample of size $n$ is fixed: $1 + 2 + \cdots + n = S_+ + S_-$. We may use any of the statistics $S_+, S_-$, or a third statistics $S_d = |S_+ - S_-|$ as a test statistic. All three have the
same information about the plausibility of $H_0$. Under $H_0$, we'd expect $S_+$ and $S_-$ to be roughly equal, and likewise $S_d$ should be close to $0$. Intermediate values of $S_+$ or $S_-$ are more supportive of $H_0$.
We can use the permutation test approach to get the
exact distribution for any of the three test statistics. We look at all the possible allocations of $+/-$ signs to the ranks $1,2, \cdots, n$. Let's take a look at the following example. Example
Heart rate (beats per minute) when standing was recorded for seven people. Assume a symmetric distribution for heart rate in the population. Continuity is questionable as heart rate is an integer, but we're okay if there are no ties, which is usually why we assume continuity.
The observed data is:
\[73, 82, 87, 68, 106, 60, 97\]
Suppose we want to test
\[\begin{aligned} &H_0: \theta = 70 \\ \text{vs. } &H_1: \theta > 70 && \text{one-sided alternative} \end{aligned}\]
First, compute $|d_i| = |x_i - \theta_0| = |x_i - 70|$, $i = 1, \cdots, 7$:
\[\begin{aligned} |d_i| && 3 && 12 && 17 && 2 && 36 && 10 && 27 \\ \text{Rank} && 2 && 4 && 5 && -1 && 7 && -3 && 6 \ \end{aligned}\]
\[\begin{aligned} \Rightarrow S_+ &= 2 + 4 + 5 + 7 + 6 = 24 \\ S_- &= 1 + 3 = 4 \\ S_d &= |S_+ - S_-| = 20 \end{aligned}\]
Sanity check: $1 + \cdots + 7 = 28 = S_+ + S_-$.
Under $H_0$ ($\theta = 70$), we'd expect $S_+ \approx S_-$ and $S_d \approx 0$. Under $H_1$ ($\theta > 70$), we'd expect $S_+ > S_-$ and $S_d$ to be larger. Here we observe $S_+$ to be appreciably larger than $S_-$, which is in support of $H_1$.
For the permutation test, we need to build the permutation distribution. Note that $S_+$ can take values from $0$ (when all ranks are negative $\Rightarrow$ all observations are less than $\theta_0$) up to $28$ (all ranks are positive). There are $2^7 = 128$ ways of allocating $+/-$ signs to the ranks. All of them are equally likely under $H_0$. Below we have a few possible configs:
1 2 3 4 5 6 7 $S_+$ $S_-$ $S_d$ - - - - - - - 0 28 28 + - - - - - - 1 27 26 $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ $\vdots$ + + + + + + + 28 0 28
With an one-sided test, the upper tail $H_1 \Rightarrow$ large values of $S_+$ (equivalently small values of $S_-$) are evidence against $H_0$. We want $P(S_+ \geq 24) = P(S_+ = 24) + P(S_+ = 25) + \cdots + P(S_+ = 128) = \frac{7}{128} \approx 0.0547$, which is pretty unlikely if $H_0$ holds. The evidence against $H_0$ is sufficiently strong to warrant a similar study with a larger group of students.
Function in R
In R, the function
wilcox.test() does the exact permutation test if $n < 50$ and there are no ties in data. Otherwise, a normal approximation is used. It should be noted that two options in
wilcox.test conflict with each other:
exact: compute exact permutation distribution. Ignored when there are ties.
correct: use continuity correction based on Normal distribution.
exactoverrides
correct, so when
exact = Tthe value of
correctdoesn't matter.
In our case,
wilcox.test(data, "greater", mu = 70) gives $S_+ = 24$ (R calls it
V) and a p-value of $0.05469$. We can also do this by hand with the following procedure.
