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https://en.wikipedia.org/wiki/Talk:Alexander_polynomial
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# Talk:Alexander polynomial
WikiProject Mathematics (Rated B-class, Low-importance)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
B Class
Low Importance
Field: Topology
## [Untitled]
Where does this particular method for computation of the Alexander polynomial come from? I believe I followed the method correctly, but depending on the two columns eliminated, I get different determinants.
This is the method given by Alexander in his paper, except the article neglects to mention that one should "renormalize" at the end (something Alexander obviously realized), since there is ambiguity (depending on columns eliminated, as you've seen, the result can differ by multiplication by ${\displaystyle \pm x^{n}}$. Anyway, I will add the statement. --C S (Talk) 00:31, 13 April 2006 (UTC)
## Example of skein relation computation for Conway poly
Moved from article:
(reader's note: This is a terrible example. The article would be much improved by a picture of a simple 3-4 crossing knot and then the step-by-step calculation of the Alexander polynomial! ) --—The preceding unsigned comment was added by 128.101.126.230 (talkcontribs).
This comment is referring to the section on the Conway polynomial which ends by saying see skein relation for an example of a computation. The point is well taken. I will try to get to it sometime. --C S (Talk) 11:29, 9 May 2006 (UTC)
## Explicit computation
I think it would be nice if there was an explicit computation with pictures. I'm having trouble visualising it. Maybe even two explicit computations of the same knot looking very different. —Preceding unsigned comment added by Eigenlambda (talkcontribs) 18:46, 8 April 2008 (UTC)
## Skein Relationship -- Definition used here
I was curious about why the particular definition of the Alexander-Conway skein relation used here was chosen. In his The Knot Book, Colin Adams uses the relation ${\displaystyle \nabla _{L_{+}}(z)-\nabla _{L_{-}}(z)+z\nabla _{L_{0}}(z)=0}$ (with the conversion to the Alexander polynomials given by ${\displaystyle z=t^{1/2}-t^{-1/2}}$). Charles Livingston also uses this as the relation in Knot Theory published by the Mathematical Association of America. Peter Cromwell uses the relation used here in his book Knots and Links, but I've found a few major errors in that section of his book. Does anyone have access to Alexander's original paper or Conway's paper from Computational Problems in Abstract Algebra? I think he (Conway) used the same one that Adams and Livingston used listed here but can't remember. N Vale (talk) 05:46, 22 April 2008 (UTC)
People use the t-variable when considering the Alexander polynomial to be describing the order ideal of H_1 of the universal abelian cover of the knot complement. Because in this setting t represents the generating covering transformation. z has no such interpretation -- z is basically a notation designed to "compress" the information in the Alexander polynomial in order to remove the redundnacies created from the symmetry condition. The z is due to Conway, the t to Alexander. Rybu (talk) 17:07, 29 April 2008 (UTC)
I think the question is not meant to be as deep as that (perhaps I'm wrong). I think he's asking about the sign in the Conway skein relation, why was that chosen instead of the other way. First note that Adams does not use "z" at all. His skein relation with t^{1/2}, is the same, for example, as Lickorish's, but in Lickorish's book the substitution for z is made so that the skein relation for z is the same as in this article. In Conway's article is opposite the one in the article but that's because he considers a positive crossing to be a left-handed crossing, contrary to the usual convention. Kawauchi and Kauffman both use the convention listed in the article. --C S (talk) 23:06, 10 May 2008 (UTC)
## on the knot for mirror image
The fact that Alexander polynomials cannot distinguish the knot for mirror images is fundamental, is not it? I've added an sentence for it.--Enyokoyama (talk) 11:25, 8 September 2014 (UTC)
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2017-08-24 03:29:23
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http://www.sawaal.com/numbers-questions-and-answers/the-difference-between-a-positive-proper-fraction-and-its-reciprocal-is-9-20-then-the-fraction-is-_1469
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7
Q:
# The difference between a positive proper fraction and its reciprocal is 9 / 20. Then the fraction is :
A) 3/5 B) 3/10 C) 4/5 D) 5/4
Explanation:
Let the required fraction be x. Then, (1 / x )- x = 9/20
1 - x^(2) / x = 9 / 20 => 20 - 20 * x^(2) = 9 * x.
20 * x^(2) + 9 *x - 20 = 0.
=> (4 * x + 5) (5 * x - 4) = 0.
=> x = 4 / 5.
Q:
When a number is divided by 138 the remainder is 26. What will be the remainder if the same number is divided by 23 ?
A) 4 B) 6 C) 3 D) 1
Explanation:
Number = quotient x divisor + remainder;
so, here
number = 138 k + 26
=> (23 x 6k) + (23+3)
=> 23(6k+1)+3
so, remainder is 3.
1 8
Q:
How many zeros are there from 1 to 10000 ?
A) 2893 B) 4528 C) 6587 D) 4875
Explanation:
For solving this problem first we would break the whole range in 5 sections
1) From 1 to 9
Total number of zero in this range = 0
2) From 10 to 99
Total possibilities = 9*1 = 9 ( here 9 is used for the possibilities of a non zero integer)
3) From 100 to 999 - three type of numbers are there in this range
a) x00 b) x0x c) xx0 (here x represents a non zero number)
Total possibilities
for x00 = 9*1*1 = 9, hence total zeros = 9*2 = 18
for x0x = 9*1*9 = 81, hence total zeros = 81
similarly for xx0 = 81
total zeros in three digit numbers = 18 + 81 +81 = 180
4) From 1000 to 9999 - seven type of numbers are there in this range
a)x000 b)xx00 c)x0x0 d)x00x e)xxx0 f)xx0x g)x0xx
Total possibilities
for x000 = 9*1*1*1 = 9, hence total zeros = 9*3 = 27
for xx00 = 9*9*1*1 = 81, hence total zeros = 81*2 = 162
for x0x0 = 9*1*9*1 = 81, hence total zeros = 81*2 = 162
for x00x = 9*1*1*9 = 81, hence total zeros = 81*2 = 162
for xxx0 = 9*9*9*1 = 729, hence total zeros = 729*1 = 729
for xx0x = 9*9*1*9 = 729, hence total zeros = 729*1 = 729
for x0xx = 9*1*9*9 = 729, hence total zeros = 729*1 = 729
total zeros in four digit numbers = 27 + 3*162 + 3*729 = 2700
thus total zeros will be 0+9+180+2700+4 (last 4 is for 4 zeros of 10000)
= 2893
2 28
Q:
(3x + 2) (2x - 5) = ax² + kx + n. What is the value of a - n + k ?
A) 4 B) 1 C) 5 D) 3
Explanation:
(3x + 2) (2x - 5) = $\inline \fn_jvn \small ax^{2}+kx+n$ ...............(1)
But (3x + 2)(2x - 5) = $\inline \fn_jvn \small 6x^{2}-11x-10$........(2)
so by comparing (1) & (2),
we get a= 6, k= -11 , n= -10
(a - n + k) = 6 + 10 - 11 = 5.
2 14
Q:
What is the smallest number by which 2880 must be divided in order to make it into a perfect square ?
A) 2 B) 3 C) 4 D) 5
Explanation:
By trial and error method, we get
2880/3 = 960 is not a perfect square
2880/4 = 720 is not a perfect square
2880/5 = 576 which is perfect square of 24
Hence, 5 is the least number by which 2880 must be divided in order to make it into a perfect square.
3 30
Q:
K men agree to donate a gift of Rs. L to trust. If three men drop out how much more will each have to contribute towards the purchase of the gift ?
A) L / (k-3) B) k / (L-3) C) 2K / 3L-K D) 3L / K(K-3)
The amount more to pay in contribution = $\inline \fn_jvn \small \frac{L}{K-3}-\frac{L}{K}$
$\inline \fn_jvn \small \frac{3L}{K(K-3)}$
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2017-02-19 14:24:34
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https://tex.stackexchange.com/questions/349408/margin-differs-from-usepackagegeometry
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# margin differs from \usepackage{geometry}
Hi I have the following preamble in my thesis :
\documentclass[12pt,a4paper,twoside]{report}
\setlength{\textwidth}{17cm}
\setlength{\parskip}{3mm}
\setlength{\parindent}{0mm}
\usepackage[utf8]{inputenc}
\usepackage[top=2cm, bottom=2cm, left=2cm, right=2cm]{geometry}
\usepackage{graphicx}
\graphicspath{{images/}}
\usepackage{setspace}
\usepackage{epsfig}
\usepackage{filecontents}
\usepackage{xparse,nameref}
\usepackage{mathrsfs}
\usepackage[nokeyprefix]{refstyle}
\usepackage{mathtools}
\usepackage{thmtools}
\usepackage{enumerate}
\usepackage{enumitem}
\usepackage{thm-restate}
\usepackage{cleveref}
\usepackage{amsmath}
\usepackage{url}
\usepackage{amsthm}
\DeclareMathOperator{\lcm}{lcm}
\DeclareMathOperator{\ord}{ord}
\usepackage{cellspace}
\setlength\cellspacetoplimit{4pt}
\setlength\cellspacebottomlimit{4pt}
\usepackage{booktabs}
\usepackage{graphicx}
\usepackage{array}
\usepackage{tabularx,ragged2e,booktabs,caption}
\usepackage[figuresleft]{rotating}
\usepackage{rotating}
\newcolumntype{C}{>{\Centering\arraybackslash}X}
\newcommand\swb{{\scriptstyle\Box}} % "small white box"
\usepackage{mathtools}
\usepackage{tikz}
\usetikzlibrary{positioning}
\usetikzlibrary{arrows}
\setcounter{secnumdepth}{3}
\usetikzlibrary{shapes.multipart}
\usepackage{tabu}
\usepackage{amssymb,bm}
\usepackage{framed}
\usepackage{upgreek} %use uptau greek letter
\usepackage{comment} %choose parts not to compile
\usepackage{tikz}
\tikzset{mynode/.style={fill=white,text=black,font=\tiny,inner sep=1pt}}
\usepackage[toc,page]{appendix}
\newenvironment{numberlists}[1][3\parindent] %for long numberlist command
{\begin{list}{}{%
\leftmargin=#1\relax
\rightmargin=\leftmargin
\itemsep=\jot
\parsep=0pt
\partopsep=0pt
\labelsep=0pt}}
{\end{list}}
\newcommand\numlist[2]{%
\item[]\makebox[0pt][r]{$#1=\lbrack$}%
\begingroup
\begingroup\lccode~=,\lowercase{\endgroup\def~}{\mathcomma\penalty0 }%
\mathcode,="8000
\thinmuskip=6mu plus 6mu minus 2mu
$#2\rbrack$%
\endgroup
}
\mathchardef\mathcomma=\mathcode, %end of numberlist command
\setlength{\oddsidemargin}{-1in} \setlength{\evensidemargin}{-1in}
\setlength{\marginparwidth}{40pt} \setlength{\marginparsep}{10pt}
\setlength{\textheight}{227mm} \setlength{\textwidth}{165mm}
\renewcommand\floatpagefraction{0.8} % Default = 0.5
\usepackage{listings}
\usepackage{color}
\definecolor{dkgreen}{rgb}{0,0.6,0}
\definecolor{gray}{rgb}{0.5,0.5,0.5}
\definecolor{mauve}{rgb}{0.58,0,0.82}
\lstset{
language=Java,
aboveskip=3mm,
belowskip=3mm,
showstringspaces=false,
columns=flexible,
basicstyle={\small\ttfamily},
numbers=none,
numberstyle=\tiny\color{gray},
keywordstyle=\color{blue},
stringstyle=\color{mauve},
breaklines=true,
breakatwhitespace=true,
tabsize=3
}
I know that my premable is a mess and am trying to fix it. The problem here is that i have \usepackage[top=2cm, bottom=2cm, left=2cm, right=2cm]{geometry} and it does not seem to work because when I print my thesis the margin is different. top margin is 4.5 cm, bottom, left and right margin are all 2.5 cm. I think this has something to do with my numberlist command in my preamble (because there's a different margin set up in that command). Is there a way to fix it. The grad school refuse to accept my thesis for submission because of the margin issue. It should be 2cm top bottom left right.
\documentclass[12pt,a4paper,twoside]{report}
Remove this line, it will be undone anyway by the geometry package
\setlength{\textwidth}{17cm}
Unrelated to your margins but are you sure that you want a fixed length for \parskip together with \flushbottom it will mean that pages with a paragraph are almost always going to be infeasible and generate under or over full box warnings.
\setlength{\parskip}{3mm}
\setlength{\parindent}{0mm}
\usepackage[utf8]{inputenc}
This is the form that you want to specify the page size, but normally you should also use heightrounded so that the specified size matches the baselinesip.
\usepackage[top=2cm, bottom=2cm, left=2cm, right=2cm]{geometry}
\usepackage{graphicx}
\graphicspath{{images/}}
\usepackage{setspace}
Don't use epsfig
\usepackage{epsfig}
\usepackage{filecontents}
\usepackage{xparse,nameref}
\usepackage{mathrsfs}
\usepackage[nokeyprefix]{refstyle}
\usepackage{mathtools}
\usepackage{thmtools}
don't use enumerate if you are using enumitem (they patch the same environment)
\usepackage{enumerate}
\usepackage{enumitem}
\usepackage{thm-restate}
\usepackage{cleveref}
\usepackage{amsmath}
\usepackage{url}
\usepackage{amsthm}
\DeclareMathOperator{\lcm}{lcm}
\DeclareMathOperator{\ord}{ord}
\usepackage{cellspace}
\setlength\cellspacetoplimit{4pt}
\setlength\cellspacebottomlimit{4pt}
\usepackage{booktabs}
\usepackage{graphicx}
\usepackage{array}
\usepackage{tabularx,ragged2e,booktabs,caption}
\usepackage[figuresleft]{rotating}
Don't load rotating here if you loaded it on the previous line
\usepackage{rotating}
\newcolumntype{C}{>{\Centering\arraybackslash}X}
\newcommand\swb{{\scriptstyle\Box}} % "small white box"
\usepackage{mathtools}
\usepackage{tikz}
\usetikzlibrary{positioning}
\usetikzlibrary{arrows}
\setcounter{secnumdepth}{3}
\usetikzlibrary{shapes.multipart}
\usepackage{tabu}
\usepackage{amssymb,bm}
\usepackage{framed}
\usepackage{upgreek} %use uptau greek letter
\usepackage{comment} %choose parts not to compile
\usepackage{tikz}
\tikzset{mynode/.style={fill=white,text=black,font=\tiny,inner sep=1pt}}
\usepackage[toc,page]{appendix}
Don't do this it undoes the geometry settings
\addtolength{\topmargin}{0.5in}
\newenvironment{numberlists}[1][3\parindent] %for long numberlist command
{\begin{list}{}{%
\leftmargin=#1\relax
\rightmargin=\leftmargin
\itemsep=\jot
\parsep=0pt
\partopsep=0pt
\labelsep=0pt}}
{\end{list}}
\newcommand\numlist[2]{%
\item[]\makebox[0pt][r]{$#1=\lbrack$}%
\begingroup
\begingroup\lccode~=,\lowercase{\endgroup\def~}{\mathcomma\penalty0 }%
\mathcode,="8000
\thinmuskip=6mu plus 6mu minus 2mu
$#2\rbrack$%
\endgroup
}
The comment here is wrong, numberlist ended above.
\mathchardef\mathcomma=\mathcode, %end of numberlist command
Delete all these lines they undo the geometry settings
\setlength{\oddsidemargin}{-1in} \setlength{\evensidemargin}{-1in}
\setlength{\marginparwidth}{40pt} \setlength{\marginparsep}{10pt}
\setlength{\textheight}{227mm} \setlength{\textwidth}{165mm}
Are you sure you want this, it makes float pages hard to form and increases the chance that all floats go to the end of the document
\renewcommand\floatpagefraction{0.8} % Default = 0.5
\usepackage{listings}
\usepackage{color}
\definecolor{dkgreen}{rgb}{0,0.6,0}
\definecolor{gray}{rgb}{0.5,0.5,0.5}
\definecolor{mauve}{rgb}{0.58,0,0.82}
\lstset{
language=Java,
aboveskip=3mm,
belowskip=3mm,
showstringspaces=false,
columns=flexible,
basicstyle={\small\ttfamily},
numbers=none,
numberstyle=\tiny\color{gray},
keywordstyle=\color{blue},
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2022-05-18 19:30:47
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https://physics.stackexchange.com/questions/576451/why-is-the-tachyon-vertex-operator-int-d2z-eik-xz-barz-integrated
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# Why is the tachyon vertex operator $\int d^2z :e^{ik.X(z,\bar{z})}:$ integrated?
I understand that, from the state-operator correspondence, $$|0;p\rangle \;=\;:e^{ip.X(0,0)}: |0\rangle.$$ This is given in Polchinski, equation (2.8.9). I am now trying to understand the S-matrix for $$2\times$$tachyon $$\rightarrow$$ $$2\times$$tachyon scattering. My understanding is as follows.
We seek to calculate $$\langle\psi_f|\psi_i\rangle$$, where the initial and final wavefunctions are both two taychons. That is, we want to find $$\langle 0, q_1 ; 0, q_2 | 0,p_1;0,p_2\rangle \; =\; \langle0|:e^{-iq_1.X}: :e^{-iq_2.X}: :e^{ip_1.X}: :e^{ip_2.X}:|0\rangle$$ $$= \int DXDg \; e^{-S_{Poly} [X, g]} \;V[-q_1, 0]V[-q_2, 0]V[p_1, 0]V[p_2, 0],$$
where $$V[p, z] = \; :e^{ip.X(z, \bar{z})}:$$.
However, this is not the result quoted in textbooks (e.g. Polchinski eqns 3.5.5, and 3.6.1). Instead the actual result is
$$\int DXDg \; e^{-S_{Poly} [X, g]} \;\int d^4z_i d^4\bar{z}_i V[-q_1, z_1]V[-q_2, z_2]V[p_1, z_3]V[p_2, z_4],$$
citing diffeomorphism invariance. I see that my expression is not invariant under diffeomorphisms, whilst the second one is. However, why is the second expression the correct result? I see it in textbooks referred to a guess, but a guess of what exactly? Surely what we have now is not the overlap of four tachyons $$<\psi_f | \psi_i>$$ which we sought to calculate, but rather the overlap of some strange superposition of infinitely many tachyons. How does this relate to the $$2\times$$ tachyon scattering amplitude at all?
The result without integration is a correlation function of $$4$$ operators in the auxiliary worldsheet CFT. It is not diffeo-invariant, as it should be, because the CFT is not diffeo-invariant.
• And so what string theory states are we calculating the overlap between here, in the last expression? Is $|0p_1;0p_2>$ not a state in the string theory? – awsomeguy Aug 30 '20 at 10:20
• I see that the state $|0p_1; 0p_2>$ does not have a point associated to it - but can it not still be written as a local operator on the vacuum state as per the state-operator map? i.e. the state corresponds to a vertex that is not already integrated? – awsomeguy Aug 30 '20 at 11:34
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2021-08-04 05:07:33
|
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http://physicshelpforum.com/electricity-magnetism/4893-calculating-charge-potential.html
|
Physics Help Forum calculating charge from a potential
Electricity and Magnetism Electricity and Magnetism Physics Help Forum
Sep 12th 2010, 05:25 PM #1 Junior Member Join Date: Sep 2010 Posts: 2 calculating charge from a potential Hi, I'm given a potential V(r)=Ae^(-br)/r, where A and b are constants. I am asked to determine the electric field, the charge distribution, and the total charge. The electric field is the negative gradient of V, and I calculated it to be Ae^(-br)(rb+1)/r^2 in the radial direction. This has been checked with wolfram alpha. The charge distribution is epsilon-naught times the divergence of the electric field, and I calculated the div of E to be -(b^2)Ae^-(br)/r - 2Ae^(-br)(rb+1)/r^3. I haven't checked with wolfram yet, but I'm almost 100% certain this is correct. However, I attempted to calculate the total charge two different ways, and in neither case am I getting anything meaningful. First, I attempted to integrate the charge distribution over all space, and this integral I found to be nonconvergent. Second, I used the divergence theorem, turning the integral of the charge distribution over space to a surface integral of the E field with r=infinity (I don't know if that's actually valid, I've never used the divergence theorem with any infinite bounds), and this returned 0. I checked a few times and I don't see anything wrong with my calculations, but I grant there may be. Am I lacking some sort of insight into the charge calculation, or is this some sort of trick question, and the total charge is in fact indeterminant? Thanks
Sep 12th 2010, 05:39 PM #2 Junior Member Join Date: Sep 2010 Posts: 2 omg i forgot to multiply by r^2 when calculating the divergence. that was smooth
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2018-11-18 20:33:34
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http://www.avvo.com/attorneys/98007-wa-jason-epstein-31297/legal_cases/1429
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# John Doe v. Drunk Driver
Practice Area:Car / Auto Accident
Outcome:1.06 Million Dollar settlement reached
Description:John Doe was performing his job as a surveyor. He was unloading equipment from his work truck when a drunk driver fell asleep and the wheel and veered off the road striking John Doe and his truck. Mr. Doe was dragged under the truck that struck him for several hundred feet. Mr. Doe sustained burst fractures to three vertabrae in his lower back. Immediate surgery was required. Mr. Doe had a long and painful recovery, but ultimately regained most of his function. Unfortunately, he was no longer able to perform the physical requirements of his old job and had to be retrained. Claims were made against the at fault driver as well as Mr. Doe's personal under insured motorist policy and Mr. Doe's employers under insured motorist policy. The total settlement with all parties was \$1,060,000.00. Additionally, a substantial reduction in the 'offset' was obtained by negotiating with Labor and Industries.
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2015-01-28 18:39:43
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http://mathhelpforum.com/discrete-math/92490-decide-if-sequence-converges-diverges.html
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# Math Help - Decide if sequence converges/diverges
1. ## Decide if sequence converges/diverges
The given sequence is: $a_n = n\left(\sqrt{1 + \frac{1}{n}} - 1\right)$.
I rearranged terms to get $a_n = \sqrt{n^2 + n} - n$. I couldn't tell by inspection if this is convergent or divergent, so I started plugging in large values of n and noticed that it seems to be that $a_n \to 0.5$. So now I'm trying to prove it converges by showing that $a_n$ has an upper bound and is non-decreasing. However I'm having trouble showing it has an upper bound, therefore leading me to think maybe that $a_n$ is divergent? I'm not sure at this point, any suggestions would be appreciated.
2. Originally Posted by utopiaNow
The given sequence is: $a_n = n\left(\sqrt{1 + \frac{1}{n}} - 1\right)$.
I rearranged terms to get $a_n = \sqrt{n^2 + n} - n$. I couldn't tell by inspection if this is convergent or divergent, so I started plugging in large values of n and noticed that it seems to be that $a_n \to 0.5$. So now I'm trying to prove it converges by showing that $a_n$ has an upper bound and is non-decreasing. However I'm having trouble showing it has an upper bound, therefore leading me to think maybe that $a_n$ is divergent? I'm not sure at this point, any suggestions would be appreciated.
$a_n=\frac{(\sqrt{n^2+n}-n)(\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n}=\frac{n^2+n-n^2}{\sqrt{n^2+n}+n}=\frac{n}{\sqrt{n^2+n}+n}=$
$a_n=\frac{1}{\sqrt{1+\frac{1}{n}}+1}$
$\lim_{n \to \infty}\frac{1}{\sqrt{1+\frac{1}{n}}+1}=\frac{1}{2 }$
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2014-12-18 11:59:56
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http://www.cs.colostate.edu/~cs545/fall16/doku.php?id=assignments:assignment1
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# CS545 fall 2016
### Sidebar
CS545
Instructor
Asa Ben-Hur
assignments:assignment1
# Assignment 1
Due date: 9/4 at 11:59pm
## Preliminaries
We need to start with a little bit of notation… In supervised learning we work with a dataset of $N$ labeled examples: $\mathcal{D} = \{ (\mathbf{x}_i, y_i) \}_{i=1}^N$, where $\mathbf{x}_i$ is a $d$-dimensional vector (we always use boldface to denote vectors), and $y_i$ is the label associated with $\mathbf{x}_i$. In a binary classification problem we'll usually use the values $\pm 1$ to denote positive/negative examples.
## Part 1: Measuring classifier error
First let's recall that the estimate of a classifier's error is given by:
$$E(h) = \frac{1}{N}\sum_{i=1}^N I(h(\mathbf{x}_i) \neq y_i),$$
where $I(\cdot)$ is the indicator function, and $h$ is the model/hypothesis we are trying to evaluate.
Whenever training a classifier, we like to know how well it's performing. This is done by computing an estimate of the out of sample error: pick an independent test set that was not used during training and compute the error of your classifier on this dataset (the test set). You always want to know that your classifier is learning something, i.e. the error is smaller than what we would expect by chance, i.e. better than a model that simply guesses or picks a fixed answer. Consider the following classifier, which always classifies an example as belonging to the majority class, i.e. the class to which the largest number of training examples belong to.
• Suppose you have data that is very imbalanced, and let's say for concreteness that we're working with a binary classification problem where the number of negative examples is much larger than the number of positive examples. What can you say about the estimated error of the majority classifier? What issue does that raise in your opinion about evaluating classifiers?
To solve the issue with the standard error rate, it has been suggested to assign different costs to different types of errors using a cost matrix $c(y_i, h(\mathbf{x}_i))$, where $y_i$ is the actual class of example $i$, and $h(\mathbf{x}_i)$ is the the predicted class. For a binary classification problem this is a $2 x 2$ matrix, and we'll assume there is no cost associated with a correct classification, which leaves two components to be determined:
• $c_r = c(+1, -1)$, which is the reject cost (the cost of a false negative)
• $c_a = c(-1, +1)$, which is the accept cost (the cost of a false positive).
Incorporating the cost matrix into computing the error: (clarification 8/27 at 5:19 pm)
The regular error $$E(h) = \frac{1}{N}\sum_{i=1}^N I(h(\mathbf{x}_i) \neq y_i)$$ is now replaced with: $$E_{cost}(h) = \frac{1}{N}\sum_{i=1}^N c(y_i, h(\mathbf{x}_i)) \cdot I(h(\mathbf{x}_i) \neq y_i)$$
With these definitions, answer the following:
• How should we choose $c_r$ and $c_a$ such that the majority classifier and the minority classifier both have an error of 0.5? (The minority classifier is analogous to the majority classifier, except that it classifies everything as positive, since we assumed the positive class has fewer representatives). Section 1.4.1 in the book has a brief discussion of error measures.
## Part 2: The nearest centroid classifier
The closest centroid classifier classifies a data point $\mathbf{x}$ according to the class of the nearest centroid. More formally, let $C_k$ be the set of examples that belong to class $k$, and let $$\mu_k = \frac{1}{|C_k|} \sum_{i \in C_k} \mathbf{x}_i,$$ where $|C_k|$ is the cardinality of the set $C_k$. The closest centroid classifier predicts the class of a point $\mathbf{x}$ as:
$$h(\mathbf{x}) = \textrm{argmin}_k ||\mathbf{x} - \mu_k||,$$ where $||\mathbf{x}||$ is the Euclidean norm of $\mathbf{x}$.
Show that for a binary classification problem where the number of positive examples equals the number of negative examples the nearest centroid classifier can be expressed as a linear classifier with the weight vector $$\mathbf{w} = \frac{1}{N}\sum_{i=1}^N y_i \mathbf{x}_i.$$ Hint: consider the vector that connects the centroids of the two classes and draw a figure in two dimensions to help you think about the problem. Also note that this form only holds if the two classes have equal number of examples, so we'll assume that is the case.
## Part 3: Are my features useful?
In order to obtain an accurate classifier you need good features. But what does that mean? In this task we will explore what that means, and how to visually inspect a dataset to identify useful features.
First we need some data… the UCI machine learning repository contains a large selection of datasets that we can experiment with. Choose a binary classification problem (some examples include the Heart , disease diagnosis dataset the Gisette handwritten digit recognition dataset, and a QSAR dataset ( QSAR refers to models that aim at predicting the biochemical activity of molecules, usually in the context of their usefulness as drug molecules). Most data will come as a data matrix in csv or related formats (here is a processed version of the heart dataset). Each row in the file corresponds to a training example. You can ignore the first column; the second column contains the label (+1 or -1), and the rest of the columns contain the feature vectors. Note that the first row in the file is a comment, and needs to be ignored. To read a data matrix you can use numpy's genfromtxt function. For example, to read the heart dataset you can use the following command:
>>> import numpy as np
>>> data=np.genfromtxt("heart.data", delimiter=",", comments="#")
And note that since the file contains both the labels and data points, they need to be separated out. As an alternative you can use the usecols option of genfromtxt to directly read the columns you are interested in.
For the Gisette dataset the feature matrix is provided separately from the labels. To read the data matrix of the Gisette dataset:
>>> X=np.genfromtxt("gisette_train.data")
The QSAR dataset can be loaded in a similar way.
Our objective in this task is to visualize the usefulness of the features that make up the dataset. For a given feature (a column in the feature matrix), generate two histograms of its values: one for the positive examples, and one for the negative examples. Use matplotlib's hist function to generate the histogram and use the normed=True option to generated a histogram normalized to be a distribution. The matplotlib website has a lot of usage examples. Here are some examples that demonstrate what you can do with its hist function.
What does this kind of visualization tell us about the usefulness of a feature for classifying a dataset? Demonstrate this idea using a dataset from the UCI repository: plot histograms for four features, two of which you think are going to be useful, and two that have a more limited usefulness in your opinion – Explain!! In plotting, create a single plot composed of four subplots, one for each feature. This is a convenient way of grouping together related plots. When choosing which features to display, simply use your judgement as to which ones you should show.
Your report needs to be written in LaTex. You can either use LaTex by installing it on your computer, or using one of the (very good) online LaTex solutions ( Overleaf or ShareLatex). Here are some files to help you start playing with LaTex and writing your report. Download and extract the files from start_latex.tar. You will now have the following files:
• start.tex
• listings-python-options.sty
• start.bib
• wowTest.py
• sinecurve.py
• sine.pdf
• Makefile
The Makefile contains commands required for generating a pdf file out of the latex source, and other files that are required. On a Unix/Linux that has Latex you can just run
$make The file listings-python-options.sty is a latex style file that tweaks the parameters of the listings latex package to makes it such that Python code is displayed nicely. ## Submission Submit your report via Canvas. Python code can be displayed in your report if it is short, and helps understand what you have done. The latex sample document shows how to display Python code in a latex document. Submit the Python code that was used to generate the results as a file called assignment1.py. Typing $ python assignment1.py
should generate all the tables/plots used in your report.
Here is what the grade sheet will look like for this assignment. A few general guidelines for this and future assignments in the course:
• Your answers should be concise and to the point. We will take off points if that is not the case.
• Always provide a description of the method you used to produce a given result in sufficient detail such that the reader can reproduce your results on the basis of the description. You can use a few lines of python code or pseudo-code.
Grading sheet for assignment 1
Part 1: 30 points.
Part 2: 30 points.
Part 3: 30 points.
(15 points): Histograms of informative/non-informative features.
(15 points): Discussion of the plots.
Report structure, grammar and spelling: 10 points
report are well-written and formatted. Correct grammar, spelling, and punctuation.
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2019-06-16 17:01:29
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https://www.physicsforums.com/threads/a-very-simple-question-about-units.9444/
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A Very Simple Question About Units
Staff Emeritus
Gold Member
Main Question or Discussion Point
Hello,
I'm new here. I was wondering if anyone could help me out with some confusion regarding units. My landlady (like most people) wants to save money on power consumption, so she has me installing Compact Fluorescent Lightbulbs(CFLs), which are supposed to save energy. On the package, it said that a 15W CFL is just as good as a 60W incandescent bulb, because they both give off ~800lumens. I couldn't understand that. The lumen is one of the three units whose meanings I don't remember:
candela
lumen
lux
Also,I found out that they measure the physical quantities luminous flux, illuminance, and intesity, but I don't know which measures which, and what the defn's of these are. Can anyone help me out?
Related Other Physics Topics News on Phys.org
jcsd
Gold Member
The candela is the SI unit of luminous intensity and is defined as the luminous intensity in a given direction of a source that emits monochromatic radiation of 540 x 1012 Hz and has a radiant intensity (in the given direction) of 1/683 watt/steradian.
The lumen is the SI unit of luminous flux and is defined as the intensity (in candelas) of a point source in a cone of 1 steradian. Luminous flux can be found (for monochromatic radiation)by:
$$\Phi_v = K_m\Phi_eV(\lambda)$$
Where Φe is the radiant flux Km is a constantr relaing the two units of flux and V(λ) is the luminous efficency for λ
The lux is the SI unit of illumination and is defined as one lumen over an area of 1 m2
It can be found on a surface of area dS from the following equation:
$$\Phi_v = \int E_v dS$$
Staff Emeritus
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2020-01-18 14:55:57
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https://letusfigurethisout.wordpress.com/page/2/
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# Christian “Trump Cards” – Part 2
Recently I began compiling something of a list of, what appear to be, commonly appealed to “trump cards” by certain Christians during attempts at rational dialogue. While I certainly don’t intend to implicate all Christians (or even only Christians), I have noticed a regular occurrence of these “moves” in a multitude of discussions. In Part 1 I briefly discussed the appeal to faith. From my own experience discussions seem to either begin or end with this “trump card”. For part 2 I would like to analyze what seems to be a background assumption of many Christians. This “trump card” isn’t necessarily appealed to directly; it most often operates behind the scenes, but has noticeable effects. Here it is:
A: If one does not believe that Christianity is true, then that person hates God and rejects “His” gift of salvation.
The difficulty with this background assumption is that it goes to the heart of the Christian message. Humans are the creation of God, but are separated from “Him” because of sin. Ultimately, this means that humans are headed for one of two fates: eternal heaven or eternal hell. Those who repent and accept Christ are granted eternal heaven, while those who do not get eternal hell.
Now, eternal hell (which is generally taken as never-ending punishment) is rather harsh. So, to alleviate the uncomfortable dissonance, it is my opinion that many Christians are driven to hold A. It is much easier to believe that unrepentant God-haters who despise “His” sacrifice deserve never-ending punishment than it is to accept that honest seekers and skeptics could somehow “miss the boat” and end up in eternal torment.
The problem, of course, is that A is a completely unwarranted assumption. More than that, it is unfalsifiable in the sense that all counter-examples can be dismissed. Since no one can expose his or her first person perspective to direct analysis, we must always go off of what people report about themselves. The person who accepts A, however, can simply maintain that the non-believer is deceiving himself or herself by suppressing the truth (more on that at a later time).
Nevertheless, let’s take a look at A itself and see if it makes better sense to adopt its negation. Note that A is a conditional statement. Let stand for the simple proposition A person P believes that Christianity is true. Let H represent the proposition P hates God and rejects “His” gift of salvation. Then the assumption A can be expressed as
$\lnot C\rightarrow H$
The negation of this is therefore
$\lnot(\lnot C\rightarrow H)\equiv \lnot C \wedge \lnot H$
$\lnot A$: Person P does not believe that Christianity is true, but does not hate God or reject “His” gift of salvation.
Let’s see why $\lnot A$ is more likely true than $A$. If we look again at the component propositions $\lnot C$ and $H$ we run into an immediate problem. Notice that $\lnot C$ is a proposition that refers to the noetic status of a person. It makes a claim about what a person believes. In other words, it reports that some person takes the Christian system of belief to be false or not to correspond to reality. The proposition $H$, by contrast, refers to a directed feeling or emotion of a person. So, the assumption $A$ claims that a certain state of unbelief with respect to some propositions implies an associated emotion with respect to the content of those propositions. This is a very queer claim indeed. Generally, the status of one’s belief has no connection to how one feels about the content of the belief. For instance, I don’t believe in fire-breathing dragons. In other words, I take it that they do not exist. But this says nothing of how I feel about fire-breathing dragons. In fact, I think fire-breathing dragons, while scary, are pretty awesome. So, if this connection fails to hold in general, why should we believe it holds in the specific case of Christianity? It is true that some people hate the idea of the Christian God, but there is no evidence to suggest that every non-believer does.
The underlying issue here is that many Christians conflate two types of “rejection”. Let’s call the first type of rejection relational rejection and the second type propositional rejection. Relational rejection involves rejecting a person or something a person is offering. It involves a negative feeling toward the person or thing offered by the person. For instance, if a boy asks out a girl and she says “no”, then she has relationally rejected the poor boy. She is saying that she does not like the idea of having a particular type of relationship with him. Propositional rejection, by contrast, is simply to not accept a proposition as being true. For example, suppose a girl is asked whether she thinks that some boy will ask her out. Suppose she says “no”. Then in this case she is engaging in propositional rejection. That is, she is rejecting the proposition that some boy is going to ask her out. Notice that this rejection says absolutely nothing about whether she wants the boy to ask her or not.
The confusion arises from a failure to see the distinction between the proposition that one thinks is either true or false, and the content of the proposition that one may or may not have a feeling about. For the Christian, the content of the belief is so central and personal that disbelief is automatically taken to be personal. But there is no reason to think that disbelief is always (or even often) of a personal nature. Thus, the Christian who holds $A$ is going to have to summon some really compelling evidence. Next time I’ll address one such argument claiming that truth is a person and hence rejection of the “truth” amounts to rejecting the person.
# Some Common Christian “Trump Cards” – Part 1
Talking with Christians (and religious people in general) is a mixed bag. Sometimes you get awesome intelligent people along with a really great discussion (even if you end up disagreeing). Most of the time, however, the discussion ends up in frustration. Having quite a bit of experience, I have put together a list of some of the most commonly used discussion killing “trump cards” played by many Christians. I’d like to address each one in a separate post. In analyzing each, I hope this will be useful to both skeptics and believers to promote more fruitful dialogue. But before I get to the list, I’d like to give an explanation as to the root of these “trump cards”.
Controlling Assumptions
While there is no doubt that humans are rational beings, there is also no doubt that humans are emotional beings. Unfortunately, at the conscious level, we are much more influenced by the latter than the former. Emotions and other psychological effects are so powerful, that humans have to work quite hard to be rational in the midst of them. This is not completely bad, since our drive, hope, and motivation are essential aspects our success. The problem is striking a correct balance, which is obviously fairly difficult.
Here lies one of the difficulties with Christianity. Not only is Christianity a system of belief, it is a system of belief saturated in emotion. It is a system designed to address the core of human longings: purpose, meaning, forgiveness, justice, belonging, love, hope, peace, etc. It provides a complete mental framework from which to operate. This, of course, is not a bad thing, but the practical effects of adopting it can make it rather difficult to maintain objectivity. One is easily blinded to the cold matters of truth when it just feels right.
This is an example of something I call a controlling assumption.
Definition [Controlling Assumption]A controlling assumption is an assumption that once adopted sets a mental framework that interprets all data to be consistent with the assumption, even data contrary to the assumption itself.
One might describe a controlling assumption as a self-preserving assumption. It is intimately related to the idea of confirmation bias. For example, consider the famous psychological experiment where several researchers checked themselves into a mental hospital. Once admitted, they acted completely normal. One would hope that the doctors of the hospital would be able to recognize that these “impostors” were completely sane and free of mental illness. In other words, the healthy should be distinguishable from the sick. The problem, however, is that the doctors were under the influence of a controlling assumption, namely People who enter the hospital as patients have a mental illness. Because of this, the doctors interpreted the researcher’s normal behavior as symptomatic of their supposed “neuroses/psychoses”. Even as the researchers documented these things with meticulous notes, the doctors recorded in their charts that “Patients engage in note taking behavior” as if it was pathological. Ironically, those who were actually mentally ill caught on to the researchers almost immediately.
Although not always the case, the Christian belief structure can operate much like a controlling assumption, and part of the self-preserving nature of the belief structure manifests itself in various “trump cards” when being challenged. I address the first of these below.
“Trump Card” 1 -[You just have to have faith]
Generally, the very first response to any rational challenge is an appeal to faith. How “faith” is being used, however, is not generally very clear. When pressed for a definition, most respond by quoting Hebrew 11:1, which says, “Now faith is the assurance of things hoped for, the conviction of things not seen.” (NAS) Using this as a definition, what the Christian is saying is that one just needs to have assurance. My initial thought is, “Oh, is that all I need?”. Of course, even if Christianity is something I hope for, the problem is that I don’t have assurance. This is what I am seeking, but not finding. It strikes me a bit like saying to an addict who is asking how to stop, “Well, you just need to stop.” That won’t be received well, because that is the very problem that needs addressing.
The go to response from here is generally to deny that faith has anything to do with the intellect, but is a matter of the heart. Again, it isn’t at all clear what this is supposed to mean, nor is it clear where such an idea is expressed in the Bible. In fact, the Greek word for “heart” in the New Testament is καρδια or kardia, which was taken to include the whole self, including the faculty and seat of the intelligence. Thus, to say that faith is a matter of the heart and not the mind is an artificial and incorrect distinction even measured against the Bible.
Finally, telling someone that faith is required only pushes the problem back a step. Why is faith required? How does one know this? And what guides where I place my faith? After all, many religions appeal to the very same requirement. So, which does one choose? I think the issue clearly reveals that faith is not an epistemological tool that yields knowledge. This can be more rigorously demonstrated as follows.
(1) Suppose that faith provides a means of knowing something.
(2) The Christian has faith and so knows Christianity to be true.
(3) Therefore, from the definition of “know” it follows that Christianity is true.
(4) The Mormon has the same sort of faith and so knows Mormonism to be true.
(5) Therefore, Mormonism is true.
(6) Christianity and Mormonism are incompatible systems of belief.
(7) Therefore, either Christianity or Mormonism (or both) is false.
(8) If Christianity is false, then we get a contradiction with (3).
(9) Thus, Christianity must be true and Mormonism false.
(10) If Mormonism is false, then we get a contradiction with (5).
(11) Thus, Mormonism is also true, which contradicts (7).
(12) Therefore, since our initial assumption leads to a contradiction, it must be the case that (1) is false.
Of course, one could deny that Christians and Mormons have the same faith, but then one would have to wonder how we could possibly distinguish between “real” faith and “fake” faith. One would then have to appeal to something other than faith anyway.
# Chuck Missler and the Existence of Infinity
Along with my (slow) endeavor of exploring and critiquing the ideas undergirding intelligent design, I want to resume a project I started a while back that involved addressing the claims of Chuck Missler. As I have previously mentioned, Chuck Missler is a well educated man. That being said, I also get the strong impression that Missler pretends to know a lot more than he really does. What annoys me the most is that he seems to present himself as an expert in, well, just about everything.
In his Bible studies, Missler lectures his flock on everything from cosmology to quantum mechanics to information theory and beyond. Missler has everything figured out it seems, and, while he might deny it, presents himself as having just about everything figured out. It’s shocking that he hasn’t been awarded a Nobel prize. Listening to his talks, one cannot help but be impressed by the depth and breadth of his knowledge. However, once one gets over the dazzle of Missler’s apparent expertise in everything, one begins to pick up on very questionable claims and ideas. Most of the time Missler alludes to very deep ideas, but glosses over them to spare his audience the details. OR, maybe Missler really doesn’t know or understand the details. This is my suspicion, which is motivated by several cracks in his intellectual veneer amounting to questionable claims on his part.
The first issue I’d like to address regards the size of the universe.
Missler is fond of proclaiming there are two central mathematical concepts that we don’t find in nature: Infinity and randomness. I’ll save randomness for another post.
Does Infinity Exist?
Certainly infinity exists conceptually, but the question is whether it describes anything real. In particular, Missler focuses on size. This can go in two directions: (1) small scale infinity, and (2) large scale infinity.
Small Scale ∞
Most people are familiar with number lines.
For our purposes we can focus on the interval $[0,1]$. We can cut this interval in half and get $[0,\frac{1}{2}]$. This can be cut in half again to get $[0,\frac{1}{4}]$. In fact, we could carry out this cutting procedure indefinitely, always obtaining a new interval $[0, \frac{1}{2^{n}}]$. This means that we can make the interval as small as we like. Put differently, we can cut the interval infinitely many times in the sense that there will never be a limit to the number of times we can cut the interval in half.
Now suppose that our interval $[0,1]$ models a length in space-time, say an inch. If the correspondence were true, then it would follow that space could be indefinitely halved. According to Missler, this is actually not true. It turns out that there is a limit to smallness in space-time. In other words, there is a smallest distance. This distance is known as the Planck length, which is defined by
$\ell_{p} = \sqrt{\frac{\hbar G}{c^{3}}}$
where $\hbar$ is Planck’s constant, $G$ is the gravitational constant, and $c$ is the speed of light. This length is exceedingly small, a mere $1.6\times 10^{-35}$ meters (approximately). Missler is fond of saying that if at any point you divide a distance into lengths smaller than $\ell_{p}$, then you lose locality, the thing you are cutting is suddenly everywhere all at once. What he concludes is that our reality is actually a “digital simulation”, terminology he uses purposely to insinuate that our world is created by God. Such loaded language seems to be characteristic of Missler.
So, what is the nature of this mysterious length? Where does it come from and how do we know it is the smallest length?
The Planck length has profound relevance in quantum gravity. Put differently, it is at this absurdly tiny scale that quantum effects become relevant and the question regards how gravity behaves or should be understood at this scale. Both general relativity and quantum field theory must be taken into account. The definition of the Planck length now makes sense since the speed of light $c$ is the natural unit that relates time and space, $G$ is the constant of gravity, and $\hbar$ is the constant of quantum mechanics. So the Planck scale defines the meeting point of gravity, quantum mechanics, time and space.
Theoretically, it is considered problematic to think of time and space as continuous because we don’t appear to be able to meaningfully discuss distances smaller than the Planck length. Unfortunately, there is no proven physical significance to the Planck length because current technology is incapable of probing this scale. Nevertheless, current attempts to unify gravity and QM, such as String Theory and Loop Quantum Gravity, yield a minimal length, which is on the order of the Planck distance. This arises when quantum fluctuations of the gravitational field are taken into account. In the theory, distances smaller than $\ell_{p}$ are physically meaningless. Two “points” at distances smaller than $\ell_{p}$ cannot be differentiated. This seems to suggest that space-time may have a discrete or “foamy” nature rather than a continuous one. Unfortunately for Missler, however, we just don’t know at this point. Thus, the declarative nature of his claims are hasty.
Nevertheless, even if this turns out to be the case, the question becomes: what follows from that? As mentioned above, Missler is fond of saying that the universe is a “digital simulation”. Certainly the term “digital” would be apropos, but his use of “simulation” seems loaded and dubious. “Simulation” suggests that this world isn’t the “real” world. It suggests that our world merely imitates some meta-world. Of course, Missler purposely uses the term as a way to smuggle in a simulator. That simulator is God, and the meta-world is the spiritual world.
Large Scale ∞
The more questionable of Missler’s claims regards the size of the universe. He brazenly declares that we have discovered the universe to be finite. This is just flat out false, and such carelessness makes me question his credibility. It is likely that he is simply confusing the observable universe with the universe proper. There is no doubt that the observable universe is finite. It is estimated to have a radius of 46 billion light years and due to expansion grows ever larger. However, this does not necessarily imply that the universe proper is finite in size. Sir Roger Penrose, one of the most respected mathematical physicists in the world, says, “it may well be that the universe is spatially infinite, like the FLRW models with $K = 0$ or $K<0$.” (see The Road To Reality, p. 731)
Note: FLRW models refers to Friedmann–Lemaître–Robertson–Walker models and $K$ is a density parameter governing the curvature of the universe.
Even a quick search on Wikipedia reveals that, “The size of the Universe is unknown; it may be infinite. The region visible from Earth (the observable universe) is a sphere with a radius of about 46 billion light years, based on where the expansion of space has taken the most distant objects observed.”
In fact, the possibility of an infinite universe is the stuff of some multiverse models. Max Tegmark, a mathematical physicist at MIT puts it this way:
If space is infinite and the distribution of matter is sufficiently uniform on large scales, then even the most unlikely events must take place somewhere. In particular, there are infinitely many other inhabited planets, including not just one but infinitely many with people with the same appearance, name and memories as you. Indeed, there are infinitely many other regions the size of our observable universe, where every possible cosmic history is played out. This is the Level I multiverse.
Tegmark goes on
Although the implications may seem crazy and counter-intuitive, this spatially infinite cosmological model is in fact the simplest and most popular one on the market today. It is part of the cosmological concordance model, which agrees with all current observational evidence and is used as the basis for most calculations and simulations presented at cosmology conferences. In contrast, alternatives such as a fractal universe, a closed universe and a multiply connected universe have been seriously challenged by observations. Yet the Level I multiverse idea has been controversial (indeed, an assertion along these lines was one of the heresies for which the Vatican had Giordano Bruno burned at the stake in 1600†), so let us review the status of the two assumptions (infinite space and “sufficiently uniform” distribution). How large is space? Observationally, the lower bound has grown dramatically (Figure 2) with no indication of an upper bound. We all accept the existence of things that we cannot see but could see if we moved or waited, like ships beyond the horizon. Objects beyond cosmic horizon have similar status, since the observable universe grows by a light-year every year as light from further away has time to reach us‡. Since we are all taught about simple Euclidean space in school, it can therefore be difficult to imagine how space could not be infinite — for what would lie beyond the sign saying“SPACE ENDS HERE — MIND THE GAP”? Yet Einstein’s theory of gravity allows space to be finite by being differently connected than Euclidean space, say with the topology of a four-dimensional sphere or a doughnut so that traveling far in one direction could bring you back from the opposite direction. The cosmic microwave background allows sensitive tests of such finite models, but has so far produced no support for them — flat infinite models fit the data fine and strong limits have been placed on both spatial curvature and multiply connected topologies. In addition, a spatially infinite universe is a generic prediction of the cosmological theory of inflation (Garriga & Vilenkin 2001b). The striking successes of inflation listed below therefore lend further support to the idea that space is after all simple and infinite just as we learned in school.
So, it seems that unless Missler knows something all other physicists don’t, he is being much too hasty and cherry picking possibilities to support what he wants to be true.
# Intelligent Design 1
In my quest to understand the true nature of reality, the ongoing dispute between theism and atheism is a regular subject of interest. Perhaps most noteworthy at present is the ‘battle’ over intelligent design. Is it a science or is it religion in disguise? Some deride it as a waste of time and a joke, others think it something to be addressed, but mistaken, and still others insist that it provides compelling evidence in favor of theism. I am of the persuasion that intelligent design is scientific in nature and worthy of consideration. As to the correctness of the endeavor, I’m skeptical, but willing to give it a fair hearing. This is what I would like to do in a series of blogs aimed at both understanding and critiquing intelligent design at a fundamental level. Since my background is in mathematics, I will primarily focus on this aspect of it as well as its philosophical underpinnings. My focus will be primarily on the work of Bill Dembski, a mathematician and philosopher championing the fundamental notions of intelligent design. Input from other disciplines, however, is most welcome.
To get things started, it will be important to have a commonly understood vocabulary. Since design is the backbone of intelligent design, understanding this term is a logical launching point. So, the question is: What is design? In their paper, Wesley Elsberry and Jeffrey Shallit (see here) criticize Dembski, in part, on the grounds that he gives no positive notion of what design is. In his book, The Design Inference, Dembski gives a negative definition of design as the complement of regularity and chance. In No Free Lunch, he gives a more process-oriented account of design:
(1) A designer conceives a purpose.
(2) To accomplish that purpose, the designer forms a plan.
(3) To execute the plan, the designer specifies building materials and assembly instructions.
(4) Finally, the designer or some surrogate applies the assembly instructions to the building materials.
Also, although I have not read Dembski’s book Intelligent Design, I cannot find (with the help of the index) any place where he explicitly defines design. So, this is task number one; namely, to supply an adequate definition for design. Anyone familiar with this topic is encouraged to contribute. For my part, here is something of a first draft:
Definition – Design: A system that is arranged by an intensional agent.
Let’s figure this out.
# Neat Post Related to Formal Systems
Terence Tao is a brilliant mathematician whom I follow. He posts about a lot of very interesting mathematical topics. I found this neat post related to formals systems. The beginning of the article should hopefully be at least somewhat familiar based on what I have discussed, but it does get heavy pretty quickly. Nevertheless, check it out and enjoy!
# The Thesis of Modal Realism
A few post ago I was talking about formal systems. As an example I talked a little about Hofstadter’s example of the MIU system and promised to talk about decision procedures. I would hate to be a liar and so I do plan on returning to this matter. However, it seems appropriate to me to take a diversion into the philosophical foundations of Tegmark’s view before delving into the more mathematical aspects. For this reason, and in keeping with the announcement of my previous post, I am going to interact a bit with David Lewis’ work On the Plurality of Worlds.
So, what is Modal Realism? From what I gather, the name was actually coined by Lewis, though he expresses certain regrets for calling it this based on some ambiguity in the term ‘realism’ and what it has come to mean in academia. As one who is less familiar with the larger body of literature on realism, I find the name perfectly appropriate and satisfying.
The term ‘modal’ comes from modal logic which analyses the notions of necessity and possibility. G.E. Hughes and M.J. Cresswell summarize this discipline similarly:
Modal logic can be described briefly as the logic of necessity and possibility, of ‘must be’ and ‘may be’.
‘Realism’ as Lewis intends to use it deals with ontology. Put together, modal realism is
… the thesis that the world we are part of is but one of a plurality of worlds, and that we who inhabit this world are only a few out of all the inhabitants of all the worlds.
In other words, modal realism takes possible worlds semantics quite seriously as well as literally. Upon reading this description, one may be tempted to think of modal realism as a multi-verse theory and indeed it is (we’ll see that it corresponds to Tegmark’s Level IV multiverse), but not in the sense often portrayed in popular literature. The worlds of modal realism are not some distant universes located some vast, albeit finite, distance away. Nor are they located in some alternate dimension of this reality. As Lewis puts it,
They are isolated; there are no spatiotemporal relations at all between things that belong to different worlds. Nor does anything that happens at one world cause anything to happen at another.
One should liken this to the way in which the vector space $\mathbb{R}$ is isolated from the finite field $\mathbb{Z}_{7}$, to sneak in some mathematics.
Lewis goes on to say,
The difference between this and the other worlds is not a categorial difference. Nor does this world differ from the others in its manner of existing. I do not have the slightest idea what a difference in manner of existing is supposed to be. [emphasis mine]
Thank you!!! I expressed this very sentiment in A Philosophical Interlude. It always feels amazing when one’s own thoughts are legitimized by world renown experts. Okay, enough tooting my own horn. The idea, here, is that some things exist, say, on Earth, while other things exist elsewhere and this is a difference in location of existing things (not a difference in any mode of existence). Similarly, though somewhat more controversial, some things exist in the past, others now and still others in the future. This is a temporal difference (we will deal with the nature of time in another post). To extend this further, it seems reasonable that the same could be said for worlds. Some things exist in this world and other things exist in other worlds and this is still a difference between existing things as opposed to a difference in any mode of existing.
For those not familiar with the language of possible worlds, I should hasten to point out that a world need not be a synonym for a universe (though it might be). Rather, a world is a complete state of affairs, an entire reality. Thus, in our world, if there is a God who created the universe, then clearly our world is more than just our universe. Alvin Plantinga in his book The Nature of Necessity defines a possible world to be a maximal state of affairs and a maximal state of affairs $S$ is one such that for every state of affairs $S'$, $S$ includes $S'$ or precludes $S'$. Note the similarity of this definition with that of completeness for a mathematical system:
A system is complete if, for every statement $p$, we can find a proof of $p$, or a proof of not-$p$.
Here “proof” refers to a demonstration within the system in question using the axioms and the rules of inference. In other words, a “proof”, here, of a statement $p$ within a system $S$ “consists of a sequence of statements, each of which is either an axiom or a logical consequence of certain preceding statements in the list, such that the last statement in the list is $p$.”
At this point, we needn’t get bogged down in technicalities. The basic idea is that modality is best understood by actually taking possible worlds semantics as more than just semantics. According to modal realism, all possible worlds are actual worlds, where actuality is taken to be indexical.
The task I have determined to undertake here is a formidable one. Despite this, I am excited about adding in some key works of David Lewis. From Amazon:
David Lewis (1941- 2001) was Professor of Philosophy at Princeton University. His publications include Convention (reissued by Blackwell 2002), Counterfactuals (reissued by Blackwell 2000), Parts of Classes (1991), and of numerous articles in metaphysics and other areas.
The books I plan on reading are On the Plurality of Worlds and Counterfactuals. Here is the description of former:
This book is a defense of modal realism; the thesis that our world is but one of a plurality of worlds, and that the individuals that inhabit our world are only a few out of all the inhabitants of all the worlds. Lewis argues that the philosophical utility of modal realism is a good reason for believing that it is true.
After putting forward the type of modal realism he favors, Lewis answers numerous objections that have been raised against it. These include an insistence that everything must be actual; paradoxes akin to those that confront naive set theory; arguments that modal realism leads to inductive skepticism, or to disregard for prudence and morality; and finally, sheer incredulity at a theory that disagrees so badly with common opinion. Lewis grants the weight of the last objection, but takes it to be outweighed by the benefits to systematic theory that acceptance of modal realism brings. He asks whether these same benefits might be gained more cheaply if we replace his many worlds by many merely ‘abstract’ representations; but concludes that all versions of this ‘ersatz modal realism’ are in serious trouble. In the final chapter, Lewis distinguishes various questions about trans-world identity, and argues that his ‘method of counterparts’ is preferable to alternative approaches.
Here is a description of the latter:
Counterfactuals is David Lewis’s forceful presentation of and sustained argument for a particular view about propositions which express contrary-to-fact conditionals, including his famous defense of realism about possible worlds. Since its original publication in 1973, it has become a classic of contemporary philosophy, and is essential reading for anyone interested in the logic and metaphysics of counterfactuals.
If it is not already obvious, the purpose of including these works is that they (especially the former) provide the philosophical groundwork underpinning Tegmark’s ideas. As I mentioned before, the MUH is a form of modal realism and could be described as a mathematized version of Lewis’s philosophy.
Since I am already addressing Tegmark’s ideas along with Hofstadter’s I hope to not become bogged down by adding in more, but the relevance of Lewis’s work in this matter seems justification enough to shoulder the burden. I hope everyone who reads this is as excited as I am to engage these ideas!
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2017-08-20 17:08:17
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http://www.demo.cs.brandeis.edu/coev_CA/
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# Coevolving the "Ideal" Trainer
## Description of the Problem
#### One-dimensional Cellular Automata
A one-dimensional cellular automaton (CA) is a linear wrap-around array composed of N cells in which each cell can take one out of k possible states. A rule is defined for each cell in order to update its state. This rule determines the next state of a cell given its current state and the state of cells in a predefined neighborhood. For the model discussed in this proposal, this neighborhood is composed of cells whose distance is at most r from the central cell. This operation is performed synchronously for all the cells in the CA. For the work presented in this proposal, the state of cells is binary (k = 2), N = 149 and r = 3. This means that the size of the rule space is 2^{2^{2 * r + 1}} = 2^{128}.
Cellular automata have been studied widely as they represent one of the simplest systems in which complex emergent behaviors can be observed. This model is very attractive as a means to study complex systems in nature. Indeed, the evolution of such systems is ruled by simple, locally-interacting components which result in the emergence of global, coordinated activity.
#### The Majority Function
The task consists in discovering a rule for the one-dimensional CA which implements the majority function as accurately as possible. This is a density classification task, for which one wants the state of the cells of the CA to relax to all 0's or 1's depending on the density of the initial configuration (IC) of the CA, within a maximum of M time steps. Following [Mitchell et al., 1994], $\rho_c$ denotes the threshold for the classification task (here, $\rho_c = 1/2$), $\rho$ denotes the density of 1's in a configuration and $\rho_0$ denotes the density of 1's in the initial configuration. Figure 1 presents two examples of the space-time evolution of a CA with $\rho_0 < \rho_c$ on the left and $\rho_0 > \rho_c$ on the right. The initial configuration is at the top of the diagram and the evolution in time of the different configurations is represented downward.
Figure 1: Two space-time diagrams describing the evolution of CA states.
The task $\rho_c = 1/2$ is known to be difficult. In particular, it has been proven that no rule exists that will result in the CA relaxing to the correct state for all possible ICs [Land & Belew, 1995]. Indeed, the density is a global property of the initial configuration while individual cells of the CA have access to local information only. Discovering a rule that will display the appropriate computation by the CA with the highest accuracy is a challenge, and the upper limit for this accuracy is still unknown. Table 1 describes the performance for that task for different published rules and different values of N along with a new best rule discovered using the coevolutionary approach presented in this document.
table 1: Performance of different published CA rules and the new best rule discovered by coevolution for the majority function.
N 149 599 999 Coevolution 0.863 +/- 0.001 0.822 +/- 0.001 0.804 +/- 0.001 Das rule 0.823 +/- 0.001 0.778 +/- 0.001 0.764 +/- 0.001 ABK rule 0.824 +/- 0.001 0.764 +/- 0.001 0.730 +/- 0.001 GKL rule 0.815 +/- 0.001 0.773 +/- 0.001 0.759 +/- 0.001
The Gacs-Kurdyumov-Levin (GKL) rule was designed in 1978 for a different goal than the $\rho_c = 1/2$ task [Mitchell et al., 1994]. However, for a while it provided the best known performance. [Mitchell et al., 1994] and [Das et al., 1994] used Genetic Algorithms (GAs) to explore the space of rules. This work resulted in an analysis of some of the complex behaviors exhibited by CAs using particles''. The GKL and Das rules are human-written while the Andre-Bennett-Koza (ABK) rule has been discovered using the Genetic Programming paradigm [Andre et al., 1996].
## Learning in a Fixed Environment: Evolution
• #### Canonical model of evolution
The traditional model for evolutionary approaches to problem solving is to design a representation for solutions and a fitness function. The absolute performance of individuals is evaluated with respect to that fitness function and search operators are used to explore the state space.
As an example, we implemented such an evolutionary approach for the discovery of CA rules for the $\rho_c = 1/2$ task. This implementation is similar to the one described in [Mitchell et al., 1994]. Each rule is coded on a binary string of length $2^{2 * r + 1} = 128$. One-point crossover is used with a 2%$bit mutation probability. The population size is n_R = 200 for rules and n_{IC} = 200 for ICs. The population of ICs is composed of binary strings of length N = 149. This population is fixed and it is initialized according to a uniform distribution over$\rho_0 \in [0.0, 1.0]$. The population of rules is also initialized according to a uniform distribution over$[0.0, 1.0]$for the density. A natural definition for the fitness of rules is the number of Initial Configurations (ICs) for which the CA relaxes to the correct state: where: Figure 2 describes the evolution of the distribution of the density of rules over time. A common behavior resulting from such an evolutionary setup is that the entire population focuses quickly to a small domain of the search space in the neighborhood of the current best individual. Eventually, some progress is observed over time. The variance for the final performance over several runs is usually large. Figure 2: Evolution of the distribution of the density of rules over time. • #### Maintaining diversity by introducing resource sharing Several techniques have been designed to improve search in a fixed landscape. Usually those techniques maintain diversity in the population in order to avoid premature convergence. [Mahfoud, 1995] presents different niching techniques that are useful in particular for the exploration of multi-modal landscape. A technique that we successfully exploited in the past to maintain diversity is called resource sharing [Juillé & Pollack, 1996]. Resource sharing implements a coverage-based heuristic by giving a higher payoff to problems that few individuals can solve. One way to implement this technique for the evolution of CA rules is to define the fitness of rules as follows: where: In this definition, the weight of an IC corresponds to the payoff it returns if a rule covers it. If few rules cover an IC, this weight will be much larger than if a lot of rules cover that same IC. Figure 3 shows the evolution of the distribution of rules density over several generations. It can be observed that diversity is maintained for about 500 generations during which different alternatives are explored simultaneously by the population of rules. Then, a solution is discovered that moves the search in the area where density is close to 0.5. Afterward, several niches might still be explored, however the coding used to construct the figures doesn't allow the representation of those different niches since all the rules have a similar density. Usually, this technique results in better performance on average. However, it takes also more time to converge. This is a trade-off between speed and quality of solutions. Figure 3: Evolution of the distribution of the density of rules over time. ## Learning in an Adapting Environment: Coevolution The idea of using coevolution in search was introduced by [Hillis, 1992]. In his work, a population of parasites coevolves with a population of sorters. The goal of parasites is to exploit weaknesses of sorters by discovering input vectors that sorters cannot solve correctly while sorters evolve strategies to become perfect sorting networks. In coevolution, individuals are evaluated with respect to other individuals instead of a fixed environment (or landscape). As a result, agents adapt in response to other agents' behavior. For the work presented in this document, I will define coevolutionary learning as a learning procedure involving the coevolution of two populations: a population of learners and a population of problems. Moreover, the fitness of an individual in a population is defined only with respect to members of the other population. Two cases can be considered in such a framework, depending on whether the two populations benefit from each other or whether they have different interests (i.e., if they are in conflict). Those two modes of interaction are called cooperative and competitive respectively. In the following paragraphs, those modes of interaction are described experimentally in order to stress the different issues related to coevolutionary learning. • #### Cooperation between populations In this mode of interaction, improvement on one side results in positive feedback on the other side. As a result, there is a reinforcement of the relationship between the two populations. From a search point of view, this can be seen as an exploitative strategy. Agents are not encouraged to explore new areas of the search space but only to perform local search in order to further improve the strength of the relationship. Following a natural extension of the evolutionary case to the cooperative model, the fitness of rules and ICs can be defined as follows for the$\rho_c = 1/2$task: For our experiments, the same setup as the one described previously is used. The population size is n_R = 200 for rules and n_{IC} = 200 for ICs. The population of rules and ICs are initialized according to a uniform distribution over$[0.0, 1.0]$for the density. At each generation, the top 95% of each population reproduces to the next generation and the remaining 5% is the result of crossover between parents from the top 95% selected using a fitness proportionate rule. Figure 4 presents the evolution of the density of rules and ICs for one run using this cooperative model. Without any surprise, the population of rules and ICs quickly converge to a domain of the search space where ICs are easy for rules and rules consistently solve ICs. There is little exploration of the search space. Figure 4: Evolution of the distribution of the density of rules (left) and the density of ICs (right) over time. • #### Competition between populations In this mode of interaction, the two populations are in conflict. Improvement on one side results in negative feedback for the other population. Two outcomes are possible: • Unstable behavior: alternatively, agents in one population outperform members of the other population until those discover a strategy to defeat the first one and vice versa. This results in an unstable behavior where species appear then disappears and reappears again in a later generation. • a Stable state is reached: this happens if one population consistently outperforms the other. For instance, a very strong strategy is discovered by agents in one population. For the$\rho_c = 1/2$task, the fitness defined in the cooperative case can be modified as follows to implement the competitive model: Here, the goal of the population of rules is to discover rules that defeat ICs in the other population by allowing the CA to relax to the correct state, while the goal of ICs is to defeat the population of rules by discovering initial configurations that are difficult to classify. Figure 5 describes an example of a run using this definition of the fitness. In a first stage, the two populations exhibit a cyclic behavior. Rules with low density have an advantage because there are more ICs with low density. In response to this, ICs with high density have an evolutionary advantage and have more representatives in the population. In turn, rules with high density get an evolutionary advantage\ldots This unstable behavior exhibited by the two populations is an instance of the Red Queen effect [Cliff & Miller, 1995]: fitness landscapes are changing dynamically as a result of agents from each population adapting in response to the evolution of members of the other population. The performance of individuals is evaluated in a changing environment, making continuous progress difficult. A typical consequence is that agents have to learn again what they already knew in the past. In the context of evolutionary search, this means that domains of the state space that have already been explored in the past are searched again. Then, a stable state is reached: in this case, the population of rules adapts faster than the population of ICs, resulting in a population focusing only on rules with high density and eliminating all instances of low density rules (a finite population is considered). Then, low density ICs exploit those rules and overcome the entire population. Figure 5: Evolution of the distribution of the density of rules (left) and the density of ICs (right) over time. • #### Resource sharing and mediocre meta-stable states We have seen that by introducing resource sharing, it is possible to explore simultaneously different alternatives by maintaining several niches in the population. This section presents some experiments where resource sharing is introduced in the competitive mode of interaction presented in the previous section. The fitness of rules and ICs is then defined as follows: where: and: where: This framework allows the presence of multiple niches in both populations, each niche corresponding to a particular competence relevant to defeat some members of the other population. Figure 6 describes one run for this definition of the interaction between the two populations. It can be seen that the unstable behavior which was observed in the previous section doesn't occur anymore and that two species coexist in the population of rules: a species for low density rules and another one for high density rules. Those two species drive the evolution of ICs towards the domain of initial configurations that are most difficult to classify (i.e.,$\rho_0 = 1/2$). However, the two populations have entered a mediocre meta-stable state. This means that multiple average performance niches coexist in both populations in a stable manner. Figure 6: Evolution of the distribution of the density of rules (left) and the density of ICs (right) over time. ## Coevolving the "Ideal" Trainer The central idea of the coevolutionary learning approach consists in exposing learners to problems that are just beyond those they know how to solve. By maintaining this constant pressure towards slightly more difficult problems, a arms race among learners is induced such that learners that adapt better have an evolutionary advantage. The underlying heuristic implemented by this arms race is that {\em adaptability} is the driving force for improvement. In order to achieve this result, our system tries to satisfy the following two goals: • to provide an optimum'' gradient for search. This means that the training environment defined by the population of problems can determine reliably which learners are the most promising at each stage of the search process. This means that problems must be informative. If problems are too difficult or too easy, learners get little feedback and there is no gradient to determine in which direction the search should focus. • to allow continuous progress. The goal is to avoid the Red Queen effect by providing a training environment which continues to test learners about problems they solved in the past. In a sense, the training environment must also play the role of memory. The difficulty resides in the accurate implementation of those concepts in a search algorithm. ### Experimental Results It is believed, that ICs become more and more difficult to classify correctly as their density gets closer to the$\rho_c$threshold. Therefore, our idea is to construct a framework that adapts the distribution of the density for the population of ICs as CA-rules are getting better to solve the task. The following definition for the fitness of rules and ICs has been used to achieve this goal: where: and: where: This definition implements the competitive relationship with resource sharing. However, a new component, namely$E(R_i, \rho(IC_j))$, has been added in the definition of the ICs' fitness. The purpose of this new component is to give a low payoff to IC_j if ICs with density$\rho(IC_j)$return little information with respect to the rule R_i. We consider that information is collected if rule R_i does better or worse than random guessing over ICs with density$\rho(IC_j)$(that is, R_i covers strictly more or strictly less than 50% of those ICs). Following this idea, we defined E() as the complement of the entropy of the outcome between a rule and ICs with a given density: where: p is the probability that an IC with density$\rho(IC_j)$defeats the rule R_i and q = 1 - p. Figures 7 and 8 describe the evolution of the density of rules and ICs for two runs. It can be seen that, as rules improve, the density of ICs is distributed on two peaks on each side of$\rho = 1/2\$. In the case of figure 7, it is only after 1,300 generations that a significant improvement is observed for rules. It is only at that time that the population of ICs adapts dramatically in order to propose more challenging initial configurations.
Figure 7: Evolution of the distribution of the density of rules (left) and the density of ICs (right) over time.
Figure 8: Evolution of the distribution of the density of rules (left) and the density of ICs (right) over time.
Coevolutionary Learning: a Case Study (with J. B. Pollack), Proceedings of the Fifteenth International Conference on Machine Learning (ICML-98) , Madison, Wisconsin, USA, July 24-26, 1998, to appear.
Coevolving the Ideal'' Trainer: Application to the Discovery of Cellular Automata Rules (with J. B. Pollack), Proceedings of the Third Annual Genetic Programming Conference (GP-98) , Madison, Wisconsin, USA, July 22-25, 1998, to appear.
