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https://sv.overleaf.com/learn/latex/Typesetting_quotations
|
## Introduction
When it comes to quotations and quotation marks, each language has its own symbols and rules. For this reason, several LaTeX packages have been created to assist in typesetting quotations in-line, in display mode or at the beginning of each chapter. It's important to remark that even if you are typing English quotes, different quotation marks used in English (UK) and English (US). Plenty of different quotation marks can be typeset with LaTeX, and there are options for almost every language (see the reference guide).
We will look at several packages suited to typesetting different types of quotation.
## dirtytalk package
dirtytalk is a small LaTeX package with only one available command: \say, as shown in the next example:
\documentclass{article}
\usepackage{dirtytalk}
\begin{document}
\section{Introduction}
Typing quotations with this package is quite easy:
\say{Here, a quotation is written and even some \say{nested} quotations
are possible}
\end{document}
This example produces the following output:
The dirtytalk package can be loaded by putting the following line in your document preamble:
\usepackage{dirtytalk}
dirtytalk supports one nested quotation and has options to redefine the characters used for the quotes. For example, in a document written in French the following code could be used:
\usepackage[french]{babel}
\usepackage[T1]{fontenc}
\usepackage[
left = \flqq{},%
right = \frqq{},%
leftsub = \flq{},%
rightsub = \frq{} %
]{dirtytalk}
The first two commands define the primary left and right quotation marks, the second pair of commands define the secondary set of quotation marks. Here is an example using the code above:
\documentclass{article}
\usepackage[french]{babel}
\usepackage[T1]{fontenc}
\usepackage[
left = \flqq{},%
right = \frqq{},%
leftsub = \flq{},%
rightsub = \frq{} %
]{dirtytalk}
\begin{document}
\section{Introduction}
Typing quotations with this package is quite easy:
\say{Here, a quotation is written and even some \say{nested} quotations are possible}
\end{document}
This example produces the following output:
This package is suitable for most situations: it's very simple, since only one command is needed, and it supports nesting quotations to one degree. If a more complex quotation mark structure is required, the options listed in the following sections may be more effective.
## csquotes package
The csquotes package provides advanced facilities for in-line and display quotations. It supports a wide range of commands, environments and user-definable quotes. Quotes can be automatically adjusted to the current language by means of the babel or polyglossia packages. This package is suitable for documents with complex quotation requirements, therefore it has a vast variety of commands to insert in-line quotes, quotes with sources, block-quotes with the support of changing language.
The following example uses the csquotes package, in conjunction with babel, within a document written in Spanish. It automatically loads the correct quotation characters "«" and "»"— known as guillemets (or "comillas angulares", in Spanish).
\documentclass{article}
\usepackage[spanish]{babel}
\usepackage{csquotes}
\begin{document}
\section{Introducción}
La siguiente frase es atribuída a Linus Torvals:
\begin{displayquote}
Sé que tengo un ego del tamaño de un planeta pequeño, pero incluso yo a veces me equivoco
\end{displayquote}
La frase revela un aspecto importante de su \textquote{jocosa} personalidad.
\end{document}
This example produces the following output:
In the example the environment displayquote prints a display quotation and the command \textquote and in-line quotation.
## epigraph package
Some authors like to write quotations at the beginning of a chapter: those quotations are known as epigraphs. The epigraph package provides a vast set of options to typeset epigraphs and epigraphs lists. To use the package, add the following line to your document preamble:
\usepackage{epigraph}
Here is an example showing an epigraph quotation typed using the command \epigraph{}{}, whose first parameter is the quotation itself and the second parameter is the quotations source (author, book, etc.):
\documentclass{book}
\usepackage{blindtext} %This package generates automatic text
\usepackage{epigraph}
\title{Epigraph example}
\author{Overleaf}
\date{August 2021}
\begin{document}
\frontmatter
\mainmatter
\chapter{Something}
\epigraph{All human things are subject to decay, and when fate summons, Monarchs must obey}{\textit{Mac Flecknoe \\ John Dryden}}
\blindtext
\end{document}
This example produces the following output:
The epigraph package can handle several quotations by means of a special environment and also has many customization options.
## fancychapters package (obsolete)
This package typesets epigraphs or quotations at the beginning of each chapter but was designed for use with LaTeX 2.09 so we no longer recommend using it.
## quotchap package
The package quotchap redefines the commands chapter, and its starred version, to reformat them. You can change the colour of the chapter number with this package. It also provides a special environment to typeset quotations and the corresponding authors.
To use this package include the following line in your document preamble:
\usepackage{quotchap}
Quotes are typed inside the environment savequote. In the example below, the parameter inside brackets, [45mm], sets the width of the quotation area. After each quote the command \qauthor{} is used to typeset and format the author's name.
\documentclass{book}
\usepackage{blindtext}
\usepackage{quotchap}
\begin{document}
\begin{savequote}[45mm]
---When shall we three meet again
in thunder, lightning, or in rain?
---When the hurlyburly’s done,
when the battle’s lost and won.
\qauthor{Shakespeare, Macbeth}
\end{savequote}
\chapter{Classic Sesame Street}
\blindtext
\end{document}
This example produces the following output:
## Reference guide
A small table of quotation marks in several languages:
Language Primary Secondary
English, UK ‘…’ “…”
English, US “…” ‘…’
Danish »…« ›…‹
Lithuanian „…“
French «…» «…»
German „…“ ‚…‘
Russian «…» „…“
Ukrainian «…»
Polish „…“ ‚…‘ or «…»
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2022-09-28 05:40:33
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https://www.electro-tech-online.com/threads/spi-to-rgb-using-ws2811-with-constant-current-amp.147612/
|
# SPI to RGB Using WS2811 with Constant Current/Amp
Status
Not open for further replies.
#### EvilGenius
##### Member
Hello
For a while I have been trying to find a pre-made board that is able to convert SPI signal (not DMX) to RGB to drive high power RGB Pixels (10W), to no avail. I had a previous post regarding this and gave up since I could not find such a board and it was cost prohibitive to manufacture. I originally gave in using WS2811 Pixel Board combined with an RGB amplifier. It was all good with the exception that I had no control over the LED currents and the pixel could easily overheat and self-destruct even with series resistors. Worth mentioning WS2811 is a 3-channel (18.5ma constant current) pixel driver that can receive SPI signal, buffers what it needs and pass on the remaining signal to the next pixel with no additional external clock. The buffer triggers 3 outputs to generate the color on the RGB pixel. This is done by a start bit, then sequence of 8bit for R, 8bit for G, 8bit for B, then a stop bit and auto latch. Its outputs can only "sink" (active low) maximum of 18.5ma per channel.
Operational Goal: Building a device that can receive SPI signal from microcontroller, or Programmable SD-RAM SPI Controller, or SPI signal from a DMX to SPI controller which utilizes DMX signal from a controller, or a PC program such as Light-O-Rama via a dongle. Device to have 3-Wires in/out (Data, +12, Gnd) for daisy chaining.
Searching on the net I came across a simple dual NPN CCR which peaked my interest yet it was not exactly what I needed since it required Positive trigger (Active High) to operate it. I reconfigured the design to use PNP transistors for Active low outputs of WS2811.
Objective: Build a device that is able to convert SPI to RGB
1- Be very low cost
2- Be very small (to fit an HP RGB metal case)
3- Be able to sink 300mA per channel (with flexibility to go a bit higher or lower by selecting Rs and Rx)
4- Use fairly common parts that are easy to find and inexpensive
5- Operate with minimum I/O wires (Single data line in/out)
6- Provide for constant current for each channel
7- Be able to drive 10W high power RGB module that runs on 12VDC
Parameters:
I-LED=300mA (under utilized vs. 350mA max)
Regards,
Rom
##### New Member
Just a note: look at this
#### EvilGenius
##### Member
Based on the simulation of the new circuit:
Vbe2=0.64 (it does not reach full saturation)
Vbe1=0.75
Vb1=Ve2-Vbe2-Vbe1=1.53
Ie2=303-289=14ma
Ib2=< 0.28ma
Ic2= 13.72ma
Ic1=284ma
Ib1=4.78ma
R2 is used to take the power stress off of Q1
Rx is used to take the slack Vf difference between Red and G/B also to bring down the voltages to take stress off Q1 and Q2
Pq1= Vq1 x Iq1 = (2.28-0.85) x (303-14+0.28)= 413mw (within specs with room for deration)
Last edited:
#### EvilGenius
##### Member
Just a note: look at this
It is an interesting device. I think this is a controller not receiver. Also at a quick glance it operates on 5v which is ideal for 5v intelligent strips but not 12v version nor the High power LED. I don't believe it can handle 300ma per channel! But I am sure you can create a interface with it to accomplish the job. Not sure if it can communicate with other cheap controllers out there.
There is a device with 27 SPI outputs (9 RGB pixels) on one board that receives DMX and converts it to SPI that a WS2811 can understand. But it is pricey and requires your pixels to be more centralized instead of spread out (5-10 feet away from each other). I don't claim I know all the products out there, but I have been searching for quite a while for such device.
Status
Not open for further replies.
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2021-06-16 21:10:41
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|
https://tex.stackexchange.com/questions/235200/reformat-my-paper-from-one-latex-format-to-another-latex-format
|
# Reformat my paper from one LaTex format to another LaTex format
I am very new to LaTex. One reason I switched to it was to have an easy way to reformat my papers based on the templates provided by different publishers.
Then I don't know yet, laTex provide that convenience for me or not. For example Suppose I have a paper based on the laTex template by IEEE. Then I want to send it to another journal named IJCA, the IJCA give me their specific laTex template.
What is the easiest way to convert my paper (in IEEE format) to XXX format? Is there any automatic conversion tool for that? I just began to use WinEDT and can edit and compile latex files, what is the next step?
I give an example of two templates
IEEE:
\author{\IEEEauthorblockN{Michael Shell}
\IEEEauthorblockA{School of Electrical and\\Computer Engineering\\
Georgia Institute of Technology\\
Atlanta, Georgia 30332--0250\\
Email: http://www.michaelshell.org/contact.html}
\and
\IEEEauthorblockN{Homer Simpson}
\IEEEauthorblockA{Twentieth Century Fox\\
Springfield, USA\\
Email: homer@thesimpsons.com}
\and
\IEEEauthorblockN{James Kirk\\ and Montgomery Scott}
San Francisco, California 96678-2391\\
Telephone: (800) 555--1212\\
Fax: (888) 555--1212}}
IJCA:
\author{
\large 1st Author \\[-3pt]
\normalsize 1st author's affiliation \\[-3pt]
\normalsize 1st line of address \\[-3pt]
\normalsize 2nd line of address \\[-3pt]
\normalsize 1st author's email address \\[-3pt]
\and
\large 2nd Author \\[-3pt]
\normalsize 2nd author's affiliation \\[-3pt]
\normalsize 1st line of address \\[-3pt]
\normalsize 2nd line of address \\[-3pt]
\normalsize 2nd author's email address \\[-3pt]
\and
\large 3rd Author \\[-3pt]
\normalsize 3rd author's affiliation \\[-3pt]
\normalsize 1st line of address \\[-3pt]
\normalsize 2nd line of address \\[-3pt]
\normalsize 3rd author's email address \\[-3pt]
}
.......
\section{USING THE ijca Article CLASS FILE}
If the file \verb ijcaArticle.cls is not already in the appropriate system directory
for \LaTeX{} files, either arrange for it to be put there or copy
it to your working directory. The \verb ijcaArticle document class is implemented
as a complete class, not a document style option. In order to
use the \verb ijcaArticle document class, replace \verb article by \verb ijcaArticle in the
\verb \documentclass command at the beginning of your document:
\vskip 6pt
\begin{centering}
\verb \documentclass{article} \end{centering}
\vskip 6pt
replace by
\vskip 6pt
\verb \documentclass{ijcaArticle} \vskip 6pt
In general, the following standard document \verb style options should
{ \itshape not} be used with the {\footnotesize \itshape article} class file:
\begin{enumerate}
\item[(1)] \verb 10pt, \verb 11pt, \verb 12pt – unavailable;
\item[(2)] \verb twoside (no associated style file) – \verb twoside is the default;
\item[(3)] \verb fleqn, \verb leqno, \verb titlepage – should not be used;
\end{enumerate}
%
I don't know much the latTex tags, then I prefer an automatic conversion tool which matches the tags and convert them to each other. At least does a preprocessing on them.
• You'll have to change document class, something like \documentclass{IEEEtrans} to \documentclass{XXXtrans} and adjust specific commands from one class to another: \IEEEkeywords to \XXXkeywords and similars. May be you'll have to adapt some formulas if one format uses two columns and the other only one. But it should be not too difficult. Mar 26 '15 at 11:56
• You just convert it and look at the result. If you didn't do anything IEEE specific, it should be just fine by changing the documentclass. What is XXX, Elsevier? Mar 26 '15 at 12:24
• @percusse XXX is IJCA, you can find the tempate there. As I checked two templates (IEEE and IJCA) the task doesn't seem easy. Isn't any tool which automatically do the conversion? Mar 26 '15 at 12:30
• @Ignasi I thought laTex is based on specific tags, then automatic conversion shouldn't be a hard task, isn't any software to do that? I am very new to the laTx codes, and expect easier solutions. Mar 26 '15 at 12:32
The conversion could be as automatic as change (manually) article by paper in the first line, or a nightmare, or really impossible. The automatic conversion, in general, is impossible.
Let go with a simple example as proof-of-concept of the problem, with the very standard class book:
\documentclass{book}
\begin{document}
\chapter{Hello}
\end{document}
Work perfectly. Now, imagine that you want change the format to another very standard class:
\documentclass{article}
\begin{document}
\chapter{Hello}
\end{document}
But that simple document cannot be compiled. The reason is a "undefined control sequence" error reading \chapter. That is, LaTeX do not know nothing about the supposedly well-know command \chapter. It is not hard coded in the TeX engines but defined in the document class book.
This is a big problem for a conversion tool. What to do if the command of one class (book) is not defined in another (article)? Just pass the definition to the new template? In this case have not sense. Well, then ... Ignore it? Down the header one level? Left as bold font?
This mean that the tool must have a predefined solution for every command in every new class.
Moreover, in this case the control sequence (command) is defined (directly or not) in the document class (.cls files), but the definition could be located also in any of the included packages (.sty files), or children .tex files (using \input{} in the preample, for example) or directly in the preamble.
Think about known only the uncountable number of control sequences of the thousands of packages in CTAN ... And what to do with user-defined control sequences like \def\myowncommand{Hello}?
A template conversion only can be compiled when all the definitions of the commands used in the old document (even the dull \myowncommand) are maintained or redefined in the new template.
Some commands works in all type of documents because they are hard coded in the LaTeX engine. These are mainly TeX primitives and some others that you can use even in a "unclassified document". For example, With pdfLaTeX document you can make a PDF with only some like:
% This is no a code chunk, but a complete MWE
\def\normalsize{\normalfont}
\begin{document}
Hello, {\em World}\par
Try me without a document class.
\end{document}
Here we must define \normalsize in some way to avoid an error in the initialization process, since there are not any \documentclass, but then we can use the commands that obviously do not pertain to any class or package.
The minimal class do not make much more than a (re)definition of \normalsize, so is hidden in minimal.cls. The the above MWE is almost equivalent to:
\documentclass{minimal}
\begin{document}
Hello, {\em World}\par
Try me without a document class.
\end{document}
This class have no utility for any real work, but show you what is not a "universally accessible" command. You may be surprised that commands as \section{} cannot be used here. Then \section{} is a control sequence specific of some type of document, without operating warranty in any template.
Said that, the conversion often is very easy. With very specific (user-defined) commands you must be careful in maintain any auxiliar file like mymacros.tex, the row of the preample with \input{mymacros} in the main document or the preamble definitions starting with \newcommand, \def and so on.
Most time, the problem of a undefined but well-known command is that you have lost in the conversion some package. That is, a undefined \includegraphics mean that you have lost \usepackage{graphicx} in the preamble, so it is a problem easy to solve.
The plight become more obscure when the package was not loaded directly but through another package or even the document class of the original template, but the after some research about the real origin, the solution is the same.
The worst problem is the command is defined at the document class level, like \chapter{}, because you cannot load a book class inside an article document, for example. Fortunately, most usual commands like \section{}are defined in almost all any document class, so the problem in practice is limited to a few commands of less standard clases, like \institution{} of paperclass, that are not defined in article and other common clases.
Then the solution for a undefined command, for example said the command \xxx{} with one argument, could be:
1. Remove, comment o replace manually every \xxx{} with another command to follow strictly the new template. Probably the best solution.
2. Copy the original definition \xxx from the old document class and use it in your custom class, package, preamble or children file. But this is only part of the problem. The other is make this without breaking anything else and ruin the layout. Do not try if you do not know what are you doing.
3. Define \xxx as something simple and inoffensive. For example:
• \newcommand\xxx{} (do nothing)
Finally, take into account that probably the editor of a journal will not be happy with the second or third solution.
In general each template writer can do pretty much whatever they want. For example I know one that forced a special paragraph command between each of their paragraphs. So in the worst case scenario it can be a pain.
In practice though I think you will find that it is mostly enough to just change document class and maybe do some other minor things.
But I don't think there are any general answers to this question.
• Even a preprocessing was good! I surprise with the lot of work on laTex, why there is not such conversion tools. Then it miss its main application! Mar 26 '15 at 12:50
• @Ahmad: The main application of LaTeX is to create documents, not to convert them. It is possible to write your documents so that they can be used without much fuss with more than one template, but due to the variety of journals and templates it is not possible to do it in a general way. Mar 26 '15 at 13:23
• @Ahmad -- ulrike has already given a good answer, but i'd like to add that, while some publishers design document classes to reuse as much as possible of the basic structure provided by "unadorned" latex, thus making transfer relatively painless, this is not true for all. so you are, unfortunately, at their mercy, and with the multiplicity of features implemented by different document classes, creating conversion tools is far from trivial. Mar 26 '15 at 16:54
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2021-09-20 17:46:52
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https://www.calculus-online.com/exercise/3852
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# Vector uses in physics – Calculate velocity and acceleration – Exercise 3852
Exercise
A particle moves according to the law of motion
$$\vec{r}(t)=R\cos(\omega t)\vec{i}+R\sin(\omega t)\vec{j}$$
Where
$$\omega>0, R>0$$
Calculate the velocity function, the acceleration function, their values (vector sizes) and unit vectors.
$$\vec{v}(t)=-R\omega\sin(\omega t)\vec{i}+R\omega\cos(\omega t)\vec{j}$$
$$|\vec{v}(t)|=R\omega$$
$$\hat{v}(t)=-\sin(\omega t)\vec{i}+\cos(\omega t)\vec{j}$$
$$\vec{a}(t)=-R\omega^2\cos(\omega t)\vec{i}-R\omega^2\sin(\omega t)\vec{j}$$
$$|\vec{a}(t)|=R\omega^2$$
$$\hat{a}(t)=-\cos(\omega t)\vec{i}-\sin(\omega t)\vec{j}$$
### Solution
Coming soon…
Share with Friends
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2023-03-21 18:33:28
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http://math.stackexchange.com/questions/9432/how-do-i-evaluate-the-clifford-product-in-dimensions-greater-than-3
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# How do I evaluate the Clifford product in dimensions greater than 3?
The Clifford product of a pair of vectors $a,b$ is an associative operation defined by
$$ab = a \cdot b + a \wedge b.$$
In sufficiently low dimensions I am used to being able to define the Clifford product on arbitrary $k$-vectors by repeatedly applying the vector definition. For instance, suppose that I build a Clifford algebra over $\mathbb{R}^3$ with the usual (positive) Euclidean inner product. Then I can easily write out the Clifford product of any pair of basis bivectors. For instance,
$$e_{12}e_{13} = e_1 e_2 e_1 e_3 = -e_2(e_1 e_1)e_3 = -e_2 (1) e_3 = -e_{23}.$$
In four dimensions I get stuck, because it's possible that two of the indices don't "cancel," and then I don't know how to apply the product:
$$e_{12}e_{34} = e_1 e_2 e_3 e_4.$$
Where do I go from here? I strongly suspect that this equals just $e_{1234}$, but I don't know how to show (in an explicit, pedantic, algebraic way) that
$$e_1 e_2 e_3 e_4 = e_1 \wedge e_2 \wedge e_3 \wedge e_4.$$
Thanks!
-
$e_i \cdot e_j = 0$ – anon Nov 8 '10 at 13:38
...and for orthogonal vectors, the Clifford product coincides with the wedge product. – Hans Lundmark Nov 8 '10 at 13:48
What you seem to be calling an "explicit, pedantic, algebraic way" is what most people would call "a proof"... – Mariano Suárez-Alvarez Nov 8 '10 at 13:52
Ok, well I want a proof that for orthogonal vectors the Clifford product coincides with the wedge product. In other words, I certainly realize that $e_i e_j = e_i \cdot e_j + e_i \wedge e_j = 0 + e_{ij}$ when $i \ne j$. But now suppose I have $e_i e_j e_k$ for distinct $i$, $j$, and $k$. Then I get $e_{ij} e_k$ but have no rule for the Clifford product between a bivector and a vector, so I am stuck! Thanks for the help. – corsecat Nov 8 '10 at 13:59
I agree with Hans; there's nothing to say until you tell us what your definition of the Clifford algebra is. – Qiaochu Yuan Nov 8 '10 at 16:30
It's not easy to give a proof without knowing exactly what your definitions of the Clifford and exterior algebras are, so the following argument is still a little handwaving. But let's say that we somehow have defined what the exterior algebra of a vector space $V$ is; we know that its elements are multivectors, and the rule which generates everything is that $x \wedge y + y \wedge x=0$ if $x$ and $y$ are vectors in $V$. The Clifford algebra has the same elements as the exterior algebra, and the same linear space structure, but the multiplication is different: it is generated by the rule $xy+yx=2 \, Q(x,y)$ where $Q$ is the inner product on $V$. Since "orthogonal" means that $Q(e_i,e_j)=0$, it shouldn't be too hard to believe that the Clifford and exterior multiplications agree for orthogonal vectors.
Maybe this section on Wikipedia can be of some help too?
-
It appears that you are thinking in terms of "geometric algebra". For those unfamliar with this doctrine, it's a way of regarding the Clifford algebra on an inner product space $(V,Q)$ and the exterior algebra on the $V$ as the same set, but with different product operations. Let's stick to a ground field of characteristic zero. Then the Clifford algebra $C$ of $(V,Q)$ is generated by the elements $v\in V$ with relations $vv=Q(v)$. Now for $v_1,\ldots,v_k\in V$ one can define wedge product $$v_1\wedge v_2\wedge\cdots\wedge v_k =\frac1{k!}\sum_{\pi\in S_k}(-1)^{\text{sgn}(\pi)}v_{\pi(1)}v_{\pi(2)}\cdots v_{\pi(k)}\in C.$$ This wedge product identifies $C$ with the exterior algebra $\bigwedge(V)$.