There are $2^{12} = 4096$ configurations in the permutation distribution. The statistic $S_+$ (or $S_-$) can take on 79 possible values - the values $0$ to $1 + 2 + \cdots + 12 = 78$. We may use
dsignrank(z, n) to build up the permutation distribution for $S_+$ (equivalently $S_-$)
[1], where
z are the possible values, and
n is the sample size.
Another useful application is to get a confidence interval (CI) for the population median (mean):
wilcox.test(x,conf.int = T, conf.level = 0.95). The test is based on
Walsh averages.
Walsh Average
Walsh averages are just pairwise averages of the observations. Set $H_0: \theta = \theta_0$ and get the order statistics $x_{(1)} <x_{(2)} < \cdots <x_{(n)}$. Now, if $x_i < \theta_0$, $X_i - \theta_0$ will have a
negative rank. In fact, the largest negative deviation will belong to $x_{(1)}$, the next largest to $x_{(2)}$, and so on. Likewise, if $x_j > \theta_0$, $x_j - \theta_0$ will have a positive rank, and the largest positive deviation will belong to $x_{(n)}$.
Now, we consider the paired averages of $x_{(1)}$ with each of $x_{(1)}, x_{(2)}, \cdots, x_{(n)}$.
Denote by $-p$ the signed rank of the deviation associated with $x_{(1)}$. Then each $\frac{1}{2}(x_{(1)} + x_{(q)})$ is less than $\theta_0$ when the deviation associated with $x_{(q)}$ also has negative rank, orpositive rank less than $p$. If $x_{(q)}$ is the smallest observation with a positive rank greater than $p$, then $|x_{(q)} - \theta_0| > |x_{(1)} - \theta_0|$, and then the average of $x_{(1)}$ and $x_{(q)} > \theta_0$. And also any $x_{(r)} > x_{(q)}$. The number of averages involving $x_{(1)}$ that are less than $\theta_0 =$ negative rank associated with $x_{(1)}$. Likewise, the number of averages involving $x_{(i)}$ less than $\theta_0$ with each of $x_{(i)}, x_{(i+1)}, \cdots, x_{(n)}$ that are less than $\theta_0 =$ negative rank associated with $x_{(i)}$. The number of
Walsh averagesless than $\theta_0 = S_-$, and the number of Walsh averages greater than $\theta_0 = S_+$.
We can use this correspondence to get point estimate and CI for the population median (mean). To do this:
Order the data from smallest to largest. Compute the pairwise Walsh averages and arrange them in a table. There are 78 unique averages. The Walsh averages increase along each row and column. Point estimator of the population median is the median of the Walsh averages, which should be the average of the $39^{th}$ and the $40^{th}$ ordered Walsh averages. In our example, they are both $9 \Rightarrow \hat{\theta} = 9$. The median of the Walsh averages is also called the
Hodges-Lehmann estimator, having been proposed by Hodges and Lehmann (1963).
To find a $95\%$ confidence interval using the Walsh averages, we select as end points values of $\theta$ that will just be acceptable if $P = 0.05$. For one tail, $0.025 \times 4096 = 102.4$, so the cumulative sum can't be more than 102.4. CI puts roughly 2.5% of density in each tail $\Rightarrow$ 14th smallest and 14th largest Walsh averages $\Rightarrow (2.5, 16.0)$. Asymptotic Results
What happens as sample size $n$ increases? The number of possible configurations (possible assignments of $+/-$ signs to the ranks) becomes large very quickly - $2^n$ increases rapidly. Therefore, it's not feasible to get the exact distribution even with the aid of software.