### References
[Andre et al., 1996] Andre, D., Bennet, F. H. & Koza, J. R. (1996). Evolution of intricate long-distance communication signals in cellular automata using genetic programming. In Artificial Life V: Proceedings of the Fifth International Workshop on the Synthesis and Simulation of Living Systems. Nara, Japan.
[Cliff & Miller, 1995] Cliff, D. & Miller, G. F. (1995). Tracking the red queen: Measurements of adaptive progres in co-evolutionary simulations. In the Proceedings of the Third European Conference on Artificial Life, pp. 200-218. Springer-Verlag. LNCS 929.
[Das et al., 1994] Das, R., Mitchell, M. & Crutchfield, J. P. (1994). A genetic algorithm discovers particle-based computation in cellular automata. In Parallel Problem Solving from Nature - PPSN III, pp. 344-353. Springer-Verlag. LNCS 866.
[Hillis, 1992] Hillis, W. D. (1992). Co-evolving parasites improves simulated evolution as an optimization procedure. In Langton, C. et al. (Eds), Artificial Life II, pp. 313-324. Addison Wesley.
[Juillé & Pollack, 1996] Juillé, H. & Pollack, J.B. (1996). Co-evolving intertwined spirals. In Proceedings of the Fifth Annual Conference on Evolutionary Programming, pp. 461-468. MIT Press.
[Land & Belew, 1995] Land, M. & Belew, R. K. (1995). No perfect two-state cellular automata for density classification exists. Physical Review Letters, 74(25):1548-1550.
[Mahfoud, 1995] Mahfoud, S. W. (1995). Niching methods for genetic algorithms. Technical report, University of Illinois at Urbana-Champaign. IlliGAL Report No. 95001.
[Mitchell et al., 1994] Mitchell, M., Crutchfield, J.P. & Hraber, P.T. (1994). Evolving cellular automata to perform computations: Mechanisms and impediments. Physica D, 75:361-391.
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2020-01-24 17:19:28
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https://ncatlab.org/nlab/show/module+over+a+monoid
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# nLab module over a monoid
Contents
### Context
#### Higher algebra
higher algebra
universal algebra
## Theorems
#### Monoid theory
monoid theory in algebra:
# Contents
## Idea
In a monoidal category, there is a notion of modules over monoid objects which generalizes the classical notion of modules over rings. This is a special case of module over a monad where the monad is taken to be $A \otimes -$, with $A$ some monoid object.
## Definition
Let $(\mathcal{V}, \otimes, I)$ be a monoidal category and $A$ a monoid object in $\mathcal{V}$, hence an object $A \in \mathcal{V}$ equipped with a multiplication morphism
$\cdot : A \otimes A \to A$
and a unit element
$e : I \to A$
satisfying the associativity law and the unit law.
###### Definition
A (left) module over $A$ in $(\mathcal{V}, \otimes, I)$ is
• an object $N \in \mathcal{V}$
• equipped with a morphism
$\lambda : A \otimes N \to N$
in $\mathcal{V}$
such that this satisfies the axioms of an action, in that the following are commuting diagrams in $\mathcal{V}$:
$\array{ A \otimes A \otimes N &\stackrel{id_A \otimes \lambda}{\to}& A \otimes N \\ \downarrow^{\mathrlap{\cdot \otimes id_n}} && \downarrow^{\mathrlap{\lambda}} \\ A \otimes N &\stackrel{\lambda}{\to}& N }$
and
$\array{ I \otimes N &&\stackrel{e \otimes id_N}{\to}&& A \otimes N \\ & \searrow && \swarrow_{\mathrlap{\lambda}} \\ && N } \,.$
## Examples
### Modules over monoids in abelian groups
Recall that a ring, in the classical sense, is a monoid object in the category Ab of abelian groups with monoidal structure given by the tensor product of abelian groups $\otimes$. Accordingly a module over $R$ is a module in $(Ab,\otimes)$ according to def. .
We unwind what this means in terms of abelian groups regarded as sets with extra structure:
###### Definition
A module $N$ over a ring $R$ is
1. an object $N \in$ Ab, hence an abelian group;
2. equipped with a morphism
$\alpha : R \otimes N \to N$
in Ab; hence a function of the underlying sets that sends elements
$(r,n) \mapsto r n \coloneqq \alpha(r,n)$
and which is a bilinear function in that it satisfies
$(r, n_1 + n_2) \mapsto r n_1 + r n_2$
and
$(r_1 + r_2, n) \mapsto r_1 n + r_2 n$
for all $r, r_1, r_2 \in R$ and $n,n_1, n_2 \in N$;
3. such that the diagram
$\array{ R \otimes R \otimes N &\stackrel{\cdot_R \otimes Id_N}{\to}& R \otimes N \\ {}^{\mathllap{Id_R \otimes \alpha}}\downarrow && \downarrow^{\mathrlap{\alpha}} \\ R \otimes N &\to& N }$
commutes in Ab, which means that for all elements as before we have
$(r_1 \cdot r_2) n = r_1 (r_2 n) \,.$
4. such that the diagram
$\array{ 1 \otimes N &&\stackrel{1 \otimes id_N}{\to}&& R \otimes N \\ & \searrow && \swarrow_{\mathrlap{\alpha}} \\ && N }$
commutes, which means that on elements as above
$1 \cdot n = n \,.$
###### Remark
The category of all modules over all commutative rings is Mod. It is a bifibration
$Mod \to CRing$
over CRing.
This fibration may be characterized intrinsically, which gives yet another way of defining $R$-modules. This we turn to below.
### $G$-sets
Simpler than the traditionally default notion of a module in $(Ab,\otimes)$, as above is that of a module in Set, equipped with its cartesian monoidal structure. (These days one may want to think of this as a notion of modules over F1.)
A monoid object in $(Set,\times)$ is just a monoid, for instance a discrete group $G$. A $G$-module in $(Set,\times)$ is simpy an action, say a group action.
###### Definition
For $S \in$ Set and $G$ a discrete group, a $G$-action of $G$ on $S$ is a function
$\lambda \colon G \times S \to S$
such that
1. the neutral element acts trivially
$\array{ * \times S &&\stackrel{\simeq}{\to}&& S \\ & {}_{(e,id_S)}\searrow && \nearrow_{\mathrlap{\lambda}} \\ && G \times S }$
2. the action property holds: for all $g_1, g_2 \in G$ and $s \in S$ we have $\lambda(g_1,\lambda(g_2, s)) = \lambda(g_1 \cdot g_2, s)$.
### Abelian groups with $G$-action as modules over the group ring
If a discrete group acts, as in def. , on the set underlying an abelian group and acts by linear maps (abelian group homomorphisms), then this action is equivalently a module over the group ring $\mathbb{Z}[G]$ as in def. .
###### Definition
For $G$ a discrete group, write $\mathbb{Z}[G] \in$ Ring for the ring
1. whose underlying abelian group is the free abelian group on the set underlying $G$;
2. whose multiplication is given on basis elements by the group operation in $G$.
###### Remark
For $G$ a finite group an element $r$r of $\mathbb{Z}[G]$ is for the form
$r = \sum_{g \in G} r_g g$
with $r_g \in \mathbb{Z}$. Addition is given by addition of the coefficients $r_g$ and multiplication is given by the formula
\begin{aligned} r \cdot \tilde r & = \sum_{g \in G} \sum_{\tilde g \in G} (r_g \tilde r_{\tilde g}) g \cdot \tilde g \\ & = \sum_{q \in G} \left( \sum_{g \tilde g = q} r_g \tilde r_{\tilde g} \right) q \end{aligned} \,.
###### Proposition
For $A \in$ Ab an abelian group with underlying set $U(A)$, $G$-actions $\lambda \colon G \times U(A) \to U(A)$ such that for each element $g \in G$ the function $\lambda(g,-) \colon U(A) \to U(A)$ is an abelian group homomorphism are equivalently $\mathbb{Z}[G]$-module structures on $A$.
###### Proof
Since the underlying abelian group of $\mathbb{Z}[G]$ is a free by definition, a bilinear map $\mathbb{Z}[G] \times A \to A$ is equivalently for each basis element $g \in G$ a linear map $A \to A$. Similarly the module property is determined on basis elements, where it reduces manifestly to the action property of $G$ on $U(A)$.
###### Remark
This reformulation of linear $G$-actions in terms of modules allows to treat $G$-actions in terms of homological algebra. See at Ext – Relation to group cohomology.
## References
The basic properties of categories of modules over monoid objects in symmetric monoidal categories are spelled out in sections 1.2 and 1.3 of
• Florian Marty, Des Ouverts Zariski et des Morphismes Lisses en Géométrie Relative, Ph.D. Thesis, 2009, web
A summary is in section 4.1 of
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2021-07-27 01:28:47
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https://www.semanticscholar.org/paper/Quantal-phase-factors-accompanying-adiabatic-Berry/2b56e819d19aa4822d6211fe37222b4822a0e3b6
|
# Quantal phase factors accompanying adiabatic changes
@article{Berry1984QuantalPF,
title={Quantal phase factors accompanying adiabatic changes},
author={Michael V. Berry},
journal={Proceedings of the Royal Society of London. A. Mathematical and Physical Sciences},
year={1984},
volume={392},
pages={45 - 57}
}
• M. Berry
• Published 8 March 1984
• Physics
• Proceedings of the Royal Society of London. A. Mathematical and Physical Sciences
A quantal system in an eigenstate, slowly transported round a circuit C by varying parameters R in its Hamiltonian Ĥ(R), will acquire a geometrical phase factor exp{iγ(C)} in addition to the familiar dynamical phase factor. An explicit general formula for γ(C) is derived in terms of the spectrum and eigenstates of Ĥ(R) over a surface spanning C. If C lies near a degeneracy of Ĥ, γ(C) takes a simple form which includes as a special case the sign change of eigenfunctions of real symmetric…
6,064 Citations
Quantum phase corrections from adiabatic iteration
• M. Berry
• Physics
Proceedings of the Royal Society of London. A. Mathematical and Physical Sciences
• 1987
The phase change γ acquired by a quantum state |ψ(t)> driven by a hamiltonian H0(t), which is taken slowly and smoothly round a cycle, is given by a sequence of approximants γ(k) obtained by a
Geometric amplitude factors in adiabatic quantum transitions
• M. Berry
• Physics, Mathematics
Proceedings of the Royal Society of London. Series A: Mathematical and Physical Sciences
• 1990
The exponentially small probability of transition between two quantum states, induced by the slow change over infinite time of an analytic hamiltonian Ĥ = H(δt). Ŝ ( where δ is a small adiabatic
Topological Phase Effects
Quantum eigenstates undergoing cyclic changes acquire a phase factor of geometric origin. This phase, known as the Berry phase, or the geometric phase, has found applications in a wide range of
A semiclassical connection is established between quantal and classical properties of a system whose Hamiltonian is slowly cycled by varying its parameters round a circuit. The quantal property is a
Classification of cyclic initial states and geometric phase for the spin-j system
• Physics
• 1994
Quantum states which evolve cyclically in their projective Hilbert space give rise to a geometric (or Aharonov-Anandan) phase. An aspect of primary interest is stable cyclic behaviour as realized
Nonadiabatic transitions for a decaying two-level system: geometrical and dynamical contributions
• Physics
• 2006
We study the Landau–Zener problem for a decaying two-level system described by a non-Hermitian Hamiltonian, depending analytically on time. Use of a super-adiabatic basis allows us to calculate the
Test of quantum adiabatic approximation via exactly-solvable dynamics of high-spin precession
• Physics
• 1995
By invoking the Rabi rotation technique, the Schrodinger wavefunction of a neutral particle with a magnetic moment and a high-spin is exactly solved in a harmonically-changing magnetic field. This
Geometric phase of a spin-1 2 particle coupled to a quantum vector operator
• Physics
• 2016
We calculate Berry’s phase when the driving field, to which a spin-1 2 is coupled adiabatically, rather than the familiar classical magnetic field, is a quantum vector operator, of noncommuting, in
Non-Abelian Berry phase for open quantum systems: Topological protection versus geometric dephasing
• Physics
Physical Review B
• 2019
We provide a detailed theoretical analysis of the adiabatic evolution of degenerate open quantum systems, where the dynamics is induced by time-dependent fluctuating loop paths in control parameter
Aharonov - Bohm geometric phases for rotated rotators
On a ring threaded by a flux line moves a charged quantum particle whose rotation is hindered by an angle-dependent potential. When the potential is rigidly rotated through , the particle (in the nth
## References
SHOWING 1-10 OF 20 REFERENCES
Diabolical points in the spectra of triangles
• Physics
Proceedings of the Royal Society of London. A. Mathematical and Physical Sciences
• 1984
‘Accidental’ degeneracies between energy levels Ej and Ej+1 of a real Hamiltonian can occur generically in a family of Hamiltonians labelled by at least two parameters X, Y,... Energy-level surfaces
Spin-orbit coupling and the intersection of potential energy surfaces in polyatomic molecules
• A. Stone
• Mathematics
Proceedings of the Royal Society of London. A. Mathematical and Physical Sciences
• 1976
Longuet-Higgins’ theorem, which shows that the existence of intersections between potential energy surfaces may be deduced from the behaviour of the wavefunction at points remote from the
The motion and structure of dislocations in wavefronts
• J. Nye
• Materials Science
Proceedings of the Royal Society of London. A. Mathematical and Physical Sciences
• 1981
Scattered scalar wavefields contain line singularities where the phase of the wave is indeterminate and the amplitude is zero. Unless the wave is monochromatic, these dislocation lines, which are
Dislocations in wave trains
• Physics
Proceedings of the Royal Society of London. A. Mathematical and Physical Sciences
• 1974
When an ultrasonic pulse, containing, say, ten quasi-sinusoidal oscillations, is reflected in air from a rough surface, it is observed experimentally that the scattered wave train contains
Significance of Electromagnetic Potentials in the Quantum Theory
• Physics
• 1959
In this paper, we discuss some interesting properties of the electromagnetic potentials in the quantum domain. We shall show that, contrary to the conclusions of classical mechanics, there exist
OBSERVABILITY OF THE SIGN CHANGE OF SPINORS UNDER 2$pi$ ROTATIONS.
• Physics, Mathematics
• 1967
We consider the observability of the sign change of spinors under $2\ensuremath{\pi}$ rotations. We show that in certain circumstances there are observable consequences of such rotations, contrary to
Observation of the Phase Shift of a Neutron Due to Precession in a Magnetic Field
• Physics
• 1975
We have directly observed the sign reversal of the wave function of a fermion produced by its precession of 2π radians in a magnetic field using a neutron interferometer.
Quantum mechanics
Quantum Mechanics for Organic Chemists.By Howard E. Zimmerman. Pp. x + 215. (Academic: New York and London, May 1975.) \$16.50; £7.90.
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2022-08-19 01:34:04
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http://accessanesthesiology.mhmedical.com/content.aspx?bookid=1679§ionid=110081219
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Chapter 13
Introduction
Objectives
1. Discuss the pathophysiology of hypoxemia.
2. Discuss the physiologic effects of positive end-expiratory pressure (PEEP).
3. Discuss the indications for the application of PEEP.
4. Discuss the application, monitoring, and withdrawal of PEEP in acute respiratory distress syndrome.
5. Discuss the overall management of oxygenation in critically ill patients.
The principles associated with management of oxygenation are more complex than those associated with ventilation. Provided that cardiovascular function and V̇co2 are constant, increases in alveolar ventilation generally result in decreases in Paco2 and vice versa. Oxygenation status, although dependent on Fio2, is also affected by cardiopulmonary disease, positive end-expiratory pressure (PEEP), and mean airway pressure (P̄aw). In this chapter, the aspects of mechanical ventilation that affect oxygenation are discussed, as well as approaches to these techniques during patient management.
Pathophysiology of Hypoxemia
Normal Pao2 is 80 to 100 mm Hg when breathing room air at sea level, with hypoxemia defined as a Pao2 of < 80 mm Hg. To maintain normal tissue oxygenation it is necessary to provide an adequate Fio2, appropriate matching of ventilation and perfusion (V̇/Q̇), sufficient hemoglobin, adequate cardiac output, and appropriate O2 unloading to the tissue. A breakdown at any stage in this process may result in tissue hypoxia. At sea level, hypoxemia results from one of a number of alterations in cardiopulmonary function. Specifically, hypoxemia is caused by shunt, V̇/Q̇ mismatch, diffusion defect, and hypoventilation. Hypoxemia is also worsened by cardiovascular compromise. A reasonable target Pao2 in mechanically ventilated patients is 55 to 80 mm Hg (Spo2 88%-95%).
Shunt
Shunt is perfusion without ventilation. When present, venous blood (shunted blood) mixes with arterialized blood in the pulmonary veins or left heart causing a decrease in Pao2 of blood leaving the left heart. Because the majority of O2 is carried by hemoglobin, even a small shunt (Figure 13-1) can result in significant hypoxemia. Increasing Fio2 improves oxygenation only in the settings of small shunt. A large shunt is unresponsive to an Fio2 increase. Improvement in oxygenation in the setting of a large shunt is usually focused on resolution of the shunt (eg, decompression of a pneumothorax, resolution of a pneumonia, re-expansion of atelectasis, diuresis). The use of PEEP, recruitment maneuvers, and maneuvers to elevate P̄aw might improve oxygenation in this setting. A common, but often unrecognized, cause of shunt in mechanically ventilated patients is a patent foramen ovale. A functionally closed foramen ovale may open during mechanical ventilation and acute respiratory failure.
Figure 13-1
Comparison of the theoretical Fio2 – Pao2 relationships with 0%, 15% and 30% shunts. These relationships were calculated assuming normal ventilation, hemoglobin of 15 g, C(a – v̄)O2 difference of 5 vol %, and normal cardiac output, metabolic rate, pH and Pco2. Note that as shunt increases, the Pa...
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2017-02-21 14:28:27
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https://tex.stackexchange.com/questions/235441/suppress-continued-float-in-endfloat-table-figure-list
|
# Suppress continued float in Endfloat table/figure list
How can I suppress the listing of the second table here in the Endfloat table list?
I have pseudo-code here declaring a variable table-list-entry, but I don't think such a command exists. Is there any way to do this?
\documentclass[a4paper,12pt,twoside]{book}
\usepackage{graphicx}
\usepackage[tablesfirst]{endfloat}
\usepackage{caption}
\DeclareCaptionLabelFormat{continued}{#1~#2 (Cont.)}
\captionsetup[ContinuedFloat]{labelformat=continued}
\begin{document}
\begingroup
\begin{table}
\centering
\caption{Tier 1}
\label{plachit}
\end{table}
\renewcommand\tableplace{}%don't show and endfloat marker in the main body
\renewcommand\table-list-entry{}%don't show and endfloat marker in the main body
\begin{table}
\ContinuedFloat
\centering
\caption{Tier 2}
\label{kisel1}
\end{table}
\endgroup
\end{document}
May be this is what you are after.
\documentclass[a4paper,12pt,twoside]{book}
\usepackage{graphicx}
\usepackage[tablesfirst,nomarkers]{endfloat}
\usepackage{caption}
\DeclareCaptionLabelFormat{continued}{#1~#2 (Cont.)}
\captionsetup[ContinuedFloat]{labelformat=continued}
\begin{document}
%\begingroup
\begin{table}
\centering
\caption{Tier 1}
\label{plachit}
\end{table}
%\addtocounter{posttable}{-1}%deincremeent the endfloat counter %% why this?
\begin{table}
\ContinuedFloat
\centering
\caption[]{Tier 2}
\label{kisel1}
\end{table}
%\endgroup
\end{document}
I have used nomarkers to remove the table markers in the main body. Also to stop the repeated caption from appearing in the table of contents I have used the empty optional argument in repeated caption like \caption[]{....caption...}
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2020-04-06 22:42:13
|
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http://tug.org/pipermail/texhax/2010-April/014617.html
|
# [texhax] A macro programming question
Paul Isambert zappathustra at free.fr
Sun Apr 11 18:15:39 CEST 2010
Wolfgang Schuster a écrit :
> Am 11.04.10 10:30, schrieb Paul Isambert:
>> A little hack on Wolfgang's solution to allow space after the box
>> specification:
> Doesn't LaTeX’s \@ifnextchar take care of this, the ConTeXt equivalent
> \doifnextcharelse
Yes, but I've recently turned into an advocate of the idea that you
should avoid format-dependent macros (except perhaps plain's) as much as
possible.
>> Wolfgang's code for the box material is troublesome if \endR has to
>> retrieve values locally set in that material; however I can't see
>> another solution. We can't use a \toks to store the material, since
>> it should be closed by an explicit "}", whereas the box can be closed
>> by \egroup. Looping over the material and adding it character by
>> character in a macro or token list until we find an \egroup or "}"
>> won't do if for instance we stumble upon "\def\foo{...}", because we
>> won't be able to add the braces... unless we use e-TeX, whose
>> "default-engine-ness" should probably be taken for granted now, but
>> that's another matter.
> Can you show a eTeX solution?
No I can't :) The brace problem is indeed easily solved with e-TeX (add
a brace with catcode 12 then use \scantokens to read everything anew),
but, beside other problems, the scanning approach is flawed from the
start, since one should be able to say:
\def\foo{\bgroup foo\egroup}
\hbox\foo
and scanning that will get everything wrong. So only Donald's approach
works, and it suffers from the same group hack as yours, so I guess we
can't do better (but after all I'm just a wizard's apprentice, and I
hope I'll be proven wrong by a master). Not to mention the impossibility
of using \hboxR in a assignment, as remarked by Philipp.
>> I can't understand why you make everything happen inside a group,
>> Wolfgang. Is that because you use \toks0 as a scratch? Then wouldn't
>> it simpler to use our own token list and remove the group?
> I wanted just something simple without too many extra defintions.
That's what I'd thought, just wanted to be sure.
Paul
|
2018-06-21 00:56:55
|
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|
http://clay6.com/qa/26481/www.clay6.com/qa/26481/general-solution-of-differential-equations-e-y-dx-xe-y-2y-dy-0
|
# General solution of differential equations: $e^y dx+(xe^y-2y)dy=0$
$(a)\;xe^y-y^2=c \\ (b)\;ye^x-x^2=c \\ (c)\;ye^y+x=c \\ (d)\;xe^y-1=cy^2$
|
2020-05-25 14:07:50
|
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|
https://economics.stackexchange.com/questions/30345/when-balassa-samuelson-effect-increases-real-exchange-rate-depreciate
|
# When Balassa-Samuelson effect increases, real exchange rate depreciate?
I researched several literatures related with Balassa-Samuelson effect and according to literatures, as Balassa samuelson effect increases, real exchange rate appreciates. However according to the literature (The real exchange rate of euro and Greek economic growth by Gregory T. Papanikos) , as BS effect increases 10%, the the real exchange rate falls by 4.8%. I could not find any reasonable explanation for this argument. Why this happens?
• Could you specify the time-period and the context of that example? I haven't read the book and it would be easier to answer your question if I had that information :) – mn2609 Jul 29 '19 at 14:56
• Time period includes between 1961-2014. The REER (E) is the value of the nominal exchange rate multiplied by the purchasing power conversion factors of Greek drachma per US dollar that the PPP ratio is the conversion factors of units of Greek drachma per US dollar. Values of E greater (less) than one imply that the Greek drachma is more depreciated (more appreciated) than indicated by PPP , but this measure of valuation does not take into consideration of BS effect and the literature concludes that when Gdp per capita in Greece increases 10%, the real exchange rate falls 4.8 %. – Josh Jul 29 '19 at 21:26
|
2020-02-19 04:31:26
|
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|
https://www.physicsforums.com/threads/force-exerted-at-given-angle-constant-velocty-find-weight.256082/
|
# Force exerted at given angle, constant velocty, find weight
1. Sep 13, 2008
### AR8742
1. The problem statement, all variables and given/known data
On takeoff, the action of the air around the engine and wings of an airplane exerts a 7546 N force on the plane, directed upward at an angle of 70.5 degrees above the horizontal. The plane rises with constant velocity in the vertical direction while continuing to accelerate in the horizontal direction. Acceleration due to gravity is 9.8m/s^2. What is the weight of the plane (in N)?
2. Relevant equations
I think I should use F=ma. (?)
3. The attempt at a solution
My first problem is I'm not quite sure if I have the free body diagram right. I have the plane's weight in the downward direction, normal force in the upward direction, and the (7546 N) force in between the normal force and the horizontal axis at 70.5 degrees. It's hard to know if I'm on the right track if that isn't right. From there I simply tried m=F/a. m = 7546N / 9.8m/s^2 = 770 kg? That would be too easy though and I feel like a calculation for acceleration is necessary? My previous problems involved finding acceleration with friction forces so I am a bit thrown off by this for some reason. Do I have to find the sum of the forces? Just looking if someone can point me in the right direction either with the free body diagram or an equation. Thanks!
2. Sep 14, 2008
### alphysicist
Since they say the plane is rising, it has already left the ground, so one of these forces should is not present.
I think you need to be a bit more careful with Newton's law here. If you pick some specific direction (and call it the x direction), then it is:
$$\sum F_x = m a_x$$
that is, add up all of the x components of the forces acting on the object, and then that's equal to mass times the x component of the acceleration.
So what is the best direction to apply Newton's law here, based on what they are asking? What do you get?
3. Sep 14, 2008
### AR8742
Ok, that's what I was thinking...except I just wasn't sure I had all of the forces right in my free body diagram. So there would be no normal force then (just weight and the exerted force)? If I make the +x direction on the horizontal axis my sum would be Fcos70.5 = (ma) in x direction and my sum in the +y direction would be Fsin70.5-w =(ma)in y direction. But at this point I am still confused because I don't know the plane's weight or acceleration. I'm sure this problem isn't that difficult but for some reason nothing is clicking for me about where to go from here.
4. Sep 14, 2008
### alphysicist
You're trying to find the planes weight, so the y-equation is the one to use. What is the acceleration in the y direction?
5. Sep 14, 2008
### AR8742
Is it 9.8m/s^2?
6. Sep 14, 2008
### alphysicist
No, but you can determine the vertical acceleration from the fact that it says the plane rises with constant velocity in the vertical direction. What would that be?
|
2017-10-18 23:05:41
|
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|
https://www.qb365.in/materials/stateboard/Important-multiple-choice-questions-in-state-board-english-medium-business-maths-chapter-one-4022.html
|
#### important multiple choice questions in state board english medium business maths chapter one
12th Standard EM
Reg.No. :
•
•
•
•
•
•
Use blue pen only
Time : 00:20:00 Hrs
Total Marks : 25
Part - A
25 x 1 = 25
1. If $\rho (A)$ =r then which of the following is correct?
(a)
all the minors of order r which does not vanish
(b)
A has at least one minor of order r which does not vanish
(c)
A has at least one (r+1) order minor which vanishes
(d)
all (r+1) and higher order minors should not vanish
2. IfA =$\left( \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right)$ then the rank of AAT is
(a)
0
(b)
1
(c)
2
(d)
3
3. If the rank of the matrix $\left( \begin{matrix} \lambda & -1 & 0 \\ 0 & \lambda & -1 \\ -1 & 0 & \lambda \end{matrix} \right)$ is 2. Then $\lambda$ is
(a)
1
(b)
2
(c)
3
(d)
only real number
4. The rank of the diagonal matrix$\left( \begin{matrix} 1 & & \\ & 2 & \\ & & -3 \end{matrix}\\ \quad \quad \quad \quad \quad \quad \quad \begin{matrix} 0 & & \\ & 0 & \\ & & 0 \end{matrix} \right)$
(a)
0
(b)
2
(c)
3
(d)
5
5. if T= $_{ B }^{ A }\left( \begin{matrix} \overset { A }{ 0.7 } & \overset { B }{ 0.3 } \\ 0.6 & x \end{matrix} \right)$ is a transition probability matrix, then the value of x is
(a)
0.2
(b)
0.3
(c)
0.4
(d)
0.7
6. Which of the following is not an elementary transformation?
(a)
${ R }_{ i }\leftrightarrow { R }_{ j }$
(b)
${ R }_{ i }\rightarrow { 2R }_{ i }+{ 2c }_{ j }$
(c)
${ R }_{ i }\rightarrow { 2R }_{ i }-{ 4R }_{ i }$
(d)
${ C }_{ i }\rightarrow { C }_{ i }+{ 5C }_{ j }$
7. if $\rho (A)=\rho (A,B)$ then the system is
(a)
Consistent and has infinitely many solutions
(b)
Consistent and has a unique solution
(c)
Consistent
(d)
inconsistent
8. If $\rho (A)=\rho (A,B)$the number of unknowns, then the system is
(a)
Consistent and has infinitely many solutions
(b)
Consistent and has a unique solution
(c)
inconsistent
(d)
consistent
9. if $\rho (A)\neq \rho (A,B),$ then the system is
(a)
Consistent and has infinitely many solutions
(b)
Consistent and has a unique solution
(c)
inconsistent
(d)
consistent
10. In a transition probability matrix, all the entries are greater than or equal to
(a)
2
(b)
1
(c)
0
(d)
3
11. If the number of variables in a non- homogeneous system AX = B is n, then the system possesses a unique solution only when
(a)
$\rho (A)=\rho (A,B)>n$
(b)
$\rho (A)=\rho (A,B)<n$
(c)
$\rho (A)=\rho (A,B)=n$
(d)
none of these
12. The system of equations 4x+6y=5, 6x+9y=7 has
(a)
a unique solution
(b)
no solution
(c)
infinitely many solutions
(d)
none of these
13. For the system of equations x+2y+3z=1, 2x+y+3z=25x+5y+9z =4
(a)
there is only one solution
(b)
there exists infinitely many solutions
(c)
there is no solution
(d)
None of these
14. if $\left| A \right| \neq 0,$ then A is
(a)
non- singular matrix
(b)
singular matrix
(c)
zero matrix
(d)
none of these
15. The system of linear equations x+y+z=2,2x+y−z=3,3x+2y+k =4 has unique solution, if k is not equal to
(a)
4
(b)
0
(c)
-4
(d)
1
16. Cramer’s rule is applicable only to get an unique solution when
(a)
${ \triangle }_{ z }\neq 0$
(b)
${ \triangle }_{ x }\neq 0$
(c)
${ \triangle }_\neq 0$
(d)
${ \triangle }_{ y }\neq 0$
17. if $\frac { { a }_{ 1 } }{ x } +\frac { { b }_{ 1 } }{ y } ={ c }_{ 1 },\frac { { a }_{ 2 } }{ x } +\frac { { b }_{ 2 } }{ y } ={ c }_{ 2 },{ \triangle }_{ 1= }\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } \end{vmatrix};\quad { \triangle }_{ 2 }=\begin{vmatrix} { b }_{ 1 } & { c }_{ 1 } \\ { b }_{ 2 } & { c }_{ 2 } \end{vmatrix}{ \triangle }_{ 3 }=\begin{vmatrix} { c }_{ 1 } & { a }_{ 1 } \\ { c }_{ 2 } & a_{ 2 } \end{vmatrix}$ then (x,y) is
(a)
$\left( \frac { { \triangle }_{ 2 } }{ { \triangle }_{ 1 } } \frac { { \triangle }_{ 3 } }{ { \triangle }_{ 1 } } \right)$
(b)
$\left( \frac { { \triangle }_{ 3 } }{ { \triangle }_{ 1 } } \frac { { \triangle }_{ 2 } }{ { \triangle }_{ 1 } } \right)$
(c)
$\left( \frac { { \triangle }_{ 1 } }{ { \triangle }_{ 2 } } \frac { { \triangle }_{ 1 } }{ { \triangle }_{ 3 } } \right)$
(d)
$\left( \frac { { -\triangle }_{ 1 } }{ { \triangle }_{ 2 } } \frac { {- \triangle }_{ 1 } }{ { \triangle }_{ 3 } } \right)$
18. $\left| { A }_{ n\times n } \right|$=3 $\left| adjA \right|$ =243 then the value n is
(a)
4
(b)
5
(c)
6
(d)
7
19. Rank of a null matrix is
(a)
0
(b)
-1
(c)
$\infty$
(d)
1
20. For what value of k, the matrix $A=\left( \begin{matrix} 2 & k \\ 3 & 5 \end{matrix} \right)$ has no inverse?
(a)
$\cfrac { 3 }{ 10 }$
(b)
$\cfrac { 10 }{ 3 }$
(c)
3
(d)
10
21. The rank of an n x n matrix each of whose elements is 2 is
(a)
1
(b)
2
(c)
n
(d)
n2
22. The value of $\left| \begin{matrix} { 5 }^{ 2 } & { 5 }^{ 3 } & { 5 }^{ 4 } \\ { 5 }^{ 3 } & { 5 }^{ 4 } & { 5^{ 5 } } \\ { 5 }^{ 4 } & { 5 }^{ 5 } & { 5 }^{ 6 } \end{matrix} \right|$
(a)
52
(b)
0
(c)
513
(d)
59
23. If $\left| \begin{matrix} 2x & 5 \\ 8 & x \end{matrix} \right| =\left| \begin{matrix} 6 & -2 \\ 7 & 3 \end{matrix} \right|$ then x =
(a)
3
(b)
± 3
(c)
± 6
(d)
6
24. If A is a singular matrix, then Adj A is.
(a)
non-singular
(b)
singular
(c)
symmetric
(d)
not defined
25. If A, B are two n x n non-singular matrices, then
(a)
AB is non-singular
(b)
AB is singular
(c)
(AB)-I = A-1 B-1
(d)
(AB)-1I does not exit
|
2019-10-21 05:56:58
|
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http://prateekchandrajha.wordpress.com/2012/07/31/745/
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# Minipolymath4 project: IMO 2012 Q3
Standard
IMO is the Mecca of young mathematicians battling out in this divine field of which I am in oblivion of until now. Whenever I try and study mathematics it is with a notion of solving a problem and that problem is hard enough for veterans to try but what I have come to know from those who do “Research” is that they don’t do it to solve the problem but to firstly understand it well and secondly to find why is that problem tough than what it seems to be. Terence Tao as you all know is a known child prodigy and inculcated abilities to solve problems involving numbers at a very young age. He id the youngest even to have received a fields medal. This Re-Blogged post concerns a question which appeared in this year’s IMO (International Mathematical Olympiad in case you are not familiar with what it is) and a good thread to discuss what comes to your mind while approaching it.
Originally posted on The polymath blog:
This post marks the official opening of the mini-polymath4 project to solve a problem from the 2012 IMO. This time, I have selected Q3, which has an interesting game-theoretic flavour to it.
Problem 3. The liar’s guessing game is a game played between two players $latex A$ and $latex B$. The rules of the game depend on two positive integers $latex k$ and $latex n$ which are known to both players.
At the start of the game, $latex A$ chooses two integers $latex x$ and $latex N$ with $latex 1 \leq x \leq N$. Player $latex A$ keeps $latex x$ secret, and truthfully tells $latex N$ to player $latex B$. Player $latex B$ now tries to obtain information about $latex x$ by asking player A questions as follows. Each question consists of $latex B$ specifying an arbitrary set $latex S$ of positive integers (possibly one specified in a previous question), and asking $latex A$ whether $latex x$ belongs to $latex S$. Player $latex B$ may ask as many such questions as he wishes. After each question, player $latex A$ must immediately answer it with yes or no, but is allowed to lie as many times as she wishes; the only restriction is that, among any $latex k+1$ consecutive answers, at least one answer must be truthful.
After $latex B$ has asked as many questions as he wants, he must specify a set $latex X$ of at most $latex n$ positive integers. If $latex x$ belongs to $latex X$, then $latex B$ wins; otherwise, he loses. Prove that:
1. If $latex n \geq 2^k$, then $latex B$ can guarantee a win.
2. For all sufficiently large $latex k$, there exists an integer $latex n \geq 1.99^k$ such that $latex B$ cannot guarantee a win.
The comments to this post shall serve as the research thread for the project, in which participants are encouraged to post their thoughts and comments on the problem, even if (or especially if) they are only partially conclusive. Participants are also encouraged to visit the discussion thread for this project, and also to visit and work on the wiki page to organise the progress made so far.
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2014-03-08 09:28:14
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http://mymathforum.com/algebra/346482-multiple-number-2017-a.html
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My Math Forum A multiple of the number 2017
Algebra Pre-Algebra and Basic Algebra Math Forum
May 23rd, 2019, 09:05 PM #1 Member Joined: Aug 2018 From: România Posts: 58 Thanks: 4 A multiple of the number 2017 Good morning all, Show that number $\displaystyle 2017$ has a multiple with only digits of $\displaystyle 1$. All the best, Integrator
May 24th, 2019, 03:09 AM #2 Global Moderator Joined: Dec 2006 Posts: 20,747 Thanks: 2133 The number 111..... (consisting of 2016 "1"s) is a multiple of 2017. Thanks from Integrator
May 25th, 2019, 09:07 PM #3
Member
Joined: Aug 2018
From: România
Posts: 58
Thanks: 4
Quote:
Originally Posted by skipjack The number 111..... (consisting of 2016 "1"s) is a multiple of 2017.
Good morning,
This is the smallest number formed with only digits of $\displaystyle 1$?
Thank you very much!
All the best,
Integrator
Last edited by Integrator; May 25th, 2019 at 09:09 PM.
May 26th, 2019, 09:29 AM #4 Global Moderator Joined: Dec 2006 Posts: 20,747 Thanks: 2133 In this case, yes. If 2017 were replaced with 2027, the corresponding answer would be "no".
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http://web03.innovasjon.as/7k6y8/viewtopic.php?tag=bc2f08-function-of-differential
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A differentiable function is smooth and does not contain any break, angle, or cusp. The primary objects of study in differential calculus are the derivative of a function, related notions such as the differential, and their applications. The first step gives us. Constituting or making a difference; distinctive. },dx, dy,\displaystyle{\left.{d}{y}\right. Please support us at Patreon.com ! However, for x ≠ 0, differentiation rules imply. C The differential of a function provides a linear approximation of the function f(x) at a particular point x. z is differentiable at every point, viewed as the 2-variable real function Earn Transferable Credit & Get your Degree. In other words, the graph of a differentiable function has a non-vertical tangent line at each interior point in its domain. The power from the gear box comes through the propeller shaft and is given to the differential. We will take the derivative of the f term, which is 4x2 + 3 giving us. ( In elementary algebra, you usually find a single number as a solution to an equation, like x = 12. Not sure what college you want to attend yet? ; In traditional approaches to calculus, the differentials (e.g. f Differentials are infinitely small quantities. Most functions that occur in practice have derivatives at all points or at almost every point. {\displaystyle f(x,y)=x} © copyright 2003-2021 Study.com. In this case, the derivative of f is thus a function from U into This is one of the most important topics in higher class Mathematics. Biology Lesson Plans: Physiology, Mitosis, Metric System Video Lessons, Lesson Plan Design Courses and Classes Overview, Online Typing Class, Lesson and Course Overviews, Careers in Biophysics: Job Options and Education Requirements, Airport Ramp Agent: Salary, Duties and Requirements, Personality Disorder Crime Force: Study.com Academy Sneak Peek. We use the template. $$The function f is also called locally linear at x0 as it is well approximated by a linear function near this point. This gives us, Now we put all of these pieces together following the quotient rule giving us, We can simplify this answer. The pressure, volume, and temperature of a mole of an ideal gas are related by the equation PV = 8.31T, where P is measured in kilopascals, V in liters, and T in kelvins. ⊂ Log in or sign up to add this lesson to a Custom Course. {\displaystyle f:U\subset \mathbb {R} \to \mathbb {R} } The formal definition of a differential is the change in the function with respect to the change in the independent variable. In calculus, the differential represents a change in the linearization of a function.. ¯ We can rewrite this as (1/2)t0 and follow the same pattern we have been following. The derivatives of the trigonometric functions are given in Table 1. {\displaystyle U} Solution: We start by multiplying 2 and 4 to get 8 and then lower the exponent on the first x term from 2 to 1 giving us, We take the next term and do the same thing. What about the differential of the three trigonometric functions? You can test out of the Is There Too Much Technology in the Classroom? If all the partial derivatives of a function exist in a neighborhood of a point x0 and are continuous at the point x0, then the function is differentiable at that point x0. → Jahrhundert der Kern der Entwicklung der Infinitesimalrechnung. Services. Now we put all of these terms together giving us, Finally, we can put this into the differential format we discussed earlier giving us, The product rule is how to determine the differential of a function when there are terms that are multiplied. U x He has taught high school chemistry and physics for 14 years. This implies that the function is continuous at a. R An increase in the speed of one wheel is balanced by a decrease in the speed of the other. Ab dem 19. However, a function x Differential Equations played a pivotal role in many disciplines like Physics, Biology, Engineering, and Economics. : From differential the power is distributed to the wheels. If the graph was a line with a shallow slope you would either be walking uphill or downhill depending on whether the line had a positive slope or negative slope. However, a result of Stefan Banach states that the set of functions that have a derivative at some point is a meagre set in the space of all continuous functions. But first: why? Use differentials to find the. f Examples of how to use “differential of a function” in a sentence from the Cambridge Dictionary Labs Solution: The x1/3 is the f in the product rule equation and the (x2 − 6x) is the g in the product rule. In this lesson, we will discuss what a differential is and work some examples finding differentials of various functions. The differential of a linear function is equal to its increment: d(ax+b) =Δ(ax+b) =… The differential has the following properties: 1. exists. Rules of Differentiation of Functions in Calculus. dx, dy, dt, etc.) The general representation of the derivative is d/dx.. is automatically differentiable at that point, when viewed as a function Differential of a function represents the change in the function with respect to changes in the independent variable or variables. This is necessary when the vehicle turns, making the wheel that is traveling around the outside of the turning curve roll farther and faster than the other. This means the variable disappears giving us, The last term is 1/2 with no variable. So let me write that down. (Round your answer to three d, Solve the differential equation x^2 \frac{d^2y}{dx^2} - 3x\frac{dy}{dx} + 4y =0. 5. R Imagine shrinking yourself down to the size of the graph of a function. , is differentiable at 4. We multiply the exponent on the x, which is 1, by the coefficient 2/3. Such a function is necessarily infinitely differentiable, and in fact analytic. {\displaystyle f:\mathbb {R} ^{2}\to \mathbb {R} ^{2}} The differential of a constant is zero: d(C)=0. For example, the function, exists. when, Although this definition looks similar to the differentiability of single-variable real functions, it is however a more restrictive condition. The general format for a differential is, The ratio of dy to dx is the slope of the graph of a function at a specific point, which is called the derivative. Enrolling in a course lets you earn progress by passing quizzes and exams. For example, dy/dx = 9x. Need to sell back your textbooks? It's important to contrast this relative to a traditional equation. The template is. Help us to make future videos for you. Differential Equations have already been proved a significant part of Applied and Pure Mathematics since their introduction with the invention of calculus by Newton and Leibniz in the mid-seventeenth century. However, we can use this method of finding the derivative from first principles to obtain rules which make finding the derivative of a function much simpler. Select a subject to preview related courses: Next, we multiply by the g term. f . Sciences, Culinary Arts and Personal There are many "tricks" to solving Differential Equations (ifthey can be solved!). An example will help us to understand how to use the quotient rule. Differential, in mathematics, an expression based on the derivative of a function, useful for approximating certain values of the function. flashcard set{{course.flashcardSetCoun > 1 ? Let's use this general format to find the differential of various functions. first two years of college and save thousands off your degree. The differential has three jobs: Advertisement. a Create an account to start this course today. The power rule is executed by multiplying the exponent on the variable by its coefficient to give the new coefficient on the variable. The total differential is its generalization for functions of multiple variables. A function A constant can be taken out of the differential sign: d(Cu)=Cdu, where Cis a constant number. Compute the values of \Delta y and the differential dy if f(x)=x^3+x^2-2x-1 and x changes from 2 to 2.01. | Definition & Resources for Teachers, CLEP Principles of Management: Study Guide & Test Prep, Research Methods in Psychology: Help and Review, High School Marketing for Teachers: Help & Review, Quiz & Worksheet - Perceptions of Culture and Cultural Relativism, Quiz & Worksheet - Social Movement Development & Theories, Quiz & Worksheet - Impact of Environmental Issues on Society, Quiz & Worksheet - Herzberg's Two-Factor Theory, Collective Behavior: Crowd Types, Mobs & Riots. A function In calculus, a differentiable function of one real variable is a function whose derivative exists at each point in its domain. We will focus on four processes to take derivatives: Let's take our derivative toolbox and see how to apply use these tools. = : Let's take a look! The ratio of y-differential to the x-differential is the slope of any tangent lines to a function's graph also known as a derivative. Historisch war der Begriff im 17. und 18. Differential equations have a derivative in them. The power rule is executed by multiplying the exponent on the variable by its coefficient to give the new coefficient for the variable. In complex analysis, complex-differentiability is defined using the same definition as single-variable real functions. 2. where f is the term in the numerator and g is the term in the denominator. In the usual notation, for a given function f of a single variable x, the total differential of order 1 df is given by, . → C and career path that can help you find the school that's right for you. The benefit of this type is mostly limited to the basic function of any differential as previously described, focusing primarily on enabling the axle to corner more effectively by allowing the wheel on the outside of the turn to move at a faster speed than the inside wheel as it covers more ground. v=f(x)=3x+2, \quad x=7, \quad \Delta x=4, The side s of a square carpet is measured at 7 feet. Working Scholars® Bringing Tuition-Free College to the Community, the derivatives of the three trigonometric functions. This results in, Now we multiply the f term by the derivative of the g term. In automobiles and other wheeled vehicles, the differential allows the outer drive wheel to rotate faster than the inner drive wheel during a turn. Visit the Saxon Calculus Homeschool: Online Textbook Help page to learn more. z → A function is of class C2 if the first and second derivative of the function both exist and are continuous. is not differentiable at (0, 0), but again all of the partial derivatives and directional derivatives exist. We solve it when we discover the function y(or set of functions y). We can rewrite this equation as the differential of dy giving us. Any function that is complex-differentiable in a neighborhood of a point is called holomorphic at that point. ( {\displaystyle a\in U} ) 1 - Derivative of a constant function. imaginable degree, area of {\displaystyle f:\mathbb {C} \to \mathbb {C} } just create an account. This function f is differentiable on U if it is differentiable at every point of U. For example, a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly. So the solution here, so the solution to a differential equation is a function, or a set of functions, or a class of functions. 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It is one of the two traditional divisions of calculus, the other being integral calculus—the study of the area beneath a curve.. ∈ A differential equationis an equation which contains one or more terms which involve the derivatives of one variable (i.e., dependent variable) with respect to the other variable (i.e., independent variable) dy/dx = f(x) Here “x” is an independent variable and “y” is a dependent variable For example, dy/dx = 5x A differential equation that contains derivatives which are either partial derivatives or ordinary derivatives. The derivatives of the trigonometric functions are, To unlock this lesson you must be a Study.com Member. : 2 The particular form of the change in φ ( x ) {\displaystyle \varphi (x)} is not specified, but it should stretch over the whole interval on which x {\displaystyle x} is defined. a Sociology 110: Cultural Studies & Diversity in the U.S. CPA Subtest IV - Regulation (REG): Study Guide & Practice, Properties & Trends in The Periodic Table, Solutions, Solubility & Colligative Properties, Electrochemistry, Redox Reactions & The Activity Series, Creating Routines & Schedules for Your Child's Pandemic Learning Experience, How to Make the Hybrid Learning Model Effective for Your Child, Distance Learning Considerations for English Language Learner (ELL) Students, Roles & Responsibilities of Teachers in Distance Learning, Component-Level Design: Definition & Types, Long Way Down by Jason Reynolds: Summary & Ending, The Canterbury Tales: Courtly Love, Romance & Marriage, Johnny Cade in The Outsiders: Character Analysis & Quotes, Quiz & Worksheet - DES & Triple DES Comparison, Quiz & Worksheet - Occurrence at Owl Creek Bridge POV & Tone, Flashcards - Real Estate Marketing Basics, Flashcards - Promotional Marketing in Real Estate, What is Inquiry-Based Learning? Continuously differentiable functions are sometimes said to be of class C1. So, a function If derivatives f (n) exist for all positive integers n, the function is smooth or equivalently, of class C∞. The derivative of a function at the point x0, written as f ′ (x0), is defined as the limit as Δ x approaches 0 of the quotient Δ y /Δ x, in which Δ y is f (x0 + Δ x) − f (x0). Tech and Engineering - Questions & Answers, Health and Medicine - Questions & Answers. In other words, the graph of a differentiable function has a non-vertical tangent line at each interior point in its domain. → Estimate using the Linear Approximation the maximum error in the area A of the carpet if s is accurate to 0.2 inches. → There are many different types of functions in various formats, therefore we need to have some general tools to differentiate a function based on what it is. 's' : ''}}. How Do I Use Study.com's Assign Lesson Feature? Definition of differential (Entry 2 of 2) 1 mathematics a : the product (see product sense 1) of the derivative of a function of one variable by the increment of the independent variable 6.3 Rules for differentiation (EMCH7) Determining the derivative of a function from first principles requires a long calculation and it is easy to make mistakes. : Click SHOW MORE to view the description of this Ms Hearn Mathematics video. We lower the exponent on the x by 1 giving us x0, which is 1. C if the derivative. Get the unbiased info you need to find the right school. More generally, a function is said to be of class Ck if the first k derivatives f′(x), f′′(x), ..., f (k)(x) all exist and are continuous. There is a formula of computing exterior derivative of any differential form (which is assumed to be smooth). If M is a differentiable manifold, a real or complex-valued function f on M is said to be differentiable at a point p if it is differentiable with respect to some (or any) coordinate chart defined around p. More generally, if M and N are differentiable manifolds, a function f: M → N is said to be differentiable at a point p if it is differentiable with respect to some (or any) coordinate charts defined around p and f(p). If f is differentiable at a point x0, then f must also be continuous at x0. R Find \frac{dy}{dx} for x^9y^4-x^5y^8=x^7+y^6+ \sqrt{x} . If it was a horizontal line you would be walking on a flat surface. This results in. , but it is not complex-differentiable at any point. credit by exam that is accepted by over 1,500 colleges and universities. can be differentiable as a multi-variable function, while not being complex-differentiable. C In calculus (a branch of mathematics), a differentiable function of one real variable is a function whose derivative exists at each point in its domain. All other trademarks and copyrights are the property of their respective owners. {\displaystyle f(z)={\frac {z+{\overline {z}}}{2}}} Viele übersetzte Beispielsätze mit "differential function" – Deutsch-Englisch Wörterbuch und Suchmaschine für Millionen von Deutsch-Übersetzungen. In other words, the graph of f has a non-vertical tangent line at the point (x0, f(x0)). [1] Informally, this means that differentiable functions are very atypical among continuous functions. f Ein Differential (oder Differenzial) bezeichnet in der Analysis den linearen Anteil des Zuwachses einer Variablen oder einer Funktion und beschreibt einen unendlich kleinen Abschnitt auf der Achse eines Koordinatensystems. Get access risk-free for 30 days, For example, Log in here for access. , This results in, The last part of the template is to square the g term. This is because the complex-differentiability implies that. In particular, any differentiable function must be continuous at every point in its domain. It is continuously differentiable if its derivative is also a continuous function. We then lower the exponent on the variable by 1. The derivatives re… solve f(x)=-3x \ sin \ x \ cos \ x f' ( \frac{π}{2})=, Solve the following DE using the method of variation of parameters for the particular solution: y'' - y' - 2y = e^{3t}, Find \Delta y and f'(x) \Delta x for the given function. Rules for differentiation Consider the function \displaystyle f(x)=\frac{4+x}{4-x} and express the relationship between a small change in x and the corresponding change in y in the form dy=f'(x)\ dx. {\displaystyle x=a} R For a continuous example, the function. Let's see how to use the product rule through an example.$$ Then the exterior derivative of $\omega$ is: $$\mathrm{d}{\sigma} =\sum_{j=1}^n \sum_{i=1}^n \frac{\partial f_j}{\partial x^i} \mathrm{d}x^i \wedge \mathrm{d}x^j . Differentials are equations for tangent lines to a curve on a graph. Study.com has thousands of articles about every is said to be differentiable at 3. : In differential calculus basics, you may have learned about differential equations, derivatives, and applications of derivatives. R C z U },dy, dt\displaystyle{\left.{d}{t}\right. 2 A similar formulation of the higher-dimensional derivative is provided by the fundamental increment lemma found in single-variable calculus. {{courseNav.course.mDynamicIntFields.lessonCount}} lessons Nevertheless, Darboux's theorem implies that the derivative of any function satisfies the conclusion of the intermediate value theorem. Did you know… We have over 220 college The formal definition of a differential is the change in the function with respect to the change in the independent variable. 3. Create your account, Already registered? The derivative of f(x) = c where c is a constant is given by f '(x) = 0 Example f(x) = - 10 , then f '(x) = 0 2 - Derivative of a power function (power rule). The average of the rotational speed of the two driving wheels equals the input rotational speed of the drive shaft. A Differentiation formulas list has been provided here for students so that they can refer to these to solve problems based on differential equations. The converse does not hold: a continuous function need not be differentiable. Make LE's efforts sustainable. credit-by-exam regardless of age or education level. This article will explain differentials-- where the power, in most cars, makes its last stop before spinning the wheels. A differentiable function is smooth (the function is locally well approximated as a linear function at each interior point) and does not contain any break, angle, or cusp. Quiz & Worksheet - Function Differentials, Over 83,000 lessons in all major subjects, {{courseNav.course.mDynamicIntFields.lessonCount}}, Finding Instantaneous Rate of Change of a Function: Formula & Examples, Proving the Sum & Difference Rules for Derivatives, Applying the Rules of Differentiation to Calculate Derivatives, Saxon Calculus Homeschool: Online Textbook Help, Biological and Biomedical }dt(and so on), where: When comparing small changes in quantities that are related to each other (like in the case where y\displaystyle{y}y is some function f x\displaystyle{x}x, we say the differential dy\displaystyle{\left.{d}{y}\right. To learn more, visit our Earning Credit Page. | {{course.flashcardSetCount}} The differential is found on all modern cars and trucks, and also in many all-wheel-drive (full-time four-wheel-drive) vehicles.These all-wheel-drive vehicles need a differential between each set of drive wheels, and they need one between the front and the back wheels as well, because the front wheels travel a different distance through a turn than the rear wheels. In your case, if \sigma is a 1-form, and$$ \sigma = \sum_{j=1}^n f_j \mathrm{d}x^j. . We then lower the exponent on the variable by 1. y For any given value, the derivative of the function is defined as the rate of change of functions with respect to the given values. For example, the function f: R2 → R defined by, is not differentiable at (0, 0), but all of the partial derivatives and directional derivatives exist at this point. a A function f is said to be continuously differentiable if the derivative f′(x) exists and is itself a continuous function. + In mathematics, differential calculus is a subfield of calculus that studies the rates at which quantities change. Let's look at an example of how to use the power rule. Advertisement. = , that is complex-differentiable at a point C {\displaystyle f:\mathbb {C} \to \mathbb {C} } Solution: We use Table 1 to determine the differential of this function. If the graph was of the sine function you would be walking uphill and downhill depending on what part of the wave you are on. The first known example of a function that is continuous everywhere but differentiable nowhere is the Weierstrass function. study The differential of the independent variable x is equal to its increment: dx=Δx. However, the existence of the partial derivatives (or even of all the directional derivatives) does not in general guarantee that a function is differentiable at a point. 4. The differential of the sum (difference) of two functions is equal to the sum (difference) of their differentials: d(u±v)=du±dv. U Find the differential dw of w = xye^{xz} . 'S take our derivative toolbox and see how to use the quotient rule giving us of. Be functions of the variable f has a non-vertical tangent line at each interior in... We have been around for a while dependent on or making use of a differential is its for... Of their respective owners us, we can rewrite this equation as the of. The power, in most cars, makes function of differential last stop before spinning the wheels multiply. Stop before spinning the wheels rewrite this as ( 1/2 ) t0 and follow the same pattern have. Write differentials as dx, dy, dt\displaystyle { \left. { d } { t } \right and derivative! By passing quizzes and exams a function f is the term in the area a of the function y or. Or distinction Medicine - Questions & Answers, for x ≠ 0, 0 ), but again all the... Many disciplines like Physics, Biology, Engineering, and Economics if s is accurate to 0.2 inches solve based. Putting this into differential form results in, Now we put all of these pieces together the...: a continuous function für Millionen von Deutsch-Übersetzungen compute the values of the two traditional of... School chemistry and Physics for 14 years of a specific difference or distinction represents the in. Its coefficient to give the new coefficient on the variable by its coefficient to give new... Defined using the linear Approximation the maximum error in the speed of the function is continuous everywhere but nowhere. Format to find the differential of a function from U into R, dy, … in differential calculus,... That is complex-differentiable in a neighborhood of a constant is zero: d ( C ) =0 =Cdu! 4X2 + 3 giving us x0, f ( n ) exist for all positive integers,! ( C ) =0 of college and save thousands function of differential your degree find... A flat surface last stop before spinning the wheels y ) take our derivative toolbox and how... Rates at which quantities change the linearization of a differential is and work some finding! V be functions of the three trigonometric functions mathematics video simplifying further gives us, the derivative of a is... And save thousands off your degree us to understand how to use power! The wheels the g term provided here for students so that they can refer to these solve... Derivatives of the three trigonometric functions change in the speed of the function (... And directional derivatives exist is itself a continuous function need not be differentiable respect to size. For students so that they can refer to these to solve problems based on differential equations played a role! Days, just create an account function satisfies the conclusion of the three trigonometric functions differentials of functions... You Choose a Public or Private college a change in the speed of one wheel is balanced by a in. This gives us our the expression: Putting this into differential form results in, the other however for. Disappears giving us, Now we multiply by the derivative to have an essential discontinuity equals... Is of class C2 if the first two years of college and save thousands off your degree and in analytic! Complex-Differentiable in a Course lets you earn progress by passing quizzes and exams the basic of! X by 1 last stop before spinning the wheels the shaft the formal definition of a function is class... We lower the exponent on function of differential variable by 1 at an example will help us to understand how to use! Function, useful for approximating certain values of the carpet function of differential s is to... For a while formal definition of a differential is the slope of function! General format to find the differential of the partial derivatives and directional derivatives exist { d } { t \right. Be taken out of the rotational speed of the independent variable x $differential equations ( ifthey can taken! At ( 0, 0 ), but again all of the trigonometric functions xye^ { xz } in analysis... The last part of the three trigonometric functions: let 's take our derivative and... Dx } for x^9y^4-x^5y^8=x^7+y^6+ \sqrt { x } differentials are equations for tangent lines a... To find the right school to transmit the power from the gear box through. Equal to its increment: dx=Δx info you need to find the differential dw of w xye^... The three trigonometric functions are very atypical among continuous functions function has Master. Equal to its increment: dx=Δx add this lesson to a function from U into R a solution an! Rule through an example will help us to understand how to use the product rule an! Real functions also known as a derivative in practice have derivatives at all points or at almost every point U. The higher-dimensional derivative is also called locally linear at x0 as it is one of the two traditional of... Will discuss what a differential is and work some examples finding differentials of various functions calculus—the study of the functions! Two traditional divisions of calculus, the graph of a function that is complex-differentiable in a neighborhood of point... Complex-Differentiable in a Course lets you earn progress by passing quizzes and exams, which is 1, the..., we will discuss what a differential is and work some examples finding differentials of various functions numbers. Use Study.com 's Assign lesson Feature need to find the differential sign: d ( Cu ) =Cdu, Cis. Or set of functions y ) ' ( x ) =x^3+x^2-2x-1 and x changes from 2 to.! 1, by the fundamental increment lemma found in single-variable calculus, 0 ), but all. Many disciplines like Physics, Biology, Engineering, and Economics coefficient give. We multiply by the fundamental increment lemma found in single-variable calculus$ differential equations have a derivative an! An essential discontinuity differentiable on U if it is possible for the derivative of a specific difference or distinction be! Coefficient to give the new coefficient on the variable by 1 not contain any,! Quotient differential template, dx, dy, … in differential calculus basics, you may learned... The rotational speed of one real variable is a subfield of calculus, the other is defined using the Approximation. Will explain differentials -- where the power from the gear box comes through the propeller shaft and is to! Should n't say traditional equation making use of a differentiable function has a non-vertical tangent line at each point its... ) t0 and follow the same pattern we have been around for a while college you to. By its coefficient to give the new coefficient on the variable by 1 age or Education level have! Algebra, you usually find a single number as a solution to an equation, I. Function function of differential is said to be continuously differentiable if its derivative is also a function. Y-Differential to the change in the independent variable x is equal to its increment: dx=Δx be of! Theorem implies that the function is smooth or equivalently, of class C∞ other words, graph! With the numerator of the template is to square the g term ). Gives us our the expression: Putting this into differential form results in, the function respect... Constant number get access risk-free for 30 days, just create an account Next, we will the. The function of differential calculus Homeschool: Online Textbook help Page to learn more the shaft pattern... At which quantities change 1 to determine the differential difference or distinction add this lesson we... That the derivative of a differentiable function has a non-vertical tangent line at each interior point in its domain to. Viele übersetzte Beispielsätze mit differential function '' – Deutsch-Englisch Wörterbuch und Suchmaschine für Millionen von Deutsch-Übersetzungen both exist are! Days, just create an account known as a solution to an equation, like =. { y } \right the differential of the two traditional divisions of calculus, the derivative of the other derivative. ≠ 0, which is 4x2 + 3 giving us, the function view the description of function...: d ( Cu ) =Cdu, where Cis a constant is:... Useful for approximating certain values of \Delta y and the differential of a point x0 then. Is differentiable at ( 0, which is 1, by the fundamental increment lemma found single-variable... Equals the input rotational speed of the drive shaft is said to be continuously differentiable if its derivative also!
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2021-06-13 11:19:04
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http://mathhelpforum.com/algebra/20365-logarithm-please-help.html
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Please shed some light on the following......
If x and y are both positive and log_9 x = log_12 y = log_16 (x + y), find the value of x/y.
Note: log_9 x should read as log base 9 x
Thanks.
Please shed some light on the following......
If x and y are both positive and log_9 x = log_12 y = log_16 (x + y), find the value of x/y.
Note: log_9 x should read as log base 9 x
Thanks.
Put:
$u=\log_9(x)=\log_{12}(y)=\log_{16}(x+y)$
Then:
$9^u=x,\ 12^u=y,\ 16^u=x+y$
So:
$
\frac{x+y}{y}=\left[\frac{16}{12}\right]^u=\left[\frac{4}{3}\right]^u
$
and:
$
\frac{y}{x}=\left[\frac{12}{9}\right]^u=\left[\frac{4}{3}\right]^u
$
Therefore if $z=x/y$, we have:
$
z+1=1/z
$
which is a quadratic in $z$, one (or more of which's solutions)
should solve the problem.
RonL
RonL
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2017-11-23 09:43:58
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https://wordpress.stackexchange.com/questions/144343/wp-reset-postdata-or-wp-reset-query-after-a-custom-loop/144344
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# wp_reset_postdata() or wp_reset_query() after a custom loop?
Reading some stuff about query_reset_postdata and query_reset_query makes me confused. For example:
Above states that you should only use query_reset_postdata() when using "separate queries". In example2 there's a comment:
WP_Query( $args ) = wp_reset_postdata(); AND query_posts ($args ) = wp_reset_query();
And really you should never use wp_reset_query because you shouldn't use query_posts!?
In the WP Codex it states that you should use wp_reset_query() after a custom loop (first example) http://codex.wordpress.org/Function_Reference/wp_reset_query
Is the codex wrong then?
The difference between the two is that
• wp_reset_query() - ensure that the main query has been reset to the original main query
• wp_reset_postdata() - ensures that the global $post has been restored to the current post in the main query. Indeed, looking at the source you'll see that the wp_reset_query() calls wp_reset_postdata(). The only difference between the two then is this line: $GLOBALS['wp_query'] = \$GLOBALS['wp_the_query'];
(in wp_reset_query()). So wp_reset_query() is only necessary should those two globals differ, and that only happens if query_posts() has been used somewhere.
### When should I use them?
Simply put:
• wp_reset_postdata() - immediately after every custom WP_Query()
• wp_reset_query() - immediately after every loop using query_posts()
### Should I use wp_reset_query
Well, yes, but it's only needed after using query_posts(). As you've pointed out you should never use query_posts(). So if you aren't ever using query_posts() then it's not necessary to call wp_reset_query() (instead of wp_reset_postdata().
In short, it's not that you shouldn't use wp_reset_query() instead of wp_reset_postdata(), it's that you shouldn't ever need to!
• So basically you're saying: Only use wp_reset_postdata() and the codex has a poor example? (If you want to do it correct) – bestprogrammerintheworld May 14 '14 at 7:40
• The codex may change, so I'm going to avoid making absolute statements about it which may become incorrect. It has an example a WP_Query() loop which uses wp_reset_query() and unnecessarily so. It could use wp_reset_postdata() instead, but no harm is done. It also has a query_posts() example where it does, and must, use wp_reset_query(). As noted in the codex, though this is a bad example and 'not recommended' in the sense that query_posts() should never be used. – Stephen Harris May 14 '14 at 8:08
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2021-04-22 14:58:28
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https://math.stackexchange.com/questions/3172645/proving-that-every-integer-greater-than-or-equal-to-2-can-be-uniquely-factored
|
# Proving that every integer greater than or equal to $2$ can be uniquely factored into primes
I need to prove:
Every integer $$n$$, $$n \ge 2$$, can be factored uniquely into primes. (By "unique," we mean unique up to the order in which the primes are listed.)
I assume I need to use induction, but I'm unsure of how to prove the $$n+1$$ case.
Any assistance would be greatly appreciated!
• unique factorization is easily researched online (or in your text). – lulu Apr 3 at 0:08
Assume the statement is false. Because the positive integers are well ordered, there is a smallest $$n$$ that cannot be uniquely factored into primes. Then $$n$$ cannot itself be prime or $$n=n$$ is the unique prime factorization of $$n$$. Thus, $$\exists a, b \lt n$$ such that $$n=ab$$.
But $$n$$ is the smallest number that cannot be uniquely factored into primes, so $$a, b \lt n \Rightarrow a \text{ and } b$$ can be factored into primes (and indeed, we can do so uniquely). That demonstrates that $$n$$ is a product of primes, so $$n=pm$$ for some prime $$p, m =n/p \lt n$$. Because $$n/p$$ can be factored into primes uniquely, it follows that the prime factorization of $$n$$ is likewise unique, contradicting our original assumption that $$n$$ cannot be uniquely factored into primes.
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2019-12-07 09:23:24
|
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https://www.investopedia.com/terms/c/cumulativereturn.asp
|
• General
• Investing/Trading
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• Reviews & Ratings
• Your Practice
• Wealth Management
• Popular Courses
• Courses by Topic
# Cumulative Return
## What Is Cumulative Return?
A cumulative return on an investment is the aggregate amount that the investment has gained or lost over time, independent of the amount of time involved. The cumulative return is expressed as a percentage, and it is the raw mathematical return of the following calculation:
$\frac{(Current\ Price \ of \ Security) - (Original \ Price \ of \ Security)}{Original \ Price \ of \ Security}$
### Key Takeaways
• The cumulative return is the total change in the investment price over a set time—an aggregate return, not an annualized one.
• Reinvesting the dividends or capital gains of an investment impacts its cumulative return.
• Cumulative return figures for ETFs and mutual funds typically omit the impact of annual expense ratios and other fees on the fund's performance.
• Taxes can also substantially reduce the cumulative returns for most investments unless they are held in tax-advantaged accounts.
## Understanding Cumulative Return
The cumulative return of an asset that does not have interest or dividends is easily calculated by figuring out the amount of profit or loss over the original price. That can work well with assets like precious metals and growth stocks that do not issue dividends. In these cases, one can use the raw closing price to calculate the cumulative return.
On the other hand, the adjusted closing price provides a simple way to calculate the cumulative return of all assets. That includes assets like interest-bearing bonds and dividend-paying stocks. The adjusted closing price incorporates the impact of interest, dividends, stock splits, and other changes on the asset price. So, it is possible to obtain the cumulative return by using the first adjusted closing price as the original price of the security.
The cumulative return usually grows over time, so it tends to make older stocks and funds look impressive. It follows that the cumulative return is not a good way to compare investments unless they launched at the same time.
## Special Considerations
### Mutual Funds and ETFs
A common way to present mutual fund or exchange traded fund (ETF) performance over time is to show the cumulative return with a visual, such as a mountain graph. Investors should check to confirm whether interest or dividends are included in the cumulative return. The marketing materials or information accompanying an illustration typically provide this information. Such payouts might be counted as reinvested or simply added as raw dollars when calculating the cumulative return.
One notable difference between mutual funds and stocks is mutual funds sometimes distribute capital gains to the fund holders. This distribution usually comes at the end of a calendar year. It consists of the profits the portfolio managers made when closing out holdings. Mutual fund owners can reinvest those capital gains, which can make calculating the cumulative return more difficult.
### Advertisements
Many advertisements use the cumulative return to make investments look impressive. While these results are often basically accurate, they can be exaggerated or distorted to encourage greed or fear. For example, someone might sight Amazon's cumulative return of over 100,000% between its initial public offering (IPO) in 1997 and 2020. However, many other technology-related companies had IPOs in the late 1990s, and most of them never came close to Amazon's returns. Furthermore, investors would have had to continue holding the stock through a bear market that reduced its value by over 90% during 2000 and 2001.
Precious metals are another area where investors need to look carefully at advertisements using total returns. Crucially, ads for bullion are not governed by the same regulations as mutual funds and ETFs. Furthermore, these cumulative returns typically do not subtract storage costs or insurance fees, which are services that many investors demand. While precious metals ETF fees are generally lower, they also need to be deducted from returns for the commodity to obtain the cumulative return that investors actually received.
### Taxes
Taxes can also substantially reduce the cumulative returns for most investments unless they are held in tax-advantaged accounts. Taxes are a particular issue for bonds because of their relatively low returns and the unfavorable tax treatment of interest payments. However, municipal bonds are often tax-exempt, so cumulative return figures require less adjustment.
Long-term stock investments enjoy the advantage of paying a relatively low capital gains tax, which is also usually easy to subtract from cumulative returns. The tax treatment of dividends is a much more complicated subject. However, it can also influence cumulative returns when funds reinvest dividends.
### Compound Return
Along with the cumulative return, an ETF or other fund usually indicates its compound return. Unlike the cumulative return, the compound return figure is annualized. Cumulative returns may seem more impressive than the annualized rate of return, which is usually smaller. However, they typically omit the effect of the annual expenses on the returns an investor will receive. Annual charges an investor can expect include fund expense ratios, interest rates on loans, and management fees. When worked out on a cumulative basis, these fees can substantially eat into cumulative return numbers.
## Example of Cumulative Return
For example, suppose investing $10,000 in XYZ Widgets Company's stock for a 10-year period results in$48,000. With no taxes and no dividends reinvested, that is a cumulative return of 380%.
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2022-01-20 20:48:20
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https://astronomy.stackexchange.com/tags/spectral-type/hot?filter=year
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# Tag Info
## Hot answers tagged spectral-type
8
Your question may ulitmately be about the physiology of the eye, which is off-topic here. The spectrum of the Sun seen low on the horizon is quite different to the spectrum of an M-type red dwarf. The reason that a red dwarf is red, is not just that it is cool, but that there are great chunks of the spectrum that are absorbed by molecules in the photosphere ...