With this definition, Hans's comments are absolutely right. If $v_1,\ldots,v_k$ are pairwise orthogonal then the Clifford product $v_1v_2\cdots v_k$ equals the exterior product $v_1\wedge v_2\wedge\cdots\wedge v_k$, and this is certainly the case for orthogonal basis vectors: $e_1 e_2 e_3 e_4 =e_1\wedge e_2\wedge e_3\wedge e_4$.
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2015-07-29 10:02:57
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https://stats.stackexchange.com/questions/376083/interpreting-a-graphed-covariance-function
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# Interpreting a graphed covariance function
I'm looking through a slide deck (slide 9) about Gaussian Processes, and I came to a slide that describes one example of a covariance function: Matérn $$\frac{3}{2}$$ Covariance.
$$C(x_1,x_2) = (1+\sqrt{6}\frac{|x_1-x_2|}{\ell})*\exp(\sqrt{6}\frac{|x_1-x_2|}{\ell})$$
where
$$\ell>0$$ is the "correlation length parameter" and
$$\sigma^2>0$$ the variance parameter (though this isn't in the formula which confuses me).
Then they show a graph like this:
You can see that changing $$\ell$$ changes the shape of the function. However, my understanding is that covariance requires two random variables/vectors as inputs. So, what is $$x$$ (on the x axis) of the graph referring to in this case?
The horizontal-axis label there is wrong, it should be $$r$$, the "radius".
The "radius" is essentially the distance between two multidimensional points $$\bf {x_1}$$ and $$\bf {x_2}$$, and this is usually written as:
$$k(x_1, x_2) = f(r) = f(||x_1 - x_2||_p)$$
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2021-01-22 04:02:54
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https://www.greaterwrong.com/posts/JB6edzY5cccrdbQxP/stepwise-inaction-and-non-indexical-impact-measures?comments=false
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# Stepwise inaction and non-indexical impact measures
Overall summary post here.
In a previous post, I asked which impact measures were vulnerable to subagents. Vika pointed out that it was not merely an issue of of the impact measure, but also of the baseline.
This is indeed the case, but the nature of the impact measure is still relevant. In this post, I’ll establish two facts: that under the stepwise inaction baseline, a subagent completely undermines all impact measures (including twenty billion questions). And for the inaction baseline, for non-indexical impact measures, a subagent will not change anything.
The next post will delve into the very interesting things that happen with the inaction baseline and an indexical impact measure.
# Stepwise inaction baseline
All impact measures are some form of distance measure between two states, of the form , where is the state the agent is actually in, and is some baseline state to compare with.
For the stepwise inaction baseline, is calculated as follows. Let be the previous state, the previous action. Then is the state that would have followed had the agent taken the noop action, , instead of .
However, in the presence of a subagent, the agent merely has to always take the action . In that case, (in a deterministic setting[1]), and the penalty collapses to nothing.
This leaves the subagent free to maximise (the positive reward for agent and subagent), without any restrictions beyond making sure that the agent can always take the action.
# Non-indexical impact penalty
The twenty billion questions is an example of a non-indexical impact penalty. The impact penalty is defined as the difference in expected value of many variables, between the current state and the baseline state.
With the inaction baseline state, is the state the system would have been in, had the agent always done nothing/had not been turned on. Thus the definition of is independent of the agent’s actions, so the inaction baseline does not collapse like the stepwise inaction baseline does.
What about subagents? Well, since the impact penalty is non-indexical, a subagent does not get around it. It matters not whether the subagent is independent, or an extension of the agent: the impact penalty remains.
1. In a non-deterministic setting, becomes a mix of an impact penalty and a measure of environment stochasticity. ↩︎
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2021-05-14 09:33:56
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https://proofwiki.org/wiki/Definition:Commutative_Ring
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# Definition:Commutative Ring
A commutative ring is a ring $\struct {R, +, \circ}$ in which the ring product $\circ$ is commutative.
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2020-04-01 17:43:38
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https://uk.pinterest.com/explore/root-mean-square/
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Pinterest • The world’s catalogue of ideas
Root mean square - Wikipedia, the free encyclopedia
### 4 Simple Ways To Maximize A Small Space
Maximize every square inch of even the most fun-sized digs.
3
2
Capacitor, Resistance, Inductor,LR DC circuit, Current growth, Current retardation, LR AC circuit, LC AC circuit, Power factor, Root mean square value of current, Root mean square value of voltage, Decay of current, LCR AC circuit across capacitor inductor resistance,
Square and Square Root Table Numbers 1 Through 30
105
12
Geometrical representation of common mathematical means. a,b-two scalars. A=Arithmetic mean of scalars 'a' and 'b'. G=Geometric mean, H=Harmonic mean, Q=Quadratic mean (also known as Root mean square)
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2016-12-05 01:56:28
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https://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-connecting-concepts-through-application/chapter-4-quadratic-functions-4-5-solving-equations-by-factoring-4-5-exercises-page-359/25
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## Intermediate Algebra: Connecting Concepts through Application
$\color{blue}{\left\{-10, 10\right\}}$
Subtract $150$ to both sides of the equation to obtain: $x^2+50-150=150-150 \\x^2-100=0$ Since $100=10^2$, the given equation can be written as: $x^2-10^2=0$ RECALL: $a^2-b^2=(a-b)(a+b)$ Factor the binomial using the formula above to obtain: $(x-10)(x+10)=0$ Use the Zero-Factor Property by equating each factor to zero to obtain: $x-10=0$ or $x+10=0$ Solve each equation to obtain: $x-10=0 \\x=0+10 \\x=10$ or $x+10=0 \\x=0-10 \\x=-10$ Therefore, the solution set is $\color{blue}{\left\{-10, 10\right\}}$.
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2019-08-19 22:50:10
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https://homework.cpm.org/category/CCI_CT/textbook/calc/chapter/6/lesson/6.3.4/problem/6-114
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Home > CALC > Chapter 6 > Lesson 6.3.4 > Problem6-114
6-114.
1. THE GREAT RACE
2. Summer and Jackie are on a floating platform on a lake and decide to race to the snack bar. Summer is a better swimmer and Jackie is a better runner. They make a bet that whoever loses the race has to buy the snacks. Summer swims at a rate of 12 feet per second and runs at a rate of 20 feet per second. Jackie swims at a rate of 7 feet per second and runs at a rate of 30 feet per second. Assuming each takes the optimal path to reach the snack bar, who is buying the snacks? Homework Help ✎
In order to optimize the time it takes to reach the Snack Bar, both girls will swim part of the way and run part of the way. Notice that the running part is horizontal but the swimming part is diagonal.
d(x) = (distance swimming) + (distance running) = (hypotenuse of triangle) + (horizontal line)
Since distance = (rate)(time),
$\text{time}=\frac{d(x)}{\text{rate}}.$
Write specific t(x) equations for each girl.
Before you apply Calculus to optimize, write a geometric equation describing the distance, d(x), of a generic journey to the snack bar. Let x represent the distance on land that each girl is NOT running.
Convert the distance equation into time equations, t(x).
Setting t'(x) = 0 will give you the optimal value of x for each girl. In other words, how far each girl should swim and run. Use that information to calculate the time it takes each girl to reach the snack bar.
Use the eTool below to explore the problem.
Click on the link to the right to view to full version of the eTool. Calc 6-114 HW eTool
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2020-01-21 02:41:20
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https://planetmath.org/ProofOfMartingaleCriterioncontinuousTime
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# proof of Martingale criterion (continuous time)
###### Proof.
1. Let $X$ be a martingale. By the optional sampling theorem we have $E(X_{c}|\mathcal{F}_{\tau})=X_{c\wedge\tau}=X_{\tau}\forall\tau\leq c$. Since conditional expectations are uniformly integrable the first direction follows.
2. Let $(\tau_{k})_{k\geq 1}$ be a local sequence of stopping times (i.e. $\tau_{k}\uparrow\infty$ a.s. and $X^{\tau_{k}}$ martingale $\forall k\in\mathbb{N}$). For each $t\in\mathbb{R}_{+}$ we have $X_{\tau_{k}\wedge t}\to X_{t},k\to\infty$ almost surely. The set
$\displaystyle\{X_{\tau_{k}\wedge t}:k\in\mathbb{N}\}$ $\displaystyle\subset\{X_{\tau}:\tau\ \text{stopping time},\tau\leq c\}$
is uniformly integrable (take $c=t$). It follows that $X_{t}^{\tau_{k}}\lx@stackrel{{\scriptstyle\begin{subarray}{c}\mathscr{L}^{1}% \end{subarray}}}{{\longrightarrow}}X_{t},k\to\infty$. Since the martingale property is stable under $\mathscr{L}^{1}$ convergence, $X$ is a martingale. ∎
Title proof of Martingale criterion (continuous time) ProofOfMartingaleCriterioncontinuousTime 2013-03-22 18:54:28 2013-03-22 18:54:28 karstenb (16623) karstenb (16623) 4 karstenb (16623) Proof msc 60G07 msc 60G48
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2023-01-30 18:11:09
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https://math.stackexchange.com/questions/2958317/determining-the-galois-group-of-splitting-field-of-cubic-polynomials
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# Determining the Galois Group of Splitting Field of Cubic polynomials
I wish to determine the Galois groups of $$L$$ over $$Q$$ when L is the splitting field of firstly, $$x^3 − 4x + 2$$ over $$Q$$ and secondly, $$x^3 − 3x + 1$$ over $$Q$$. For another example, I know $$x^3 + x^2 − 1$$. is irreducible by the rational root theorem. And the discriminant of the polynomial is $$−23$$, we have the group $$S_3$$ but I can't determine the Galois group for the above cubic polynomials in the same way, can I
• Can you work out the discriminants of your polynomials? – Lord Shark the Unknown Oct 16 '18 at 18:46
• opps yeah 148 and 81 respectively~ – Homaniac Oct 16 '18 at 19:03
• $148$ is not a square, but $81$ is. So the first has group $S_3$, and the second $A_3$. – Lord Shark the Unknown Oct 16 '18 at 19:44
In general: Let $$d$$ is discriminants polynomials.
if $$\sqrt d \in \Bbb Q$$ , then $$Gal(L, \Bbb Q)\cong A_3$$, while if $$\sqrt d \notin \Bbb Q$$ , then $$Gal(L, \Bbb Q)\cong S_3$$
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2019-09-21 00:33:20
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http://bibli.cirm-math.fr/listRecord.htm?list=link&xRecord=19249724146910679069
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# Documents 05C80 | enregistrements trouvés : 48
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## Self-interacting walks and uniform spanning forests Peres, Yuval | CIRM H
Post-edited
2 y
Research talks
In the first half of the talk, I will survey results and open problems on transience of self-interacting martingales. In particular, I will describe joint works with S. Popov, P. Sousi, R. Eldan and F. Nazarov on the tradeoff between the ambient dimension and the number of different step distributions needed to obtain a recurrent process. In the second, unrelated, half of the talk, I will present joint work with Tom Hutchcroft, showing that the component structure of the uniform spanning forest in $\mathbb{Z}^d$ changes every dimension for $d > 8$. This sharpens an earlier result of Benjamini, Kesten, Schramm and the speaker (Annals Math 2004), where we established a phase transition every four dimensions. The proofs are based on a the connection to loop-erased random walks. In the first half of the talk, I will survey results and open problems on transience of self-interacting martingales. In particular, I will describe joint works with S. Popov, P. Sousi, R. Eldan and F. Nazarov on the tradeoff between the ambient dimension and the number of different step distributions needed to obtain a recurrent process. In the second, unrelated, half of the talk, I will present joint work with Tom Hutchcroft, showing that the ...
## Non-backtracking spectrum of random graphs: community detection and non-regular Ramanujan graphs Massoulié, Laurent | CIRM H
Post-edited
2 y
Research talks
A non-backtracking walk on a graph is a directed path such that no edge is the inverse of its preceding edge. The non-backtracking matrix of a graph is indexed by its directed edges and can be used to count non-backtracking walks of a given length. It has been used recently in the context of community detection and has appeared previously in connection with the Ihara zeta function and in some generalizations of Ramanujan graphs. In this work, we study the largest eigenvalues of the non-backtracking matrix of the Erdos-Renyi random graph and of the Stochastic Block Model in the regime where the number of edges is proportional to the number of vertices. Our results confirm the "spectral redemption" conjecture that community detection can be made on the basis of the leading eigenvectors above the feasibility threshold. A non-backtracking walk on a graph is a directed path such that no edge is the inverse of its preceding edge. The non-backtracking matrix of a graph is indexed by its directed edges and can be used to count non-backtracking walks of a given length. It has been used recently in the context of community detection and has appeared previously in connection with the Ihara zeta function and in some generalizations of Ramanujan graphs. In this work, we ...
## Bootstrap percolation on Erdos-Renyi graphs Angel, Omer | CIRM H
Post-edited
2 y
Research talks
We consider bootstrap percolation on the Erdos-Renyi graph: given an initial infected set, a vertex becomes infected if it has at least $r$ infected neighbours. The graph is susceptible if there exists an initial set of size $r$ that infects the whole graph. We identify the critical threshold for susceptibility. We also analyse Bollobas's related graph-bootstrap percolation model.
Joint with Brett Kolesnik.
## Séminaire Bourbaki. Vol. 2011/2012:exposés 1043-1058 | Société Mathématique de France 2013
Congrès
V
- xi; 556 p.
ISBN 978-2-85629-371-3
Astérisque , 0352
Localisation : Périodique 1er étage
Algorithme d'approximation # carte brownienne # cartes planaires # champ libre gaussien # champ moyen # choix social # concentration-compacité # condition nulle # configuration polynomiale # courbe elliptique # D-module holonome # difficulté d'approximation # équation aux dérivées partielles # équations d'Einstein # équations différentielles partielles # équations non-linéaires dispersives # espaces adiques # espaces de Berkovich # espaces homogènes # espaces métriques # espaces normés # espaces perfectoïdes # existence globale # fibré de Higgs # fibré holomorphe plat # forme quartique binaire # formule de KPZ # gravité quantique # groupe de Galois motivique # groupe de Selmer # groupes de Lie # groupes quasi-fuchsiens # hamiltonien # marches aléatoires # mélange exponentiel du fibré des repères # mesures de Liouville # mesures stationnaires # métrique harmonique # modération topologique # monodromie-poids # motifs de Tate mixtes # multizêtas # nonlinéaire # norme d'uniformité # orbites coadjointes # plongement métrique # principe de transfert # programmation semi-définie # Programme de Ribe # progression arithmétique # pureté # rang # réarrangement # Relativité générale # représentations des groupes algébriques réductifs # représentations des groupes de Lie compacts # résonances en espace temps # rigidité # singularités irrégulières # stabilité orbitale # surfaces enfermées # système stellaire auto-gravitant # théorème de Lefschetz difficile # théorie de Hodge # théorie géométrique des invariants # topologie étale # trous noirs # types stablement dominés # variétés de drapeaux # variétés hyperboliques de dimension 3 # Vlasov-Poisson Algorithme d'approximation # carte brownienne # cartes planaires # champ libre gaussien # champ moyen # choix social # concentration-compacité # condition nulle # configuration polynomiale # courbe elliptique # D-module holonome # difficulté d'approximation # équation aux dérivées partielles # équations d'Einstein # équations différentielles partielles # équations non-linéaires dispersives # espaces adiques # espaces de Berkovich # espaces ...
## Random graphs'85.Based on lectures presented at the 2nd international seminar on random graphs and probabilistic methods in combinatoricsPoznan # august 5-9, 1985 Karonski, Michal ; Palka, Zbigniew | North-Holland 1987
Congrès
V
- vii; 354 p.
ISBN 978-0-444-70265-4
North-Holland mathematics studies , 0144
Localisation : Colloque RdC
## Surveys in combinatorics 2011.Papers from the 23rd British combinatorial conferenceExeter # july 3-8, 2011 Chapman, Robin | Cambridge University Press 2011
Congrès
V
- vii; 437 p.
ISBN 978-1-107-60109-3
London mathematical society lecture note series , 0392
Localisation : Collection 1er étage
combinatoires # graphe topologique # hypergraphe # graphe aléatoire
## Combinatorial stochastic processes :école d'été de probabilités de Saint-Flour XXXII#2002 Pitman, Jim ; Picard, Jean | Springer 2006
Congrès
V
- 256 p.
ISBN 978-3-540-30990-1
Lecture notes in mathematics , 1875
Localisation : Collection 1er étage
arbre aléatoire # mouvement brownien # probabilité combinatoire # processu stochastique # combinatoire asymptotique # position aléatoire
## Random discrete structuresproceedings of a workshop of the 1993-94 IMA program on emerging applications of probability Aldous, David ; Pemantle, Robin | Springer 1996
Congrès
V
ISBN 978-0-387-94623-8
The IMA volumes in mathematics and its applications , 0076
Localisation : Colloque RdC
approximation normale par méthode de Stein # arbre aléatoire # couverture universelle de graphe # distribution aléatoire de masse # distribution de probabilités sur cladogramme # ensemble régénératif # environnement aléatoire # grande déviation # graphe libre de triangle # intersection et limite # marche aléatoire transitoire # matrice positive complètement # méthode du second moment # métrique sur composition et coïncidence # processus aléatoire # recurrence amenabilité # stabilité de processus auto-organisant # structure discrète aléatoire # suite de renouvellement # théorème du cycle impaire long # tresse de jeux de minimax aléatoire # énergie et intersection de chaîne de Markov approximation normale par méthode de Stein # arbre aléatoire # couverture universelle de graphe # distribution aléatoire de masse # distribution de probabilités sur cladogramme # ensemble régénératif # environnement aléatoire # grande déviation # graphe libre de triangle # intersection et limite # marche aléatoire transitoire # matrice positive complètement # méthode du second moment # métrique sur composition et coïncidence # processus ...
## Expanding graphs :proceedings of a DIMACS workshop#May 11-14 Friedman, Joel | American Mathematical Society 1993
Congrès
V
- 142 p.
ISBN 978-0-8218-6602-3
DIMACS series in discrete mathematics and theoretical computer science , 0010
Localisation : Collection 1er étage
approche corps de fonction de graphe et diagramme de Ramanuj # construction algébrique de graphe dense de grand contour et # graphe de Cayley aléatoire et expanseur # graphe demi-plan supérieur fini et Ramanujan # graphe en expansion # graphe en expansion hautement tiré de groupe diédral # groupe et expanseur # géométrie spectral et constante de Cheeger # investigation numérique du spectre pour famille de graphe de # laplacien d'hypergraphe # seconde valeur propre et développement linéaire de graphe ré # simulation de chaîne de Markov # échantillonnage uniforme modulo ou groupe de symétrie approche corps de fonction de graphe et diagramme de Ramanuj # construction algébrique de graphe dense de grand contour et # graphe de Cayley aléatoire et expanseur # graphe demi-plan supérieur fini et Ramanujan # graphe en expansion # graphe en expansion hautement tiré de groupe diédral # groupe et expanseur # géométrie spectral et constante de Cheeger # investigation numérique du spectre pour famille de graphe de # laplacien d'hypergraphe # ...
## Ecole d'été de probabilités de Saint-Flour XXI - 1991cours donnés à l'école d'été de calcul des probabilités de Saint-Flour du 18 Août au 4 sept., 1991 Hennequin, P. L. | Springer-Verlag 1993
Congrès
V
ISBN 978-3-540-56622-9
Lecture notes in mathematics , 1541
Localisation : Collection 1er étage
branchement à valeur de mesure # calcul stochastique # distribution de Palm # fonctionnelle de Log-Laplace # mesure aléatoire # mesure de Campbell # probabilité # problème de martingale # processus de Markov naissance # processus de Markov à valeur de mesure # processus de construction à valeur de mesure et interaction # regénération # représentation d'amas de Poisson # représentation de De Finetti # retournement # structure de famille # super mouvement Brownien branchement à valeur de mesure # calcul stochastique # distribution de Palm # fonctionnelle de Log-Laplace # mesure aléatoire # mesure de Campbell # probabilité # problème de martingale # processus de Markov naissance # processus de Markov à valeur de mesure # processus de construction à valeur de mesure et interaction # regénération # représentation d'amas de Poisson # représentation de De Finetti # retournement # structure de famille # super ...
## Probabilistic combinatorics and its applicationsproceedings of a symposia held in San Francisco, CaliforniaJanv.14-15 Bollobas, Bela ; Chung, Fan R. K. ; Diaconis, Persi | American Mathematical Society 1991
Congrès
V
ISBN 978-0-8218-5500-3
Proceedings of symposia in applied mathematics , 0044
Localisation : Collection 1er étage
calcul du volume des corps convexes # chaîne de Markov se mélangeant rapidement # combinatoire probabiliste # graphe aléatoire # inégalité isopérimétrique discrète # méthode de Fourier finie
## Probabilistic problems of discrete mathematics Kolchin, V. F. | American Mathematical Society 1989
Congrès
V
- 217 p.
ISBN 978-0-8218-3123-6
Proceedings of the Steklov institute of mathematics , 0177
Localisation : Collection 1er étage
mathematique discrete # probabilite # probleme probabiliste # statistique
## Graph theory and combinatoricsproceedings of the cambridge combinatorial conference in honour of paul erdos,cambridge,trinity college,21-25 march,1983 Bollobas, Bela | Academic Press 1984
Congrès
V
ISBN 978-0-12-111760-3
Localisation : Colloque RdC
## Random hyperbolic graphs Kiwi, Marcos | CIRM H
Multi angle
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Research talks
Random hyperbolic graphs (RHG) were proposed rather recently (2010) as a model of real-world networks. Informally speaking, they are like random geometric graphs where the underlying metric space has negative curvature (i.e., is hyperbolic). In contrast to other models of complex networks, RHG simultaneously and naturally exhibit characteristics such as sparseness, small diameter, non-negligible clustering coefficient and power law degree distribution. We will give a slow pace introduction to RHG, explain why they have attracted a fair amount of attention and then survey most of what is known about this promising infant model of real-world networks. Random hyperbolic graphs (RHG) were proposed rather recently (2010) as a model of real-world networks. Informally speaking, they are like random geometric graphs where the underlying metric space has negative curvature (i.e., is hyperbolic). In contrast to other models of complex networks, RHG simultaneously and naturally exhibit characteristics such as sparseness, small diameter, non-negligible clustering coefficient and power law degree ...
## Connected chord diagrams, bridgeless maps, and perturbative quantum field theory Yeats, Karen | CIRM H
Multi angle
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Research talks
Rooted connected chord diagrams can be used to index certain expansions in quantum field theory. There is also a nice bijection between rooted connected chord diagrams and bridgeless maps. I will discuss each of these things as well as how the second sheds light on the first. (Based on work with Nicolas Marie, Markus Hihn, Julien Courtiel, and Noam Zeilberger.)
## Random cubic planar graphs revisited Rué, Juanjo | CIRM H
Multi angle
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Research School
We analyze random labelled cubic planar graphs according to the uniform distribution. This model was analyzed first by Bodirsky et al. in a paper from 2007. Here we revisit and extend their work. The motivation for this revision is twofold. First, some proofs where incomplete with respect to the singularity analysis and we provide full proofs. Secondly, we obtain new results that considerably strengthen those known before. For instance, we show that the number of triangles in random cubic planar graphs is asymptotically normal with linear expectation and variance, while formerly it was only known that it is linear with high probability.