But we also have symmetry, and as $n$ increases, the number of possible values of $S_+$ also increases ($\frac{1}{2}n(n+1)$). There is a
normal approximation to the exact distribution of the test statistic! We can easily show that
\[\begin{aligned} E(S_+) &= \frac{1}{4}n(n+1) \\ Var(S_+) &= \frac{1}{24}n(n+1)(2n+1)\end{aligned}\]
since the sum of the integers from $1$ to $n$ is $\frac{n(n+1)}{2}$ and the sum of squares is $\frac{n(n+1)(2n+1)}{6}$. For large enough $n$, we can define a test statistic $Z$
\[Z = \frac{S_+ - E(S_+)}{\sqrt{Var(S_+)}} \dot\sim N(0,1)\]
We can also improve the effectiveness of the approximation by using a
continuity correction, which comes to account for the fact that we approximate a discrete quantity $S_+$ with a continuous distribution (in this case the normal). The idea is if $S$ is the smaller of the sums of the positive of negative ranks, we replace $S$ by $S + \frac{1}{2}$. If $S$ is the larger sum it is replaced by $S - \frac{1}{2}$. In R, this is called with the option
correct = T in
wilcox.test.
Another thing we could do when $2^n$ is too large to enumerate all of the possible configurations (e.g. in the millions) is to take a random sample of them instead (e.g. $\approx 10,000$).
Wilcoxon with Ties
We've been assuming the underlying population distribution is continuous, so in theory in the data there should be no ties. In practice, however, ties can of course happen. Real observations never have a distribution that is strictly continuous either because of their nature, or as a result of rounding errors or limited measurement precision.
The other in theory impossible but in practice often observed are deviations of $0$. We're talking about values in the sample that are exactly equal to the median hypothesis under $H_0$, $\theta_0$. There's no one agreed upon best approach for handling these cases!
Ties in the Observations
One suggestion is to replace the ranks for the tied values with their
mid-ranks. For example, if we had a sample
\[12, 18, 24, 26, 37, 40, 42, 47, 49, 49, 78, 108\]
We take $H_0: \theta = 30$. The deviations are then
\[\begin{aligned} -18 && -12^{\ast} && -6 && -4 && 7 && 10 && 12^{\ast} && 17 && 19 && 19 && 48 && 78 \end{aligned}\]
The *s have the same magnitude $\Rightarrow$ ranks would be tied. They would be ranked $5$ and $6$, so we give them both $5.5$. Similarly the $19$s are both ranked $9.5$:
\[\begin{aligned} -8 && -5.5 && -2 && -1 && 3 && 4 && 5.5 && 7 && 9.5 && 9.5 && 11 && 12 \end{aligned}\]
We can now calculate $S_+$ and $S_-$ same as before. But the exact distribution changes because of using these mid-ranks. In fact, the exact permutation distribution depends on
the number of ties and where they fall in the rank sequence. It's much harder to work out. The distribution with no ties is unimodal, with values of the statistic confined to ingegral values increasing in steps, while the distribution with ties is heavily multimodal with $S$ taking unevenly spaced and not necessarily integer values. Discontinuities are also more marked.
Streitberg and Rohmel
[2] came up with an algorithm called the
shift algorithm to handle this situation. The R package
exactRankTests implements this algorithm under the function
wilcox.exact. Works for ties or no ties!
Another suggestion is to modify the normal approximation. The idea is to consider each of the signed ranks as a score $S_i$ for observation $x_i$. Under $H_0$, as before each score has equal probability of being positive or negative, so the expected value and the variance of $S_+$ or $S_-$ is
\[\begin{aligned} E(S_i) &= \frac{1}{2}\sum\limits_{i=1}^{n} |S_i| \\ Var(S_i) &= \frac{1}{4}\sum\limits_{i=1}^n S_i^2 \end{aligned}\]
These work out to be the same as before for the particular choice of score $\Leftrightarrow$ rank. The
score representation is thus
\[Z = \frac{S_+ - \frac{1}{2}\sum{|S_i|}}{\sqrt{\frac{1}{4}\sum{S_i^2}}} \dot\sim N(0,1)\]
Deviations of 0
Again, opinions differ for data points that are equal to the median hypothesized under $H_0$. A standard advice is to drop such points from the calculation of $S_+$ (this is equivalent to assigning them rank 0), but this decreases the sample size and we lose data!