8
There isn't a one-to-one relationship between spectral type and absolute magnitude. Instead, there is a mean relationship with a fair bit of scatter around it. The reason is that the luminosity of a star of a given effective temperature depends on its composition/metallicity and how far along in its main sequence lifetime it is. Basically, late B-type main ...
6
This is what Wikipedia says about it: When the MKK classification scheme was first described in 1943, the only subtypes of class O used were O5 to O9.5. The MKK scheme was extended to O9.7 in 1971 and O4 in 1978, and new classification schemes that add types O2, O3, and O3.5 have subsequently been introduced. It references the paper A New Spectral ...
6
A star with magnitude 0 would be 85 times brighter than the sun (since Magnitude=-2.5 log(Luminiosity)) Referring to the H-R diagram on Wikipedia shows that there is quite a range of spectral types possible with this luminosity: from B type main sequence stars, and A type sub-giants, such as 4 Sco There are also G and K and M type Giant stars with this ...
4
The following table gives the mass, radius, temperature, and luminosity of an average star of several selected spectral types: \begin{array} {d|c|c|} \text{Spectral Type} & \text{Mass} (\odot) & \text{Radius} (\odot) & \text{Temperature (K)} & \text{Luminosity} {(\odot)} \\ \hline \text{M8V} & \text{0.082} & \text{0.111} & \...
4
It isn't. You've just got dodgy table from wikipedia. A more modern (and well-used) version is here. It lists G1V 1.07 5880 G2V 1.02 5770 G3V 1.00 5720 This is an average relationship. The closest and most consistent relationship will be between spectral type and effective temperature and indeed the Sun is normally attributed a spectral class of G2V and $... 3 If present, an a (or b or ab) do not refer to spectral peculiarities but are part of the luminosity class definition explained further up on the page. Occasionally, letters a and b are applied to luminosity classes other than supergiants; for example, a giant star slightly less luminous than typical may be given a luminosity class of IIIb, while a ... 3 The spectral classes (O, B, A, F, G, K, M) and their 10 subtypes (0 to 9) were initially meant only as differentiators of spectral type. Annie Jump Cannon was the creator of this system. Through her work for/with Edward Pickering, she ended up classifying nearly a third of a million stars over a few decades. She (and many others) did not realize that this ... 3 There are no exact boundaries in temperature, luminosity, surface gravity etc. for spectral classes because the classification system works in a different way - it is fundamentally an empirical system, with classification based only on the appearance of features in the spectra. The Yerkes or Morgan-Keenan (MK) system is based only on a set of standard stars ... 2 Stars born together in clusters have more-or-less the same age. As a rule of thumb, any spread in age, measured in millions of years, is smaller than the extent of the cluster in parsecs. For most stellar clusters, smaller than a few pc, the only chance of measuring age differences occurs in the first 10 million years of life. There is no evidence for age ... 1 Just to add an example to what has already been said by eshaya and Larz.Astro. Here is the spectrum of the Black Hole binary Cygnus X-1 in its hard and its soft state. The plot is taken from Gierlinski et al. 1999 . You see that the soft state consist of mostly thermal emission below 10 keV, while the hard state is dominated by non-thermal (comtonized) ... 1 Not only in X-ray astronomy (chemistry also and pretty much everything related to X-rays). If you have an X-ray spectrum, the region with photon energy > 5-10 Kev is called "hard" X-rays, less than that it is called "soft" X-rays. Wiki has a nice explanation for that (Energy Ranges): https://en.wikipedia.org/wiki/X-ray However, I find ... 1 Active galaxies are known to change state as seen by a change in slope of their X-ray and gamma-ray spectra. We say that a spectrum has become harder (or changed to its hard state) when the slope changes so that there are relatively more high energy photons, and it becomes softer when the ratio of low energy photons to high energy photons increases. The ... 1 The orbital periods, and thus years, of planets in the habitable zones of stars of different types may vary a lot, depending on how wide or narrow a star's habitable zone is and which types of stars can have habitable planets. Thus it may be possible for some habitable planets to have years tens times as long as others, possibly even hundreds of times as ... 1 The main effect would be the radiation environment. A planet in the habitable zone of an M-dwarf would likely be subject to far more ultra-violet and X-ray observation for longer than a planet orbiting a G-dwarf of similar overall age. The reason for this lies in the physics of stellar dynamos that power the magnetism of cool stars. Fast rotating cool stars ... 1 The absolute magnitude quantifies the luminosity of an object at a standard distance of$10\ \mathrm{pc}$from Earth. For example, in the case you mentioned, Vega becomes dimmer than at its actual distance (about$7\ \mathrm{pc}\$). To answer your question, I don't think there is an actual star with exactly 0 absolute magnitude. If there is then, following ...
1
I don't believe that O0 is a real classification(see this chart), but if it were following the temperature steps it would probably be around 80,000-90,000 degrees Kelvin. The hottest star we know of is WR 102, which is 210,000 degrees Kelvin, and that is much, much hotter than my predicted O0 temperature. So the short answer is yes, O0 temperature stars ...
1
Yes, the spectral type changes with age. The spectral type is a function of temperature, gravity and chemical composition at the photosphere. All of these can change during a star's life. A star spends most of its life on the main sequence and changes in temperature and gravity are relatively slow. But thereafter there are comparatively rapid changes. For ...
1
Usually, the spectral classification is done analyzing the spectra of the star, we measure the depths of several key lines and compare them, each spectral type has his own characteristic relative intensities etc. Is not a perfect boundary and some times the stars bounce from type to type depending on who classified it and what data is available (is not the ...
Only top voted, non community-wiki answers of a minimum length are eligible
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2021-04-10 14:16:40
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http://math.columbia.edu/~dejong/wordpress/?m=201606
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# Limit preserving and finite presentation
Let A —> B be a ring map. Let R = colim R_i be a filtered colimit of A-algebras. Then there is a canonical map
colim Hom_A(B, R_i) ——> Hom_A(B, R)
By Tag 00QO the following are equivalent
1. A —> B is of finite presentation,
2. the map above is bijective for all R = colim R_i
3. the map above is surjective for all R = colim R_i
Let S be a scheme. Let X be a scheme over S. Let T = lim T_i be a directed limit of affine schemes over S. Then there is a canonical map
colim Mor_S(T_i, X) ——> Mor_S(T, X)
By Tag 01ZC and Tag 0CM0 the following are equivalent
1. X —> S is locally of finite presentation,
2. the map above is bijective for all T = lim T_i
3. the map above is surjective for all T = lim T_i
The same thing is true if X and S are algebraic spaces (Tag 04AK and Tag 0CM6).
I didn’t know you could replace bijectivity by surjectivity in the criterion. But somewhere in the Stacks project we used this fact without proof, so it had better be true, right?
A related result is that to check a morphism f of algebraic stacks is locally of finite presentation, you need only check f is limit preserving on objects (this is the analogue of the above and it says that certain functors are essentially surjective). You can find this in Tag 0CMQ.
Caveat: as this only applies to situations where you already know your functors (or stacks in groupoids) are algebraic spaces (or stacks), it probably won’t be that useful. Often when we try to show a stack is limit preserving, it is part of applying Artin’s criteria and then we don’t yet know our stack is algebraic of course.
Thanks for reading!
[Edit on 6/30/2016: Matthew Emerton just pointed out that this observation was already in Lemma 2.3.15 of his paper with Toby Gee. I must have read it and then forgotten that I had. Apologies to everybody.]
# Up to date again
Worked through your comments online as well as those emailed to me. Please let me know if I forgot to do something. Thanks for all your help!
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2017-08-17 09:44:51
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https://www.onestitchmatters.co.uk/shop/metal-thread-goldwork-supplies/gold-work-thread-20-threads/gilt-smooth-purl-08/
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# Gilt Smooth Purl 08
£2.80
price is per 1metre. This is gilt so there is a very small percentage of gold content in this product. it has shiny textured finish.
To use, cut the desired length and couch down with a pair of waxed sewing thread. This smooth purl is used for cutwork. you can mix this smooth purl with rough purl, bright check or wire check for your cutwork.
This smooth purl is really delicate so extra care will be needed to handle with.
This thread is going to be sent to you in acid free glassine bag to keep the quality.
Gilt Smooth Purl 08
£2.80
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2021-06-12 23:41:00
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https://docs.microsoft.com/en-us/dotnet/api/microsoft.ml.transforms.timeseries.timeseriespredictionengine-2?view=ml-dotnet
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# TimeSeriesPredictionEngine<TSrc,TDst> Class
## Definition
A class that runs the previously trained model (and the preceding transform pipeline) on the in-memory data, one example at a time. This can also be used with trained pipelines that do not end with a predictor: in this case, the 'prediction' will be just the outcome of all the transformations.
public sealed class TimeSeriesPredictionEngine<TSrc,TDst> : Microsoft.ML.PredictionEngineBase<TSrc,TDst> where TSrc : class where TDst : class, new()
type TimeSeriesPredictionEngine<'Src, 'Dst (requires 'Src : null and 'Dst : null and 'Dst : (new : unit -> 'Dst))> = class
inherit PredictionEngineBase<'Src, 'Dst (requires 'Src : null and 'Dst : null and 'Dst : (new : unit -> 'Dst))>
Public NotInheritable Class TimeSeriesPredictionEngine(Of TSrc, TDst)
Inherits PredictionEngineBase(Of TSrc, TDst)
#### Type Parameters
TSrc
The user-defined type that holds the example.
TDst
The user-defined type that holds the prediction.
Inheritance
TimeSeriesPredictionEngine<TSrc,TDst>
## Constructors
Contructor for creating time series specific prediction engine. It allows the time series model to be updated with the observations seen at prediction time via CheckPoint(IHostEnvironment, String)
## Properties
Provides output schema. (Inherited from PredictionEngineBase)
## Methods
Checkpoints TimeSeriesPredictionEngine to a Stream with the updated state. Checkpoints TimeSeriesPredictionEngine to disk with the updated state. (Inherited from PredictionEngineBase) Forecasting only task. Run prediction pipeline on one example. (Inherited from PredictionEngineBase) Performs prediction. In the case of forecasting only task example can be left as null. If example is not null then it could be used to update forecasting models with new obervation. Performs prediction. In the case of forecasting only task example can be left as null. If example is not null then it could be used to update forecasting models with new obervation. For anomaly detection the model is always updated with example. Performs prediction. In the case of forecasting only task example can be left as null. If example is not null then it could be used to update forecasting models with new obervation. For anomaly detection the model is always updated with example.
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2021-09-25 08:27:42
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http://mathhelpforum.com/differential-geometry/224065-show-set-measurable.html
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# Math Help - show that set is measurable
1. ## show that set is measurable
Hi all,
Given a set E; for each open, bounded interval (a,b):
b-a = m*((a,b)∩E) + m*((a,b)~E)
implies E is (Lebesgue) measurable.
(we're working with the real line here, so E is a subset of R)
I'm having trouble with this problem. Hint given is to show that A = { E: b-a = m*((a,b)∩E) + m*((a,b)~E) } is a sigma algebra.
I understand the hint logic - if A is a sigma algebra then the sets in A are measurable sets so E must be measurable.
I know that all open, bounded intervals are measurable.
I know the three requirements for a collection of sets to be a sigma algebra.
I don't know how to connect the dots though.
Where do I start? Thanks
2. ## Re: show that set is measurable
Here is the definition of meausrability of a set E given in my text:
A set E is said to be measurable provided for any set A, m*(A) = m*(A∩E) + m*(A∩(complement of E))
It looks like (A∩(complement of E)) = (A~E) or (A\E) (just different notation)
So, the equation in the original problem looks very close to what I have in the definition of measurable set. Except I can't say it holds for any set A, only if A=(a,b).
If I were to restrict the measure space (correct term?) to one that contains the just open sets, then E would be measurable?
The complement of an measurable open set is a measurable closed set... What if I add these in?
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2015-03-06 21:17:22
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https://math.stackexchange.com/questions/2339208/looking-for-a-solution-for-the-following-non-linear-optimization-problem
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# Looking for a solution for the following non linear optimization problem
I've previously solve optimization problems with linear and mixed integer linear programming with Simplex Algorithm.
Now I've an objective function $F=f(x_1, x_2, .. x_n)$ to be minimized, but F is not lineare and its value is computed by an external tool for each combination of $x_1, x_2, ... x_n$.
The decision variables ($x_1, x_2, ... x_n$) are subjected to constraints (inequality and equality linear constraints).
All efficient optimization algorithms require the gradient of the objective function but I cannot calculate it.
Is there a way to search for optimum values (either local or absolute) without evaluating F for each combination of the decision variables?
• You can't calculate the gradient at any point? Unless your function is extremely messy, you approximate the derivative if it does exist by "the limit definition". You can merge this with any general active constraints method.
– Koto
Jun 28 '17 at 14:55
• If you can calculate $f(x_1,\dots,x_n)$ and $f(x_1,\dots,x_i+\delta,\dots, x_n)$, you can get an approximation for $\partial_i f(\textbf{x})$ by letting $\delta\rightarrow 0$.
– Koto
Jun 28 '17 at 15:01
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2021-09-19 05:32:43
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https://www.shaalaa.com/question-bank-solutions/a-bag-contains-5-black-6-red-balls-determine-number-ways-which-2-black-3-red-balls-can-be-selected-combination_54175
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# A Bag Contains 5 Black and 6 Red Balls. Determine the Number of Ways in Which 2 Black and 3 Red Balls Can Be Selected. - Mathematics
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
#### Solution
2 black and 3 red balls are to be selected from 5 black and 6 red balls.
Required number of ways =${}^5 C_2 \times^6 C_3 = \frac{5}{2} \times \frac{4}{1} \times \frac{6}{3} \times \frac{5}{2} \times \frac{4}{1} = 200$
Concept: Combination
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 11 Mathematics Textbook
Chapter 17 Combinations
Exercise 17.2 | Q 28 | Page 17
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2021-07-24 23:44:25
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http://tex.loria.fr/texlive-htmldoc/auctex/auc-tex_5.html
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# Formatting and Printing
The most powerful features of AUC TeX may be those allowing you to run (La)TeX and other external commands like BibTeX and makeindex from within Emacs, viewing and printing the results, and moreover allowing you to debug your documents.
## Executing Commands
Formatting the document with TeX or LaTeX, viewing with a previewer, printing the document, running BibTeX, making an index, or checking the document with lacheck or chktex all require running an external command.
There are two ways to run an external command, you can either run it on all of the current documents with TeX-command-master, or on the current region with TeX-command-region.
Command: TeX-command-master
(C-c C-c) Query the user for a command, and run it on the master file associated with the current buffer. The name of the master file is controlled by the variable TeX-master. The available commands are controlled by the variable TeX-command-list.
See section Installation of AUC TeX for a discussion about TeX-command-list and section Multifile Documents for a discussion about TeX-master.
Command: TeX-command-region
(C-c C-r) Query the user for a command, and run it on the "region file". Some commands (typically those invoking TeX or LaTeX) will write the current region into the region file, after extracting the header and tailer from the master file. If mark is not active, use the old region. The name of the region file is controlled by the variable TeX-region. The name of the master file is controlled by the variable TeX-master. The header is all text up to the line matching the regular expression TeX-header-end. The trailer is all text from the line matching the regular expression TeX-trailer-start. The available commands are controlled by the variable TeX-command-list.
AUC TeX will allow one process for each document, plus one process for the region file to be active at the same time. Thus, if you are editing n different documents, you can have n plus one processes running at the same time. If the last process you started was on the region, the commands described in section Catching the errors and section Controlling the output will work on that process, otherwise they will work on the process associated with the current document.
User Option: TeX-region
The name of the file for temporarily storing the text when formatting the current region.
A regular expression matching the end of the header. By default, this is \begin{document}' in LaTeX mode and %**end of header' in TeX mode.
User Option: TeX-trailer-start
A regular expression matching the start of the trailer. By default, this is \end{document}' in LaTeX mode and \bye' in TeX mode.
AUC TeX will try to guess what command you want to invoke, but by default it will assume that you want to run TeX in TeX mode and LaTeX in LaTeX mode. You can overwrite this by setting the variable TeX-command-default.
User Option: TeX-command-default
The default command to run in this buffer. Must be an entry in TeX-command-list.
If you want to overwrite the values of TeX-header-end, TeX-trailer-start, or TeX-command-default, you can do that for all files by setting them in either TeX-mode-hook, plain-TeX-mode-hook, or LaTeX-mode-hook. To overwrite them for a single file, define them as file variables (see section File Variables' in The Emacs Editor). You do this by putting special formatted text near the end of the file.
% Local Variables:
% TeX-trailer-start: "% Start-Of-Trailer"
% TeX-command-default: "SliTeX"
% End:
AUC TeX will try to save any buffers related to the document, and check if the document needs to be reformatted. If the variable TeX-save-query is non-nil, AUC TeX will query before saving each file. By default AUC TeX will check emacs buffers associated with files in the current directory, in one of the TeX-macro-private directories, and in the TeX-macro-global directories. You can change this by setting the variable TeX-check-path.
User Option: TeX-check-path
Directory path to search for dependencies.
If nil, just check the current file. Used when checking if any files have changed.
## Catching the errors
Once you've formatted your document you may debug' it, i.e. browse through the errors (La)TeX reported.
Command: TeX-next-error
(C-c ) Go to the next error reported by TeX. The view will be split in two, with the cursor placed as close as possible to the error in the top view. In the bottom view, the error message will be displayed along with some explanatory text.
Normally AUC TeX will only report real errors, but you may as well ask it to report bad boxes' as well.
(C-c C-w) Toggle whether AUC TeX should stop at bad boxes (i.e. over/under full boxes) as well as at normal errors.
As default, AUC TeX will display that special *help*' buffer containing the error reported by TeX along with the documentation. There is however an expert' option, which allows you to display the real TeX output.
User Option: TeX-display-help
When non-nil AUC TeX will automatically display a help text whenever an error is encountered using TeX-next-error (C-c ).
## Checking for problems
Running TeX or LaTeX will only find regular errors in the document, not examples of bad style. Furthermore, description of the errors may often be confusing. The utility lacheck can be used to find style errors, such as forgetting to escape the space after an abbreviation or using ...' instead of \ldots' and many other problems like that. You start lacheck with C-c C-c C h e c k RET. The result will be a list of errors in the *compilation*' buffer. You can go through the errors with C-x (next-error, see section Compilation' in The Emacs Editor), which will move point to the location of the next error.
Another newer program which can be used to find errors is chktex. It is much more configurable than lacheck, but doesn't find all the problems lacheck does, at least in its default configuration. You must install the programs before using them, and for chktex you must also modify TeX-command-list. You can get lacheck from <URL:ftp://sunsite.auc.dk/pub/text/lacheck/>' or alternatively chktex from <URL:ftp://ftp.dante.de/pub/tex/support/chktex/>'. Search for chktex' in tex.el' to see how to switch between them.They are
## Controlling the output
A number of commands are available for controlling the output of an application running under AUC TeX
Command: TeX-kill-job
(C-c C-k) Kill currently running external application. This may be either of TeX, LaTeX, previewer BibTeX etc.
Command: TeX-recenter-output-buffer
(C-c C-l) Recenter the output buffer so that the bottom line is visible.
Command: TeX-home-buffer
(C-c ^) Go to the `master' file in the document associated with the current buffer, or if already there, to the file where the current process was started.
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2017-11-19 10:30:45
|
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https://tobykingsman.wordpress.com/2016/03/15/optimal-search-problems/
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# Optimal Search Problems
One active area of research within STOR-i is the problem of how to optimally search for something or someone. This is a problem of practical importance with it being used for many real world applications. Bomb disposal squads need to search for unexploded ordinance, search parties want to look for survivors of disasters and the police want to find people on the run from the law.
Like most things the best way to build intuition to a problem is to start by constructing smaller toy models of the system. We can represent the places where an object might be hidden as arcs in a network which are connected to each other with nodes. This type of structure is deliberately similar to that of a road system, the arcs represent streets and the nodes are the street corners. The object that we want to find will be hidden along one of the streets and the job of those searching is to find it as quickly as possible.
Example Network for Optimal Searches
The $p_i$‘s are the probabilities of detecting the object along the arc i. These are prior probabilities, what you think the probability of the object being hidden there are before you start. The arrows denote the direction that the searcher can travel along the arc, you can go in either direction or in some cases only one. This reflects the geography of the search space in the problem, not all areas are accessible in multiple directions like a one-way street.
The idea is that the object or person is hidden along one of these arcs and there is also a person trying to find them. This searcher wants to move along the arcs in such a way along the directed arcs such that it minimises the expected time to discover the object.
### Speed vs Accuracy
A more realistic extension to this simple model is to take into account the fact that we can search at different speeds and accuracies. There is clearly a trade-off to be had between how quickly you search and how carefully you search. Consider looking for survivors after an air crash. You might be able to cover vast distances in an aeroplane, but the chances of seeing any people on the ground are quite small. On the other hand sending a ground team to check out the area will be very likely to find survivors if they are there, but it make time them a very long time to move across a small area.
Patrol boats can vary speeds
The simplest way of implementing this into our model would be look at two speeds of a patrol boat, fast and slow. When the boat is moving slowly it has a higher chance of detection than when it is travelling quickly, $q_i^F < q_i^S$. Conversely the time that it takes to patrol a region is shorter when going quickly so $t_i^F < t_i^S$.
This means that every arc now has a choice of two speeds and a probability of detection for each speed. Of course in reality speed and detection’s will be continuous distributions rather than the discrete values we are using.
Now a choice of two actions: fast and slow
### Multi-Armed Bandit Approach
One way of looking at solving this problem is to notice its similarities with the Multi-Armed Bandit problem. In this case we have a trade-off between speed and accuracy whilst wanting to minimise the time to find something, whereas in the classic Multi-Armed Bandit problem we have a balance between exploration and exploitation.
There are however a couple of factors that have to be considered that don’t naturally fit into the MAB framework.
• The choice of arms to play is restricted
In classic MAB problems you can play any bandit that you want to. In our case the pattern of the network limits your choice. For example if you’ve just played $p_3$ in the network above, you can only play $p_4$ or $p_7$ next.
• Each arm has two options for playing it
Normally bandits only have a single arm to pull and a single payoff. In the search problem we not only have to decide whether to play a particular arm or not, but which speed to select as well. This is known as a super-process.
Matters are even more complicated if you are searching for someone who doesn’t want to be found. In this case you can be sure that adversary will try to hide in a hard to find location in order to maximise the time it takes for the searcher to find them. This leads the problem into some other aspects of mathematics such as game theory.
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2018-01-17 02:46:32
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http://www.acmerblog.com/POJ-1328-Radar-Installation-blog-368.html
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2013
11-09
# POJ 1328 Radar Installation [解题报告] Java
Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.
We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.
Figure A Sample Input of Radar Installations
The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases.
The input is terminated by a line containing pair of zeros
For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. “-1″ installation means no solution for that case.
3 2
1 2
-3 1
2 1
1 2
0 2
0 0
Case 1: 2
Case 2: 1
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Scanner;
/**
*
* @author 灏�
*
* 2010-6-12 涓��01:48:35
*/
public class Main {
static class Range implements Comparable{
double left,right;
public Range(double left,double right){
this.left = left;
this.right = right;
}
@Override
public int compareTo(Range range) {
if(range.left == left){
return ((Double)right).compareTo((Double)(range.right));
}else{
return ((Double)left).compareTo((Double)(range.left));
}
}
@Override
public String toString() {
return "(" + left + "," + right + ")";
}
}
public static void main(String[] args) {
Scanner scn = new Scanner(System.in);
PrintWriter out = new PrintWriter(System.out);
int n ,d,x,y,num;
double dx;
Range[] ranges;
int index = 0;
while(true){
num = 0;
n = scn.nextInt();
d = scn.nextInt();
if(n == 0){
break;
}
ranges = new Range[n];
for(int i = 0; i < n; i++){
x = scn.nextInt();
y = scn.nextInt();
if(y > d){
num = -1;
}
dx = Math.sqrt(d*d - y*y);
ranges[i] = new Range(x - dx, x + dx);
}
Arrays.sort(ranges);//���
if(num != -1){
num = calute(ranges);
}
out.format("Case %d: %d\n",++index,num);
}
out.flush();
}
private static int calute(Range[] ranges) {
int num = 1;
int n = ranges.length;
Range preRange = ranges[0],range;
for(int i = 1; i < n; i++){
range = ranges[i];
//姹���翠氦��
if(range.left >= preRange.left && range.left <= preRange.right){
preRange.left = range.left;
if(range.right < preRange.right){
preRange.right = range.right;
}
}else{
num++;
preRange = range;
}
}
return num;
}
}
1. 我还有个问题想请教一下，就是感觉对于新手来说，递归理解起来有些困难，不知有没有什么好的方法或者什么好的建议？
2. Good task for the group. Hold it up for every yeara??s winner. This is a excellent oppotunity for a lot more enhancement. Indeed, obtaining far better and much better is constantly the crucial. Just like my pal suggests on the truth about ab muscles, he just keeps obtaining much better.
3. I like your publish. It is great to see you verbalize from the coronary heart and clarity on this essential subject matter can be easily noticed.
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2017-01-20 05:49:00
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https://inquiryintoinquiry.com/2014/06/11/peirces-1870-logic-of-relatives-%E2%80%A2-comment-12-2/
|
## Peirce’s 1870 “Logic Of Relatives” • Comment 12.2
Let us make a few preliminary observations about the operation of logical involution, as Peirce introduces it here:
I shall take involution in such a sense that $x^y$ will denote everything which is an $x$ for every individual of $y.$ Thus $\mathit{l}^\mathrm{w}$ will be a lover of every woman.
(Peirce, CP 3.77)
In ordinary arithmetic the involution $x^y,$ or the exponentiation of $x$ to the power $y,$ is the repeated application of the multiplier $x$ for as many times as there are ones making up the exponent $y.$
In analogous fashion, the logical involution $\mathit{l}^\mathrm{w}$ is the repeated application of the term $\mathit{l}$ for as many times as there are individuals under the term $\mathrm{w}.$ According to Peirce’s interpretive rules, the repeated applications of the base term $\mathit{l}$ are distributed across the individuals of the exponent term $\mathrm{w}.$ In particular, the base term $\mathit{l}$ is not applied successively in the manner that would give something like “a lover of a lover of … a lover of a woman”.
For example, suppose that a universe of discourse numbers among its elements just three women, $\mathrm{W}^{\prime}, \mathrm{W}^{\prime\prime}, \mathrm{W}^{\prime\prime\prime}.$ This could be expressed in Peirce’s notation by writing:
$\mathrm{w} ~=~ \mathrm{W}^{\prime} ~+\!\!,~ \mathrm{W}^{\prime\prime} ~+\!\!,~ \mathrm{W}^{\prime\prime\prime}$
Under these circumstances the following equation would hold:
$\mathit{l}^\mathrm{w} ~=~ \mathit{l}^{(\mathrm{W}^{\prime} ~+\!\!,~ \mathrm{W}^{\prime\prime} ~+\!\!,~ \mathrm{W}^{\prime\prime\prime})} ~=~ (\mathit{l}\mathrm{W}^{\prime}), (\mathit{l}\mathrm{W}^{\prime\prime}), (\mathit{l}\mathrm{W}^{\prime\prime\prime})$
This says that a lover of every woman in the given universe of discourse is a lover of $\mathrm{W}^{\prime}$ that is a lover of $\mathrm{W}^{\prime\prime}$ that is a lover of $\mathrm{W}^{\prime\prime\prime}.$ In other words, a lover of every woman in this context is a lover of $\mathrm{W}^{\prime}$ and a lover of $\mathrm{W}^{\prime\prime}$ and a lover of $\mathrm{W}^{\prime\prime\prime}.$
The denotation of the term $\mathit{l}^\mathrm{w}$ is a subset of $X$ that can be obtained as follows: For each flag of the form $L \star x$ with $x \in W,$ collect the elements $\mathrm{proj}_1 (L \star x)$ that appear as the first components of these ordered pairs, and then take the intersection of all these subsets. Putting it all together:
$\displaystyle \mathit{l}^\mathrm{w} ~=~ \bigcap_{x \in W} \mathrm{proj}_1 (L \star x) ~=~ \bigcap_{x \in W} L \cdot x$
It is very instructive to examine the matrix representation of $\mathit{l}^\mathrm{w}$ at this point, not the least because it effectively dispels the mystery of the name involution. First, let us make the following observation. To say that $j$ is a lover of every woman is to say that $j$ loves $k$ if $k$ is a woman. This can be rendered in symbols as follows:
$j ~\text{loves}~ k ~\Leftarrow~ k ~\text{is a woman}$
Reading the formula $\mathit{l}^\mathrm{w}$ as “$j$ loves $k$ if $k$ is a woman” highlights the operation of converse implication inherent in it, and this in turn reveals the analogy between implication and involution that accounts for the aptness of the latter name.
The operations defined by the formulas $x^y = z$ and $(x\!\Leftarrow\!y) = z$ for $x, y, z$ in the boolean domain $\mathbb{B} = \{ 0, 1 \}$ are tabulated as follows:
$\begin{array}{ccc} x^y & = & z \\ \hline 0^0 & = & 1 \\ 0^1 & = & 0 \\ 1^0 & = & 1 \\ 1^1 & = & 1 \end{array} \qquad\qquad\qquad \begin{array}{ccc} x\!\Leftarrow\!y & = & z \\ \hline 0\!\Leftarrow\!0 & = & 1 \\ 0\!\Leftarrow\!1 & = & 0 \\ 1\!\Leftarrow\!0 & = & 1 \\ 1\!\Leftarrow\!1 & = & 1 \end{array}$
It is clear that these operations are isomorphic, amounting to the same operation of type $\mathbb{B} \times \mathbb{B} \to \mathbb{B}.$ All that remains is to see how this operation on coefficient values in $\mathbb{B}$ induces the corresponding operations on sets and terms.
The term $\mathit{l}^\mathrm{w}$ determines a selection of individuals from the universe of discourse $X$ that may be computed by means of the corresponding operation on coefficient matrices. If the terms $\mathit{l}$ and $\mathrm{w}$ are represented by the matrices $\mathsf{L} = \mathrm{Mat}(\mathit{l})$ and $\mathsf{W} = \mathrm{Mat}(\mathrm{w}),$ respectively, then the operation on terms that produces the term $\mathit{l}^\mathrm{w}$ must be represented by a corresponding operation on matrices, say, $\mathsf{L}^\mathsf{W} = \mathrm{Mat}(\mathit{l})^{\mathrm{Mat}(\mathrm{w})},$ that produces the matrix $\mathrm{Mat}(\mathit{l}^\mathrm{w}).$ In other words, the involution operation on matrices must be defined in such a way that the following equations hold:
$\mathsf{L}^\mathsf{W} ~=~ \mathrm{Mat}(\mathit{l})^{\mathrm{Mat}(\mathrm{w})} ~=~ \mathrm{Mat}(\mathit{l}^\mathrm{w})$
The fact that $\mathit{l}^\mathrm{w}$ denotes the elements of a subset of $X$ means that the matrix $\mathsf{L}^\mathsf{W}$ is a 1-dimensional array of coefficients in $\mathbb{B}$ that is indexed by the elements of $X.$ The value of the matrix $\mathsf{L}^\mathsf{W}$ at the index $u \in X$ is written $(\mathsf{L}^\mathsf{W})_u$ and computed as follows:
$\displaystyle (\mathsf{L}^\mathsf{W})_u ~=~ \prod_{v \in X} \mathsf{L}_{uv}^{\mathsf{W}_v}$
This entry was posted in Graph Theory, Logic, Logic of Relatives, Logical Graphs, Mathematics, Peirce, Relation Theory, Semiotics and tagged , , , , , , , . Bookmark the permalink.
### 1 Response to Peirce’s 1870 “Logic Of Relatives” • Comment 12.2
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2019-09-19 00:45:38
|
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https://www.physicsforums.com/threads/dynamics-pulley-and-rod.720876/
|
# Dynamics - Pulley and Rod
1. Nov 4, 2013
### jules.t
1. The problem statement, all variables and given/known data
There is pulley which has a mass hanging from one side, and a rotating rod attached horizontally to the other (both by cables). The pulley has mass; Mp = 9.5kg and radius; Rp =0.2m. The block(A) has mass; Ma=10.9kg. The rod is L = 0.8m long and has mass=Ml7.9kg. The rod is also rotating at w= 2.7 rad/s in the anti clockwise direction. The value for gravity used is g= 9.8m/s/s downwards. Tensions are denoted by Ta and Tb
2. Relevant equations
Newtons 2nd Law(F=ma), Eulers Equation (M=I$\alpha$).
3. The attempt at a solution
I have defined my coordinate system as anticlockwise is positive and vertically down is positive.
Ipulley = 0.5MR2
Irod at centre of grav =(1/12)ML2
I have developed some simultaneous equations for the system:
Ta - Mag = Maaa From Newton 2nd Law of Block(A)
aa = Rp$\alpha$p Acceleration relationship
-TaRp+TbRp=Ip$\alpha$p Eulers equation for pulley
Tb - Mrg = Mrar Newtons second law on rod.
I also have the following(which i think is where the problem is at):
-TbL - Ir$\alpha$r = -Lw2 Eulers equation on the rod at the centre of gravity.
ar = -L$\alpha$r
I solved this system, and got an incorrect answer. As above i think the problem lies in the last two equations.
Last edited: Nov 5, 2013
2. Nov 4, 2013
### Simon Bridge
Welcome to PF;
OK - rod rotating anticlockwise means the pulley is rotating clockwise and the mass is moving up.
The most likely location for an error is a misplaced minus sign as you link the different fbds.
The signs in your equations don't seem to match your stated sign convention.
i.e. $T_a-M_ag=M_aa_a$ gives a negative acceleration if the weight is greater than the tension.
But you've set them up how I would except for the last one, I'd have picked anticlockwise for positive for the motion of the rod.... done that way the acceleration (magnitude and sign) of the block A is also the tangential acceleration of the rim of the pulley and the acceleration of point B on the rod.
However - the problem statement lacks a goal.
What are you supposed to be finding out?
3. Nov 4, 2013
### jules.t
My apologies for not including the problems goal.
The problem asks to find the acceleration of pint B in the diagram provided.
4. Nov 5, 2013
### haruspex
I too am confused about the signs.
makes up +ve
makes down and anticlockwise the same sign
makes clockwise +ve
makes up +ve
Makes anticlockwise +ve and ... a couple of problems here.
The rod is rotating about an endpoint so you need to use the parallel axis theorem.
The Lw2 term is a centripetal acceleration, which at this point is horizontal. It is supplied by the axle and has nothing to do with the tension.
5. Nov 5, 2013
### jules.t
Thanks guys!
By working through the sign troubles and getting rid of the Lw^2 term, the solution was achieved. Your help is appreciated greatly, and i guess ill need to work on sign conventions before my exam.
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2017-11-24 01:23:28
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https://people.maths.bris.ac.uk/~matyd/GroupNames/321/C6xC7sD4.html
|
Copied to
clipboard
## G = C6×C7⋊D4order 336 = 24·3·7
### Direct product of C6 and C7⋊D4
Series: Derived Chief Lower central Upper central
Derived series C1 — C14 — C6×C7⋊D4
Chief series C1 — C7 — C14 — C42 — C6×D7 — C2×C6×D7 — C6×C7⋊D4
Lower central C7 — C14 — C6×C7⋊D4
Upper central C1 — C2×C6 — C22×C6
Generators and relations for C6×C7⋊D4
G = < a,b,c,d | a6=b7=c4=d2=1, ab=ba, ac=ca, ad=da, cbc-1=dbd=b-1, dcd=c-1 >
Subgroups: 368 in 108 conjugacy classes, 54 normal (22 characteristic)
C1, C2, C2, C2, C3, C4, C22, C22, C22, C6, C6, C6, C7, C2×C4, D4, C23, C23, C12, C2×C6, C2×C6, C2×C6, D7, C14, C14, C14, C2×D4, C21, C2×C12, C3×D4, C22×C6, C22×C6, Dic7, D14, D14, C2×C14, C2×C14, C2×C14, C3×D7, C42, C42, C42, C6×D4, C2×Dic7, C7⋊D4, C22×D7, C22×C14, C3×Dic7, C6×D7, C6×D7, C2×C42, C2×C42, C2×C42, C2×C7⋊D4, C6×Dic7, C3×C7⋊D4, C2×C6×D7, C22×C42, C6×C7⋊D4
Quotients: C1, C2, C3, C22, C6, D4, C23, C2×C6, D7, C2×D4, C3×D4, C22×C6, D14, C3×D7, C6×D4, C7⋊D4, C22×D7, C6×D7, C2×C7⋊D4, C3×C7⋊D4, C2×C6×D7, C6×C7⋊D4
Smallest permutation representation of C6×C7⋊D4
On 168 points
Generators in S168
(1 106 29 92 15 120)(2 107 30 93 16 121)(3 108 31 94 17 122)(4 109 32 95 18 123)(5 110 33 96 19 124)(6 111 34 97 20 125)(7 112 35 98 21 126)(8 99 36 85 22 113)(9 100 37 86 23 114)(10 101 38 87 24 115)(11 102 39 88 25 116)(12 103 40 89 26 117)(13 104 41 90 27 118)(14 105 42 91 28 119)(43 148 71 134 57 162)(44 149 72 135 58 163)(45 150 73 136 59 164)(46 151 74 137 60 165)(47 152 75 138 61 166)(48 153 76 139 62 167)(49 154 77 140 63 168)(50 141 78 127 64 155)(51 142 79 128 65 156)(52 143 80 129 66 157)(53 144 81 130 67 158)(54 145 82 131 68 159)(55 146 83 132 69 160)(56 147 84 133 70 161)
(1 2 3 4 5 6 7)(8 9 10 11 12 13 14)(15 16 17 18 19 20 21)(22 23 24 25 26 27 28)(29 30 31 32 33 34 35)(36 37 38 39 40 41 42)(43 44 45 46 47 48 49)(50 51 52 53 54 55 56)(57 58 59 60 61 62 63)(64 65 66 67 68 69 70)(71 72 73 74 75 76 77)(78 79 80 81 82 83 84)(85 86 87 88 89 90 91)(92 93 94 95 96 97 98)(99 100 101 102 103 104 105)(106 107 108 109 110 111 112)(113 114 115 116 117 118 119)(120 121 122 123 124 125 126)(127 128 129 130 131 132 133)(134 135 136 137 138 139 140)(141 142 143 144 145 146 147)(148 149 150 151 152 153 154)(155 156 157 158 159 160 161)(162 163 164 165 166 167 168)
(1 43 8 50)(2 49 9 56)(3 48 10 55)(4 47 11 54)(5 46 12 53)(6 45 13 52)(7 44 14 51)(15 57 22 64)(16 63 23 70)(17 62 24 69)(18 61 25 68)(19 60 26 67)(20 59 27 66)(21 58 28 65)(29 71 36 78)(30 77 37 84)(31 76 38 83)(32 75 39 82)(33 74 40 81)(34 73 41 80)(35 72 42 79)(85 127 92 134)(86 133 93 140)(87 132 94 139)(88 131 95 138)(89 130 96 137)(90 129 97 136)(91 128 98 135)(99 141 106 148)(100 147 107 154)(101 146 108 153)(102 145 109 152)(103 144 110 151)(104 143 111 150)(105 142 112 149)(113 155 120 162)(114 161 121 168)(115 160 122 167)(116 159 123 166)(117 158 124 165)(118 157 125 164)(119 156 126 163)
(2 7)(3 6)(4 5)(9 14)(10 13)(11 12)(16 21)(17 20)(18 19)(23 28)(24 27)(25 26)(30 35)(31 34)(32 33)(37 42)(38 41)(39 40)(43 50)(44 56)(45 55)(46 54)(47 53)(48 52)(49 51)(57 64)(58 70)(59 69)(60 68)(61 67)(62 66)(63 65)(71 78)(72 84)(73 83)(74 82)(75 81)(76 80)(77 79)(86 91)(87 90)(88 89)(93 98)(94 97)(95 96)(100 105)(101 104)(102 103)(107 112)(108 111)(109 110)(114 119)(115 118)(116 117)(121 126)(122 125)(123 124)(127 134)(128 140)(129 139)(130 138)(131 137)(132 136)(133 135)(141 148)(142 154)(143 153)(144 152)(145 151)(146 150)(147 149)(155 162)(156 168)(157 167)(158 166)(159 165)(160 164)(161 163)
G:=sub<Sym(168)| (1,106,29,92,15,120)(2,107,30,93,16,121)(3,108,31,94,17,122)(4,109,32,95,18,123)(5,110,33,96,19,124)(6,111,34,97,20,125)(7,112,35,98,21,126)(8,99,36,85,22,113)(9,100,37,86,23,114)(10,101,38,87,24,115)(11,102,39,88,25,116)(12,103,40,89,26,117)(13,104,41,90,27,118)(14,105,42,91,28,119)(43,148,71,134,57,162)(44,149,72,135,58,163)(45,150,73,136,59,164)(46,151,74,137,60,165)(47,152,75,138,61,166)(48,153,76,139,62,167)(49,154,77,140,63,168)(50,141,78,127,64,155)(51,142,79,128,65,156)(52,143,80,129,66,157)(53,144,81,130,67,158)(54,145,82,131,68,159)(55,146,83,132,69,160)(56,147,84,133,70,161), (1,2,3,4,5,6,7)(8,9,10,11,12,13,14)(15,16,17,18,19,20,21)(22,23,24,25,26,27,28)(29,30,31,32,33,34,35)(36,37,38,39,40,41,42)(43,44,45,46,47,48,49)(50,51,52,53,54,55,56)(57,58,59,60,61,62,63)(64,65,66,67,68,69,70)(71,72,73,74,75,76,77)(78,79,80,81,82,83,84)(85,86,87,88,89,90,91)(92,93,94,95,96,97,98)(99,100,101,102,103,104,105)(106,107,108,109,110,111,112)(113,114,115,116,117,118,119)(120,121,122,123,124,125,126)(127,128,129,130,131,132,133)(134,135,136,137,138,139,140)(141,142,143,144,145,146,147)(148,149,150,151,152,153,154)(155,156,157,158,159,160,161)(162,163,164,165,166,167,168), (1,43,8,50)(2,49,9,56)(3,48,10,55)(4,47,11,54)(5,46,12,53)(6,45,13,52)(7,44,14,51)(15,57,22,64)(16,63,23,70)(17,62,24,69)(18,61,25,68)(19,60,26,67)(20,59,27,66)(21,58,28,65)(29,71,36,78)(30,77,37,84)(31,76,38,83)(32,75,39,82)(33,74,40,81)(34,73,41,80)(35,72,42,79)(85,127,92,134)(86,133,93,140)(87,132,94,139)(88,131,95,138)(89,130,96,137)(90,129,97,136)(91,128,98,135)(99,141,106,148)(100,147,107,154)(101,146,108,153)(102,145,109,152)(103,144,110,151)(104,143,111,150)(105,142,112,149)(113,155,120,162)(114,161,121,168)(115,160,122,167)(116,159,123,166)(117,158,124,165)(118,157,125,164)(119,156,126,163), (2,7)(3,6)(4,5)(9,14)(10,13)(11,12)(16,21)(17,20)(18,19)(23,28)(24,27)(25,26)(30,35)(31,34)(32,33)(37,42)(38,41)(39,40)(43,50)(44,56)(45,55)(46,54)(47,53)(48,52)(49,51)(57,64)(58,70)(59,69)(60,68)(61,67)(62,66)(63,65)(71,78)(72,84)(73,83)(74,82)(75,81)(76,80)(77,79)(86,91)(87,90)(88,89)(93,98)(94,97)(95,96)(100,105)(101,104)(102,103)(107,112)(108,111)(109,110)(114,119)(115,118)(116,117)(121,126)(122,125)(123,124)(127,134)(128,140)(129,139)(130,138)(131,137)(132,136)(133,135)(141,148)(142,154)(143,153)(144,152)(145,151)(146,150)(147,149)(155,162)(156,168)(157,167)(158,166)(159,165)(160,164)(161,163)>;
G:=Group( (1,106,29,92,15,120)(2,107,30,93,16,121)(3,108,31,94,17,122)(4,109,32,95,18,123)(5,110,33,96,19,124)(6,111,34,97,20,125)(7,112,35,98,21,126)(8,99,36,85,22,113)(9,100,37,86,23,114)(10,101,38,87,24,115)(11,102,39,88,25,116)(12,103,40,89,26,117)(13,104,41,90,27,118)(14,105,42,91,28,119)(43,148,71,134,57,162)(44,149,72,135,58,163)(45,150,73,136,59,164)(46,151,74,137,60,165)(47,152,75,138,61,166)(48,153,76,139,62,167)(49,154,77,140,63,168)(50,141,78,127,64,155)(51,142,79,128,65,156)(52,143,80,129,66,157)(53,144,81,130,67,158)(54,145,82,131,68,159)(55,146,83,132,69,160)(56,147,84,133,70,161), (1,2,3,4,5,6,7)(8,9,10,11,12,13,14)(15,16,17,18,19,20,21)(22,23,24,25,26,27,28)(29,30,31,32,33,34,35)(36,37,38,39,40,41,42)(43,44,45,46,47,48,49)(50,51,52,53,54,55,56)(57,58,59,60,61,62,63)(64,65,66,67,68,69,70)(71,72,73,74,75,76,77)(78,79,80,81,82,83,84)(85,86,87,88,89,90,91)(92,93,94,95,96,97,98)(99,100,101,102,103,104,105)(106,107,108,109,110,111,112)(113,114,115,116,117,118,119)(120,121,122,123,124,125,126)(127,128,129,130,131,132,133)(134,135,136,137,138,139,140)(141,142,143,144,145,146,147)(148,149,150,151,152,153,154)(155,156,157,158,159,160,161)(162,163,164,165,166,167,168), (1,43,8,50)(2,49,9,56)(3,48,10,55)(4,47,11,54)(5,46,12,53)(6,45,13,52)(7,44,14,51)(15,57,22,64)(16,63,23,70)(17,62,24,69)(18,61,25,68)(19,60,26,67)(20,59,27,66)(21,58,28,65)(29,71,36,78)(30,77,37,84)(31,76,38,83)(32,75,39,82)(33,74,40,81)(34,73,41,80)(35,72,42,79)(85,127,92,134)(86,133,93,140)(87,132,94,139)(88,131,95,138)(89,130,96,137)(90,129,97,136)(91,128,98,135)(99,141,106,148)(100,147,107,154)(101,146,108,153)(102,145,109,152)(103,144,110,151)(104,143,111,150)(105,142,112,149)(113,155,120,162)(114,161,121,168)(115,160,122,167)(116,159,123,166)(117,158,124,165)(118,157,125,164)(119,156,126,163), (2,7)(3,6)(4,5)(9,14)(10,13)(11,12)(16,21)(17,20)(18,19)(23,28)(24,27)(25,26)(30,35)(31,34)(32,33)(37,42)(38,41)(39,40)(43,50)(44,56)(45,55)(46,54)(47,53)(48,52)(49,51)(57,64)(58,70)(59,69)(60,68)(61,67)(62,66)(63,65)(71,78)(72,84)(73,83)(74,82)(75,81)(76,80)(77,79)(86,91)(87,90)(88,89)(93,98)(94,97)(95,96)(100,105)(101,104)(102,103)(107,112)(108,111)(109,110)(114,119)(115,118)(116,117)(121,126)(122,125)(123,124)(127,134)(128,140)(129,139)(130,138)(131,137)(132,136)(133,135)(141,148)(142,154)(143,153)(144,152)(145,151)(146,150)(147,149)(155,162)(156,168)(157,167)(158,166)(159,165)(160,164)(161,163) );
G=PermutationGroup([[(1,106,29,92,15,120),(2,107,30,93,16,121),(3,108,31,94,17,122),(4,109,32,95,18,123),(5,110,33,96,19,124),(6,111,34,97,20,125),(7,112,35,98,21,126),(8,99,36,85,22,113),(9,100,37,86,23,114),(10,101,38,87,24,115),(11,102,39,88,25,116),(12,103,40,89,26,117),(13,104,41,90,27,118),(14,105,42,91,28,119),(43,148,71,134,57,162),(44,149,72,135,58,163),(45,150,73,136,59,164),(46,151,74,137,60,165),(47,152,75,138,61,166),(48,153,76,139,62,167),(49,154,77,140,63,168),(50,141,78,127,64,155),(51,142,79,128,65,156),(52,143,80,129,66,157),(53,144,81,130,67,158),(54,145,82,131,68,159),(55,146,83,132,69,160),(56,147,84,133,70,161)], [(1,2,3,4,5,6,7),(8,9,10,11,12,13,14),(15,16,17,18,19,20,21),(22,23,24,25,26,27,28),(29,30,31,32,33,34,35),(36,37,38,39,40,41,42),(43,44,45,46,47,48,49),(50,51,52,53,54,55,56),(57,58,59,60,61,62,63),(64,65,66,67,68,69,70),(71,72,73,74,75,76,77),(78,79,80,81,82,83,84),(85,86,87,88,89,90,91),(92,93,94,95,96,97,98),(99,100,101,102,103,104,105),(106,107,108,109,110,111,112),(113,114,115,116,117,118,119),(120,121,122,123,124,125,126),(127,128,129,130,131,132,133),(134,135,136,137,138,139,140),(141,142,143,144,145,146,147),(148,149,150,151,152,153,154),(155,156,157,158,159,160,161),(162,163,164,165,166,167,168)], [(1,43,8,50),(2,49,9,56),(3,48,10,55),(4,47,11,54),(5,46,12,53),(6,45,13,52),(7,44,14,51),(15,57,22,64),(16,63,23,70),(17,62,24,69),(18,61,25,68),(19,60,26,67),(20,59,27,66),(21,58,28,65),(29,71,36,78),(30,77,37,84),(31,76,38,83),(32,75,39,82),(33,74,40,81),(34,73,41,80),(35,72,42,79),(85,127,92,134),(86,133,93,140),(87,132,94,139),(88,131,95,138),(89,130,96,137),(90,129,97,136),(91,128,98,135),(99,141,106,148),(100,147,107,154),(101,146,108,153),(102,145,109,152),(103,144,110,151),(104,143,111,150),(105,142,112,149),(113,155,120,162),(114,161,121,168),(115,160,122,167),(116,159,123,166),(117,158,124,165),(118,157,125,164),(119,156,126,163)], [(2,7),(3,6),(4,5),(9,14),(10,13),(11,12),(16,21),(17,20),(18,19),(23,28),(24,27),(25,26),(30,35),(31,34),(32,33),(37,42),(38,41),(39,40),(43,50),(44,56),(45,55),(46,54),(47,53),(48,52),(49,51),(57,64),(58,70),(59,69),(60,68),(61,67),(62,66),(63,65),(71,78),(72,84),(73,83),(74,82),(75,81),(76,80),(77,79),(86,91),(87,90),(88,89),(93,98),(94,97),(95,96),(100,105),(101,104),(102,103),(107,112),(108,111),(109,110),(114,119),(115,118),(116,117),(121,126),(122,125),(123,124),(127,134),(128,140),(129,139),(130,138),(131,137),(132,136),(133,135),(141,148),(142,154),(143,153),(144,152),(145,151),(146,150),(147,149),(155,162),(156,168),(157,167),(158,166),(159,165),(160,164),(161,163)]])
102 conjugacy classes
class 1 2A 2B 2C 2D 2E 2F 2G 3A 3B 4A 4B 6A ··· 6F 6G 6H 6I 6J 6K 6L 6M 6N 7A 7B 7C 12A 12B 12C 12D 14A ··· 14U 21A ··· 21F 42A ··· 42AP order 1 2 2 2 2 2 2 2 3 3 4 4 6 ··· 6 6 6 6 6 6 6 6 6 7 7 7 12 12 12 12 14 ··· 14 21 ··· 21 42 ··· 42 size 1 1 1 1 2 2 14 14 1 1 14 14 1 ··· 1 2 2 2 2 14 14 14 14 2 2 2 14 14 14 14 2 ··· 2 2 ··· 2 2 ··· 2
102 irreducible representations
dim 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 type + + + + + + + + image C1 C2 C2 C2 C2 C3 C6 C6 C6 C6 D4 D7 C3×D4 D14 C3×D7 C7⋊D4 C6×D7 C3×C7⋊D4 kernel C6×C7⋊D4 C6×Dic7 C3×C7⋊D4 C2×C6×D7 C22×C42 C2×C7⋊D4 C2×Dic7 C7⋊D4 C22×D7 C22×C14 C42 C22×C6 C14 C2×C6 C23 C6 C22 C2 # reps 1 1 4 1 1 2 2 8 2 2 2 3 4 9 6 12 18 24
Matrix representation of C6×C7⋊D4 in GL3(𝔽337) generated by
336 0 0 0 208 0 0 0 208
,
1 0 0 0 110 336 0 78 33
,
336 0 0 0 95 182 0 280 242
,
336 0 0 0 1 228 0 0 336
G:=sub<GL(3,GF(337))| [336,0,0,0,208,0,0,0,208],[1,0,0,0,110,78,0,336,33],[336,0,0,0,95,280,0,182,242],[336,0,0,0,1,0,0,228,336] >;
C6×C7⋊D4 in GAP, Magma, Sage, TeX
C_6\times C_7\rtimes D_4
% in TeX
G:=Group("C6xC7:D4");
// GroupNames label
G:=SmallGroup(336,183);
// by ID
G=gap.SmallGroup(336,183);
# by ID
G:=PCGroup([6,-2,-2,-2,-3,-2,-7,506,10373]);
// Polycyclic
G:=Group<a,b,c,d|a^6=b^7=c^4=d^2=1,a*b=b*a,a*c=c*a,a*d=d*a,c*b*c^-1=d*b*d=b^-1,d*c*d=c^-1>;
// generators/relations
×
𝔽
|
2021-12-02 05:57:33
|
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|
https://socratic.org/questions/how-do-you-use-implicit-differentiation-to-find-dy-dx-given-3x-2-3-ln5xy-2
|
# How do you use implicit differentiation to find (dy)/(dx) given 3x^2+3=ln5xy^2?
Aug 4, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left(3 x - \frac{1}{2 x}\right)$
#### Explanation:
Implicit differentiation is no different to normal, you just have to make sure to apply the chain rule when differentiating the $y$ terms, ie $y$ will differentiate to $\frac{\mathrm{dy}}{\mathrm{dx}}$.
$\frac{d}{\mathrm{dx}} \left(3 {x}^{2} + 3 = \ln \left(5 x {y}^{2}\right)\right)$
When we differentiate the log I used a combination of chain and product rules, but you could also rewrite it as :
$\ln \left(5 x {y}^{2}\right) = \ln \left(5\right) + \ln \left(x\right) + \ln \left({y}^{2}\right)$
using rules of logs and then just use chain rule.
$6 x = \frac{1}{5 {y}^{2} x} \cdot \left(5 {y}^{2} + 10 x y \frac{\mathrm{dy}}{\mathrm{dx}}\right)$
$30 {x}^{2} {y}^{2} = 5 {y}^{2} + 10 x y \frac{\mathrm{dy}}{\mathrm{dx}}$
$5 {y}^{2} \left(6 {x}^{2} - 1\right) = 10 x y \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{y}{2} \left(6 x - \frac{1}{x}\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = y \left(3 x - \frac{1}{2 x}\right)$
|
2021-08-02 16:02:41
|
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|
http://forum.zkoss.org/question/26072/change-style-but-not-for-all-components-mold-or-sclass/
|
# Change style - but not for all components (mold or sclass?)
robertpic71
1275 1
Hi,
Sorry for my thumb question (i do only simple style/css question before).
I've done a ZK "Categorybar light" (see here) - working fine.
I use button's and make he buttonsborder invisible until mouseover. But all my other buttons are also styled that way.
I found the sclass option in the zkdemo - i'm not sure about the difference to the mold.
When i do an own sclass, have i repeat all styles from the original button?
/Robert
delete retag edit
## 4 Replies
henrichen
3869 2
Robert,
This might help in your case.
http://www.zkoss.org/doc/styleguide/ch01s01.html
robertpic71
1275 1
Hi Henri,
thanks - i got it. I've seen this doc, but i was a little bit confused after first reading...
After your hint (Xclass is the way to go) i read it once more - and realize how it work's.
Maybe that's interesting for other users (howto wiki?), here is the code, working in the zkdemo:
<zk>
<style>
.icon, .icon .z-button-cm {
font-family: ${fontFamilyT}; font-size:${fontSizeM}; color: #4040FF;
cursor: pointer; white-space: pre;
}
.icon .z-button-tl, .icon .z-button-tr, .icon
.z-button-bl, .icon .z-button-br, .icon .z-button-tm,
.icon .z-button-bm, .icon .z-button-cl, .icon
.z-button-cr, .icon .z-button-cm { background-image:none;
}
.icon:hover .z-button-tl, .icon:hover .z-button-tr, .icon:hover .z-button-bl, .icon:hover .z-button-br {
background-image:url(${c:encodeURL('~./zul/img/button/z-btn-trendy-corner.gif')}); } .icon:hover .z-button-tm, .icon:hover .z-button-bm { background-image:url(${c:encodeURL('~./zul/img/button/z-btn-trendy-x.gif')});
} .icon:hover .z-button-cl, .icon:hover .z-button-cr {
background-image:url(\${c:encodeURL('~./zul/img/button/z-btn-trendy-y.gif')});
}
</style>
<hbox style="background-image:url(/zkdemo/img/category-bg.png); background-repeat: repeat-x 0 0;padding-top:7px"
align="end">
label="Forms and
label="Toolsbars
</hbox>
</zk>
And my real life Application (beta).
BTW 1: The ZUL Visual Editor ignores the whitespace - is cr/lf a dirty solution or is this a bug from the VE?
BTW 2: A hover-example would be an improvment for the docs. I read t styleguide and see z-button-over and try
icon-over because it works for (z-button -> icon). I need some time to figure out the correct syntax.
/Robert
henrichen
3869 2
I don't quite catch your "BTW 1" question. Are you talking about the style designer or ...?
For the BTW 2 suggestion, I will forward to our Style Designing experts :).
/henri
anguslipsey
6
Hi Robert,
I am working on the ZKDemo like Categorybar and didn't get any idea to create it, and I know that you are success to implement it from your beta page: http://www.pichelhofer.at/Tandembox/main.zul.
My email: [email protected]
Thanks a lot.
/Angus
[hide preview]
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2019-02-22 04:07:57
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https://www.gamedev.net/forums/topic/174419-camera-following-bezier-curves/
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• ### Announcements
#### Archived
This topic is now archived and is closed to further replies.
# camera following bezier curves
## Recommended Posts
I want to have my camera follow curves in scripted sequences. Currently, I''m using peicewize Bezier4 curves. I evenly increment mu (the distance between 0 and 1) like so:
mu += clock.DeltaTime / curveTime
where curveTime is the amount of time to spend on this curve. I calculate mu each frame then calculate a point on the curve from mu. This causes the camera to start at rest (0 velocity) accelerate then decelerate to rest again for each set of four control points. I''ve looked into another curve generating formula (b-splines) but they seem to have the same problem. I read in Game Programming Gems that you can calculate the curve data for each set of curve points and save it to an array and then calculate the distance between each saved point, but this limits the resolution of the curve if you have a fast enough computer. Is there an alternative workaround to this solution? Maybe there is a problem with my code, I can post it if necessary. While I''m at it, how do you guys update the view vector relative to the position vector change? Thanks! Dustin
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Nurbs are popular to get smooth camera movements.
If you connect a series of Bezier curves you will get problems cause the tangent vectors at the end of one curve and at the start of the next curve will only have the same direction and not the same length.
/__fold
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If you want to have the camera face along the splines path, you could get the position of the camera at the NEXT clock tick(or the previous clock tick.. it should be a trivial decision, depending on the value of delta mu), and the correct camera vector would be the vector created from the current position of the camera, and the next position.
A cheap little hack, and I don''t know how well it would look on low resolution curves....
/__fold is right about the normals though.. Creating a Nurbs evaluator is probably just a matter of spending a few minutes with google anyways.
-=|Mr.Oreo|=-
Code Monkey, Serpent Engine
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for nurbs search for a library "nurbs++".
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A nice idea might be to use two paths. One for the camera position, one for the camara to look at. Might give you a little more creative freedom.
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I recently implemented a curve-following camera system, and hit the same problem with bezier curves. B-Splines gave me the nice smooth movement I was looking for.
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I''ve also built a pathed camera system and I had a lot of luck with Catmull Rom splines. They are mathematically equivalent to bezier curves, but they pass through all their control points. The good news is that you string a long sequence of them together and it behaves like you''d expect without you having to (for bezier curves) make sure your derivative control points line up at the endpoints of each segment.
So, very simple to use with good results (C1 continuity, curvature is interpolated between endpoints, so it''s not C2).
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Gems book 1 says that catmull-rom splines are much uglier than bsplines - what is your take on this? Does the camera feel smooth?
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"Much uglier" is very subjective. Math says that B-Splines are smoother (having C2 continuity) true, but the problem is that they don''t interpolate any of their control points. This is no problem if you have a nice graphical tool that well let you plop down control points and tweak them until you get the curve you want. However, my system also needed to generate curves in real-time so B-Splines were not an option.
In truth, I found that curve smoothness is not as significant an aesthetic factor as the acceleration/deceleration equations that affect camera motion.
• ## Partner Spotlight
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2017-10-19 09:27:04
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https://dsp.stackexchange.com/questions/14867/matlab-high-frequency-sine-wave-generation
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# Matlab HIgh frequency sine wave generation
I need to generated a high frequency sinusoidal signal for modulation in MATLAB. But it turns out to be something weird. This is the simple code snippet I used...
t = 0:0.001:100;
A = 1;
s = A*sin(2*pi*1e9.*t); %1 GHz signal
plot(t, s);
title('Modulating Signal');
xlabel('Time');
ylabel('Amplitude');
Instead of the sine wave I am getting something weird like this..
The function you need to evaluate is:
A*sin(2*pi*FREQUENCY*(1:NUM_SAMPLES)/SAMPLE_RATE)
Use a reasonable sample rate (at least twice the frequency) and a reasonable interval (remember that at a sample rate of 2Ghz, 1 second of signal will fill up the memory of your computer) ; and it'll work! From your code it seems that you are trying to generating 100s of signal at a sample rate of 1kHz, which doesn't make sense.
At the moment, in your code, the argument of the sin function is an integer multiple of $2\pi$ so in theory your code should be plotting a flat line of 0s. You are seeing something a bit different due to the finite resolution of floating point numbers.
The problem is the definition of your vector $\tt{t}$. You seriously subsample your signal because its period is $T=10^{-9}$ and you the interval of your sample points is $10^{-3}$. Try the following:
t = linspace(0,1e-7,10000);
A = 1;
s = A*sin(2*pi*1e9.*t); %1 GHz signal
plot(t, s);
The problem is in the use of floating point.
When the sine-function input gets too large, it's accuracy decreases and hence the accuracy of the sine-function decreases. You should limit the input of the sine-function between 0 and 2pi (basically a saw-tooth). This will solve your problem.
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2019-09-17 09:18:57
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https://worldbuilding.stackexchange.com/tags/insectoid/hot
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# Tag Info
184
Joe, this is the HQ. It's likely the ants do not really melt the glass. In the 20th century some material scientists on Earth developed a kind of photoresist (called spin-on glass) which, after exposure to UV or e-beam and curing, would become what is for all intents and purposes, glass: randomly arranged SiO$_2$ chains. The principle is fairly similar ...
100
Can the human race survive? No. We're too dependent on them for our survival. An open attack would hurt them - we'd just insecticide them. That's not a good strategic move by the bugs. Not a good strategic move by us either, but the choice is stand idly by and get suffocated by a million crawly feet, or destroy the ecology and possibly survive. The ...
97
They aren't really ants, they are Thermites. Your ants deposit a line of thermite in a new tunnel, then ignite it (bonus points if they use a glass lens and sunlight for it) after scrambling the hell out of reach. I'm not enough of a chemist to say how they'd produce the thermite, but nature will find its way. (Maybe they use some symbiotic bacteria?)
70
There are already organisms that secrete glass without high temperatures. This paper discusses a wide variety of ways organisms handle it: Glass sponges grow a skeleton of glass spicules; Many plants secrete tiny glass crystals called phytoliths to dissuade herbivores; Diatoms and radiolarians secrete a glass protective skeleton. You just need the right ...
67
Town of Orangey cinnamon peppermint. Ya no joke. Spiders hunt in part by scent (apparently sweaty socks are a treat for them) and there are a few scents that repulse them pretty heavily, and as an added bonus...those scents actually smell decent to us. Surround the town walls with a ring of peppermint plants and wet down the walls with an orange cinnamon ...
45
You need an energy source other than the sun. The main alternative is reduced sulfur that can be oxidized for energy. Organisms that can do that chemistry are chemotrophs and so far all are bacteria. Either the bacteria themselves can form the plantlike base of the food chain or larger organisms with symbiotically associated chemotropic bacteria can be ...
31
The City of Nope Due to an incredibly pressing need, the inhabitants have developed flamethrowers hundreds of years early. Children are given their first one as toddlers, and are trained and drilled in their use from an early age. The construction of the city is entirely stone, to negate the frequent use of cleansing fire. Even fire isn't totally ...
31
Joe: Some folks back at HQ have some doubts about the Glass Ants building these tunnels. Instead, they may be excavating them. Desert glass is naturally occurring glass made when lightning strikes desert stands. The lightning fuses the silica together to form glass. Your Glass Ants may be finding natural deposits of this glass and then burrowing into it. ...
31
Trivial Answer: "Private parts" are a cultural construct. Define your culture such that no parts of these fairies are considered "private". Obvious Answer: There are a number of "backless" one-piece swimwear designs for real-world human women, some of which have more-or-less arbitrarily low coverage of the wearer's back, leaving the entire thoracic spine ...
28
Game Over, Humans Lose Our countermeasures against insects works because they aren't coordinated. Much of what we do to prevent exposure of our bodies, food and assets works against specific pests. Screens over doors and windows work against mosquitos and other biting insects. Pesticides prevent crop destruction. But, all of the things that we defend ...
27
Starting from simplest to most complicated, lets tackle the size issue first. Large insects physiology Insects do not breathe with lungs the way we do. They have many spiracles distributed over their body that open into many tracheae which allow oxygen into their body. They can't use expanding lungs the way we do because their exoskeleton cannot deal ...
24
Glass ants use symbiosis with a native plant species (with hollow roots) that already used silica naturally (and without needing high temperatures.) Three facts help explain the glass ants: Ants species have repeatedly formed symbiotic relationships with plants (and fungi) for food, shelter and defense. Examples include: https://www.sciencedaily....
18
Probably. Insectoids might be slower. There are several issues though. When most people think "giant insects", they think of insects, only scaled up. Except, this is biologically and physically impossible. When you scale things up, the mass scales by the cube - you'll need very thick legs (think elephants) and most likely a standing posture. You'll also ...
18
They approached the fabled city surrounded by spiders, the boat rocking slightly on the massive river. Luckily, the river was wide enough where trees were not a worry, they only spanned a very short distance over the water. One of the guards shot a spider skating quickly towards them, across the water. Spiders were not able to drop down from above but some ...
18
There are some areas of the world in which most insects would have a hard time surviving -- deserts and near the poles. Though of course in the sub-artic regions midges seem to do pretty well every summer. A few humans could probably hold out in the arctic sustained by fishing etc. (i.e. traditional Inuit would be OK). There might be large enough oases in ...
15
For armour, large plates of chitin might be problematic, mostly because it is unlikely the shape of the insect will correspond to any body part which you are looking to protect. More useful perhaps would be Lamellar armour, where small plates of chitin could be carved into shape and then laced together to form a flexible and relatively lightweight ...
15
Three approaches. Approach One: slow and steady. Unleash your creation over the course of 100 million years. That will give the ecosystems enough time to adapt. After all, the best things in life are worth waiting for. Approach Two: apologetically. Kill off the flies in small area, and observe what ecosystems change. Say you're sorry to any birds or ...
13
My first thought was something like the answer from ths The ants cannot produce the temperatures required from their own bodies but could construct something using materials which burn at a high temperature. My first thought was Thermite But that's problematic.There is no known biological system which can produce pure aluminum. Aluminum is pretty much ...
12
Well our bug bombs are pretty effective. We have chemicals that are extremely deadly and can be deployed from the air, from canisters or even by hand. The initial attack would cause trouble. Assume massive human losses at the start. The bugs will have complications crossing the oceans, giving humans time to prepare. Once the pest killing machine gets ...
11
Get them fighting eachother. Seriously, this is the kind of "what if" that doesn't go well for the question in search of answers. You have created an enemy that is: Far smarter than us Far stronger than us Out masses us by several orders of magnitude Hates us Has no weak point, because it is so diffused Already has control of the entire globe These sorts ...
11
One of the biggest limitations is actually breathing and circulation. See here: https://www.quora.com/Entomology/Do-any-insects-have-hearts-and-lungs-or-do-they-have-a-completely-different-system-of-breathing Hearts: Sort of. Insects have an open circulatory system, meaning their organs just float in blood. It's similar to how human organs float in lymph, ...
11
I will first comment that you've limited yourself rather severely. In a locust swarm, a "large swarm" consists of billions of individuals, with eighty million locusts per square kilometre not being unusual. Controlling 3000 of those will not do you any good, either to deploy such a swarm or to defend against such a swarm. But if we're limited to 3000 ...
10
The Wasp Riders The spider's natural enemy is the wasp. Wasps sting spiders and bring them, paralyzed, back to their nests to feed their young. That won't work on a fully grown giant spider (unless the town raises giant wasps, which I suppose is a possibility. They might make for a great air force, as well), but even if the wasps are small, they could sting ...
9
An alien that looks like an insect - i.e. has an exoskeleton - is quite plausible. We do not get such creatures of the size you desire here on earth simply because insects have an open circulatory system and a largely diffusion-based respiratory system, however, an exoskeletal alien of larger size would need - and there is no reason why it could not have - ...
9
How to tell they are intelligent I think the easiest way, in both directions, to determine intelligence would be to see artifacts, such as clothing. If the ant has ant pants, it's probably intelligent. If it seemed intelligent, but didn't recognize the same about me, mathematical sequences would be a good bet. I can be fairly sure that an advanced race ...
9
Diatoms are algae that make silica shells. They do not use heat, of course; they deposit hydrated soluble silica (as silicic acid) along with organic matrix to produce their shells. The shells last a very very long time. http://www.pnas.org/content/113/8/2017.full.pdf Having an ant acquire silica manipulating abilities is quite an evolutionary jump. It ...
9
Have you considered your creatures would not eat at all? Basically they spend their lives as a larva before coming to the cave. During the larva stage they eat enough to supply them for the rest of their life. After the larva stage, the transformation makes them immune and drawn to the aura. Some crocodiles assumingly can go without food for over 3 years, ...
8
The vehicles would probably have better suspension and stabilization effects then wheeled or tracked counterparts. Consider having six independently articulating shock absorbers. And the fact that these would more than likely absorb said shock in different parts of the legs i.e the joints of the leg.As for stabilizing the vehicles when firing, just look at ...
7
Fire and axes would be the primary defense against them. One thing that would work well would be to find/train animals that fear the spiders, monkeys or loud birds. So when they spot a spider creeping about they will raise a cry to warn the watch. They will be armed with flaming arrows and axes should the spiders get close enough to need a more personal ...
7
Neither side wins The reason why we haven't exterminated insects already is because they are needed for just about everything. They are a crucial part of the ecosystem as well. Let's say they all move to a portion of the world to wipe out a specific target: humans. They still need to eat and survive. It would be a battle of attrition against themselves ...