This is based on a joint work with Marc Noy (UPC) and Clément Requilé (FU Berlin - BMS).
We analyze random labelled cubic planar graphs according to the uniform distribution. This model was analyzed first by Bodirsky et al. in a paper from 2007. Here we revisit and extend their work. The motivation for this revision is twofold. First, some proofs where incomplete with respect to the singularity analysis and we provide full proofs. Secondly, we obtain new results that considerably strengthen those known before. For instance, we show ...
## Recurrence of half plane maps Angel, Omer | CIRM H
Multi angle
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Research talks
On a graph $G$, we consider the bootstrap model: some vertices are infected and any vertex with 2 infected vertices becomes infected. We identify the location of the threshold for the event that the Erdos-Renyi graph $G(n, p)$ can be fully infected by a seed of only two infected vertices. Joint work with Brett Kolesnik.
## Random walk on random digraph Salez, Justin | CIRM H
Multi angle
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Research talks
A finite ergodic Markov chain exhibits cutoff if its distance to equilibrium remains close to its initial value over a certain number of iterations and then abruptly drops to near 0 on a much shorter time scale. Originally discovered in the context of card shuffling (Aldous-Diaconis, 1986), this remarkable phenomenon is now rigorously established for many reversible chains. Here we consider the non-reversible case of random walks on sparse directed graphs, for which even the equilibrium measure is far from being understood. We work under the configuration model, allowing both the in-degrees and the out-degrees to be freely specified. We establish the cutoff phenomenon, determine its precise window and prove that the cutoff profile approaches a universal shape. We also provide a detailed description of the equilibrium measure. A finite ergodic Markov chain exhibits cutoff if its distance to equilibrium remains close to its initial value over a certain number of iterations and then abruptly drops to near 0 on a much shorter time scale. Originally discovered in the context of card shuffling (Aldous-Diaconis, 1986), this remarkable phenomenon is now rigorously established for many reversible chains. Here we consider the non-reversible case of random walks on sparse ...
## Anchored expansion in the hyperbolic Poisson Voronoi tessellation Paquette, Elliot | CIRM H
Multi angle
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Research talks
We show that random walk on a stationary random graph with positive anchored expansion and exponential volume growth has positive speed. We also show that two families of random triangulations of the hyperbolic plane, the hyperbolic Poisson Voronoi tessellation and the hyperbolic Poisson Delaunay triangulation, have 1-skeletons with positive anchored expansion. As a consequence, we show that the simple random walks on these graphs have positive speed. We include a section of open problems and conjectures on the topics of stationary geometric random graphs and the hyperbolic Poisson Voronoi tessellation. We show that random walk on a stationary random graph with positive anchored expansion and exponential volume growth has positive speed. We also show that two families of random triangulations of the hyperbolic plane, the hyperbolic Poisson Voronoi tessellation and the hyperbolic Poisson Delaunay triangulation, have 1-skeletons with positive anchored expansion. As a consequence, we show that the simple random walks on these graphs have positive ...
## Spectral measures of factor of i.i.d. processes on the regular tree Backhausz, Ágnes | CIRM H
Multi angle
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Research talks
We prove that a measure on $[-d,d]$ is the spectral measure of a factor of i.i.d. process on a vertex-transitive infinite graph if and only if it is absolutely continuous with respect to the spectral measure of the graph. Moreover, we show that the set of spectral measures of factor of i.i.d. processes and that of $\bar{d}_2$-limits of factor of i.i.d. processes are the same.
##### Codes MSC
Nuage de mots clefs ici
Ressources Electroniques
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2017-11-21 02:44:21
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https://math.stackexchange.com/questions/2918467/proof-verification-sum-k-0n-binomnk-2n-spivaks-calculus
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# Proof Verification $\sum_{k=0}^n \binom{n}{k} = 2^n$ (Spivak's Calculus)
$$\sum_{k=0}^n \binom{n}{k} = 2^n$$ I'll use induction to solve prove this. Then
$$\sum_{k=0}^n \binom{n}{k} = 2^n = \binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n - 1} + \binom{n}{n}$$
First prove with n = 1
$$\binom{1}{0} + \binom{1}{1} = 2^1$$
Since $$\binom{1}{0} = \binom{1}{1} = 1$$
it's true.
Now suppose that is true with $n$ if is true with $n + 1$
Then, multiply both sides by two
$$2(2^n) = 2(\binom{n}{0} + \binom{n}{1} + ... + \binom{n}{n - 1} + \binom{n}{n})$$
$$2^{n+1} = 2\binom{n}{0} + 2\binom{n}{1} + ... + 2\binom{n}{n - 1} + 2\binom{n}{n}$$
$$2\binom{n}{0} + 2\binom{n}{1} + ... + 2\binom{n}{n - 1} + 2\binom{n}{n} = \binom{n}{0} + \binom{n}{0} + \binom{n}{1} + \binom{n}{1} + ... + \binom{n}{n} + \binom{n}{n}$$
The first term have two equal term, then, you sum the last one with the first one of the next term, and you'll get this
$$\binom{n}{0} + \binom{n}{1} +... + \binom{n}{n-1} + \binom{n}{n}$$
If we use this equation (Already proved)
$$\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}$$
Of course, we'll have two term without sum, one $\binom{n}{0}$ and $\binom{n}{n}$
We can write these two term like this
$$\binom{n}{0} = \binom{n}{n} = \binom{n+1}{0} = \binom{n+1}{n+1}$$
Then, we get
$$2^{n+1} = \binom{n+1}{0} + \binom{n+1}{1} + ... + \binom {n+1}{n+1}$$
And it's already proved. Note: Just if we take $0! = 1$
I have to prove these too.
$\sum_{k}^n \binom{n}{m} = 2^{n-1}$ If $m$ is even. And $\sum_{j}^n \binom{n}{j} = 2^{n-1}$ If $j$ is odd. Then, I just said that
If $$\sum_{m}^n \binom{n}{m} + \sum_{j}^n \binom{n}{j} = \sum_{k=0}^n \binom{n}{k}$$
Then
$$2^{n - 1} + 2^{n - 1} = 2^n$$
Which is true, then, I already prove this. And I have a last one.
$$\sum_{i=0}^n (-1)^i\binom{n}{i} = 0$$
if n is odd. Then
$$\binom{n}{0} - \binom{n}{1} + ... + \binom{n}{n-1} - \binom{n}{n} = 0$$
And that's can be solve knowing that $$\binom{n}{k} = \binom{n}{n - k}$$
And if n is even
$$\binom{n}{0} - \binom{n}{1} + ... - \binom{n}{n-1} + \binom{n}{n} = 0$$
That means that every negative term if when n is odd, then, we can use our two last prove to prove it
If $$\sum_{m}^n \binom{n}{m} - \sum_{j}^n \binom{n}{j} = 0$$
Then
$$2^{n-1} - 2^{n-1} = 0$$
Which is true.
And that's it, I want to know if my proves are fine and are rigorous too and what is the meaning of every combinatorics prove .
I want to know too better approaches to prove these (Or forms more intuitive)
• Please only ask one question at a time. Your first proof looks fine. I don't follow your second one. Your third one looks fine, but it could be written concisely as a consequence of the second result since the value of $(-1)^i$ depends on the parity of $i$. Also, if you are allowed to do so, the first problem can be answered via the binomial theorem when expanding $(1 + 1)^n$. – theyaoster Sep 16 '18 at 1:12
These are corollaries of the Binomial Theorem, which states that $$(x + y)^n = \sum_{k = 0}^{n} \binom{n}{k} x^{n - k}y^k$$
If we set $x = y = 1$, we obtain $$2^n = (1 + 1)^n = \sum_{k = 0}^{n} 1^{n - k}1^k = \sum_{k = 0}^{n} \binom{n}{k}$$ which means that the number of subsets of a set with $n$ elements is $2^n$.
If we set $x = 1$ and $y = -1$, we obtain $$0^n = [1 + (-1)]^n = \sum_{k = 0}^{n} 1^{n - k}(-1)^{k} = \sum_{k = 0}^{n} (-1)^k\binom{n}{k}$$ Notice that each term in which $k$ is even is positive and each term in which $k$ is odd is negative. Hence, $$\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} - \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1} = 0 \tag{1}$$ which means the number of subsets with an even number of elements is equal to the number of subsets with an odd number of elements.
Since every subset has an even number of elements or an odd number of elements, $$\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} + \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1} = \sum_{k = 0}^{n} \binom{n}{k} = 2^n \tag{2}$$ Adding equations 1 and 2 yields $$2\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} = 2^n \implies \sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} = 2^{n - 1}$$ which means the number of subsets with an even number of subsets is $2^{n - 1}$.
Since the number of subsets with an odd number of elements is equal to the number of subsets with an even number of elements, the number of subsets with an odd number of elements is $$\sum_{k = 0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k + 1} = 2^{n - 1}$$
A better mathematician than you or I once struggled as mightily with this question as you have, and I told him to expand $(1+1)^n$ with the binomial expansion.
here is another approach:
Consider a set with $n$ objects. There are $\binom nk$ ways to choose $k$ of them. That way gives $\sum_{i=0}^n\binom ni$ ways to choose objects from the set. On the other hand, we can choose to choose each item or not.That gives $2^n$. And of course they are equal, since they both represent the number of ways to choose some objects from the set.
The second and the third one are actually the same (can you see why?). Using Pascal's triangle $\sum_{2|i}^n\binom ni=\sum_{2\nmid i}^n\binom ni$ since they both equal to the sum of the line above them. (which is $2^{n-1}).$
• I don't get your second statement – Enigsis Sep 16 '18 at 1:09
• If we want to choose some elements from a set of size $n$, then for each element we must decide to choose or not to choose.There are 2 ways of choosing to a single element. Therefore to a set S with n elements there are $2^n$ ways. – abc... Sep 16 '18 at 1:16
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2019-08-18 18:16:09
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https://www.cram.com/subjects/calibration/5
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# Calibration
Page 5 of 33 - About 328 Essays
• ## An Analysis Of Pin Hole Camera Model
${\bf n}$ calibration points we have ${\bf 2n}$ equations:\\ $${\bf AP=0}$$ Camera calibration can be done by known patterns. Chess pattern is one of the standard pattern used for calibration. We know the number of rectangles and size of rectangles in the chess board pattern and pattern is a plane. The units that we use in calibration will be same for all other steps in SFM, using these information and $zang's$ method we can calibrate the camera. After calibration we…
Words: 2288 - Pages: 10
• ## Estimation Of Raloxifene Hydrochloride
standard solutions of concentrations in the range of 3-60µg/mL were obtained by further dilution of the aliquots with methanol. All the solutions were stored at -20oC and were equilibrated to room temperature prior to use. To prepare the standard calibration samples, 50µL of standard solution and 100µL of Acetonitrile were added to 100µL of blank plasma. The mixture was vortex-mixed for 10 min, followed by centrifugation at 4000 rpm for 15 min. 20µLof the supernatant was taken, then filtered…
Words: 777 - Pages: 4
• ## Solubility Analysis Of Ketoprofen
was found to be 96˚C by open capillary tube using microcontroller based melting point apparatus which is within the reported range of 94˚C to 97˚C. It complies with the standards thus indicating the purity of the drug sample. CALIBRATION CURVE OF KETOPROFEN: • Calibration curve of Ketoprofen in Ethanol Table 8: Absorbance values of ketoprofen in ethanol at 256nm CONCENTRATION(µg/ml) ABSORBANCE…
Words: 1273 - Pages: 6
• ## Biochemistry Lab Report
DIAGNOSIS OF ANAEMIA: DETERMINATION OF HAEMOGLOBIN CONCENTRATION Introduction Red blood cells contain the protein molecule haemoglobin (Hb); it is made up of four polypeptide chains and contains iron. Its function is to carry oxygen from the lungs to the bodies tissue and to carry carbon dioxide from tissue back to the lungs. Anaemia can be caused when there is a lack of iron in the body which leads to a reduction in the number of red blood cells. Iron is used to carry oxygen in the blood so If…
Words: 1430 - Pages: 6
• ## Importance And Methodology Of Palaeoclimatology
(Bradley, 2015) Data collection and calibration Geological sediments Geological evidence is the most widely distributed and commonly used approach to reconstruct the past climate. One of the essential way is the isotopic test. For example, the proportion of O18 in relation to O16 is temperature…
Words: 977 - Pages: 4
• ## Optokinetic Research Paper
However use of this device requires a trained operator. Calibration also needs to be done before using the device. Flexible electrogoniometers are strain-gage-based systems that consists of two light-weight end blocks fixed to the twelfth thoracic vertebrae(T12) and sacral spine(S2)spinal processes. They are separated…
Words: 992 - Pages: 4
• ## Internal Validity Essay
effect causes a threat to internal validity due to changes on the measuring instruments. 4). Instrumentation error is a threat to internal validity due instability or variations in the measuring instruments. These may include such as changes in calibration of the instrument. 5). Statistical regression threat may occur when research participants selected by relying on extreme indicators and assessments exhibit unforeseen extreme outcomes. 6). Selection of subject’s threat occurs where research…
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• ## Pah Lab
Determination of Most Advantageous Sequestering Agent By: Kayla Jeeter and Ashley Driscoll Chemistry 112-508 Abstract The creation and execution of this project served to determine what material is a better sequestering agent of the polycyclic aromatic hydrocarbons (PAHs) that are found in Texas’ water. Charcoal is currently the most typical agent used to sequester PAHs. In this experiment, two different Bentonite clays (zeolites) were synthesized, one magnetized and one not, to determine if…
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• ## Unit 3 Physics Lab 3
Lab 3: Technology Name: Section: General Physic Lab 3 Date: 10/03/2017 Purpose The goal of this experiment was to learn how to apply computerized instruments in physics. In addition, this experiment aimed at enhancing practice on how to use three equipment in the lab: the rotary motion sensor, high resolution force sensor, and motion sensor. Further, this experiment sought to clarify the circumstances under which the instruments fail to comply with expectations. Results/Data Part 1: The…
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• ## Research Paper Chromatography
Introduction Chromatographic process is a separation technique which has to do with the mass transfer of test sample between a stationary and mobile phase. HPLC (High Performace Liqud Chromatography) is a vast system which involves the use of a mobile and stationary phase to separate materials. The stationary phase is usually a column packed with solids (usually silica gel) while the mobile phase is usually a solvent or a mixture of various solvents. The mobile phase usually, is the carrier of…
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https://www.goconqr.com/flashcard/5251067/wjec-core-2-maths-key-facts
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# WJEC Core 2 Maths - Key Facts
Flashcards by , created about 3 years ago
## Key facts and formulae which must be known for the WJEC Core 2 examination.
78
2
0
Created by Daniel Cox about 3 years ago
Core 2 AS level maths formulae OCR
Pure Mathematics with FP1 and FP2
WJEC Mechanics 1 - Key Facts
Plano de Revisão Geral
THE CASES - GERMAN
WJEC Core 1 Maths - Key Facts
AS Pure Core 1 Maths (AQA)
Further Mathematics
WJEC Core 4 Maths - Key Facts
WJEC Core 3 Maths - Key Facts
Question Answer $\log_a x+\log_a y = ?$ $\log_a x+\log_a y = \log_a(xy)$ $\log_a x-\log_a y = ?$ $\log_a x-\log_a y = \log_a\left (\frac{x}{y} \right )$ NOT $$\frac{\log_a x}{\log_a y}$$ $k \log_a x = ?$ $k \log_a x = \log_a\left (x^k \right )$ State the sine rule $$\frac{a}{\sin A}=\frac{b}{\sin B}$$ or $$\frac{\sin A}{a}=\frac{\sin B}{b}$$ True or false? $\log_a\left (xy^k \right )=k \log_a\left ( xy \right )$ FALSE \begin{align*} \log_a\left (xy^k \right )&=\log_a x +\log_a \left (y^k \right )\\ &=\log_a x + k \log_a y \end{align*} What is the trigonometric formula for the area of a triangle? $Area=\frac{1}{2} ab \sin C$ Here, the sides $$a$$ and $$b$$ surround the angle $$C$$ What is the Pythagorean trigonometric identity? (Hint: it involves $$\sin^2 x$$ and $$\cos^2 x$$ $\sin^2 x + \cos^2 x = 1$ If $$y=a^x$$, then $$x=?$$ If $$y=a^x$$, then $$x=\log_a y$$ State the cosine rule $a^2=b^2+c^2-2bc \cos A$ $\log_a a =?$ $\log_a a =1$ $\log_a 1 =?$ $\log_a 1 =0$ State an identity relating $$\sin x$$, $$\cos x$$ and $$\tan x$$ $\frac{\sin x}{\cos x}=\tan x$ How many degrees is $$\pi$$ radians? $$\pi$$ radians is $$180^{\circ}$$ Formula for the area of a sector? 246e5789-f0a7-4bdd-adf3-6c292cebdf77.png (image/png) $Area=\frac{1}{2}r^2 \theta$ d9908c47-eea7-4a82-8040-ff473cf307cf.png (image/png) Formula for the length of an arc? 7b0c27d1-8771-4950-8671-09fd560d29b0.gif (image/gif) $s=r \theta$ 33c8b397-c996-4ddc-be92-21aa0cb46497.gif (image/gif) How would you find the area of a segment of a circle? 3f2f7ddf-b1c4-480f-86b7-8009148d4c6a.gif (image/gif) \begin{align*} \mathrm{Segment}&= \mathrm{Sector}-\mathrm{Triangle}\\ &=\frac{1}{2}r^2\theta-\frac{1}{2}r^2 \sin\theta\\ &=\frac{1}{2}r^2\left ( \theta - \sin\theta \right ) \end{align*} Formula for the $$n$$th term of an arithmetic sequence... [given in the formulae booklet] $u_n=a+(n-1)d$ Formula for the sum of the first $$n$$ terms of an arithmetic sequence... [given in the formulae booklet] $S_n=\frac{n}{2}\left ( 2a+(n-1)d \right )$ or $S_n=\frac{n}{2}\left ( a+l \right )$ where $$l$$ is the last term Formula for the $$n$$th term of a geometric sequence... [given in the formulae booklet] $u_n=ar^{n-1}$ Formula for the sum of the first $$n$$ terms of a geometric sequence... [given in the formulae booklet] $S_n=\frac{a\left ( 1-r^n \right )}{1-r}$ Formula for the sum to infinity of a convergent geometric series (one where $$\left | r \right |<1$$) [given in the formulae booklet] $S_\infty=\frac{a}{1-r}$ $\int ax^n \, dx=\, ?$ $\int ax^n \, dx= \frac{ax^{n+1}}{n+1}+c$ How would you find this shaded area? 0a3690b2-1482-449a-aae6-dff7cd33d632.png (image/png) Work out $$\int_{a}^{b} f(x) \, dx$$ General equation of a circle, centre $$\left ( a,b \right )$$ and radius $$r$$ $\left ( x-a \right )^2+\left ( y-b \right )^2=r^2$ If we are given $$\frac{dy}{dx}$$ or $$f'(x)$$ and told to find $$y$$ or $$f(x)$$, we need to... Integrate [remember to include $$+c$$] What is the angle between the tangent and radius at $$P$$? 836b171c-39e7-4991-8c29-be862870372c.jpg (image/jpg) $90^{\circ}$ This is always true at the point where a radius meets a tangent 7b9e0001-91f1-483e-9305-f4ace70cf194.png (image/png) What does the graph of $$y=a^x$$ look like? Where does it cross the axes? It goes through the $$y$$-axis at $$\left ( 0,1 \right )$$. It does not cross the $$x$$-axis. The $$x$$-axis is an asymptote. 16493cc7-79d3-45f7-b601-63787610188f.jpg (image/jpg) This is a triangle inside a semicircle, where one side of the triangle is the diameter of the circle. What is the size of angle $$C$$? 7528b419-763e-4ed3-95f2-3fd66f8da729.gif (image/gif) $90^{\circ}$ Draw the graph of $$y=\sin x$$ for $$0\leq x \leq 2\pi$$ c5195b8d-e25a-460f-a4a3-3189515f3468.gif (image/gif) Draw the graph of $$y=\cos x$$ for $$0\leq x \leq 2\pi$$ eda70f8f-b1c3-4e65-9f56-d06139c3c3ad.gif (image/gif) Draw the graph of $$y=\tan x$$ for $$0\leq x \leq 2\pi$$ The lines $$x=\frac{\pi}{2}$$ and $$x=\frac{3\pi}{2}$$ are asymptotes 9e44c61a-ff49-4279-92e9-50bbaf346ee4.png (image/png) Differentiation is the reverse of ...? Integration Integration is the reverse of ...? Differentiation What is the condition for two circles to touch externally? The distance between their centres is the sum of their radii d5d1fe08-2a7a-4f54-a678-1df57dcfee1c.png (image/png) What is the condition for two circles to touch internally? The distance between their centres is the difference in their radii c9d8250c-de62-48e6-b82e-3bbde32d1de6.png (image/png) If we draw the perpendicular bisector of any chord on a circle, which point will it definitely go through? The perpendicular bisector of a chord always passes through the centre of the circle 0c4a576d-58b4-4667-8408-927571967d0f.png (image/png)
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2019-05-19 16:27:28
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https://www.physicsforums.com/threads/greens-theorem.797126/
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# Green's Theorem
1. Feb 10, 2015
### Amy Marie
1. The problem statement, all variables and given/known data
Use, using the result that for a simple closed curve C in the plane the area enclosed is:
A = (1/2)∫(x dy - y dx) to find the area inside the curve x^(2/3) + y^(2/3) = 4
2. Relevant equations
Green's Theorem:
∫P dx + Q dy = ∫∫ dQ/dx - dP/dy
3. The attempt at a solution
I solved the equation of the curve for x:
x = (4 - y^(2/3))^(3/2)
Also, from the original curve equation x^(2/3) + y^(2/3) = 4, when x = 0, y = +/- 8 because 4^(3/2) = 8.
But when I plug x = +/- (4 - y^(2/3))^(3/2) in for the x bounds and y = +/- 8 in for the y bounds in the resulting double integral
(1/2)∫∫ 2 dxdy
I have trouble integrating x = (4 - y^(2/3))^(3/2) it with respect to y.
Does anybody happen to know if there is a more correct way to solve this problem?
Thank you for your help!
2. Feb 10, 2015
### LCKurtz
The question does not want you to work the integral out by doing a double integral. It wants you to find a nice parameterization of the curve and do$$A = \frac 1 2 \int_C x~dy - y~dx$$ So your first job is to find a nice parameterization. As a hint think about a way to parameterize it so that you can use the identity $(r\cos\theta)^2 + (r\sin\theta)^2 = r^2$.
3. Feb 12, 2015
### Amy Marie
Thank you for your help!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook
Have something to add?
Draft saved Draft deleted
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2017-11-20 02:34:31
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http://artificial-intelligence-class.org/r2d2_assignments/hw1/homework1.html
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Skip to main content
Robot Excercise 1: Using Python to Control R2D2 [0 points]
Instructions
In this assignment, you’ll learn how to write Python code to control your robot. A skeleton file homework1.py containing empty definitions for each question has been provided. Since portions of this assignment will be graded automatically, none of the names or function signatures in this file should be modified. However, you are free to introduce additional variables or functions if needed.