Sprent and Smeeton (in part 3.3.4) proposed a slight alternative: temporarily assign such points a rank of 1 (or the appropriate tied rank if there's more than one zero), then sign-rank all the other observations as usual. Finally, switch the rank(s) associated with zero deviation to 0. This keeps all the data and uses the ranks up to $n$. However, the effect on the exact distribution is unclear.
Summary The
sign test, unlike the
Wilcoxon signed rank test, does not require symmetry of the underlying distribution. When the data come from a skewed distribution (e.g. income), both the
t-testand the
Wilcoxonmay be inappropriate in the sense that they may not give us valid inference. This depends in part on how skewed the population distribution is. The
sign testwill still be valid. Confidence intervals based on the
t-test/
Wilcoxonmay also be misleading. Those based on the
sign test(and
Binomial distribution) will still be fine. In other situations, all three will lead to similar conclusions.
A suggestion: try different analyses and see if your conclusions are consistent. When the symmetry assumption is violated, the
sign testmay have higher efficiency - higher power in tests and shorter confidence intervals for a given confidence level. We can compare the asymptotic relative efficiency of the sign test, Wilcoxon and t-test:
ARE of the Wilcoxon compared to the t-test is at least 0.864, and can go up to infinity under some circumstances. The Wilcoxon is never "too bad" and can be very good. When the data are actually normally distributed (the situation where the t-test is optimal), the ARE of Wilcoxon is 0.955, so very little loss here.
Up to this point, we've been talking about location inference - questions about the mean or median of the population distribution, but we can do a lot more. One particularly useful application is studying whether our data are consistent with having been drawn from some specified distribution.
For each distribution, there's a series of functions in R to get the density
dsignrank, distribution function
psignrank, quantile function
qsignrankand random numbers
rsignrank. ↩︎
Streitberg, Bernd, and Joachim Röhmel. "Exakte Verteilungen für Rang-und Randomisierungstests im allgemeinen c-Stichprobenfall." EDV in Medizin und Biologie 18.1 (1987): 12-19. ↩︎ |
Submitted by Lanowen on Sun, 05/24/2015 - 02:15
A friend asked me about this problem, on how to calculate the distance an object needs to be from the camera to appear a certain height on the screen.
The solution is simpler when dealing with an object centered in the in the screen where a right angle can be created from the camera to the object.
Fig. 1: right angle triangle diagram. $\figLbl{1}{tanSol}$
Solving the triangle in \ref{fig:tanSol}:
\begin{equation}$\vdots$
\tan{\left( \frac{\theta}{2} \right)} = \frac{h}{2d} \label{eq:tanTriangle} \end{equation} |
I saw a really great study guide that got passed around before the first Engineering 195 test, so I thought it would be a great idea to create a collaborative study guide for MA 181.
Convergent and Divergent Series Tests
1. The nth-Term Test:
Unless $ a_{n}\rightarrow 0 $, the series diverges
2. Geometric series:
$ \sum_{n=1}^\infty ar^n $ converges if $ |r| < 1 $; otherwise it diverges.
3. p-series:
$ \sum_{n=1}^\infty 1/n^p $ converges if $ p>1 $; otherwise it diverges
4. Ratio Test(for series with non-negative terms):
$ \lim \limits_{n \to \infty }{\frac{a_{n+1}}{a_{n}}}=p $ The series converges if p<1, diverges if p>1, and is inconclusive if p=1
5. Root Test(for series with non-negative terms):
$ \lim \limits_{n \to \infty }{\sqrt[n]{a_{n}} }=p $ The series converges if p<1, diverges if p>1, and is inconclusive if p=1
More to come...
I am looking to check my answers to the practice problems that Dr. Bell posted on our home page, so if someone would please post them I would really appreciate it. Gleenhee 20:10, 17 November 2008 (UTC)
That's a lot to post. Are there any that you are just not sure about? --Ctuchek 19:31, 18 November 2008 (UTC) |
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