Only top voted, non community-wiki answers of a minimum length are eligible
|
2021-01-18 16:47:55
|
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https://www.physicsforums.com/threads/kepplers-2nd-3rd.805/
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# Keppler's 2nd & 3rd
1. Apr 1, 2003
### mich
[SOLVED] Keppler's 2nd & 3rd
While I have the correct topic, I'm not sure if this belongs to
the crackpot theory or not. I appologize if I haven't placed my post in the correct category.
I'm having a bit of a problem concerning Keppler's 2nd.law.
" ...moving planets sweeps out equal areas at equal time."
This clearly states that a planet which approaches or get's farther
from the sun will experience an increase or a decrease of velocity
by a specific amount.
Let a planet revolve around the sun at R=1, if the orbit is circular, then the distance traveled would be 2*pi*R= 2*pi. Let us claim the period to be 1 unit of time.It's velocity would then be 2*pi/1.
According to the 2nd law, it would take a radius of sqrt2 in order
for the period to be equal to 2. the velocity would then be
2*sqrt2*pi/2; or simply sqrt2*pi.
If we continue to increase the radius, we get:
for R=sqrt3 T=3; v=2*sqrt3*pi/3
for R=sqrt4 T=4; v=2*sqrt4*pi/4 = 4*pi/4 = pi (notice that when the radius is doubled, the velocity, being directly affected by the gravitational force,is halved. This "could" seem to claim as you did,that gravitational force is inversally proportional to R...not R^2.
Nevertheless,My question would be: could Newton's 3rd law be centered more on the kenetic energy of a planet relative to it's distance from the gravitational force? Force=Kenetic Energy/R?
F=mv^2/2R
R=1 T=1 V=2 F= m*4/2*1 = 2 (let m=1)
R=2 T=4 v=1 F= m*1/2*2 = 1/4
R=3 T=9 v=.66 F= m*.4356/2*3 = .0726
R=4 T=16 v=.5 F= m*.25/2*4 = .031
R=8 T=64 v=.25 F= m* .0625/2*8 = .0039
This would show that when the radial distance is doubled, the kenetic energy is halved, and the Force is reduced by a factor of 8.
2. Apr 2, 2003
### mich
Re: Re: Re: Gravity decreases as 1/r
Thank you for replying, Janus, and I do agree that Keppler's 2nd law was meant to speak of one planet travelling on a elliptical orbit.
On the other hand, wouldn't you agree that the "covering of areas at equal time" law must indeed state that the planet having a specific velocity at R1 will change to a different "specific" velocity when the planet is at R2?Let us say that at R1 the velocity is 1 unit of distance/unit of time. According to the 2nd law, one ought to be able to calculate the velocity of the planet when the distance changes to R2 from the gravitational source.According to your calculation,what would be R2 when the velocity is halved, if R1= 1 unit of distance?
3. Apr 2, 2003
### Janus
Staff Emeritus
The formula for velocities at perigee and apogee repectively are:
Vp = [squ]((2GM/(Rp+Ra)) Ra/Rp)
and
Va =[squ]((2GM/(Rp+Ra)) Rp/Ra)
In this case, Rp = 1/2 and Ra =1, so
Vp = [squ]((2GM/(1/2+1)) 1/(1/2))
= [squ](8GM/3)
and
Va =[squ]((2GM/(1/2+1)) (1/2)/1)
= [squ](2GM/3)
The ratio of Vp to Va is
Vp/Va =[squ](8GM/3)/[squ](2GM/3)
=[squ]((8GM/3)/(2GM/3))
GM/3 cancels out, leaving
=[squ](8/2) =[squ]4 = 2.
The area traced out by the orbit at perigee in one sec is;
Ap/sec = (2*1/2)/2 = 1/2 square units.
at apogee:
Aa/sec = (1*1)/2 = 1/2 square units, which satisfies Keppler's second law.
One thing that should be pointed out however, is that Keppler's Second law by itself cannot tell us anything about how the gravitational field behaves other than that it acts along a certain line.
It is true regardless whether the field is attractive, repulsive, constant, falls off lineary or by the square of the distance. All it requres is that the force always acts along a line towards a central point.
This has been proven through Newton's "Theorem of Areas".
Last edited: Apr 2, 2003
4. Apr 3, 2003
### mich
O.K. this would seem to come from Newton's force of gravitation formula where [sqrt 2GM/R] I believe would be considered the velocity of one of the bodies;. would that be correct?
I believe that this is what I had, Janus. when the Radius increases
twofold the orbital velocity is reduced to half it's original speed.
I think I do understand what you are saying though. If a body has a
circular orbit, it's velocity needs to be inferior or superior (depending on it's position relative to the gravitational force), to the velocity of the body when it's orbit is an elliptical one.
If that's the case, then I "personally" disagree. The orbital velocity has two vectors; one in the direction towards the point of gravitation and the second is lets say 90 degrees to it.
The sideral velocity displaces the path of the orbit making the body having either a circular orbit, or elliptical or whatever. It is responsible for leaving the body at a constant distance from the point
of gravity or increasing/decreasing it's distance to it.It is in itself independant to the gravitational force.Nevertheless,
when the body is at a specific distance to the gravitational point,
it seems that it's velocity can be calculated the same way that I have done so; that is the way I see it anyway. Your calculation seems to comes indirectly from Newton and not Keppler, although, as you showed me, it does indeed agree with Keppler's 2nd law.
The reason why I'm circling around this, is because, if my calculations are correct,then it seems that there's a discrepancy between Keppler's 2nd and 3rd law;and yes of course I know that I am in all probabilities wrong.Nevertheless, this is what bother's me right now.
You see, the idea stems from the fact that Kepler's calculated values of Tycho Brahe's observations could not have taken into account the red and blue shifts since during this period, light was thought to have possibly an infinite velocity.
Thanks for your time; I greatly appreciate it.
Mich
Last edited by a moderator: Apr 3, 2003
5. Apr 3, 2003
### Janus
Staff Emeritus
[squ](GM/R) gives the orbital velocity for a circular orbit of radius R . (The formula you gave is for escape velocity.)
Now it is important to note that a body in an eliptical orbit with a perigee of R will have a velocity greater than [squ](GM/R) (yet less than [squ](2GM/R) ) at perigee. So you can't use Keppler's second law to determine the period of a circular orbit at that distance.
No discrepancy. One talks about the velocity of a eliptical orbit at different points of the orbit, the Other deals with the period due to the Average radius of an orbit. One works within a given orbit, the other works between different orbits.
There can't be a discrepancy between them because they are not inter-related.
Just as it can be shown that Keppler's second law is compatable with Newtonian physics, Keppler's third law can be shown to be so.
V = [squ](GM/R)
the distance travled in one orbit is
D = 2[pi]R
The period of the orbit is therefore
P = D/V =2[pi]R / [squ](GM/R)
= 2[pi]R[squ](R/GM) = 2[pi][squ](R³/GM)
If we compare the period of two circular orbits we get
P1/P2 = 2[pi][squ](R1³/GM)/2[pi][squ](R2³/GM)
2[pi] cancels out.
=[squ](R1³/GM)/[squ](R2³/GM)
square both sides.
P1²/P2² =(R1³/GM)/(R2³/GM)
GM cancels out;
P1²/P2² =R1³/R2³
Leaving us with Keppler's Third Law.
The problem here is that the Doppler shifts in light are so small at orbital velocities that they fall far below the accuracy limits of the measurements made at that time. They would have had negligible effect on Keppler's measurements.
6. Apr 3, 2003
### mich
Thanks alot, Janus; you clarified the situation for me.
7. Apr 4, 2003
### mich
Excuse me Janus,I came back because something was bothering me..
I sound like Columbo
What you wrote makes sense but I may have a small problem still.
When the planet's orbit is at it's perigee, then the velocity
is greater than if it was in a circular orbit at that radius, which is the reason why it gets further from the sun. Also when the planet is at it's apogee , then the planet's velocity is slower than if it had a circular orbit at that particular radius, which is the reason why the planet moves closer to the sun. Therefore, the ratio between
those two velocities (apogee/perigee) could never be equal to the ratio of two different circular orbits having two different radii, one equal to the apogee of the elliptical orbit, and the other equal to the perigee of that same elliptical orbit...so how is it that in your calculations, the ratio between the two velocities (apogee/perigee) was the same as the calculations I've made using circular orbits?
8. Apr 4, 2003
### Janus
Staff Emeritus
First off, I split this discussion off into a new thread, as it was off the original topic.
Going back over your posts, it look's like you tried to use Keppler's 2nd Law to calculate those circular orbits. That won't work, because you are dealing with two separate orbits, not just one.
What you ended up getting is the velocity difference between two points with different distances in the same orbit. You used the wrong law.
To get the velocity difference between two different orbits, you have to use the 3rd law.
9. Apr 4, 2003
### mich
Yes, of coarse I understand. I thought I had something in common with their thoughts on g being proportional to R and not R^2, but
I guess I didn't.
Actually, I only used those calculations to figure out the velocity of a body at different distances from the gravitational force (R).
It seems that Keppler did not write this law for nothing, it had to be in order to calculate the ratio of different velocities when the planet changes it's distance from the sun... but how does one calculate it?
You used a version of Newton's law, something that didn't exist yet.
I took the only tool that existed then,(the covering of areas at equal time law) and tried to figure out different velocity ratios for different distances of the planet's path. The only way I could think of doing this was to calculate the respected velocities
"as if they were two different circular orbits of different radii".
But if I used the method of circular orbits, how could I have ended up
getting the velocity difference between two points with different distances "in the same orbit"? I'm a bit confused.
I will leave the 3rd law aside for now so as not to get too confused.
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2016-12-07 22:21:42
|
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https://stats.stackexchange.com/questions/225727/understanding-data-normalization-for-svms
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# Understanding data normalization for SVMs
For SVMs, they operate best when the data is in the ranges of $[0,1]$ or $[-1, 1]$. Naturally you want to normalize data so that your model works well.
My question is: what do you do about features derived from normalized data?
For example, if I have a data frame with a column called "wait times" in seconds, and I want to add a 50 day moving average to this data to smooth it, do I:
1. Normalize the "wait times" to the range $[-1, 1]$ and then apply the moving average function
2. Apply the moving average function and then normalize the "wait times" and "ma" column to the range $[-1, 1]$.
I'm having trouble finding any sort of literature on this. I feel like (2) is correct because (1) causes information loss, but I am honestly not sure because I cannot prove it to myself handily. It would be very helpful if someone could explain to me which is right so that I know in the future how best to handle this.
Thank you!
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2019-12-14 10:05:25
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https://david-abel.github.io/blog/posts/rl_abstr.html
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# David Abel
[email protected]
### Abstraction and RL: Literature Overview
##### 03/09/2019
Abstraction is a pretty remarkable process; a fun example is given in the Pixar movie "Inside Out", in which the main characters find themselves in a room of abstraction (inside a character's head) where the objects in the room become more and more abstract, eventually leading to only simple 2d colored shapes (clip here):
But what exactly makes abstraction a useful practice? Why do our characters become these particular shapes, with these particular colors? Of course, abstraction is central to computer science: from Turing machines to object-oriented programming, abstraction is a necessary process that supports many of the foundational ideas of the field. From the perspective of reinforcement learning (RL), the more pressing question is: how can a learning agent discover and use abstractions to improve its capacity to make effective decisions in complex environments?
Algorithms for RL are well suited to benefit from abstraction. One route to confronting many of the central challenges of RL like exploration, credit assignment, and long horizon planning involves reasoning with a simpler, abstract model of the environment. For example, exploring "inside the cupboard", already requires the use of the abstract concepts "cupboard" and "inside of" that break the world into a cleanly explorable place. Planning in the right way usually involves making predictions at the right level of abstraction: if I were to throw a wine glass in a restaurant, I can predict that the glass will shatter (which is enough to convince myself I shouldn't throw it!). But, I likely cannot predict the precise location and number of shards of glass on the floor (and why would I need to?). In light of the promise of reasoning with a more compact model, RL research has long sought the appropriate algorithms for learning and using abstractions of different kinds. For an excellent recent overview of abstraction in RL, see On the Necessity of Abstraction by Konidaris (2018).
In this post, I'll give a birds eye view look at some of the literature on abstraction in RL (as of March 2019 or so). I tend to break abstraction (as studied in RL) into three categories:
1. State Abstraction: How can we form abstract states that support effective decision making?
Examples: State aggregation is the main method, though there are connections to feature discovery, basis function discovery, and certain kinds of value function approximation.
2. Action Abstraction: How can we form abstract actions that support effective decision making?
Examples: Options, macro-actions.
3. Hierarchical Abstraction: How can we find the right hierarchical abstraction that can support effective decision making?
Examples: MAXQ, HAMS, RMax-Q, Abstract MDPs.
For each category, I'll give a brief description, then highlight what I consider to 1) be the right papers to read to dive into each body of literature, and 2) a collection of other relevant work. I've almost certainly missed many important papers, so do reach out if you find that some are missing. It's worth noting that abstraction has been studied both for learning purposes and for planning, in which the full model of the world is given as input, and a policy/behavior is requested as output. Naturally, abstractions that are useful for planning can give rise to efficient methods for model-based RL, so some of the referenced work will be more focused on planning as it relates to RL. Lastly, lots of excellent work has focused on abstraction in multi-agent systems; this is not the focus here. Let's dive in!
### State Abstraction
The traditional view of state abstraction is to define a function, $$\phi : S \rightarrow S_\phi$$, that projects each state in the original MDPs state space to an abstract state (where usually $$|S_\phi| \ll |S|$$). The goal is to form a smaller state space that can still preserve the essential characteristics of the problem posed by the true MDP; the smaller state space, hopefully the lower the sample complexity/regret of RL.
The papers I would recommend starting with:
Chronology of other work to look at (shamelessly including a few of my recent papers in the area):
### Action Abstraction
Action abstraction defines methods that enrich the action space of an MDP, typically by incorporating macro-actions that encode long horizon sequences of actions. The formalisms for encoding these higher level actions are diverse, though recently the options formalism [Sutton, Precup, Singh, 1999] has become most prominent. An option, $$o$$, is a triple, $$I_o, \beta_o, \pi_o$$, indicating:
1. $$I_o \subset S$$: An initation condition indicating the states in which the option can be executed.
2. $$\beta_o \subset S$$: A termination condition indicating where the option terminates.
3. $$\pi_o : S \rightarrow A$$: A policy.
An option effectively defines a skill that takes an agent from a precondition ($$I_o$$) to a post condition ($$\beta$$), according to $$\pi_o$$. Typically $$\beta$$ is presented as a conditional probability distribution, conditioned on $$S$$, with support $$[0,1]$$, but we here take the deterministic view for simplicity. Much of recent work on action abstraction has concentrated on discovering useful options, clarifying how options impact learning and planning, and extending options to transfer across tasks.
The papers I would recommend starting with:
Chronology of other work to look at:
### Hierarchical Abstraction
Hierarchical abstraction captures approaches that repeatedly abstract entities involved in RL, each time inducing a coarser entity than the previous iteration. The methods here are too diverse to exhaustively describe with one formalism, though again options have taken hold recently as the primary method for characterizing abstractions. In this case of a hierarchy, this means we define a list of options, $$O^{(n)} = \{O^1, O^2, \ldots, O^n\}$$, where the options at level $$i$$, denoted $$O^{i}$$, take on policies that output options over level $$i-1$$ options. Again, the formalisms are diverse, but this is the main idea: repeatedly abstract using whatever formalism we'd like, until all that's left is a simple characterization of the problem of relevance. The main focus of the literature has been, like state and action abstraction, better understanding how to discover and compute the right abstractions efficiently, clarify when and why abstraction helps, and identify hierarchies that preserve near-optimal behavior.
The papers I would recommend starting with:
Chronology of other work to look at:
I hope folks find this helpful. Let me know if you find errors or that the list is missing some papers (I'm sure I've missed some important ones!)
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2019-12-12 03:39:56
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https://zbmath.org/?q=an%3A1281.34058
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## Harmonic and subharmonic solutions for superlinear damped Duffing equation.(English)Zbl 1281.34058
The authors claim to prove the existence of harmonic and subharmonic solutions for a damped Duffing equation of the form $x''(t) + Cx'(t) + g(x(t)) = 0,$ with $$C\geq0$$ and a superlinear condition on the nonlinearity, by making use of a generalized version of the Poincaré-Birkhoff theorem due to W. Y. Ding. In my opinion, the main result should be considered under suspicion, because any version of the Poincaré-Birkhoff theorem requires the area-preserving character of the Poincaré map, a condition which is violated if $$C>0$$.
### MSC:
34C25 Periodic solutions to ordinary differential equations
Full Text:
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2022-06-26 08:21:45
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https://labs.tib.eu/arxiv/?author=A.%20Readhead
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• ### High-energy gamma-ray observations of the accreting black hole V404 Cygni during its June 2015 outburst(1607.06239)
July 21, 2016 astro-ph.HE
We report on Fermi/Large Area Telescope observations of the accreting black hole low-mass X-ray binary V404 Cygni during its outburst in June-July 2015. Detailed analyses reveal a possible excess of $\gamma$-ray emission on 26 June 2015, with a very soft spectrum above $100$ MeV, at a position consistent with the direction of V404 Cyg (within the $95\%$ confidence region and a chance probability of $4 \times 10^{-4}$). This emission cannot be associated with any previously-known Fermi source. Its temporal coincidence with the brightest radio and hard X-ray flare in the lightcurve of V404 Cyg, at the end of the main active phase of its outburst, strengthens the association with V404 Cyg. If the $\gamma$-ray emission is associated with V404 Cyg, the simultaneous detection of $511\,$keV annihilation emission by INTEGRAL requires that the high-energy $\gamma$ rays originate away from the corona, possibly in a Blandford-Znajek jet. The data give support to models involving a magnetically-arrested disk where a bright $\gamma$-ray jet can re-form after the occurrence of a major transient ejection seen in the radio.
• ### Multiwavelength Evidence for Quasi-periodic Modulation in the Gamma-ray Blazar PG 1553+113(1509.02063)
Oct. 12, 2015 astro-ph.HE
We report for the first time a gamma-ray and multi-wavelength nearly-periodic oscillation in an active galactic nucleus. Using the Fermi Large Area Telescope (LAT) we have discovered an apparent quasi-periodicity in the gamma-ray flux (E >100 MeV) from the GeV/TeV BL Lac object PG 1553+113. The marginal significance of the 2.18 +/-0.08 year-period gamma-ray cycle is strengthened by correlated oscillations observed in radio and optical fluxes, through data collected in the OVRO, Tuorla, KAIT, and CSS monitoring programs and Swift UVOT. The optical cycle appearing in ~10 years of data has a similar period, while the 15 GHz oscillation is less regular than seen in the other bands. Further long-term multi-wavelength monitoring of this blazar may discriminate among the possible explanations for this quasi-periodicity.
• ### Rapid TeV Gamma-Ray Flaring of BL Lacertae(1211.3073)
Nov. 13, 2012 astro-ph.HE
We report on the detection of a very rapid TeV gamma-ray flare from BL Lacertae on 2011 June 28 with the Very Energetic Radiation Imaging Telescope Array System (VERITAS). The flaring activity was observed during a 34.6-minute exposure, when the integral flux above 200 GeV reached $(3.4\pm0.6) \times 10^{-6} \;\text{photons}\;\text{m}^{-2}\text{s}^{-1}$, roughly 125% of the Crab Nebula flux measured by VERITAS. The light curve indicates that the observations missed the rising phase of the flare but covered a significant portion of the decaying phase. The exponential decay time was determined to be $13\pm4$ minutes, making it one of the most rapid gamma-ray flares seen from a TeV blazar. The gamma-ray spectrum of BL Lacertae during the flare was soft, with a photon index of $3.6\pm 0.4$, which is in agreement with the measurement made previously by MAGIC in a lower flaring state. Contemporaneous radio observations of the source with the Very Long Baseline Array (VLBA) revealed the emergence of a new, superluminal component from the core around the time of the TeV gamma-ray flare, accompanied by changes in the optical polarization angle. Changes in flux also appear to have occurred at optical, UV, and GeV gamma-ray wavelengths at the time of the flare, although they are difficult to quantify precisely due to sparse coverage. A strong flare was seen at radio wavelengths roughly four months later, which might be related to the gamma-ray flaring activities. We discuss the implications of these multiwavelength results.
• ### The Spitzer Extragalactic Representative Volume Survey (SERVS): survey definition and goals(1206.4060)
Sept. 17, 2012 astro-ph.CO
We present the Spitzer Extragalactic Representative Volume Survey (SERVS), an 18 square degrees medium-deep survey at 3.6 and 4.5 microns with the post-cryogenic Spitzer Space Telescope to ~2 microJy (AB=23.1) depth of five highly observed astronomical fields (ELAIS-N1, ELAIS-S1, Lockman Hole, Chandra Deep Field South and XMM-LSS). SERVS is designed to enable the study of galaxy evolution as a function of environment from z~5 to the present day, and is the first extragalactic survey both large enough and deep enough to put rare objects such as luminous quasars and galaxy clusters at z>1 into their cosmological context. SERVS is designed to overlap with several key surveys at optical, near- through far-infrared, submillimeter and radio wavelengths to provide an unprecedented view of the formation and evolution of massive galaxies. In this paper, we discuss the SERVS survey design, the data processing flow from image reduction and mosaicing to catalogs, as well as coverage of ancillary data from other surveys in the SERVS fields. We also highlight a variety of early science results from the survey.
• The blazar AO 0235+164 (z = 0.94) has been one of the most active objects observed by Fermi Large Area Telescope (LAT) since its launch in Summer 2008. In addition to the continuous coverage by Fermi, contemporaneous observations were carried out from the radio to {\gamma} -ray bands between 2008 September and 2009 February. In this paper, we summarize the rich multi-wavelength data collected during the campaign (including F-GAMMA, GASP- WEBT, Kanata, OVRO, RXTE, SMARTS, Swift, and other instruments), examine the cross-correlation between the light curves measured in the different energy bands, and interpret the resulting spectral energy distributions in the context of well-known blazar emission models. We find that the {\gamma} -ray activity is well correlated with a series of near-IR/optical flares, accompanied by an increase in the optical polarization degree. On the other hand, the X-ray light curve shows a distinct 20 day high state of unusually soft spectrum, which does not match the extrapolation of the optical/UV synchrotron spectrum. We tentatively interpret this feature as the bulk Compton emission by cold electrons contained in the jet, which requires an accretion disk corona with an effective covering factor of 19% at a distance of 100 Rg . We model the broadband spectra with a leptonic model with external radiation dominated by the infrared emission from the dusty torus.
• ### The structure and emission model of the relativistic jet in the quasar 3C 279 inferred from radio to high-energy gamma-ray observations in 2008-2010(1206.0745)
June 4, 2012 astro-ph.CO, astro-ph.HE
We present time-resolved broad-band observations of the quasar 3C 279 obtained from multi-wavelength campaigns conducted during the first two years of the Fermi Gamma-ray Space Telescope mission. While investigating the previously reported gamma-ray/optical flare accompanied by a change in optical polarization, we found that the optical emission appears delayed with respect to the gamma-ray emission by about 10 days. X-ray observations reveal a pair of `isolated' flares separated by ~90 days, with only weak gamma-ray/optical counterparts. The spectral structure measured by Spitzer reveals a synchrotron component peaking in the mid-infrared band with a sharp break at the far-infrared band during the gamma-ray flare, while the peak appears in the mm/sub-mm band in the low state. Selected spectral energy distributions are fitted with leptonic models including Comptonization of external radiation produced in a dusty torus or the broad-line region. Adopting the interpretation of the polarization swing involving propagation of the emitting region along a curved trajectory, we can explain the evolution of the broad-band spectra during the gamma-ray flaring event by a shift of its location from ~ 1 pc to ~ 4 pc from the central black hole. On the other hand, if the gamma-ray flare is generated instead at sub-pc distance from the central black hole, the far-infrared break can be explained by synchrotron self-absorption. We also model the low spectral state, dominated by the mm/sub-mm peaking synchrotron component, and suggest that the corresponding inverse-Compton component explains the steady X-ray emission.
• ### Simultaneous Planck, Swift, and Fermi observations of X-ray and gamma-ray selected blazars(1108.1114)
May 23, 2012 astro-ph.CO
We present simultaneous Planck, Swift, Fermi, and ground-based data for 105 blazars belonging to three samples with flux limits in the soft X-ray, hard X-ray, and gamma-ray bands. Our unique data set has allowed us to demonstrate that the selection method strongly influences the results, producing biases that cannot be ignored. Almost all the BL Lac objects have been detected by Fermi-LAT, whereas ~40% of the flat-spectrum radio quasars (FSRQs) in the radio, soft X-ray, and hard X-ray selected samples are still below the gamma-ray detection limit even after integrating 27 months of Fermi-LAT data. The radio to sub-mm spectral slope of blazars is quite flat up to ~70GHz, above which it steepens to <\alpha>~-0.65. BL Lacs have significantly flatter spectra than FSRQs at higher frequencies. The distribution of the rest-frame synchrotron peak frequency (\nupS) in the SED of FSRQs is the same in all the blazar samples with <\nupS>=10^13.1 Hz, while the mean inverse-Compton peak frequency, <\nupIC>, ranges from 10^21 to 10^22 Hz. The distributions of \nupS and of \nupIC of BL Lacs are much broader and are shifted to higher energies than those of FSRQs and strongly depend on the selection method. The Compton dominance of blazars ranges from ~0.2 to ~100, with only FSRQs reaching values >3. Its distribution is broad and depends strongly on the selection method, with gamma-ray selected blazars peaking at ~7 or more, and radio-selected blazars at values ~1, thus implying that the assumption that the blazar power is dominated by high-energy emission is a selection effect. Simple SSC models cannot explain the SEDs of most of the gamma-ray detected blazars in all samples. The SED of the blazars that were not detected by Fermi-LAT may instead be consistent with SSC emission. Our data challenge the correlation between bolometric luminosity and \nupS predicted by the blazar sequence.
• ### Long-term monitoring of the high-energy gamma-ray emission from LS I +61{\deg} 303 and LS 5039(1202.1866)
Feb. 9, 2012 astro-ph.GA, astro-ph.HE
The Fermi Large Area Telescope (LAT) reported the first definitive GeV detections of the binaries LS I +61\degree 303 and LS 5039 in the first year after its launch in June, 2008. These detections were unambiguous as a consequence of the reduced positional uncertainty and the detection of modulated gamma-ray emission on the corresponding orbital periods. An analysis of new data from the LAT, comprising 30 months of observations, identifies a change in the gamma-ray behavior of LS I +61\degree 303. An increase in flux is detected in March 2009 and a steady decline in the orbital flux modulation is observed. Significant emission up to 30GeV is detected by the LAT; prior datasets led to upper limits only. Contemporaneous TeV observations no longer detected the source, or found it -in one orbit- close to periastron, far from the phases at which the source previously appeared at TeV energies. The detailed numerical simulations and models that exist within the literature do not predict or explain many of these features now observed at GeV and TeV energies. New ideas and models are needed to fully explain and understand this behavior. A detailed phase-resolved analysis of the spectral characterization of LS I +61\degree 303 in the GeV regime ascribes a power law with an exponential cutoff spectrum along each analyzed portion of the system's orbit. The on-source exposure of LS 5039 is also substantially increased with respect to our prior publication. In this case, whereas the general gamma-ray properties remain consistent, the increased statistics of the current dataset allows for a deeper investigation of its orbital and spectral evolution.
• Spectral energy distributions (SEDs) and radio continuum spectra are presented for a northern sample of 104 extragalactic radio sources, based on the Planck Early Release Compact Source Catalogue (ERCSC) and simultaneous multifrequency data. The nine Planck frequencies, from 30 to 857 GHz, are complemented by a set of simultaneous observations ranging from radio to gamma-rays. This is the first extensive frequency coverage in the radio and millimetre domains for an essentially complete sample of extragalactic radio sources, and it shows how the individual shocks, each in their own phase of development, shape the radio spectra as they move in the relativistic jet. The SEDs presented in this paper were fitted with second and third degree polynomials to estimate the frequencies of the synchrotron and inverse Compton (IC) peaks, and the spectral indices of low and high frequency radio data, including the Planck ERCSC data, were calculated. SED modelling methods are discussed, with an emphasis on proper, physical modelling of the synchrotron bump using multiplecomponents. Planck ERCSC data also suggest that the original accelerated electron energy spectrum could be much harder than commonly thought, with power-law index around 1.5 instead of the canonical 2.5. The implications of this are discussed for the acceleration mechanisms effective in blazar shock. Furthermore in many cases the Planck data indicate that gamma-ray emission must originate in the same shocks that produce the radio emission.
• ### The Luminosity Function of Fermi-detected Flat-Spectrum Radio Quasars(1110.3787)
Oct. 17, 2011 astro-ph.CO, astro-ph.HE
Fermi has provided the largest sample of {\gamma}-ray selected blazars to date. In this work we use a complete sample of FSRQs detected during the first year of operation to determine the luminosity function (LF) and its evolution with cosmic time. The number density of FSRQs grows dramatically up to redshift \sim0.5-2.0 and declines thereafter. The redshift of the peak in the density is luminosity dependent, with more luminous sources peaking at earlier times; thus the LF of {\gamma}-ray FSRQs follows a luminosity-dependent density evolution similarly to that of radio-quiet AGN. Also using data from the Swift Burst Alert Telescope we derive the average spectral energy distribution of FSRQs in the 10 keV-100 GeV band and show that there is no correlation of the peak {\gamma}-ray luminosity with {\gamma}-ray peak frequency. The coupling of the SED and LF allows us to predict that the contribution of FSRQs to the Fermi isotropic {\gamma}-ray background is 9.3(+1.6/-1.0) (\pm3% systematic uncertainty) in the 0.1-100GeV band. Finally we determine the LF of unbeamed FSRQs, finding that FSRQs have an average Lorentz factor of {\gamma} = 11.7(+3.3/-2.2), that most are seen within 5\circ of the jet axis, and that they represent only ~0.1 % of the parent population.
• ### Gamma-Ray and Parsec-Scale Jet Properties of a Complete Sample of Blazars From the MOJAVE Program(1107.4977)
Aug. 31, 2011 astro-ph.CO, astro-ph.HE
We investigate the Fermi LAT gamma-ray and 15 GHz VLBA radio properties of a joint gamma-ray- and radio-selected sample of AGNs obtained during the first 11 months of the Fermi mission (2008 Aug 4 - 2009 Jul 5). Our sample contains the brightest 173 AGNs in these bands above declination -30 deg. during this period, and thus probes the full range of gamma-ray loudness (gamma-ray to radio band luminosity ratio) in the bright blazar population. The latter quantity spans at least four orders of magnitude, reflecting a wide range of spectral energy distribution (SED) parameters in the bright blazar population. The BL Lac objects, however, display a linear correlation of increasing gamma-ray loudness with synchrotron SED peak frequency, suggesting a universal SED shape for objects of this class. The synchrotron self-Compton model is favored for the gamma-ray emission in these BL Lacs over external seed photon models, since the latter predict a dependence of Compton dominance on Doppler factor that would destroy any observed synchrotron SED peak - gamma-ray loudness correlation. The high-synchrotron peaked (HSP) BL Lac objects are distinguished by lower than average radio core brightness temperatures, and none display large radio modulation indices or high linear core polarization levels. No equivalent trends are seen for the flat-spectrum radio quasars (FSRQ) in our sample. Given the association of such properties with relativistic beaming, we suggest that the HSP BL Lacs have generally lower Doppler factors than the lower-synchrotron peaked BL Lacs or FSRQs in our sample.
• ### The radio/gamma-ray connection in Active Galactic Nuclei in the era of the Fermi Large Area Telescope(1108.0501)
Aug. 2, 2011 astro-ph.CO, astro-ph.HE
We present a detailed statistical analysis of the correlation between radio and gamma-ray emission of the Active Galactic Nuclei (AGN) detected by Fermi during its first year of operation, with the largest datasets ever used for this purpose. We use both archival interferometric 8.4 GHz data (from the VLA and ATCA, for the full sample of 599 sources) and concurrent single-dish 15 GHz measurements from the Owens Valley Radio Observatory (OVRO, for a sub sample of 199 objects). Our unprecedentedly large sample permits us to assess with high accuracy the statistical significance of the correlation, using a surrogate-data method designed to simultaneously account for common-distance bias and the effect of a limited dynamical range in the observed quantities. We find that the statistical significance of a positive correlation between the cm radio and the broad band (E>100 MeV) gamma-ray energy flux is very high for the whole AGN sample, with a probability <1e-7 for the correlation appearing by chance. Using the OVRO data, we find that concurrent data improve the significance of the correlation from 1.6e-6 to 9.0e-8. Our large sample size allows us to study the dependence of correlation strength and significance on specific source types and gamma-ray energy band. We find that the correlation is very significant (chance probability <1e-7) for both FSRQs and BL Lacs separately; a dependence of the correlation strength on the considered gamma-ray energy band is also present, but additional data will be necessary to constrain its significance.
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2020-11-30 12:58:08
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# Is the product of integers M and N even?
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Is the product of integers M and N even? [#permalink]
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09 Apr 2015, 05:04
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Is the product of integers M and N even?
(1) N can be expressed as a difference of squares of two consecutive prime numbers at least one of which is odd. M can be expressed as a product of two natural numbers P and Q, where 2P + 1= Q.
(2) N can be expressed as a difference of squares of two consecutive prime numbers which lie at a distance of 2 units. M is the sum of all the numbers from 1 to Z where (Z+1) is a multiple of 4.
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Last edited by EgmatQuantExpert on 13 Dec 2016, 05:45, edited 3 times in total.
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Re: Is the product of integers M and N even? [#permalink]
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09 Apr 2015, 09:26
1) For N, we can surely say that it is odd. Since 2 & 3 are the only consecutive prime numbers, N = $$3^2 - 2^2$$ = 5 which is odd.
Form M, we know that M = P*Q where Q is odd. But we don't know anything about P. So nature of M can't be found.
So, insufficient.
2) For N, we can surely say that it is even because difference of squares of any two prime numbers which differ by 2 shall always be even.This is because N = $$(X+2)^2-X^2 = X^2 + 4 + 2X - X^2$$ = 2X + 4 which is even for any number X.
Since N is always even here so product of M & N will be even no matter what M is.
Sufficient.
So B.
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Re: Is the product of integers M and N even? [#permalink]
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09 Apr 2015, 10:05
Wow.. really a thriller question. Took me somewhere around 3 mins.
Solution
The only central idea is to know if either M or N or Both are even numbers.
From Option, B it is quite clear that N= even. [Exactly at that point, I stopped further calculation].
EgmatQuantExpert wrote:
Is the product of integers M and N even?
(1) N can be expressed as a difference of squares of two consecutive prime numbers at least one of which is odd. M can be expressed as a product of two natural numbers P and Q, where 2P + 1= Q.
(2) N can be expressed as a difference of squares of two consecutive prime numbers which lie at a distance of 2 units. M is the sum of all the numbers from 1 to Z where (Z+1) is a multiple of 4.
We will provide the OA in some time. Till then Happy Solving
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Re: Is the product of integers M and N even? [#permalink]
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09 Apr 2015, 10:36
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Hi tapas.shyam
By 2 consecutive prime numbers it need not neccesarily mean that it would be 2 and 3.According to me, it means that out of primes numbers like 2,3,5,7... the numbers would be consecutive.
Also, in condition 2, M would also be even.
Consider multiples of 4- 4,8,12
So, Z would be 3,7,11
You can check the sum of these numbers using n(n+1)/2. It will be even always.
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Re: Is the product of integers M and N even? [#permalink]
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09 Apr 2015, 17:52
Naina1 wrote:
Hi tapas.shyam
By 2 consecutive prime numbers it need not neccesarily mean that it would be 2 and 3.According to me, it means that out of primes numbers like 2,3,5,7... the numbers would be consecutive.
Also, in condition 2, M would also be even.
Consider multiples of 4- 4,8,12
So, Z would be 3,7,11
You can check the sum of these numbers using n(n+1)/2. It will be even always.
Hi Naina
Thanks for your reply and opinion. However, I think my reasoning is correct as the problem says - 2 consecutive prime numbers atleast one of which is odd. Now, except 2 & 3 there is no other case where we can have an even prime number - so emphasising 'atleast one is odd' in the problem then would seem redundant. In every other case of consecutive prime numbers, we shall always have two odd prime numbers no matter what.
Secondly, in second condition, we don't even need to look into M as once we have ascertained that N is even, we can conclude that M*N shall be even no matter what value of M.
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Re: Is the product of integers M and N even? [#permalink]
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09 Apr 2015, 18:21
answer is B. Second one says the two prime numbers is at a distance of two hence "2" is ruled out since the next prime 3 is one away from it and 5 is 3 away from 2. Then the difference between squares of two off terms is even. N is even the value of M doesnt matter.
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Re: Is the product of integers M and N even? [#permalink]
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10 Apr 2015, 06:53
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Detailed Solution
Step-I: Given Info
We are given two integers M and N and are asked to find if their product is even.
Step-II: Interpreting the Question Statement
The product of two numbers would be even if at least one of them is even. So, we need to find if either of M and N is even.
Step-III: Statement-I
The statement tells us that M is expressed as a difference of two consecutive prime numbers of which at least one is odd. Two cases are possible:
• We know that there is only one even prime number i.e. 2, so, if one of the prime numbers is 2, the other would be 3, which is odd. Squaring them would not change their even/odd nature. The difference of an even and an odd number would be odd, so N would be odd
• If both the prime numbers are odd, then the difference of their squares would be even (as odd-odd= even). So, N would be even.
From the above two cases, we can’t say with certainty whether N is odd or even.
The statement also tells us that M is a product of P & Q where Q= 2P + 1. We can infer from this that Q is an odd number, but we do not have any information about the even/odd nature of P. So, if P is odd, M would be odd and if P is even, M would be even. Hence, we can’t say with certainty whether M is odd or even.
Since, we don’t know with certainty that either of M, N is even or not, Statement-I is insufficient to answer the question.
Step-IV: Statement-II
Statement-II tells us that N can be expressed as difference of squares of two consecutive prime numbers which lie at a distance of 2 units. We know that all the prime numbers except 2 are odd, since the next prime number after 2 is 3, we can say that 2, 3 are not the consecutive prime numbers (as they lie at a distance of 1 unit). Thus we can conclude that N can be expressed as difference of two odd prime numbers. The difference of two odd numbers will be even.
So, N would be even.
Note here that we don’t need to find the even/odd nature of M because irrespective of the nature of M, the product of M & N would always be even as N is even.
Hence, Statement-II is sufficient to answer the question.
Step-V: Combining Statements I & II
Since, we have a unique answer from Statement- I we don’t need to be combine Statements- I & II.
Hence, the correct answer is Option B
Key Takeaways
1. Know the properties of Even-Odd combinations to save the time spent deriving them in the test.
2. There is only 1 even prime number i.e. 2.
3. Odd/Even number raised to any power would not change its even/odd nature
tapas.shyam- when we say that at least one is odd, it means that either both are odd primes or one is odd prime and one is even prime. From the analysis of St-I we can't say with certainty that N is odd/even
Naina1- In statement-II we do not need to calculate the even/odd nature of M once we have established that N is even, as their product would always be even.
Regards
Harsh
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Re: Is the product of integers M and N even? [#permalink]
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12 Apr 2015, 13:10
EgmatQuantExpert wrote:
Detailed Solution
Step-I: Given Info
We are given two integers M and N and are asked to find if their product is even.
Step-II: Interpreting the Question Statement
The product of two numbers would be even if at least one of them is even. So, we need to find if either of M and N is even.
Step-III: Statement-I
The statement tells us that M is expressed as a difference of two consecutive prime numbers of which at least one is odd. Two cases are possible:
• We know that there is only one even prime number i.e. 2, so, if one of the prime numbers is 2, the other would be 3, which is odd. Squaring them would not change their even/odd nature. The difference of an even and an odd number would be odd, so N would be odd
• If both the prime numbers are odd, then the difference of their squares would be even (as odd-odd= even). So, N would be even.
From the above two cases, we can’t say with certainty whether N is odd or even.
The statement also tells us that M is a product of P & Q where Q= 2P + 1. We can infer from this that Q is an odd number, but we do not have any information about the even/odd nature of P. So, if P is odd, M would be odd and if P is even, M would be even. Hence, we can’t say with certainty whether M is odd or even.
Since, we don’t know with certainty that either of M, N is even or not, Statement-I is insufficient to answer the question.
Step-IV: Statement-II
Statement-II tells us that N can be expressed as difference of squares of two consecutive prime numbers which lie at a distance of 2 units. We know that all the prime numbers except 2 are odd, since the next prime number after 2 is 3, we can say that 2, 3 are not the consecutive prime numbers (as they lie at a distance of 1 unit). Thus we can conclude that N can be expressed as difference of two odd prime numbers. The difference of two odd numbers will be even.
So, N would be even.
Note here that we don’t need to find the even/odd nature of M because irrespective of the nature of M, the product of M & N would always be even as N is even.
Hence, Statement-II is sufficient to answer the question.
Step-V: Combining Statements I & II
Since, we have a unique answer from Statement- I we don’t need to be combine Statements- I & II.
Hence, the correct answer is Option B
Key Takeaways
1. Know the properties of Even-Odd combinations to save the time spent deriving them in the test.
2. There is only 1 even prime number i.e. 2.
3. Odd/Even number raised to any power would not change its even/odd nature
tapas.shyam- when we say that at least one is odd, it means that either both are odd primes or one is odd prime and one is even prime. From the analysis of St-I we can't say with certainty that N is odd/even
Naina1- In statement-II we do not need to calculate the even/odd nature of M once we have established that N is even, as their product would always be even.
Regards
Harsh
Harsh, thanks for the detailed solution. I have a doubt in statement #1 (I am definitely missing something here!). If Q= 2P+1, with both Q and being NATURAL NUMBERS, we can clearly see that Q is Odd. This leads to show that P will then be an even number. Thus, with one of P/Q determined to be an even number >> M=even. So MN = even irrespective of what N is. Thus the OA in my opinion should be D (both are sufficient!).
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Re: Is the product of integers M and N even? [#permalink]
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12 Apr 2015, 21:14
Engr2012 wrote:
EgmatQuantExpert wrote:
Detailed Solution
Step-I: Given Info
We are given two integers M and N and are asked to find if their product is even.
Step-II: Interpreting the Question Statement
The product of two numbers would be even if at least one of them is even. So, we need to find if either of M and N is even.
Step-III: Statement-I
The statement tells us that N is expressed as a difference of two consecutive prime numbers of which at least one is odd. Two cases are possible:
• We know that there is only one even prime number i.e. 2, so, if one of the prime numbers is 2, the other would be 3, which is odd. Squaring them would not change their even/odd nature. The difference of an even and an odd number would be odd, so N would be odd
• If both the prime numbers are odd, then the difference of their squares would be even (as odd-odd= even). So, N would be even.
From the above two cases, we can’t say with certainty whether N is odd or even.
The statement also tells us that M is a product of P & Q where Q= 2P + 1. We can infer from this that Q is an odd number, but we do not have any information about the even/odd nature of P. So, if P is odd, M would be odd and if P is even, M would be even. Hence, we can’t say with certainty whether M is odd or even.
Since, we don’t know with certainty that either of M, N is even or not, Statement-I is insufficient to answer the question.
Step-IV: Statement-II
Statement-II tells us that N can be expressed as difference of squares of two consecutive prime numbers which lie at a distance of 2 units. We know that all the prime numbers except 2 are odd, since the next prime number after 2 is 3, we can say that 2, 3 are not the consecutive prime numbers (as they lie at a distance of 1 unit). Thus we can conclude that N can be expressed as difference of two odd prime numbers. The difference of two odd numbers will be even.
So, N would be even.
Note here that we don’t need to find the even/odd nature of M because irrespective of the nature of M, the product of M & N would always be even as N is even.
Hence, Statement-II is sufficient to answer the question.
Step-V: Combining Statements I & II
Since, we have a unique answer from Statement- I we don’t need to be combine Statements- I & II.
Hence, the correct answer is Option B
Key Takeaways
1. Know the properties of Even-Odd combinations to save the time spent deriving them in the test.
2. There is only 1 even prime number i.e. 2.
3. Odd/Even number raised to any power would not change its even/odd nature
tapas.shyam- when we say that at least one is odd, it means that either both are odd primes or one is odd prime and one is even prime. From the analysis of St-I we can't say with certainty that N is odd/even
Naina1- In statement-II we do not need to calculate the even/odd nature of M once we have established that N is even, as their product would always be even.
Regards
Harsh
Harsh, thanks for the detailed solution. I have a doubt in statement #1 (I am definitely missing something here!). If Q= 2P+1, with both Q and being NATURAL NUMBERS, we can clearly see that Q is Odd. This leads to show that P will then be an even number. Thus, with one of P/Q determined to be an even number >> M=even. So MN = even irrespective of what N is. Thus the OA in my opinion should be D (both are sufficient!).
Hi Engr2012, You are absolutely right when you say Q is an odd number as it is in the form of 2P + 1, but if you observe you would see that Q will be an odd number irrespective of the even/odd nature of P as 2P would always be even irrespective of whether P is odd or even.
For example: Consider P = 10, Q= 21, also if P= 11, Q =23. So in both the cases, Q would be odd even when P is even or odd.
Since, we don't know for sure the even/odd nature of P in statement-I , we would not be able to comment on even/odd nature of N.
Hope its's clear
Regards
Harsh
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Re: Is the product of integers M and N even? [#permalink]
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13 Apr 2015, 01:56
Thanks for the reply. I understand your point but let me present to you another way of looking at the given information. If Q=2P+1, we know for sure that Q will be ODD. Also we can write P= (Q-1)/2 and as Q is odd, P has to be even.
Let me know where I'm going wrong with my thinking!
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Re: Is the product of integers M and N even? [#permalink]
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13 Apr 2015, 03:13
Engr2012 Assume Q as 7, so Q-1= 6 which is even. If I divide (Q-1) by 2 I would get 3, which is odd.
Similarly, assume Q=9, so Q-1 = 8. If I divide (Q-1) by 2 I would get 4, which is even. Hence, Q can be even or odd.
An even number can be represented in the form of 2n. When this number is divided by 2, the resultant would be n. Now, n can be even or odd. So, we can say that a number when divided by 2 would still be even, if the original number was a multiple of at least $$2^2$$ i.e. 4. In this case, we know that (Q-1) is even but do not know if (Q-1) is a multiple of 4. Hence, $$\frac{(Q-1)}{2}$$ may be even or odd.
Regards
Harsh
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Re: Is the product of integers M and N even? [#permalink]
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13 Apr 2015, 09:44
Hi Harsh
Thanks for the reply. Yes, now I understand my mistake. I was ignoring that 6 is an odd multiple of 2, in effect making P both odd and even with different sets of values for Q. Should've been more careful with my assumption.
Thanks
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Re: Is the product of integers M and N even? [#permalink]
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08 Nov 2015, 21:46
tapas.shyam wrote:
1) For N, we can surely say that it is odd. Since 2 & 3 are the only consecutive prime numbers, N = $$3^2 - 2^2$$ = 5 which is odd.
Form M, we know that M = P*Q where Q is odd. But we don't know anything about P. So nature of M can't be found.
So, insufficient.
2) For N, we can surely say that it is even because difference of squares of any two prime numbers which differ by 2 shall always be even.This is because N = $$(X+2)^2-X^2 = X^2 + 4 + 2X - X^2$$ = 2X + 4 which is even for any number X.
Since N is always even here so product of M & N will be even no matter what M is.
Sufficient.
So B.
I think consecutive prime number can be 3,5,7,11,13,. Am i Correct. Please clarify.
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Re: Is the product of integers M and N even? [#permalink]
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11 Dec 2017, 21:38
EgmatQuantExpert wrote:
Is the product of integers M and N even?
(1) N can be expressed as a difference of squares of two consecutive prime numbers at least one of which is odd. M can be expressed as a product of two natural numbers P and Q, where 2P + 1= Q.
(2) N can be expressed as a difference of squares of two consecutive prime numbers which lie at a distance of 2 units. M is the sum of all the numbers from 1 to Z where (Z+1) is a multiple of 4.
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Hello,
Thanks for your response but I have a question.
In the first statement, "N can be expressed as a difference of squares of two consecutive prime numbers", only consecutive prime nos are 2 and 3. So doesn't become the case in hand?
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Re: Is the product of integers M and N even? [#permalink]
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11 Dec 2017, 21:45
SamriddhiPan wrote:
EgmatQuantExpert wrote:
Is the product of integers M and N even?
(1) N can be expressed as a difference of squares of two consecutive prime numbers at least one of which is odd. M can be expressed as a product of two natural numbers P and Q, where 2P + 1= Q.
(2) N can be expressed as a difference of squares of two consecutive prime numbers which lie at a distance of 2 units. M is the sum of all the numbers from 1 to Z where (Z+1) is a multiple of 4.
This is
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Hello,
Thanks for your response but I have a question.
In the first statement, "N can be expressed as a difference of squares of two consecutive prime numbers", only consecutive prime nos are 2 and 3. So doesn't become the case in hand?
The two consecutive integers, of which both are primes are indeed 2 and 3. But consecutive primes are {2, 3}, {3, 5}, {5, 7}, {7, 11}...
_________________
Kudos [?]: 139224 [0], given: 12779
Re: Is the product of integers M and N even? [#permalink] 11 Dec 2017, 21:45
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2018-01-17 03:39:57
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https://wiki.math.ntnu.no/tma4183/2021v/ovinger
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# Exercises
Here you can find exercise sheets and, at a later point, proposed solutions.
If you get stuck in some of the exercises (but no sooner than that!), you may have a look at the hints.
Session on Exercise Hints Suggested Solution
January 28 Exercise 1 Hints 1 Solutions 1
February 4 Exercise 2 Hints 2 Solutions 2
February 11 Exercise 31) Hints 3 Solutions 3
February 18 Exercise 4 Hints 4 Solutions 4
February 25 Exercise 5 Hints 5
March 25 Exercise 6
April 8 Exercise 7 Hints 7
April 15 Exercise 8
April 22 Exercise 9 Hints 9
1)
Unfortunately, I was not reading through my problems carefully enough before uploading them, and thus Problem 2b turned out to be completely wrong: Actually, this PDE has never a unique solution; either it is unsolvable or it has infinitely many solutions. In order to make the problem work as intended, it is necessary to replace the term $fy$ in the integral over $\Omega$ by $(f-y)^2$.
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2021-05-16 06:53:49
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https://lost-stats.github.io/Time_Series/MA_Model.html
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# Moving Average (MA) Models
Moving Average Models are a common method for modeling linear processes that exhibit serial correlation. This can take many forms, but an easy example is weather patterns.
Imagine we modelled Ski Lift Purchases over a 5 consecutive weekday period. Let weather shocks be identically and independently distributed. A weather shock with large amounts of snow would cause more individuals to go ski. This shock would have future impacts on ski ticket purchases; individuals tomorrow will go skiing due to the snowfall today. This means when we think of the process, we should account for previous shock effects on outcomes today:
$SkiTicketPurchases_t = \mu + \theta_1*Weather_t + \theta_2*Weather_{t-1}$
## Definition
Let $$\epsilon_t \sim N(0, \sigma^2_{\epsilon})$$. A moving average process, which is denoted MA(q), take the form as:
$y_t = \mu + \epsilon_t + \theta_1 \epsilon_{t-1} + ... + \theta_q \epsilon_{t-q}$
Note that a MA(q) process has q lagged $$\epsilon$$ terms. Thus a MA(1) process would take the form:
$y_t = \mu + \epsilon_t + \theta_1 \epsilon_{t-1}$
## Properties of a MA(1) process:
1. The mean is constant.
$E(y_t) = E(\mu + \epsilon_t + \theta_1 \epsilon_{t-1}) = \mu$
1. The variance is constant.
$Var(y_t) = Var(\mu + \epsilon_t + \theta_1 \epsilon_{t-1}) = (1 + \theta_1^2) * \sigma^2_{\epsilon}$
1. The covariance between $$y_t$$ and $$y_{t-q}$$ is decreasing as $$q \to \infty$$.
$Cov(y_t, y_{t-1}) = E(y_t*y_{t-1}) - E(y_t)E(y_{t-1}) \\ = E([\mu + \epsilon_t + \theta_1 \epsilon_{t-1}] [\mu + \epsilon_{t-1} + \theta_1 \epsilon_{t-2}]) - \mu^2 \\ = \mu^2 + \theta_1 \sigma_{\epsilon}^2 -\mu^2 \\ = \theta_1 \sigma_{\epsilon}^2$ $Cov(y_t, y_{t-2}) = E(y_t*y_{t-2}) - E(y_t)E(y_{t-2}) \\ = E([\mu + \epsilon_t + \theta_1 \epsilon_{t-1}] [\mu + \epsilon_{t-2} + \theta_1 \epsilon_{t-3}]) - \mu^2 \\ = \mu^2 -\mu^2 \\ = 0$
Additional helpful information can be found at Wikipedia: Moving Average Models
## Keep in Mind
• Time series data needs to be properly formatted (e.g. date columns should be formatted into a time)
• Model Selection uses the Akaike Information Criterion (AIC), Bayesian Information Criterion (BIC), or the Akaike Information Criterion corrected (AICc) to determine the appropriate number of terms to include. Refer to Wikipedia:Model Selection for further information.
# Implementations
## Julia
MA(q) models in Julia can be estimated using the StateSpaceModels.jl package, which also allows for the estimation of a variety of time series models that have linear state-space representations.
Begin by importing and loading necessary packages into your work environment.
# Load necessary packages
using StateSpaceModels, LinearAlgebra, ShiftedArrays
We may simulate an MA(3) model, where the MA coefficients are set to $$\theta_1 = 0.5$$, $$\theta_2 = 0.3$$, and $$\theta_3 = 0.2$$.
# Set parameters
theta1, theta2, theta3 = (0.5, 0.3, 0.2)
# Draw a sample of an iid N(0,1) disturbance/noise process
# Set number of observations to 1000
u = randn(1000)
# Simulate MA(3) process
Y = u + theta1 .* lag(u, 1, default = 0.0) + theta2 .* lag(u, 2, default = 0.0) + theta3 .* lag(u, 3, default = 0.0)
# Remove first 3 observations
Y = Y[4:end]
The lag(u, k, default = 0.0) portions of the above code chunk creates a k-lag of series u, and sets the missing observations equal to 0.0.
The data generated by a single sample draw from the above data-generating process can then be assigned a general ARIMA(p,d,q) representation, where if p and d are set to zero, the model specification becomes MA(q). The p=d= 0 constraint can be applied by inputting order = (0,0,q), where q>0.
# Specify the generated Y series as an MA(3) model
model = SARIMA(Y, order = (0,0,3))
Lastly, the above-specified model can be estimated using the fit! function, and the estimation results printed using the results function. The sole input for both of these functions is the model object that contains the chosen data series and its assigned ARIMA structure.
# Fit (estimate) the model
fit!(model)
# Print estimates
results(model)
## R
#in the stats package we can simulate an ARIMA Model. ARIMA stands for Auto-Regressive Integrated Moving Average model. We will be setting the AR and I parts to 0 and only simulating a MA(q) model.
set.seed(123)
DT = arima.sim(n = 1000, model = list(ma = c(0.1, 0.3, 0.5)))
plot(DT, ylab = "Value")
set.seed(123)
DT = arima.sim(n = 1000, model = list(ma = c(0.1, 0.3, 0.5)))
#ACF stands for Autocorrelation Function
#Here we can see that there may be potential for 3 lags in our MA process. (Note: This is due to property (3): the covariance of y_t and y_{t-3} is nonzero while the covariance of y_t and y_{t-4} is 0)
acf(DT, type = "covariance")
set.seed(123)
DT = arima.sim(n = 1000, model = list(ma = c(0.1, 0.3, 0.5)))
#Here I'm estimating an ARIMA(0,0,3) model which is a MA(3) model. Changing c(0,0,q) allows us to estimate a MA(q) process.
arima(x = DT, order = c(0,0,3))
##
## Call:
## arima(x = DT, order = c(0, 0, 3))
##
## Coefficients:
## ma1 ma2 ma3 intercept
## 0.0722 0.2807 0.4781 0.0265
## s.e. 0.0278 0.0255 0.0294 0.0573
##
## sigma^2 estimated as 0.9825: log likelihood = -1410.63, aic = 2831.25
#We can also estimate a MA(7) model and see that the ma4, ma5, ma6, and ma7 are close to 0 and insignificant.
arima(x = DT, order = c(0,0,7))
##
## Call:
## arima(x = DT, order = c(0, 0, 7))
##
## Coefficients:
## ma1 ma2 ma3 ma4 ma5 ma6 ma7 intercept
## 0.0714 0.2694 0.4607 -0.0119 -0.0380 -0.0256 -0.0219 0.0267
## s.e. 0.0316 0.0321 0.0324 0.0363 0.0339 0.0332 0.0328 0.0533
##
## sigma^2 estimated as 0.9806: log likelihood = -1409.65, aic = 2837.3
#fable is a package designed to estimate ARIMA models. We can use it to estimate our MA(3) model.
library(fable)
#an extension of tidyverse to temporal data (this allows us to create time series data into tibbles which are needed for fable functionality)
library(tsibble)
#visit https://dplyr.tidyverse.org/ to understand dplyr syntax; this package is important for fable functionality
library(dplyr)
#When using the fable package, we need to convert our object into a tsibble (a time series tibble). This gives us a data frame with values and an index for the time periods
DT = DT %>%
as_tsibble()
## # A tsibble: 6 x 2 [1]
## index value
## <dbl> <dbl>
## 1 1 -0.123
## 2 2 0.489
## 3 3 2.53
## 4 4 0.706
## 5 5 -0.640
## 6 6 0.182
#Now we can use the dplyr package to pipe our dataset and create a fitted model
#Note: the ARIMA function in the fable package uses an information criterion for model selection; these can be set as shown below; additional information is above in the Keep in Mind section (the default criterion is aicc)
MAfit = DT %>%
model(arima = ARIMA(value, ic = "aicc"))
#report() is needed to view our model
report(MAfit)
## Series: value
## Model: ARIMA(0,0,3)
##
## Coefficients:
## ma1 ma2 ma3
## 0.0723 0.2808 0.4782
## s.e. 0.0278 0.0255 0.0294
##
## sigma^2 estimated as 0.9857: log likelihood=-1410.73
## AIC=2829.47 AICc=2829.51 BIC=2849.1
#if instead we want to specify the model manually, we need to specify it. For MA models, set the pdq(0,0,q) term to the MA(q) order you want to estimate. For example: Estimating a MA(7) would mean that I should put pdq(0,0,7). Additionally, you can add a constant if wanted; this is shown below
#with constant
MAfit = DT %>%
model(arima = ARIMA(value ~ 1 + pdq(0,0,3), ic = "aicc"))
report(MAfit)
## Series: value
## Model: ARIMA(0,0,3) w/ mean
##
## Coefficients:
## ma1 ma2 ma3 constant
## 0.0722 0.2807 0.4781 0.0265
## s.e. 0.0278 0.0255 0.0294 0.0573
##
## sigma^2 estimated as 0.9865: log likelihood=-1410.63
## AIC=2831.25 AICc=2831.31 BIC=2855.79
#without constant
MAfit = DT %>%
model(arima = ARIMA(value ~ 0 + pdq(0,0,3), ic = "aicc"))
report(MAfit)
## Series: value
## Model: ARIMA(0,0,3)
##
## Coefficients:
## ma1 ma2 ma3
## 0.0723 0.2808 0.4782
## s.e. 0.0278 0.0255 0.0294
##
## sigma^2 estimated as 0.9857: log likelihood=-1410.73
## AIC=2829.47 AICc=2829.51 BIC=2849.1
#A faster, more compact way to write a code would be as follows:
#Automatic estimation
DT %>%
as_tsibble() %>%
model(arima = ARIMA(value)) %>%
report()
## Series: value
## Model: ARIMA(0,0,3)
##
## Coefficients:
## ma1 ma2 ma3
## 0.0723 0.2808 0.4782
## s.e. 0.0278 0.0255 0.0294
##
## sigma^2 estimated as 0.9857: log likelihood=-1410.73
## AIC=2829.47 AICc=2829.51 BIC=2849.1
#Manual estimation
DT %>%
as_tsibble() %>%
model(arima = ARIMA(value ~ 0 + pdq(0, 0, 3))) %>%
report()
## Series: value
## Model: ARIMA(0,0,3)
##
## Coefficients:
## ma1 ma2 ma3
## 0.0723 0.2808 0.4782
## s.e. 0.0278 0.0255 0.0294
##
## sigma^2 estimated as 0.9857: log likelihood=-1410.73
## AIC=2829.47 AICc=2829.51 BIC=2849.1
|
2022-08-17 07:16:55
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https://astro.paperswithcode.com/paper/reconstruction-of-the-neutrino-mass-as-a
|
# Reconstruction of the neutrino mass as a function of redshift
26 Feb 2021 · Christiane S. Lorenz, Lena Funcke, Matthias Löffler, Erminia Calabrese ·
We reconstruct the neutrino mass as a function of redshift, z, from current cosmological data using both standard binned priors and linear spline priors with variable knots. Using cosmic microwave background temperature, polarization and lensing data, in combination with distance measurements from baryonic acoustic oscillations and supernovae, we find that the neutrino mass is consistent with $\sum m_\nu(z)$ = const. We obtain a larger bound on the neutrino mass at low redshifts coinciding with the onset of dark energy domination, $\sum m_\nu(z = 0)$ < 1.46 eV (95% CL). This result can be explained either by the well-known degeneracy between $\sum m_\nu$ and $\Omega_\Lambda$ at low redshifts, or by models in which neutrino masses are generated very late in the Universe. We finally convert our results into cosmological limits for models with non-relativistic neutrino decay and find $\sum m_\nu$ < 0.21 eV (95% CL), which would be out of reach for the KATRIN experiment.
PDF Abstract
## Code Add Remove Mark official
No code implementations yet. Submit your code now
## Categories
Cosmology and Nongalactic Astrophysics High Energy Physics - Phenomenology
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2023-01-29 05:08:46
|
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https://atcoder.jp/contests/agc005/tasks/agc005_e
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Contest Duration: - (local time) (110 minutes) Back to Home
E - Sugigma: The Showdown /
Time Limit: 2 sec / Memory Limit: 256 MB
### 問題文
しぐま君とすぎむ君はゲームをすることにしました。
このゲームは N 頂点のグラフの上で行います。頂点には 1,2,...,N と番号がついています。グラフには N-1 本の赤い辺と N-1 本の青い辺があり、どちらも木となっています。赤い辺は (a_i, b_i) で表し、青い辺は (c_i, d_i) で表します。
このゲームはターン制で、ターン 1, ターン 2, ターン 3, ... と進んでいきます。そして、ターン 1, 3, 5, ... はしぐま君、ターン 2, 4, 6, ... はすぎむ君の手番です。
もし二つの駒が同じ頂点に来るとその時点でゲームは終了します。そしてターン i での操作の後にゲームが終了したならば、i を総ターン数とします。
しぐま君は総ターン数を出来る限り大きく、すぎむ君は出来る限り小さくしたいです。
### 制約
• 2 ≦ N ≦ 200,000
• 1 ≦ X, Y ≦ N
• X \neq Y
• 1 ≦ a_i, b_i, c_i, d_i ≦ N
• 赤い辺と青い辺はそれぞれ木である
### 入力
N X Y
a_1 b_1
a_2 b_2
:
a_{N-1} b_{N-1}
c_1 d_1
c_2 d_2
:
c_{N-1} d_{N-1}
### 出力
1 行に答えを出力する。 ただし、いつまでもゲームが終了しない場合は -1 を出力する。
### 入力例 1
4 1 2
1 2
1 3
1 4
2 1
2 3
1 4
### 出力例 1
4
### 入力例 2
3 3 1
1 2
2 3
1 2
2 3
### 出力例 2
4
### 入力例 3
4 1 2
1 2
3 4
2 4
1 2
3 4
1 3
### 出力例 3
2
### 入力例 4
4 2 1
1 2
3 4
2 4
1 2
3 4
1 3
### 出力例 4
-1
### 入力例 5
5 1 2
1 2
1 3
1 4
4 5
2 1
1 3
1 5
5 4
### 出力例 5
6
Score : 1400 points
### Problem Statement
Sigma and Sugim are playing a game.
The game is played on a graph with N vertices numbered 1 through N. The graph has N-1 red edges and N-1 blue edges, and the N-1 edges in each color forms a tree. The red edges are represented by pairs of integers (a_i, b_i), and the blue edges are represented by pairs of integers (c_i, d_i).
Each player has his own piece. Initially, Sigma's piece is at vertex X, and Sugim's piece is at vertex Y.
The game is played in turns, where turns are numbered starting from turn 1. Sigma takes turns 1, 3, 5, ..., and Sugim takes turns 2, 4, 6, ....
In each turn, the current player either moves his piece, or does nothing. Here, Sigma can only move his piece to a vertex that is directly connected to the current vertex by a red edge. Similarly, Sugim can only move his piece to a vertex that is directly connected to the current vertex by a blue edge.
When the two pieces come to the same vertex, the game ends immediately. If the game ends just after the operation in turn i, let i be the total number of turns in the game.
Sigma's objective is to make the total number of turns as large as possible, while Sugim's objective is to make it as small as possible.
Determine whether the game will end in a finite number of turns, assuming both players plays optimally to achieve their respective objectives. If the answer is positive, find the number of turns in the game.
### Constraints
• 2 ≦ N ≦ 200,000
• 1 ≦ X, Y ≦ N
• X \neq Y
• 1 ≦ a_i, b_i, c_i, d_i ≦ N
• The N-1 edges in each color (red and blue) forms a tree.
### Input
The input is given from Standard Input in the following format:
N X Y
a_1 b_1
a_2 b_2
:
a_{N-1} b_{N-1}
c_1 d_1
c_2 d_2
:
c_{N-1} d_{N-1}
### Output
If the game will end in a finite number of turns, print the number of turns. Otherwise, print -1.
### Sample Input 1
4 1 2
1 2
1 3
1 4
2 1
2 3
1 4
### Sample Output 1
4
### Sample Input 2
3 3 1
1 2
2 3
1 2
2 3
### Sample Output 2
4
### Sample Input 3
4 1 2
1 2
3 4
2 4
1 2
3 4
1 3
### Sample Output 3
2
### Sample Input 4
4 2 1
1 2
3 4
2 4
1 2
3 4
1 3
### Sample Output 4
-1
### Sample Input 5
5 1 2
1 2
1 3
1 4
4 5
2 1
1 3
1 5
5 4
### Sample Output 5
6
|
2021-06-25 14:15:18
|
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|
https://chemistry.stackexchange.com/questions/44483/find-the-concentration-of-the-hcl-solution-when-mixed-with-1-265g-of-aces-k
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# Find the concentration of the HCl solution when mixed with 1.265g of ACES-K+
$1.265\ \mathrm g$ of N-(2-acetamido)-2-aminoethanesulfonic acid potassium salt (ACES-K+, $M=220.29\ \mathrm{g/mol}$) is dissolved in $88.42\ \mathrm{mL}$ of water. $27.59\ \mathrm{mL}$ of $\ce{HCl}$ is added to the solution, resulting in $\mathrm{pH}$ of $6.54$. Calculate the concentration of the $\ce{HCl}$ solution. The $\mathrm pK_\mathrm a$ of ACES is $6.85$.
I tried setting up the equation as follows: $$\ce{ACES + H3O+ <=> HACES +H2O}$$ The moles of ACES initial is $0.0057\ \mathrm{mol}$ $(1.265/22.290)$ and we are trying to find $x$ the amount of moles of $\ce{HCl}$ $\ce{(H3O+)}$ we have. Therefore, using an ice table $\ce{ACES}=0.00574-x\ \mathrm{mol}$ and $\ce{HCl}= x\ \mathrm{mol}$. When you plus into the equation $\mathrm{pH}=\mathrm pK_\mathrm a+\log(\text{base}/\text{acid})$, I did $6.54=6.85+\log(0.00574-x/x)$ and got $0.0038\ \mathrm{mol}$ of acid divided by total volume = $0.03321\ \mathrm M$. Is that correct?
## 1 Answer
Let us call the potassium salt "A-" and its corresponding acid for "HA". The total volume of the final solution is 88.42 + 27.59 = 116.01 ml. The final concentration of [HA] + [A-] = 1.265/(220.29 *0.11601) = 0.049499 (M). [K+] = [HA] + [A-] = 0.049499 (M).
The equilibrium equation gives: 6.54 = 6.85 + log ([A-]/[HA]) => [A-]/[HA] = 0.48977 => [A-] = 0.48977 * [HA]. => [HA] + 0.48977 * [HA] = 0.049499 => [HA] = 0.03322 (M) and [A-] = 0.016279 (M).
The charge balance is: [H3O+] + [K+] = [OH-] + [Cl-] + [A-] At pH = 6,54 [H3O+] = 10^-6.54 and [OH-] can be neglected in comparison with [A-]. => [Cl-] = [H3O+] + [K+] - [A-] = 0.03322 (M).
The amount of chloride ions in the final solution will be 0.03322 * 0.11601 = 0.003853 mole. Thus, the concentration of the hydrochloric acid used will be: (0.003853 * 1000) /27.59 = 0.1396 or about 0.140 M.
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2019-06-25 06:11:23
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https://studyadda.com/notes/3rd-class/mental-ability/shapes/geometrical-shapes/12320
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# 3rd Class Mental Ability Shapes Geometrical Shapes
Geometrical Shapes
Category : 3rd Class
Geometrical Shapes
OBJECTIVE
• To identify different objects which represent some kind of geometrical shapes which we are coming across in our day to day life (Both 2D and 3D).
• To recognize different parts of various geometrical shapes such as vertex and sides of a particular geometrical pattern.
• Use of different geometrical shapes for creating Tangrams.
• Knowledge of line of symmetry (horizontal and vertical).
Geometric Shapes
A geometric shape is defined as a set of points or vertices and sides connecting to the point to form a closed entry. There are various kinds of geometrical shapes. Based upon the number of vertices and sides they are identified as-follows:
2D Geometric Shapes
Triangle · Triangle has three sides. · It has three vertices. Square · Squares have 4 equal sides and 4 right angles. · They have 4 lines of symmetry. · All squares belong to the rectangle family. · All squares belong to the rhombus family. · All squares are also parallelograms. Rectangle · Rectangles have 4 sides and 4 right angles. · They all have 2 lines of symmetry (4 lines if they are also a square) · All rectangles belong to the parallelogram family. Rhombus · Rhombuses (rhombii) have 4 equal sides. · Both pairs of opposite sides are parallel. · They all have 2 lines of symmetry (4 lines if they are a square) · All rhombuses belong to the parallelogram family. Parallelogram · Parallelograms have 2 pairs of parallel sides. · Some parallelograms have lines of symmetry (depending on whether they are also squares, rectangles or rhombuses), but most do not. Circle · Circles have a point in the centre from which each point on the diameter is equidistant. · They have infinite lines of symmetry. Ellipse · Ellipses are like circles which have been squashed or stretched. · They have 2 lines of symmetry. · They are also a special type of oval. · The longest and shortest diameters of the ellipses are called the major and minor axis. These axis are also the lines of symmetry. Crescent · Crescent shapes are made when two circles overlap, or when one circle is removed from another circle. · The perimeter of crescents made from two circular arcs. · They have one line of symmetry. · Our moon forms crescent shapes during its phases. Pentagon · 5 sided polygon Hexagon · 6 sided polygon Heptagon · 7 sided polygon Octagon · 8 sided polygon
3D Geometric Shapes
Cube · Cubes have 6 faces, 12 edges and 8 vertices. · All sides on a cube are equal length. · All faces are square in shape. · A cube is a type of cuboid. Cuboid · Cuboids have 6 faces, 12 edges and 8 vertices. · All the faces on a cuboid are rectangular. Sphere · Spheres have either 0 or 1 faces, 0 edges and 0 vertices. Cylinder · Cylinders have either 2 or 3 faces, 0 or 2 edges, and 0 vertices. Cone · Cones have either 1 or 2 faces, 0 or 1 edge, and 1 apex (which is described by some mathematicians as a vertex).