Unless explicitly stated otherwise, you may not import any of the standard Python modules, meaning your solutions should not include any lines of the form import x or from x import y. Accordingly, you may find it helpful to refresh yourself on Python’s built-in functions and data types.
You will find that in addition to a problem specification, each programming question also includes a pair of examples from the Python interpreter. These are meant to illustrate typical use cases, and should not be taken as comprehensive test suites.
You may submit as many times as you would like before the deadline, but only the last submission will be saved.
You are strongly encouraged to follow the Python style guidelines set forth in PEP 8, which was written in part by the creator of Python. However, your code will not be graded for style.
Once you have completed the assignment, you should submit your file on Gradescope.
You may submit as many times as you would like before the deadline, but only the last submission will be saved.
Set Up
For now, we have setup instructions for Mac. We are working on creating instructions for Windows and Linux.
To get started, we are going to download a set of APIs for controling the robots via Bluetooth. This step is a little bit complicated, so we recommend that you start early, so that you can ask for help if you get stuck.
1. Download and install the current version of Python from https://www.python.org/downloads/.
2. After you have installed it, you will need to Double click on the “Install Certificates.command” file in your /Applications/Python 3.7 folder.
3. The Mac installer will put it at /usr/local/bin/python3 which you can verify by checking the date after typing
ls -la /usr/local/bin/py*
You should see today’s date.
4. Make sure you have brew installed. You can find out whether you’ve got it by typing
which brew
in ther terminal. If you don’t have it, then install it from here: https://brew.sh
5. Download the repo
git clone https://github.com/josephcappadona/sphero-project.git
cd sphero-project
6. Create a virtual environment
/usr/local/bin/python3 -m venv r2d2
source r2d2/bin/activate
python -m pip install --upgrade pip
7. Set your Path
export PATH=/usr/bin:/bin:/usr/sbin:/sbin:/usr/local/bin:/opt/X11/bin
8. Set up Node in your virtual environment
python -m pip install nodeenv
nodeenv -p --node=10.15.3
brew install yarn
9. Install python dependencies
python -m pip install numpy pygame
10. Compile the server library and dependencies
cd spherov2.js
rm yarn.lock
sudo yarn install
cd lib/
yarn rebuild
11. Start the server, and leave it running in its own Terminal window.
cd ../examples/
sudo yarn server
12. Open a new Terminal window, and change into your sphero-project director. Then activate your virual environment.
cd sphero-project/
source r2d2/bin/activate
1. Change into the src directory and run python:
cd src/
python
2. Try copying and pasting these commands into the Python environment:
from client import DroidClient
droid = DroidClient()
droid.scan() # Scan the area for droids.
# Connect to your robot.
droid.connect_to_droid('D2-55A2') # Replace D2-55A2 with your ID
droid.animate(5)
If you hear a happy chirp, you’re ready to go!
1. Let’s get rolling
You can start a Python session on your terminal by typing python3. This will launch a “REPL” environment. REPL stands for read, eval, print, loop. That means you can interactively write and test code in the terminal.
First launch the R2D2 server. This will manage the bluetooth connection between your computer and the robot, and will relay commands from your python code to the robot. You should launch the server in its own terminal window by running the commands:
cd sphero-project/spherov2.js/examples
sudo yarn server
It will ask for your password, and then if everything is working properly, you’ll see this:
yarn run v1.16.0
$ts-node src/server Listening... Next start a new terminal window and launch python by typing: python3 You’ll now see a REPL environment with a >>> prompt that lets you interactively enter python commands. Try copying and pasting these commands: from client import DroidClient droid = DroidClient() droid.scan() # Scan the area for droids. # Connect to your robot. droid.connect_to_droid('D2-55A2') # This is not the droid you're looking for. Replace D2-55A2 with your droid's ID droid.animate(5) This should cause your robot to make a friendly chirping sound. Next you can have it roll by passing it 3 arguments, the speed (ranging from 0.0 to 1.0), the heading (0 to 360) and the amount of time to roll (in seconds). droid.roll(0.3, 90, 3) The heading variable is relative to the robot’s orientation when you first connect to it. 0° is straight ahead, 90° is to its right, 180° is behind it, and 270° is to its left. droid.turn(90) # turn right droid.turn(180) # turn to face backwards from the initial orientation droid.turn(270) # turn to face left from the initial orientation droid.turn(0) # return to the initial orientation The robot has a gyroscope, which allows it to remember its original orientation, even if you pick it up and manually reposition it. Turning back to 0° will cause it to face the same direction as when it woke up. If you’d like to explore what commands a Python class supports, you can use the dir(ClassName) or help(ClassName) functions. Type help(DroidClient) in the Python REPL environment to see what methods your robots supports. 2. For Loops in Python Let’s drive in a square by dividing forward, turning 90 degrees and then driving forward again. We can use a for loop to roll forward and change heading 4 times: heading = 0 for i in range(4): droid.roll(100, heading, 2) heading = heading + 90 For loops in Python are done differently than for loops in Java or C++. In a for loop in Java, we initialize a variable, test a truth condition, and then increment (or decrement a variable) like so: for(int i=1; i<=10; i++){ System.out.println(i); } Whereas in Python, we actually have a variable that gets assigned a value based on each element in a list. In our Python for statement that caused the robot to drive in a square, we used the range() function, which generates a list of numbers. The python expression for i in range(num): is very common. 3. Lists Python’s for loop also allows us to execute a series of roll commands based on a list of headings. Creating a list in python is easy. We can initalize a list with a bunch of vables like like this: headings = [0, 90, 180, 270] Or we can start with an empty list, and then add values to it using the append() method. headings = [] # initalize an empty list headings.append(0) # add an item to the end of the list headings.append(90) headings.append(180) headings.append(270) Once we’ve got that list, we could use a for loop to drive the robot in a square. We can do this in a few ways. We could use a for loop with range() again and giving it the length of the list as its argument, which we can get via len(headings). for i in range(len(headings)): heading = headings[i] droid.roll(100, heading, 2) But there is a cooler way to do it in Python without a variable i: for heading in headings: droid.roll(100, heading, 2) 4. Tuples and multiple return variables Our roll command takes three arguments speed, heading, and duration. We can encode all three of those into a Python type called a tuple. A tuple is an ordered list of values. In Python a tuple is immutable, meaning the its elements cannot be changed (unlike a list). In Python tuples are written with round brackets, and their elements can be accessed with an index in square brackets (just like accessing an element of a list). roll_command = (100, 0, 3) speed = roll_command[0] heading = roll_command[1] duration = roll_command[2] Let’s create a list of (speed, heading, duration) tuples. roll_commands = [] # create an empty list roll_commands.append((200, 0, 1)) # add an item to the end of the list roll_commands.append((200, 72, 1)) roll_commands.append((200, 144, 1)) roll_commands.append((200, 216, 1)) roll_commands.append((200, 288, 1)) roll_commands.append((0, 0, 0)) Or equivalently: roll_commands = [ (200, 0, 1), (200, 72, 1), (200, 144, 1), (200, 216, 1), (200, 288, 1), (0, 0, 0), ] You could use a for loop to execute each of the roll commands in turn like this: for i in range(len(roll_commands)): command = roll_commands[i] speed = command[0] heading = command[1] duration = command[2] droid.roll(speed, heading, duration) Python also allows multiple return types, which means that we can assign several variables simulatenously. So we can say: speed, heading, duration = roll_commands[0] This allows us to write very concise for loops: for speed, heading, duration in roll_commands: droid.roll(speed, heading, duration) 5. Python functions Instead of manually specifying the commands to have the robot drive in a square or a pentgon, let’s write a function that will let it drive in the shape of any polygon. Let’s see how to write a python function. Here’s how we can write a function for driving in a square. def trace_square(speed=100, duration=2): heading = 0 for i in range(4): print("Heading: %i" % heading) droid.roll(speed, heading, duration) heading += 90 Now let’s generalize it to be any regular polygon. For a polygon with$n\$ sides, we’ll need to compute what angle to turn to turn instead of 90° in a square.
Here’s how to compute the interior angle of a polygon: $\frac{(n-2) \cdot 180}{n}$ The angle that you want to turn the droid is the exterior angle. The exterior angle is 180° minus the interior angle. Try to implement this function:
def trace_polygon(n, speed=100, duration=2):
interior_angle = # todo
exterior_angle = # todo
heading = 0
for i in range(n):
droid.roll(speed, heading, duration)
heading += # todo
Did you get it? If you want to see the solution check here.
In Python, you can set default values to arguments that you pass into a function. In this case, we have set default values for speed and duration. This means that you can call the function just by specifying the number of sides in the polygon.
trace_polygon(3) # triangle
trace_polygon(3, speed=255) # faster, and therefore larger triangle
trace_polygon(3, speed=255, duration=1)
trace_polygon(4) # square
trace_polygon(5) # pentagon
trace_polygon(8, duration=1) # octogon, making it smaller by setting the duration value to be lower
6. Python Dictionaries
Python dictionaries are hash tables that let us store key-value pairs. Let’s use a dictionary to map from color names (Strings) onto their corresponding RGB values. We’ll store the RGB values as (red, green, blue) triples that indiciate the intesity of each of those colors (ranging from 0 to 255).
Here’s how we create an empty dictionary in Python:
color_names_to_rgb = {}
We can add elements like this:
color_names_to_rgb['black'] = (0,0,0)
color_names_to_rgb['white'] = (255,255,255)
color_names_to_rgb['red'] = (255,0,0)
color_names_to_rgb['blue'] = (0,0,255)
color_names_to_rgb['yellow'] = (255,255,0)
color_names_to_rgb['cyan'] = (0,255,255)
color_names_to_rgb['aqua'] = (0,255,255) # There can be two keys with the same value, but not one key with multiple values.
color_names_to_rgb['magenta'] = (255,0,255)
color_names_to_rgb['gray'] = (192,192,192)
We can look up values like this:
color_name = "blue"
color_code = color_names_to_rgb[color_name]
We can iterate through keys like this:
for color_name in color_names_to_rgb.keys():
r,g,b = color_names_to_rgb[color_name]
print("The color %s has %i parts red, %i parts green, and %i parts blue." % (color_name,r,g,b)) # This is. a fancy print statement.
We can set the color on our droid’s front and back lights with the set_front_LED_color and set_back_LED_color functions, which take RGB values as input.
droid.set_front_LED_color(255,0,255)
droid.set_back_LED_color(255,0,255)
Why don’t you try to write a function that will set the color of the droid’s lights to the same value using the color name?
def set_lights(color_name, which_light='both'):
r,g,b = color_names_to_rgb[color_name]
if(which_light=='both'):
# TODO
Now, let’s write a function to walk through a list of colors, and set the light to each color in turn for a certain number of seconds. We’ll use Python’s time library to sleep between color changes.
import time
time.sleep(2) # Example of how to wait for 2 seconds
rainbow = ['red', 'orange', 'yellow', 'green', 'blue', 'indigo', 'violet']
def init_color_names_to_rgb():
# TODO create a color names to RGB code dictionary
# Here's a web page with color names to RGB codes
# https://www.rapidtables.com/web/color/RGB_Color.html
def flash_colors(colors, seconds=1):
# TODO call the set_lights method on each color in the colors list
# wait for the specified number of seconds in between
color_names_to_rgb = init_color_names_to_rgb()
If you want to see the solution check here.
7. Sorting and Lambda Functions
Let’s create a list of roll commands:
roll_commands = [ (20, 0, 1),
(40, 72, 1),
(60, 144, 1),
(80, 216, 1),
(100, 288, 1),
(120, 0, 1),
(140, 72, 1),
(160, 144, 1),
(180, 216, 1),
(200, 288, 1),
(220, 0, 1),
(240, 72, 1),
(260, 144, 1),
(280, 216, 1),
(300, 288, 1),
(0, 0, 0), ]
Here’s a quick function to exectute them in order:
def roll_list(roll_commands):
for speed, heading, duration in roll_commands:
droid.roll(speed, heading, duration)
OK, so we’ve a list of a bunch of headings. You can think of these as vectors. We should be able add these in any order, and the resulting end point will be the same. Is that right? I can’t remember. Let’s just try it out and see.
Let’s try sorting the roll_commands in different orders, and see whether the droid ends up at the same location, as when the commands are executed in the initial order.
You can user the shuffle command to put the items in a list in random order.
from random import shuffle
shuffle(roll_commands)
print(roll_commands)
You can sort them with the sort method on lists:
roll_commands.sort()
print(roll_commands)
You can also sort them in reverse order with the reverse keyword argument:
roll_commands.sort(reverse=True)
print(roll_commands)
You can also sort a list by specifying a custom function to call on the objects within the list:
def sortSecond(val):
return val[1]
roll_commands.sort(key=sortSecond, reverse=False)
print(roll_commands)
Python supports lambda functions, which allows us to rewrite the previous snippet without the need to explicitly define a sortSecond function:
roll_commands.sort(key=lambda e: e[1], reverse=False)
print(roll_commands)
8. Driving with the keyboard arrow keys
Let’s design a video game style controler for the robot, where we can use the arrow keys to change its speed (by pressing up or down) and its orientation (by pressing left or right)
First, we’ll give you a function for reading in a keystroke from the keyboard. Here it is:
import sys,tty,os,termios
def getkey():
old_settings = termios.tcgetattr(sys.stdin)
tty.setcbreak(sys.stdin.fileno())
try:
while True:
b = os.read(sys.stdin.fileno(), 3).decode()
if len(b) == 3:
k = ord(b[2])
else:
k = ord(b)
key_mapping = {
127: 'backspace',
10: 'return',
32: 'space',
9: 'tab',
27: 'esc',
65: 'up',
66: 'down',
67: 'right',
68: 'left'
}
return key_mapping.get(k, chr(k))
finally:
termios.tcsetattr(sys.stdin, termios.TCSADRAIN, old_settings)
Next, you can write a function to continuously read in the keyboard input and use it to drive the robot.
def drive_with_keyboard(speed_increment=30, heading_increment=45, duration=0.1):
speed = 0
heading = 0
max_speed = 255
while True:
key = getkey()
if key == 'esc':
break
elif key == 'up':
# TODO - finish this function
9. Sending a message
In Star Wars, R2-D2 delivers a message from Princess Leia to Obi-Wan Kenobi. Our robots can only play pre-programmed sounds, so we will use the robot’s lights to blink out the message “Help me, Obi-Wan Kenobi. You’re my only hope.” in Morse Code.
Here we will use the Python concept of a generator.
Generators behave similar to iterators like lists, so they can be used in Python’s for loops. They have the added nice property of creating the next item on-demand, which means that they can often be more efficient than the equivalent operation of generating a list and returning its iterator. That property is nice in this example, since the message that Leia sent to Obi-Wan is actually quite a bit longer than everyone remembers:
General Kenobi. Years ago, you served my father in the Clone Wars. Now he begs you to help him in his struggle against the Empire. I regret that I am unable to present my father’s request to you in person, but my ship has fallen under attack and I’m afraid my mission to bring you to Alderaan has failed. I have placed information vital to the survival of the Rebellion into the memory systems of this R2 unit. My father will know how to retrieve it. You must see this droid safely delivered to him on Alderaan. This is our most desperate hour. Help me, Obi-Wan Kenobi. You’re my only hope.
Try writing these methods:
def encode_in_morse_code(message):
# TODO - Create a generator that walks through each
# character in a message string.
# Skip over any non-alphanumeric characters.
# Encode numbers and letters as morse code strings like
# A = '.-'
# B = '-...'
# ...
# Return a string for the next character.
def play_message(message, droid, short_length=0.1, long_length=0.3, length_between_blips=0.1, length_between_letters=0.5):
# TODO - blink out the message on the holo projector.
|
2019-07-21 13:31:43
|
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|
https://stats.stackexchange.com/questions/144190/probability-for-selecting-centroids-k-means
|
# Probability for selecting centroids - K-means++
K-means++ selects centroids one by one, where each point has the chance to become next centroid with probability proportional to distance to closest centroid already selected.
I implemented it like this (for selecting one centroid):
• Calculate distance for each point to existing centroids and save distance to closest centroid
• Divide distance to closest centroid by total distance in cluster of that centroid (sum of distances from each point in cluster to centroid)
• Sort distances for all points (in this step we consider that given distances that were divided by total distance represent probability for selecting the point as centroid)
• Create an array of cummulative probabilities (for example, array 0.2, 0.3, 0.5 gives array 0.2, 0.5, 1)
• Generate a random number between 0 and 1
• New centroid is the point that is represented by the interval to which the generated number belongs
In this way third point from the example has the greatest probability being selected since it has the greatest interval.
Is this a good way to implement K-means++ like initialization?
The problem is that I'm not getting the expected result, so I'm not sure if I misunderstood the concept or I got something wrong with implementation.
The problem occurs with a relatively large dataset (5000 points) with 15 clusters. Concretely - it happens that multiple points from the same cluster, that are relatively close to each other, get selected as centroids.
I'm guessing that intervals get relatively small since there are 5000 points and then the concept of greater probabilities gets lost. For example, if there's 800 points in some cluster we get probabilities of about 1/800, so probability for one point could be for example 0.00125 and for the other one 0.00130, so it seems like I'm getting away from weighted probability and getting to uniform probability.
Am I missing something?
The paper (see section 2.2) suggests that you use the squared distance when computing probabilities. In fact, you can try distance$^{\ell}$ using any exponent $\ell$ greater than 1. It stands to reason that as $\ell$ increases, the likelihood of initial centroids being close together will go to zero.
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2019-11-22 00:09:09
|
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|
https://www.turito.com/ask-a-doubt/if-2pi-7-then-tan-alpha-tan2alpha-tan2alpha-tan4alpha-tan4alpha-tan-alpha-7-5-3-1-q265c77
|
Maths-
General
Easy
### Hint:
In this question, we have to find the value of tan, if . For this we will solve the function using common trigonometric identity and simplify it and later substitute the value of in the equation.
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2022-09-27 23:04:49
|
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https://www.physicsforums.com/threads/does-more-angular-monentum-mean-faster-spin.874282/
|
# B Does more angular monentum mean faster spin?
1. Jun 3, 2016
### avito009
If a top has angular momentum of 12 units and the earth has angular momentum of 100. Does this mean that Earth is spinning faster than the top since it has more angular momentum? The answer is there at the back of my head but cant articulate it.
2. Jun 3, 2016
### A.T.
No. Angular velocity tells you what spins faster.
3. Jun 3, 2016
### Aniruddha@94
No. Angular momentum ( its magnitude) is the product of moment of inertia and the angular velocity. In your case the rotational inertia of the earth and the top aren't equal, so you can't make the conclusion as you did.
Last edited: Jun 3, 2016
4. Jun 3, 2016
### sophiecentaur
A typo, there, it's Angular Momentum that you meant.
5. Jun 3, 2016
### Aniruddha@94
Aah yes, sorry.. I'll edit it. Thanks.
6. Jun 3, 2016
### DrewD
As everyone else has said, the answer is no. You could only draw the conclusion you draw if the objects are the same (same mass and distribution of mass). To be more precise, the moment of inertia needs to be the same. I think you could draw this conclusion from other posts, but I just wanted to be explicit.
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2017-08-22 06:58:47
|
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https://tednote.com/post/2020-01-29-salesforce-business-date-formula/
|
Just putting this here so I remember it if I ever need it in the future.
There was a Trailblazer community post I came across and thought “I can do that” - and it looks like I did. The OP had a formula he was using to calculate 5 business days before an event date and now needed to change the calculation to 9 days prior. The OP wasn’t sure how to do it.
The issue was that with the date calculation, the OP needed to adjust the number of weekend days for some of the days of the week as they would now cross two weekends (4 days) and not just one weekend (two days). I will admit that I had to do some searching to help figure out how the Salesforce date formulas worked, but it was easily found and here’s the result. Sunday is “0” and you can figure out the rest. It then ends with the last two days of the week using the (2 + 9) formula since there’s no point in repeating it for them. And you get an error if there are an even number of arguments in the CASE formula.
Big_Event_date__c -
CASE(
MOD( Big_Event_Date__c - DATE( 1900, 1, 7 ), 7 ) ,
0, (3 + 9),
1, (4 + 9),
2, (4 + 9),
3, (4 + 9),
4, (4 + 9),
(2 + 9))
If you have stumbled across this post for a date calculation, I hope it helps.
Resources:
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2022-01-26 08:53:04
|
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|
http://hal.in2p3.fr/in2p3-00003488
|
# Study of Neutral-Current Four-Fermion and ZZ Production in $e^+ e^-$ Collisions at $\sqrt{s}$= 183 GeV
3 CMS
IP2I Lyon - Institut de Physique des 2 Infinis de Lyon
Abstract : Study of Neutral--Current Four--Fermion and ZZ \\ Production in $\rm e^+ e^-$ Collisions at $\rm \sqrt{s}=$ 183 GeV A study of neutral--current four--fermion processes is performed using a data sample corresponding to 55.3~pb$^{-1}$ of integrated luminosity collected by the L3 detector at LEP at an average centre--of--mass energy of $183~\GeV$. The neutral--current four--fermion cross sections for final states with a pair of charged leptons plus jets and with four charged leptons are measured to be consistent with the Standard Model predictions. Events with fermion pair masses close to the Z boson mass are selected in all observable final states and the ZZ production cross section is measured to be %\begin{center} $\rm \sigma_{ZZ} = 0.30 ^{+0.22\,\,+0.07}_{-0.16\,\,-0.03}\,\mathrm{pb},$ %\end{center} in agreement with the Standard Model expectation. No evidence for the existence of anomalous triple gauge boson ZZZ and ZZ$\gamma$ couplings is found and limits on these couplings are set.
Document type :
Journal articles
Cited literature [19 references]
http://hal.in2p3.fr/in2p3-00003488
Contributor : Sylvie Flores <>
Submitted on : Tuesday, May 25, 1999 - 4:23:39 PM
Last modification on : Friday, December 4, 2020 - 1:22:16 PM
Long-term archiving on: : Friday, May 29, 2015 - 4:11:38 PM
### Citation
M. Acciarri, P. Achard, O. Adriani, M. Aguilar-Benitez, J. Alcaraz, et al.. Study of Neutral-Current Four-Fermion and ZZ Production in $e^+ e^-$ Collisions at $\sqrt{s}$= 183 GeV. Physics Letters B, Elsevier, 1999, 450, pp.281-293. ⟨10.1016/S0370-2693(99)00103-3⟩. ⟨in2p3-00003488⟩
Record views
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2021-05-10 09:28:36
|
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http://ergodicity.net/tag/signal-processing/
|
I’m in Austin right now for the first GlobalSIP conference. The conference has a decentralized organization, with semi-independent day-long workshops (“symposia”) scheduled in parallel with each other. There are 8 of these, with 6 running in parallel per day, with 1 session of “plenary” talks and 2 poster sessions. Each workshop is scheduled in AAB, ABA, or BAA, where A = posters and B = plenary, so there are 2 talk sessions and 4 poster sessions running in parallel.