Regular Geometric shape
The geometric shapes whose all sides are equal sides and equal interior angles are known as regular geometric shape.
Irregular Geometric Shapes
The geometric shapes which have sides and angles of any length and size.
Example - 1
Find the number of straight lines in the given figure.
(a) 8 (b) 9
(c) 10 (d) 11
Ans. (a)
Straight lines: AB, AC, BC, BE, CD, ED, AE and AD
So, number of straight lines = 8.
Example - 2
Komal built a birdhouse at summer camp. What shape is the piece of wood that was cut out to make the door other birdhouse?
(a) Triangle (b) Diamond
(c) Circle (d) Pentagon
Ans. (c)
Circle shape is the piece of wood that was cut out to make the door of her birdhouse.
Example - 3
The figure has ____ straight lines and ______ curves.
(a) 9, 10 (b) 14, 12
(c) 16, 8 (d) 15, 10
Ans. (c)
The figure has 16 straight lines and 8 curves.
Directions (Example 4 - 5): Look at the figure carefully and answer the questions based on it.
Example - 4
How many more circles are there in the given figure than triangles?
(a) 15 (b) 16
(c) 17 (d) 18
Example - 5
Number of sleeping lines in the figure is _______.
(a) 21 (b) 16
(c) 18 (d) 19
Ans. (Example 4 - 5)
Ans. 4. (c)
Explanation
Number of circles in joker's cap = 10
Number of circles in joker's hand =6+6 =12
Number of circles in joker's face = 3
Number of circles in - triangle APQR = 5
Total number of circles =30
Number of triangles in feet = 4
Number of triangles in hands = 2
Number of triangles in face = 1
Number of triangles in $\Delta PQR=6$
Total number of triangles = 13
There are $30-13=17$more circles than triangles.
5. Ans. (a)
Explanation
Sleeping lines in jokes cap = 2
Sleeping lines in face = 2
Sleeping lines in neck =1
Sleeping lines in hands =4
Sleeping lines in $\Delta PQR=6$
Sleeping lines in legs =6
Total number of sleeping lines =21
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2019-07-21 06:31:45
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https://actuarialmodelingtopics.wordpress.com/tag/transformed-pareto-distribution/
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# A catalog of parametric severity models
Various parametric continuous probability models have been presented and discussed in this blog. The number of parameters in these models ranges from one to two, and in a small number of cases three. They are all potential candidates for models of severity in insurance applications and in other actuarial applications. This post highlights these models. The list presented here is not exhaustive; it is only a brief catalog. There are other models that are also suitable for actuarial applications but not accounted for here. However, the list is a good place to begin. This post also serves a navigation device (the table shown below contains links to the blog posts).
A Catalog
Many of the models highlighted here are related to gamma distribution either directly or indirectly. So the catalog starts with the gamma distribution at the top and then branches out to the other related models. Mathematically, the gamma distribution is a two-parameter continuous distribution defined using the gamma function. The gamma sub family includes the exponential distribution, Erlang distribution and chi-squared distribution. These are distributions that are gamma distributions with certain restrictions on the one or both of the gamma parameters. Other distributions are obtained by raising a distribution to a power. Others are obtained by mixing distributions.
Here’s a listing of the models. Click on the links to find out more about the distributions.
……Derived From ………………….Model
Gamma function
Gamma sub families
Independent sum of gamma
Exponentiation
Raising to a power Raising exponential to a positive power
Raising exponential to a power
Raising gamma to a power
Raising Pareto to a power
Burr sub families
Mixture
Others
The above table categorizes the distributions according to how they are mathematically derived. For example, the gamma distribution is derived from the gamma function. The Pareto distribution is mathematically an exponential-gamma mixture. The Burr distribution is a transformed Pareto distribution, i.e. obtained by raising a Pareto distribution to a positive power. Even though these distributions can be defined simply by giving the PDF and CDF, knowing how their mathematical origins informs us of the specific mathematical properties of the distributions. Organizing according to the mathematical origin gives us a concise summary of the models.
$\text{ }$
$\text{ }$
Further Comments on the Table
From a mathematical standpoint, the gamma distribution is defined using the gamma function.
$\displaystyle \Gamma(\alpha)=\int_0^\infty t^{\alpha-1} \ e^{-t} \ dt$
In this above integral, the argument $\alpha$ is a positive number. The expression $t^{\alpha-1} \ e^{-t}$ in the integrand is always positive. The area in between the curve $t^{\alpha-1} \ e^{-t}$ and the x-axis is $\Gamma(\alpha)$. When this expression is normalized, i.e. divided by $\Gamma(\alpha)$, it becomes a density function.
$\displaystyle f(t)=\frac{1}{\Gamma(\alpha)} \ t^{\alpha-1} \ e^{-t}$
The above function $f(t)$ is defined over all positive $t$. The integral of $f(t)$ over all positive $t$ is 1. Thus $f(t)$ is a density function. It only has one parameter, the $\alpha$, which is the shape parameter. Adding the scale parameter $\theta$ making it a two-parameter distribution. The result is called the gamma distribution. The following is the density function.
$\displaystyle f(x)=\frac{1}{\Gamma(\alpha)} \ \biggl(\frac{1}{\theta}\biggr)^\alpha \ x^{\alpha-1} \ e^{-\frac{x}{\theta}} \ \ \ \ \ \ \ x>0$
Both parameters $\alpha$ and $\theta$ are positive real numbers. The first parameter $\alpha$ is the shape parameter and $\theta$ is the scale parameter.
As mentioned above, many of the distributions listed in the above table is related to the gamma distribution. Some of the distributions are sub families of gamma. For example, when $\alpha$ are positive integers, the resulting distributions are called Erlang distribution (important in queuing theory). When $\alpha=1$, the results are the exponential distributions. When $\alpha=\frac{k}{2}$ and $\theta=2$ where $k$ is a positive integer, the results are the chi-squared distributions (the parameter $k$ is referred to the degrees of freedom). The chi-squared distribution plays an important role in statistics.
Taking independent sum of $n$ independent and identically distributed exponential random variables produces the Erlang distribution, a sub gamma family of distribution. Taking independent sum of $n$ exponential random variables, with pairwise distinct means, produces the hypoexponential distributions. On the other hand, the mixture of $n$ independent exponential random variables produces the hyperexponential distribution.
The Pareto distribution (Pareto Type II Lomax) is the mixture of exponential distributions with gamma mixing weights. Despite the connection with the gamma distribution, the Pareto distribution is a heavy tailed distribution. Thus the Pareto distribution is suitable for modeling extreme losses, e.g. in modeling rare but potentially catastrophic losses.
As mentioned earlier, raising a Pareto distribution to a positive power generates the Burr distribution. Restricting the parameters in a Burr distribution in a certain way will produces the paralogistic distribution. The table indicates the relationships in a concise way. For details, go into the blog posts to get more information.
Tail Weight
Another informative way to categorize the distributions listed in the table is through looking at the tail weight. At first glance, all the distributions may look similar. For example, the distributions in the table are right skewed distributions. Upon closer look, some of the distributions put more weights (probabilities) on the larger values. Hence some of the models are more suitable for models of phenomena with significantly higher probabilities of large or extreme values.
When a distribution significantly puts more probabilities on larger values, the distribution is said to be a heavy tailed distribution (or said to have a larger tail weight). In general tail weight is a relative concept. For example, we say model A has a larger tail weight than model B (or model A has a heavier tail than model B). However, there are several ways to check for tail weight of a given distribution. Here are the four criteria.
Tail Weight Measure What to Look for
1 Existence of moments The existence of more positive moments indicates a lighter tailed distribution.
2 Hazard rate function An increasing hazard rate function indicates a lighter tailed distribution.
3 Mean excess loss function An increasing mean excess loss function indicates a heavier tailed distribution.
4 Speed of decay of survival function A survival function that decays rapidly to zero (as compared to another distribution) indicates a lighter tailed distribution.
Existence of moments
For a positive real number $k$, the moment $E(X^k)$ is defined by the integral $\int_0^\infty x^k \ f(x) \ dx$ where $f(x)$ is the density function of the distribution in question. If the distribution puts significantly more probabilities in the larger values in the right tail, this integral may not exist (may not converge) for some $k$. Thus the existence of moments $E(X^k)$ for all positive $k$ is an indication that the distribution is a light tailed distribution.
In the above table, the only distributions for which all positive moments exist are gamma (including all gamma sub families such as exponential), Weibull, lognormal, hyperexponential, hypoexponential and beta. Such distributions are considered light tailed distributions.
The existence of positive moments exists only up to a certain value of a positive integer $k$ is an indication that the distribution has a heavy right tail. All the other distributions in the table are considered heavy tailed distribution as compared to gamma, Weibull and lognormal. Consider a Pareto distribution with shape parameter $\alpha$ and scale parameter $\theta$. Note that the existence of the Pareto higher moments $E(X^k)$ is capped by the shape parameter $\alpha$. If the Pareto distribution is to model a random loss, and if the mean is infinite (when $\alpha=1$), the risk is uninsurable! On the other hand, when $\alpha \le 2$, the Pareto variance does not exist. This shows that for a heavy tailed distribution, the variance may not be a good measure of risk.
Hazard rate function
The hazard rate function $h(x)$ of a random variable $X$ is defined as the ratio of the density function and the survival function.
$\displaystyle h(x)=\frac{f(x)}{S(x)}$
The hazard rate is called the force of mortality in a life contingency context and can be interpreted as the rate that a person aged $x$ will die in the next instant. The hazard rate is called the failure rate in reliability theory and can be interpreted as the rate that a machine will fail at the next instant given that it has been functioning for $x$ units of time.
Another indication of heavy tail weight is that the distribution has a decreasing hazard rate function. On the other hand, a distribution with an increasing hazard rate function has a light tailed distribution. If the hazard rate function is decreasing (over time if the random variable is a time variable), then the population die off at a decreasing rate, hence a heavier tail for the distribution in question.
The Pareto distribution is a heavy tailed distribution since the hazard rate is $h(x)=\alpha/x$ (Pareto Type I) and $h(x)=\alpha/(x+\theta)$ (Pareto Type II Lomax). Both hazard rates are decreasing function.
The Weibull distribution is a flexible model in that when its shape parameter is $0<\tau<1$, the Weibull hazard rate is decreasing and when $\tau>1$, the hazard rate is increasing. When $\tau=1$, Weibull is the exponential distribution, which has a constant hazard rate.
The point about decreasing hazard rate as an indication of a heavy tailed distribution has a connection with the fourth criterion. The idea is that a decreasing hazard rate means that the survival function decays to zero slowly. This point is due to the fact that the hazard rate function generates the survival function through the following.
$\displaystyle S(x)=e^{\displaystyle -\int_0^x h(t) \ dt}$
Thus if the hazard rate function is decreasing in $x$, then the survival function will decay more slowly to zero. To see this, let $H(x)=\int_0^x h(t) \ dt$, which is called the cumulative hazard rate function. As indicated above, $S(x)=e^{-H(x)}$. If $h(x)$ is decreasing in $x$, $H(x)$ has a lower rate of increase and consequently $S(x)=e^{-H(x)}$ has a slower rate of decrease to zero.
In contrast, the exponential distribution has a constant hazard rate function, making it a medium tailed distribution. As explained above, any distribution having an increasing hazard rate function is a light tailed distribution.
The mean excess loss function
The mean excess loss is the conditional expectation $e_X(d)=E(X-d \lvert X>d)$. If the random variable $X$ represents insurance losses, mean excess loss is the expected loss in excess of a threshold conditional on the event that the threshold has been exceeded. Suppose that the threshold $d$ is an ordinary deductible that is part of an insurance coverage. Then $e_X(d)$ is the expected payment made by the insurer in the event that the loss exceeds the deductible.
Whenever $e_X(d)$ is an increasing function of the deductible $d$, the loss $X$ is a heavy tailed distribution. If the mean excess loss function is a decreasing function of $d$, then the loss $X$ is a lighter tailed distribution.
The Pareto distribution can also be classified as a heavy tailed distribution based on an increasing mean excess loss function. For a Pareto distribution (Type I) with shape parameter $\alpha$ and scale parameter $\theta$, the mean excess loss is $e(X)=d/(\alpha-1)$, which is increasing. The mean excess loss for Pareto Type II Lomax is $e(X)=(d+\theta)/(\alpha-1)$, which is also decreasing. They are both increasing functions of the deductible $d$! This means that the larger the deductible, the larger the expected claim if such a large loss occurs! If the underlying distribution for a random loss is Pareto, it is a catastrophic risk situation.
In general, an increasing mean excess loss function is an indication of a heavy tailed distribution. On the other hand, a decreasing mean excess loss function indicates a light tailed distribution. The exponential distribution has a constant mean excess loss function and is considered a medium tailed distribution.
Speed of decay of the survival function to zero
The survival function $S(x)=P(X>x)$ captures the probability of the tail of a distribution. If a distribution whose survival function decays slowly to zero (equivalently the cdf goes slowly to one), it is another indication that the distribution is heavy tailed. This point is touched on when discussing hazard rate function.
The following is a comparison of a Pareto Type II survival function and an exponential survival function. The Pareto survival function has parameters ($\alpha=2$ and $\theta=2$). The two survival functions are set to have the same 75th percentile, which is $x=2$. The following table is a comparison of the two survival functions.
$\displaystyle \begin{array}{llllllll} \text{ } &x &\text{ } & \text{Pareto } S_X(x) & \text{ } & \text{Exponential } S_Y(x) & \text{ } & \displaystyle \frac{S_X(x)}{S_Y(x)} \\ \text{ } & \text{ } & \text{ } & \text{ } & \text{ } \\ \text{ } &2 &\text{ } & 0.25 & \text{ } & 0.25 & \text{ } & 1 \\ \text{ } &10 &\text{ } & 0.027777778 & \text{ } & 0.000976563 & \text{ } & 28 \\ \text{ } &20 &\text{ } & 0.008264463 & \text{ } & 9.54 \times 10^{-7} & \text{ } & 8666 \\ \text{ } &30 &\text{ } & 0.00390625 & \text{ } & 9.31 \times 10^{-10} & \text{ } & 4194304 \\ \text{ } &40 &\text{ } & 0.002267574 & \text{ } & 9.09 \times 10^{-13} & \text{ } & 2.49 \times 10^{9} \\ \text{ } &60 &\text{ } & 0.001040583 & \text{ } & 8.67 \times 10^{-19} & \text{ } & 1.20 \times 10^{15} \\ \text{ } &80 &\text{ } & 0.000594884 & \text{ } & 8.27 \times 10^{-25} & \text{ } & 7.19 \times 10^{20} \\ \text{ } &100 &\text{ } & 0.000384468 & \text{ } & 7.89 \times 10^{-31} & \text{ } & 4.87 \times 10^{26} \\ \text{ } &120 &\text{ } & 0.000268745 & \text{ } & 7.52 \times 10^{-37} & \text{ } & 3.57 \times 10^{32} \\ \text{ } &140 &\text{ } & 0.000198373 & \text{ } & 7.17 \times 10^{-43} & \text{ } & 2.76 \times 10^{38} \\ \text{ } &160 &\text{ } & 0.000152416 & \text{ } & 6.84 \times 10^{-49} & \text{ } & 2.23 \times 10^{44} \\ \text{ } &180 &\text{ } & 0.000120758 & \text{ } & 6.53 \times 10^{-55} & \text{ } & 1.85 \times 10^{50} \\ \text{ } & \text{ } \\ \end{array}$
Note that at the large values, the Pareto right tails retain much more probabilities. This is also confirmed by the ratio of the two survival functions, with the ratio approaching infinity. Using an exponential distribution to model a Pareto random phenomenon would be a severe modeling error even though the exponential distribution may be a good model for describing the loss up to the 75th percentile (in the above comparison). It is the large right tail that is problematic (and catastrophic)!
Since the Pareto survival function and the exponential survival function have closed forms, We can also look at their ratio.
$\displaystyle \frac{\text{pareto survival}}{\text{exponential survival}}=\frac{\displaystyle \frac{\theta^\alpha}{(x+\theta)^\alpha}}{e^{-\lambda x}}=\frac{\theta^\alpha e^{\lambda x}}{(x+\theta)^\alpha} \longrightarrow \infty \ \text{ as } x \longrightarrow \infty$
In the above ratio, the numerator has an exponential function with a positive quantity in the exponent, while the denominator has a polynomial in $x$. This ratio goes to infinity as $x \rightarrow \infty$.
In general, whenever the ratio of two survival functions diverges to infinity, it is an indication that the distribution in the numerator of the ratio has a heavier tail. When the ratio goes to infinity, the survival function in the numerator is said to decay slowly to zero as compared to the denominator.
It is important to examine the tail behavior of a distribution when considering it as a candidate for a model. The four criteria discussed here provide a crucial way to classify parametric models according to the tail weight.
severity models
math
Daniel Ma
mathematics
$\copyright$ 2017 – Dan Ma
# Transformed Pareto distribution
One way to generate new probability distributions from old ones is to raise a distribution to a power. Two previous posts are devoted on this topic – raising exponential distribution to a power and raising a gamma distribution to a power. Many familiar and useful models can be generated in this fashion. For example, Weibull distribution is generated by raising an exponential distribution to a positive power. This post discusses the raising of a Pareto distribution to a power, as a result generating Burr distribution and inverse Burr distribution.
Raising to a Power
Let $X$ be a random variable. Let $\tau$ be a positive constant. The random variables $Y=X^{1/\tau}$, $Y=X^{-1}$ and $Y=X^{-1/\tau}$ are called transformed, inverse and inverse transformed, respectively.
Let $f_X(x)$, $F_X(x)$ and $S_X(x)=1-F_X(x)$ be the probability density function (PDF), the cumulative distribution function (CDF) and the survival function of the random variable $X$ (the base distribution). The goal is to express the CDFs of the “transformed” variables in terms of the base CDF $F_X(x)$. The following table shows how.
Name of Distribution Random Variable CDF
Transformed $Y=X^{1 / \tau}, \ \tau >0$ $F_Y(y)=F_X(y^\tau)$
Inverse $Y=X^{-1}$ $F_Y(y)=1-F_X(y^{-1})$
Inverse Transformed $Y=X^{-1 / \tau}, \ \tau >0$ $F_Y(y)=1-F_X(y^{-\tau})$
If the CDF of the base distribution, as represented by the random variable $X$, is known, then the CDF of the “transformed” distribution can be derived using $F_X(x)$ as shown in this table. Thus the CDF, in many cases, is a good entry point of the transformed distribution.
Pareto Information
Before the transformation, we first list out the information on the Pareto distribution. The Pareto distribution of interest here is the Type II Lomax distribution (discussed here). The following table gives several distributional quantities for a Pareto distribution with shape parameter $\alpha$ and scale parameter $\theta$.
Pareto Type II Lomax
Survival Function $S(x)=\displaystyle \biggl( \frac{\theta}{x+\theta} \biggr)^\alpha \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x >0$
Cumulative Distribution Function $F(x)=1-\displaystyle \biggl( \frac{\theta}{x+\theta} \biggr)^\alpha \ \ \ \ \ \ \ \ \ \ \ \ \ x >0$
Probability Density Function $\displaystyle f(x)=\frac{\alpha \ \theta^\alpha}{(x+\theta)^{\alpha+1}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x >0$
Mean $\displaystyle E(X)=\frac{\theta}{\alpha-1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \alpha>1$
Median $\displaystyle \theta \ 2^{\frac{\alpha}{2}}-\theta$
Mode 0
Variance $\displaystyle Var(X)=\frac{\theta^2 \ \alpha}{(\alpha-1)^2 \ (\alpha-2)} \ \ \ \ \ \ \ \alpha>2$
Higher Moments $\displaystyle E(X^k)=\frac{k! \ \theta^k}{(\alpha-1) \cdots (\alpha-k)} \ \ \ \ \ \ \alpha>k \ \ \ k$ is integer
Higher Moments $\displaystyle E(X^k)=\frac{\theta^k \ \Gamma(k+1) \Gamma(\alpha-k)}{\Gamma(\alpha)} \ \ \ \ \alpha>k$
The higher moments in the general case use $\Gamma(\cdot)$, which is the gamma function.
The Distributions Derived from Pareto
Let $X$ be a random variable that has a Pareto distribution (as described in the table in the preceding section). Assume that $X$ has a shape parameter $\alpha$ and scale parameter $\theta$. Let $\tau$ be a positive number. When raising $X$ to the power $1/\tau$, the resulting distribution is a transformed Pareto distribution and is also called a Burr distribution, which then is a distribution with three parameters – $\alpha$, $\theta$ and $\tau$.
When raising $X$ to the power $-1/\tau$, the resulting distribution is an inverse transformed Pareto distribution and it is also called an inverse Burr distribution. When raising $X$ to the power -1, the resulting distribution is an inverse Pareto distribution (it does not have a special name other than inverse Pareto).
The paralogistic family of distributions is created from the Burr distribution by collapsing two of the parameters into one. Let $\alpha$, $\theta$ and $\tau$ be the parameters of a Burr distribution. By equating $\tau=\alpha$, the resulting distribution is a paralogistic distribution. By equating $\tau=\alpha$ in the corresponding inverse Burr distribution, the resulting distribution is an inverse paralogistic distribution.
Transformed Pareto = Burr
There are two ways to create the transformed Pareto distribution. One is to start with a base Pareto with shape parameter $\alpha$ and scale parameter 1 and then raise it to $1/\tau$. The scale parameter $\theta$ is added at the end. Another way is to start with a base Pareto distribution with shape parameter $\alpha$ and scale parameter $\theta^\tau$ and then raise it to the power $1/\tau$. Both ways would generate the same CDF. We take the latter approach since it generates both the CDF and moments quite conveniently.
Let $X$ be a Pareto distribution with shape parameter $\alpha$ and scale parameter $\theta^\tau$. The following table gives the distribution information on $Y^{1/\tau}$.
Burr Distribution
CDF $F_Y(y)=\displaystyle 1-\biggl( \frac{1}{(y/\theta )^\tau+1} \biggr)^\alpha$ $y >0$
Survival Function $S_Y(x)=\displaystyle \biggl( \frac{1}{(y/\theta )^\tau+1} \biggr)^\alpha$ $y >0$
Probability Density Function $\displaystyle f_Y(y)=\frac{\alpha \ \tau \ (y/\theta)^\tau}{y \ [(y/\theta)^\tau+1 ]^{\alpha+1}}$ $y >0$
Mean $\displaystyle E(Y)=\frac{\theta \ \Gamma(1/\tau+1) \Gamma(\alpha-1/\tau)}{\Gamma(\alpha)}$ $1 <\alpha \ \tau$
Median $\displaystyle \theta \ (2^{1/\alpha}-1)^{1/\tau}$
Mode $\displaystyle \theta \ \biggl(\frac{\tau-1}{\alpha \tau+1} \biggr)^{1/\tau}$ $\tau >1$, else 0
Higher Moments $\displaystyle E(Y^k)=\frac{\theta^k \ \Gamma(k/\tau+1) \Gamma(\alpha-k/\tau)}{\Gamma(\alpha)}$ $-\tau
The distribution displayed in the above table is a three-parameter distribution. It is called the Burr distribution with parameters $\alpha$ (shape), $\theta$ (scale) and $\tau$ (power).
To obtain the moments, note that $E(Y^k)=E(X^{k/\tau})$, which is derived using the Pareto moments. The Burr CDF has a closed form that is relatively easy to compute. Thus percentiles are very accessible. The moments rely on the gamma function and are usually calculated by software.
Inverse Transformed Pareto = Inverse Burr
One way to generate inverse transformed Pareto distribution is to raise a Pareto distribution with shape parameter $\alpha$ and scale parameter 1 to the power of -1 and then add the scale parameter. Another way is to raise a Pareto distribution with shape parameter $\alpha$ and scale parameter $\theta^{-\tau}$. Both ways derive the same CDF. As in the preceding case, we take the latter approach.
Let $X$ be a Pareto distribution with shape parameter $\alpha$ and scale parameter $\theta^{-\tau}$. The following table gives the distribution information on $Y^{-1/\tau}$.
Inverse Burr Distribution
CDF $F_Y(y)=\displaystyle \biggl( \frac{(y/\theta)^\tau}{(y/\theta )^\tau+1} \biggr)^\alpha$ $y >0$
Survival Function $S_Y(x)=\displaystyle 1-\biggl( \frac{(y/\theta)^\tau}{(y/\theta )^\tau+1} \biggr)^\alpha$ $y >0$
Probability Density Function $\displaystyle f_Y(y)=\frac{\alpha \ \tau \ (y/\theta)^{\tau \alpha}}{y \ [1+(y/\theta)^\tau]^{\alpha+1}}$ $y >0$
Mean $\displaystyle E(Y)=\frac{\theta \ \Gamma(1-1/\tau) \Gamma(\alpha+1/\tau)}{\Gamma(\alpha)}$ $1 <\tau$
Median $\displaystyle \theta \ \biggl[\frac{1}{ 2^{1/\alpha}-1} \biggr]^{1/\tau}$
Mode $\displaystyle \theta \ \biggl(\frac{\alpha \tau-1}{\tau+1} \biggr)^{1/\tau}$ $\alpha \tau >1$, else 0
Higher Moments $\displaystyle E(Y^k)=\frac{\theta^k \ \Gamma(1-k/\tau) \Gamma(\alpha+k/\tau)}{\Gamma(\alpha)}$ $-\alpha \tau
The distribution displayed in the above table is a three-parameter distribution. It is called the Inverse Burr distribution with parameters $\alpha$ (shape), $\theta$ (scale) and $\tau$ (power).
Note that both the moments for Burr and inverse Burr distributions are limited, the Burr limited by the product of the parameters $\alpha$ and $\tau$ and the inverse Burr limited by the parameter $\tau$. This is not surprising since the base Pareto distribution has limited moments. This is one indication that all of these distributions have a heavy right tail.
The Paralogistic Family
With the facts of the Burr distribution and the inverse Burr distribution established, paralogistic and inverse paralogistic distributions can now be obtained. A paralogistic distribution is simply a Burr distribution with $\tau=\alpha$. An inverse paralogistic distribution is simply an inverse Burr distribution with $\tau=\alpha$. In the above tables for Burr and inverse Burr, replacing $\tau$ by $\alpha$ gives the following table.
Paralogistic Distribution
CDF $F_Y(y)=\displaystyle 1-\biggl( \frac{1}{(y/\theta )^\alpha+1} \biggr)^\alpha$ $y >0$
Survival Function $S_Y(x)=\displaystyle \biggl( \frac{1}{(y/\theta )^\alpha+1} \biggr)^\alpha$ $y >0$
Probability Density Function $\displaystyle f_Y(y)=\frac{\alpha^2 \ \ (y/\theta)^\alpha}{y \ [(y/\theta)^\alpha+1 ]^{\alpha+1}}$ $y >0$
Mean $\displaystyle E(Y)=\frac{\theta \ \Gamma(1/\alpha+1) \Gamma(\alpha-1/\alpha)}{\Gamma(\alpha)}$ $1 <\alpha^2$
Median $\displaystyle \theta \ (2^{1/\alpha}-1)^{1/\alpha}$
Mode $\displaystyle \theta \ \biggl(\frac{\alpha-1}{\alpha^2+1} \biggr)^{1/\alpha}$ $\alpha >1$, else 0
Higher Moments $\displaystyle E(Y^k)=\frac{\theta^k \ \Gamma(k/\alpha+1) \Gamma(\alpha-k/\alpha)}{\Gamma(\alpha)}$ $-\alpha
Inverse Paralogistic Distribution
CDF $F_Y(y)=\displaystyle \biggl( \frac{(y/\theta)^\alpha}{(y/\theta )^\alpha+1} \biggr)^\alpha$ $y >0$
Survival Function $S_Y(x)=\displaystyle 1-\biggl( \frac{(y/\theta)^\alpha}{(y/\theta )^\alpha+1} \biggr)^\alpha$ $y >0$
Probability Density Function $\displaystyle f_Y(y)=\frac{\alpha^2 \ (y/\theta)^{\alpha^2}}{y \ [1+(y/\theta)^\alpha]^{\alpha+1}}$ $y >0$
Mean $\displaystyle E(Y)=\frac{\theta \ \Gamma(1-1/\alpha) \Gamma(\alpha+1/\alpha)}{\Gamma(\alpha)}$ $1 <\alpha$
Median $\displaystyle \theta \ \biggl[\frac{1}{ 2^{1/\alpha}-1} \biggr]^{1/\alpha}$
Mode $\displaystyle \theta \ (\alpha-1)^{1/\alpha}$ $\alpha^2 >1$, else 0
Higher Moments $\displaystyle E(Y^k)=\frac{\theta^k \ \Gamma(1-k/\alpha) \Gamma(\alpha+k/\alpha)}{\Gamma(\alpha)}$ $-\alpha^2
Inverse Pareto Distribution
The distribution that has not been discussed is the inverse Pareto. Again, we have the option of deriving it by raising to a base Pareto with just the shape parameter to -1 and then add the scale parameter. We take the approach of raising a base Pareto distribution with shape parameter $\alpha$ and scale parameter $\theta^{-1}$. Both approaches lead to the same CDF.
Inverse Pareto Distribution
CDF $F_Y(y)=\displaystyle \biggl( \frac{y}{\theta+y} \biggr)^\alpha$ $y >0$
Survival Function $S_Y(x)=\displaystyle 1-\biggl( \frac{y}{\theta+y} \biggr)^\alpha$ $y >0$
Probability Density Function $\displaystyle f_Y(y)=\frac{\alpha \ \theta \ y^{\alpha-1}}{[\theta+y ]^{\alpha+1}}$ $y >0$
Median $\displaystyle \frac{\theta}{2^{1/\alpha}-1}$
Mode $\displaystyle \theta \ \frac{\alpha-1}{2}$ $\alpha >1$, else 0
Higher Moments $\displaystyle E(Y^k)=\frac{\theta^k \ \Gamma(1-k) \Gamma(\alpha+k)}{\Gamma(\alpha)}$ $-\alpha
The distribution described in the above table is an inverse Pareto distribution with parameters $\alpha$ (shape) and $\theta$ (scale). Note that the moments are even more limited than the Burr and inverse Burr distributions. For inverse Pareto, even the mean $E(Y)$ is nonexistent.
Remarks
The Burr and paralogistic families of distributions are derived from the Pareto family (Pareto Type II Lomax). The Pareto connection helps put Burr and paralogistic distributions in perspective. The Pareto distribution itself can be generated as a mixture of exponential distributions with gamma mixing weight (see here). Thus from basic building blocks (exponential and gamma), vast families of distributions can be created, thus expanding the toolkit for modeling. The distributions discussed here are found in the appendix that is found in this link.
$\copyright$ 2017 – Dan Ma
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2020-05-27 01:45:11
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https://socratic.org/questions/how-do-you-solve-the-following-system-4x-6y-8-x-5y-4
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How do you solve the following system?: -4x+6y=8 , x-5y=4
Jan 10, 2016
$x = - \frac{32}{7}$ $y = - \frac{12}{7}$
Explanation:
To solve by elimination you will need to make coefficients of x or y "match" so that when you add or subtract the equations one of the variables will be eliminated.
If we multiply the second equation by 4 we get:
$- 4 x + 6 y = 8$
$4 x - 20 y = 16$
Now if we add the two new equations, we will eliminate x and will be able to solve an equation in y.
$- 14 y = 24 \implies y = - \frac{12}{7}$
Substitute this back into one of the original equations to find x.
$x - 5 \cdot - \frac{12}{7} = 4$
$x + \frac{60}{7} = 4$
$x = 4 - \frac{60}{7} = - \frac{32}{7}$
$x = - \frac{32}{7}$
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2019-08-18 01:38:06
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https://geant4-forum.web.cern.ch/t/radionuclide-decay/8658
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Hello, I’ve this geometry
and I can simulate a particle source using the GPS, for example
########## Particles ##########
/gps/particle gamma
/gps/energy 1808.65 keV
/gps/pos/type Volume
/gps/pos/shape Sphere
/gps/pos/centre .0 .0 3.67 cm
/gps/pos/halfz 16.4 cm
/gps/ang/type iso
/gps/source/intensity 1
/gps/pos/confine sample
########## Number of particles ##########
/run/printProgress 10
/run/beamOn 100
Now instead of using the single photons (as I do in in the sketch of the macro), my source has to be radionuclide dacay. In particular I’ve to simulate the decays of Am-241, Co-60, Cs-137, Cs-134, Pb-210.
What should I do to use the radionuclide decays as source?
Hi @faca87 ,
Change first two lines of your mac to
/gps/particle ion
/gps/ion 27 60 0 0
/gps/energy 0. 0. eV
Rest remain same.
Thank you @drvijayraj
Regarding the physics list, in my main file I’ve
``````// Physics list
G4VModularPhysicsList* physicsList = new QBBC;
physicsList->SetVerboseLevel(1);
// runManager->SetUserInitialization(physicsList);
runManager->SetUserInitialization(new gem90PhysicsList);
``````
where gem90PhysicsList calls the external file
gem90PhysicsList.cc (9.4 KB)
How to include also the radioactive physics?
You should try running the macro.
I get some deposited energy, and I see photons production …do you think I can be sure it is from radioactive decay?
Yeah !! You are good to go with more events.
Well,thank you @drvijayraj !
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2022-09-26 00:37:28
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https://lintr.r-lib.org/reference/parse_exclusions.html
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read a source file and parse all the excluded lines from it
## Usage
parse_exclusions(
file,
exclude = settings$exclude, exclude_start = settings$exclude_start,
exclude_end = settings$exclude_end, exclude_linter = settings$exclude_linter,
exclude_linter_sep = settings\$exclude_linter_sep,
lines = NULL,
linter_names = NULL
)
## Arguments
file
R source file
exclude
regular expression used to mark lines to exclude
exclude_start
regular expression used to mark the start of an excluded range
exclude_end
regular expression used to mark the end of an excluded range
exclude_linter
regular expression used to capture a list of to-be-excluded linters immediately following a exclude or exclude_start marker.
exclude_linter_sep
regular expression used to split a linter list into indivdual linter names for exclusion.
lines
a character vector of the content lines of file
linter_names
Names of active linters
## Value
A possibly named list of excluded lines, possibly for specific linters.
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2023-01-31 19:54:42
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https://forum.allaboutcircuits.com/threads/55-sec-time-delay-in-assembly-language.9993/
|
55 Sec Time Delay in Assembly Language
pauliewalnuts
Joined Mar 9, 2008
5
Hi,
I'm new to assembly language, I have read a couple of tutorials and am beginning to get the fundamentals.
I have a fairly simple project to do, whereby I am using a PIC and need it to produce a time delay between 55-60 sec when an input goes high. after this delay, an output will go high.
I would really appreciate some ideas, as I am unsure how to produce a delay of this length.
Many thanks in advance for any info.
Cheers, Paulie.
John Luciani
Joined Apr 3, 2007
477
Use a counter-timer and an interrupt routine.
For example in one of my uC boards I have 32768Hz crystal. The counter is set to interrupt
when the count = 32768 (which is one second). You could change the counter
to interrupt at a 60 second interval by using different settings for the counter
clock or a larger counter.
(* jcl *)
Papabravo
Joined Feb 24, 2006
12,389
I'll give you the carpenter's answer. If you want to drive a BAN(big a** nail), you use a BAH(big a** hammer). If you want to time a long delay with a fast clock you need a LAC (long a** counter). In other words you keep making the counter longer until it has enough bits to exceed the interval of interest.
Example: How many cycles of an 8 MHz. clock do you need to count in one day?
Answer: 8e6 cycles/sec * 60 seconds/min * 60 Minutes/hour * 24 hours/day = 6.912 e 11
Now log(6.912 e 11)/log(2) = 39.33 which means a 40 bit or 5 byte counter will do the trick. Just to check my math 2^40 = 1.099 e 12 which is > 6.912 e 11
Wasn't that a dizzying mental math exercise?
To implement the multibyte or multiword counter in the assembly language of your choice you need to locate and understand the instructions which produce and use the carry flag. For the least significant part of an addition you do an ordinary "ADD" instruction. For each of the higher order parts you use the "ADD with Carry" instruction that adds two things together AND adds the carry bit from the previous operation. With the carry flag you can make the number of bytes or words anything you desire within the limits of memory on your particular machine.
Your machine might also have subtract and subtract with borrow instructions where the borrow flag is the complement of the carry flag.
0xFF
Joined Feb 26, 2008
12
I think the answer depends on what you're trying to do with it, and how accurate you need your count to be. For me, I have a project where I use Timer1 to set a 10ms interrupt. Accordingly, I set a counter at 100 and decrement it with each interrupt. Once the counter = 0, 1 second has passed (10ms x 100 = 1000ms == 1 Second). Add that to a seconds counter and you can keep track of the seconds.
However, this is an ESTIMATED interrupt rollover! It WILL NOT be accurate over time (ie. if you plan on using it for a clock, etc.). If you need the accuracy, set up your Timer to use an external crystal (32768Hz as pointed out above), which will keep accurate time.
The reason I choose this method is that I can keep an "estimated" time of how long the unit has been running (and I don't need it to the second -- even days is fine for what I need), but the main reason is, I can handle as many counters as I need. My timers are for time delay -- you must wait a certain amount of time before being able to do a certain task. I, on occasion, have 3 timeout timers running at the same time, all from the same interrupt / timer setup. Here's how I do it;
NOTE: this code was pulled from an 18F series chip -- 18F4620 to be precise. Please make sure to check your settings for your particular chip (ie. INTCON, T1CON, etc. -- just make sure to look them up in your chips datasheet and adjust as required.
First, setup the timer to rollover every 10ms, which creates an interrupt.
Rich (BB code):
; setup Timer1
movlw b'00000000' ; TMR1 prescaler 1:1, internal clock source
movwf T1CON ;
movlw 0xD8 ; load timer 1 with 0xD8EF
movwf TMR1H ; 10ms rollover @ 4 MHz
movlw 0xEF ;
movwf TMR1L ;
Next, turn on the interrupts.
Rich (BB code):
; enable/disable interrupts
bcf PIR1, TMR1IF ; clear the TMR1 interrupt flag
bsf INTCON, PEIE ; enable peripheral interrupts
bsf PIE1, TMR1IE ; enable TMR1 interrupts
bsf INTCON, GIE ; enable global interrupts
Then, in your ISR (Interrupt Service Routine);
Rich (BB code):
ISR:
bcf INTCON, GIE ; turn off global interrupts
btfsc INTCON, GIE ; test the flag to ensure it's off
bra \$-4 ; still on, try again
ISR_TMR1:
btfss PIR1, TMR1IF ;
bra IRestore ; jump if not TMR1 interrupt
; handle the TMR1 event
bcf PIR1, TMR1IF ; clear the TMR1 interrupt flag
movlw 0xD8 ; reload timer 1 with 0xD8EF
movwf TMR1H ; 10ms rollover @ 4 MHz
movlw 0xEF ;
movwf TMR1L ;
decfsz second_count, F ; decrement the seconds count
; interrupt every 10ms x 100 = 1000ms (1 second)
bra IRestore ;
movlw .100 ; reset the seconds count
movwf second_count ;
ITime_Delay:
decf delay_md1_sec, F ; decrement seconds
incfsz delay_md1_sec, W ; check for underflow
bra ICheck_Zero ;
movlw .59 ; reset seconds to 59
movwf delay_md1_sec ;
decf delay_md1_min, F ; decrement MIN
incfsz delay_md1_min, W ; check for underflow
nop ;
ICheck_Zero:
movf delay_md1_sec, F ; test SEC for zero
bnz IRestore ; NOT ZERO
movf delay_md1_min, F ; test MIN for zero
bnz IRestore ; NOT ZERO
ITimeout:
{DO WHAT YOU NEED TO DO AFTER A TIMEOUT HERE}
IRestore:
bcf PIR1, TMR1IF ; clear the TMR1 interrupt flag
retfie FAST ; enable global interrupt (INTCON,GIE)
So, basically, what happens is this -- you set a 10ms interrupt -- with every interrupt, you decrement the count -- when the count hits zero, 1 second has elapsed. When 1 second has elapsed, you send the ISR to the next function, which decrements the time out period you have set (in the example I have provided, delay_md1_sec and delay_md1_min are your seconds and minutes counters (respectively)). For example, if I wanted a 55 second timeout, I would set delay_md1_sec = 55 and delay_md1_min = 0. Each time you decrement the seconds, you check them for zero -- if they are zero, you set them back to 59 (assuming you haven't hit your timeout) and decrement the minutes counter. If they're both = 0, you have hit your timeout.
I have removed some of the other checks for other timeouts (ITime_Delay2, ITime_Delay3, etc.) -- but it's easy to have multiple counts running at the same time.
I have to run, so I don't have much time to proof read this -- hopefully there aren't many mistakes!
If I haven't explained something well, please ask and I'll try to explain it better.
SAE140
Joined May 1, 2008
4
Hi,
I'm new to assembly language, I have read a couple of tutorials and am beginning to get the fundamentals.
I have a fairly simple project to do, whereby I am using a PIC and need it to produce a time delay between 55-60 sec when an input goes high. after this delay, an output will go high.
I would really appreciate some ideas, as I am unsure how to produce a delay of this length.
Many thanks in advance for any info.
Cheers, Paulie.
Providing the processor isn't required to do anything else during this delay period, it's possible to achieve this kind of delay without the use of interrupts (to keep things as simple as possible). The key word here is 'nested-delay'.
So - create a loop which keeps checking for your desired input state. On change, it enters a nested loop, then after the long delay it alters the output, then whatever .... (program ends, re-starts ... ?)
The nested loop structure itself takes the form of an innermost loop with a down counter, starting from max (255), and exitting on zero. With a 1 uSec clock period (say), that gives you a delay of 255 or 256 uSecs (depending on whether you decrement the counter before or after the decision to exit) - so you've now created a 255 uSec delay.
If this inner loop is then placed within an outer loop, also with a down-counter set at 255, then you'll have a 255x255x1 uSec delay (approximately). If you need even more delay, then arrange for this lot to sit (nest) within yet another loop, and so on ....
You could make a timer to delay for many years like this.
Couple of points:
a) this is extremely wasteful of processing power, but for a demonstration it doesn't matter. Using interrupts is certainly better, but adds complexity if you've just started programming.
b) consider toggling an LED from an output line when your delay gets up to around the 0.5 - 5 second mark, so that you have some indication that the program is still running and the processor hasn't gone off picking daisies ...
c) In the loops there will probably be a timing difference depending on whether the program keeps looping, or whether it exits. This may be small difference, but is obviously magnified if it occurs in the innermost loop.
Adjustment will be needed if precision is sought.
Hope this helps. Try doing a Google for 'nested delay loops'.
'best
Colin
rathishag007
Joined Jun 12, 2008
2
Hi,,,,,
I need a subprogram of 1 SECOND delay.......I want to call this routine frequently..Because i am doing my project traffic light.so i want to CALL the delay of 1 second ,20 times for highway road and 5 times for pocket road. please anyone can help me..I need my delay pgm in assembly language.................
nanovate
Joined May 7, 2007
665
Hi,,,,,
I need a subprogram of 1 SECOND delay.......I want to call this routine frequently..Because i am doing my project traffic light.so i want to CALL the delay of 1 second ,20 times for highway road and 5 times for pocket road. please anyone can help me..I need my delay pgm in assembly language.................
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2019-11-14 21:42:28
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https://discourse.mcneel.com/t/where-is-text/59397
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# Where is Text?
Where is the Text box ? i normally type text , then a dialog appears and i can put in … text , now a dialog appears and i can not do anything ? where is it
latest update btw
Have you tried pulling the bottom of the dialog downward to see if a text box becomes visible? I know there was a bug to this effect - the dialog does not look resizable, but it is…
Bummer thats it , anoying …
Thanks had not think of that , and it was there always…
Do you have the latest service release? I don’t know if/when the bug got fixed… It also might be that doing that once is all you need, it will remember for the next time…
yes latest update that did not show the text field , but problem solved now
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2020-09-22 18:37:07
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http://mathhelpforum.com/geometry/18573-equation-set-up.html
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1. ## equation set up
I am having a problem setting up this problem and being successful at solving it.
A rectangular garden is to be surrounded by a walkway of constant width. The garden's dimensions are 30 ft by 40 ft. The total area, plus garden walkway, is to be 1800ft^2. What must be the width of the walkway to the nearest thousandth.
I thought A= L X W.
But I am still stuck and do not get a successful answer.
2. Hello, lizard4!
A rectangular garden is to be surrounded by a walkway of constant width.
The garden's dimensions are 30 ft by 40 ft.
The total area, plus garden walkway, is to be 1800 ft².
What must be the width of the walkway to the nearest thousandth.
I hope you made a sketch . . .
Code:
: x : - -40 - - : x :
*-------------------* - -
x | | :
| *-----------* | :
| | | | :
30| |30 | | 30+2x
| | 40 | | :
| *-----------* | :
x | | :
- *-------------------* - -
: - - - 40+2x - - - :
The garden itself is 30 × 40 ft.
The walkway is $x$ feet wide . . . all the way around.
The entire region has length $(40 + 2x)$ and width $(30 + 2x)$.
There is our equation: . $(40 + 2x)(30 + 2x) \;=\;1800$
3. A = L x W is formula for area of a rectangle, alright.
Which rectangle here?
One rectangle is the 30ft by 40ft garden.
Another rectangle is that garden that is surrouded by walkways of constant width.
If c = constant width of the walkways, the other rectangle is
(30 +2c) by (40 +2c)
whose area is 1800 sq.ft.
So,
(30 +2c)x(40 +2c) = 1800
30*40 +30*2c +2c*40 +2c*2c = 1800
("*" is x or times or multipllied by.)
1200 +60c +80c +4c^2 = 1800
4c^2 +140c +1200 -1800 = 0
4c^2 +140c -600 = 0
Divide both sides of the equation by 4,
c^2 +35c -150 = 0
Cannot factor that. Use the Quadtratic formula,
c = {-35 +,-sqrt[35^2 -4(1)(-150)]} / 2(1)
c = {-35 +,-sqrt[1825]} / 2
c = {-35 +,-42.72001873} /2
c = -38.860 ft, or 3.860 ft.
Reject the -38.860 ft because there are no negative dimensions.
Therefore, the constant width of the walkways is 3.860 ft. ------answer.
check,
(3.860)[(30+3.860+3.860)*2 +(40*2)] + 30*40 =? 1800
600 +1200 =? 1800
1800 =? 1800
Yes, so, OK.
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2013-12-09 23:46:24
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https://lachozademigue.com/2lwj5t/d3aa38-is-so2-polar-or-nonpolar
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Explain why carbon dioxide… 01:24. • Why Isn’t Carbon Dioxide Polar? The greater the difference in electronegativity more will be the polarity of the molecule. Top. It's a large one since they have a big electronegativity. Ej25 to jdm ej20 swap Posting Lebih Baru Posting Lama Beranda. Since … Hypothesize why long, nonp… Watch More Solved Questions in Chapter 5. Categories . The two bond dipoles vectors only partially cancel out, leaving a NET dipole moment. II) … Anonymous (not verified) Sat, 11/14/2009 - 19:37. SO2 is polar and it is because of the difference in electronegativity between sulfur and oxygen atoms. The difference in electronegativity between the two atoms is 1.7. Data from NIST Standard Reference Database 69: NIST Chemistry WebBook. Log in or register to post comments; Similar Questions. Problem 8. SO 3 is a nonpolar molecule. Therefore, the molecule is not completely symmetrical. Active 1 year, 2 months ago. With other molecules, like PF 3 Cl 2, it's not possible to divide it into non-polar shapes. They have “poles”, just like the opposite poles on the Earth, or like the … The electronegativity of sulfur is 2.58 and the... See full answer below. Lone pairs repel more than bonding pairs so the lone pairs cause a bent shape. Polar or Nonpolar? is so2 polar or nonpolar. If oxygen is highly electr… 00:29. Science Is CO2 polar or nonpolar: Check carbon dioxide polarity. Nonpolar molecules: A nonpolar molecule either formed due to an equal sharing of electrons among the two atoms of a diatomic molecule or due to the symmetrical array of polar bonds in mainly complex molecules. A polar covalent bond is an unequal sharing of electrons between two atoms with different electronegativities (χ). Priyanka 1 week ago. Keeping all that aside, it is a colorless gas that is used in several industries today, right from … NO2 (Nitrogen dioxide) is Polar. One easy way to … What about nonpolar? Combine that will lower … N2O is linear but polar. SO2 is a bent molecule. water is polar due to the dipole moment created. Polar molecules must contain polar bonds due to a difference in electronegativity between … Due to the molecules relatively strong dipole, the compound a melting point of -72˚C and a boiling point of -10˚C. Click to see full answer. 0 0 posted by bobpursley Apr 10, 2010 so N20 is polar SO2 is polar and N3- is ionic? Is no2 … Is SO2 Polar or Nonpolar? ICl4- is Nonpolar I'll tell you the polar or nonpolar list below. 22 Related Question Answers Found What is the bond angle of co2? Definition of Polarity As It Relates To CO2. CO2 is non polar but SO2 is polar why Share with your friends. Due to the molecules relatively strong dipole, the compound a melting point of -72˚C and a boiling point of -10˚C. Conclusion. Nonpolar Molecule: Compounds or molecules can be classified as nonpolar or polar, depending on the presence of the dipole moment in the structure. a) all three have a trigonal planar electron group geometry b) all three are polar c) NO2- is the only polar one d) SO2 and NO2- are the only polar ones e) CO2 and SO2 are polar while NO2- is non polar Here, we have … Which one of the following molecules is nonpolar? Which means S-O forms polar covalent bond. By bagus amin - Januari 23, 2018 - Add Comment. In the case of water, it is polar. Let’s look at the structure of carbon dioxide: As you can clearly see, the molecule has a carbon atom sharing two double bonds with oxygen. Post by Aya Ghoneum 1H » Thu Oct 22, 2015 3:13 pm . And after bonding in the SO2 molecule, … CO 2 is a linear molecule so the resultant dipole moment is zero since the individual dipole moments of CO bond cancel each other.Whereas, SO 2 is an angular molecule due to which it has a net dipole moment.Hence SO 2 is polar and CO 2 is non- polar. Sure … However, molecules are not necessarily polar just because there are polar bonds present in them. Polarity alone doesn't determine if a molecule is polar. There are covalent bonds between one sulfur atom and three oxygen atoms. So, the conclusion is, SO2 is a Polar molecule. (The electrons on the oxygen aren't pictured, but each has two pairs, just like above, sulfur has no electrons). 0 0 posted by … Is so2 polar or nonpolar? When comparing a polar and nonpolar molecule with similar molar masses, the polar molecule in general has a higher boiling point, because the dipole–dipole interaction between polar molecules results in stronger intermolecular attractions. Neither exist as a discrete molecules at room temperature, but as a network solids with alternating Ge and O atoms. Ask Question Asked 4 years, 9 months ago. Why is carbon dioxide nonpolar? Problem 10. Therefore, it would be good to contextualize carbon dioxide’s non-polar attributes with water polar molecules and to go into detail about how a molecule’s polarity is decided. carbon dioxide looks like this O=C=O in that exact plane. Emoticon Emoticon. So SO3 is nonpolar, and SO2 is polar because of substituent differences, but especially because of geometry. Viewed 21k times -1 $\begingroup$ Is $\ce{SiO2}$ is a polar molecule or not? The latter has no charge either formal or dipole moment since it is a nonpolar molecule with a trigonal planar configuration (i.e. Molecular Sieves Water Adsorbent Zeolite. It has no angles (180 degrees across from eachother). Problem 3. One common form of polar interaction is the hydrogen bond, which is also known as the H-bond. A. BeCl2 -+labels Is BeCl2 polar or nonpolar? This negates pull on the carbon either way causing no vector. CO2, SO2, NO2-Which of the statements is correct? Is NO2 (Nitrogen dioxide) polar or nonpolar ? Answer: SO2 is a polar molecule because the lone pair of electrons on the central sulfur atom cause electron-electron repulsion that forces a bent structure leading to an unequal distribution of charge throughout the molecule. So, shouldn’t carbon dioxide, which contains a positive carbon and two partially negative oxygens, be polar? Aya Ghoneum 1H Posts: 22 Joined: Fri Sep 25, 2015 10:00 am. ADVERTISEMENT. SO2 polar N2O polar N3- non polar? Polar or non-polar: In molecules, how can you tell if it's polar or nonpolar? Published by at December 2, 2020. Sulfur has 2 double bonds between the oxygens, with 2 lone pairs on the sulfur. Re: SiO2 Polar or Nonpolar SiO2 has a linear shape, and since the elements at each end are the same, the pull is canceled out, making the overall compound non-polar. 0 0 posted by Johannie Apr 10, 2010 SO2 polar N2O polar N3- non polar ? The basic adsorbates are polar CF4 and non-polar CF3Cl and CF3Br molecules. Silicon tetrachloride is non-polar because the four chemical bonds between silicon and chlorine are equally distributed.. Share 4. Is SO2 covalent or nonpolar covalent? (b) The C=O bond is the most polar, with O at the negative end. SO3(2-) has three resonance structures with one being shown in the diagram below. Share this. If a molecule has polar bo… 02:39. Glavni izbornik Therefore, most non-polar molecules form volatile compounds. By analyzing the Lewis structure of SO2, we can see that the SO2 is asymmetrical because it contains a region with different sharing. Pretraži. Is ICl4- polar or nonpolar ? ClF3 IOF3 PF5 SO2 NOCl. Besides, is SiO2 polar? Re: Polar vs Nonpolar. 14 ; View Full Answer Sulfur dioxide O - S - 0 has two valence pairs on one side … H2O has a 109.5 degree bond angle, but CO2 has exactly 180 degrees. AND because the bond angles created a vector pull on the oxygen. Ex: – Pentane, Hexane. The reason that these polar bonds make SO2 polar is that the molecule is also bent, similarly to H2O, and is positioned with sulfur in the middle. … The other structures move the double bond to another part of the atom with all three positions … Share on Facebook Tweet on Twitter Plus on Google+. Sportska akademija Vunderkid Vaše dijete, čudo od pokreta! For example, water forms H-bonds and has a molar mass M = 18 and a … Explain why CO2 and CCl4 a… 00:52. The molecular geometry of SO2 has a bent shape which means the top has less electronegativity , and the bottom placed atoms of Oxygen have more of it. Problem 9. Difference Between Polar and Nonpolar Molecules Net Dipole. polar vs. non-polar: Identifying Nonpolar molecules: Polar or … A. BeI2 -+labels Is BeI2 polar or nonpolar? Polar or Nonpolar. Polar Molecules: Net dipole is present due to electronegativity differences of participating atoms or asymmetrical arrangement of the molecule. However, it would be good to contextualize carbon dioxide’s non-polar attributes with other polar molecules and to go into detail about how a molecule’s polarity is decided. 0 0 posted by DBob222 Apr 10, 2010; it tells me that it is incorrect.. 0 0 posted by Johannie Apr 10, 2010 N3- is an ion. … Is the molecule TiO2 polar? If … What does polar mean? Is SO2 polar or nonpolar? Roll20 5etools / Water is a polar molecule because its oxygen is strongly electronegative and, as such, pulls the electron pair towards itself (away from the two hydrogen atoms) Carbon dioxide is a great example of how the geometry of a molecule plays a crucial role in determining whether it's polar or nonpolar. There is a slight negative charge at the oxygen atoms, and a slight positive charge at the sulfur atom. SO2 and H2O are polar because of their bent shape. Carbon Dioxide, or CO2, is one of the most known gases due to its contribution to the greenhouse effect and global warming. CS2 and CO2 are non‐polar because of their linear shape. Popular Posts. This means that SO2 is a … (bond angle 105 degrees.) Individual Ge-O bonds, based on the electronegativity difference (1.43), are polar with a 40% ionic character. Problem 4. 3 g kg NaHCO3 Sep 05 2015 Is CH3COOH Polar or Nonpolar NaHCO3 HC2H3O2 sodium bicarbonate baking soda and vinegar acetic acid. The repulsion between the Fluorine atoms is greater than that between the Chlorine, and that has to be accounted for when positioning them. Explain why $\mathrm{CO}_{… 01:45. 120˚ between the outer molecules). Problem 6. The molecules$\mathrm{CH}… 01:16. SO2 has polar bonds but is a polar molecule as it is bent and the bond dipoles do not cancel one another out. So, the conclusion is, SO2 is a Polar molecule. The fact that it is bent, and asymmetrical, makes it polar. Next we need to look at the Lewis structure SO2. Problem 7. Similarly, it is asked, is SiO2 dipole dipole? Problem 1. The bent shape of SO2 is because of the repulsion between the unbonded electrons present on the sulfur and oxygen atoms. The former is described by the description right above. The molecular geometry of SO2 has a bent shape which means the top has less electronegativity, and the bottom placed atoms of Oxygen have more of it. Is SO2 Polar or Non-Polar? Problem 2. Let’s understand this scenario better by considering the carbon dioxide molecule. The side that has both oxygen atoms pointing toward it is partially negative, and the side that has a … Is silicon dioxide a polar molecule or not? Explanation Relates to CO2. C. SCl2 -+labels Is SCl2 polar or nonpolar? (a) BFCl2 is trigonal planar. Home/Science/ Is CO2 polar or nonpolar: Check carbon dioxide polarity. In general, a bond between two atoms with an χ difference … I know there is a difference in electronegativity but do they cancel each other out since there are 2O's? Is this a polar or non-polar molecule? Since there are 4 pairs of electron density around the sulfur it is sp3 hybridized. Polar molecules are molecules that have regions/areas of both positive and negative charge. By analyzing the Lewis structure of SO2, we can see that the SO2 is asymmetrical because it contains a region with different sharing. 0 0 posted by Johannie Apr 10, 2010 Those look ok to me. GeO2 is germanium dioxide, compared to germanium monoxide (GeO). Problem 5. B. SO2 -+labels Is SO2 polar or nonpolar? There are many things that determine whether something is polar or nonpolar, such as the chemical structure of the molecule. If a molecule consists of more than one bond, then the combined effect of all these bonds must be considered. 0 61 1 minute read. is Silicon polar or nonpolar? It has two resonance structures with a double and single bond and a lone pair on sulfur. Is ICl4- polar or nonpolar ? CS2 is linear, symmetrical, and therefore nonpolar. Is described by the description right above its contribution to the molecules $\mathrm { }. Polarity alone does n't determine if a molecule consists of more than bonding pairs so the lone pairs cause bent... No2-Which of the most known gases due to its contribution to the molecules relatively dipole... Slight negative charge at the sulfur and oxygen atoms that between the oxygens, be polar form polar... That determine whether something is polar because of their linear shape cs2 and CO2 non‐polar., is one is so2 polar or nonpolar the statements is correct with other molecules, how can tell. Polar due to the molecules$ \mathrm { CH } … 01:16 the combined effect of these... Co2 has exactly 180 degrees ok to me this negates pull on the oxygen is the hydrogen bond, contains... One being shown in the case of water, it is because of the molecule GeO2 is germanium dioxide which. In Chapter 5 \begingroup $is$ \ce { SiO2 } $is a … is silicon dioxide polar... Chemical structure of the molecule like this O=C=O in that exact plane non-polar because four... Just because is so2 polar or nonpolar are 4 pairs of electron density around the sulfur$ \mathrm CO. Equally distributed baking soda and vinegar acetic acid two resonance structures with one being shown the. Big electronegativity basic adsorbates are polar CF4 and non-polar CF3Cl and CF3Br molecules the chemical of... Molecule, … is silicon dioxide a polar molecule or not and O atoms be! H2O has a 109.5 degree bond angle of CO2 Add Comment all these bonds must be considered or! A bent shape and chlorine are equally distributed an unequal sharing of electrons between two atoms is greater than between. Bonds between the chlorine, and therefore nonpolar the greenhouse effect and global warming not possible to divide into...: NET dipole moment created, makes it polar the carbon either way causing no vector repulsion... By … water is polar and it is bent, and that has to be accounted for when them... Sulfur and oxygen atoms, and asymmetrical, makes it polar, the conclusion is, SO2 is and... Boiling point of -72˚C and a boiling point of -72˚C and a boiling point of -10˚C resonance with! Is greater than that between the two atoms is 1.7. is SO2 polar N2O polar N3- non polar Check! 4 years, 9 months ago so SO3 is nonpolar i 'll tell the! Compared to germanium monoxide ( GeO ) degrees across from eachother ) three resonance with. ; Similar Questions … is SO2 polar or nonpolar 2.58 and the... see full answer below since there many! Lewis structure of SO2 is polar because of their linear shape it 's not possible to divide into! N20 is polar and it is bent, and therefore nonpolar unequal sharing electrons. Whether something is polar Tweet on Twitter Plus on Google+ common form of interaction... Bobpursley Apr 10, 2010 SO2 polar or nonpolar explain why $\mathrm { CO } {. It contains a region with different sharing both positive and negative charge at the.... More will be the polarity of the repulsion between the chlorine, and asymmetrical, makes it.! It is bent, and asymmetrical, makes it polar the SO2 is because! Be accounted for when positioning them former is described by the description right above the statements is correct on... Vinegar acetic acid combined effect of all these bonds must be considered BeCl2... Between two atoms with different electronegativities ( χ ) … 01:16 Check carbon dioxide or. And asymmetrical, makes it polar is polar Vaše dijete, čudo od pokreta as a network solids with Ge! Joined: Fri Sep 25, 2015 10:00 am … CO2, SO2, can. Tell if it 's polar or non-polar bent shape of SO2, we see..., then the combined effect of all these bonds must be considered NaHCO3 Sep 2015! Does n't determine if a molecule consists of more than one bond, which is known... Sulfur has 2 double bonds between silicon and chlorine are equally distributed be polarity. Tell you the polar or non-polar pair on sulfur has exactly 180 degrees CO2 are non‐polar because of most! Od pokreta with alternating Ge and O atoms then the combined effect of all these bonds be. Many things that determine whether something is polar SO2 is because of the most known due!, how can you tell if it 's a large one since they have a big electronegativity and therefore.! By … water is polar because of the molecule know there is a polar covalent is! Bond angle of CO2 t carbon dioxide looks like this O=C=O in that exact plane, we can see the. And vinegar acetic acid, … is SO2 polar or nonpolar: Check dioxide! Electrons between two atoms is 1.7. is SO2 polar N2O polar N3- polar... Or CO2, SO2 is a … is silicon dioxide a polar covalent bond is unequal! Are equally distributed or asymmetrical arrangement of the most known gases due to the relatively. Different sharing since … Sportska akademija Vunderkid Vaše dijete, čudo od pokreta because!: in molecules, like PF 3 Cl 2, it is because of substituent differences but. Are 2O 's is also known as the chemical structure of the statements is correct oxygen. Two partially negative oxygens, with 2 lone pairs on the oxygen the chemical structure of statements. Pair on sulfur _ { … 01:45 either way causing no vector be the of. One of the repulsion between the chlorine, and that has to be accounted for when positioning them at... Than bonding pairs so the lone pairs on the electronegativity difference ( 1.43 ) are. Different sharing out, leaving a NET dipole is present due to electronegativity differences of participating atoms or asymmetrical of. A polar covalent bond is an unequal sharing of electrons between two is!, leaving a NET dipole is present due to the molecules relatively strong dipole, the is... Polarity alone does n't determine if a molecule consists of more than bond. A NET dipole moment created the lone pairs cause a bent shape molecules are not necessarily polar just because are! Science is CO2 polar or nonpolar ( GeO ) a slight negative charge Add Comment …... Bonding pairs so the lone pairs cause a bent shape has to be accounted for when positioning them is …... Neither exist as a discrete molecules at room temperature, but as a discrete molecules at temperature! -1$ \begingroup $is$ \ce { SiO2 } $is$ \ce SiO2. If … Home/Science/ is CO2 polar or nonpolar alternating Ge and O atoms and asymmetrical, it... 0 0 posted by Johannie Apr 10, 2010 Those look is so2 polar or nonpolar to.! Sulfur has 2 double bonds between the chlorine, and asymmetrical, it... Nahco3 Sep 05 2015 is CH3COOH polar or non-polar: in molecules, how can tell. Share on Facebook Tweet on Twitter Plus on Google+ since they have a big electronegativity greenhouse effect and global.... Network solids with alternating Ge and O atoms posted by … water polar! Present on the electronegativity of sulfur is 2.58 and the... see full answer below log in or register post... 69: NIST Chemistry WebBook fact that it is polar and N3- is?... Bond, which is also known as the chemical structure of the known. The sulfur atom and three oxygen atoms and vinegar acetic acid linear, symmetrical, asymmetrical... See that the SO2 is a polar molecule is so2 polar or nonpolar not times -1 $\begingroup$ is \ce! $\begingroup$ is $\ce { SiO2 is so2 polar or nonpolar$ is $\ce { }! Into non-polar shapes therefore nonpolar the repulsion between the chlorine, and nonpolar., molecules are molecules that have regions/areas of both positive and negative at... Post comments ; Similar Questions chemical bonds between one sulfur atom and three oxygen atoms are not necessarily polar because! Sodium bicarbonate baking soda and vinegar acetic acid 4 years, 9 months ago character. Or not … CO2, SO2 is polar there is a … is SO2 N2O. Than that between the Fluorine atoms is 1.7. is SO2 polar N2O polar non... Is correct eachother ) Reference Database 69: NIST Chemistry WebBook bonds between silicon and are... Of both positive and negative charge at the sulfur atom and three oxygen atoms, and is. Watch more Solved Questions in Chapter 5 sure … CO2, SO2 is because of their shape... Viewed 21k times -1$ \begingroup $is a polar molecule or not double between. They cancel each other out since there are 2O 's acetic acid melting point of -10˚C consists of more bonding. Becl2 -+labels is BeCl2 polar or nonpolar -1$ \begingroup $is$ \ce { SiO2 } is. 2015 3:13 pm 1H » Thu Oct 22, 2015 10:00 am between sulfur and atoms. Into non-polar shapes 11/14/2009 - 19:37 A. BeCl2 -+labels is BeCl2 polar non-polar! Hydrogen bond, then the combined effect of all these bonds must be considered amin... Molecule consists of more than bonding pairs so the lone pairs repel more than bond... Hc2H3O2 sodium bicarbonate baking soda and vinegar acetic acid 3 g kg NaHCO3 Sep 05 is... Polar molecules are molecules that have regions/areas of both positive and negative charge at the sulfur atom Vunderkid dijete... So2, we can see that the SO2 molecule, … is silicon dioxide a polar molecule or not have. … Sportska akademija Vunderkid Vaše dijete, čudo od pokreta post by aya Ghoneum 1H Posts: Joined!
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2021-05-08 01:37:39
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https://www.physicsforums.com/threads/setting-up-formulas-algebra-how-to-help-needed-tonight-please.213822/
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# Setting up formulas, algebra how to? Help needed tonight, please
1. Feb 7, 2008
### Saturnfirefly
1. The problem statement, all variables and given/known data
A:
1
To play tennis at the local tennis club costs 150 kr/hour for non-members and 90 kr/hour for members. A membership costs 2100 kr a year. How many hous of tennis does one have to play to make it wothwhile to be a member? (kr is the swedish currency)
2
In a factory, goods are packed in both large and small boxes. The total mass of 3 large and 7 small boxes is 48.3 kg. The total mass of 4 large and 5 small boxes is 46.2 kg. Find the mass of each type of box.
B:
1
The final bill for a recent party was $400. There were two guests of hounour who did not pay anything so the cost was spread equally amongst the rest. If tehy each paid$10 mor than they would otherwise have done, how many were at the party altogether?
2
A shopkeeper has two brands of coffe, A at the price 13.50 kr/kg and B at 10.50 kr/kg. He wants to make a 50 kg mixture of these two, that can be sold for 11.30 kr/kg. How much of each brand shall he take? (simultaneous equation)
3. Evariste bought n oranges for 70p. Later he bough (n+1) apples for 60p. Each orange costs 4p more than each apple. Find n.
4. Two consecutive positive even numbers are squared and added together. The answer is 340. Find the two numbers.
2. Relevant equations
3. The attempt at a solution
I'm having some problems with setting up formulas... I just don't understand how I'm supposed to. I can't do it at all. Please tell me how to do it easy. :) Thank you!!
Pleasee... at least one from a and one from b.... it really means a lot to me.
2. Feb 7, 2008
### symbolipoint
For A2 and B2, you need two unknowns and you will derive two equations (for each exercise). For the large and small boxes exercise, say x=mass of small box, y=mass of large box. The rest of this problem is just simple translation.
3. Feb 7, 2008
### dlgoff
Hint for 1A. First you know you are going to be a member. So what operation do you need to do with the membership cost and the hourly rate inorder to get an answer in units of hours?
4. Feb 7, 2008
### Staff: Mentor
You know that we will not provide you with the answers. In your other threads, you showed your work and were told where you were making mistakes. You must show your work here as well.
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2016-12-08 04:00:19
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https://encyclopediaofmath.org/wiki/Functional
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Functional
A mapping \$f\$ of an arbitrary set \$X\$ into the set \$\mathbb R\$ of real numbers or the set \$\mathbb C\$ of complex numbers. If \$X\$ is endowed with the structure of a vector space, a topological space or an ordered set, then there arise the important classes of linear, continuous and monotone functionals, respectively (cf. Linear functional; Continuous functional; Monotone mapping).
References
[1] A.N. Kolmogorov, S.V. Fomin, "Elements of the theory of functions and functional analysis" , 1–2 , Graylock (1957–1961) (Translated from Russian)
How to Cite This Entry:
Functional. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Functional&oldid=29314
This article was adapted from an original article by V.I. Sobolev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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2020-07-06 05:36:24
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https://nickhigham.wordpress.com/2018/06/11/how-to-program-log-z/?shared=email&msg=fail
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## How to Program log z
While Fortran was the first high-level programming language used for scientific computing, Algol 60 was the vehicle for publishing mathematical software in the early 1960s. Algol 60 had real arithmetic, but complex arithmetic had to be programmed by working with real and imaginary parts. Functions of a complex variable were not built-in, and had to be written by the programmer.
I’ve written a number of papers on algorithms to compute the (principal) logarithm of a matrix. The problem of computing the logarithm of a complex scalar—given a library routine that handles real arguments—might appear trivial, by comparison. That it is not can be seen by looking at early attempts to provide such a function in Algol 60.
The paper
J. R. Herndon (1961). Algorithm 48: Logarithm of a complex number. Comm. ACM, 4(4), 179.
presents an Algol 60 code for computing $\log z$, for a complex number $z$. It uses the arctan function to obtain the argument of a complex number.
The paper
A. P. Relph (1962). Certification of Algorithm 48: Logarithm of a complex number. Comm. ACM, 5(6), 347.
notes three problems with Herndon’s code: it fails for $z$ with zero real part, the imaginary part of the logarithm is on the wrong range (it should be $(-\pi,\pi]$ for the principal value), and the code uses log (log to the base 10) instead of ln (log to the base $e$). The latter error suggests to me that the code had never actually been run, as for almost any argument it would produce an incorrect value. This is perhaps not surprising since Algol 60 compilers must have only just started to become available in 1961.
The paper
M. L. Johnson and W. Sangren, W. (1962). Remark on Algorithm 48: Logarithm of a complex number. Comm. CACM, 5(7), 391.
contains more discussion about avoiding division by zero and getting signs correct. In
D. S. Collens (1964). Remark on remarks on Algorithm 48: Logarithm of a complex number. Comm. ACM, 7(8), 485.
Collens notes that Johnson and Sangren’s code wrongly gives $\log 0 = 0$ and has a missing minus sign in one statement. Finally, Collens gives in
D. S. Collens (1964). Algorithm 243: Logarithm of a complex number: Rewrite of Algorithm 48. Comm. CACM, 7(11), 660.
a rewritten algorithm that fixes all the earlier errors.
So it took five papers over a three year period to produce a correct Algol 60 code for the complex logarithm! Had those authors had the benefit of today’s interactive computing environments that period could no doubt have been shortened, but working with multivalued complex functions is necessarily a tricky business, as I have explained in earlier posts here and here.
This entry was posted in research. Bookmark the permalink.
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2019-05-22 14:20:26
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https://stats.stackexchange.com/questions/562811/mixed-models-if-the-interaction-is-significant-but-the-main-effect-is-not-shou
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# Mixed models: if the interaction is significant but the main effect is not, should I remove the factor from the fixed effects or the random slope?
I run a glmer model using two factors both as fixed effects and as random slopes. Here is the formula:
maximal_RTs.model = glmer(RTs ~ FT1* FT2+ (1+ FT1* FT2|Num_part)
, data = data_RTs_hands
, control=glmerControl(optimizer="bobyqa"
, optCtrl=list(maxfun=1e6))
)
The model converges and the summary() functions show that the main effect of FT2 is significant and the interaction is significant, but the simple effect of FT1 is not significant.
Should I reduce the model by removing FT1? And if yes, should I remove it from the fixed effects or the random slopes?
Thank you,
• Don't trust "significance" tests of "main effects" for predictors involved in interactions. They test whether the "main effect" coefficient is different from 0 when all of the interacting predictors are at 0 or at reference levels. Simply re-centering the F2 predictor can lead to an apparent change in the "significance" of the F1 "main effect." See this page.
– EdM
Feb 2 at 20:50
• Thank you for your response, that was very clear! I created the model following Barr, 2013 "maximal random structure allowed by the design" and given that the model converged and the anova() function showed that the BIC was lower compared to reduced models, I decided to use it. F1 has 3 levels/condition and represent my primary research question, while F2 represents the sequence of trials that were randomly generated (Sequence effects, show that the response changes depending on the type of the previous trial, e.g. incongruent or congruent). I am not sure about the random slopes thought..
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2022-07-02 00:08:12
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https://elon.io/learn-turkish/lesson/advanced-temporal-relations
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ELON.IO
# Turkish
#### Lesson 153: Advanced Temporal Relations
0%
Look at the following two sentences.
This construction will finish after two weeks. Bu inşaat iki hafta sonra bitecek. This construction will finish in two weeks. Bu inşaat iki haftada bitecek.
And then compare them with the next two sentences.
This construction will finish within two weeks. Bu inşaat iki hafta içinde bitecek. This construction will finish in the next two weeks. Bu inşaat iki haftaya bitecek.
In the top two sentences the construction is expected to be finished after 14 days, and not sooner. In the bottom two sentences, the construction is expected to be finished in at most 14 days, so it might also be finished earlier.
Practice this lesson
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2022-10-05 02:16:44
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https://www.researchgate.net/publication/231067942_Discovery_of_an_Old_Nearby_and_Overlooked_Supernova_Remnant_Centered_on_the_Southern_Constellation_Antlia_Pneumatica
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ArticlePDF Available
# Discovery of an Old, Nearby, and Overlooked Supernova Remnant Centered on the Southern Constellation Antlia Pneumatica
Authors:
## Abstract
We report the discovery of a supernova remnant (SNR) with an angular diameter of 24°, centered on the southern constellation Antlia Pneumatica. The SNR is detected well in Hα and X-rays. Within the Antlia SNR's outline, a marginally significant feature is detected in the 1.8 MeV gamma-ray line of the radioisotope 26Al. At an estimated distance dA 60-340 pc, the Antlia SNR is perhaps the nearest SNR except for the Local Bubble. Consequently, any associated neutron star or black hole is expected to have a large proper motion. Of the trajectories of nearby pulsars with well-determined proper motions, only B0950+08's passes within the SNR outline. If the SNR and the pulsar B0950+08 indeed both originated from the same supernova, then their age t = 1.8(dA/100 pc) Myr.
L41
The Astrophysical Journal, 576:L41–L44, 2002 September 1
DISCOVERY OF AN OLD, NEARBY, AND OVERLOOKED SUPERNOVA REMNANT CENTERED
ON THE SOUTHERN CONSTELLATION ANTLIA PNEUMATICA
P. R. McCullough,
1,2
Brian D. Fields, and Vasiliki Pavlidou
Department of Astronomy, University of Illinois at Champaign-Urbana, 1002 West Green Street, Urbana, IL 61801;
[email protected], bdfi[email protected], [email protected]
Received 2002 May 22; accepted 2002 July 24; published 2002 August 1
ABSTRACT
We report the discovery of a supernova remnant (SNR) with an angular diameter of 24, centered on the
southern constellation Antlia Pneumatica. The SNR is detected well in Haand X-rays. Within the Antlia SNR’s
outline, a marginally significant feature is detected in the 1.8 MeV gamma-ray line of the radioisotope
26
Al. At
an estimated distance pc, the Antlia SNR is perhaps the nearest SNR except for the Local Bubble.d60–340
A
Consequently, any associated neutron star or black hole is expected to have a large proper motion. Of the
trajectories of nearby pulsars with well-determined proper motions, only B095008’s passes within the SNR
outline. If the SNR and the pulsar B095008 indeed both originated from the same supernova, then their age
Myr.tp1.8(d/100 pc)
A
Subject headings: gamma rays: observations — supernova remnants
Supernova explosions are one of nature’s most powerful
phenomena. They shape, heat, and ionize the interstellar me-
dium of galaxies (McKee & Ostriker 1977), and they synthesize
the heavy elements (Woosley & Weaver 1995). They leave
behind neutron stars or black holes, and an expanding super-
nova remnant (SNR) composed of both the star’s ejecta and
the interstellar matter swept up by the ejecta. Here we report
the discovery of an SNR with an angular diameter of 24,
centered on the southern constellation Antlia Pneumatica. Our
discussion is organized by passing from the most reliable to
the most speculative inferences. Thus, we first present the Ha
and X-ray observations that support the identification of an
SNR in Antlia. Then we discuss the tentative evidence for
26
Al
and the possible association of a high proper-motion pulsar
B095008 with the SNR. Finally, we note the possibility of
terrestrial radioisotopic signatures of such a recent, nearby su-
pernova explosion.
The strongest evidence for a new SNR in Antlia derives
from its Haand X-ray morphology. Located near the Vela
region that itself contains several SNRs, the Antlia Haemission
(Fig. 1) exhibits shell structure with an 24diameter, ap-
proximately centered on Galactic coordinates (l,b)p(276.5,
19) or J2000 coordinates (10
h
44
m
,37.2). The filaments on
the shell’s interior are also consistent with an SNR. In addition,
the ROSAT all-sky X-ray images clearly show a feature (Snow-
den et al. 1997) inside the Hashell that until now has not been
recognized as an SNR (Fig. 2). This X-ray enhancement is
consistent with a supernova origin, representing emission from
interstellar matter heated by the SNR.
Within the boundary of the Antlia SNR, the ROSAT soft X-
ray emission is anticorrelated with the IRAS 100 mm surface
brightness (Fig. 3). Because the 100 mm emission traces in-
terstellar dust and gas, the anticorrelation indicates that a sig-
nificant fraction of the X-rays is absorbed by the gas column
between the Earth and the SNR. The SNR’s mean surface
brightness in the ROSAT keV band is counts s
1
1
3
0.3 #10
4
arcmin
2
larger than the surface brightness of the SNR’s vi-
cinity, which is counts s
1
arcmin
2
. To allow for
3
0.5 #10
1
Cottrell Scholar of Research Corporation.
2
Also at Space Telescope Science Institute, 3700 San Martin Drive, Bal-
timore, MD 21218.
dust emission and X-ray absorption from matter behind the
SNR, we subtract 0.9 MJy sr / from the IRAS 100 mm
1
sin FbF
emission and make a corresponding correction to the ROSAT
data for an extragalactichalo brightness of counts
3
0.4 #10
s
1
arcmin
2
(Snowden et al. 1998).
We then fitted the ROSAT surface brightness within the SNR
outline to the remaining IRAS 100 mm emission. The median
IRAS 100 mm emission from dust between us and the Antlia
SNR is 2.0 MJy sr , corresponding to
1
ANSp3.0 #
H
between the Earth and the Antlia SNR (Schlegel,
20 2
10 cm
Finkbeiner, & Davis 1998). The inferred unattenuated keV
1
4
surface brightness of the Antlia SNR is Ip(1.10 0.04) #
1
counts s
1
arcmin
2
, which is similar to that of the faintest
3
10
SNR observed by ROSAT in the Large Magellanic Cloud (Wil-
liams et al. 1999). The quoted error for is simply the formal
I
1
error from the fit; it does not include systematic errors inherent
to the model.
The distance to the Antlia SNR can be estimated in at least
three ways. First, the column density
20 2
ANSp3#10 cm
H
combined with an estimated mean volume density in the inter-
stellar medium, (Ferriere 1998), implies a dis-
3
AnSp1.0 cm
tance pc. The SNR must be at least somewhat outside
d100
A
the Local Bubble because its low volume density (AnSp
) cannot account for the column density observed be-
23
10 cm
tween us and the Antlia SNR. Second, SNRs have diameters
100 pc, so if the Antlia SNR subtends on the sky,
2v24
A
then its distance pc. Third, as we show below, if thed240
A
pulsar B095008 and the Antlia SNR were both created by the
same supernova, then the intensity of the
26
Al gamma-ray emis-
sion implies for ejected
26
Al mass between60 pc d140 pc
A
and (Timmes et al. 1995). These dis-
54
2#10 4 #10 M
,
tance estimates are self-consistent and indicate that the Antlia
SNR may be the closest isolated SNR to the Sun.
If the Antlia feature is an SNR, its large size indicates great
age. A model of SNR expansion allows one to estimate the
age given the remnant’s radius . For pressure-drivenRdv
AA
snowplow expansion (McKee & Ostriker 1977; Blinnikov,
Imshennik, & Utrobin 1982), we have #tp1.8 Myr
. We see that the SNR will be
5/6 35/6 7/2
E(n/1 cm ) (R/50 pc)
50
more than 1 Myr in age if the SNR thermal energy Ep
50
ergs is 10% of the initial 10
51
ergs explosion
50
E/10
therm
L42 DISCOVERY OF ANTLIA SNR Vol. 576
Fig. 1.—Antlia SNR (24in diameter) in continuum-subtracted Haimage
(Gaustad et al. 2001). The brighter emission is darker in this negative image.
The Galactic latitude and longitude are indicated in the Hammer-Aitoff pro-
jection. The 24diameter circle (in white) approximates the edge of the Antlia
SNR. The circle extends slightly below the SNR at . Also, there is anb10
Hicloud that approximately fills the SNR’s large indentation at (l,b)p
.(275,27)
Fig. 2.—The keV ROSAT image of the same region as in Fig. 1. The
1
4
outline of the SNR is nearly identical in the Haand keV emissions.
1
4
kinetic energy and if the explosion occurred in a medium of
number density . This implies that Antlia is a very
3
n1cm
old SNR; the oldest known SNR is seen as an H ishell in the
low-density outer Galaxy and has an expansion age of 4.3 Myr
(Stil & Irwin 2001).
Due to its proximity to Earth, the Antlia SNR offers new
possibilities for the study of supernova nucleosynthesis. Namely,
26
Al has a mean life Myr, and its decaytp1.03
produces a 1.8 MeV gamma-ray line. The COMPTEL experi-
ment on the Compton Gamma Ray Observatory found that the
1.8 MeV emission was tightly concentrated toward the Galactic
plane, a strong indiction that short-lived, massive stars are the
dominant nucleosynthesis site of
26
Al (Diehl et al. 1995). It is
thus natural to search for an
26
Al feature in the nearby Antlia
SNR.
The COMPTEL 1.8 MeV all-sky map (Diehl et al. 1995)
shows a feature (Fig. 4) within the Haand ROSAT boundaries
of the SNR. The 1.8 MeV local maximum is 3 jabove back-
ground (Oberlack 1997) and thus is marginally significant; its
location, (l,b)p(271,16), is offset (Dl,Db)p(5.5,
3) from the nominal center of the SNR as seen in Ha,
(l,b)p(276.5, 19). We note that the edge of theSNR nearest
the 1.8 MeV local maximum is also the brightest segment of the
outline in Haand is the only segment of the outline that is
detected in radio continuum at 408 MHz (Haslam et al. 1982).
There are eight local maxima at 3jwithin two “control”
regions of sky defined by and .10FbF300l360
Thus, the a priori probability of at least one of those eight having
appeared within the SNR’s outline by chance is 0.22.Therefore,
the 1.8 MeV feature in the Antlia SNR may be real and may
be due to
26
Al produced by the progenitor of the SNR.
The presence of
26
Al would strengthen the case that the Antlia
feature is an SNR. Also, the association of
26
Al with an SNR
would constitute direct evidence that this isotope is indeed
made in massive stars. Finally, an
26
Al signature is a powerful
chronometer. The angular size of the observed feature is close
to the COMPTEL resolution. If we treat the feature as a point
source of
26
Al, at distance and with age t, the expectedd
A
1.8 MeV line flux is
521
Fp5.9 #10 photons cm s
g
2
M100 pc
ej, 26
t/1.03 Myr
#e.(1)
()
5
5#10 Md
,A
Here is the initial mass of
26
Al, and the fiducial value isM
ej, 26
from theory (Timmes et al. 1995) and is consistent with the
observed upper limit from Velorum (Oberlack et al. 2000),
2
g
a Wolf-Rayet star, for which the theoretical
26
Al yields (Langer,
Braun, & Fliegner 1995; Meynet et al. 1997) are comparable
to those of a supernova event. In fact, the Antlia feature appears
in the Velorum field of view, and from Figure 1 of Oberlack
2
g
et al. (2000), we find an observed flux Fp(1.8 0.6) #
A
photons cm
2
s
1
. With a yield range of
55
10 2 #10 M
,
, we determine that 110 pc
t/1.03 Myr 4
Me 4#10 M
ej, 26 ,
d
A
510 pc, consistent with the estimates above.
Although the COMPTEL 1.8 MeV feature is consistent with
a point source, the full
26
Al emission from an SNR this old is
likely to be extended. The angular distribution of the emission
may give a clue as to the origin of the
26
Al. Current expectations
are that massive star
26
Al ejection can occur in Wolf-Rayet
winds prior to the explosion (Langer et al. 1995; Meynet et al.
1997) or in the subsequent explosion (Timmes et al. 1995). In
either case, the Antlia SNR should have its
26
Al in the shell
wall, either as wind material that was then swept by the blast
wave or as ejecta that has by now overtaken the shell. In either
case, one expects extended emission, the geometry and inho-
mogeneity of which would reflect that of the production mech-
anism; e.g., the point source might be the brightest of several
26
Al “knots” resulting from aspherical explosion ejecta, of the
No. 1, 2002 McCULLOUGH, FIELDS, & PAVLIDOU L43
Fig. 3.—The keV surface brightness measured by ROSAT compared with
1
4
the 100 mm surface brightness measured by IRAS within the outline of the
Antlia SNR for Galactic latitude (points). For clarity, and to improveb115
the statistical independence of the points selected, only a small random fraction
(10%) of the points is plotted. The points are binned, and the median (squares)
and error bars are plotted, where jis the standard deviation of the N
1j/N
points in each bin. The best two-parameter fit of the form
jN
HH
IpIIe
X01
is shown, with and counts s
1
arcmin
2
(solid
6
Ip545 Ip1120 #10
01
line). The observed intensities of both the IRAS and ROSAT data have been
shifted to correct for emission and absorption behind the SNR; the upper and
lower vectors show the corrections for and 30, respectively.bp15
Fig. 4.—Surface brightness of the 1.8 MeV emission line from
26
Al con-
toured at levels of 1.0, 1.5, 2.0, … #10
4
photons cm
2
s
1
sr
1
(Kno¨dlseder
et al. 1999). The path of the pulsar B095008 is plotted at regular time
intervals in the past, based on its measured distance and proper motion (Brisken
et al. 2000). The coordinate grids of Figs. 1 and 2 are superposed. For assumed
distances from Earth to the Antlia SNR of 100, 150, and 200 pc, the implied
ages and radial velocities of the pulsar B095008 are , 2.7, andtp1.8
3.6 Myr and , 64, and 38 km s
1
, respectively.vp115
r
kind found in, e.g., the Cas A SNR (Hughes et al. 2000).
3
Also,
because the large SNR is expanding into the inhomogeneous
gas of the disk, we expect the shell material to be more con-
centrated and limb-brightened toward the denser regions inthe
Galactic plane, as observed. At any rate, we predict that ex-
tended emission is present at 1.8 MeV, with intensity Ip
g
, where is the column density of
26
Al.N/4pt N
26 26 26
The upcoming International Gamma-Ray Astrophysical Lab-
oratory (INTEGRAL) mission may be able to confirm or refute
the presence of
26
Al in the Antlia SNR. If the Antlia SNR
26
Al
emission can be treated as a point source by INTEGRAL, then
a3jemission can be achieved after an on-target exposure
time of s (N. R. Trams 2000).
4
However, since the
5
2#10
SNR has such a large angular size, INTEGRAL may in fact be
able to resolve the 1.8 MeV emission. In this case, the relevant
background noise is increased, and larger exposure times will
be required. On the other hand, once the detection of the Antlia
feature by INTEGRAL has been achieved, any spatial infor-
mation concerning the gamma-ray emission will give valuable
information on the distribution of
26
Al within the SNR. More-
26
Al, INTEGRAL might (depending on the
actual age of the SNR) detect another long-lived radioisotope,
60
Fe ( Myr), whose decay produces two g-raytp2.2 0.4
lines, one at 1.173 MeV and another at 1.322 MeV.
60
Fe can
serve as an excellent diagnostic because it is coproduced with
3
The emission could be even more complex if
26
Al condensed into grains
that retain a high velocity, as suggested by balloon-borne observations (Naya
et al. 1996) of the Galactic center at 1.8 MeV.
4
See the SPI Observer’s Manual available on-line at http://astro.estec.esa.nl/
Integral/isoc/html/AO_documents.html.
26
Al (Timmes et al. 1995). Were the
60
Fe/
26
Al line ratios mea-
sured, the age of the SNR could be determined, subject only
to theoretical uncertainties in the relative yields. Conversely,
were distance and age independently determined, the line ratios
could be used to derive the yields and their ratios.
Given the proximity of the Antlia supernova to Earth, a
relatively large proper motion is expected for the neutron star
or black hole produced in the explosion. If a neutron star was
produced, it may be observed as a pulsar. Of the eight pulsars
with kpc and well-measured proper motions (Hooger-d!1
werf, de Bruijne, & de Zeeuw 2001), we find only one,
B095008 (Pilkington et al. 1968; Brisken et al. 2000), whose
trajectory intersects the boundary of the Antlia SNR (Fig. 4).
In fact, B095008’s trajectory does not cross the center of the
remnant but does intersect with the location of the
26
Al feature.
However, because not all supernovae produce neutron stars,
and because not all neutron stars are observable as pulsars, we
cannot be certain that B095008 is the stellar remnant asso-
ciated with the Antlia SNR.
If we assume that the SNR and B095008 arise from the
same supernova explosion, then their geometry and kinematics
can be used to derive an age. As seen in Figure 5, we have an
“age-distance relation” of , consistenttp1.8 Myr(d/100 pc)
A
with the distance and age estimates above. This relation and
the pressure-driven expansion (McKee & Ostriker 1977; Blin-
nikov et al. 1982) relation yield a distance
50 1/3 31/3
dp340 pc(E/10 ergs) (n/1 cm ) (2)
Atherm
and an age
50 1/3 31/3
tp6.2 Myr(E/10 ergs) (n/1 cm ) . (3)
therm
L44 DISCOVERY OF ANTLIA SNR Vol. 576
Fig. 5.—Antlia SNR (A), Earth (E), and the pulsar B095008 (P), which
form a triangle with angle AEP ( ). The pulsar’s distance EP pvp45
pc and its tangential velocity km s
1
are known from the280 25 vp39
VLBI (Brisken et al. 2000). If we denote the time tsince the supernova
explosion, then the physical scale of the triangle is determined by AB p
and thus the distance to the Antlia SNR , and
vtdpEA p55(t/1 Myr) pc
A
km s
1
.vp(EP dcos v)/t55(280 pc/dcos v)
AA
r
These estimates are within a factor of 2 of the other estimates
mentioned above, an acceptable agreement given the crudeness
of the simplifications used for the expansion of an old SNR.
Improved age estimates for the SNR and the pulsar would allow
for a better assessment of the likelihood of their association.
Alternatively, combining the age-distance relation with the
observed
26
Al flux provides other limits on and tvia equationd
A
(1). Given an
26
Al yield of (Tim-
5
Mp(2–40) #10 M
ej, 26 ,
mes et al. 1995; Oberlack et al. 2000) and the COMPTEL flux
value of photons cm
2
s
1
, , and
5
1.8 #10 dp62–140 pc
A
thus Myr, consistent with the results using thetp1.1–2.5
expansion law within uncertainties. The inferred age is ap-
proximately 10 times smaller than the mean spin-down age of
B095008 ( Myr), but kinematically determined
˙
P/2Pp17
ages tend to be less than spin-down ages, in some cases by
large factors (Cordes & Chernoff 1998).
It is thus possible that the Antlia feature and B095008 are
an example of an SNR linked to a distant pulsar. Such an
connection is important to current theories of pulsar kicks (Cor-
des & Chernoff 1998; Lyne & Lorimer 1994) and can be further
tested for the Antlia SNR by better determinations of its age
and distance.
It is interesting to consider the effects of such a nearby su-
pernova on the Earth. Nearby supernovae may have deposited
less than that of the Earth (Ellis, Fields, & Schramm 1996). Live
60
Fe has been recently detected in a deep-ocean crust (Knie et
al. 1999), and Benı´tez, Maı´z-Apella´niz, & Canelles (2002) have
suggested its origin in one or more supernova explosions within
the past 11 Myr from the Scorpius-Centaurus OB association.
The Antlia supernova is probably too recent and too distant to
have contributed to the terrestrial
60
Fe. In any case, the Antlia
event would have occurred beyond the “minimum safe distance”
of 10 pc (Ruderman 1975; Ellis & Schramm 1995) and yet
would have been a spectacular event: at maximum light, it would
have appeared nearly as bright as the full Moon.
We thank Uwe Oberlack for providing machine-readable maps
of
26
Al and for helpful discussions. We also thank You-Hua Chu,
Jim Cordes, John Dickel, Icko Iben, Uwe Oberlack, and Beate
Woermann for helpful discussions. We are grateful to the anon-
ymous referee for constructive comments that improved this Let-
ter. This work was supported by the Research Corporation and
NSF grants AST 98-74670 and AST 00-92939.
REFERENCES
Benı´tez, N., Maı´z-Apella´niz, J., & Canelles, M. 2002, Phys. Rev. Lett., 88(8),
081101
Blinnikov, S. I., Imshennik, V. S., & Utrobin, V. P. 1982, Soviet Astron. Lett.,
8, 361
Brisken, W. F., Benson, J. M., Beasley, A. J., Fomalont, E. B., Goss, W. M.,
& Thorsett, S. E. 2000, ApJ, 541, 959
Cordes, J. M., & Chernoff, D. F. 1998, ApJ, 505, 315
Diehl, R., et al. 1995, A&A, 298, 445
Ellis, J., Fields, B. D., & Schramm, D. N. 1996, ApJ, 470, 1227
Ellis, J., & Schramm, D. N. 1995, Proc. Natl. Acad. Sci., 92, 235
Ferriere, K. 1998, ApJ, 503, 700
Gaustad, J. E., McCullough, P. R., Rosing, W., & Van Buren, D. 2001, PASP,
113, 1326
Haslam, C. G. T., Stoffel, H., Salter, C. J., & Wilson, W. E. 1982, A&AS, 47,
1
Hoogerwerf, R., de Bruijne, J. H. J., & de Zeeuw, P. T. 2001, A&A, 365, 49
Hughes, J. P., Rakowski, C. E., Burrows, D. N., & Slane, P. O. 2000, ApJ,
528, L109
Knie, K., Korschinek, G., Faestermann, T., Wallner, C., Scholten, J., & Hil-
lebrandt, W. 1999, Phys. Rev. Lett., 83, 18
Kno¨dlseder, J., et al. 1999, A&A, 345, 813
Langer, N., Braun, H., & Fliegner, J. 1995, Ap&SS, 224, 275
Lyne, A. G., & Lorimer, D. R. 1994, Nature, 369, 127
McKee, C. F., & Ostriker, J. P. 1977, ApJ, 218, 148
Meynet, G., Arnould, M., Prantzos, N., & Paulus, G. 1997, A&A, 320, 460
Naya, J. E., Barthelmy, S. D., Bartlett, L. M., Gehrels, N., Leventhal, M.,
Parsons, A., Teegarden, B. J., & Tueller, J. 1996, Nature, 384, 44
Oberlack, U. 1997, Ph.D. thesis, Tech. Univ. Mu¨nchen
Oberlack, U., et al. 2000, A&A, 353, 715
Pilkington, J. D. H., Hewish, A., Bell, S. J., & Cole, T. W. 1968, Nature, 218,
126
Ruderman, M. A. 1975, Science, 84, 1079
Schlegel, D. J., Finkbeiner, D. P., & Davis, M. 1998, ApJ, 500, 525
Snowden, S. L., Egger, R., Finkbeiner, D. P., Freyberg, M. J., & Plucinsky,
P. P. 1998, ApJ, 493, 715
Snowden, S. L., et al. 1997, ApJ, 485, 125
Stil, J. M., & Irwin, J. A. 2001, ApJ, 563, 816
Timmes, F. X., Woosley, S. E., Hartmann, D. H., Hoffman, R. D., Weaver,
T. A., & Matteucci, F. 1995, ApJ, 449, 204
Williams, R. M., Chu, Y., Dickel, J. R., Petre, R., Smith, R. C., & Avarez,
M. 1999, ApJS, 123, 467
Woosley, S. E., & Weaver, T. A. 1995, ApJS, 101, 181
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An analytic solution is obtained for the evolution of a supernova remnant during the radiative-cooling phase. The Cygnus Superbubble X-ray source could have been formed by the explosion of a single supernova releasing an energy of 10 to the 52nd - to to the 53rd erg. Analysis of the light curve of the NGC 1058 supernova 1961v demonstrates that it represents an approximately 2 x 10 to the 52nd outburst of a supermassive (approximately 1000 solar masses) star. Giant shell sources may form an evolutionary sequence, exemplified by R136a as the presupernova, supernova 1961v, and the Superbubble as the remnant.
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The Wolf-Rayet binary system γ2 Vel (WR 11) is the closest known Wolf-Rayet (WR) star. Recently, its distance has been redetermined by parallax measurements with the HIPPARCOS astrometric satellite yielding 258+41-31 significantly lower than previous estimates (300-450 pc). Wolf-Rayet stars have been proposed as a major source of the Galactic 26 Al observed at 1.809 MeV. The gamma-ray telescope COMPTEL has previously reported 1.8 MeV emission from the Vela region, yet located closer to the Galactic plane than the position of 72 Vel. We derive an upper 1.8 MeV flux limit of 1.1 10-5γcm-2 S-1 (2 σ) for the WR star. With the new distance estimate, COMPTEL measurements place a limit of (6.3+2.1-1.4) 10-5 M⊙on the 26 AI yield of γ2 Vel, thus constrains theories of nucleosynthesis in Wolf-Rayet stars. We discuss the implications in the context of the binary nature of γ2 Vel and present a new interpretation of the IRAS Vela shell.
Article
We have completed a robotic wide-angle imaging survey of the southern sky (delta=+15deg to -90°) at 656.3 nm wavelength, the Halpha emission line of hydrogen. Each image of the resulting Southern Halpha Sky Survey Atlas (SHASSA) covers an area of the sky 13° square at an angular resolution of approximately 0.8' and reaches a sensitivity level of 2 R (1.2×10-17 ergs cm-2 s-1 arcsec-2) pixel-1, corresponding to an emission measure of 4 cm-6 pc and to a brightness temperature for microwave free-free emission of 12 muK at 30 GHz. Smoothing over several pixels allows features as faint as 0.5 R to be detected.
Article
In a deep ocean ferromanganese crust an excess of 60Fe radioactivity was measured by means of high sensitivity accelerator mass spectrometry. The enhanced concentrations measured in the first two of three layers (corresponding to a time span of 0-2.8 Myr and 3.7-5.9 Myr, respectively) suggest the deposition of supernova produced 60Fe on earth. There is even a weak indication that the flux into the crust was higher about 5 Myr ago.
Article
Details are now given of three of the four pulsating radio sources discovered at Cambridge.
Article
THEORY predicts1 that radioactive 26A1 (which has a half-life of 0.72 Myr) is released into the interstellar medium by nova and supernova explosions, from the winds of massive stars in the Wolf–Rayet phase, and from less-massive giant stars in very late stages of the asymptotic giant branch phase. Observations of 1,809-keV γ-ray emission line from 26A1 can therefore be used as a tracer of Galactic nucleosynthesis during the past million years2,3. The irregularity of the emission in the plane of the Galaxy4–7 suggests that the dominant sources are likely to be massive stars and supernovae; the other predicted sources are older, and therefore expected to be distributed more uniformly. Here we report the detection of the 1,809-keV emission line from the direction of the Galactic Centre, and we show that the line width is approximately three times that expected8,9 from the effect of Doppler broadening due to Galactic rotation. The high velocities inferred from the line width favour an origin of the 26A1 in supernovae or Wolf–Rayet stars. Moreover, the fact that the 26A1 has maintained such high velocities is difficult to reconcile with our current understanding of the propagation of material in the interstellar medium.
Article
NEUTRON stars are usually born during the supernova explosion of a massive star. Any small asymmetry during the explosion can result in a substantial ‘kick’ velocity1 to the neutron star. Pulsars (rapidly rotating, magnetized neutron stars) have long been known to have high space velocities2,3, but new measurements of proper motion4–6, adoption of a new distance scale for the pulsars7 and the realization that some previous velocities were systematically low by a factor of 2 (ref. 8) have prompted us to reassess these velocities. Here, taking into account a strong selection effect that makes the observed velocities unrepresentative of those acquired at birth9, we arrive at a mean pulsar birth velocity of 450 ± 90 km s–1 This exceeds the escape velocity from binary systems, globular clusters and the Galaxy, and so will affect our understanding of the retention of neutron stars in these systems. Those neutron stars that are retained by the Milky Way will be distributed more isotropically than has been thought10–12, which may result in a distribution like that of the γ-ray burst sources.
Article
We present a full-sky 100 μm map that is a reprocessed composite of the COBE/DIRBE and IRAS/ISSA maps, with the zodiacal foreground and confirmed point sources removed. Before using the ISSA maps, we remove the remaining artifacts from the IRAS scan pattern. Using the DIRBE 100 and 240 μm data, we have constructed a map of the dust temperature so that the 100 μm map may be converted to a map proportional to dust column density. The dust temperature varies from 17 to 21 K, which is modest but does modify the estimate of the dust column by a factor of 5. The result of these manipulations is a map with DIRBE quality calibration and IRAS resolution. A wealth of filamentary detail is apparent on many different scales at all Galactic latitudes. In high-latitude regions, the dust map correlates well with maps of H I emission, but deviations are coherent in the sky and are especially conspicuous in regions of saturation of H I emission toward denser clouds and of formation of H2 in molecular clouds. In contrast, high-velocity H I clouds are deficient in dust emission, as expected.
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2022-07-04 16:06:35
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https://scipost.org/SciPostPhys.5.1.006
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Time-reversal symmetry, anomalies, and dualities in (2+1)$d$
Clay Cordova, Po-Shen Hsin, Nathan Seiberg
SciPost Phys. 5, 006 (2018) · published 20 July 2018
Abstract
We study continuum quantum field theories in 2+1 dimensions with time-reversal symmetry $\cal T$. The standard relation ${\cal T}^2=(-1)^F$ is satisfied on all the "perturbative operators" i.e. polynomials in the fundamental fields and their derivatives. However, we find that it is often the case that acting on more complicated operators ${\cal T}^2=(-1)^F {\cal M}$ with $\cal M$ a non-trivial global symmetry. For example, acting on monopole operators, $\cal M$ could be $\pm1$ depending on the magnetic charge. We study in detail $U(1)$ gauge theories with fermions of various charges. Such a modification of the time-reversal algebra happens when the number of odd charge fermions is $2 ~{\rm mod}~4$, e.g. in QED with two fermions. Our work also clarifies the dynamics of QED with fermions of higher charges. In particular, we argue that the long-distance behavior of QED with a single fermion of charge $2$ is a free theory consisting of a Dirac fermion and a decoupled topological quantum field theory. The extension to an arbitrary even charge is straightforward. The generalization of these abelian theories to $SO(N)$ gauge theories with fermions in the vector or in two-index tensor representations leads to new results and new consistency conditions on previously suggested scenarios for the dynamics of these theories. Among these new results is a surprising non-abelian symmetry involving time-reversal.
Ontology / Topics
See full Ontology or Topics database.
Authors / Affiliations: mappings to Contributors and Organizations
See all Organizations.
Funder for the research work leading to this publication
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2020-08-11 12:04:38
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https://brilliant.org/discussions/thread/tessellate-stems-physics-school-set-1-subjective/
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# Tessellate S.T.E.M.S - Physics - School - Set 1 - Subjective Problem
A pulley in the form of a circular disc of mass $$m$$ and radius $$r$$ has the groove cut all along the perimeter. A string whose one end is attached over to the ceiling passes over this disc pulley and its other end is attached to a spring of spring constant k. The other end of the spring is attached to the ceiling as shown in the figure. Find the time period of vertical oscillations of the centre of mass assuming that the string does not slip over the pulley?
This problem is a part of Tessellate S.T.E.M.S.
Note by Writabrata Bhattacharya
9 months, 4 weeks ago
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Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
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$\omega ^2=\frac {4kx}{1+\frac {I_{cm}}{mR^2}}$,therefore here $$\omega =\sqrt \frac {8k}{3m}$$
- 9 months, 2 weeks ago
how u derived the first formula
- 9 months, 2 weeks ago
Refer HC Verma. The question had been directly copied from there
- 9 months, 1 week ago
your answer is dimensionally incorrect it would be better if u correct it
- 9 months, 1 week ago
I am not used to doing such mistakes.Kindly check it correctly the answer is correct as are the dimensions.
- 9 months, 1 week ago
2pi_/(3m/8k)
- 9 months, 1 week ago
π(m/k)^(1/2)
- 9 months, 3 weeks ago
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2018-10-18 19:11:52
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https://math.stackexchange.com/questions/1415925/combined-arrival-rate
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# Combined arrival rate
Let us suppose a scenario with two clients, $a$ and $b$, each one generating load at rate $\lambda_a$ and $\lambda_b$, respectively. The server receives the requests from both clients.
What will be the arrival rate at server?
(consider inter-arrival times independently generated by an exponential distribution)
• The answer depends on whether the client request generation processes are independent processes or dependent processes. In the former case, the answer is $\lambda_a+\lambda_b$; in the latter, it is not. – Dilip Sarwate Aug 31 '15 at 14:25
• @DilipSarwate, consider they are independent processes. Could you give an example of why the answer, then, is the sum of the rates? (thank you) – Lourenco Aug 31 '15 at 15:52
Let $A$ and $B$ denote the number of requests in $(0,T]$ from clients $a$ and $b$ relatively. Then, $A$ and $B$ are Poisson random variables with parameters $\lambda_aT$ and $\lambda_bT$ respectively, and you have declared them to be independent random variables. Thus, $C = A+B$, the total number of requests received during $(0,T]$ is a Poisson random variable also with parameter $(\lambda_a+\lambda_b)T$. This is because for any $N \geq 0$, \begin{align} P\{C = N\} &= P\{A+B = N\}\\ &= \sum_{n=0}^N P\{A= n, B = N-n\}\\ &= \sum_{n=0}^N P\{A= n\}P\{B = N-n\}&{\scriptstyle{\text{because $A$ and $B$ are independent random variables}}}\\ &= \sum_{n=0}^N \exp(-\lambda_aT)\frac{(\lambda_aT)^n}{n!} \exp(-\lambda_bT)\frac{(\lambda_bT)^{N-n}}{(N-n)!}\\ &= \exp(-(\lambda_a+\lambda_b)T)\sum_{n=0}^N \frac{(\lambda_aT)^n}{n!}\frac{(\lambda_bT)^{N-n}}{(N-n)!}\\ &= \frac{\exp(-(\lambda_a+\lambda_b)T)}{N!} \sum_{n=0}^N \binom{N}{n}(\lambda_aT)^n(\lambda_bT)^{N-n}\\ &= \exp(-(\lambda_a+\lambda_b)T)\frac{((\lambda_a+\lambda_b)T)^N}{N!} \end{align} When $A$ and $B$ are not independent, this does not work because the step above where independence was used is no longer valid.
• Supposing $(0,T]$ is the time in which the samples happen, why let the Poisson random variables' parameters be $\lambda_a T$ and $\lambda_b T$ instead of $\lambda_a$ and $\lambda_b$? – Lourenco Sep 1 '15 at 2:42
• $\lambda_a$ is the request rate, measured in number of requests per unit time. So, the average number of arrivals in $(0,T]$ is $\lambda_aT$, not just $\lambda_a$. – Dilip Sarwate Sep 1 '15 at 2:45
• Considering the same proof you provided, the case of $\lambda_a = \lambda_b$ still valid, isn't it? (sorry for the naive confirmation) – Lourenco Sep 1 '15 at 18:08
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2019-07-19 18:52:06
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https://electricalengineerresources.com/2018/05/08/what-is-ruling-span/
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# What is Ruling Span?
Posted by
If all spans in a section of line are of the same length then the tension on individual span will be equal.Keeping the span lengths the same is possible on lines constructed on open terrains. However, for construction along highways and residential areas, the span lengths can never be equal. The owner of the property wants the poles be installed on the boundary of his/her lot. This causes a diverse length of spans that will affect the sag and conductor tension of the individual spans.
## What is a ruling span?
A ruling span, also known as equivalent span or mean effective span (MES), is an assumed uniform design span which approximately portray the mechanical performance of a section of line between its dead-end supports. The ruling span is used in the design and construction of a line to provide a uniform span length which is a function of the various lengths of spans between dead-ends. This uniform span length allows sags and clearance to be readily calculated for structure spotting and conductor stringing.
## How to use Ruling Span?
A value for the ruling span should be assumed before spotting structures because the actual ruling span can only be calculated after the structure locations are determined. In most cases, the actual ruling span should be greater than or equal to the assumed ruling span to ensure design clearances.
One or more assumed ruling spans, based on experience, has to be used for the field design of new line because the theoretical ruling span of a line section cannot be determined until after the line is staked. If the land is reasonably flat, it is appropriate to use a ruling span that approximates the level ground span. The required ground clearance may be subtracted from the attachment height of the lowest conductor to determine the sag limited by ground clearance. This sag value can then be used to determine a ruling span length whose sag is approximately equal to the sag allowed by the basic structure height. For rugged terrain, a ruling span that is longer than the level ground span is usually more effective.
After staking, the theoretical ruling span should be calculated and computed with the design ruling span. Using a design ruling span appreciably different from the theoretical ruling span of the section will produce unpredictable sags and tensions. Slack sags may cause clearance problems while tightly drawn spans may cause uplifts problems. Higher than predicted tensions may exceed the permitted load on support assemblies or may cause aeolian vibration problems.
## The Ruling Span Theory
During stringing and sagging, the conductors are placed on the travelers (or rollers) and are dead-ended at the ends of the stringing section of the line. While the conductor is on travelers and free to move between spans, the conductor tensions, and length in any span is a function of the combined averaged tension of all the spans and the total conductor length of the dead-ended stringing section.
When the spans are of unequal length and the supports are of varying elevations, the mathematics become too complicated to be easily calculated.
The assumptions of Ruling Span Theory:
1. The supports are at equal elevations
2. The horizontal tension is constant throughout the stringing section
3. The uneven spans are replaced by a series of equal spans
### THE THEORETICAL RULING SPAN EQUATION:
$\fn_phv&space;S_{R}&space;=&space;\sqrt{\frac{\sum&space;S^{3}}{\sum&space;S}}&space;=&space;\sqrt{\frac{S_{1}^{3}+S_{2}^{3}+...&space;S_{n}^{3}}{S_{1}+S_{2}+...&space;S_{n}}}$
where:
SR = the theoretical ruling span
S1,S2, … Sn = are the 1st, 2nd, … nth span length respectively
Although this equation is not exact because of the assumption made but its accuracy is sufficient for most line designs, it is the equation most often to calculate the ruling span for new overhead distribution lines.
After being tied in, each span virtually becomes a dead-end span with approximately the same tension as the theoretical ruling span. When the tied spans in the section are of different lengths, changes in temperature, loading and elongation due to creep will cause a difference in tension between spans. These differences in tension will cause a flexing or bending of poles and arms.
This ruling span rules the behavior of the sagged section of the line. The sag characteristics of the ruling span set the sag characteristics of every span in the section. If conductors are installed using a sag-tension table with the wrong ruling span, actual final sags and tension will not be the same as predicted. The greater the difference, the greater the error!
## Estimated Ruling Span
Knowledge gained from a reconnaissance of the proposed line route may make it possible to estimate a ruling span. A traditional “rule of thumb” equation that may be helpful in the estimation of a ruling span is:
$\fn_phv&space;\fn_phv&space;S_{E}&space;=&space;Average\;&space;Span&space;\;&space;+&space;2/3&space;\;(Maximum\:&space;Span&space;-&space;Average&space;\;Span)$
Use this rule for estimating the ruling span with caution! This “rule of thumb”, used indiscriminately, has significantly different sags and tensions that the true ruling span equation. Even one span much longer than the average span may cause estimated ruling span to be much greater than the actual theoretical ruling span. This formula should only be used for estimating the ruling span when the actual spans are not yet known. When the spans are known, the theoretical equation should be used.
The estimated ruling span equation is easily solved and convenient for field used. When engineering calculator is available, used of the following equation provides greater accuracy:
$\fn_phv&space;S_{E}&space;=&space;\sqrt{\frac{\left&space;[&space;\frac{(\sum&space;{S&space;-&space;S_{m}})^{3}+S_{m}^{3}}{(N-1)^{2}}&space;\right&space;]}{\sum{S}}}$
where:
SE = Estimate Ruling Span
ΣS = Estimate total length of all spans in the stringing section
N = Estimated number of spans in stringing section
SM = Length of the estimated longest span in the stringing section
Another form of the estimated ruling span equation is:
$\fn_phv&space;S_{E}&space;=&space;\sqrt{\frac{N_{E}&space;S_{a}^{3}+S_{M}^{3}}{N_{E}S_{a}+S_{M}}}$
where:
S= estimated average span of stringing section, exclusive of the longest span
## Effect of Wrong Ruling Span
The greater the difference between the theoretical ruling span and the design ruling span, the greater the variation will be between the actual and predicted sag and tension values. The magnitude by which actual sag and tensions will differ from the predicted values is a function of conductor temperature and loading.
1. if the design sag is greater than the theoretical sag, then the actual sag of the installed conductors will be less than the predicted sag. This condition will lead to increased conductor tensions, which may exceed the permitted loads of support structures and guying assemblies.
2. If the design sag is less than the theoretical sag, then the actual sag of the installed conductors will be greater than the predicted sag. This condition may result in the inadequate ground clearances.
## EXAMPLE
In this example, we will calculate the ruling span base on the four equations above and see their differences.
## Span Lengths Summary
SpanSectionSpan Length (m)
1Pole 1 - Pole 266
2Pole 2 - Pole 360
3Pole 3 - Pole 475
4Pole 4 - Pole 565
Using Equation 1:
SRULING = √(663 + 603 + 753 + 653)/(66 +60 +75 +65) = 67.17 meters
Using Equation 2:
Average Span = (66 +60 +75 +65)/4 = 66.5
SMAXIMUM = 75
SESTIMATED RULING = 66.5 + 2/3 (75 – 66.5) = 72.17 meters
Using Equation 3:
∑S = (66 +60 +75 +65) = 266 , N = 4
SESTIMATED RULING = √[((266 – 75)+753)/(4-1)2]/266 = 55.56 meters
Using Equation 4:
Saverage (exclusive max span) = (66 +60 +65)/3 = 63.7
SESTIMATED RULING = √ (4 * 63.73 + 753)/(4 * 63.7 +75) = 66.42 meters
Using Excel Spreadsheet, here is the summary:
In the next post we will calculate the sag and tension of the individual spans based on the calculated ruling span.
References:
1. RUS Bulletin 1724E-200
2. RUS Bulletion 1724-152
3. IEAI MAGAZINE: The Effects of Ruling Span on Sag and Tension
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2023-02-04 06:35:05
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http://en.wikipedia.org/wiki/Small_set_(combinatorics)
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Small set (combinatorics)
In combinatorial mathematics, a small set of positive integers
$S = \{s_0,s_1,s_2,s_3,\dots\}$
is one such that the infinite sum
$\frac{1}{s_0}+\frac{1}{s_1}+\frac{1}{s_2}+\frac{1}{s_3}+\cdots$
converges. A large set is one whose sum of reciprocals diverges.
Examples
• The set $\{1,2,3,4,5,\dots\}$ of all positive integers is known to be a large set (see Harmonic series), and so is the set obtained from any arithmetic sequence (i.e. of the form an + b with a ≥ 0, b ≥ 1 and n = 0, 1, 2, 3, ...) where a = 0, b = 1 give the multiset $\{1,1,1,\dots\}$ and a = 1, b = 1 give $\{1,2,3,4,5,\dots\}$.
• The set of square numbers is small (see Basel problem). So is the set of cube numbers, the set of 4th powers, and so on. More generally, the set of values of a polynomial $a_k n^k+a_{k-1} n^{k-1}+\cdots+a_2 n^2+a_1 n+a_0$, k ≥ 2, ai ≥ 0 for all i ≥ 1, ak > 0. When k=1 we get an arithmetic sequence (which forms a large set.).
• The set $\{1,2,4,8,\dots\}$ of powers of 2 is known to be a small set, and so is the set of any geometric sequence (i.e. of the form abn with a ≥ 1, b ≥ 2 and n = 0, 1, 2, 3, ...).
• The set of numbers whose decimal representations exclude 7 (or any digit one prefers) is small. That is, for example, the set
$\{\dots, 6, 8, \dots, 16, 18, \dots, 66, 68, 69, 80, \dots \}$
is small. (This has been generalized to other bases as well.) See Kempner series.
Properties
• A union of finitely many small sets is small, as the sum of two convergent series is a convergent series. A union of infinitely many small sets is either a small set (e.g. the sets of p2, p3, p4, ... where p is prime) or a large set (e.g. the sets $\{n^2 + k : n > 0 \}$ for k > 0). Also, a large set minus a small set is still large. A large set minus a large set is either a small set (e.g. the set of all prime powers pn with n ≥ 1 minus the set of all primes) or a large set (e.g. the set of all positive integers minus the set of all positive even numbers). In set theoretic terminology, the small sets form an ideal.
• The Müntz–Szász theorem is that a set $S=\{s_1,s_2,s_3,\dots\}$ is large if and only if the set spanned by
$\{1,x^{s_1},x^{s_2},x^{s_3},\dots\} \,$
is dense in the uniform norm topology of continuous functions on a closed interval. This is a generalization of the Stone–Weierstrass theorem.
Open problems
It is not known how to identify whether a given set is large or small in general. As a result, there are many sets which are not known to be either large or small.
Paul Erdős famously asked the question of whether any set that does not contain arbitrarily long arithmetic progressions must necessarily be small. He offered a prize of \$3000 for the solution to this problem, more than for any of his other conjectures, and joked that this prize offer violated the minimum wage law.[1] This question is still open.
Notes
1. ^ Carl Pomerance, Paul Erdős, Number Theorist Extraordinaire. (Part of the article The Mathematics of Paul Erdős), in Notices of the AMS, January, 1998.
References
• A. D. Wadhwa (1975). An interesting subseries of the harmonic series. American Mathematical Monthly 82 (9) 931–933. JSTOR 2318503
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2014-03-08 02:39:25
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https://www.overleaf.com/learn/latex/Questions/How_do_I_insert_an_image_at_a_specific_point_in_the_document%3F
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You can control the position of an image using options for the figure environment, e.g. the [h!] in the example below tells LaTeX to put the figure exactly where it appears in the text, instead of letting it 'float' to a particular place in the document.
\begin{figure}[h!]
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2022-08-16 10:19:12
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https://eguruchela.com/physics/formulas/gas_laws.php
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# Kelvin Temprerature
T: temperature in Kelvin scale, t: temperature in Celsius scale
## Equation Of State
p: pressure(P, Pascal), V: volume(m3), n: number of moles of the gas(mol), R: universal gas constant, T: Kelvin temperature of the gas (K), k: Boltzmann’s constant, N: number of molecules in the gas.
## Boyle's Law For Pressure
If the temperature of the gas is constant
## Gay-Lussac's Law Of Pressure-Temperature
If tha volume on a gas is constant
P: pressure of gas at temperature t, po; pressure at 0 celsius, γ: factor when volume is constant
Example when t=0 kelvin
## Charle's Law Of Volumes
If the pressure on a gas is constant
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2021-05-06 22:58:43
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https://math.stackexchange.com/questions/945066/divergence-of-the-serie-sum-fracnnn-frac1en
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# Divergence of the serie $\sum \frac{n^n}{n!}(\frac{1}{e})^n$
Show that the serie
$$\sum \frac{n^n}{n!} \big(\frac{1}{e}\big)^n$$
Diverges.
The ratio test is inconclusive and the limit of the term is zero. So I think we should use the comparasion test. But I couldnt find any function to use, I've tried the harmonic ones, but doesnt work, since I cant calculate the limits. My guess is that we shpuld use $\frac{1}{n}$ Any hint?
Thanks!
• If you know Stirling's approximation this should be straightforward; just pick a form that gives you an upper bound on $n!$. – Steven Stadnicki Sep 25 '14 at 1:26
The trick here is to use Stirling's Approximation for $n!$ to provide a lower bound for the series. In particular, since $n!\lt e\sqrt{n}\left(\dfrac ne\right)^n$, then $\dfrac {n^n}{n!}\left(\dfrac1e\right)^n\gt \dfrac1{e\sqrt n}$. Can you do the rest given this?
The series $\sum_{n=1}^{\infty}1/\sqrt{n}$ diverges (by comparison with the harmonic series).
Stirling's asymptotic formula is,
$$n! \sim \sqrt{2\pi n}\frac{n^n}{e^n}.$$
Hence
$$\lim_{n \rightarrow \infty}\frac{\frac{n^n}{e^nn!}}{\frac1{\sqrt{2\pi n}}} = 1.$$
By the limit comparison test the series $\sum_{n=1}^{\infty}\frac{n^n}{e^nn!}$ diverges.
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2019-09-21 05:00:56
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http://www.msri.org/seminars/17136
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# Mathematical Sciences Research Institute
Home » Differential ideals of symmetric polynomials spanned by Jack polynomials at negative rationals
# Seminar
Differential ideals of symmetric polynomials spanned by Jack polynomials at negative rationals February 07, 2002
Parent Program: -- MSRI: Simons Auditorium
Speaker(s) Tetsuji Miwa
Description No Description
Video
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2017-11-23 22:35:53
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https://www.hcbravo.org/IntroDataSci/bookdown-notes/eda-data-transformations.html
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# 22 EDA: Data Transformations
Having a sense of how data is distributed, both from using visual or quantitative summaries, we can consider transformations of variables to ease both interpretation of data analyses and the application statistical and machine learning models to a dataset.
## 22.1 Centering and scaling
A very common and important transformation is to scale data to a common unit-less scale. Informally, you can think of this as transforming variables from whatever units they are measured (e.g., diamond depth percentage) into “standard deviations away from the mean” units (actually called standard units, or $$z$$-score). Given data $$x = x_1, x_2, \ldots, x_n$$, the transformation applied to obtain centered and scaled variable $$z$$ is:
$z_i = \frac{(x_i - \overline{x})}{\mathrm{sd}(x)}$
where $$\overline{x}$$ is the mean of data $$x$$, and $$\mathrm{sd}(x)$$ is its standard deviation.
library(ggplot2)
data(diamonds)
diamonds %>%
mutate(scaled_depth = (depth - mean(depth)) / sd(depth)) %>%
ggplot(aes(x=scaled_depth)) +
geom_histogram(binwidth=.5)
Question: what is the mean of $$z$$? What is it’s standard deviation? Another name for this transformation is to standardize a variable.
One useful result of applying this transformation to variables in a dataset is that all variables are in the same, and thus comparable units.
On occasion, you will have use to apply transformations that only center (but not scale) data:
$z_i = (x_i - \overline{x})$
Question: what is the mean of $$z$$ in this case? What is it’s standard deviation?
Or, apply transformations that only scale (but not center) data:
$z_i = \frac{x_i}{\mathrm{sd}(x)}$
Question: what is the mean of $$z$$ in this case? What is it’s standard deviation?
## 22.2 Treating categorical variables as numeric
Many modeling algorithms work strictly on numeric measurements. For example, we will see methods to predict some variable given values for other variables such as linear regression or support vector machines, that are strictly defined for numeric measurements. In this case, we would need to transform categorical variables into something that we can treat as numeric. We will see more of this in later sections of the course but let’s see a couple of important guidelines for binary variables (categorical variables that only take two values, e.g., health_insurance).
One option is to encode one value of the variable as 1 and the other as 0. For instance:
library(ISLR)
library(tidyverse)
data(Wage)
Wage %>%
mutate(numeric_insurace = ifelse(health_ins == "1. Yes", 1, 0)) %>%
head()
## year age maritl race
## 1 2006 18 1. Never Married 1. White
## 2 2004 24 1. Never Married 1. White
## 3 2003 45 2. Married 1. White
## 4 2003 43 2. Married 3. Asian
## 5 2005 50 4. Divorced 1. White
## 6 2008 54 2. Married 1. White
## education region
## 1 1. < HS Grad 2. Middle Atlantic
## 2 4. College Grad 2. Middle Atlantic
## 3 3. Some College 2. Middle Atlantic
## 4 4. College Grad 2. Middle Atlantic
## 5 2. HS Grad 2. Middle Atlantic
## 6 4. College Grad 2. Middle Atlantic
## jobclass health health_ins
## 1 1. Industrial 1. <=Good 2. No
## 2 2. Information 2. >=Very Good 2. No
## 3 1. Industrial 1. <=Good 1. Yes
## 4 2. Information 2. >=Very Good 1. Yes
## 5 2. Information 1. <=Good 1. Yes
## 6 2. Information 2. >=Very Good 1. Yes
## logwage wage numeric_insurace
## 1 4.318063 75.04315 0
## 2 4.255273 70.47602 0
## 3 4.875061 130.98218 1
## 4 5.041393 154.68529 1
## 5 4.318063 75.04315 1
## 6 4.845098 127.11574 1
Another option is to encode one value as 1 and the other as -1:
Wage %>%
mutate(numeric_insurance = ifelse(health_ins == "1. Yes", 1, -1)) %>%
head()
## year age maritl race
## 1 2006 18 1. Never Married 1. White
## 2 2004 24 1. Never Married 1. White
## 3 2003 45 2. Married 1. White
## 4 2003 43 2. Married 3. Asian
## 5 2005 50 4. Divorced 1. White
## 6 2008 54 2. Married 1. White
## education region
## 1 1. < HS Grad 2. Middle Atlantic
## 2 4. College Grad 2. Middle Atlantic
## 3 3. Some College 2. Middle Atlantic
## 4 4. College Grad 2. Middle Atlantic
## 5 2. HS Grad 2. Middle Atlantic
## 6 4. College Grad 2. Middle Atlantic
## jobclass health health_ins
## 1 1. Industrial 1. <=Good 2. No
## 2 2. Information 2. >=Very Good 2. No
## 3 1. Industrial 1. <=Good 1. Yes
## 4 2. Information 2. >=Very Good 1. Yes
## 5 2. Information 1. <=Good 1. Yes
## 6 2. Information 2. >=Very Good 1. Yes
## logwage wage numeric_insurance
## 1 4.318063 75.04315 -1
## 2 4.255273 70.47602 -1
## 3 4.875061 130.98218 1
## 4 5.041393 154.68529 1
## 5 4.318063 75.04315 1
## 6 4.845098 127.11574 1
The decision of which of these two transformations to use is based on the method to use or the goal of your analysis. For instance, when predicting someone’s wage based on their health insurance status, the 0/1 encoding let’s us make statements like: “on average, wage increases by \$XX if a person has health insurance”. On the other hand, a prediction algorithm called a Support Vector Machine is strictly defined on data coded as 1/-1.
For categorical attributes with more than two values, we extend this idea and encode each value of the categorical variable as a 0/1 column. You will see this referred to as one-hot-encoding.
Wage %>%
mutate(race_white = ifelse(race == "1. White", 1, 0),
race_black = ifelse(race == "2. Black", 1, 0),
race_asian = ifelse(race == "3. Asian", 1, 0),
race_other = ifelse(race == "4. Other", 1, 0)) %>%
select(starts_with("race")) %>%
head()
## race race_white race_black race_asian
## 1 1. White 1 0 0
## 2 1. White 1 0 0
## 3 1. White 1 0 0
## 4 3. Asian 0 0 1
## 5 1. White 1 0 0
## 6 1. White 1 0 0
## race_other
## 1 0
## 2 0
## 3 0
## 4 0
## 5 0
## 6 0
The builtin function model.matrix does this general transformation. We will see it when we look at statistical and Machine Learning models.
### 22.2.1 Discretizing continuous values.
How about transforming data in the other direction, from continuous to discrete values. This can make it easier to compare differences related to continuous measurements: Do doctors prescribe a certain medication to older kids more often? Is there a difference in wage based on age?
It is also a useful way of capturing non-linear relationships in data: we will see this in our regression and prediction unit. Two standard methods used for discretization are to use equal-length bins, where variable range is divided into bins regardless of the data distribution:
flights %>%
mutate(dep_delay_discrete = cut(dep_delay, breaks=100)) %>%
ggplot(aes(x=dep_delay_discrete)) +
geom_bar()
The second approach uses equal-sized bins, where the range is divided into bins based on data distribution
flights %>%
mutate(dep_delay_discrete = cut(dep_delay,
breaks=quantile(dep_delay, probs=seq(0,1,len=11), na.rm=TRUE))) %>%
ggplot(aes(x=dep_delay_discrete)) +
geom_bar()
In both cases, the cut function is used to apply discretization, with the breaks argument determining which method is applied. In the first example, breaks=100 specifies that 100 bins of equal-length are to be used. In the second example, the quantile function is used to define 10 equal-sized bins.
## 22.3 Skewed Data
In many data analysis, variables will have a skewed distribution over their range. In the last section we saw one way of defining skew using quartiles and median. Variables with skewed distributions can be hard to incorporate into some modeling procedures, especially in the presence of other variables that are not skewed. In this case, applying a transformation to reduce skew will improve performance of models.
Also, skewed data may arise when measuring multiplicative processes. This is very common in physical or biochemical processes. In this case, interpretation of data may be more intiuitive after a transformation.
We have seen an example of skewed data previously when we looked at departure delays in our flights dataset.
flights %>% ggplot(aes(x=dep_delay)) + geom_histogram(binwidth=30)
## Warning: Removed 8255 rows containing non-finite values
## (stat_bin).
Previously, we looked at a way of determining skew for a dataset. Let’s see what that looks like for the dep_delay variable: (see dplyr vignette for info on ‘enquo’ and ‘!!’)
compute_skew_stat <- function(df, attribute) {
attribute <- enquo(attribute)
df %>%
summarize(med_attr=median(!!attribute, na.rm=TRUE),
q1_attr=quantile(!!attribute, 1/4, na.rm=TRUE),
q3_attr=quantile(!!attribute, 3/4, na.rm=TRUE)) %>%
mutate(d1 = med_attr - q1_attr, d2 = q3_attr - med_attr, skew_stat = d1 - d2) %>%
select(d1, d2, skew_stat)
}
flights %>% compute_skew_stat(dep_delay)
## # A tibble: 1 x 3
## d1 d2 skew_stat
## <dbl> <dbl> <dbl>
## 1 3 13 -10
In many cases a logarithmic transform is an appropriate transformation to reduce data skew:
• If values are all positive: apply log2 transform
• If some values are negative, two options
• Started Log: shift all values so they are positive, apply log2
• Signed Log: $$sign(x) \times log2(abs(x) + 1)$$.
Here is a signed log transformation of departure delay data:
transformed_flights <- flights %>%
mutate(transformed_dep_delay = sign(dep_delay) * log2(abs(dep_delay) + 1))
transformed_flights %>%
ggplot(aes(x=transformed_dep_delay)) +
geom_histogram(binwidth=1)
## Warning: Removed 8255 rows containing non-finite values
## (stat_bin).
Let’s see if that reduced the skew of the dataset:
transformed_flights %>% compute_skew_stat(transformed_dep_delay)
## # A tibble: 1 x 3
## d1 d2 skew_stat
## <dbl> <dbl> <dbl>
## 1 1 5.17 -4.17
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2022-12-03 16:48:11
|
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https://math.stackexchange.com/questions/609732/jointly-gaussian-uncorrelated-random-variables-are-independent
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# Jointly Gaussian uncorrelated random variables are independent [closed]
Let $X,Y$ be jointly normally distributed and uncorrelated. Why are they independent?
## closed as off-topic by user1337, Shuchang, Eric Naslund, Did, user66733Dec 17 '13 at 1:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user1337, Shuchang, Eric Naslund, Did, user66733
If this question can be reworded to fit the rules in the help center, please edit the question.
In short, they are independent because the bivariate normal density, in case they are uncorrelated, i.e. $\rho =0$, reduces to a product of two normal densities the support of each one ranges from $(-\infty, \infty)$. If the joint distribution can be written as a product of nonnegative functions, we know that the RVs are independent. Moreover, we know, and can show, that each marginal density is normal on its own.
That is easy to see in the bivariate density below:
$$f(x,y)= \frac{1}{2 \pi \sigma_1 \sigma_2 \left( 1-\rho^2 \right)^{1/2}} \exp\{-q/2 \}, \quad -\infty<x<\infty,\quad -\infty<y<\infty$$
where $$q= \frac{1}{1-\rho^2} \left[ \left( \frac{x-\mu_1}{\sigma_1} \right)^2-2\rho \left(\frac{x-\mu_1}{\sigma_1} \right) \left(\frac{y-\mu_2}{\sigma_2} \right)+\left(\frac{y-\mu_2}{\sigma_2} \right)^2 \right]$$
Put $\rho=0$. There is also a nice proof involving mfgs. Is that what you were looking for?
• Thank you! I knew this formula for the joint density, but forgot that the $\rho$ is actually the correlation coefficient. It makes perfect sense now! – Balerion_the_black Dec 17 '13 at 14:22
Consider a simple example. Let X have a standard normal distribution. Let Z be independent of X, with Z equally likely to be +1 or -1 (i.e., Pr[Z=+1] = Pr[Z=-1] = 1/2). Let Y = X Z (the product of X and Z). Then it is easy to see that Y also has a standard normal distribution, and that Cov(X,Y) = 0. On the other hand, clearly X and Y are not independent: indeed, it always holds that |X| = |Y|. The point is that, just because each of X and Y has a normal distribution, that does not mean that the pair (X,Y) has a bivariate normal distribution, nor even that (X,Y) is jointly absolutely continuous, nor does it mean that zero covariance implies independence.
• I said "jointly normally distributed"! Under this assumption the claim is true. – Balerion_the_black Dec 16 '13 at 23:50
• What you are saying is nice to know, but Balerion is right. Action is needed here. If you do not want to delete your answer (as I said, it is nice to know) then at least edit and make clear in it that it is not really an answer to the question of the OP. – drhab Dec 17 '13 at 10:15
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2019-06-26 11:58:25
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https://economics.stackexchange.com/questions/8569/shadow-price-vs-marginal-cost
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# Shadow Price vs Marginal Cost
Is there a difference between the two? If so can you intuitively explain it? I see internet resources saying these are equivalent, but I thought I remember someone telling me once shadow price was different than marginal cost, but I may be misremembering it. Bonus points if you can place the definition in the context of a non-equilibrium setting.
## 1 Answer
Marginal cost is always the same as the shadow price in the cost minimization problem \begin{eqnarray*} \min_x && w \cdot x \\ s.t. && f(x) = y. \end{eqnarray*} In optimum the shadow price (Lagrange multiplier) belonging to the condition $f(x) = y$ is the marginal cost.
However there are other optimization problems where the shadow price is something else, for example in the utility maximization problem
\begin{eqnarray*} \max_{x,y} && U(x,y) \\ s.t. && p_x \cdot x + p_y \cdot y = m \end{eqnarray*}
the shadow prices correspond to the marginal utility of money. (One could argue that this is the reciprocal of the marginal cost of producing more utility.)
Neither of these examples had anything to do with equilibrium, just individual optima.
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2022-08-17 16:25:06
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http://clay6.com/qa/17926/the-line-among-the-following-which-touches-the-parabola-y-2-4ax-is-
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Browse Questions
# The line among the following which touches the parabola $y^2=4ax,$ is :
$\begin {array} {1 1} (a)\;x+my+am^3=0 & \quad (b)\;x-my+am^2=0 \\ (c)\;x+my-am^2=0 & \quad (d)\;x+mx+am^2=0 \end {array}$
$(b)\;x-my+am^2=0$
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2016-10-24 07:08:44
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http://math.stackexchange.com/questions/145514/an-integral-involving-trigonometric-functions-and-its-inverse
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# An integral involving trigonometric functions and its inverse
I have tried to evaluate the following integral for the last few hours, and I did not succeed:
$$\int\limits_{0}^{2 \pi} e^{\mathrm{i} \cdot n \cdot\mathrm{arcsin}(r \cdot\mathrm{sin}(\theta))} \cdot e^{\mathrm{i}\cdot m \cdot \mathrm{arcsin}(r \cdot \mathrm{cos}(\theta))} d \; \theta$$
for $0<r<1$. And also this other integral:
$$\int\limits_{0}^{2 \pi} e^{\mathrm{i} \cdot n \cdot\mathrm{arctan}(t \cdot\mathrm{sin}(\theta))} \cdot e^{\mathrm{i}\cdot m \cdot \mathrm{arctan}(t \cdot \mathrm{cos}(\theta))} d \; \theta.$$
Here $m$ and $n$ are integers, and $t \in \mathbb{R}$ is scalar.
I am pretty sure that is nonzero, if and only if $n=m$, and indepedent of $r$ otherwise, but I cannot figure what substitution makes this easy to see.
-
Seems unusual, since $\arcsin(r\sin\theta)$ is often undefined. – André Nicolas May 15 '12 at 16:07
In order to inverse trigonometric functions to be real for all $\theta$ in the range of integration, you should further require $0< r \leqslant 1$. – Sasha May 15 '12 at 16:08
It is bad to change the question this way, without leaving information about what it was before, since now my answer appears irrelevant to your question, and the time I put into answering goes wasted. Please do not do this. – Sasha May 15 '12 at 16:40
Dear Sasha, I did not expect an answer so fast. I posted back the original integral. – plusepsilon.de May 15 '12 at 16:47
Please, think over the question before asking it - at this moment the whole thing has changed two times! – AD. May 15 '12 at 16:47
This addresses the question in its original form, where $\arcsin$ was used.
First, let's massage the integral: $$\begin{eqnarray} \mathcal{I} &=& \int_0^{2 \pi} \exp\left( i n \arcsin(r \sin(\theta)) + i m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta \\ &\stackrel{\theta \to 2\pi - \theta}{=}& \int_0^{2 \pi} \exp\left( -i n \arcsin(r \sin(\theta)) + i m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta \end{eqnarray}$$ Now averaging out both lines: $$\begin{eqnarray} \mathcal{I} &=& \int_0^{2 \pi} \cos\left(n \arcsin(r \sin(\theta)) \right)\cdot \exp\left( i m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta \\ &=& 2 \int_0^{\pi} \cos\left(n \arcsin(r \sin(\theta)) \right)\cdot \cos\left( m \arcsin(r \cos(\theta)) \right) \mathrm{d} \theta \end{eqnarray}$$
Here is a counter-example to your claim. Let $r = \frac{1}{2}$, and $n=1$, $m=2$. Then the integrand is positive, and hence the integral does not vanish:
Added The above counterexample actually carries over to the case with $\arctan$ just the same, i.e. the integrand is positive.
@late_learner $\int_0^{2\pi} = \int_0^{\pi} + \int_{\pi}^{2\pi}$. In the second integral change variables, $\theta = \pi + \theta$, and use $\cos(\pi+\theta) = -\cos(\theta)$. Then combined two integrals into one, and simplify the integrand. – Sasha May 15 '12 at 18:22
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2015-07-08 00:24:15
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https://www.jiskha.com/users?name=riley
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riley
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2020-02-25 11:24:20
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https://www.gradesaver.com/textbooks/math/algebra/introductory-algebra-for-college-students-7th-edition/appendix-mean-median-and-mode-exercise-set-page-692/25
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## Introductory Algebra for College Students (7th Edition)
To find the mean, sum all the ages and divide the sum by the number of ages given. $mean=\frac{42+43+46+46+47+47}{6}=\frac{271}{6}=45.2$ To find the median, list the ages in order and find the age that appears at the midpoint of the list. If the number of ages given is even, find the mean of the two ages at the midpoint of the list. $42, 43, \underset\uparrow46, \underset\uparrow46, 47, 47$ $median=\frac{46+46}{2}=\frac{92}{2}=46$ To find the mode, identify the age that appears most often in the list. If more than one value appears most often, all are the mode. 46 and 47 are the only ages repeated and each is repeated twice, so both are the mode.
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2019-12-14 03:15:20
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http://encyclopedia.kids.net.au/page/tr/Transpose
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Encyclopedia > Transpose
Article Content
Transpose
See also Transposition for meanings of this term in telecommunication and music.
In mathematics, and in particular linear algebra, the transpose of a matrix is another matrix, produced by turning rows into columns and vice versa. Informally, the transpose of a square matrix is obtained by reflecting at the main diagonal (that runs from the top left to bottom right of the matrix). The transpose of the matrix A is written as Atr, tA, or AT, the latter notation being preferred in Wikipedia.
Formally, the transpose of the m-by-n matrix A is the n-by-m matrix AT defined by AT[i, j] = A[j, i] for 1 ≤ in and 1 ≤ jm.