Fortunately, there are a wide range of topics covered in the workshops, from biology to controlled sensing, to financial signal processing. The downside is that the actual papers in each workshop often fit well with other workshops. For example, the distributed optimization posters (in which I am interested), were sprinkled all over the place. This probably has a lot to do with the decentralized effects.
In terms of the “results” at the conference, it seems from my cursory view that many people are presenting “extra” results from other conference papers, or preliminary work for future papers. This actually works well in the poster format: for the former, the poster contains a lot of information about the “main result” as well, and for the latter, the poster is an invitation to think about future work. In general I’m a little ambivalent about posters, but if you’re going to have to do ‘em, a conference like this may be a better way to do it.
I recently saw that Andrew Gelman hasn’t really heard of compressed sensing. As someone in the signal processing/machine learning/information theory crowd, it’s a little flabbergasting, but I think it highlights two things that aren’t really appreciated by the systems EE/algorithms crowd: 1) statistics is a pretty big field, and 2) the gulf between much statistical practice and what is being done in SP/ML research is pretty wide.
The other aspect of this is a comment from one of his readers:
Meh. They proved L1 approximates L0 when design matrix is basically full rank. Now all sparsity stuff is sometimes called ‘compressed sensing’. Most of it seems to be linear interpolation, rebranded.
I find such dismissals disheartening — there is a temptation to say that every time another community picks up some models/tools from your community that they are reinventing the wheel. As a short-hand, it can be useful to say “oh yeah, this compressed sensing stuff is like the old sparsity stuff.” However, as a dismissal it’s just being parochial — you have to actually engage with the use of those models/tools. Gelman says it can lead to “better understanding one’s assumptions and goals,” but I think it’s more important to “understand what others’ goals.”
I could characterize rate-distortion theory as just calculating some large deviations rate functions. Dembo and Zeitouni list RD as an application of the LDP, but I don’t think they mean “meh, it’s rebranded LDP.” For compressed sensing, the goal is to do the inference in a computationally and statistically efficient way. One key ingredient is optimization. If you just dismiss all of compressed sensing as “rebranded sparsity” you’re missing the point entirely.
There’s an opening in Professor Madhow’s group at UC Santa Barbara:
We are looking for a postdoctoral researcher with a strong background in communications/signal processing/controls who is interested in applying these skills to a varied set of problems arising from a number of projects. These include hardware-adapted signal processing for communications and radar, neuro-inspired signal processing architectures, and inference in online social networks. In particular, familiarity with Bayesian inference is highly desirable, even if that is not the primary research area for his/her PhD. There are also opportunities to work on problems in next generation communication systems, including millimeter wave networking and distributed communication. While the researcher will be affiliated with Prof. Madhow’s group in the ECE Department at UCSB, depending on the problem(s) chosen, he/she may need to interact with faculty collaborators in other disciplines such as circuits, controls, computer science and neuroscience, as well as with colleagues with expertise in signal processing and communications. Thus, in addition to technical depth and talent, a flexible attitude and openness to interdisciplinary collaboration is essential.
Interested candidates should send a brief statement of research experience and interests and a CV (including the names and contact info for at least three references) to Prof. Upamanyu Madhow.
I’m on the program committee for the Cyber-Security and Privacy symposium, so I figured I would post this here to make more work for myself.
GlobalSIP 2013 – Call for Papers
IEEE Global Conference on Signal and Information Processing
December 3-5, 2013 | Austin, Texas, U.S.A.
GlobalSIP: IEEE Global Conference on Signal and Information Processing is a new flagship IEEE Signal Processing Society conference. The focus of this conference is on signal and information processing and up-and-coming signal processing themes.
GlobalSIP is composed of symposia selected based on responses to the call-for-symposia proposals. GlobalSIP is composed of symposia on hot topics related to signal and information processing.
The selected symposia are:
Paper submission will be online only through the GlobalSIP 2013 website Papers should be in IEEE two-column format. The maximum length varies among the symposia; be sure to check each symposium’s information page for details. Authors of Signal Processing Letters papers will be given the opportunity to present their work at GlobalSIP 2013, subject to space availability and approval by the Technical Program Chairs of GlobalSIP 2013. The authors need to specify in which symposium they wish to present their paper. Please check conference webpage for details.
Important Dates:
*New* Paper Submission Deadline – June 15, 2013
Review Results Announce – July 30, 2013
Camera-Ready Papers Due – September 7, 2013
*New* SPL request for presentation – September 7, 2013
Assumptionless consistency of the Lasso
Sourav Chatterjee
The title says it all. Given $p$-dimensional data points $\{ \mathbf{x}_i : i \in [n] \}$ the Lasso tries to fit the model $\mathbb{E}( y_i | \mathbf{x_i}) = \boldsymbol{\beta} \mathbf{x}_i$ by minimizing the $\ell^1$ penalized squared error
$\sum_{i=1}^{n} (y_i - \boldsymbol{\beta} \mathbf{x}_i)^2 + \lambda \| \boldsymbol{\beta} \|_1$.
The paper analyzes the Lasso in the setting where the data are random, so there are $n$ i.i.d. copies of a pair of random variables $(\mathbf{X},Y)$ so the data is $\{(\mathbf{X}_i, Y_i) : i \in [n] \}$. The assumptions are on the random variables $(\mathbf{X},Y)$ : (1) each coordinate $|X_i| \le M$ is bounded, the variable $Y = (\boldsymbol{\beta}^*)^T \mathbf{X} + \varepsilon$, and $\varepsilon \sim \mathcal{N}(0,\sigma^2)$, where $\boldsymbol{\beta}^*$ and $\sigma$ are unknown constants. Basically that’s all that’s needed — given a bound on $\|\boldsymbol{\beta}\|_1$, he derives a bound on the mean-squared prediction error.
On Learnability, Complexity and Stability
Silvia Villa, Lorenzo Rosasco, Tomaso Poggio
This is a handy survey on the three topics in the title. It’s only 10 pages long, so it’s a nice fast read.
Adaptivity of averaged stochastic gradient descent to local strong convexity for logistic regression
Francis Bach
A central challenge in stochastic optimization is understanding when the convergence rate of the excess loss, which is usually $O(1/\sqrt{n})$, can be improved to $O(1/n)$. Most often this involves additional assumptions on the loss functions (which can sometimes get a bit baroque and hard to check). This paper considers constant step-size algorithms but where instead they consider the averaged iterate $\latex \bar{\theta}_n = \sum_{k=0}^{n-1} \theta_k$. I’m trying to slot this in with other things I know about stochastic optimization still, but it’s definitely worth a skim if you’re interested in the topic.
On Differentially Private Filtering for Event Streams
Jerome Le Ny
Jerome Le Ny has been putting differential privacy into signal processing and control contexts for the past year, and this is another paper in that line of work. This is important because we’re still trying to understand how time-series data can be handled in the differential privacy setting. This paper looks at “event streams” which are discrete-valued continuous-time signals (think of count processes), and the problem is to design a differentially private filtering system for such signals.
Gossips and Prejudices: Ergodic Randomized Dynamics in Social Networks
Paolo Frasca, Chiara Ravazzi, Roberto Tempo, Hideaki Ishii
This appears to be a gossip version of Acemoglu et al.’s work on “stubborn” agents in the consensus setting. They show similar qualitative behavior — opinions fluctuate but their average over time converges (the process is ergodic). This version of the paper has more of a tutorial feel to it, so the results are a bit easier to parse.
I’ve been trying to get a camera-ready article for the Signal Processing Magazine and the instructions from IEEE include the following snippet:
*VERY IMPORTANT: All source files ( .tex, .doc, .eps, .ps, .bib, .db, .tif, .jpeg, …) may be uploaded as a single .rar archived file. Please do not attempt to upload files with extensions .shs, .exe, .com, .vbs, .zip as they are restricted file types.
While I have encountered .rar files before, I was not very familiar with the file format or its history. I didn’t know it’s a proprietary format — that seems like a weird choice for IEEE to make (although no weirder than PDF perhaps).
What’s confusing to me is that ArXiV manages to handle .zip files just fine. Is .tgz so passé now? My experience with RAR is that it is good for compressing (and splitting) large files into easier-to-manage segments. All of that efficiency seems wasted for a single paper with associated figures and bibliography files and whatnot.
I was trying to find the actual compression algorithm, but like most modern compression software, the innards are a fair bit more complex than the base algorithmic ideas. The Wikipedia article suggests it does a blend of Lempel-Ziv (a variant of LZ77) and prediction by partial matching, but I imagine there’s a fair bit of tweaking. What I couldn’t figure out is if there is a new algorithmic idea in there (like in the Burrows-Wheeler Transform (BWT)), or it’s more a blend of these previous techniques.
Anyway, this silliness means I have to find some extra software to help me compress. SimplyRAR for MacOS seems to work pretty well.
Venkatesh Saligrama sent out a call for an ICML workshop he is organizing:
I wanted to bring to your attention an ICML workshop on “Machine Learning with Test-Time Budgets” that I am helping organize. The workshop will be held during the ICML week. The workshop will feature presentations both from data-driven as well as model-based perspectives and will feature researchers from machine learning and control/decision theory.
We are accepting papers related to these topics. Please let me know if you have questions about the workshop or wish to submit a paper.
I’m sick today so here are some links.
Click That Hood, a game which asks you to identify neighborhoods. I was lousy at San Diego, but pretty decent at Chicago, even though I’ve lived here for half the time. Go figure.
For those who care about beer, there’s been some news about the blocked merger of Inbev and Modelo. I recommend Erik’s podcast post on the structure of the beer industry (the three-tier system) for those who care about craft beer, and (with reservations) Planet Money’s show on the antitrust regulatory framework that is at work here.
Remember step functions from your signals and systems course? We called them Heaviside step functions after Oliver Heaviside — you can read more about him in this Physics Today article.
I need this album, since I love me some Kurt Weill. I can also live vicariously through NPR’s list of SXSW recommendations.
I just wanted to write a few words about the workshop at the Bellairs Research Institute. I just returned from sunny Barbados to frigid Chicago, so writing this will help me remember the sunshine and sand:
The beach at Bathsheba on the east coast of Barbados
Mike Rabbat put on a great program this year, and there were lots of talks on a range of topics in machine learning, signal processing, and optimization. The format of the workshop was to have talks with lots of room for questions and discussion. Talks were given out on the balcony where we were staying, and we had to end at about 2:30 because the sunshine would creep into our conference area, baking those of us sitting too far west.
I promised some ITA blogging, so here it is. Maybe Alex will blog a bit too. These notes will by necessity be cursory, but I hope some people will find some of these papers interesting enough to follow up on them.
A Reverse Pinsker Inequality
Daniel Berend, Peter Harremoës , Aryeh Kontorovich
Aryeh gave this talk on what we can say about bounds in the reverse direction of Pinsker’s inequality. Of course, in general you can’t say much, but what they do is show an expansion of the KL divergence in terms of the total variation distance in terms of the balance coefficient of the distribution $\beta = \inf \{ P(A) : P(A) \ge 1/2 \}$.
Unfolding the entropy power inequality
Mokshay gave a talk on the entropy power inequality. Given vector random variables $X_1$ and $X_2$ is there a term we know that $h(X_1 + X_2) \ge h(Z_1 + Z_2)$ where $Z_1$ and $Z_2$ are isotropic Gaussian vectors with the same differential entropy as $X_1$ and $X_2$. The question in this paper is this : can we insert a term between these two in the inequality? The answer is yes! They define a spherical rearrangement of the densities of $X_1$ and $X_2$ into variables $X_1^{\ast}$ and $X_2^{\ast}$ with spherically symmetric decreasing densities and show that the differential entropy of their sum lies between the two terms in the regular EPI.
This talk was on tradeoffs in caching. If there are $N$ files, $K$ users and a size $M$ cache at each user, how should they cache files so as to best allow a broadcaster to share the bandwidth to them? More simply, suppose there are three people who may want to watch one of three different TV shows, and they can buffer the content of one TV show. Since a priori you don’t know which show they want to watch, the idea might be to buffer/cache the first 3rd of each show at each user. They show that this is highly suboptimal. Because the content provider can XOR parts of the content to each user, the caching strategy should not be the same at each user, and the real benefit is the global cache size.
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2014-04-23 11:48:06
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http://pldml.icm.edu.pl/pldml/element/bwmeta1.element.bwnjournal-article-doi-10_4064-am32-2-7?q=bwmeta1.element.bwnjournal-number-zm-2005-32-2;6&qt=CHILDREN-STATELESS
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Pełnotekstowe zasoby PLDML oraz innych baz dziedzinowych są już dostępne w nowej Bibliotece Nauki.
Zapraszamy na https://bibliotekanauki.pl
PL EN
Preferencje
Język
Widoczny [Schowaj] Abstrakt
Liczba wyników
• # Artykuł - szczegóły
## Applicationes Mathematicae
2005 | 32 | 2 | 195-223
## On existence of solutions for the nonstationary Stokes system with boundary slip conditions
EN
### Abstrakty
EN
Existence of solutions for equations of the nonstationary Stokes system in a bounded domain Ω ⊂ ℝ³ is proved in a class such that velocity belongs to $W_p^{2,1}(Ω × (0,T))$, and pressure belongs to $W_p^{1,0}(Ω × (0,T))$ for p > 3. The proof is divided into three steps. First, the existence of solutions with vanishing initial data is proved in a half-space by applying the Marcinkiewicz multiplier theorem. Next, we prove the existence of weak solutions in a bounded domain and then we regularize them. Finally, the problem with nonvanishing initial data is considered.
195-223
wydano
2005
### Twórcy
autor
• Faculty of Mathematics, and Information Sciences, Warsaw University of Technology, Pl. Politechniki 1, 00-661 Warszawa, Poland
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2022-08-16 13:28:43
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http://ufdc.ufl.edu/UFE0021259/00001
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<%BANNER%>
State Estimation
Material Information
Title: State Estimation A Decision Theoretic Approach
Physical Description: 1 online resource (114 p.)
Language: english
Creator: Levinbook, Yoav Nir
Publisher: University of Florida
Place of Publication: Gainesville, Fla.
Publication Date: 2007
Subjects
Subjects / Keywords: bayes, estimation, filter, kalman, minimax, restricted, risk, state
Electrical and Computer Engineering -- Dissertations, Academic -- UF
Genre: Electrical and Computer Engineering thesis, Ph.D.
bibliography ( marcgt )
theses ( marcgt )
government publication (state, provincial, terriorial, dependent) ( marcgt )
born-digital ( sobekcm )
Electronic Thesis or Dissertation
Notes
Abstract: The problem of state estimation with stochastic and deterministic (set membership) uncertainties in the initial state, model noise, and measurement noise is approached from a statistical decision theory point of view. The problem is initially treated within a general framework in which the state estimation problem is a special case. General existence results such as the existence of a minimax estimator and a least favorable a priori distribution are derived for the state estimation problem. Then, attention is restricted to two important cases of the state estimation problem. In the first case uncertainties in the initial state, model noise, and observation noise are considered. It is assumed that the a priori distributions of the initial state and the noises are not perfectly known, but that some a priori information may be available. The restricted risk Bayes approach, which incorporates the available a priori information, is adopted. When attention is restricted to affine estimators based on a quadratic loss function, a systematic method to derive restricted risk Bayes solutions is proposed. When the filtering problem is considered, the restricted risk Bayes approach provides us with a robust method to calibrate the Kalman filter, considering the presence of stochastic uncertainties. This method is illustrated with an example in which Bayes, minimax, and restricted risk Bayes solutions are derived and their performance is compared. In the second case only the initial state uncertainty is considered. The initial state is regarded as deterministic and unknown. It is only known that the initial state vector belongs to a specified parameter set. The (frequentist) risk is considered as the performance measure and the minimax approach is adopted. The search of estimators is done within the class of all estimators. If the parameter set is bounded, a method of finding estimators whose maximum risk is arbitrarily close to that of a minimax estimator is provided. This method is illustrated with an example in which an estimator whose maximum risk is at most 3% larger than that of a minimax estimator is derived.
General Note: In the series University of Florida Digital Collections.
General Note: Includes vita.
Bibliography: Includes bibliographical references.
Source of Description: Description based on online resource; title from PDF title page.
Source of Description: This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Statement of Responsibility: by Yoav Nir Levinbook.
Thesis: Thesis (Ph.D.)--University of Florida, 2007.
Record Information
Source Institution: UFRGP
Rights Management: Applicable rights reserved.
Classification: lcc - LD1780 2007
System ID: UFE0021259:00001
Material Information
Title: State Estimation A Decision Theoretic Approach
Physical Description: 1 online resource (114 p.)
Language: english
Creator: Levinbook, Yoav Nir
Publisher: University of Florida
Place of Publication: Gainesville, Fla.
Publication Date: 2007
Subjects
Subjects / Keywords: bayes, estimation, filter, kalman, minimax, restricted, risk, state
Electrical and Computer Engineering -- Dissertations, Academic -- UF
Genre: Electrical and Computer Engineering thesis, Ph.D.
bibliography ( marcgt )
theses ( marcgt )
government publication (state, provincial, terriorial, dependent) ( marcgt )
born-digital ( sobekcm )
Electronic Thesis or Dissertation
Notes
Abstract: The problem of state estimation with stochastic and deterministic (set membership) uncertainties in the initial state, model noise, and measurement noise is approached from a statistical decision theory point of view. The problem is initially treated within a general framework in which the state estimation problem is a special case. General existence results such as the existence of a minimax estimator and a least favorable a priori distribution are derived for the state estimation problem. Then, attention is restricted to two important cases of the state estimation problem. In the first case uncertainties in the initial state, model noise, and observation noise are considered. It is assumed that the a priori distributions of the initial state and the noises are not perfectly known, but that some a priori information may be available. The restricted risk Bayes approach, which incorporates the available a priori information, is adopted. When attention is restricted to affine estimators based on a quadratic loss function, a systematic method to derive restricted risk Bayes solutions is proposed. When the filtering problem is considered, the restricted risk Bayes approach provides us with a robust method to calibrate the Kalman filter, considering the presence of stochastic uncertainties. This method is illustrated with an example in which Bayes, minimax, and restricted risk Bayes solutions are derived and their performance is compared. In the second case only the initial state uncertainty is considered. The initial state is regarded as deterministic and unknown. It is only known that the initial state vector belongs to a specified parameter set. The (frequentist) risk is considered as the performance measure and the minimax approach is adopted. The search of estimators is done within the class of all estimators. If the parameter set is bounded, a method of finding estimators whose maximum risk is arbitrarily close to that of a minimax estimator is provided. This method is illustrated with an example in which an estimator whose maximum risk is at most 3% larger than that of a minimax estimator is derived.
General Note: In the series University of Florida Digital Collections.
General Note: Includes vita.
Bibliography: Includes bibliographical references.
Source of Description: Description based on online resource; title from PDF title page.
Source of Description: This bibliographic record is available under the Creative Commons CC0 public domain dedication. The University of Florida Libraries, as creator of this bibliographic record, has waived all rights to it worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law.
Statement of Responsibility: by Yoav Nir Levinbook.
Thesis: Thesis (Ph.D.)--University of Florida, 2007.
Record Information
Source Institution: UFRGP
Rights Management: Applicable rights reserved.
Classification: lcc - LD1780 2007
System ID: UFE0021259:00001
Full Text
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STATE ESTIMATION: A DECISION THEORETIC APPROACH
By
YOAV N. LEVINBOOK
A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL
OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT
OF THE REQUIREMENTS FOR THE DEGREE OF
DOCTOR OF PHILOSOPHY
UNIVERSITY OF FLORIDA
2007
@ 2007 Yoav N. Levinbook
To the memory of my father, Benjamin Levinbook
ACKNOWLEDGMENTS
I would like to thank Professor Tan F. Wong, my advisor and the chair of the supervisory
committee, for his guidance, useful advice, and, in particular, for the freedom and encouragement
he gave me for pursuing my own research interests. I feel that I significantly evolved as an
electrical engineer in the four years I worked closely with him. I have no doubt that without his
help I would not have completed this work.
I wish to take this opportunity to thank all the members of the supervisory committee for
I would also like to thank Professor Paul Robinson and Professor Sergei Shabanov from the
Department of Mathematics, who were my instructors in several courses. The knowledge I have
gained from them proved to be very valuable for this work.
I would like to express my deepest gratitude to my beloved mother and late father, which
have always supported me and encouraged me. I hope I lived up to their expectations.
Finally, I am indebted to my dear wife, Eliane, for her support, encouragement, and
patience. Without her, I could not have confronted all the difficulties of the last four years.
page
ACKNOWLEDGMENTS ......... . . .. 4
LISTOFFIGURES ............. .............. 7
LIST OF ABBREVIATIONS ......... . .. .. 8
ABSTRACT.............. ......... ...... 9
CHAPTER
1 INTRODUCTION ......... ... .. 11
2 GENERAL NOTATION AND CONVENTIONS .... .... .. 17
3 DECISION THEORETIC FORMULATION ... .. . .. 19
4 GENERAL DECISION THEORETIC RESULTS ... .. .. 24
5 THE CASE THAT THE RISK IS SPECIFIED BY A LOSS FUNCTION .. .. .. 34
6 THE CASE OF CONVEX LOSS FUNCTION .... .... .. 40
7 FINDING A MINIMAX ESTIMATOR AND THE DUAL PROBLEM .. .. .. .. 49
8 APPROXIMATING A MINIMAX ESTIMATOR .... .... .. 52
9 THE RESTRICTED RISK BAYES PROBLEM AS A MINIMAX PROBLEM .. 61
10 ESTIMATION WITH A RESTRICTION ON THE OBSERVATIONS THAT CAN
BEUSED ............. ............... 66
11 THE STATE ESTIMATION PROBLEM . ... .. 73
12 AFFINE STATE ESTIMATION BASED ON QUADRATIC LOSS FUNCTIONS .. 77
12.1 Finding a Restricted Risk Bayes Solution ... .... .. 77
12.2 Finding a Maximizer of the Risk . ... .. .. .. 84
12.3 Connection to the Kalman Filter and E-Minimax Approach . . 87
12.4 Numerical Example . .. .... .. 90
13 STATE ESTIMATION WITH INITIAL STATE UNCERTAINTY .. . 96
13.1 Conditional Mean Estimators ....... .. .. 97
13.2 Approximations to Minimax Estimators ..... .... .. 98
13.3 Numerical Example ... . ..... .. .00
14 CONCLUSIONS ......... . ... .. 106
APPENDIX
A PROOF OFLEMMA 5.2 . . . 1..07
B PROOFOFLEMMA9.1 ............. ...........109
C PROOF OFLEMMA 9.2 . . .. .... .10
REFERENCES . .. . ... .111
BIOGRAPHICAL SKETCH ......... ... .. .. 114
LIST OF FIGURES
Figure page
12-1 Achieved Bayes risk vs. maximum risk . .... .. .. 94
12-2 The maximum risk over O, of the Bayes, minimax, and restricted risk Bayes solu-
tionsvs.e ............. ............... 95
13-1 A full view of 31o', which is a finite (6, V)-dense subset of 31o .. .. .. .. .. .. 103
13-2 A zoom-in view of the bottom left comer of 31o', which is a finite (6, V)-dense subset
of31o. ........ ... ......... .......104
13-3 The risk of .i-(To), the derived -,-optimal estimator, as a function of .ro E 31o' .. .. 104
13-4 The a priori distribution -ro, which is defined on 31o' .. .. .. .. .. 105
13-5 The maximum risk of the Kalman Filter initialized with zero mean and covariance
O.2I as a function of o . ..... .. 105
LIST OF ABBREVIATIONS
CM: conditional mean ......... . ... .. 97
KF: Kalman Filter ......... . .. .. 11
LMMSE: linear minimum mean squared error . ... .. 11
MSE: mean squared error ......... . . .. 45
Abstract of Dissertation Presented to the Graduate School
of the University of Florida in Partial Fulfillment of the
Requirements for the Degree of Doctor of Philosophy
STATE ESTIMATION: A DECISION THEORETIC APPROACH
By
Yoav N. Levinbook
August 2007
Chair: Tan F. Wong
Major: Electrical and Computer Engineering
The problem of state estimation with stochastic and deterministic (set membership)
uncertainties in the initial state, model noise, and measurement noise is approached from
a statistical decision theory point of view. The problem is initially treated within a general
framework in which the state estimation problem is a special case. General existence results such
as the existence of a minimax estimator and a least favorable a priori distribution are derived
for the state estimation problem. Then, attention is restricted to two important cases of the state
estimation problem. In the first case uncertainties in the initial state, model noise, and observation
noise are considered. It is assumed that the a priori distributions of the initial state and the noises
are not perfectly known, but that some a priori information may be available. The restricted
risk Bayes approach, which incorporates the available a priori information, is adopted. When
attention is restricted to affine estimators based on a quadratic loss function, a systematic method
to derive restricted risk Bayes solutions is proposed. When the filtering problem is considered,
the restricted risk Bayes approach provides us with a robust method to calibrate the Kalman filter,
considering the presence of stochastic uncertainties. This method is illustrated with an example
in which Bayes, minimax, and restricted risk Bayes solutions are derived and their performance
is compared. In the second case only the initial state uncertainty is considered. The initial state is
regarded as deterministic and unknown. It is only known that the initial state vector belongs to a
specified parameter set. The (frequentist) risk is considered as the performance measure and the
minimax approach is adopted. The search of estimators is done within the class of all estimators.
If the parameter set is bounded, a method of finding estimators whose maximum risk is arbitrarily
close to that of a minimax estimator is provided. This method is illustrated with an example in
which an estimator whose maximum risk is at most 3%b larger than that of a minimax estimator is
derived.
CHAPTER 1
INTRODUCTION
The problem of state estimation for linear dynamical systems has received considerable
attention in signal processing, controls, communications, econometrics, and a wealth of other
fields. The usual formulation of the problem assumes that the initial state, model noise, and
measurement noise are random vectors with perfectly known a priori distribution or at least with
known covariance and mean. It is well known that if these assumptions, together with other usual
assumptions regarding the noises, are valid, the Kalman filter (KF) [1] is the linear minimum
mean squared error (LMMSE) estimator. If in addition all the stochastic quantities are Gaussian,
the KF is the minimum mean squared error estimator.
Since the assumption of complete knowledge of the a priori distribution is seldom satisfied,
a Bayesian approach is used in practice. The a priori distributions of the initial state, model
noise, and measurement noise are learned from past experience and used as approximations of
the corresponding true distributions. Nevertheless, even if extensive past experience is available,
the estimated distributions may still deviate from the true ones. The effect of such errors in the
a priori information on the performance of the KF is studied in [2]-[4]. The effect of the errors
in the a priori information of the initial state, model noise, and measurement noise may be very
significant and a KF updated based on erroneous a priori information may perform poorly. Thus
it is necessary to consider other approaches that are robust against uncertainties in the a priori
distribution of the initial state, model noise, and measurement noise.
The state estimation literature deals extensively with the general problem of linear systems
with stochastic or deterministic uncertainties using game theory and the minimax approach (cf.
[5]-[14] and the references therein). Usually the so-called E-minimax approach is adopted. The
F-minimax approach [15] regards the parameter as random with its distribution lies in a class F.
However, the exact distribution in the class is unknown. A F-minimax estimator is an estimator
that minimizes the supremum of the Bayes risk, where the supremum is taken over all elements
of F. When the F-minimax approach is used, the class of available estimators is usually restricted
to linear or affine estimators. As a result, an element of the class 0 is specified by a first-order
statistic (mean) and second-order statistic covariancee) pair.
There are several other approaches in statistical decision theory that seem suitable in this
context. Among the most prominent approaches is the restricted risk Bayes approach. The
restricted risk Bayes approach, proposed by Hodges and Lehmann [16], is a compromise between
the Bayes approach and the minimax approach. A restricted risk Bayes estimator minimizes
the Bayes risk with respect to an a priori distribution suggested based on some past experience
subject to the restriction that the maximum risk does not exceed the minimax risk by more than a
given amount. This approach utilizes available a priori information but at the same time provides
a safeguard in case this information is not accurate. If the a priori information is fairly accurate,
a restricted risk Bayes estimator has good Bayes risk properties. Other work considering the
restricted risk Bayes approach or closely related approaches include [17]-[20]. Despite the
appealing formulation of the restricted risk Bayes approach, it has not been utilized in the context
of state estimation.
Although the problem of state estimation with stochastic uncertainties has been approached
from the F-minimax approach, we believe that approaching this problem from the restricted risk
Bayes approach also has a considerable merit. If a state estimation problem can be regarded as
a zero-sum two-person game (henceforth to be referred as a game) against a rational opponent,
then the F-minimax approach seems very attractive. However, in most applications, if we regard
the state estimation problem as a game, the game is against Nature. Using the F-minimax
approach in this case corresponds to a very pessimistic viewpoint that regards Nature as a rational
opponent that wishes to cause us the largest possible loss. The F-minimax approach may still
seem reasonable in the case that there is no a priori information that enables us to regard certain
distributions in C as more likely than others. However, in many applications, some a priori
information regarding the true distribution may be available. This a priori information may be in
the form of a nominal distribution, which is suggested based on some past experience. It is well
known that under certain conditions, a F-minimax estimator is a KF relative to a least favorable a
priori distribution. If our a priori information suggests that the true distribution is very different
from a least favorable a priori distribution, then using the F-minimax approach may result in
very conservative estimators. The restricted risk Bayes approach enables us to adopt a less
conservative but still rather robust approach, which gives us a guaranteed safeguard in terms of
the risk. We assume the nominal distribution is an approximation of the true a priori distribution
and search for an estimator that minimizes the Bayes risk relative to the nominal distribution
subject to the constraint that the Bayes risk relative to any distribution in 0 is less than a given
value. We can determine this value based on the amount of past experience that we have. The
stronger the available past experience, the more we can trust the approximated distribution and
the larger the value we can allow.
In this work, we consider the restricted risk Bayes approach. Under our formulation, the
F-minimax approach is a special case. We consider the risk, based on quadratic loss functions,
as our performance measure. In this case, we restrict ourselves to affine estimators in order
to derive estimators that are attractive in terms of computational complexity. We provide a
systematic method for solving restricted risk Bayes and E-minimax solutions. In some important
cases, this method can be easily used to calibrate the KF, considering the presence of stochastic
uncertainties.
While in most applications, the observation noise can be indeed modeled as random, or
even Gaussian (e.g., the thermal noise in communications systems), in some applications one
can argue whether the initial state can be better modeled as an unknown constant, rather than as
random, and the same may be also argued for the model noise. In this work, we also consider the
case that there is a deterministic uncertainty in the initial state as a special case of the problem
of state estimation with deterministic uncertainties. The distributions of the model noise and
measurement noise are assumed known. As mentioned above, it may be reasonable to assume
the model noise is deterministic and unknown as well. We do not pursue this approach here
since we believe that this would obscure the results that we derive for the initial state uncertainty
due to the technical difficulties of considering deterministic uncertainties in the model noise.
In addition, if we are concerned with estimation problems in which it is necessary to estimate
fast, using a small number of samples, the uncertainty in the initial state may a have a much
more significant effect than the uncertainty in the model noise. The reason is that if the estimated
signal changes slowly (which is the case in many applications), the uncertainty in the model
noise may be small whereas the uncertainty in the initial state may be very large. For instance,
in tracking problems the changes of the velocity or acceleration of the target between adjacent
samples are usually small relative to the range of their possible values at the initial time. When
deterministic uncertainty is considered in the initial state, the minimax approach may be the
preferable approach. A minimax estimator minimizes the supremum of the (frequentist) risk.
The minimax approach seems a reasonable approach when it is only known that the initial
state vector belongs to a certain parameter set. The minimax approach is especially suitable for
applications in which we may interpret the estimation problem as a game against an opponent.
This is certainly the case in many military applications.
The state estimation literature has traditionally focused on the Bayesian approach and the
F-minimax approach when dealing with uncertainty in the initial state vector. The classical
approach that regards the initial state vector as deterministic, has been mostly ignored. This is
despite the fact that in many important applications (see for example [21] and [22]), it seems
more reasonable to model the initial state vector as deterministic and unknown. One example that
falls within the classical estimation framework is Danyang and Xuanhuang [23], where the state
estimation problem was considered from a least squares viewpoint and the best linear unbiased
estimator was derived. In [24], the authors consider the problem of state estimation with initial
state uncertainty from a decision theoretic point of view. The initial state vector is regarded as
deterministic and unknown. It is only known that the initial state vector belongs to a parameter
set. The risk, based on quadratic loss functions, is considered as the performance measure. The
search for a minimax estimator is done within the class of all possible estimators. Minimax
estimators are derived for the case of unbounded parameter set and approximations of minimax
estimators are derived for the case of bounded parameter set. In this work, we will repeat some
of the results in [24] that deal with the bounded parameter set case in order to illustrate how
minimax estimators can be approximated with arbitrarily prescribed accuracy.
While we are mainly interested in the state estimation problem, a large part of this work will
be concerned with a more general estimation problem. In fact, some of the existence results such
as the existence of a minimax estimator and a restricted risk Bayes solution hold in a very general
setting and may have applicability not only in the state estimation problem.
The rest of this work is organized as follows. In Chapter 2, we present notation and
conventions that are used throughout this work. In Chapter 3, we present a general decision
theoretic formulation that is needed in order to address the problem of state estimation with
stochastic and deterministic uncertainties and also to derive other, more general, results. In
Chapter 4, we derive several general decision theoretic results, which are based on well known
results from decision theory and game theory; the applicability of these results is not limited
only to the problem of state estimation with stochastic and deterministic uncertainties. In
Chapter 5, we consider the case that the risk function is specified by a loss function. We derive
rather weak conditions that guarantee the existence of a minimax estimator and a restricted risk
Bayes solution. In Chapter 6, we restrict ourselves to convex loss functions, in general, and the
quadratic loss function, in particular. In Chapter 7, we discuss how a minimax estimator can
be found by solving the dual problem of finding a least favorable a priori distribution. Since,
in general, finding a minimax estimator may be an extremely difficult task, in Chapter 8 we
discuss how one can derive approximations to minimax estimators, where the approximation
can be made as accurate as desired. In Chapter 9, we consider the general restricted risk Bayes
estimation problem and show that this problem is equivalent to a sequence of minimax estimation
problems. In some estimation problems there are restrictions on the observations that can be
used in order to estimate the parameters. This is the case in the state estimation problem in which
each state can be estimated using only certain observations. We consider this type of estimation
problems in Chapter 10. In Chapters 11, 12, and 13, we restrict ourselves to the problem of state
estimation with uncertainties in the initial state, model noise, and measurement noise, which
is the main problem considered in this work. In Chapter 11, we derive some general existence
results that are based on the results of the previous chapters. In Chapter 12, we consider the case
of stochastic uncertainties in the initial state, model noise, and measurement noise, and restrict
ourselves to affine estimators. We propose a method that can be easily used to derive a restricted
risk Bayes solution in many important cases. In Chapter 13, we consider the case of deterministic
initial state uncertainty, and search for estimators within the class of all estimators. We conclude
this work in Chapter 14.
CHAPTER 2
GENERAL NOTATION AND CONVENTIONS
Let R"N denote the N~-dimensional Euclidean space, and let RW = RIW. Let R"~ = {a e
R"N : a(i) > 0 for i = 1, 2, ..., N}. We use RWNxM to denote the space of NV-by-M~ real
matrices. Let S" denote the space of n-by-n real symmetric matrices. Let S" denote the cone
of positive semi-definite matrices in S". Let N = {0, 1, .. .}. Let a denote an NV dimensional
vector. We use a(i) to denote the ith element of a and |a| to denote the Euclidean norm of a.
We let ||a|| = CE |a(i)|. Let Ai and B denote arbitrary matrices. We use N~(A) to denote
the nullspace of A, At to denote the Moore-Penrose pseudo inverse of A, tr(A) to denote the
trace of A, and ||IA||2 to denote the 2-norm of A. We use A > 0 (A > 0) to denote that A
is positive definite (positive semi-definite) and symmetric and A > B (A > B) to denote
that A B > 0 (A B > 0). We use A 0 B to denote the Kronecker product of A and
B. Let A denote an arbitrary set. We use | A| to denote the cardinality of A. If AC R ", we
let A = { x E R"N : x~y = 0, for ally E A}.Let A1,. ,AS be arbitrary sets. Let
A = n0 Ali, where product on sets means cartesian product. We use vri : A Ai to denote
the projection onto the ith factor. Given a s A, let a(i) = xri(a). If Al,. ,AS are subsets of
vector spaces V1, .. respectively, a' and a" are in A, and as R I, then a = a' + a" means
that a(i) = a'(i) + a"(i) for i = 1,. ., n, and a = caa' means that a(i) = caa'(i) for i = 1, .., n.
Let f : X Y and let Z be a subset of X. Then f (Z) = {f (z) : z E Z}. Let lni denote the
NVx NV identity matrix. Let ONxM/ denote the NVx M~ zero matrix. When the dimensions of the
zero matrix and identity matrix are clear from the context, they will simply be denoted 0 and I,
respectively. For any topological space X, we use B(X) to denote the o--algebra of Borel subsets
of X. Given a measurable space (X, FT), where X is a subset of R"N and FT = B(X), and a
probability measure -r defined on (X, FT), by the mean vector and autocorrelation matrix of -r, we
mean the mean vector and autocorrelation matrix of a random vector that is distributed according
to -r, i.e., the mean vector of -r is fx xd-r(x) and the autocorrelation matrix of -r is fx xx~d-r(x).
Let S denote an index set, possibly uncountable, and (Xs, F,), for each s e S, be a measurable
space. Then nses Fs, denotes the o--algebra on nses X, generated by subsets of the form
nses A,, where A, E T, for each s ES and A, = X, for all but a finite number of s E S. In
particular, given measurable spaces (X1, Fi), (X2, F2,),. ., (XN, Fy), FI~x T~ 2 x FTN
denotes the o--algebra on X1 x X2 x x XN generated by subsets of the form Al x A2 x x AN,
where Ai E Fei for i = 1,. ., NV. Given a measurable space (X, FT), let m(FT) denote the set of all
real-valued measurable functions on (X, FT) and m(FT) denote the set of all extended real-valued
measurable functions on (X, FT). Given real numbers a and b, a Vb denotes max {a, b} and a Ab
denotes mina, b}. Some of the derivations in the sequel require the use of the extended real
number system. We use the usual conventions for arithmetic operations in the extended reals.
When we take the supremum or infimum of a subset AC R we regard A as a subset of the
extended reals, i.e., the supremum and infimum always exist and may take the values +oo and
-oo, respectively.
CHAPTER 3
DECISION THEORETIC FORMULATION
Let (R, &~, Q) be a probability space, where R is the sample space of an experiment and
elements of -TA are the events of the experiment. Suppose there are given a set 8, called the set of
possible states of nature, and a family P = { Po : 0 E 8}) of probability measures defined on the
measurable space (R, &~). The probability measure Q is unknown to the experimenter; however
it is known that Q is an element of P. Suppose there is given a measurable space (Y, FTy). The
experimenter observes the value y taken by a random element (or a Y-valued random variable )
Y : R (f FTy); this value is called an observation.
In Wald's interpretation [25] of the statistical decision problem as a zero sum two-person
game, Nature chooses an element 8 E 8, called the true state of nature. The experimenter needs
to reach a certain decision based on the observation y without knowing the true state of nature.
The experimenter reaches such a decision by choosing a decision rule from a class of decision
rules. Here we are mainly interested in estimation problems, although some of the results of this
work apply also to decision problems.
In classical estimation, the experimenter estimates the true state of nature 8 E 8 or more
generally a function of 0. In this case the space (R, -T2, Q) can be taken to be (Y, FTy, Q), where
Q is the distribution of Y. In this work, we are interested in a more general case. Suppose
there is given an additional measurable space (X, 6T) and a random element X : R
(X, 8Tx). Suppose Y and X have a joint distribution defined on the product measurable space
(y x X, FTy x 5T). Suppose the experimenter wants to estimate the value x taken by X. In this
case, the space (R, -T2, Q) can be taken to be the space (y x X, FTy x 8x, Q), where Q is the
joint distribution of Y and X and is an element of P. Note that the previous mentioned case can
be regarded as a special case, in which the distribution of X assigns mass 1 to the true state of
nature 8 E 8.
Suppose the experimenter is allowed to choose estimates for x from a class D. The class
D, equipped with a topology, is referred to as the space of possible estimates. Often D coincides
with X. However, it is convenient not to make this restriction. Given a topological space A, let
My/1 denote the class of all probability measures on (A, B(A)). A nonrandomized estimator i-
is simply a ~T-measurable mapping from Y into D. A mapping :i : Y M zD, y H 9(I)
is said to be a randomized estimator if y H -(D'| y) is a 8 y-measurable function for all
D' E B(D). A randomized estimator (from now on just an estimator) can be used to determine
uniquely a procedure for choosing an estimate for :r. The estimate of :r, given y is observed
and i- is used, is an element of D selected according to the probability measure i-(-|y). A
nonrandomized estimator i- can be regarded as a randomized estimator that assigns to each
ye E a Dirac measure, i.e., a measure that assigns a mass 1 to a single point in D. Let 2'
denote a class of estimators for :r. Since we adopt a decision theoretic approach, the merit of
an estimator will be judged based on a risk function. The risk function is the expected loss
incurred to the experimenter when using an estimator :i and 8 is the true state of Nature. Let
R : 8 x X R WU {+oo} denote a risk function. In the formulation of this work, an estimation
problem may be specified by a triplet (8, 2', R).
For the convenience of the reader, we present here some basic decision-theoretic definitions
taken from [26] and [25], with slight modifications. These definitions are given for an estimation
problem (8, 2 R).
Definition 1. An estimator :i* E 2' is said to be a nzinintax estimator if
sup R(8, i-*) inf sup R(0,;i).
Definition 2. Let e > 0. An estimator :i* E X is said to be an e-nzinintax estimator if
sup R(8, i-*) < inf sup R(8, i-) +
BOe *E~ aee
Definition 3. An estintator i-' is said to be as good as an estintator i-" if R(0, i-') < R(0, i-") for
all 8 E 8. An estintator i-' is said to be better than an estintator i-" if R(0,;i') < R(0,;i") for all
8 E 8 and R(0, i-') < R(0, i-") for at least one 8 E 8. An estintator i-' is said to be equivalent to
an estintator i-" if R(0, i-') = R(0, i-") for all 8 E 8.
Definition 4. An estimator & is said to be admissible if there exists no estimator better than 2. An
Definition 5. A class A of estimators is said to be essentially complete relative to a class B of
estimators if given any estimator & in B, there exists an estimator i* E A that is as good as 2.
If AC c is essentially complete relative to X, we simply say that A is essentially
complete.
Definition 6. An estimator & is said to be an essentially unique minimax estimator if any
minimax estimator is equivalent to 2.
Let V(8, X, R) = infecx supeoe R(0, 2) be the minimax risk. The following assumptions
will be assumed throughout this work, unless mentioned otherwise:
Assumption 3.1. V(8, K, R) < +oo.
The case V(8, .[, R) = +oo is not very interesting. Clearly in this case the minimax
approach is not suitable since any estimator is a minimax estimator.
Assumption 3.2. 8 is a metric space with metric a.
Assumption 3.3. The risk function R is nonnegative.
Assumption 3.3 can be weakened to the assumption that infeee int ,x R(0, 2) > -oo, but
for the sake of presentation it is advantageous to assume that R is nonnegative.
We need additional definitions. If & is such that R(-, 2) is bounded from below and is a
B(8)-measurable function, we denote the Bayes risk relative to -r Me Zl of &
~(~,i (3J8
In this work, we consider only estimation problems (8, X, R) for which R(-, 2) is bounded from
below and is a B(8)-measurable function for all ie E Hence the Bayes risk is always defined.
Definition 7. An estimator i* E X is a Bayes solution relative to -r Me Zl ifr (-, i*)=
infEx gr (-, 2). An estimator i* E X is a Bayes solution if it is a Bayes solution relative to some
distribution -r Ms/l.
Definition 8. An estimator & is said to be an essentially unique Bayes solution relative to -r if any
Bayes solution relative to -r is equivalent to x.
Note that we may regard a point 8 E 8 as an element of Me/l by regarding it as the
probability measure that assigns mass 1 to 0. Thus 8 may be considered as a subset of Mel. With
this viewpoint, r(-, 2) is an extension of R(-, ) from 8 to Me/l, i.e., r(0, 2) = R(0, 2) for all
S8.O
We also use the notion of Bayes solution in the wide sense:
Definition 9. Let {n }r be a sequence of a priori distributions in Me/l and i* be an estimator
Then i* is a Bayes solution relative to the sequence {n }) if
lim [r(ni, i*) -- inf r(i, 2)] = 0.
An estimator i* is a Bayes solution in the wide sense if there exists a sequence {nri } EZ/e such
that i* is a Bayes solution relative to the sequence {8 }).
It iseasy to show that for each re M 2e, inf ~xr-,~: I ~,2 V(8, K, R). Thus by
Assumption 3.1, infe,x r (-, 2) < +oo, and the term r (-r, i*) infa,x r (-r, 2), in the above
definition, is well defined.
Definition 10. An a priori distribution -ro E Me/l is said to be least favorable if
1111T T, Z Sup inf r 7r, x.
Definition 11. The estimation problem (8, I R) is said to be strictly determined if
inf sup r(-r,i)= sup inf r(-r, ).
For any class AC c of estimators, we let A(M~, R) = { E A : supose R(0, 2) < M}).
Definition 12. Let -r Me Zl and Co ERIWU {+oo }. An estimator io is said to be a restricted risk
Bayes solution relative to (-r, Co) if
T T,0 111 T T I
Note that if Qo = V(8, K, R), then for any -r Me Zl, a restricted risk Bayes solution
relative to (-r, Qo) is a minimax estimator. In addition, a restricted risk Bayes solution relative to
(-r, +oo) is a Bayes solution relative to -r. Thus the problems of finding a minimax estimator and
a Bayes solution may be regarded as two extreme cases of the problem of finding a restricted risk
Bayes solution.