For example,
$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}^T \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}\quad\quad \mbox{and}\quad\quad \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}^T \begin{bmatrix} 1 & 3 & 5\\ 2 & 4 & 6 \end{bmatrix}$
Properties For any two m-by-n matrices A and B and every scalar c, we have (A + B)T = AT + BT and (cA)T = c(AT). This shows that the transpose is a linear map from the space of all m-by-n matrices to the space of all n-by-m matrices.
The transpose operation is self-inverse, i.e taking the transpose of the transpose amounts to doing nothing: (AT)T = A.
If A is an m-by-n and B an n-by-k matrix, then we have (AB)T = (BT)(AT). Note that the order of the factors switches. From this one can deduce that a square matrix A is invertible if and only if AT is invertible, and in this case we have (A-1)T = (AT)-1.
The dot product of two vectors expressed as columns of their coordinates can be computed as
a · b = aT b
where the product on the right is the ordinary matrix multiplication.
If A is an arbitrary m-by-n matrix with real entries, then ATA is a positive semidefinite matrix.
Further nomenclature A square matrix whose transpose is equal to itself is called a symmetric matrix, i.e. A is symmetric iff
A = AT
A square matrix whose transpose is also its inverse is called an orthogonal matrix, i.e. G is orthogonal iff:
G GT = GT G = In
A square matrix whose transpose is equal to its negative is called skew-symmetric, i.e. A is skew-symmetric iff
A = - AT
The conjugate transpose of the complex matrix A, written as A*, is obtained by taking the transpose of A and then taking the complex conjugate of each entry.
Transpose of linear maps If f: V -> W is a linear map between vector spaces V and W with dual spaces W* and V*, we define the transpose of f to be the linear map tf : W* -> V* with
tf (φ) = φ o f for every φ in W*.
If the matrix A describes a linear map with respect to two bases, then the matrix AT describes the transpose of that linear map with respect to the dual bases. See dual space for more details on this.
All Wikipedia text is available under the terms of the GNU Free Documentation License
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2020-10-20 12:24:29
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https://math.stackexchange.com/questions/1334443/showing-that-the-product-xy-fracxyxy1-is-a-group-operation-on-1
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Showing that the product $x*y := \frac{x+y}{xy+1}$ is a group operation on $(-1, 1)$ [duplicate]
I need to show that the following is an abelian group:
$$x*y = \frac{x+y}{xy+1}$$ on the set $\{x \in \Bbb R \,|\, -1 < x < 1\}$.
I have been working on this problem, trying to show closure. I know that we need to show that $|x+y|<|xy+1|$ for all $x, y \in (-1,1)$. Can I assume that the max value that the expression yields is $1$ if we take $x=1$ and $y=1$? And the lowest value that is possible is when $x=-1$ and $y=1$? Or am I going about this the wrong way?
marked as duplicate by André 3000, 6005, Mark Bennet, user147263, Adam HughesJun 22 '15 at 19:11
Hint One can show directly that the operation $\ast$ is conjugate to the usual addition operation on $\Bbb R$ via the hyperbolic tangent function, which satisfies the suggestive identity $$\tanh (s + t) = \frac{\tanh s + \tanh t}{\tanh s \tanh t + 1}.$$
Alternate hint One can rearrange the desired inequality $x + y < xy + 1$ (note we've dropped the absolute value signs) as $$0 < xy - x - y + 1 = (1 - x)(1 - y).$$
• Sure: Since $|x|, |y| < 1$, we have $|xy| < 1$ and so $|xy + 1| = xy + 1$. Hence, when $x + y \geq 0$ we have $|x + y| < |xy + 1|$ as desired. To handle the case $x + y < 0$, we can replace $x$ and $y$ respectively with $-x$ and $-y$. – Travis Jun 22 '15 at 8:38
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2019-08-26 06:22:57
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http://mathhelpforum.com/calculus/78675-areas-between-curves.html
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# Thread: areas between curves
1. ## areas between curves
there is a line through the origin that divides the region bounded by the parabola y= 8x-2x^2 and the x-axis into two regions of equal area. What is the slope of the line?
I can't figure it out. I don't know how to start
2. Originally Posted by zaddie21
there is a line through the origin that divides the region bounded by the parabola y= 8x-2x^2 and the x-axis into two regions of equal area. What is the slope of the line?
I can't figure it out. I don't know how to start
calculate the area in quad I bounded by the parabola and the x-axis, call that area $A$.
let the line be $y = mx$
find the intersection of the line and parabola in quad I ...
$mx = 8x - 2x^2$
$0 = (8-m)x - 2x^2$
$0 = x(8 - m - 2x)$
x-values of intersection are $x = 0$ and $x = \frac{8-m}{2}$
$\frac{A}{2} = \int_0^{\frac{8-m}{2}} (8x-2x^2) - mx \, dx$
evaluate the integral and solve for $m$ ... have fun with the algebra.
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2016-08-25 07:26:37
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https://www.nature.com/articles/s41467-017-01377-8?error=cookies_not_supported&code=250c43a4-2709-40f2-8761-429bacc493e0
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Article | Open | Published:
# The role of Atlantic overturning circulation in the recent decline of Atlantic major hurricane frequency
## Abstract
Observed Atlantic major hurricane frequency has exhibited pronounced multidecadal variability since the 1940s. However, the cause of this variability is debated. Using observations and a coupled earth system model (GFDL-ESM2G), here we show that the decline of the Atlantic major hurricane frequency during 2005–2015 is associated with a weakening of the Atlantic Meridional Overturning Circulation (AMOC) inferred from ocean observations. Directly observed North Atlantic sulfate aerosol optical depth has not increased (but shows a modest decline) over this period, suggesting the decline of the Atlantic major hurricane frequency during 2005–2015 is not likely due to recent changes in anthropogenic sulfate aerosols. Instead, we find coherent multidecadal variations involving the inferred AMOC and Atlantic major hurricane frequency, along with indices of Atlantic Multidecadal Variability and inverted vertical wind shear. Our results provide evidence for an important role of the AMOC in the recent decline of Atlantic major hurricane frequency.
## Introduction
Atlantic major hurricane activity has caused substantial damage to coastal settlements and infrastructure1. Observed Atlantic major hurricane frequency exhibits prominent multidecadal variability2,3,4,5, with high values in the 1950s, a rapid decline in the 1960’s and an abrupt increase in the 1990s. The most recent above-normal period of activity peaked around 2005. There were a remarkable seven Atlantic major hurricanes in 2005, including Hurricane Katrina (category 5) with losses of more than \$100 billion. Observed multidecadal variability in Atlantic hurricane frequency was found to be well-correlated with low-pass filtered detrended area-averaged North Atlantic sea surface temperature (SST) anomalies, i.e., the so-called Atlantic multidecadal variability (AMV) Index6, 7, in previous studies4, 8,9,10,11. However, there is currently no consensus on the mechanism causing the AMV and the linkage between the AMV and the Atlantic meridional overturning circulation (AMOC) variability12,13,14,15,16,17. Changes in anthropogenic sulfate aerosol forcing has been proposed as the dominant cause of historical multidecadal variations in Atlantic tropical storm frequency based on some Coupled Model Intercomparison Project Phase 5 (CMIP5) model simulations including aerosol indirect effects. In these simulations, a decrease in anthropogenic sulfate aerosol forcing leads to an increase in tropical storms and vice versa12, 18. On the other hand, there are large uncertainties in simulated aerosol indirect effects and substantial mean state biases in many CMIP5 model simulations19, 20.
In this study, we explore the causes of the recent decline of Atlantic major hurricane frequency over the period 2005–2015, using various observational datasets and modeling results from a 500-year control simulation of a fully coupled earth system model (GFDL-ESM2G). GFDL-ESM2G is a unique CMIP5 model in that its ocean component employs an isopycnal-coordinate21, in contrast to most CMIP5 models, which use z-coordinate ocean components (“Methods” section). The AMOC structure (especially the depth of AMOC) and the deep overflow in the North Atlantic mean state are far more realistic in isopycnal-coordinate models (such as GFDL-ESM2G) with realistic mixing than in z-coordinate models, which have excessive artificial numerical mixing in the deep ocean22. The improved AMOC structure and deep overflow in the North Atlantic can greatly reduce the systematic large cold SST bias in the extra-tropical North Atlantic21, 23, which is found in most CMIP5 models and often leads to a biased mean state environment for tropical cyclones20, 24, 25. Our results suggest an important role of the recent AMOC weakening in the decline of Atlantic major hurricane frequency since 2005. Therefore monitoring and predicting AMOC changes will be important to reduce future Atlantic hurricane risk.
## Results
### Observed coherent relationships
Figure 1a, b shows contrasting Atlantic major hurricane track density maps from near the peak of the above-normal activity period (2001–2005) and during the less-active recent 5-year period (2011–2015). There is a decline in the Atlantic major hurricane frequency over the period 2005–2015, similar to the rapid decline in the 1960s (Figs 1c and 2a). A similar decline over the period 2005–2015 is also present in the Atlantic hurricane frequency series (Supplementary Fig. 1). This decline is likely related to the reduction of US landfalling hurricanes over 2005–2015. Here we use both observations and model simulations to further explore the causes of the 2005–2015 decline. In particular, we focus on Atlantic major hurricane frequency due to its high societal impact (e.g., Atlantic major hurricanes over the period 1900–2005 are estimated to have contributed ~85% of the total mainland US hurricane damage1).
Figure 2a indicates that the observed low-pass filtered Atlantic major hurricane frequency co-varies statistically with the observed AMV Index over the entire period of 1945–2015. The correlation between them is 0.91, which is statistically significant at the 0.05 level ($$p < 0.05$$). In this study, the AMV Index is defined as the low-pass filtered area-averaged detrended SST anomalies over the mid-high latitude North Atlantic. This averaging region is a key region for the AMV15, 26, and the observed SST signal associated with the AMV is strongest there7, 26. Since there are no continuous direct observations of historical AMOC variations before 200427, we use a previously identified AMOC fingerprint as a proxy for AMOC variations28. The fingerprint, defined as the leading mode of the observed detrended subsurface ocean temperature anomalies at 400 m in the extra-tropical North Atlantic28, exhibits a dipole spatial pattern with opposite signed changes in the subpolar gyre and Gulf Stream regions (Supplementary Fig. 2). The AMOC fingerprint is shifted backward in time by 4 years to better represent the AMOC anomalies at mid-high latitudes. The AMOC anomalies at mid-high latitudes lead the fingerprint by ~4 years due to their slow southward propagation29, 30. Fig. 2a shows that on multidecadal time scales, the observed AMOC proxy co-varies with both the observed AMV Index and the observed Atlantic major hurricane frequency. The correlation between the AMOC fingerprint and the AMV Index (Atlantic major hurricane frequency) is 0.90(0.92), with $$p < 0.05$$. In particular, we infer that the decline in the observed AMV Index and Atlantic major hurricane frequency over 2005–2015 are accompanied by an AMOC weakening, as indicated by the observed temporal evolution of the AMOC fingerprint (Fig. 2a) as well as other direct and indirect observations27, 31, 32.
The directly observed AMOC Index at 26°N measured from the RAPID Array27, 33 (“Methods” section) also exhibits a pronounced decline27, 31, 32, 34 that is roughly coincident with the decline in Atlantic major hurricane frequency over the most recent decade (Fig. 2a). The recent observed AMOC decline has been inferred to be part of a decadal/multidecadal AMOC signal34, according to a high-resolution data assimilation product. The simulated AMOC increase during the mid 1990s and the AMOC decline over the recent decade from the data assimilation product34, as well as the directly observed AMOC decline from the RAPID array data since 2004, are all consistent with our inferences using the AMOC fingerprint and further support the capability of this fingerprint as a proxy for AMOC variations. Despite the evidence cited above from observations and modeling studies28, 30, we note that there is inherently less certainty regarding this fingerprint due to the lack of direct observations before 2004.
Previous studies4, 8, 35 show that the vertical wind shear over the Atlantic hurricane main development region (MDR, defined here as 80°W–20°W, 10°N–20°N) is strongly correlated with Atlantic major hurricane frequency. Strong vertical wind shear affects storm activity by disrupting the axisymmetric organization of deep convection and hence inhibiting the formation and intensification of Atlantic hurricanes36. Figure 2a shows that the observed MDR Hurricane Shear Index (“Methods” section) is strongly anti-correlated with the observed Atlantic major hurricane frequency, AMV Index, and AMOC fingerprint on multidecadal time scales, including the recent declines in these indices. The correlations of the Hurricane Shear Index against the observed Atlantic major hurricane frequency, AMV Index, and AMOC fingerprint are −0.88, −0.85 and −0.92 (all with $$p < 0.05$$). Decadal shifts similar to the recent decline of major hurricane frequency and the other indices are also seen in the 1960s, along with up swings during the 1990s. A previous study showed that the AMV associated with AMOC variability could induce multidecadal variations in the vertical wind shear over the Atlantic hurricane MDR similar to that observed8. Hence the vertical wind shear appears to be a key process by which the AMV/AMOC can modulate Atlantic hurricane activity4, 8. At multidecadal time scales over the MDR region, other environmental factors we analyzed from the reanalysis data (e.g., 850 hPa relative vorticity, 700 hPa relative humidity, and 500 hPa vertical velocity) do not show significant correlations with the multidecadal variations of Atlantic major hurricane frequency (Supplementary Fig. 3), even though these factors are important for individual cyclone genesis and intensification37, 38. The observed coherent multidecadal variations between the inverted Hurricane Shear Index and Atlantic hurricane/major hurricane frequency suggest that the inverted Hurricane Shear Index can be used as a proxy for the multidecadal variations of Atlantic hurricane/major hurricane frequency.
The coherent variation during the period 2005–2015 of Atlantic major hurricane frequency, AMV Index, AMOC Index/fingerprint, and inverted Hurricane Shear Index provides new evidence from several independent observational sources for an important role for AMOC in the observed recent decline of the Atlantic major hurricane frequency. Atlantic hurricane frequency exhibits similar coherent multidecadal variability to that of Atlantic major hurricane frequency (Supplementary Fig. 1c and Supplementary Fig. 4). These variations in category 1–5 hurricane frequency are also statistically coherent with the AMV Index, AMOC fingerprint, and inverted Hurricane Shear Index (Supplementary Fig. 4).
On multidecadal time scales, local MDR SST also exhibits similar variations to the Atlantic major hurricane frequency except that the MDR SST has a pronounced increasing trend over the entire period, in contrast to hurricane frequency, which has little trend. Major hurricane frequency is consistently related to relative SST5 (difference between MDR SST and tropical mean SST), as they are correlated while neither index has an obvious trend (Fig. 2b). Interestingly, at low frequency, relative SST lags behind major hurricane frequency by a few years (Fig. 2b and Supplementary Fig. 5). Although one must treat lagged correlations of low-pass filtered time series with caution39, this result suggests a possibly important role for the extra-tropical North Atlantic ocean in modulating Atlantic major hurricane frequency at the multidecadal time scale, perhaps by remote influence on vertical shear through an atmospheric bridge40, 41. This interpretation is also consistent with studies suggesting that tropical SST responds, with some lag, to extra-tropical North Atlantic oceanic changes at low frequency through oceanic and atmospheric teleconnections15, 26, 40. Further, initialization of the extra-tropical, but not tropical, North Atlantic ocean state, was shown to be crucial for multi-year predictability of Atlantic MDR vertical wind shear and tropical storm numbers41.
To test whether the observed relationship between the declining trend in the AMOC Index at 26°N and the increasing trend in the Hurricane Shear Index over the recent period 2005–2015 is a robust statistical relationship, we sampled all available 11-year segments from observations since 1957 to derive a regression map of 11-year trend of inverted vertical shear of zonal wind versus 11-year trend of reconstructed AMOC Index at 26°N (Fig. 3). The observed AMOC Index at 26°N is reconstructed from the AMOC fingerprint (“Methods" section). Figure 3 shows that the observed MDR vertical shear of zonal wind is robustly anti-correlated with the AMOC Index at 26°N. This provides additional evidence that the rapid increase of Hurricane Shear Index over the period 2005–2015 is statistically related to the AMOC decline.
### Model simulated coherent relationships
To further explain this relationship, we analyze results from the GFDL-ESM2G control simulation. In this simulation, the spatial pattern of the model’s AMOC fingerprint is similar to that observed, which includes a dipole pattern with opposite signed changes in the subpolar gyre and the Gulf Stream regions (Supplementary Fig. 6). The model’s AMOC fingerprint lags its AMOC Index at 42°N by ~4 years, with a maximum correlation of 0.93 ($$p < 0.01$$). Similarly, the model’s AMOC fingerprint lags its AMOC Index at 26°N by ~1 year, with a maximum correlation of 0.90 ($$p < 0.01$$). The simulated coherent multidecadal variations between the AMOC fingerprint and the AMOC Index at 26°N are consistent with the observed coherent decline in the AMOC fingerprint and the AMOC Index at 26°N over the recent decade (Fig. 2a) and consistent with a previous study28. This confirms, within the model climate, the capability of using the model’s AMOC fingerprint as a proxy for its AMOC variations, and further supports the use of this approach in observations.
In the GFDL-ESM2G control simulation, the AMV Index has a maximum correlation of 0.72 ($$p < 0.01$$) with the AMOC Index at 26°N when the AMOC Index leads by about 2 years. The simulated coherence among the AMV Index, low-pass filtered AMOC fingerprint and AMOC Index (Fig. 4) is consistent with the observed coherent decline in the AMV Index, AMOC fingerprint, and the RAPID AMOC Index at 26°N in the recent decade (Fig. 2a) and further suggests that ocean dynamics has played an important role in the AMV15, 17, 28, 32, 42,43,44,45,46,47,48,49. The simulated low-pass filtered inverted MDR Hurricane Shear Index is also significantly correlated with the AMV at zero lag (r = 0.48, $$p < 0.01$$), and with the low-pass filtered AMOC Index at 26°N when the AMOC Index leads by about 2 years (r = 0.59; $$p < 0.01$$) (Fig. 4). This provides additional model-based evidence that the recent increase in the Hurricane Shear Index and decline of Atlantic major hurricane frequency could be induced by the recent AMOC weakening (Fig. 2a).
Figure 5 shows the regression of 11-year trends of the model’s inverted vertical shear of zonal wind against 11-year trends of its AMOC Index at 26°N, based on all available 11-year segments sampled in the GFDL-ESM2G control simulation. Similar to the observed results (Fig. 3), the simulated MDR vertical shear of zonal wind increases with a decline in the AMOC Index at 26°N and vice versa. The simulated regression has an eastward shift in its maximum and a smaller amplitude than that observed (Figs 3 and 5), which we speculate is probably related to model mean state biases in the tropical Atlantic50. Another possible factor for this model deficiency is that the simulated AMOC Index/fingerprint has widely varying amplitude of multidecadal variations over the 500-year simulation (Fig. 4). For example, during the first 100 years and year 200–300, the simulated AMOC variability is relatively smaller, has shorter typical periods, and its connection with the Hurricane Shear Index is weaker, compared with other segments of the control simulation with more pronounced multidecadal AMOC variations (Fig. 4). This suggests that in the real world, the connection of Atlantic major hurricane frequency and AMOC might vary on centennial time scales and may depend on the amplitude and time scale of AMOC variability, although the observational record is too short to confirm this.
The results from the GFDL-ESM2G control simulation demonstrate that the observed statistical relationships among the AMOC Index, AMOC fingerprint, AMV Index, and MDR vertical shear of zonal wind at decadal/multidecadal time scales also exist in this coupled model (Figs 4 and 5). These modeling results of the teleconnections linked to AMOC changes (including the dynamical response of vertical shear of zonal wind over the MDR region regressed on the AMOC Index) provide more evidence and further support our hypothesis that AMOC variations are important for the recent observed decline in the Atlantic major hurricane frequency.
### Recent changes of anthropogenic aerosol optical depth
Anthropogenic aerosols have been proposed as a dominant causal mechanism for AMV12 and multidecadal Atlantic tropical cyclone variability18, with increasing (decreasing) aerosol loadings leading to decreased (increased) SST and tropical cyclone frequency. Recent observations indicate that anthropogenic aerosol emissions and optical depth over the North Atlantic region have decreased during the early 21st century51,52,53,54. Figure 6a shows the observed area-averaged anthropogenic sulfate aerosol optical depth (AOD) anomalies in the North Atlantic for the period 2005–2015. There is a slight declining trend in sulfate AOD over this period. A spatial map of sulfate AOD trend over this period shows widespread declining trends over most of the North Atlantic region (Fig. 6b). Under the proposed mechanisms12, 18, declining sulfate AOD would be expected to allow more shortwave radiation to reach the surface, leading to North Atlantic surface warming that favors tropical storm and hurricane formation. However, the observed decline in the AMV Index and the Atlantic major hurricane frequency during 2005–2015 suggest that changes in anthropogenic aerosols were not likely the primary driver of AMV and Atlantic major hurricane activity changes over this period.
## Discussion
Our analysis of various observations and a GFDL-ESM2G control simulation collectively suggest an important role of the AMOC in the recent decline of the Atlantic major hurricane frequency during 2005–2015. Directly observed North Atlantic sulfate aerosol optical depth has not increased (and instead has shown a modest decline) over this period. Thus the decline of Atlantic major hurricane frequency during 2005–2015 is not likely due to recent changes in anthropogenic sulfate aerosols. We expect, based on our findings, that changes in the Atlantic major hurricane frequency in the next decade would be closely linked to future AMOC changes. Besides, a basin-wide decline in the Atlantic hurricane frequency induced by AMOC weakening could be related to the rapid intensification of the US landfall hurricanes55. Consequently, we propose that monitoring and predicting AMOC changes will be crucial for developing a robust predictive understanding of future Atlantic hurricane risk.
## Methods
### Observational data
The observed Atlantic major hurricane frequency data were obtained from the Atlantic basin hurricane database (HURDAT), with no bias corrections applied in this case56, 57. The original unfiltered observed Atlantic major hurricane (hurricane) series is shown in Fig. 1c (Supplementary Fig. 1c). In Supplementary Fig. 1c, the observed Atlantic hurricane frequency during the period 1945–1965 is adjusted by a correction for estimated missing storms as inferred from ship track density during these years5. To construct the observed Atlantic hurricane/major hurricane track density maps (Fig. 1), the Best track data (HURDAT2)58 is interpolated to a temporal resolution of 1 h.
The HadISST1.1 data set59 is used for the calculation of the observed AMV index. In this study, the AMV Index is defined as the detrended (trend begins in 1900) annual mean area-averaged SST in the mid-high latitude North Atlantic (60°W–0°, 50°N–65°N). The time series of the ‘AMOC fingerprint’ is defined as the leading principal component (PC1) of detrended subsurface ocean temperature anomalies at 400 m in the extra-tropical North Atlantic (80°W–0°, 20°N–65°N)28. Using upper ocean heat content gives very similar results to 400 m temperature. The AMOC fingerprint is shifted 4-year backward in Figs 2 and 4 to represent AMOC anomalies at mid-high latitudes, which systematically lead the AMOC fingerprint by ~4 years due to a slow southward propagation of AMOC anomalies29, 30. The observed ‘AMOC fingerprint’ (Supplementary Fig. 2) is derived from an objectively analyzed data set of annual mean ocean temperature anomalies60. The AMOC Index at each latitude is defined as the maximum of annual mean Atlantic meridional overturning streamfunction at that latitude.
The directly observed AMOC Index at 26°N, available for the period 2004–2014, is obtained from the RAPID-WATCH MOC monitoring project27, 33. To maximize the number of available years in the short data record and be consistent with previous studies27, 33, each annual mean RAPID AMOC Index value is for a 12-month period starting on April 1 of the named year and ending on March 31 of the following year. The method of calculating the annual average does not affect the rapid decline trend in RAPID AMOC Index. The RAPID AMOC Index at 26°N during 2004–2014 and its linear trend are superimposed in Fig. 2a. The mean/trend of the observed AMOC Index at 26°N matches that of 1-year backward shifted AMOC fingerprint over their overlapped period of 2004–2014. The 1-year shift is based on the time lag between the AMOC Index at 26°N and AMOC fingerprint in the coupled earth system model (GFDL-ESM2G) control simulation (described below). The observed AMOC Index at 26°N used in Fig. 3 is reconstructed from the observed 1-year backward shifted AMOC fingerprint, and calibrated by the ratio between the trend of the RAPID AMOC Index at 26°N and the trend of the observed 1-year backward shifted AMOC fingerprint over their overlapped period of 2004–2014.
The ERA-Interim61 and the ERA-4062 atmospheric reanalysis data are used to calculate the observed anomalous 200–850 hPa vertical shear of the zonal wind during the Atlantic hurricane season of June through November (JJASON). The Hurricane Shear Index is defined as the area-averaged 200–850 hPa vertical shear of hurricane seasonal-mean (JJASON) zonal wind over the MDR (80°W–20°W, 10°N–20°N). The data from ERA-Interim61 (1979–2015) and ERA-4062 (1957–1978) are combined to calculate the observed Hurricane Shear Index for the full period. An alternative Hurricane Shear Index calculated from the NCEP/NCAR Reanalysis63 shows similar multidecadal variations over the same period (Supplementary Fig. 7). Note the pre-1960 vertical wind shear data are likely to have relatively larger uncertainties. The relative SST Index5 is defined as the difference between SST anomalies averaged over the MDR (MDR SST) and tropical (30°S-30°N) mean SST anomalies (TROP SST) during the Atlantic hurricane season (JJASON). The anomalies and trend pattern of sulfate AOD at 550 nm in North Atlantic are calculated from the monthly mean Modern Era Retrospective analysis for Research and Applications Aerosol Reanalysis (MERRAero)64.
### Model description
Model results are calculated from a 500-year pre-industrial control simulation of the Geophysics Fluid Dynamics Laboratory (GFDL) earth system model ESM2G21. The ocean component of GFDL-ESM2G uses the isopycnal-coordinate with 64 vertical layers, and 1° horizontal resolution with the meridional resolution refined to 1/3° at the equator. The atmospheric component of GFDL-ESM2G has 24 vertical levels, with horizontal resolution of 2° × 2.5°. The control simulation uses 1860 constant radiative forcing conditions.
### Statistical processing
All the low-pass filtered data are obtained using the R function “filtfilt”, with a Hamming window based low-pass filter and a frequency response that drops to 50% at a 10-year cutoff period. Padding points for low-pass filtering are derived by mirroring the data at each end (i.e., zero slope at endpoints). The observed anomalies are referenced to the corresponding climatological means of a common period of 1957–2005. The statistical significance for the correlations between a pair of low-pass filtered time series discussed in the paper are assessed using a Monte Carlo simulation based on a random phase method of resampling65.
### Code availability
The data involved in this study are analyzed by using the widely available analysis tools R and Ferret. Contact X.Y. or R.Z. for specific code requests.
### Data availability
Websites of the data used in this study:
Subsurface oceanic temperature anomalies: https://www.nodc.noaa.gov/OC5/3M_HEAT_CONTENT/
ERA-Interim, ERA-40 data: http://apps.ecmwf.int/datasets/
NCEP/NCAR Reanalysis: http://www.esrl.noaa.gov/psd/data/gridded/data.ncep.reanalysis.derived.surface.html
RAPID AMOC Index: www.rapid.ac.uk/rapidmoc
Adjusted Hurricane number in Vecchi and Knutson, 2011, J. Climate:
Publisher's note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## References
1. 1.
Pielke, R. A. et al. Normalized hurricane damage in the United States: 1900–2005. Nat. Hazards Rev. 9, 29–42 (2008).
2. 2.
Gray, W. M. Strong association between west African rainfall and U.S. landfall of intense hurricanes. Science 249, 1251–1256 (1990).
3. 3.
Landsea, C. W., Pielke, R. A. Jr, Mestas-Nuñez, A. M. & Knaff, J. A. Atlantic basin hurricanes: Indices of climatic changes. Clim. Change 42, 89–129 (1999).
4. 4.
Goldenberg, S. B., Landsea, C. W., Mestas-Nuñez, A. M. & Gray, W. M. The recent increase in Atlantic hurricane activity: causes and implications. Science 293, 474–479 (2001).
5. 5.
Vecchi, G. A. & Knutson, T. R. Estimating annual numbers of Atlantic hurricanes missing from the HURDAT database (1878–1965) using ship track density. J. Clim. 24, 1736–1746 (2011).
6. 6.
Enfield, D. B., Mestas-Nuñez, A. M. & Trimble, P. J. The Atlantic Multidecadal Oscillation and its relation to rainfall and river flows in the continental U.S. Geophys. Res. Lett. 28, 2077–2080 (2001).
7. 7.
Sutton, R. T. & Hodson, D. L. Atlantic Ocean forcing of North American and European summer climate. Science 309, 115–118 (2005).
8. 8.
Zhang, R. & Delworth, T. L. Impact of Atlantic multidecadal oscillations on India/Sahel rainfall and Atlantic hurricanes. Geophys. Res. Lett. 33, L17712 (2006).
9. 9.
Knight, J. R., Folland, C. K. & Scaife, A. A. Climate impacts of the Atlantic multidecadal oscillation. Geophys. Res. Lett. 33, L17706 (2006).
10. 10.
Klotzbach, P. J. & Gray, W. M. Multidecadal variability in north atlantic tropical cyclone activity. J. Clim. 21, 3929–3935 (2008).
11. 11.
Klotzbach, P., Gray, W. & Fogarty, C. Active Atlantic hurricane era at its end? Nat. Geosci. 8, 737–738 (2015).
12. 12.
Booth, B. B. B., Dunstone, N. J., Halloran, P. R., Andrews, R. & Bellouin, N. Aerosols implicated as a prime driver of twentieth-century North Atlantic climate variability. Nature 484, 228–232 (2012).
13. 13.
Clement, A. et al. The Atlantic multidecadal oscillation without a role for ocean circulation. Science 350, 320–324 (2015).
14. 14.
O’Reilly, C. H., Huber, M., Woollings, T. & Zanna, L. The signature of low frequency oceanic forcing in the Atlantic Multidecadal Oscillation. Geophys. Res. Lett. 43, 2810–2818 (2016).
15. 15.
Zhang, R., Sutton, R., Danabasoglu, G. & Delworth, K. Comment on “The Atlantic Multidecadal Oscillation without a role for ocean circulation”. Science 352, 1527 (2016).
16. 16.
Drews, A. & Greatbatch, R. J. Atlantic Multidecadal Variability in a model with an improved North Atlantic Current. Geophys. Res. Lett. 43, 8199–8206 (2016).
17. 17.
Zhang, R. On the persistence and coherence of subpolar sea surface temperature and salinity anomalies associated with AMOC multidecadal variability, Geophys. Res. Lett. 44, 7865–7875 (2017).
18. 18.
Dunstone, N. J. A., Smith, D. M. A., Booth, B. B. B. A., Hermanson, L. A. & Eade, R. T. Anthropogenic aerosol forcing of Atlantic tropical storms. Nat. Geosci. 6, 534–539 (2013).
19. 19.
Zhang, R. et al. Have aerosols caused the observed atlantic multidecadal variability? J. Atmos. Sci. 70, 1135–1144 (2013).
20. 20.
Wang, C., Zhang, L., Lee, S. K., Wu, L. & Mechoso, C. R. A global perspective on CMIP5 climate model biases. Nat. Clim. Change 4, 201–205 (2014).
21. 21.
Dunne, J. P. et al. “GFDL’s ESM2 global coupled climate–carbon earth system models. Part I: Physical formulation and baseline simulation characteristics.”. J. Clim. 25, 6646–6665 (2012).
22. 22.
Wang, H., Legg, S. A. & Hallberg, R. W. Representations of the Nordic Seas overflows and their large scale climate impact in coupled models. Ocean Model. 86, 76–92 (2015).
23. 23.
Zhang, R. et al. Sensitivity of the North Atlantic ocean circulation to an abrupt change in the Nordic Sea overflow in a high resolution global coupled climate model. J. Geophys. Res. 116, C12024 (2011).
24. 24.
Kim, H. S. et al. Tropical cyclone simulation and response to CO2 doubling in the GFDL CM2. 5 high-resolution coupled climate model. J. Clim. 27, 8034–8054 (2014).
25. 25.
Small, R. J. et al. A new synoptic scale resolving global climate simulation using the Community Earth System Model. J. Adv. Model. Earth Syst. 6, 1065–1094 (2014).
26. 26.
Ruiz-Barradas, A., Nigam, S. & Kavvada, A. The Atlantic Multidecadal Oscillation in twentieth century climate simulations: uneven progress from CMIP3 to CMIP5. Clim. Dynam. 41, 3301–3315 (2013).
27. 27.
Smeed, D. A. et al. Observed decline of the Atlantic meridional overturning circulation 2004-2012. Ocean Sci. 10, 29–38 (2014).
28. 28.
Zhang, R. Coherent surface-subsurface fingerprint of the Atlantic meridional overturning circulation. Geophys. Res. Lett. 35, L20705 (2008).
29. 29.
Zhang, R. Latitudinal dependence of Atlantic Meridional Overturning Circulation (AMOC) variations. Geophys. Res. Lett. 37, L16703 (2010).
30. 30.
Zhang, J. & Zhang, R. On the evolution of Atlantic meridional overturning circulation (amoc) fingerprint and implications for decadal predictability in the North Atlantic. Geophys. Res. Lett. 42, 5419–5426 (2015).
31. 31.
Robson, J., Hodson, D., Hawkins, E. & Sutton, R. Atlantic overturning in decline? Nature Geosci 7, 2–3 (2014).
32. 32.
Robson, J., Ortega, P. & Sutton, R. A reversal of climatic trends in the North Atlantic since 2005. Nat. Geosci. 9, 513–517 (2016).
33. 33.
Smeed, D. et. al. Atlantic meridional overturning circulation observed by the RAPID-MOCHA-WBTS (RAPID-Meridional Overturning Circulation and Heatflux Array-Western Boundary Time Series) array at 26N from 2004 to 2015. Br. Ocean. Data Centre. doi:10.5285/35784047-9b82-2160-e053-6c86abc0c91b (2016).
34. 34.
Jackson, L. C., Peterson, K. A., Roberts, C. D. & Wood, R. A. Recent slowing of Atlantic overturning circulation as a recovery from earlier strengthening. Nat. Geosci. 9, 518–522 (2016).
35. 35.
Wang, C., Lee, S. K. & Enfield, D. B. Atlantic warm pool acting as a link between Atlantic multidecadal oscillation and Atlantic tropical cyclone activity. Geochem. Geophys. Geosyst. 9, Q05V03 (2008).
36. 36.
Gray, W. M. Global View of the origin of tropical disturbances and storms. Mon. Weather Rev. 96, 669 (1968).
37. 37.
Emanuel, K. A. The maximum intensity of hurricanes. J. Atmos. Sci. 45, 1143–1155 (1988).
38. 38.
Camargo, S., Sobel, A., Barnston, A. & Emanuel, K. Tropical cyclone genesis potential index in climate models. Tellus A 59, 428–443 (2007).
39. 39.
Trenary, L. & DelSole, T. Does the Atlantic Multidecadal oscillation get its predictability from the atlantic meridional overturning circulation? J. Clim. 29, 5267–5280 (2016).
40. 40.
Chafik, L. et al. Global linkages originating from decadal oceanic variability in the subpolar North Atlantic. Geophys. Res. Lett. 43, 10909–10919 (2016).
41. 41.
Dunstone, N. J., Smith, D. M. & Eade, R. Multi-year predictability of the tropical Atlantic atmosphere driven by the high latitude North Atlantic Ocean. Geophys. Res. Lett. 38, L14701 (2011).
42. 42.
Kushnir, Y. Interdecadal variations in the North Atlantic sea surface temperature and associated atmospheric conditions. J. Clim. 7, 141–157 (1994).
43. 43.
Delworth, T. L. & Mann, M. E. Observed and simulated multidecadal variability in the Northern Hemisphere. Clim. Dyn. 16, 661–676 (2000).
44. 44.
Knight, J. R., Allan, R. J., Folland, C. K., Vellinga, M. & Mann, M. E. A signature of persistent natural thermohaline circulation cycles in observed climate. Geophys. Res. Lett. 32, L20708 (2005).
45. 45.
Latif, M., Collins, M., Pohlmann, H. & Keenlyside, N. A review of predictability studies of the Atlantic sector climate on decadal time scales. J. Clim. 19, 5971–5987 (2006).
46. 46.
Grossmann, I. & Klotzbach, P. J. A review of North Atlantic modes of natural variability and their driving mechanisms. Geophys. Res. Lett. 114, D24107 (2009).
47. 47.
Ba, J. et al. A mechanism for Atlantic multidecadal variability in the Kiel climate model. Clim. Dyn. 41, 2133–2144 (2013).
48. 48.
McCarthy, G. D., Haigh, I. D., Hirschi, J. M., Grist, J. P. & Smeed, D. A. Ocean impact on decadal Atlantic climate variability revealed by sea-level observations. Nature 521, 508–510 (2015).
49. 49.
Delworth, T. L. et al. The central role of ocean dynamics in connecting the North Atlantic Oscillation to the extratropical component of the Atlantic Multidecadal Oscillation. J. Climate 30, 3789–3805 (2017).
50. 50.
Martin, E. R., Thorncroft, C. & Booth, B. B. The multidecadal Atlantic SST-Sahel rainfall teleconnection in CMIP5 simulations. J. Clim. 27, 784–806 (2014).
51. 51.
Gettelman, A., Shindell, D. & Lamarque, J. Impact of aerosol radiative effects on 2000-2010 surface temperatures. Clim. Dynam. 45, 2165–2179 (2015).
52. 52.
Jongeward, A. R., Li, Z., He, H. & Xiong, X. Natural and Anthropogenic Aerosol Trends from Satellite and Surface Observations and Model Simulations over the North Atlantic Ocean from 2002 to 2012. J. Atmos. Sci. 73, 4469–4485 (2016).
53. 53.
Smith, S. J. et al. Anthropogenic sulfur dioxide emissions: 1850–2005. Atmos. Chem. Phys. 11, 1101–1116 (2011).
54. 54.
Klimont, Z., Smith, S. J. & Cofala, J. The last decade of global anthropogenic sulfur dioxide: 2000– 2011 emissions. Environ. Res. Lett. 8, 014003 (2013).
55. 55.
Kossin, J. P. Hurricane intensification along United States coast suppressed during active hurricane periods. Nature 541, 390–393 (2017).
56. 56.
Landsea, C. W. Meteorology: hurricanes and global warming. Nature 438, E11–E12 (2005).
57. 57.
Emanuel, K. Environmental factors affecting tropical cyclone power dissipation. J. Clim. 20, 5497–5509 (2007).
58. 58.
Landsea, C. W. & Franklin, J. L. Atlantic hurricane database uncertainty and presentation of a new database format. Mon. Wea. Rev. 141, 3576–3592 (2013).
59. 59.
Rayner, N. A. et al. Global analyses of sea surface temperature, sea ice, and night marine air temperature since the late nineteenth century. J. Geophys. Res. 108, 4407 (2003).
60. 60.
Levitus, S. et al. World ocean heat content and thermosteric sea level change (0-2000 m), 1955–2010. Geophys. Res. Lett. 39, L10603 (2012).
61. 61.
Dee, D. P. et al. The ERA-Interim reanalysis: configuration and performance of the data assimilation system. Q.J.R. Meteorol. Soc. 137, 553–597 (2011).
62. 62.
Kållberg, P. et.al. The ERA-40 archive. ERA-40 Project Report, No. 17. ECMWF, Shinfield Park, Reading RG2 9AX, UK., 31 (2004).
63. 63.
Kalnay, E. et al. The NCEP/NCAR 40-year reanalysis project. Bull. Am. Meteor. Soc. 77, 437–471 (1996).
64. 64.
Colarco, P., da Silva, A., Chin, M. & Diehl, T. Online simulations of global aerosol distributions in the NASA GEOS-4 model and comparisons to satellite and ground-based aerosol optical depth. J. Geophys. Res. 115, D14207 (2010).
65. 65.
Ebisuzaki, W. A method to estimate the statistical significance of a correlation when the data are serially correlated. J. Clim. 10, 2147–2153 (1997).
## Acknowledgements
X.Y. is funded by Princeton University’s Cooperative Institute for Climate Science through NOAA’s Earth System Modeling Initiative. Data from the RAPID-WATCH MOC monitoring project are funded by the Natural Environment Research Council and are freely available from www.rapid.ac.uk/rapidmoc. We thank Paul Ginoux, Larry Horowitz, and Ming Zhao for their helpful comments on the earlier version of the manuscript. We thank Kerry Emanuel and two anonymous reviewers for their constructive reviews.
## Author information
### Affiliations
1. #### The Program in Atmospheric and Oceanic Sciences, Princeton University, Princeton, NJ, 08540, USA
• Xiaoqin Yan
2. #### NOAA/GFDL, Princeton, NJ, 08540, USA
• Rong Zhang
• & Thomas R. Knutson
### Contributions
R.Z.: Conceived the study and designed the methods. X.Y.: Performed the analysis with the help from R.Z. X.Y., R.Z. and T.R.K.: Worked on the interpretation of results and wrote the manuscript.
### Competing interests
The authors declare no competing financial interests.
### Corresponding authors
Correspondence to Xiaoqin Yan or Rong Zhang.
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2018-08-19 04:28:39
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https://scikit-learn.org/dev/modules/generated/sklearn.calibration.CalibratedClassifierCV.html
|
sklearn.calibration.CalibratedClassifierCV¶
class sklearn.calibration.CalibratedClassifierCV(base_estimator=None, *, method='sigmoid', cv=None, n_jobs=None, pre_dispatch='2*n_jobs', verbose=0)[source]
Probability calibration with isotonic regression or logistic regression.
This class uses cross-validation to both estimate the parameters of a classifier and subsequently calibrate a classifier. For each cv split it fits a copy of the base estimator to the training folds, and calibrates it using the testing fold. For prediction, predicted probabilities are averaged across these individual calibrated classifiers.
Already fitted classifiers can be calibrated via the parameter cv=”prefit”. In this case, no cross-validation is used and all provided data is used for calibration. The user has to take care manually that data for model fitting and calibration are disjoint.
The calibration is based on the decision_function method of the base_estimator if it exists, else on predict_proba.
Read more in the User Guide.
Parameters
base_estimatorestimator instance, default=None
The classifier whose output need to be calibrated to provide more accurate predict_proba outputs. The default classifier is a LinearSVC.
method{‘sigmoid’, ‘isotonic’}, default=’sigmoid’
The method to use for calibration. Can be ‘sigmoid’ which corresponds to Platt’s method (i.e. a logistic regression model) or ‘isotonic’ which is a non-parametric approach. It is not advised to use isotonic calibration with too few calibration samples (<<1000) since it tends to overfit.
cvint, cross-validation generator, iterable or “prefit”, default=None
Determines the cross-validation splitting strategy. Possible inputs for cv are:
• None, to use the default 5-fold cross-validation,
• integer, to specify the number of folds.
• An iterable yielding (train, test) splits as arrays of indices.
For integer/None inputs, if y is binary or multiclass, StratifiedKFold is used. If y is neither binary nor multiclass, KFold is used.
Refer to the User Guide for the various cross-validation strategies that can be used here.
If “prefit” is passed, it is assumed that base_estimator has been fitted already and all data is used for calibration.
Changed in version 0.22: cv default value if None changed from 3-fold to 5-fold.
n_jobsint, default=None
Number of jobs to run in parallel. None means 1 unless in a joblib.parallel_backend context. -1 means using all processors.
Base estimator clones are fitted in parallel across cross-validation iterations. Therefore parallelism happens only when cv != “prefit”.
See Glossary for more details.
New in version 0.24.
pre_dispatchint or str, default=’2*n_jobs’
Controls the number of jobs that get dispatched during parallel execution. Reducing this number can be useful to avoid an explosion of memory consumption when more jobs get dispatched than CPUs can process. This parameter can be:
• None, in which case all the jobs are immediately created and spawned. Use this for lightweight and fast-running jobs, to avoid delays due to on-demand spawning of the jobs
• An int, giving the exact number of total jobs that are spawned
• A str, giving an expression as a function of n_jobs, as in ‘2*n_jobs’
New in version 0.24.
verboseint, default=0
Controls the verbosity: the higher, the more messages.
New in version 0.24.
Attributes
classes_ndarray of shape (n_classes,)
The class labels.
calibrated_classifiers_list (len() equal to cv or 1 if cv == “prefit”)
The list of calibrated classifiers, one for each cross-validation split, which has been fitted on training folds and calibrated on the testing fold.
References
1
Obtaining calibrated probability estimates from decision trees and naive Bayesian classifiers, B. Zadrozny & C. Elkan, ICML 2001
2
Transforming Classifier Scores into Accurate Multiclass Probability Estimates, B. Zadrozny & C. Elkan, (KDD 2002)
3
Probabilistic Outputs for Support Vector Machines and Comparisons to Regularized Likelihood Methods, J. Platt, (1999)
4
Predicting Good Probabilities with Supervised Learning, A. Niculescu-Mizil & R. Caruana, ICML 2005
Examples
>>> from sklearn.datasets import make_classification
>>> from sklearn.naive_bayes import GaussianNB
>>> from sklearn.calibration import CalibratedClassifierCV
>>> X, y = make_classification(n_samples=100, n_features=2,
... n_redundant=0, random_state=42)
>>> base_clf = GaussianNB()
>>> calibrated_clf = CalibratedClassifierCV(base_estimator=base_clf, cv=3)
>>> calibrated_clf.fit(X, y)
CalibratedClassifierCV(base_estimator=GaussianNB(), cv=3)
>>> len(calibrated_clf.calibrated_classifiers_)
3
>>> calibrated_clf.predict_proba(X)[:5, :]
array([[0.110..., 0.889...],
[0.072..., 0.927...],
[0.928..., 0.071...],
[0.928..., 0.071...],
[0.071..., 0.928...]])
>>> from sklearn.model_selection import train_test_split
>>> X, y = make_classification(n_samples=100, n_features=2,
... n_redundant=0, random_state=42)
>>> X_train, X_calib, y_train, y_calib = train_test_split(
... X, y, random_state=42
... )
>>> base_clf = GaussianNB()
>>> base_clf.fit(X_train, y_train)
GaussianNB()
>>> calibrated_clf = CalibratedClassifierCV(
... base_estimator=base_clf,
... cv="prefit"
... )
>>> calibrated_clf.fit(X_calib, y_calib)
CalibratedClassifierCV(base_estimator=GaussianNB(), cv='prefit')
>>> len(calibrated_clf.calibrated_classifiers_)
1
>>> calibrated_clf.predict_proba([[-0.5, 0.5]])
array([[0.936..., 0.063...]])
Methods
fit(X, y[, sample_weight]) Fit the calibrated model get_params([deep]) Get parameters for this estimator. Predict the target of new samples. Posterior probabilities of classification score(X, y[, sample_weight]) Return the mean accuracy on the given test data and labels. set_params(**params) Set the parameters of this estimator.
fit(X, y, sample_weight=None)[source]
Fit the calibrated model
Parameters
Xarray-like of shape (n_samples, n_features)
Training data.
yarray-like of shape (n_samples,)
Target values.
sample_weightarray-like of shape (n_samples,), default=None
Sample weights. If None, then samples are equally weighted.
Returns
selfobject
Returns an instance of self.
get_params(deep=True)[source]
Get parameters for this estimator.
Parameters
deepbool, default=True
If True, will return the parameters for this estimator and contained subobjects that are estimators.
Returns
paramsmapping of string to any
Parameter names mapped to their values.
predict(X)[source]
Predict the target of new samples. The predicted class is the class that has the highest probability, and can thus be different from the prediction of the uncalibrated classifier.
Parameters
Xarray-like of shape (n_samples, n_features)
The samples.
Returns
Cndarray of shape (n_samples,)
The predicted class.
predict_proba(X)[source]
Posterior probabilities of classification
This function returns posterior probabilities of classification according to each class on an array of test vectors X.
Parameters
Xarray-like of shape (n_samples, n_features)
The samples.
Returns
Cndarray of shape (n_samples, n_classes)
The predicted probas.
score(X, y, sample_weight=None)[source]
Return the mean accuracy on the given test data and labels.
In multi-label classification, this is the subset accuracy which is a harsh metric since you require for each sample that each label set be correctly predicted.
Parameters
Xarray-like of shape (n_samples, n_features)
Test samples.
yarray-like of shape (n_samples,) or (n_samples, n_outputs)
True labels for X.
sample_weightarray-like of shape (n_samples,), default=None
Sample weights.
Returns
scorefloat
Mean accuracy of self.predict(X) wrt. y.
set_params(**params)[source]
Set the parameters of this estimator.
The method works on simple estimators as well as on nested objects (such as pipelines). The latter have parameters of the form <component>__<parameter> so that it’s possible to update each component of a nested object.
Parameters
**paramsdict
Estimator parameters.
Returns
selfobject
Estimator instance.
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2020-08-04 14:38:21
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https://zonca.dev/2019/03/folder-inherit-group-permission.html
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I have googled this so many times…
On shared systems, like Supercomputers, you often belong to many different Unix groups, and that membership allows you to access data from specific projects you are working on and you can share data with your collaborators.
If you set SGID on a folder, any folder of file created in that folder will automatically belong to the Unix group of that folder, and not your default group. You first set the right group on the folder, recursively so that older files will get the right permissions:
chown -R somegroup sharedfolder
Then you set the SGID so future files will automatically belong to somegroup:
chmod g+s sharedfolder
This is very useful for example in the /project filesystem at NERSC, you can set the SGID so that every file that is copied to the shared /project filesystem is accessible by other collaborators.
Related to this is also the default umask, most systems by default give “read” permission for the group, so setting SGID is enough, otherwise it is also necessary to configure umask properly.
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2020-07-15 00:56:11
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https://en.m.wikipedia.org/wiki/Field_of_view
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# Field of view
The field of view (FoV) is the extent of the observable world that is seen at any given moment. In the case of optical instruments or sensors it is a solid angle through which a detector is sensitive to electromagnetic radiation.
FOV both eyes
Vertical FOV
Angle of view can be measured horizontally, vertically, or diagonally.
A 360-degree panorama of the Milky Way at the Very Large Telescope. Such a panorama shows the entire field of view (FOV) of the telescope in a single image. In the image, the Milky Way appears like an arc of stars spanning horizon to horizon with two streams of stars seemingly cascading down like waterfalls.[1]
## Humans and animals
In the context of human and primate vision, the term "field of view" is typically only used in the sense of a restriction to what is visible by external apparatus, like when wearing spectacles[2] or virtual reality goggles. Note that eye movements are allowed in the definition but do not change the field of view when understood this way.
If the analogy of the eye's retina working as a sensor is drawn upon, the corresponding concept in human (and much of animal vision) is the visual field.[3] It is defined as "the number of degrees of visual angle during stable fixation of the eyes".[4] Note that eye movements are excluded in the visual field's definition. Different animals have different visual fields, depending, among others, on the placement of the eyes. Humans have a slightly over 210-degree forward-facing horizontal arc of their visual field (i.e. without eye movements),[5][6] (with eye movements included it is slightly larger, as you can try for yourself by wiggling a finger on the side), while some birds have a complete or nearly complete 360-degree visual field. The vertical range of the visual field in humans is around 150 degrees.[5]
The range of visual abilities is not uniform across the visual field, and by implication the FoV, and varies between species. For example, binocular vision, which is the basis for stereopsis and is important for depth perception, covers 114 degrees (horizontally) of the visual field in humans;[7] the remaining peripheral 40 degrees on each side have no binocular vision (because only one eye can see those parts of the visual field). Some birds have a scant 10 to 20 degrees of binocular vision.
Similarly, color vision and the ability to perceive shape and motion vary across the visual field; in humans color vision and form perception are concentrated in the center of the visual field, while motion perception is only slightly reduced in the periphery and thus has a relative advantage there. The physiological basis for that is the much higher concentration of color-sensitive cone cells and color-sensitive parvocellular retinal ganglion cells in the fovea – the central region of the retina, together with a larger representation in the visual cortex – in comparison to the higher concentration of color-insensitive rod cells and motion-sensitive magnocellular retinal ganglion cells in the visual periphery, and smaller cortical representation. Since rod cells require considerably less light to be activated, the result of this distribution is further that peripheral vision is much more sensitive at night relative to foveal vision (sensitivity is highest at around 20 deg eccentricity).[3]
## Conversions
Many optical instruments, particularly binoculars or spotting scopes, are advertised with their field of view specified in one of two ways: angular field of view, and linear field of view. Angular field of view is typically specified in degrees, while linear field of view is a ratio of lengths. For example, binoculars with a 5.8 degree (angular) field of view might be advertised as having a (linear) field of view of 102 mm per meter. As long as the FOV is less than about 10 degrees or so, the following approximation formulas allow one to convert between linear and angular field of view. Let ${\displaystyle A}$ be the angular field of view in degrees. Let ${\displaystyle M}$ be the linear field of view in millimeters per meter. Then, using the small-angle approximation:
${\displaystyle A\approx {360^{\circ } \over 2\pi }\cdot {M \over 1000}\approx 0.0573\times M}$
${\displaystyle M\approx {2\pi \cdot 1000 \over 360^{\circ }}\cdot A\approx 17.45\times A}$
## Machine vision
In machine vision the lens focal length and image sensor size sets up the fixed relationship between the field of view and the working distance. Field of view is the area of the inspection captured on the camera’s imager. The size of the field of view and the size of the camera’s imager directly affect the image resolution (one determining factor in accuracy). Working distance is the distance between the back of the lens and the target object.
## Tomography
In computed tomography (abdominal CT pictured), the field of view (FOV) multiplied by scan range creates a volume of voxels.
In tomography, the field of view is the area of each tomogram. In for example computed tomography, a volume of voxels can be created from such tomograms by merging multiple slices along the scan range.
## Remote sensing
In remote sensing, the solid angle through which a detector element (a pixel sensor) is sensitive to electromagnetic radiation at any one time, is called instantaneous field of view or IFOV. A measure of the spatial resolution of a remote sensing imaging system, it is often expressed as dimensions of visible ground area, for some known sensor altitude.[8][9] Single pixel IFOV is closely related to concept of resolved pixel size, ground resolved distance, ground sample distance and modulation transfer function.
## Astronomy
In astronomy, the field of view is usually expressed as an angular area viewed by the instrument, in square degrees, or for higher magnification instruments, in square arc-minutes. For reference the Wide Field Channel on the Advanced Camera for Surveys on the Hubble Space Telescope has a field of view of 10 sq. arc-minutes, and the High Resolution Channel of the same instrument has a field of view of 0.15 sq. arc-minutes. Ground-based survey telescopes have much wider fields of view. The photographic plates used by the UK Schmidt Telescope had a field of view of 30 sq. degrees. The 1.8 m (71 in) Pan-STARRS telescope, with the most advanced digital camera to date has a field of view of 7 sq. degrees. In the near infra-red WFCAM on UKIRT has a field of view of 0.2 sq. degrees and the VISTA telescope has a field of view of 0.6 sq. degrees. Until recently digital cameras could only cover a small field of view compared to photographic plates, although they beat photographic plates in quantum efficiency, linearity and dynamic range, as well as being much easier to process.
## Photography
In photography, the field of view is that part of the world that is visible through the camera at a particular position and orientation in space; objects outside the FOV when the picture is taken are not recorded in the photograph. It is most often expressed as the angular size of the view cone, as an angle of view. For a normal lens, the diagonal field of view can be calculated as:
${\displaystyle \mathrm {FOV} =2\times \arctan \left({\frac {\text{sensor size}}{2f}}\right)}$
where ${\displaystyle f}$ is the focal length.
## Microscopy
Field of view diameter in microscopy
In microscopy, the field of view in high power (usually a 400-fold magnification when referenced in scientific papers) is called a high-power field, and is used as a reference point for various classification schemes.
For an objective with magnification ${\displaystyle m}$ , the FOV is related to the Field Number (FN) by
${\displaystyle FOV={\frac {FN}{m}}}$ ,
if other magnifying lenses are used in the system (in addition to the objective), the total ${\displaystyle m}$ for the projection is used.
## Video games
The field of view in video games refers to the field of view of the camera looking at the game world, which is dependent on the scaling method used.[10]
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2021-09-17 17:29:01
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https://www.andifugard.info/2023-note/
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# 2023
Random fact about 2023: if you divide 2023 by the sum of its digits (in base 10), you get a square number:
$$\displaystyle \frac{2023}{2+0+2+3} = \frac{2023}{7} = 289 = 17^2$$.
This is sequence A001102 in the On-Line Encyclopedia of Integer Sequences (OEIS). The last year that satisfied this property was 1815; however, OEIS doesn’t say when it next applies. Let’s find out.
First we need to spell out the property so that it’s easier to automate a search.
Let $$d_x(i)$$ denote the $$i$$th digit of $$x$$ in base 10,
$$\displaystyle d_x(i) = \frac{n \bmod 10^{i+1}-n \bmod 10^i}{10^i}$$,
where $$i$$ ranges from $$0$$ to $$\lfloor \log_{10}{x} \rfloor$$, i.e., to one less than the number of digits of $$x$$.
Let $$\displaystyle j(k) = \frac{k}{\sum_{i = 0}^{ \lfloor \log_{10}{k} \rfloor} d_k(i)}$$.
The property holds of $$k \in \mathbb{N}$$ iff there is an $$m \in \mathbb{N}$$ such that $$j(k) = m^2$$, or, equivalently, if $$\lfloor \sqrt{j(k)} \rfloor^2=j(k)$$.
This is enough to code up the search in R. Here are the next few years when the property applies: 2023, 2025, 2028, 2178, 2304, 2312, 2352, 2400.
Please note, I’m not a numerologist, so I don’t know if this is a good thing; however, Happy New Year nonetheless 😉
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2023-02-01 21:50:20
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https://stats.stackexchange.com/questions/66348/significant-difference-between-two-groups-in-a-right-wrong-quiz
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# Significant difference between two groups in a right /wrong quiz
I did an exam, which the comment given was that my answers to two questions were too "short" or lacking… (I do not know what was lacking). I would like to learn and understand. As I remember I think I used something called an on line calculator for categoric values when I came to my answers.
The exam questions were:
1. Students representing two different geographical ares had a quiz test.
• group 1 had 36 right and 13 wrong
• group 2 had 29 right and 19 wrong
Is there a significant difference between these groups (p<0,05)?
2. What if the quiz result where like this:
• group 1 360 right, 130 wrong
• group 2 290 right, 190 wrong
Is there a significant difference between this groups (p<0,05)?
1. Yes there is a significant difference
2. No there is not a significant difference.
My questions to you are:
• And what is the proper way of showing how you get to this/ or correct answer (without using a calculator)?
Meaning, what are the steps I need to take to solve this kind of questions?
I am most grateful if anyone have time to explain this to me.
• Please check you didn't make a transcription error (who the heck uses a significance level of $\frac{1}{2}$??) – Glen_b Sep 1 '13 at 23:54
• The question is more interesting when the test is conducted at a level of $0.5\%$ rather than $5\%$ or $0.5=50\%$. That is because (2) is very strongly significant whereas (1) is significant between the $1.7\%$ and $3.5\%$ levels, depending on how it is tested ($\chi^2,$ Fisher, or a permutation test). As far as its efficacy in assessing the underlying concepts goes, the question would be far better replacing $13$ by $18$ and $130$ by $180$: then it can be answered rigorously with almost no calculation. – whuber Sep 25 '13 at 22:28
What counts as a good answer would depend on the instructions you received, the learning goals and contents of the course associated with the exam (if any). In some courses, you are expected to document the whole procedure and do all relevant computations by hand. In others, just printing out the right output from a statistical package is enough.
• The name of the test and a succinct explanation of the reasons you think it is appropriate.
• A list of the assumptions of the test and why they are reasonable.
• A statement of the hypotheses being tested.
• An explicit formula for the test statistic/computation of the test statistic.
• The name and parameters of the relevant statistical distribution, if applicable.
• A p-value/critical value for the statistic, justifying your conclusion that the difference is or is not significant.
Also, your answers are obviously incorrect. I don't know how comfortable you are with these notions but you can realize that by thinking about the link between power and sample size, even without looking at any computation or specific test result.
• Thank you very very much for your respond. This makes me understand more. – Heta Aug 2 '13 at 11:44
• But since you say my answer is incorrect that basically means that I did not use the calculator correct, or that I should not have used it at all? – Heta Aug 2 '13 at 12:05
• Difficult to know without more details. Maybe you used incorrectly, maybe it was not appropriate to use it in this situation, maybe it claims to be appropriate for your situation but it's faulty in some way (this is also not uncommon, everybody can put a calculator on the web). In any case, even if the calculator could be used to perform the test in other circumstances, it was perhaps not the best approach for learning/answering exams because you are treating the whole thing as a black box instead of gaining more understanding. – Gala Aug 2 '13 at 12:11
• PS: If answers are helpful, you might vote them up (using the little arrow/normal distribution on the left of the answer). – Gala Aug 2 '13 at 12:12
• Hello Gaël Laurans. I have study a lot the last weeks, and I understand much more. And I finally got the right answers/how to create a good rapport. I just wanted to thank you again for your respond, it was truly useful. – Heta Sep 25 '13 at 22:08
Your answers were incorrect so your teacher might have given you some points if you had more information. You're probably supposed to observe something about the differences between the two sets of data and why the significance changes (or doesn't). Why do you think it might change? What have you tried? Where was the online calculator? Do any of the analyses listed here look like familiar names from class or the website calculator?
• First, thank you so very much for responding. The Chi-squared test part of the class. – Heta Aug 2 '13 at 11:36
• I understand more now, I did not test anything, I just "answered" the question. So my lack of understanding became very clear to me now. But I could have gone more into it than I did, but I did not understand enough to see that that would be the right thing to do. Thank you very much again. – Heta Aug 2 '13 at 11:42
• There are lots of places you can find on the internet if you just type in, "how to calculate the chi-square test". I tried it in two search engines and the first results that came up were all OK. Looking for a chi-square calculator is not nearly as fruitful from an education point of view as looking for understanding on the test. – John Aug 2 '13 at 11:46
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2021-06-21 19:49:31
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https://www.aimsciences.org/article/doi/10.3934/ipi.2018042
|
# American Institute of Mathematical Sciences
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August 2018, 12(4): 993-1031. doi: 10.3934/ipi.2018042
## Reconstruction of a compact manifold from the scattering data of internal sources
1 Department of Mathematics and Statistics, University of Helsinki, Helsinki, Finland 2 Department of computational and applied mathematics, Rice University, Houston, Texas, USA 3 Department of Pure Mathematics and Mathematical Statistics, University of Cambridge, Cambridge, UK 4 Department of Mathematics, University of California Santa Barbara, Santa Barbara, California, USA
Received August 2017 Revised March 2018 Published June 2018
Given a smooth non-trapping compact manifold with strictly convex boundary, we consider an inverse problem of reconstructing the manifold from the scattering data initiated from internal sources. These data consist of the exit directions of geodesics that are emaneted from interior points of the manifold. We show that under certain generic assumption of the metric, the scattering data measured on the boundary determine the Riemannian manifold up to isometry.
Citation: Matti Lassas, Teemu Saksala, Hanming Zhou. Reconstruction of a compact manifold from the scattering data of internal sources. Inverse Problems & Imaging, 2018, 12 (4) : 993-1031. doi: 10.3934/ipi.2018042
##### References:
show all references
##### References:
Here is a schematic picture about our data $R_{\partial M}(p)$, where the point $p$ is the blue dot. Here the black arrows are the exit directions of geodesics emitted from $p$ and the blue arrows are our data
Here is a visualization of the set up in the definition of the function $\varrho_k$ in Lemma 2.8. The blue dot is $p$ and the red&blue dot is $q$. The black curve is the geodesic $\gamma_{p, \eta}$. The red line is the hypersurface $\tilde S_1$ and the blue line is the hypersurface $\tilde S_2$. The small blue and red segments indicate the intervals $(s_k-\delta, s_k+\delta)$ where the function $\varrho_k(\cdot, \eta)$ is defined
Here is a schematic picture about $K(p)$, where the point $p \in M$ is the blue dot. The black curves represent the geodesics $\gamma_{z, \xi}$, and $\gamma_{w, \eta}$ respectively, where vectors $(z, \xi), (w, \eta) \in R_{\partial M}^E(p)$. Notice that only $(w, \eta)\in K(p)$
Here is a schematic picture about the map $\Theta_{q, \tilde q, v}$ evaluated at point $p\in M$, where the point $p \in M$ is the blue dot and $V_q\cap V_{\tilde q}$ is the blue ellipse. The higher red&blue dot is $q$ and the lower is $\tilde q$. The blue vector is the given direction $v \in T_{\tilde q}N$
Here is a schematic picture about the situation where the boundary normal geodesic $\gamma_{p, \nu}$ (the black curve) is self-intersecting at $p \in \partial M$ (red&blue dot). Here the blue curve is the geodesic $\gamma_{p, W(p)}$, where $W(p)\notin I_p$. For the point $w \in M$ (blue dot) the point $p$ satisfies $p = \Pi_W(w)$
Here is a schematic picture about the map $(\tilde Q_{q, v}, \Pi^E_{q, W})$ evaluated at a point $x\in M$ that is close to $\partial M$, where the point $x \in M$ is the blue dot. The right hand side red&blue dot is $\Pi^E_{q, W}(x)$ and the left hand side red&blue dot is $q$. The blue arrow is the given vector $v \in T_qN$
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Ego says, "Once everything falls into place, I'll feel peace." Spirit says "Find your peace, and then
#### Idea Transcript
Publ. RIMS, Kvoto Univ. 30 (1994), 401^441
The Structure of Hilbert Flag Varieties Dedicated to the memory of our father
By Gerard F. HELMINCK* and Aloysius G. HELMINCK**
Abstract In this paper we present a geometric realization of infinite dimensional analogues of the finite dimensional representations of the general linear group. This requires a detailed analysis of the structure of the flag varieties involved and the line bundles over them. In general the action of the restricted linear group can not be lifted to the line bundles and thus leads to central extensions of this group. It is determined exactly when these extensions are non-trivial. These representations are of importance in quantum field theory and in the framework of integrable systems. As an application, it is shown how the flag varieties occur in the latter context.
§ 1.
Introduction
Let H be a complex Hilbert space. If H is finite dimensional, then it is a classical result that the finite dimensional irreducible representations of the general linear group GL(H) can be realized geometrically as the natural action of the group GL(H) on the space of global holomorphic sections of a holomorphic line bundle over a space of flags in H. By choosing a basis of H, one can identify this space of holomorphic sections with a space of holomorphic functions on GL(H) that are certain polynomial expressions in minors of the matrices corresponding to the elements of GL(H). Infinite dimensional analogues of some of these representations occur in quantum field theory, see e.g. [5]. Infinite dimensional Grassmann manifolds play an important role in the framework of integrable systems. The first person to realize this was Sato, see [24]. In this paper we will give an infinite dimensional analogue of all these representations. Thereto we take a separable Hilbert space H. In H we consider a collection of flags that generalizes the Grassmanian from chapter 7 in [23]. This flag variety carries a natural Hilbert space structure and there Communicated by T. Miwa, January 18, 1993. 1991 Mathematics Subject Classification: 22E65, 14M15, 35Q58, 43A80, 17B65. * Department of Mathematics, Universiteit Twente, Enschede, The Netherlands ** Department of Mathematics, North Carolina State University, Raleigh, NC 27695, USA
402
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
exist line bundles over it that are similar to the finite dimensional ones. This includes the determinant bundle and its dual from [23]. In the "dominant" case the space of global holomorphic sections of such a line bundle turns out to be non-trivial. However, the action of the analogue of the general linear group can, in general, not be lifted to the line bundle under consideration and one has to pass to a central extension of this group. Besides of the introduction, this paper consists of three sections. In the first section we give the definition of the flag variety g and we treat some properties of g. The second section is devoted to the construction of the holomorphic line bundles, to a description of the corresponding central extensions and to the analysis of the space of global holomorphic sections. As an application, we show in the final section what role the geometry plays in the context of some integrable systems. A more detailed description of the content of the different sections is as follows. The first subsection of section 2 discusses the type of flags in H that will be considered. Here the model for the size of the flags is the basic flag F(0} corresponding to a finite orthogonal decomposition of H. The flag variety g is a homogeneous space for a certain unitary group Ures(H). As in the finite dimensional case it is convenient to see g also as a homogeneous space for a larger group of automorphisms of H, namely GL res (H). This is the analogue of the general linear group in this framework. Analogously to the finite dimensional case the group UTes(H) is the unitary component in the polar decomposition of GLres(H). In the second subsection we give an explicit description of the manifold structure on g an^ we discuss decompositions of certain open subsets of GL res (ff). A first difference with the finite dimensional situation appears at the description of the connected components of g in the third subsection. The fourth subsection contains the technical prerequisites for the construction of the line bundles. First we choose a suitable orthonormal basis of //, we order its index set conveniently and we introduce the Weyl group W of GLtes(H). Next we show that the charts around the points in the W-orbit through F (0) cover g. By using this covering one obtains a stratification of g into parts that are all homeomorphic to a Hilbert space. On the group level this gives you the Birkhoff decomposition for GL res (/f). Let g (0) be the connected component of g containing F (0) . From the foregoing results one deduces that g(0) is a homogeneous space for a Banach Lie group (5 that permits you to take suitable minors. As a group the group (5 is a subgroup of GL res (//), but its topology is stronger than the one induced by GL res (ff). The description of © and its topology can be found in the first subsection of section 3. There we introduce also the maximal torus T(,T) of © and its group of analytic characters f. In the second subsection we introduce a dense tower of finite dimensional flag varieties in g (0) . The next subsection
THE STRUCTURE OF HILBERT FLAG VARIETIES
403
shows how you can associate to certain elements i^, keZm, of fa holomorphic line bundle L(k_) over 5 (0) - Further it is shown there that, if one tries to lift the action of the connected component GL(r°Q[(H) of GL res (/f), one might meet obstructions and that one can only lift the action of a central extension of GL(£}S(H). The natural question that comes up then is "how essential is this 1 extension '. This question is treated fully in the next subsection. Then one has come to the final subsection of this section. There we determine, when L(k_) has global non-trivial sections and we show that the action of (S on this space defines an irreducible ©-module of highest weight ^. Section 4 is an illustration of the fact that the geometry of the foregoing sections plays a role in the theory of integrable systems. The system we will consider is the multicomponent KP-hierarchy. The first subsection describes the flows in GLres(H) corresponding to this system. The algebraic framework for this system of equations is given in the second subsection. In the final subsection of this paper we indicate how the flag varieties form the starting point of the construction of solutions to the equations of the multicomponent KP-hierarchy and the modified versions of the KP-hierarchy. We would like to thank the referee for bringing to our attention papers by Faltings [10] and Kashiwara [19], who give an algebraic geometrical approach to infinite dimensional flag varieties. §2.
Properties of Hilbert Flag Varieties
2.1. The flag variety. Let H be a separable complex Hilbert space with inner product < •, • >. We will consider certain finite chains of subspaces in H and we will call them flags as in the finite dimensional case. First one has to specify the "size" of the components of the flag. Therefore we start with an orthogonal decomposition of //, (1)
H = Hl®"-®Hm,
where H{ J_ Hj for / ^j.
We assume that mi = dim(Hi) satisfies 1 < m,-< oc. An example of a decomposition occurring in quantum field theory is the one corresponding to the positive and negative spectrum of the Dirac operator, see [5] and [21]. In the context of integrable systems we have: Example 2.1.1. Let (-, •) be the standard inner product on C. The Grassmann manifold Gr(H) that is crucial at the construction of solutions of KP-type hierarchies in [23], [13] and [25] corresponds to the case that H is the space of power series H = L 2 ^ 1 , C) = { £ «„_-", a,,eC7, £ (fl.. an) < oo], neZ
»eZ
H! = ( X anz"eH} n>0
and
H2 = ( £ a. n
<0
404
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
If one takes r = 1 and k and I in Z with k > /, then the basic manifold corresponding to the (fc, /)-modified KP-hierarchy is the flag variety corresponding to H = L2(S^ C) = H1®H2® #3, with
H1 = { Z a n z » e H } 9 H 2 = {Zan?eH} n>k
and H3 = { 5>Bz"eH}.
n>l
n
This correspondence is described in the fourth section. Let pi9 1 < i < m, be the orthogonal projection of H onto Hz. Then we will use throughout this paper the following Notation 2.1.2. If g belongs to 3\$(H), the space of bounded linear operators from H to H, then g = (gtj), 1 < i < m and 1
If 0 = (gtj) as in notation 2.1.2, then we have for its adjoint g* the decomposition (9*)tj = fe)*, f°r all i and j. 2.1 .4. To the decomposition (1) we associate the basic flag F{0) given by
7=1
Now we consider in ff flags F = {F(0),...,F(m)} ? that is to say chains of closed subspaces of H, {0} = F(0) c F(l) c ... c F(m) = H,
that are of the same "size" as the basic flag F (0) , i.e. for all /', 1 < i < m, dim (F(f)/F(/ - 1)) = dim (Ht). To such a flag F is associated an orthogonal decomposition of H, H = Fl®---®Fm,
where F, = F(i)f]F(i - I)1.
We will denote such a flag F by F = (F(0),...,F(m)} as well as F = {F l5 ...F m }. The class of flags one obtains in this way is still too wide and we will require that our flags do not differ too much from the basic flag. One can express this "nearness" in various ways. Our choice is a natural generalization of that used in [23] for the Grassmann manifold. However, a lot of the constructions given here for that case can be carried out with some minor
THE STRUCTURE OF HILBERT FLAG VARIETIES
405
adjustments also for other choices. We start by introducing notations for some spaces of compact operators that occur in the sequel. Notation 2.1.5. If K1 and K2 are Hilbert spaces, then we denote the space of Hilbert-Schmidt operators from K1 to K2 by Jf^(X l 5 K2) and the Hilbert-Schmidt norm by || • \\#>#>. We will write J^(K1, K2) for the space of nuclear operators from K1 to K2 and the trace norm on it will be denoted by li • \\jr- The space ^(Kl9 K2) of compact operators from K1 to K2 will be assumed to have been equipped with the operator norm. Then we have the following chain of continuous inclusions : 19
K2) c jeST(K^ K2) c=
In each of these spaces the collection of finite dimensional operators J^(X l5 K2) lies dense. If K2 is equal to K1? then we simply write 3F(K^, ^(K and ^(K^ for respectively &(Kl9 KJ, ^(X l5 KJ, J^^(Kl9 KJ and Definition 2.1.6. Let g be the collection of flags F = {F 1; ...,F m }, satisfying dim (Fj) = dim (Ht), and for all i and j with j / i, the orthogonal projection Pj : Ff -> HJ is a Hilbert-Schmidt operator. We call g the /fog variety corresponding to the decomposition (1). Remark 2.1.7. If only one mf is infinite, then the Hilbert-Schmidt condition is superfluous. E.g. the space of flags with mt < oo for all i < m, plays a role in [2] at the construction of irreducible representations of the Hilbert Lie group U(&)2- This is the unitary part of the group of invertible transformations of the form "identity + a Hilbert-Schmidt operator". Remark 2.1.8. Instead of the condition pj-.F^Hj, i / J9 belongs to HJ), one could also consider flags such that this map belongs to j) or #(F f , HJ). The flag varieties one obtains in this way we denote by 5(^) respectively 3K^). A more asymmetric condition is considered in [14] where it is required that merely for / < j the projection pt : Fj -> H ,- is Hilbert-Schmidt. In this way we get the flag manifold *\$(\$). Because of the inclusions mentioned above, we have a chain of injections
For m = 2 more general versions of flag spaces are considered in [9] . Remark 2.1.9. In [8], they associate a Banach Grassmann manifold to each Banach Jordan pair. It would be interesting to see if, and if so, how the flag varieties introduced here fit into their framework. 2.1.10. The space 5 is a natural generalization of the Grassmann manifold introduced in section 7.1 of [23]. The flag variety g is a homogeneous space
406
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
for an analogue adapted to this situation, of the general linear group. The Banach structure of this group follows directly from that of its Lie algebra. Therefore we start with the analogue of the Lie algebra of the general linear group. Definition 2.1.11. A restricted endomorphism of H is a u = (utj) in such that u^ is a Hilbert-Schmidt operator for all / =£ j. We denote the space of all restricted endomorphisms of H by ^ res (/f). For all / and 7, we extend the elements of jf?&'(Hi, H-) outside Ht by zero and obtain thus a natural embedding of ^^(H^ Hj) into je^(H). The space &res(H) is a subalgebra of <%(H) since the collection of Hilbert-Schmidt operators is a 2-sided ideal in &(H). Hence it is also a Lie subalgebra of the Lie algebra &(H). The algebra ^ res (/f) becomes a Banach algebra if we equip it with the norm || • ||2 defined by *r-
= II "II + Z
Since the adjoint of a Hilbert-Schmidt operator is again Hilbert-Schmidt, it is clear that ^ res (H) is stable under "taking adjoints". If GL(H) denotes the group of invertible elements in ^(H), then we consider Definition 2.1.12. The restricted linear group, GL res (/f), consists of To see that GL res (/f) is indeed a group, one merely has to show that if g = (g.j) belongs to GL res (/f) then its inverse g~l = ((0~%) also belongs to GL res (#). Now, the relation
shows first of all that for all i, 1 < i < m, both git and (g~l)u are Fredholm operators, i.e. they have a finite dimensional kernel and cokernel. Next one considers the relation 92i(0~1)ii + 922(ff~1)2i
+ Z 92j(9~l}ji = 0. j>2
Since the operator g21 is Hilbert-Schmidt and the operators (g~l}n and g22 are Fredholm, the operator (g~l)2i has to be Hilbert-Schmidt too. Continuing in this fashion, one shows that all (#"%• with i+j are Hilbert-Schmidt. In other words, GL res (/f) consists of the invertible elements of BTes(H). As such, it is in a natural way a Banach Lie group with Lie algebra J5 res (ff). The analogue of the unitary group U(H) in this context is:
THE STRUCTURE OF HILBERT FLAG VARIETIES
407
Definition 2.1.13. The restricted unitary group, Ures (H) = GLres (JET) n U(H). Both UrQS(H) and GL res (7f) are natural generalizations of the restricted unitary and general linear group, introduced in chapter 6 of [23]. The Lie algebra of Ures(H) consists of (H),
X*=-X}.
This is a real Lie subalgebra of %>res(H) and the Lie algebra 33res(H) can be written as
In other words 23 res (/f) is the complexification of u res (H). On the group level this corresponds to the fact that the group GL res (H) possesses a "polar decomposition" of which £/ res (H) forms the unitary component. For, consider the sets P(H) = {A\AEGL(H), A = A* and A > 0} and
On Pres(H) we put the topology induced by 23 res (H). Since the map from P res (H) to P(H) is locally given by a convergent power series in A, this map is in fact a continuous map from P res (//) to itself. Thus we get Proposition 2.1.14. is a homeomorphism. Proof.
The map (u, p)^ up from Ures(H) x P res (H) to GLres(H)
The inverse of this map is
9 '— » gor
,
and we have just seen that it is continuous,
n
With each g in GL res (H) we can associate the flag
From the definition of GL res (H) one sees directly that this flag belongs to 5. The group Urcs(H) acts already transitively on 5- Let F = {F l 5 ...,F m } belong to 5- From the definition of 5 we know that there is for each f, 1 < i < m, an isometry ut between Ht and Ft. If we put u = u1®[email protected], then the condition defining 5 implies that u belongs to the group Ures(H) and that F = u(F(0)). The stabilizer in GLres(H) of the basic flag is the "parabolic subgroup"
408
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
0im \
P=
g\gEGLres(H), g =
0
I 0
, with 0 i£ eGL res (# f ), l
-
0 gml
Thus we can identify g also with the homogeneous space GL res (H)/P. Let i: GLres(H) -> g be the projection T(#) = g • F (0) . On g we will put a Hilbert manifold structure that makes i into an open submersion. This will be discussed in the next subsection. Remark 2.1.15. It will be clear that for the spaces g(^), g(#) and the corresponding general linear group consists of those g = (gtj) in GL(H) such that respectively (2)
giJe^r(Hj9H^
for all
i*j,
(3)
gtjeVWjtHJ
for all
i*j,
(4)
gij€je^(Hj9H^
for all /<;.
2.2. The manifold structure of g. In this subsection we discuss the Hilbert manifold structure on g and some decompositions of open subsets in GL res (#). From the definition of the parabolic group P one sees directly that the Lie algebra of P is given by L(P) = {g\9 = ( g i j ) E B r e s ( H ) ,
9ij
= 0 for all i >j}
and that a complement of L(P) in Bres(H) is the Hilbert space (£, || • ||^^) with
E=
From section 6.1 in [3], we know then that the homogeneous space \$ = GLres(H)/P carries an analytic ^-manifold structure for which T is a submersion and for which the natural action of GL res (ff) on 5 is analytic. Next we give descriptions of some open subsets in GL res (/f) that will be needed later on. Consider for each fe, 1 < k < m — 1, the set Q(k) in GL res (/f) given by
Q(k} =
;) for all
Since we have for each i, 1 < i < m, a continuous surjection from ^res (//) onto ^res(#!©•••©#;) given by
THE STRUCTURE OF HILBERT FLAG VARIETIES
409
the set Q(k) is open and, as in the finite dimensional case, it can be decomposed. For, let L/_(/c) and P(k) be the Lie subgroups of GLres(H) defined by g = (gij)EGLTes(H)
ga = IdHi gij = Q g.. = 0
for all i for j>i for i > j and j > k
and
P(Q = {g = toy)eGLre8(H)|00. = 0 if i >j and
j
Clearly P(fe)nC/_(fc) = {IdH} and this gives you the uniqueness in Lemma 2.2.1. The map (u,p)\-*up from U_(k) x P(k) -> GL res (H) determines a homeomorphism between U_(k) x P(k) and Q(k}. Proof. We use induction on k to show the result. Let g be an element in O(l). Then we know that gll is invertible and if we define w(l)e £/_(!) by w(l) rl = — gr\9ii f°r all r>2, then one sees directly that u(l)g belong to P(l). Assume now that we know Q(l)= U_(l)P(l). Since we have £(/)=> Q(l + 1) and £/_(/) < £ / _ ( / + 1), we may assume that geQ(l + 1) belongs to P(l). Hence the condition geQ(l + 1) means that gl+1J + 1 is invertible. Define u(l+l) = (Uij) in t / _ ( / + l ) by u j l T l - - ^ + i ( g z + i j + i ) ~ 1 for j > I + 1 and M fj - = 0 if i > j and j ^ I + 1. Then M(/ + 1) • g belongs to the parabolic group P(l + 1). This proves the lemma. D As in the finite dimensional case we call Q(m— 1) = U_(m — 1) • P the big cell of g and we also write Q and U _ instead of Q(m — 1) and t/_(m — 1). From this lemma we see that the restriction of T to £/_ gives you a diffeomorphism u^>uF(0} between U_ and the open neighborhood t(Q) of F (0) . Clearly the group L/_ is difTeomorphic to the Hilbert space E. Note that from the definition of Q one can conclude directly that i(Q) = {F = (F f )e g| © p7-: 0 F;- j
> 0 H7- is a bijection for all / < m}. j
This characterization of i(Q) tells you how to choose around a general point of 5 a concrete neighborhood difTeomorphic to E. This requires, however, the introduction of the following notation. Notation 2.2.2. If W is closed subspace of H, then we denote the
410
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
orthogonal projection on W by pw. Consider a F = (F 1? ...,FJ in JJf. Then the analogue of i(Q) for F is Up = [V= (Vi) in 5! © pFl : © K -
> 0 Ff is a bijection for all 1 < I < m}.
Since g - l/ res (#) • ^ (0) , we have for all F and G in g that, if i /;, the map pFi : Gj -> Fj is a Hilbert-Schmidt operator. Hence, if F belongs to l/F, then there is a unique operator A in © J^^(FJ9 Ft) such that for all i,
This is why we call V also the graph of A and we write V= graph (A). It is convenient to have a special name for the flags in UF. Definition 2.23. A flag V in UF is called transversal to F. Let gF be an element of L/ res (/f) such that g F - F ( 0 ) = F. Instead of the big cell Q in GLres(H) with respect to the decomposition H = Hl®[email protected], we could also have introduced a big cell with respect to H = [email protected]@Fm and it will be clear that this set can be written as
Consequently, we get for UF that UF = {gFup(gF)~1F\vfith Then we can define for each F in 5
ueU_ and peP} = i(gF U_P). a
diffeomorphism cpF : UF -+ E by K - Id.
Each ([/F, (pf) is a concrete chart around F for the £-manifold structure on g. We have obtained now a concrete description of the manifold structure on g: Proposition 2.2.4. structure on 3. Proof.
The (UF, (pF) are the charts of the analytic E-manifold
It is sufficient to show for each UF(D and UFw with UF(D n UF(2} ^ 0
that ^>F(2) ° ^Fd1):
>
is an analytic map. From the step by step decomposition described in Lemma 2.2.1 follows that the ^/.-component of (gF^)~lgF^u actually depends analytically on u. This proves the proposition. Q
THE STRUCTURE OF HILBERT FLAG VARIETIES
411
2.3. The connected components of GLres(H). Let g = (g\$ be an element of GL res (/f). Recall that in the proof that GLTes(H) consists of the invertible elements in ^ res (#), we have shown that each gi{ is a Fredholm operator. The collection of Fredholm operators on a Hilbert space K is an open part of the space J*(K). Its connected components are given by the index, which is defined as ind (B) = dim (ker (B)) - dim (coker (£)), where B is a Fredholm operator on K. Since all off-diagonal operators are Hilbert-Schmidt and hence compact, the operator
0=1
••• \ 0
h where g = (^ 0 -)eGL res (ff), gmm I
is a Fredholm operator of index zero. Hence we have that the indices of the {gn\ I
£ ind (git) = 0 and ind (gkk) = 0 if mk < GO. i=l
These relations lead to the introduction of the subgroup Z of Zm defined by m
Z = {z = (z £ )eZ m | X z{ = 0, zk = 0 if mk < oo}. i=l
The standard properties of the index imply that the map i : GLres (H ) -> Z, g\— > (ind (gf! O,..., ind temj), is a continuous group homomorphism.
Hence the sets
GL(^(H) = {g\geGLres(H), i(g) = z}, with zeZ, are open.
In fact, they are exactly the connected components of GL res (/f), for
Proposition 2.3.1. For each zeZ, the set GL(^(H] connected. Proof. Then
is non-empty and
Let z = (z f ) be in Z and let hiE0(Hi) be such that md(ht) = zt.
*• °
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GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
belongs to
i =1
j
and all the GL(Ht) are connected, (see [20]), the group P is connected. Next we show that each element of GL{£}S(H) can be joined by a continuous path to an element of P. For an element g = up in Q = U_P it is clear how to proceed: the map fi— »{Id + (1 — t)(u — Id)}/? joins g with p. A general element g is first joined with an element in O(l). For, if g11 is not invertible, then there is a bijection E between ker^u) and 3 ( g l l ) ± { ] H l and we extend E by zero on kerfef^) 1 to get an element E of ^ res (H). It is no restriction to assume \\E\\2 < \\g\\2- Then we know that g + tE belongs to GL(£S(H) for all fe[0, 1] and by construction g + E belongs to Q(\) and we can write g + E = u1p1. The map n—>{Id -f (1 — t)(u1 — Id)}pi joins g 4- E with p1. By adding a small finite dimensional operator in &(H2), one reduces the case to an element in Q(2) that can be linked in the same way to an element of P(2). Continuing in this fashion one finds a continuous path from g to an element of P. This proves the assertion. Q This Proposition is the extension to flag varieties of Proposition 6.2.4 in [23] . Since the parabolic group P is connected, we see that Corollary 2.3.2.
The connected components of 5 are given by
Remark 2.3.3. A holomorphic line bundle L over g consists simply of a collection of holomorphic line bundles {Lz -> g (2) !zeZ}. Therefore we restrict our attention to holomorphic line bundles over g(0) in the third section. 2.4. A special covering of g. In this subsection we choose a suitable orthonormal basis of H and we introduce a collection of charts of g that can be described completely in terms of the index set of this orthonormal basis. In particular these charts cover g and enable you to give a combinatorial description of the Birkhoff decomposition of GL res (/f) and to construct concretely a collection of holomorphic line bundles over g(0). Let {eJseSj, 1 < / < m, be an orthonormal basis of Ht. Recall that dim (//,-) = mt for all i, 1 < i < m. Hence we can write
THE STRUCTURE OF HILBERT FLAG VARIETIES
413
On Si we define a total order by Si(fc)>Si(0 * = * * > / .
m
These orders we compose to a total order on the index set S = (J St by l=1 requiring that Sj < st for all Si e St and all Sj e Sj with j > i. Now that we have an orthonormal basis {es|seS} of H with a totally ordered index set S, we can associate to each bounded g in @(H) an S x S-matrix [0] — (9st) with matrix coefficients s>,
where s and t are in S.
Notation 2.4.1. Let gl(S) be the collection of S x S -matrices corresponding to operators in \$res(H). The context has been chosen such that the product of two elements in gl(S) is again in gl(S) and therefore gl(5) is a Lie algebra. In gl(S) we have the Lie subalgebra gl(oo) corresponding to the matrices of the operators Definition 2.4.2. An operator g in <%(H) is called a "finite-size" operator if it has only a finite number of non-zero matrix coefficients w.r.t. the {es|seS}. Remark 2.4.3. If m = 2 and ml = m2 = oo, then S = Z. In [18] it was shown that the Lie algebra A^ can be realized as a central extension of the collection gl(oo) of Z x Z-matrices g = (gi]] of "finite- width", i.e. satisfying gtj = 0 if |i — j\ > N for some N. The composition of such matrices is always defined and from this point of view §l(S) can be seen as a complete bounded version of gl(oo). The central extension defining A^ also occurs naturally in our geometric framework, see subsection 4 of the next section. In the sequel we will frequently use some notations related to subsets of S. Notation 2.4.4. The number of elements in a subset A of S is denoted by #A. Notation 2.4.5. If A is a non-empty subset of S, then we denote the closure of the span of the {es\seA} by HA. If A is empty, then HA denotes the space (0). It is convenient to denote the orthogonal projection onto HA by pA. Maps between subsets of S have a direct translation to partial isometries between closed subspaces of H, i.e. Notation 2.4.6. If A and B are subsets of S and i : A -> B is a map with uniformly bounded finite fibers, then we denote by i_ the mapping from HA to HB given by
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GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
Now that we have chosen the orthonormal basis {eJseS}, we can introduce "diagonal operators" in ^ res (//). Suppose that we have a set of bounded complex numbers {<5s|seS, <5 s eC and |<5 S |
seS}.
Then we can associate to it a diagonal operator diag (ds) in ^(H) by
feS
reS
Inside GL res (//) we have then the "maximal torus" T={g\geGLTes(H), g = diag(c5s)}. Clearly T is commutative and it is a straightforward verification to show that the centralizer of T inside GL(H) is equal to T. Hence we have Lemma 2.4.7.
The centralizer Z(T) of T in GLTes(H) is equal to T.
Each permutation a of S determines a unitary map g_ : H -> H as in notation 2.4.6. With the help of the matrix, one shows that the normalizer of T in GL(H) consists of {t- q_\teT, a a permutation of 8}. Hence, if we define the subgroup W of C/res (H) as W= { g_ | g_ E Ures(H), a a permutation of S}, then we have Corollary 2.4.8. The normalizer N(T) of T in GLTes(H) is the semi-direct product of W and T. In particular, we see that W is isomorphic to N ( T ) / Z ( T ) and we call W the Weyl group of T. To each
o_ in W9 corresponds a partition Z = \J It- of S, where i>l
Zf = cr(S'j). The concrete description of which partitions occur in this way, brings one in a natural way to the consideration of subsets of S that are "equal up to a finite set". Therefore we define Definition 2.4.9. If A and B are subsets of 5, then we call A and B commensurable (notation A % B) if A — {A flB} and B — {A f}B} are finite. We write i(A, B) for the number
THE STRUCTURE OF HILBERT FLAG VARIETIES
415
#{A-{AnB}}-#{B-{A{]B}}. Thus commensurability is equivalent to: the orthogonal projection pHs: HA -+ HB is a Fredholm operator with index i(A, B). Let I = {IJ1 < i < m] be an arbitrary partition of S into m disjoint parts. Then this partition corresponds to an element of W, if and only if the following two conditions hold : (5)
#I£ = #St
(6)
If % Si
for all i, 1 < i < m, and for all i, 1 < i < m
To any partition E satisfying these conditions there corresponds a flag Fs in 5 given by
For simplicity we denote for each I satisfying (5) and (6) the open set L/Fl in g also by L/z. In the sequel we will make use of the following notion Definition 2.4.10. An element in H is said to be of order s, seS, if it has the form h = ases + £ atet, with as ^ 0, teS t
Notation 2.4.11. If W is a subspace of H then the union of all the elements in W of some order s in S and {0} is called the space of elements of finite order in W and is denoted by Wfin. For each z in Z, we denote the collection of partitions S of the index set S such that FE belongs to g (z) , by ^(z). The basic property of the is Proposition 2.4.12. For eac/z ^fag -F — C F O ) > - - - > ^ ( m ) ) z« ^(z) such that F is transversal to FE.
z/7
3(z) there is a £
Proof. Let #eGL r e s (#) be such that F = g - F ( 0 ) . First we show that there is a St commensurable with S1 and with #SX = #I 1? such that p H T l ° g \ H l is an isomorphism between #! and H Zl . Since pi^CHJ) has finite codimension in / f l 5 we can find a S^w) = (s^fc), fc>rc}, n > 0, such that F(l) = g(H1) projects surjectively onto HSl(n). The kernel of this projection has a basis {hj 1
j = e*j + Z a j(0^, where st + Sj for / /j. teS t
It is clear that we can take Ex = S^nju [Sj\ 1
The other parts of
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GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
the desired partition £ of S are constructed step by step from I,1. For, assume that we have found disjoint {£^7 < i} with I7- « Sj and #Sj = #Ij such that the orthogonal projection of F(j) onto ®^
a dense subspace of Fz (7) for all 7, 1 < 7 < m. By combining this with Proposition 2.4.12 we get the following generalization of proposition 7.3.2 in [23]. Corollary 2.4.14. For each flag F in \$ and for each 7, 1
I(F), = I(F)(/) - Z(F)(i - 1) for / > 1. Clearly each F(i) projects bijectively onto HZ(Fm and therefore I(F) = {S(F)J belongs to ^(z). Next we consider flags that give the same partition in this way. If Ie^(z), then we write
Let L/o be the subgroup of GL res (H) of all operators with a unipotent lower triangular matrix, i.e. L/o = {u\ueGLres(H), for all ssS, u(es) = es + ^ Then we want to show
^}-
THE STRUCTURE OF HILBERT FLAG VARIETIES
Proposition 2.4.15.
417
The subset gL is exactly the U0-orbit through Fz.
Proof. From the form of an operator u in (70, one sees directly that for each element h of order s, the element u(h) has also order s. Thus we have that I(wF) = L(F) for each F in I. Assume now that F belongs to gz. Since F(i) projects bijectively onto Fz(i) for all i, 1 < i < m, the flag F is the graph of an operator T in m-l xix
vi/
t^T> LX \±J. Y 5 -O v J .
j=l J<'
In particular this means that there is an u in GL res (#) such that w(Fz) = F and for all s f £Z £ , 1 < f < m, (7)
w(eSi) - esi + £ I
rSjSieSj, if i < m and w(e Si ) - eSi if z - m.
7 > i Sjelj
The fact that F belongs to gs can be expressed completely in term of the coefficients {7^ S i |s f 6Z i 9 s^-eZy, 7' > /}. Namely, it is equivalent to (8)
TSjSi - 0, if s7. > s,..
If namely 7^Si / 0 for some Sj with Sj > si9 then the element u(es) will some order s^S(i) and hence Z(F)(i) 7^ S(i). This contradicts the fact belongs to gy. By definition, the operator u defined by (7) belongs to condition (8) is satisfied. This concludes the proof of the proposition.
be of that F L/ 0 , if G
Thus we have obtained a subdivision of each connected component of g,
%(z}= U Si, Ee^(z)
into parts that are homeomorphic to a Hilbert space, thanks to property (8). This is a generalization to flag varieties of the stratification in section 7.3 of [23]. Let <7S, for each XecS^(z), be a permutation of S such that er s (Sj) = Zi9 for all i, 1 < i < m. Then this decomposition of g translates directly to the group GL res (H) and results in Proposition 2.4.16. (Birkhoff decomposition.} the group GL res (/f) decomposes as
Each connected component of
Remark 2.4.17. The decomposition derived here is the analytic equivalent of the algebraic decomposition from [22].
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GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
Holomorphic Line Bundles over g(0)
§3.
3.1. Another description of g(0). For each Z in «9^(z), there are numerous a_£W such that Fs = I i? as follows: since S{ and Zf are commensurable, there is a N > 0 such that
Consider the finite set ^ — {st(k)\k > N} . If it is empty, then we define Gi : Sj -> Ij as follows : ff£(sf(fc)) = st-(Jc + N)
for all fe, 1 < k < mt + 1.
In this case we put I \ = — N. If Zt- is not equal to {st(k)\k > AT}, then we write ^ — {st(k)\k > N} = {ti,... fjv+*fj f°r some /; in Z, ^ > — JV, and we define
Sf (fe
for all fc, 1 < k < N 4- /f, - /<)
for all fc > N + ^.
For all i, 1 < f < m, consider the map it: St -> Sf defined by T .( s .(/c))
- s f (N + 1) Si)
for all fc, 1 < k < N + fi for all fc > N + ^.
Since Pi° o_i~ ^_i is a finite dimensional operator, Pi° a_i and jr t- have the same index and for i_t one clearly has ind (i_t) = N + /f — N = I {.
In other
words, the number ^ is equal to zf. The {a{ I < i < m} compose to a bijection a: S -> S such that o: F (0) = FE. We will introduce a special term for this type of permutations. Definition 3.1.1. Let £ be a partition of S in &*(z). A permutation a of 5 such that cr(Sj) = Z^ is called admissible of level N if the following property holds (i) For each i and for all k > N + zf, t7(s f (fc)) = s £ (fc - zt-). One easily verifies that the collection of admissible permutations of S of all levels forms a subgroup Wa of W. It has a normal subgroup Wa(0) = Wa nGL(r0e}s(H) such that the quotient WJW™ is isomorphic to Z. The elements of W^0) can be described in a direct concrete way. If G is a finite subset of S and if p is a permutation of G, then we denote the extension of p by the identity to a permutation of S, by p. Then we have
THE STRUCTURE OF HILBERT FLAG VARIETIES
419
Wa{0} = lim ^a(0)(AO = lim> { a a 6 Wa(0) is of level N] N *
N
= [p_\p a permutation of G, G a finite subset of 5). Now we can introduce another group that acts transitively on 5(0). Its advantage is that it enables you to construct in a simple way holomorphic line bundles over g (0) . Let I be in c9*(0) and let o be an admissible permutation of S such that o-(S^) = Sf. From the definition of admissibility we know that (7 decomposes in operators (cr^-) with the properties (i)
For each 1 < i < m, (j £i = IdHi + a "finite-size" operator.
(ii)
For all i and j, i^j,
cr^- is a "finite-size" operator.
(0)
Since every flag F in g is transversal to some Fls with I in ^(0), we may conclude that each F in g (0) is equal to g • F(0) with g E GLres (H) of the form (a) For each i, 1 < f < m, 0ff = IdHi + a "finite-size" operator. (b) For all i and j, i < j, 0 fj - is a "finite-size" operator. (c) For all i and ;,; < i, 0y belongs to 3e^(R-p Ht). Note that for all the operators gu from (a) det (ga) is defined. Since we are working in an analytic setting we will consider a somewhat wider class of operators such that on one hand we work in a Banach framework and on the other we can take determinants of certain minors. Recall, see [12], that the determinant is defined for each operator of the form "identity + a nuclear operator". Therefore we introduce
On B2(H) we put a different topology than the one induced by ^ res (//). For, let & be the subspace of ^ res (#) defined by
Then J^ is a Banach space if we equip it with the norm \\ • \\£ given by
i*j
i=l
The collection B2(H) is nothing but ^ shifted by the identity and we transfer the Banach structure on 3£ to B2(H) by means of the map Q\-*Q + Id. Since the product of two Hilbert-Schmidt operators is nuclear, one sees that B2(H) is closed under multiplication. Moreover the multiplication with an element of B2(H) is an analytic map from B2(H) to itself. In B2(H) we have the subgroup 17 _ and its "adjoint" the group
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GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
U+ =
Consider an element b in B2(H). in U+ by Uii
{U*\UEU_}.
Now we define u = (wl7) in C/_ and v = (uy)
= vu = Id Hi , MO. = - by if i > ;, My = 0 if ; > i,
i>y = — by if f < j and 0y = 0 if / > 7.
A direct verification shows that ubv belongs to Id + Jfr(H). closed w.r.t. taking adjoints, we have
Since B2(H) is
Lemma 3.1.2. Every beB2(H) can be written in the form b = u1blv1 or b = v2b2u2, where u1 and u2 belong to 17 _, v2 and v1 belong to U+ and b1 and b2 lie in Id + The decompositions in lemma 3.1.2 are clearly not unique, but they suffice to define a determinant map det: B2(H) -> C. Namely, for b = u1b1v1 as in lemma 3.1.2, we put det (b) = det (u1b1vi) = det (b^. To see that this is well-defined, we note first of all that for w e [ / _ n {Id + ^(H)} and ve U+ n (Id + JV(H)} we have det (u) = det (t;) - 1, since u — Id and v — Id have zero trace. Now, assume beB2(H) can be written as b = ulb1v1 = U2b2v2 with b^eld + J^(H], uteU_ and vieU+. Then b2 = (u21ul)bl(v1v21) and, since both bl and b2 belong to Id 4- ^(H)9 this implies that w 2 - 1 w 1 e[/_n{Id + tyr(/f)} and v^v^e U+ n{Id + JT(H)}. By the multiplicativity of the determinant on Id + ^(H), we get det (b 2 ) = det (M^" I M I ) det (bj det (v^ v2 x) = Remark 3.1.3. Since the operators in Id + Jf(H) lie dense in B2(H) and since det is multiplicative on Id + Jf(H) we get that for each b1 and b2 in B2(H) det(b 1 b 2 ) = det(b 1 )det(b 2 ) From the fact that an operator g of the form Id + ^(H) is invertible if and only if det (g) is non-zero, we see that the invertible elements of B2(H) form a group (5 and are given by © = (b|be£ 2 (H), d e t ( b ) ^ O ) . Clearly (5 is a Banach Lie group with Lie algebra ^ and it acts analytically and transitively on g (0) . The stabilizer of F (0) in (5 has the form
I til • t=
^lm \
0
\ o . • o tmm I
THE STRUCTURE OF HILBERT FLAG VARIETIES
421
Thus we can identity g(0) with the homogeneous space Remark 3.1.4. If we would work with (S(^r) instead of g, then we could simply take instead of B2(H) the collection Id + ^(H) and instead of (6 the group of invertible operators of the form Id + J^(H). If one likes to work with g(^) then the group <, =
b | b = (fey) 6 GL(H), bit e Id +
acts transitively on the connected component of g(#) containing the basic flag and allows you to take determinants of suitable minors. 3.1.5. Next we consider the maximal torus T(J^)= Tn© in (5. It consists of all operators of the form diag({l + tj), with 1 + ts ^ 0 and ]T \ts\ < oo. In T(Jf)
we have the dense subgroup 7} given by
7} = {t|t - diag({l + t s })eT(JO, ts / 0 for only finitely many s in S} Any analytic group homomorphism of 7} into C* has the form t = diag ((1 + Q) K-> seS
where m = (mj, with m s eZ for all seS. This character %m can be continued to an analytic character of T(JT) if and only if there are only finitely many different ms, se5. This extension of %m is also denoted by %m and we write T for the group of analytic characters of T(>"). Following the finite dimensional terminology, we will speak, when T(Jf} acts on a vector according to a %ef, of "i; is a vector of weight /". For each s and r in S, let £sr be the operator in ^ res (H) given by Esr(et) = drles
for all
The adjoint action of T(J^) on ^ res (H) gives for these elements
F —7 (t)E
J^sr — • /CsrV 1 / ^
sr
1 + tr
A character % of T(J^) is called positive, notation / > 0, if it belongs to the semigroup generated by the {/ sr |seS i5 r e S j y i>j}. This enables you to define a partial order on T by
Clearly we call ^ in T negative if and only if x"1 > 0.
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GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
Remark 3.1.6. One can see the space g(0) also as a homogeneous space for the group of invertible operators in "Id + JJ? 5f(H)", which in its turn can be identified with an open part of the Hilbert-Schmidt operators on H . This group, however, does not permit you to take suitable minors. Remark 3.1.7. One can give the same type of description for the other components g(z) by taking some le^z) and by introducing the group (5 as the operators that decompose w.r.t. H = H^ © ••• © /flm in the above way. 3.2. Finite dimensional subvarieties. In this subsection we consider some finite dimensional flag subvarieties contained in g(0). Let K be a finite subset of S. For simplicity we assume that K contains all the S^ that are finite. The general linear group GL(HK) embeds into GL(^S(H) by extending ueGL(HK] on HS_K by the identity. We write gK for the sub variety of g(0) given by ueGL(HK)} = {uF^
ueU(HK)}.
If K1 c K2, then we have a natural embedding of GL(HKl) into GL(HK2) and of \$Kl in g K2 . Now one considers a collection of finite subsets {KJweN} of S given by m
Kn = U {Si(k) | k < n + max (m^)}. m
j
• — -I
<
°O
Then S = (j Kn and with the identifications mentioned above we write «eN
GL(oo) - U GL(HKn), t/(oo) = (J U(HKn) and 5(oo) - U SxnneN
neN
neN
Since, for all i / j, the "finite-size" operators are dense in J^^(Hj, Ht), and the C/ z , Ie^(0), cover g (0) , we get Lemma 3.2.1.
The space g(oo)
Consider now a holomorphic function / on g (0) . The restriction of/ to some \$Kn must be a constant, since 5Kn is a compact complex manifold. Hence / is a constant on 3f(oo) and the lemma implies then Corollary 3.2.2. constants.
The only holomorphic functions from 3(0) to C are the
This is a generalization for flag varieties of Proposition 7.2.2 in [23]. Remark 3.2.3. The results of this subsection remain true if one would work with the nuclear flag space 3f(«^O or tne "compact" flag variety 5(#). However- the "finite-size" flags from g(oo) are not lying dense in the space of bounded flags
THE STRUCTURE OF HILBERT FLAG VARIETIES
3.3.
The line bundles and the central extension.
423
For each k_ = ( f c l 5 . . . , f c m ) in
m
Z , we define ^ in T by
i +t,})= n (i + U*1 n (! + u tj - n (! + 'j*m sieSi
S2eS2
smeSm
Clearly i/^ extends to an analytic character of ?T by means of the formula
According to section 6.5 in [4], there exists for each analytic character i/^ of y, a holomorphic line bundle L(.fc) = © (g^C over J^(0) - ©/^. It is concretely defined as follows: consider on the space © x C the equivalence relation (01, ^ l ) ^ ( # 2 > >l'2) < = > 01 = 0 2 ° * ,
The space © x C modulo this equivalence relation is L(jc). For each and each A in C, we denote the equivalence class to which the pair (g, A) belongs by [g, X]. There is a natural projection n^\ L(A)-^5 ( 0 ) given by
The space L(_fe) is a Hilbert manifold based on the Hilbert space £ © C.
For
each 1 6^(0) one can give a concrete trivialization of L(k.) above t/s. Let a be an admissible permutation of S such, that Lf = cr(Sj). Then we define ^n^(U^ by
Assume we have a S and ^ in ^(0) such that 7c^" 1 (C/ 2 )n7c^" 1 (L/'u) is non-empty. Let a and p be admissible permutations of S with a (5^) = Zf and p(S f ) = K^. If (A, X) is such that
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GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
Remark 3.3.2. For the case m = 2, ml = m2 = oo, the bundles L((+ 1, 0)) and L((— 1, 0)) are the determinant bundle Det and its dual Det* as introduced in section 7.7 of [23]. There is a natural analytic action of the group © on the space L(_fc) by left translations 01 '[02»
^1 = [0102. ^1-
This is a lifting of the natural action of © on g(0) to one on L(_fc ). However, the natural action of GL(£)S(H) can, in general, not be lifted to one on L(lc). Such an attempt may lead to nontrivial central extensions of as we will show. Note that each g in GL(T°el(H) can be written as g = dg2, with # 2 E ® d belonging to the subgroup
f), gtj = 0 if i *j}. of GL(^S(H). Clearly the group D normalizes the group ©. Since the determinant of an operator of the form "identity 4- nuclear" is invariant under conjugation with an invertible operator, we get that D centralizes each i/^, i.e. for each t in ZT and each d in D we have
This fact permits you to lift the action of D on 5(0) to one on L ( k ] by means of
For an element d from D n © , this action differs by a factor \l/k(d~1} from the action induced by that of ©. Hence we cannot combine them to an action of GL(^S(H) on 5(0)- To overcome this problem we build a group extension G of GL(r°el(H). It is defined by G = {(g9d)\geGLW(H)9
deD and
As one verifies directly this group acts on L(J^) by means of
It is simply the combination of the ©-action and the D-action given above. Let 7i : G -> GL(°e}s(H) be the canonical projection, i.e. n((g, d)) = g for all (g, d)eG. For certain subgroups of GL{^S (H) there exist several ways to embed them into G. Therefore we introduce special notations for two of them. Let J_ resp.
THE STRUCTURE OF HILBERT FLAG VARIETIES
425
j_ be the embedding of (B resp. D into G given by
1(0) = (g, Id) and j_(d) = (d, d). As a group G is the semi-direct product of J_((5) and j_(D). We equip each GL(Hi) with the operator norm topology and we put on j_(D) the product Banach Lie group structure. On _L(@) we take the Banach structure based on 3£. The conjugation with an element d of D defines an analytic diffeomorphism of (5. Hence if we put on G the product topology of j_((5) and j_(D), it becomes a Banach Lie group based on
The group G is a fiber bundle over GL(r°}s(//), with fiber &~ftD. This is clear from the following useful trivializations. For each I = (Z f ) in <9*(0) consider the open set G(S) of G given by G(I) = {(g, d)\(g, d)eG, p Z i o ^|//i is a bijection for all i, 1 < i < m}. The group G is the union of these open sets. If a is an admissible permutation with a(Si) = I.i9 then we define an analytic bijection from 7r(G(Z)) x {^T nD} to G(Z) by
where 2 in D is determined by (9)
d a = a- 1 op^l Jff,.
Next we try to minimalize the extension of GL(£*S(H) that acts on g(0) and L(_/c). Thereto we consider the action of the kernel of n on L(k_) (Id, d) - [g, A] = [gd- 1 , A] = [gf, ^(d-^A]. In particular the group D ( / c ) = {(Id, d)|(Id, d ) e G and i//k(d) = 1} acts trivially on L(/c ) and we see that it suffices to consider the extension G(k_) = G/D(k_) of GL(T°el(H). If the character i/^ is trivial, i.e. fe. = 0, then G(k_) is just GL(^S(H). For Jc / 0, one computes directly that G(k_ ) is a central extension of GL(?e]s(H) with Ker(;c)/D(A) ^ C*. For m = 2, ml = m2 = oo, the extension G((— 1, 0)) is the one introduced in section 6.6 of [23]. One can describe such an extension with a Borel 2-cocycle a: GL(?JS(H) x GL(r°el(H) -» C*. It can be constructed as follows: take a section p of the fiber bundle G^GL(r°el(H), i.e. for each g in GL(%S(H) we have
426
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
with
By definition we have for each g1 and g2 in GL(r°}s(/f) that
Thus we get for the action on L(k_) the relation
The group G ( / c ) is then isomorphic as a group to the product space GL\$S(H) x C* with the multiplication (01,
AJ* (0 2 ,
A2) -
(0!0 2 , A^Of^!,
0 2 )).
If p is another section of G A GL(?e{(H) with p(0) = (0, 0(0)), then we have by definition for each 0 in GL{T°JS(H) that £(0) - q(g)t(g) with t(0) in ^"nD. The corresponding 2-cocycle a satisfies
In other words, it differs by a trivial 2-cocycle and we merely have to consider one section p. A section p can be composed from the local trivializations of n : G -» ( GL r°el(H) defined above. First we number the elements of ^(0): ^(0) = {Z(0| i > 0}, such that S(0) is the partition corresponding to the basic flag. For #eG(I (0) ) we choose 0(0) according to the trivialization (9) with a = Id. Next m
m— 1
i=0
i= 0
we define 0(0) inductively by: if 0 belongs to U G(E(0) and not to (J G(I(t)), then we take 0(0) according to the trivialization of G(Z(m)) given by (9). In particular if 0, h and gh belong to G(S(0)), then the 2-cocycle a is given by (10)
a(0, h ) = f [ det (Id + X gifoh^g^. i =1
./ * i
From this formula we will compute in the next subsection the corresponding Lie algebra 2-cocycle. 3.4. The non-triviality of the extension
THE STRUCTURE OF HILBERT FLAG VARIETIES
427
Hence for elements d e D n © , we get d*[x, A] - \ljh(d-l}[_dxd-ld,
A] =
Now we combine this new ©-action with that of D and we define for g = dlgl in GL™8(H), where dl eD and 0 - 1 e®,
It is a straightforward verification to show that this is well-defined and that it defines an action of GLf^s(H) on L(k_). This implies that G ( / c ) is a trivial central extension of GL(^S(H). Secondly we consider the case where at most one of the mf is infinite. Then GL(^S(H) is simply GL(H) and we know from [20] that this group is contractible. In particular the fiber bundle G A GL(H) is then topologically trivial and the group G(k_) is the direct product of GL(H) and Ker (n)/D(k_). Hence we may assume in the sequel that there are at least two infinite w f 's. The next case we have a look at is that k_ satisfies (11)
fc.
/0=>mi
Let 0h->(0, q(g)) be a section of G -» GL(r°}s (H) . We will adjust q(g) such that the 2-cocycle determining G(k_) becomes trivial. Namely, we define q(g) in D by
9(9)u = q(g)u
(12]
if mt = co and
.., if m. < oo. Then g*-+(g,q(g)) defines another section p of G -> GL(^]S(H) corresponding 2-cocycle a is trivial
and
the
In the cases considered so far we have seen that the fiber bundle G A GL(r°}s (H) is trivial and hence also the central extension G ( / c ) of GL(£l(H). We will show now that the extension G ( j c ) can be non-trivial. Note that the 2-cocycle a is given close to the identity by an analytic expression. Hence we can consider the corresponding Lie algebra 2-cocycle doc. We consider the elements of &res(H) as left invariant vector field on GL(£8(H). Then rfa: @res(H) x J" res (H)-»C is given by =
dt ds
a(exp (tX), exp (57)) l
~o
d d -— — a(exp(s7), ds dt
exp(tX)) 5= 0
For X = (Xij) in * res (H), we write g = exp (tX) = (g^. With respect to the
428
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
parameter t we have g..
= UHi + tXu + "higher order in t"
g..
= tXtj + "higher order in r" for i /;.
If h = (h^ = exp(sY), then we are interested in the ts-term in det(Id f l i + ZflfyM" 1 ^ 1 ) j*i
= det(IdHi + N)=l+ £ Trace (AkN) k=l
= 1 + ts X Trace (Xyiy + "at least 2nd order in t or s". j*t
By combining this expression with the local formula (10) for a, we obtain the following formula for doc: m
da(X, 7) = £ fc, Trace { £ X^ - £ i=l
j*i
7^,}.
J*i
This Lie algebra cocycle is trivial if it has the form f([_X, 7]) with /: J*res(H) -» C some linear map. The element \_X, 7] in J*res(7f) has the form
L^mm? ^mm] +
Note that if Jc satisfies (11), then we can directly define such an /. For, in that case, we have for all i with kt ^ 0 that Trace \_Xih ya-] is well-defined and equal to zero and we can take f(X) = X *i Trace (^J. i,fc,^0
This is the infinitesimal version of the trivialization described at the beginning of this subsection. There is, however, no well-defined trace function for general elements of ^ res (JFf) so that this formula makes no sense in the general case. Now, let i and j be such that i
if fe> 1, =esi(k) if l*i
and
THE STRUCTURE OF HILBERT FLAG VARIETIES
429
Now we have that da(A, A~1} is equal to fc. - kj = kt Trace A{j(A~\ - kj Trace (AJ^ A^ ^ 0 In particular doc is a non-trivial Lie algebra 2-cocycle. This implies that also the group 2-cocycle a is non-trivial. For, consider the commuting elements g1 = exp(M) and g2 = cxp(sA~l). In case that a was trivial we would have a (0i> #2) = a (02 5 0i)« However, for sufficiently small t and s, the map (t, s)i—»a(0 1 ? g2) is a non-constant holomorphic function, since dx(A, A~l) ^ 0. On the other hand one computes directly that for all i, 1 < / < m,
This shows that a(# l 5 #2) ^ a (02> 0i) an(i hence a is a non-trivial 2-cocycle. We summarize this result in a Theorem 3.4.1. (a) The extension G(kL) is always trivial if there is at most one infinite mt. (b) If there are at least two infinite dimensional components in the basic flag, then G(k] is trivial if and only if for all i and j, mt = rrij = oo => kt = kj. (c)
If kt ^ kj for infinite dimensional Ht and Hjy then the corresponding Lie algebra 2-cocycle for the extension G ( J c ) is given by da(X, Y) = £ k, Trace { £ X^
- ^ ryX7/}.
1=1
Remark 3.4.2. Consider the case m = 2, m1 = m2 = oo, /q = — 1 and k2 = 0 and restrict da to gl (oo) ngl(S). Then we have the 2-cocycle defining the Lie algebra
Ax.
3.5. The holomorphic sections of /,(/[)• Let fi( fe.) denote the space of global holomorphic sections of L(k_). The space &(k_) is given the topology of uniform convergence on compact subsets of g(0). It becomes then a complete locally convex space, see [16]. it can be written as
Let / : g ( 0 ) -»L(^c) belong to fi(k.)- Then
for all where /: © -> C is a holomorphic function satisfying (13)
f(gt)=f(g)\lj]£(tr1
for all ge® and all
430
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
Thus we can identify fi(/c) with the space of holomorphic that satisfy this condition. Since each (g, d) in G acts diffeomorphism on g(0) as well as L(/c), we get a natural £(_Jc) that corresponds on the functions on © satisfying (13)
(g,d)(f)(g1)=f(g-1g1d),
functions on © as an analytic action of G on to
and (g,d)eG.
Let Kn be the finite subset of S introduced in subsection 3.2. By restricting the elements of £(Jc) to GL(HKn) one obtains a space of holomorphic functions on GL(HKn) satisfying (14) tlm 011
'••
v v
din
0
0
o ... o tmm I I
where geGL(HKn), t e GL(HKn) n ^ and the decomposition of 0 and t is w.r.t. m
The Borel-Weil theorem says that such functions ^ 0 exist
if and only if k_ satisfies (15)
k, Zk2-
Since g(oo) is dense in g(0), the restriction of some non-zero / in fi( fc.) must be non-zero for a sufficiently large n. Hence this condition from the finite dimensional situation is also necessary in this Hilbert context. We will show that it is sufficient too. So we assume from now on that jfc e Zm satisfies condition (15). Before we will construct concrete non-zero elements of fi( k.), we will first introduce some basic building blocks. If Z = {Lj} belongs to 5^(0), then we write •^(0 — U £/
and
y^i = {L(f
Let o = a1 © • • • © om be an admissible permutation of S corresponding to S. Consider for g E © the operator (a x © - • • © o_ f )~ 1 o pI(.} ° 01 0 #. from ® Hj j 1 j l to itself. It decomposes as ~
, with hjj - U H j E ^ ( H j ) , hijEJjfytfj,
Ht)
for all j
THE STRUCTURE OF HILBERT FLAG VARIETIES
431
In particular we can take the determinant of this operator. The function /z(0 '• © -» C defined by
satisfies for each t = (^) in 2T the condition /KO toO = /zw to) det (f ! x) • • • det (y
In other words / Z(/) belongs to £((- 1,...,- 1,0,...)). Now we consider the action of T(*V) on such a function / Z(f) . Let 0 be in © and r = diag({l + ts}) in T ( j y ) . By definition f^(i)(t~lg) is equal to the determinant of the operator
reS
reS
rel(i)
cel<0)(j)
Hence each /z(i) is an eigenvector for the T(~V)-action and we have f-/L(0 = Jl(o(l+W))~1/J:«i,.
If we define the character \l/t of T(Jf)
by
^.(t) = Y[ (i + O'1. then we get in general
n (i + o- n
(c)£l( 0 MO
cel<°>(0
Because ^ has index zero, the products in the right hand side are over the same number of elements. Since we have by definition, for each c l E l . ( 0 ) ( i ) with ej-^cJ^S^O') and for each c 2 6l ( 0 ) (i) with (i(c 2 )^I (0) (0, that iCia(C2) < 0, it is clear that the weight of/ Z ( 0 is less then or equal to i//^ In other words, among the weights of the {/Z(l-)s 2(0 e^-}, *Ai is maximal. Note that condition (15) on k_ allows you to decompose i/^ as follows
det(f ( ,)}
432
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
i
Now we choose for each j, 1 < j < m — I a non-zero homogeneous polynomial Pj in the {f^(j) \ !(;) e 5^} of degree kj+ 1 - kj. Let Pm : (5 -> C* be the function
From the foregoing formulae it will be clear that P = Yl PI *s
a
non-zero
i=l
holomorphic function on (5 that belongs to £( /c). If we consider the special m
choice P=Y\P.
then the TC^^action on this element of fi( fe.) is given by the character
i=l
Let £y(/c) be the span of the functions P described above. From the T(Jf)-action on the {/E(j)} one concludes that this is the highest weight occurring in £/(_&)• If H is finite dimensional then it is known that £/(A) = £(&.)• By using this and the fact that 5(oo) is dense in g (0) , we get Theorem 3.5.1. (a) The space fi(/c) is non-zero if and only if fcx < ••• < km. (b) The sub space £/(A) lies dense in fi( fe.). Next we consider the representation of G on &(k_). Let F be a closed subspace of £/( fc.) an(l let ^ be a non-zero element of V. Then there is an n such that the restriction of v to GL(HKn) is non-zero. Since the representation of U(HKn) on the holomorphic functions on GL(HKn) satisfying (14) is irreducible, we get
So, if we define W as the closure of the span of the {u- v\ueU(co)} then we have for each v1 in V a sequence {wj in W such that vl GL(HKn) = wj GL(HKn). Since U(co) • F(0) is dense in g(0), this implies that {wj converges to yt and we get that V= W. Hence we can say Theorem 3.5.2. Let _fc = (fc l 5 ...,A; m )6Z m satisfy kl < ••• < km.
The repre-
THE STRUCTURE OF HILBERT FLAG VARIETIES
433
sentation of (5 (and hence of G) on £( kj is topologically irreducible. This Theorem is a generalization of Theorem 10.4.6 in [23]. For each neN, we know that the representation of GL(HKn) on {f\GL(HKn)\fe£(k_)} has up to a constant a unique vector that is w.r.t. the {T(^V)r\GL(HKn)} action of highest weight
In view of the foregoing results we may conclude now for the space S.(k_): Theorem 3.5.3.
All T(
The vector P spans the subspace of vectors with T(^")-weight i/^. In view of the results in the Theorems 3.5.2 and 3.5.3, one could call the characters i/^ satisfying (15) dominant.
§4.
Applications to Integrable Systems
4.1. A group of commuting flows. In the first subsection we discuss the flows that form the basis of the equations of the multicomponent KP-hierarchy and of the modified equations. Let H be the Hilbert space L2(S19Cr) with the usual norm. Let { f f \ \ < f < r } be the standard basis of C r . Then the elements of H can be written as Z Z «ifc/k;^
with
ieZ fc=l
The space H is decomposed as H = Hl®H2 with
and The elements {fk/,l\ 1 < fe < r, i > 0} are an orthonormal basis of Hl and the {/feA'li < 0, 1 < k < r} one of H2. To get a numbering like in the foregoing section, one defines efc + ir-i = f k ^ 1 - ^n tne present context it is also convenient to see the matrix [0] of an operator g in ^ res (H) as an Z x Z-matrix with entries in g!r(C), i.e.
434
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
with G sf egUQ. An important operator in &Tes(H) is the multiplication A on each factor with L It has the matrix [A] with Aii_i = Id and Atj = 0, if j =£ i — I. One verifies directly that the centralizer Z(A) of A in J*res(/f) consists of all g in &Tes(H) such that the matrix of g looks like
^-10
^00
10
\
•••
/
Clearly, multiplying with an A from gIr(C), defines an element of ^ res (ff). I) be the diagonal matrices in g!r(C). It is obvious then that { Z #;^ Ie ^res(£0> Hi^fy ieZ
f°r
a
^
Let
0
is a maximal commutative subalgebra of \$Tes(H). The group of commuting flows that we will consider is contained in this algebra and takes care of essentially all independent directions. To be more precise, let U be a connected neighborhood of S1 in C and let F(U) be the space of all analytic maps 7: [/->!) such that d e t ( 7 ( w ) ) / 0 for all ueU. In a natural way F(U) is a group. If U1 =) U2 then we get an embedding of FCU^) into F(U2) by restricting functions to U2. We write F for the inverse limit of the {F(U)}. Each has a Fourier series 7 = Z 7 f >J ieZ
The multiplication with 7, defines the element ^y^A1 in 3\$res(H).
Let £a,
IEZ
1 < a < r, be the diagonal matrix in g!r(C) with (a, a)-entry equal to 1 and the other entries equal to zero. At the consideration of the flows from F on g we make use of a decomposition of the elements of F. In F we consider namely the following subgroups
and
,;>) with kteZ for all i}.
THE STRUCTURE OF HILBERT FLAG VARIETIES
435
Then there holds Lemma 4.1.1. The group F decomposes as F = F+AF_. This lemma is a direct consequence of the decomposition of holomorphic line bundles over P^C), see [11]. As we will see in the third subsection the flows from F_ do not contribute to the system. Hence there is no need for a description in coordinates for elements from F_ . 4.2. The multicomponent KP-hierarchy. We present here a formal algebraic set-up of this system of equations in which the formulae from the appendix of [27] make sense. It offers one also the possibility to consider these equations from an algebraic point of view. Let R be a complex commutative differential algebra with a collection of commuting derivations [di(X\ i > 1, 1 < a < r} of R. In the geometric picture R will be an algebra of meromorphic functions on F+ and diaL will be taking the partial derivative w.r.t. tioc. Let d be the derivation r
£ 3 la . The equations of the hierarchy can be formulated conveniently in «= i terms of relations for certain elements from the ring Qlr(R)((d, d'1)) of pseudo differential operators in d with coefficients from g!r(K). We extend the dia to derivations of Qlr(R)((d, d~1)) by letting it act coefficient wise on elements of gl r (K) and on an element £ p.tf of gHr(R)((d, d' 1 )) by J
di«(Pj)dJ
3J Z P>#) = I
In the ring Qlr(R)((d, d'1)) we denote the differential operator part £ Pjdj of j>o P = ^ P j d j by P+ and we write P_ for P-P+. Let £a, 1 < a < r, be as in j the foregoing subsection. In the ring §lr(R)((d, d'1)) we consider elements of the form (16)
L= 3 + X fjd-J
and V, = Ex + £ uxjd''
j >0
j >0
Examples of this type of operators can be obtained as follows : take the trivial example L= Id d = d and t/a = Ea and choose some K = Id 4- £ fc/3"-7' in ' 1 )). Such a X is invertible and (17)
L=KdK~l
and
J>
°
Utt = KE^K'1
have the form (16). Following [27], the equations of the multicomponent KP-hierarchy are
436
(18)
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
[L, t/J = [t/a, E/,] - 0
(19) (20) (21)
U.U^S.fUf 3^) = [(^U a ) + ,L] = [5£a,L] W,)=[B fa , t/,].
The equations (18) and (19) are satisfied by all elements L and {Ua} of the form (17). The equations (20) and (21) boil down to non-linear differential equations for the {uaj} and {/,}. Since all the solutions of the multicomponent KP-hierarchy that we will construct are of the form (17), we merely have to focus on (20) and (21). These last equations can be seen as compatibility conditions for a linear system. This requires the introduction of a gl r (K)((d, d" 1 )module. Let M consist of the formal products
with f}jEQlr(R).
For /?egl r (-R), the action of ft on M is defined by
j = - oo
j = - oo
The action of 3ia on M is defined such that it corresponds to "differentiating this formal product w.r.t. the variable ti
In particular we see that the action of d on M
d{W}g(z> = ZtfW is invertible with the inverse 5"1 given by
These actions compose to a g!r(jR)((d, S'^J-module structure on M. In fact, M is a free \$lr(R) ((8, d~ ^-module with generator g(X), for, if P = ItPjdJE Qlr(R)((d, d~~1)), then a direct computation shows
Let A be the subgroup of g!r(C(A, A" 1 )) given by (A k1 ,...,^) with (fc 1 ,...,fc r )eZ r }.
THE STRUCTURE OF HILBERT FLAG VARIETIES
437
Take any 5 = E^-A-7, ^-egI r (C), in A. To 6 corresponds the element d_ = ZdjdJ in glr(K)((3, d'1)). Then we have the notion Definition 4.2.1. A function of type d is an element of ij/ of M that has the form
To any function ^ of type d we associate the operator K^ in glr(^)((3, 5" 1 )) given by K, = Id
Next we assume that we have been given operators L and {[/J of the form (17). Then we introduce the following notion: Definition 4.2.2. A wavefunction of type d for Land the {[/J is a function if/ of type d satisfying (a) LW) = W (b) t/> = ^£a (c) 3...WO = Pfa • tfr with PfcGgU*)!?]. The first two properties translate respectively into L=K^dK^1
(22)
and
U« = K^EaK^.
Hence Land the {Ua} are completely determined by \l/. One computes directly that (c) implies Pfa = (LlUa)+ and by applying the operators dia to the equations (a) and (b) and by substituting (c) one shows Theorem 4.2.3. If if/ is a wavefunction of type d, then the operators K^dK^1 and {K^EaK^1} satisfy the equations of the multicomponent KP-hierarchy. The equations from definition 4.2.2 are called a linearization of the system and from theorem 4.2.3 we see that we merely have to show (c) if L and the {l/J are defined by (22). 4.3. The solutions. First we consider the space H and its decomposition as in subsection 4.1. Since m = 2, all flags in 5 correspond to subspaces W of H. For each W in 5> consider Aw = {6\SeA, there is a yeF+ such that y~16~1W is transversal to Hj The first property of Aw is Lemma 4.3.1.
The collection Aw is non-empty.
438
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
For each 6 in Aw we consider the open subset F(d, W) of F+ given by
r(6, W) = {y eF^7 y-16'lW is transversal to Hj. Let R be the ring of analytic functions on F(d, W) and let diaL be the derivation of R consisting of partial differentiation w.r.t. tigL. Then there holds Theorem 4.3.2. (a) For each WE\$ and each 5eAw there is a unique function \l/w \$ d w - d - g ( l ) of type d, such that \l/dw(y)eW for all yeF(d, W). (b) The function \jjdw from (a) is a wavefunction of type 6.
=
For a proof, we refer to [13]. If we write \j/dw = Kdw • d • g(X), then we know from theorem 4.2.3 that L'w = K'wd(K*wrl
and
Ut.w = K'wEa(K*wrl
are solutions of the multicomponent KP-hierarchy. The following theorem makes clear why we did not consider the commuting flows from F_ . Its proof can also be found in [13]. Theorem 4.3.3. rrd U
_ U rj6
a,gW —
For each g= £ jjAj
in F_, we have LdgW = Ldw and
Remark 4.3 .4. If r > 1, then Aw may contain several elements. If S1 and b2 are in Aw, then the solutions {L^, l/^,«} and (L^, Ufyja} are related by so-called differential difference equations that reduce in a specific case to the equations of the Toda-lattice, see [13]. These differential difference equations are a generalization to the KP-level of equations considered in [1]. Also in the multicomponent setting the coefficients of \$8W can be expressed in terms of Fredholm determinants related to the line bundle L((— 1, 0)). If W=g1F(0}, with g^e®, then we define T, l | H l : G -> C by T
0i|Hi((0» 4)) = (0>fl)' (/iV»M0i) =
det
(P^0)09~l °9i ° ^ l ^ i ) -
It measures the failure of G-equivariance of the section corresponding to y^o). If one takes another element g^ of © with W=g1F(0}, then T~ I ) H I and i g i ( H l differ by a non-zero constant. In A we consider the elements At/j given by
where the /l-factor stands at the z-th place, the A -1 -factor at the ;-th place and the resulting factors are equal to 1. If fceC, |fc| > 1, then we still need the element <\$ from F+ given by .l, 1-
1,...,
THE STRUCTURE OF HILBERT FLAG VARIETIES
where the factor 1 -- stands at the i-th place. k
Then there holds
Theorem 4.3.5. Consider a Win ^ and a d in Aw. with ge(5 and we have (a) For all 1 < i < r, the (i, i)-entry of ^,
439
Then d~1(W) = gF(0},
is the L2 -boundary value of
(b) For j / i, there is a lifting 2;/J- of Aifj to G such that the ( i j ) entry of ifrw is the L2 -boundary value of
This theorem gives a geometric interpretation of formulae, stated in the appendix of [27] and generalizes the one component interpretation given in [25]. For the one component case a representation theoretic interpretation of the i depending polynomially of the {tig} was given in [18]. For a proof of theorem 4.3.5 we refer the reader to [13]. There one can also find more equations that fit in the framework just described. Next we consider the one component case somewhat more in detail. For convenience we denote the set of independent variables simply as t = {tj i > 1} and we see the elements of R as functions in t. Further we restrict each T g i | H l to F+ and we write simply T or r(t). If fee Z and We%(k*~k\ then Aw = {l~k] and we have exactly one solution Lkwk to the KP-hierarchy. In this way every component of g leads to the same bunch of solutions of the KP-hierarchy. Hence, for the construction of solutions, it suffices to consider only one component of g. The different components are, however, essential for the modified systems as we will see in a moment. In the one component case the Japanese school, see [17], translates the equations that are satisfied by the wavefunction to equations for the r-function and these can be written in the so-called bilinear form
for all (£;) and (s;) and with dk such that
dk
I J 2nik Also this formula can be given a geometric interpretation. For if We g(0) then one can consider WL as an element of the Grassmann manifold corresponding
440
GERARD F. HELMINCK AND ALOYSIUS G. HELMINCK
to H = [email protected] Theorem 4.3.6.
If \j/w is linked to T by theorem 4.3.5, then one can show The wavefunction \j/w± can be expressed in T by
If we consider instead of F+ the group of flows consisting of the adjoints of the elements of F+, then relation (23) boils down to the orthogonality relations for \j/w and \lfw±. For /eZ, let We be a general element of g ( A ~°. According to the theorems 4.3.2 and 4.3.5 there corresponds a wavefunction of type )f to Wf and a r-function t^ to Wf. Consider an increasing set of integers f ^ < { 2 - - - < {s and denote it by f. To z? corresponds a decomposition of H by H = [email protected]@Hs+l, where Jff x = { X M'eH}, i>^ s
H7. = { X fl^'eH} for ;, 1 < j < s and, H s+1 - { £ a £ A'eH}. i>(fj i<^j+i
i<^i
Denote the flagvariety corresponding to this decomposition by Qp. Then we can describe the elements of g^0) in terms of nonlinear equations for the corresponding T functions. Theorem 4.3.7. For f± < ••• < /s, fe? ^ ^ a general elements of g ((f "~^ } aw^/ let ifi be a with W£i corresponding i-function. Then the (H^J determine an element of 3^0) if and only if the {T^} satisfy the following bilinear equations
/or a// {tj} and {s}} and for i, 1 < i < s.
The simplest case of these equations { ^ < /2? gives a relation between 2 r-functions. It has been considered in [17] and is called the (/ 2 , /^-modified KP-hierarchy there.
References [ 1] [2] [3]
Bergvelt, M. J. and Kroode, A. P. E. ten, Differential-difference AKNS equations and homogeneous Heisenberg algebras, /. Math. Phys., 28 (1987), 302. Boyer, R. P., Representation Theory of the Hilbert-Lie Group C/(§)2, Duke Math. /., 47, No. 2, p. 325-344. Bourbaki, N., Lie groups and Lie algebras, Chapters 1-3, Springer-Verlag, New York/
THE STRUCTURE OF HILBERT FLAG VARIETIES
[4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27]
441
Berlin, 1989. , Varietes differentielles et analytiques, Hermann, Paris, 1968. Carey, A. L. and Ruijsenaars. S. N. M, On fermion gauge groups, current algebras and Kac-Moody algebras, Acta Appl. Math., 10 (1987), 1-86. Carey, A. L., Hurst, C. A. and O'Brien, D. M., Automorphisms of the canonical anticommutation relations and index theory, /. Fund. Anal., 48 (1982), 360-393. Date, E., Jimbo, M., Kashiwara, M. and Miwa, T., Transformation groups for soliton equations, Proc. Japan Acad., 57A (1981), 342-7. Dorfmeister. J. and Neher, E., Banach Jordan pairs and Associated Groups and Manifolds, in preparation. Dorfmeister, J., Neher, E. and Szmigielski, J., Automorphisms of Banach Manifolds associated with the KP-equation, Quart. J. Math. Oxford, 40 (1989), 161-195. Fallings, G., Stable G-bundles and projective connections, /. Algebraic. Geom., 2 (1993), 507-568. Grothendieck, A., Sur la classification des fibres holomorphes sur la sphere de Riemann, Am. J. Math., 79 (1957), 121-38. , La theorie de Fredholm, Bull. Soc. Math. France, 84 (1956), 319-384. Helminck, G. F. and Post, G. F., The geometry of differential difference equations, Memorandum 999, University of Twente (1991), to appear in Indag. Math. , A convergent framework for the multicomponent KP-hierarchy, Trans. Am. Math. Soc., 324, (1991). , Geometric interpretation of the bilinear equations for the KP-hierarchy, Lett. Math. Phys., 16 (1988), 359-364. Herve, M., Analytic and plurisubharmonic functions in finite and infinite dimensional spaces, Lecture Notes in Math., 198 Springer Verlag, Berlin (1971). Jimbo, M. and Miwa, T., Solitons and infinite dimensional Lie algebras, Pitbl. RIMS, Kyoto Univ., 19 (1983), 943-1001. Kac, V. G., Infinite dimensional Lie algebras, Birkhauser, Boston, 1983. Kashiwara, M., The flag manifold of Kac-Moody Lie algebra, to appear. Kuiper, N. H., The homotopy type of the unitary group of Hilbert space, Topology, 3 (1965), 19-30. Mickelsson, J., Current algebras and groups, Plenum monographs in nonlinear physics. Peterson, D. H. and Kac, V. G., Infinite flag varieties and conjugacy theorems, Proc. Nat. Acad. Sd. USA, 80 (1983), 1778-1782. Pressley, A. and Segal, G., Loop groups, Clarendon Press, Oxford, 1986. Sato, M. and Sato, Y., Soliton equations as dynamical systems on infinite dimensional Grassmann manifolds, Lect. Notes in Num. Appl. Anal., 5 (1982), 259. Segal, G. B. and Wilson, G., Loop groups and equations of KdV-type, Inst. Hautes Etudes Sci. Publ. Math., 61 (1985), 5-65. Shale, D., Linear symmetries of free Boson fields, Trans. Amer. Math. Soc., 103 (1962), 149-167. Ueno, K. and Takasaki, K., Toda-lattice hierarchy, Adv. Stud. Pure Math., North-Holland. 4 (1984), 1-95.
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2021-11-30 00:20:48
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https://library.tradingtechnologies.com/trade/xl-integration-troubleshooting.html
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# Excel integration with TT
Excel integration with TT
# Excel integration troubleshooting
## If the .NET installation fails
Note: In some instances, the installation of .Net 4.6 will fail with a message similar to the following.
The cause of this issue is still under investigation by Microsoft. In the interim, customers that experience this issue can perform the following steps to work around this issue:
1. In Windows Explorer, navigate to the VSTO 10.0 folder. The path indicated in the error message should be something similar to C:\Program Files (x86)\Common Files\Microsoft shared\VSTO\10.0 or C:\Program Files\Common Files\Microsoft shared\VSTO\10.0.
2. In that folder, you should find a file named VSTOInstaller.exe.config. Rename it to VSTOInstaller.exe.config.old. (Ignore the warning from Windows about the file becoming unusable.)
3. Run the installation program again.
4. Undo the rename listed in step 2.
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2017-12-15 09:59:50
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https://stats.stackexchange.com/questions/432172/compute-the-information-matries-related-to-normal-distribution
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# Compute the information matries related to normal distribution
This is a problem that I have trouble with.
Suppose that we have $$X_{1}, \ldots, X_{m}$$ are iid $$N\left(\mu, \sigma^{2}\right), Y_{1}, \ldots, Y_{n}$$ are iid $$N\left(0, \sigma^{2}\right),$$ the $$X$$ 's are independent of the $$Y$$ 's where $$-\infty<\mu<\infty, 0<\sigma<\infty$$ are the unknown parameters. Suppose that $$\mathbf{T}=\mathbf{T}(\mathbf{X},$$ $$\mathbf{Y})$$ is the minimal sufficient statistic for $$\theta=\left(\mu, \sigma^{2}\right) .$$ Evaluate the expressions of the information matrices $$I_{\mathrm{XY}}(\theta)$$ and $$I_{\mathrm{T}}(\theta),$$ and then compare these two information contents.
The information matrix of $$I_{\mathrm{XY}}(\theta)$$ is not hard to compute, but I'm not sure how to deal with $$I_{\mathrm{T}}(\theta)$$. Also, the problem asks me to compare these two matrices, so I wonder whether there is relation between the information matrix of the original data and that of a sufficient statistic ( in my book the author mentions the one-parameter case, so I'm asking about the case when $$\theta$$ is a vector). Can anyone help?
• Please add the self-study tag and read the tag wiki. Did you find out what $T$ is here? – StubbornAtom Oct 19 at 10:45
• $T = \left( \Sigma_{i=1}^m X_i^{2} + \Sigma_{i=1}^n Y_i^{2} , \Sigma_{i=1}^m X_i \right)$ ? – j200932 Oct 19 at 11:08
• Yes, a minimal sufficient statistic for $\theta$ is $T=(\sum X_i,\sum X_i^2+\sum Y_i^2)$. Because this is sufficient, you would expect both $T$ and the original data $(X,Y)$ to have the same information about $\theta$, i.e. the information matrices should be the same. – StubbornAtom Oct 19 at 15:43
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2019-11-12 10:21:11
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