CHAPTER 4
GENERAL DECISION THEORETIC RESULTS
In this section, we present several decision theoretic results for an estimation problem
(8, X2, R). These results hold in a much more general case than the estimation problem
considered in this work and may be used in general decision theoretic problems. These results are
based on well known results from decision theory and game theory.
First, we note that it is well known that (cf. [26, Exercise 2.2.1])
sup R (B, i) sup r (-r, .). (4-1)
Consider the space Me/l equipped with the topology of weak convergence [27, pp. 236]. The
topology of weak convergence makes Me/l a Hausdorff space.
Lemma 4.1. Suppose 8 is compact. Then
1. Me/l is nzetrizable and compact.
2. Let~i e E be such that R(0,.i-) < +oc for each Ie 8 and R(-,.i) is continuous on 8.
Then r (-, i) is continuous on M sl.
3. Let~i e E X be such that R(0,.i-) < +oc for each Ie 8 and R(-,.i) is upper senicontinu-
ous on 8. Then r (-, i) is upper senticontinuous on M sl.
Proof 1) By the hypothesis of the lemma, 8 is a compact metric space and hence is separable
(cf. [28, Exercise 2.25]). This implies that Me/l is metrizable [29, pp. 122]. In addition, [29,
Theorem 3.1.9] furnishes that Me/l is compact.
2) By [29, Theorem 3.1.5], J, fdvr is a continuous function of -r for any bounded and
continuous f. Since R(-,.i) is a continuous real-valued function on a compact set, it is bounded.
Hence r(-,.i) is continuous on Ms/l.
3) By [29, Theorem 3.1.5], J, fdvr is an upper semicontinuous function of -r for any up-
per semicontinuous f that is bounded from above. Since R(-,.i) is an upper semicontinuous
real-valued function on a compact set, it is bounded from above. Hence r(-,.i) is upper semicon-
tinuous on Me/l for any .i E C. O
Given a set Z, let & (Z) denote the space of nonnegative extended real-valued functions on
Z equipped with the topology of pointwise convergence. For u', u" E W (Z), we write u' < u"
to denote that u'(z) < u"(z) for all z E Z. Consider the space W (8). Let A denote an arbitrary
set and f : 8 x A [0, +oo] an arbitrary function. Then for each a s A, f (-, a) E W (8). Let
A' and A" denote arbitrary sets, and let f' : 8 x A' [0, +oo] and f" : 8 x A" [0, +oo]
denote arbitrary functions. It is convenient to use the notation (A', f') -4 (A", f") to denote that
for each element a" E A", there exists an element a' E ~A' such that f'(-, a') < f"(-, a"). Let
(A', f') ~ (A", f") denote that (~A', f') -4 (A", f") and (A", f") -4 (~A', f'). Clearly ~ is an
equivalence relation. Now, consider an estimation problem (8, .[, R), and let A' and A" denote
classes of estimators (i.e., subsets of X). In this case, (A', R) -4 (A", R) simply states that the
class A' is essentially complete relative to ~A". In this case, the notation can be simplified, and we
can write A' -4 A" to denote that A' is essentially complete relative to A". Note that the relation
-4 is a preorder on the collection of all classes of estimators. Clearly A -4 A, and it can be easily
verified that A -4 B and B -4 C imply A -4 C. Similarly, if A and C are classes of estimators, we
use A ~ C instead of (A, R) ~ (C, R).
Definition 13. A subset U of W (Z) is said to be half-closed if for each a in the closure of &,
there exists a u* E U such that u* < u.
Note that given an estimation problem (8, X, R), for each is E r (-, ) E W (Mel) and
R(-, 2) E & (8). Let M(A, Me/l) = {r(-, 2) : S E A} and M(A, 8) = {R(-, ) : f E A}. The
following definition is due to LeCam [30].
Definition 14. Given an estimation problem (8, K, R), a subset A of K is said to have the
property (W) if& (A, 8) is half-closed.
We need also the following closely related definition.
Definition 15. Given an estimation problem (8, K, R), a subset A of K is said to have the
property (W*) if&(A,Me/l) is half-closed.
The preceding definition can be reformulated as follows: Given an estimation problem
(8, K, R), a subset A of .[ is said to have the property (W*) if for each net {idaeA n A,
there exists a subnet {:ib bEB and an element i-* E A such that lim inf r(-, i'b) > r( ~).
Indeed, suppose #'(A, Me/l) is half-closed. Let {:i'o,,A denote a net in A. Let U denote the set
of limit points of this net. The set U is not empty since the space ~(Mel) is compact [31]. Since
~(Mel) is compact, M(A, Me/l) is relatively compact. Thus there exists a subnet {ib bEB that
converges to a point it in the closure of d(A, Me/l). Hence lim inf R(-, ib) = lim R( *,;b) = IL.
Since '(A, Me/l) is half-closed, there exists an element t' E M(A, Me/l) such that < Is.
Hence there exists an element i-* E X such that R(-,;i*) < liminf R(-,;ib). Suppose
for each net {:i )aEA in A, there exists a subnet {:ib bEB and an element i-* E A such that
lim inf r (-, ib) > r (-,;i*). Let it* belong to the closure of s'(A, Me/l). Then there exists a net
{u,),eA Ed 8(,iZ) that converges to u*. Clearly for each element u, in this net, there is
an element i's E A such that u, = R(-,;i). Thus lim R(-, is) = u*. It follows that there
exists a subnet {:ib bEB and an element i-* E A such that lim inf R(-, i'b) > R(-,;i*), whence
u* > R(-,;i*) and M(A, Me/l) is half-closed. In entirely analogous way, A has the property (W)
if and only if for each net {:i' )aEA in A, there exists a subnet {:ib bEB and an element i-* E A
such that lim inf R(-, ib) > R(-,;i*). Since each 8 can be identified as an element in Me/l, as
discussed previously, it is clear that the property (W*) implies the property (W). The property
(W*), as formulated with nets, is closely related to Wald's weak compactness [25, pp. 53].
The only difference is that in the current definition sequences are replaced by nets. Hence the
property (W*) is weaker than Wald's weak compactness. Sufficient conditions for a subset A of
X to have the property (W) are given in [31]. A simple sufficient condition is that there exists
a Hausdorff topology J for A such that A is compact and R(0, -) is lower semicontinuous on A
for each 8 E 8. Similarly, a sufficient condition for A to have the property (W*) is that there
exists a Hausdorff topology J for A such that A is compact and r(-r, -) is lower semicontinuous
on A for each -r Me Zl. Suppose there exists a topology J for A such that A is compact. Then
if the topological space (A, J) satisfies the first axiom of countability, by Fatou's Lemma, the
requirement that r(-r, -) is a lower semicontinuous on A for each -r Me Zl can be replaced by the
requirement that R(0, -) is a lower semicontinuous on A for each 8 E 8.
The following definition appears in [31].
Definition 16. A class A of estimators is said to be subconvex if for any i', i" E A and
0 < a~ < 1, there exists E A such that caR(0, i') + (1 ~) R(0, i") > R(0, 2) for all 8 E 8.
Clearly if A is a convex subset of a certain vector space and R(0, -) is convex over A for
each 8 E 8, then A is subconvex.
We are now ready to state several theorems some of which will be used throughout this work
and some of which are presented for their own sake.
Theorem 4.1. Given an estimation problem (8, X, R). Suppose that K has the property (W).
Then there exists a minimax estimator
Proof The proof is very similar to the proof of Wald's Theorem 3.7 [25]. The main differences
is that sequences are replaced by nets. Recall that we assume that V(8, X9, R) < +oo. Let {x,}
be a sequence such that supose R(8, in) converges to infc,x supose R(0, 2). Then there exists a
subnet {idaeA and an element i* E X such that lim inf R(-, is) > R(-, i*). Since {idaeA iS a
subnet of {x,}, lim sup,, R(-, i) > lim inf R(-, fe). In addition, lim,,, supose R(8, in)
lim sup,, R(0, in) for all 8 E 8. Thus in f a supose R(0, i) > supose R(0, i*), and i* is a
minimax estimator. O
After addressing the existence of a minimax estimator, we want to address the existence of a
restricted risk Bayes solution.
Theorem 4.2. Given an estimation problem (8, X9, R). Suppose .9 has the property (W*).
Then if V(8, K, R) < Qo, there exists a restricted risk Bayes solution relative to (v, Qo).
Proof Suppose V(8, X9, R) < Q. Note that the class X9(Qo, R) is not empty even if
Qo = V(8, K, R) since by Theorem 4.1i, there exists a minimax estimator.
Let {i,} be a sequence in X(Qo, R) such that limes, r(v, 2,) = infiEX(co,R) doI, 2).
Then since .9 has the property (W*), there exists a subnet {idaeA Of (Xn} and an element
i* E K such that liminf r(-r, i) > r(-r, *) for each r E Mel. Certainly r(-r, ,) < Qofor
all n and each -r E Mo/. Since {idaeA is a subnet of {x,}, it is clear that lim inf r(-r, 2,) < Co
for each -r Me Zl, whence R(0, i*) < Co for each 8 E 8. Thus i* E X(Qo, R). It is also
clear that infier(co,,) r(v, 2) > r(v, i*). Thus i* is a restricted risk Bayes solution relative to
Putting Qo = +oo in the above theorem, we get that for each -r Me Zl, there exists a Bayes
solution relative to -r (provided that X has the property (W*)).
The following Theorem, a version of Wald's complete class theorem, appears in [31].
Theorem 4.3. Suppose X is subconvex and has the property (W). Then the class of Bayes
solutions in the wide sense is essentially complete.
The following theorem is essentially a corollary to a very well known game-theoretic result
[32, Theorem 4.2].
Theorem 4.4. Suppose 8 is compact, K is subconvex, R(-, 2) < +oo for each & E K, and
R(-, 2) is upper semicontinuous on 8 for each is K Then the estimation problem (8, X, R)
is strictly determined and there exists a least favorable a priori distribution.
Proof Fix is E Since R(-, 2) is an upper semicontinuous real-valued function on a compact
set 8, it is bounded. Thus r (-, ) is bounded on 8, and hence a real-valued function. By Lemma
4.1, Me/l is compact and r(-, 2) is upper semicontinuous on Me/l for each is E It is easy to
verify that since X is subconvex, for each i', i" E X and 0 < a~ < 1, there exists i s such
that r (-, 2) < ~r (-r, i') + (1 a~)r (-, ") for all 8 E 8. Certainly r (-, ) is concave on Me/l for
all ie E Applying [32, Theorem 4.2],
1111 Sup r 7r, x sup inf r 7r, x.
Thus (8, X, R) is strictly determined. Since inf~Ex r(-, 2) is upper semicontinuous on a
compact set Me/l, there exists -ro E Me/l such that
III T TOZ) Sup inf r (r, x).
Thus -ro is a least favorable a priori distribution. O
The assumption in Theorem 4.4 that R(-, 2) < +oo for each i s and R(-, 2) is upper
semicontinuous on 8 for each i s is obviously too strong. In the following theorem this
assumption is considerably weakened, but at the price of a stronger assumption on X. Let
& C K denote the class of Bayes solutions relative to -r Me Zl. Before stating the theorem, we
need the following lemma.
Lemma 4.2. Suppose .F has the property (W*), then
infr~~i =inf r(-r, i) for all y E Me-lo
Proof In Theorem 4.2 it is shown that if X has the property (W*), there exists a Bayes solution
relative to -r for all -r Me Zl. The lemma follows easily. O
Theorem 4.5. Suppose 8 is compact, K is subconvex and has the property (W*), R(-, 2) <
+oo for each is E and R(-, ) is upper semicontinuous on 8 for each is E Then the
estimation problem (8, X, R) is strictly determined and there exists a least favorable a priori
distribution.
Proof Let G and Go denote the convex hull of (.[, Me/l) and &(9, Me/l), respectively. It
is easy to see that each g EG is a concave upper semicontinuous real-valued function on the
compact set Me/l. Thus by [32, Theorem 4.2],
inf sup g(-r) =sup inf g(-r). (4-2)
Fix -r Me Zl. On one hand, since &(9, Me/l) C Ga,
inf g(-r) < inf gr)=if(, )
On the other hand, if we fix g E G'a, then there exist an integer n > 0, real numbers
ai1, an > 0, and elements ul, ., an e (9, Mel) such that g = CE asse and
I~ as = 1. Let 1 < i < n be such that a () < Up (7) for j = 1,. ., n. Then g(-r) > a ().
Thus
inf g(-r) > inf gr)
It follows that
inf g(r) = inf r (-, 2). (4-3)
Since G > Ga,
inf sup g (r) < inf sup g (-). (4-4)
g6G 76MenL g6GSB 76Men
Fix g E G. Then since X is subconvex, there exists i s such that r (r, i) < g(-r) for all
-r MeiZ Hence
inf sup r (-, ) < inf sup g (-). (4-5)
26X 76enL g6G 76MenL
By (4-2)-(4-5) and Lemma 4.2, (8, X, R) is strictly determined. Since info,EG g is upper
semicontinuous on the compact set Me/l, there exists -ro E Me/l such that
inf g(-ro) = sup inf g(-r).
g6GSB 76Men g6GgB
By (4-3) and Lemma 4.2, -ro is a least favorable a priori distribution. O
If the compactness of 8 is dropped in the above theorem, then there may not be a least
favorable a priori distribution. If there exists no least favorable a priori distribution, but the
estimation problem (8, X, R) is strictly determined, a minimax estimator can be found as a
Bayes solution relative to a least favorable sequence of a priori distributions, i.e., a sequence
{7r,} E Me/ that satisfies lim,,, infie g r(-r,, i) = supe~Me inf~Ec r(r, i). We are not going
to deal with the question of how such a sequence can be constructed. The following theorem is
essentially Wald's Theorem 3.9 in [25].
Theorem 4.6. Suppose the estimation problem (8, X, R) is strictly determined. Then if -r is a
least favorable a priori distribution, any minimax estimator is also a Bayes solution relative to
Proof It can be verified that the proof of [25, Theorem 3.9] applies without any changes. O
When 8 is compact, we have the following version to Wald's complete class theorem.
Theorem 4.7. Suppose 8 is compact, K is subconvex and has the property (W*), R(-, 2) is
lower semicontinuous on 8 for each is K R(-, 2) < +oo for each is E and R(-, ) is
continuous on 8 for each is E Then & is essentially complete.
Proof Since the class of Bayes solutions in the wide sense is essentially complete, we are done
if we show that any Bayes solution in the wide sense is a Bayes solution. Suppose that i* is a
Bayes solution in the wide sense. Then there exists a sequence {n}i E Me/l such that
lim [ n n 2 *)] = 0 (4-6)
By Lemma 4. 1 part 1), there exists -rs Me Z and a subsequence {4}i ) Me Zl such that -ri
converges weakly to -r. By the hypothesis of the theorem, r(-, 2) is continuous on Me/l for any
is E Thus infies r(-, 2) is upper semicontinuous on Me/l [33, Proposition 1.5.12]. By Lemma
4.2, ini~,x r(-,2) is upper semicontinuous on Me/l. Since r(-,i*) is lower semicontinuous on
M el, inf a r (-, ) r (-, *) is upper semicontinuous on Me/l and
lim sup[ inf r(i,,) Tirq,,i*)] < inf r(O,i) -- r(q0, *). (4-7)
By (4-6) and (4-7), inf~Ex r(-r, 2) > r(-r, i*), whence i* is a Bayes solution relative to Ire. O
Remark 4.1. Suppose C7 is a metric space and 8 is a subset of W. Let 8 denote the closure
of 8. Then if R(-, 2) is lower semicontinuous on 8 for each is K a minimax estimator for
(8, K R) is also a minimax estimator for (8, X R). In addition, if A is essentially complete
in the estimation problem (8, X, R), it is also essentially complete in the estimation problem
(8, K, R). Thus we can solve (8, X, R) instead of (8, X. R). Indeed, suppose A is essentially
complete in the estimation problem (8, X, R). Fix is E Then there exists i' E A such that
R(0, i') < R(0, 2) for all 8 E 8. Since 8 C 8, R(0, i') < R(0, 2) for all 8 E 8 and A is also
essentially complete in the estimation problem (8, X, R). Fix is E It is easy to show that
the lower semicontinuity of R(-, 2) on 8 implies that supeoe R(0, i) = supose R(0, 2). Since
& is arbitrary, infe, g supose R8, i) = infie g supose R(0, 2). Thus a minimax estimator for
(8, K R) is also a minimax estimator for (8, X R). In particular if C7 is a finite-dimensional
normed space and 8 is a bounded subset of 7, then, II idustI~, loss of generality, we may assume
that 8 is compact since if thri\ is not the case, we can consider the estimation problem (8, X, R)
In this work it will be sufficient to impose the following conditions:
Condition 4.1. X is subconvex.
Condition 4.2. There exists a compact metrizable space 9 and a function R* : 8 x X*
[ 0, +oo] such that (X9, R) ~ ( 9 ", R*), R* (0, -) is lower semicontinuous on 9 for each 8 E 8,
and R*(-, a) E m(B(8)) for each as K *.
Condition 4.3. R(-, 2) is lower semicontinuous on 8 for each is K R(-, 2) < +oo for each
is 9 and R(-, ) is continuous on 8 for each a E .
Condition 4.4. If~ E X is a Bayes solution relative to -r M Z/e, & is an essentially unique
Bayes solution relative to -r.
In the following corollary, we summarize all the results of this chapter that are needed in the
subsequent chapters.
Corollary 4.1. Suppose Conditions 4.1 and 4.2 hold. Then X has the property (W*), there
exists a minimax estimator and there exists a restricted risk Bayes solution relative to (-r, Co) for
each -r M Z/e and Co > V(8, X9, R). If in addition, 8 is compact and Condition 4.3 holds,
(8, K, R) is strictly determined, there exists a least favorable a priori distribution -ro E Me/l,
any minimax estimator is a Bayes solution relative to -ro, and the class of Bayes solutions is
essentially complete. If in addition, Condition 4.4 holds, then the (essentially unique) Bayes
solution relative to a least favorable a priori distribution -rs Me Z is an essentially unique
Proof Although, the results of this chapter were formulated for an estimation problem
(8, 2', R), there is no use whatsoever of the fact that .9 is a class of estimators. Therefore,
the results also hold for the triplet (8, 2'*, R*), which is, in fact, a zero sum two- person game
(cf. [25]). For the sake of simplicity, we assume that (8, 9 ', R*) is an estimation problem. The
proof can be trivially modified to the case that 9 is not a class of estimators, but an arbitrary
set. Let r*(-r, a) = je R*(0, a)d-r, where -r Me Zl and as E *. It can be verified that since
2'* is a compact metrizable space and R*(0, -) is lower semicontinuous on 2 for each 8 E 8,
{r*(-, a) : as E X} is half-closed, i.e., 2'* has the property (IT*). It can be verified that
since ( 2, R) ~ ( 2*, R*), .9 has the property (IT*). By Theorem 4.1, there exists a minimax
estimator. By Theorem 4.2, there exists a restricted risk Bayes solution relative to (-r, Qo) for
each -r M Z/e and Qo > V(8, 2', R). If, in addition, Condition 4.3 holds, it follows from the
preceding results, that (8, 2', R) is strictly determined, there exists a least favorable a priori
distribution -roE Me/l, any minimax estimator is a Bayes solution relative to -r, and the class of
Bayes solutions is essentially complete. Suppose Condition 4.4 holds as well. Let -r denote a
least favorable a priori distribution. Since any minimax estimator is a Bayes solution relative to
I-r (Theorem 4.6), the (essentially unique) Bayes solution relative to -r is an essentially unique
minimax estimator. By [26, Theorem 2.3.1], this Bayes solution is admissible. O
While sometimes it is possible to verify Conditions 4.1-4.4 directly, other times, especially
when there is no closed from expression for R, it may be difficult to check whether these
conditions hold. In the next chapter, we consider the case that R is based on a loss function L.
In this case, it is possible to derive conditions that can be more easily checked when there is no
closed form expression for R.
CHAPTER 5
THE CASE THAT THE RISK IS SPECIFIED BY A LOSS FUNCTION
A risk function R is usually chosen by first specifying a loss function. A loss function
L : E x D R WU {+oo} specifies that L(:r, d) is the loss incurred to the experimenter
when using the estimate d and :r is the true value of the parameter. Let 9 denote the class of all
(randomized) estimators. Ignoring measurability considerations for now, let L (:r, i) denote the
mapping y H D L(:r, d)d~i(dly). The risk function R : 8 x 9 R WU {+oo} is specified as
follows:
R(0,;i- = x L(:,I-)d~/o U,;r)L. (5-1)
We impose several condition on the family p, the space Y, the space D of possible esti-
mates, and the loss function L.
Condition 5.1. The space Y is a Borel subset of a separable complete metric space and
FTy = B(y), where B(Y) is meant in the sense of the relative topology.
Recall that we assume there is given a probability space (R, Fo2, Q), where R = Y x X,
Fo~ = FTy x FTx, and Q2 is the joint distribution of Y and X. The marginal distribution of X,
denoted Qx, is an element of a family ~Px = (Pox : 8 E 8} of probability measures on (X, FTx),
where Pox(A) = Po(Y x A) for each Ae F x and 8 E 8. Similarly, the marginal distribution
of Y is an element of a family ~Py = {Po' : 0 E 8} of probability measures on (Y, FTy), where
Po ~ (A = Po(A x CL) I fo eac Aey andIC V t 8. Let denote a a-subalgebra of Fo.2 Let
f denote a nonnegative random variable. Let E( f |) denote the conditional expectation of f
with respect to the a-subalgebra W. Let Fa~x denote the a-subalgebra generated by the random
element X. Let E( f |X) = E( f |Fo~x). Let Ae F o.2 The conditional probability of the event
A given the random element X is denoted Pr(A|X) and is defined as E(I;|X), where Ig is the
indicator function of A. The conditional probability of the event A given that X = :r, which is
denoted Pr(A|X = :r), is any FTx-measurable function g for which
/ Ig()dQ)() g/~(:r)ilxdrx(r)foreah Ce (5-2)
To see that such a function indeed exists, the reader is referred to [34, pp. 220]. If y is a version
of Pr(A|X), then according to [34, pp. 221], a conditional probability of the event A given that
X = :r, g(:r), can be constructed as follows: g(X(w)) = g(w) (i.e., g(:r) = g(Lo), where w is an
element in R such that X (w) = :r).
Since in this work the distribution Q is unknown, but is known to be an element of P, the
conditional probability of Ae F o~ given X and the conditional probability of A given X = :r
depend on the true state of nature 8 E 8.
Definition 17. A faction Q (- |X) : FTy x R [0, 1], (A, w) H Q (A|X) (Lo) is said to be a
regular conditional distribution of Y given X if
1. For each we E Q(-|IX) (w) is a probability measure on (f Fy).
2. For each Ae F y, Q (A|X) is a version of the conditional probability Pr (Axx | X).
Fix 0 E 8. Condition 5.1 furnishes that there exists a regular conditional distribution of Y
give X wen Q= P [34, Theorem 2.7.5]. Let Po(-|X)l denote a regular conditionall disriutionvl
of Y given X when Q2 = Pa. For each r E and AeIC F y, we defineI Po(cA|:) = Po(A|X)(w),
where w E is such that X(w) = :r. We call Po(-|:r a regular conditional distribution of Y given
X = :r when Q = Po (or 8 is the true state of nature). Note that since Po(-|X) is regular, Po(-|:r
is a probability measure for each E X. In addition, for each Ae F y, Po(A|:r) is a version of
the conditional probability Pr (Axx | X = :r). Let pv x = { Po(- |:) : xrE X, 8 E 8 }.
Condition 5.2. There exist a o--Jinite measure p on (f FTy) and for each I E 8 a regular
conditional distribution of Y given X = :r when Q = Po, denoted Po (- |:r, such that Po (- |:r is
absolutely continuous with respect to ftfor each Ie 8 and xre X.
Let p~lo| I e) denote a density of Po (-|:r) with respect to p.
Condition 5.3. The loss faction Le E (FTx x B(D)) and is nonnegative.
LICondition 5.4. ThIe mpingLYIL 1 Po(A) E m(B(8)) for each Ae F y x FTx.
Conditions 5.3 and 5.4 were added to guarantee that the integration necessary in the
calculation of the risk function are well defined. It can be verified that if Conditions 5.1- 5.4
hold, R(-,;i) is an extended real-valued nonnegative B(8)-measurable function and hence the
Bayes risk relative to any distribution TrE Me/l is also well defined. We have the following
alternative expression for the risk function:
R(0, i) =~ LS (x.:)pol/ll)dP B] dox. (5-3)
Condition 5.5. The space D is a locally compact metrizable space and is o--compact.
We need the following condition for the case that D is not compact.
Condition 5.6. If D is not compact, then for each sequence of compact sets D, such that
U" ,D, = D and each element d, Sf D, (n = 1, 2, .. .), lim inf,,, L(x, d,) = supdED L(X, d)
for all x E X.
Condition 5.7. For each x e X, L(x, -) is lower semicontinuous on D.
It is easy to see that 9~ is subconvex. Indeed, given i:', i:" E 9~ and 0 < a~ < 1, let i:*
be an element in 9 that satisfies i*(-|y) = ak~'(-|y) + (1 a~)&"(-|y) for all ye E Then
R(-, i*) = a~R(-, i') + (1 a~)R(-, i"). Thus 9 is subconvex.
The space 9 can be identified with a subset of a certain vector space; this was shown
by LeCam in [30]. It is useful to discuss the properties of this vector space. The following
discussion essentially appears in [30] and [31]. Let C,(D) denote the class of continuous real-
valued functions on D with compact support, and let ||lu|| = supdED |U(d)|I for U E Oc(D).
Let LI1 denote the Banach space of equivalence classes of integrable functions on (Y, Ty, p)
with norm || f |1 = | f |dlp Denote by Lpv x the linear subspace of EL spanned by py lx. Let
the product space Co(D) x L~pv be equipped with the norm | | (u, f) | | = | |u| | V | | fl | for
ne E (D) and fe La x~. Let # denote the vector space of bounded linear functionals on
Oc(D) x L~pv The weak*-topology turns # into a locally convex topological vector space
[33]. Call gl, g2 EE Y|Tj) pX-eyUIValent if fr, |yl g2| fdy = 0 for all f e py x. Call
two estimators it,~ 2 ~ Y|X-equivalent if for each D' E B(D), Az(D'|-) and 2(D'|-)
are Py~lx-equivalent. L~et 9 denote the class of equivalence classes so obtained. A functional
e E is said to be positive if a > 0 and f > 0 imply ~(u, f) > 0. According to LeCam if
Conditions 5.1-5.5 hold, every positive linear functional of norm 1 can be represented by an
integral Q(u f ) = @~ u(d)di.~(dl~y) f (y~d y, where is E Certainly the converse is also true, i.e.,
each is E (or more precisely the equivalence class containing 2) is a positive linear functional
on C,(D) x LIv x of norm 1. Thus the class 9 is the class of positive linear functionals of
norm 1. In the sequel, we are not going to distinguish between an estimator and the equivalence
class containing this estimator. Whether a class A of estimators refers to the estimators or the
corresponding equivalence classes can be understood from the context. Let J denote the relative
topology for 9 induced by the weak*-topology.
Lemma 5.1. Suppose Conditions 5.1-5.5 and 5. 7 hold. Then the topological space (9, J) is
metrizable and R(0, -) is lower semicontinuous on 9 for all 8 E 8. If~ in addition, D is compact,
9 is compact.
Proof Under Condition 5.5, the space C,(D) is separable [30]. Since by Condition 5.1, Y
is a separable metric space, the space LI1 is separable [35, pp. 92]. Since LIv x is a subspace
of a separable normed space, it is also separable. It follows that the space C,(D) x L1pv x is
separable. By Theorem [36, Theorem 3.16], if @oc C is weak*-compact, then @o is metrizable.
By Banach-Alaoglu Theorem [33, Theorem 2.5.2], the set Be = {~ E # : || || < 1} is
weak*-compact, where | |~ | denotes the operator norm of a E Thus Be is metrizable. Since ]
is the relative topology, 9 is metrizable.
Let R(x, o, 2) = f L(x, 2)pa (1i| I )dpl. Then by (5-3), R(0, 2) = f R(x, 0, 2)dlox (x).
Using the results of LeCam [30], it can be shown that R(x, 8, -) is lower semicontinuous on 9
for each (0, x) E 8 x X. Since 9 is metrizable, we have by Fatou's Lemma that R(0, -) is lower
semicontinuous on 9 for each 8 E 8.
Suppose D is compact. LeCam [30] showed that a class A of estimators is J-compact if it
is J-closed and if the following conditions holds: For each e > 0 and each (0, x) E 8 x X, there
exists a nuE C:(D) satisfying 0 < Ir < 1 and fy u(d)dit(d y)dYo(,,I|, > 1 e unifo~rmly for all
i E A. The preceding condition certainly holds for 9 when D is compact. Since 9 is 1-closed,
it is compact. O
The following auxiliary lemma is needed in order to prove that Condition 4.2 holds under
very weak conditions.
Lemma 5.2. Let (A, F4~) be a measurable space. Let (C, w) be a topological space that is
locally compact, o--compact, and metrizable, but not compact. Let f : A x C R WU {+oo}
be a nonnegative function. Suppose f (a, -) is lower semicontinuous on C for each a s A.
Suppose for each sequence of compact sets C, such that U" ,C, = C and each element c, ( C,
(n = 1, 2, .. .), lim inf us, f (a, en) = supeec f (a, c) for all as A.~ Let (C*, 0*) denote the
one-point compactification ofC, and let oo denote the added point. Let f : A x C* R WU {+oo},
be defined as follows: For each as A f (a, c) = f (a, c) if ce C, and f (a, 00) = supeec f (a, c).
Then C* is compact and metrizable, f is nonnegative, f (a, -) is lower semicontinuous on C*
for each as A and a subset Co of C is in B(C) if and only if it is in B(C*). In addition, if
fe m(F x B(C)), fe m(Fa x B(C*)), and if f (, c) E m(A) for each Ee C,
f (-, c) EM (F~) for each c E C*.
Proof See Appendix A. O
Theorem 5.1. Suppose Conditions 5.1-5. 7 hold. Then Condition 4.1 and 4.2 hold for (8, 9, R).
As a consequence, there exists a minimax estimator and a restricted risk Bayes solution relative
to (-r, Co) for each -r Me Zl and Co > V(8, 9, R).
Proof We already showed that Condition 4.1 holds. Let us show that Condition 4.2 holds.
Suppose D is compact. Then by Lemma 5.1, Condition 4.2 holds. Suppose D is not compact.
We will use the one-point compactification of D to prove the theorem. The idea to use the
one-point compactification of the class of possible estimates to prove results of the type of this
theorem seems to appear first in [30]. Let D* denote the one-point compactification of D, and
let oo a D* denote the point that is added to D. Let L* : E x D* R WU {+oo} be defined
as follows: For each x E X, L*(x, d) = L(x, d) if d e D, and L*(x, 00) = supdeD L(x, d).
Let 9* denote the class of all estimators with D* as their space of possible estimates. By Lemma
5.2, L* E m(FTx x B(D*)) and D* is compact. Let L*(x, 2)(y) = JD* L*(x, d)di(dly),
and let R* (0, i) = fy xx L*(x, 2) (y)dYo(y, x). Then Conditions 5.1-5.5 and 5.7 hold f~or
(8, 9*, R*). Then by Lemma 5.1, / is compact and R*(0, -) is lower semicontinuous on 9*
for each 8 E 8. It can also be verified that R*(-, a) E m(B(8)) for each as E *. Let 9*(D)
denote the class of estimators & in 9* such that 2(Dly) = 1 for all ye E We claim that
(9*(D), R*) -4 (9*, R*). Indeed, fix i' E 9* such that i'(ooly) > 0 for some ye E Let
y' = {ye Y : i'(Dly) = 0}. Clearly Y' is measurable. Let (Aly) = i'(A n Dly)/i'(Dly)
for y ( y' (A E B(D)), and let 2(-|ly) be a Dirac measure with respect to a point d' E D
for all ye Y'. It can be verified that 2(D'|y) is B(y)-measurable for each D' E B(D).
Now if ye Y', then L* (x, 2) = L* (x, d') < L* (x, 00). If y ( y', then SD* L* (x, d) di(d|y)=
JD L*(x, d)/i'(Dly)di'(dly) < x'(Dly) JD L*(x, d)/i'(Dly)di'(dly)+ (1-i'(Dly))L*(x, 00) =
JD* L*(x, d)di'(dly). Thus L*(x, i) < L*(x, i'). It follows that & is as good as i'. This proves
that ( / "(D), R*) -4 ( / R*). Since 9* (D) C / ', we have that (9* (D), R*) ~ ( / R*).
Clearly ( /, R) ~ (9*(D), R*). Thus ( /. R) ~ ( /', R*). It follows that Condition 4.2
holds. O
As a consequence of Theorem 5.1, under the rather weak Conditions 5.1-5.7, there exists a
minimax estimator and a restricted risk Bayes solution relative to (-r, Qo) for each -r Me Zl and
Qo > V(8. /. R). In order to get the stronger results when 8 is compact, namely that (8, 9, R)
is strictly determined and that there exists a least favorable a priori distribution, we need to
prove that Condition 4.3 holds. Unfortunately, this seems to require rather strong conditions
on the loss function and family P. A set of such conditions is well known for the case that the
loss function is uniformly bounded and P is dominated by a o--finite measure. However, we
are mainly interested in the case that the loss function is unbounded (e.g., the quadratic loss
function). Moreover, in many cases P is not necessarily dominated by a o--finite measure.
CHAPTER 6
THE CASE OF CONVEX LOSS FUNCTION
In this chapter we consider the special case in which D is a convex subset of a finite
dimensional normed space and L(x, -) is convex over D for each x E X. Throughout this chapter
we will assume that L is a real-valued function, Y = RA~, X = WN, Ty = a(IN), and
-T = B(RWN) even if it is not implicitly stated. We are mainly interested in the case that the loss
function is quadratic, i.e., L (x, d) = | V(x d) | 2, where Ve R N, X Nz We will need the following
conditions.
Condition 6.1. for each x E X there exists an e > 0 and a c such that L(x, d) > eld| + c for all
dE D.
Condition 6.2. The measures {PY : 0 E 8 } are mutually absolutely continuous, i.e., for each
8, 8' E 8, PY is absolutely continuous with respect to Pf.
Clearly nonrandomized estimators are more attractive than randomized estimators in terms
of implementation. In general, randomized estimators can outperform nonrandomized estimators.
However, if D is a convex subset of R"N, and if for all x E X, L(x, -) is convex over D, it may
be sufficient to consider nonrandomized estimators. Let ~D denote the class of nonrandomized
estimators. The following lemma is closely related to the Rao-Blackwell theorem [37, Theorem
1.6.4].
Lemma 6.1. Suppose D is a convex subset of R", L(x, -) is convex over D for all x E X, and
Condition 5.3 holds. Then if Conditions 6.1 and 6.2 hold, ~D -4 9.
Proof Let is E Let Oo = {0 E 8 : R(0, 2) < +oo}. If 80 is empty, any estimator in ~D is
as good as 2. Suppose then that 0o is nonempty. Clearly the lemma is proved if we can show that
there exists an estimator i' E D such that R(0, i') < R(0, 2) for each 8 E 0o. Fix 0 E 80. Since
R(0, 2) < +oo, there exists a set Ae F y x &T such that L(x, 2)(y) < +oo for each (y, x) E A
and Po(A) = 1. Let C = {ye Y : (y, x) E A}. It follows that JD Id \$dy) < +OO for 811 y 6 C.
Thus the integral JD d di(dly) is well defined for all y e C. It is well known that C E FTy. Clearly
Po(C x X) = 1. Thus Po'(C) = 1. Since the measures {Po' : 0 E 8} are mutually absolutely
continuous, Po'(C) =1 fort eac n 8~. Thu Po(C x y) 1 for each 8 E 8.
Let i' denote a nonrandomized estimator such that i'(y) = JD d di \$ly) for y 6 C
and i'(y) = 0 otherwise. That i' is indeed a nonrandomized estimator, i.e., a FTy-measurable
function, follows from the fact that C E FTy By the Jensen inequality,
L~~x,2)(y)= L~~~idy ~~'y)) onC X (6-1)
Integrlating (6-1) with respectL to Po, we have R(0, 2) > R(0, i'). Since i' E ~D, ~D -4 9. O
Since ~D -4 9, we consider in the rest of this chapter the estimation problem (8, ~D, R)
instead of the estimation problem (8, '/, R). Clearly, ~D is subconvex by Jensen inequality.
Hence Condition 4. 1 holds for (8, ~D, R). Since ~DC 9, ~D ~ 9. Hence it is clear that Condition
4.2 holds for (8, D, R) if it holds for (8, 9, R). Thus if Conditions 5.1-5.7 hold, Condition
4.2 holds for (8, D, R). In Lemma 6.2 below, we show that under weak conditions, Condition
4.4 also holds for (8, ~D, R). In order to prove that Condition 4.3 holds for (8, ~D, R), it seems
necessary to make additional assumptions on the loss function and the family P.
Sometimes it is convenient to restrict the class of estimators that are available to the experi-
menter to the class of affine estimators. An estimator x is said to be affine if it is nonrandomized
and is an affine function on y. Since we consider only the case that Y = RWN and X = RAN, & is
affine exactly when & = Ay + b for some Ae R N X N and be R AN, and the space of possible es-
timates D is then RWN. Let L denote the class of affine estimators. The space L can be identified
with the space RWNXN x RWN where (A, b) E RWNX~y x RWN, is the estimator & = Ay + b E
and vice versa. Thus the space L can be identified with a finite-dimensional vector space with the
following addition and multiplication by a scalar: If & = (A, b), i' = (A', b') and a~ is a scalar,
& + i' = (A + A', b + b') and ai~ = (a~A, a~b). Let the space be equipped with the norm
|| "'| |A- '|| + b b'|i, where & = (A, b) and i' = (A', b'). Clearly is convex.
It is easy to see that Jensen's inequality furnishes that R(0, -) is convex over L for each 8 E 8
if L(x, -) is convex over D for each x E X. Thus if L(x, -) is convex over D for each x E X, L
is subconvex and hence Condition 4. 1 holds for (8, L, R). In Theorem 6.1 below, we show that
under rather weak conditions, Condition 4.2 holds for (8, L, R). As in the estimation problem
(8, ~D, R), it seems necessary to make further assumptions on the loss function and the family
p in order to prove that Condition 4.3 holds for (8, L, R). In Lemma 6.2 below, we show that
under weak conditions, Condition 4.4 also holds for (8, L, R).
Theorem 6.1. Suppose D = RWN, and Conditions 5.3, 5.4, 5.6, and 5.7 hold. Then Condition
4.2 holds for (8, L, R). As a consequence, there exists a minimax estimator and a restricted risk
Bayes solution relative to (Qo, r) for each Qo > V (8, 9, R) and -r E eZl.
Proof Clearly if {i,} E is a sequence that converges to an element i* in the sense of the
norm |I | | |, it converges pointwise on y. Let in = (A,, b,) for n = 1, 2, ... Suppose in
converges to i* = (A*, b*). Fix x E X. By Condition 5.7, lim inf,,, L(x, 2,(y)) > L(x, i* (y))
for each ye E By Fatou's lemma, lim inf,,, R(0, 2,) > R(0, i*) for each 8 E 8. Thus
R(0, -) is lower semicontinuous on for each 8 E 8. Let us show that for each sequence of
compact subsets C, of L such that U" zC, = L and each element in Sf C, (n = 1, 2, ...),
lim inf,,, R(8, in) = supy,~ R(0, i) for all 8 E 8. Fix a sequence of compact subsets C,
of L such that U" zC, = L and a sequence {i,} E L such that in ( C, (n =1,2...)
Fix 0 E 8. Certainly lim inf,,, R(8, in) < supper R(0, 2). Thus it is left to prove that
lim inf,,, R(0, in) > supy,~ R(0, 2). By Fatou's lemma,
lim inf R(0, 2,) > lim inf L(x, 2, (y))dPo(y, x). (6-2)
Fix (y, x) E Y x X. Let D, = {i,(y) : in E C,} (n = 1, 2, .. .). We claim that D, is a
compact subset of D. Indeed, let {di} be a sequence in D,. Then there exists a sequence {ij} in
C, such that (~(y) = di. Since C, is compact, there exists a subsequence {i~ } of the sequence
{ii} and an element i'* E C, such that &~ i '*. This implies that &~ (y) i '*(y), whence
{di, } i '*(y). Since i'*(y) E D,, D, is compact. We claim that U" zD, = D. Fix d E D.
Clearly there exists i s such that 2(y) = d (e.g., & = (A, b), where A = 0 and b = d).
Since U" ,C, = C, i s C, for a sufficiently large. This implies that d e D, for a sufficiently
large. Thus U" zD, = D. By Condition 5.6, lim inf,,, L(x, d,) = supdeD L(x, d), where
d, = 2,(y). It is easy to verify that supdeD L(x, d) = sup,,, L(x, 2(y)). Thus for each i s
we have
lim inf L(x, in(y)) > L(x, 2(y)). (6-3)
Since y and x are arbitrary,
Since x is arbitrary,
lim inf R(8, in) > sup R(0, 2). (6-5)
Since is a finite dimensional normed space, it is locally compact, o--compact, and
metrizable. Let denote the one-point compactification of L and let oo denote the added point.
For each 8 E 8 let R*(0, 2) = R(0, 2) if & E L, and let R*(0, 00) = sup,,, R(0, 2). By Lemma
5.2, is compact and metrizable, R*(0, -) is lower semicontinuous on for each 8 E 8, and
R*(-, a) E m(B(8)) for each as E *. Clearly L C *. Since oo is the only element in \ L and
R*(-, 00) > R(-, 2) for each is L (, R) -4 (*, R*). Since C *, (, R) ~ (*, R*). Thus
Condition 4.2 holds. O
The following lemma is concerned with the essential uniqueness of Bayes solutions.
Lemma 6.2. Consider the estimation problem (8, K, R), where X is either ~D or L. Suppose
D is a convex subset of R"N, L(x, -) is strictly convex over D for all x E ./, Condition 5.3 holds,
and V (8, K R) < +oo. Then if Condition 6.2 holds, a Bayes solution relative to -r Me Zl is an
essentially unique Bayes solution relative to -r.
Proof We prove the lemma for the case X = ~D. The proof for the case X = L is entirely
analogous. Fix -r Me Zl. It can be verified that our assumption that V(8, ~D, R) < +oo
implies that infeez, r (-, 2) < +oo. Suppose i', i" E D are Bayes solutions relative to -r.
The lemma is proved once we show that is equivalent to i'. Since inf~Er r(-r, i) < +oo,
r (-, i') = r (-, i") < +oo. Let i* = 1/22' + 1/22". Then by Jensen inequality, L (x, i* (y)) <
1/2L(x, i'(y)) + 1/2L(x, i"(y)) with strict inequality whenever i'(y) / i"(y). It follows that
R(0, 1i*) < 1/2R(0, 1i') + 1/2R(0, i"). Then clearly r(-r, i*) < 1/2r(r, i') + 1/2r(r, i"). Since
i' and i" are Bayes solutions relative to -r, we must have r (-, i*) = 1/2r (-, i') + 1/2r (-, i").
Thus R(0, i*) = 1/2R(0, i') + 1/2R(0, i") (a.e. -r). Clearly there exists a set 0o E B(8) such
that R(0, i') < +oo and R(0, i") < +oo for each 8 E 80 and -r(80) = 1. Thus there exists
an element 8o E Oo such that R(00, i*) = 1/2R(00, i') + 1/2R(00, i"). It follows that there
exists a set Ae F y x FTx such that L(x, i*(y)) = 1/2L(x, i'(y)) + 1/2L(x, i"(y)) for each
(y, x)It EAIC an Poo(,A) =. Let C = {ye Y : (y, x) E A}. Then Ce F y, ~I = ~I" on C and
Poo(C x X) = 1. Since the measures {Po' : 0 E 8, x E X} are mutually absolutely continuous,
Po(C x X) = 1 for each 8 E 8. It follows easily that i" is equivalent to i'. O
In the rest of this chapter, we assume the loss function is quadratic, i.e., L : (x, d) H
|V(x d)|12, where Ve R N"X A. We assume VTV > 0. The extension to the case VTV > 0 is
discussed later. In this case it can be verified that Conditions 5.1, 5.3, 5.5-5.7, and 6.1 hold. Thus
if D is convex and Condition 6.2 holds, ~D -4 9. Certainly ~D is subconvex and Condition 4.1
holds. In addition, if Conditions 5.2 and 5.4 hold, Conditions 4.2 holds for (8, ~D, R). Similarly,
Condition 4.1 holds for (8, L, R), and if Conditions 5.2 and 5.4 hold, Condition 4.2 holds for
(8, L, R). Suppose, in addition, that P is a Gaussian family of distributions, i.e., Y and X are
jointly Gaussian, when 8 is the true state of nature, for each 8 E 8. Then py and pv x are also
Gaussian families of distributions. Suppose the family pylx is dominated by the Lebesgue-
Borel measure on (RW~y x B(RIWy)), which is denoted p. It can be easily verified that py is
also dominatedLL by p andC thatL sinlce, in1 addUition, for echLI t V po~y), the densitLy of1, Po:,+, 1~,., 1,,,,:,~+, ,,,,Dr/\,/. 1,~,:,, kwith
respect to p, is positive, the measures {Po' : 0 E 8} are mutually absolutely continuous. Thus
Condition 6.2 holds. We would like to check under what conditions Condition 4.3 holds. Under
the current assumptions, it is well known that there exists a regular conditional distribution of X
given Y =. y,~, when is the true state of ntreLt Po(-|y) denote thi conditional distributor,:
which is well known to be Gaus sian. Let is : Ya R Nz y H Ea(X |y), where Ea (X | y
denotes the conditional expectation of X given Y = y, when 8 is the true state of nature. That
is, is is the conditional mean estimator for x based on the observation y when Q = Pa. Let
Fe = Ea(|V(fo(Y) X)|2), i.e., ~e iS the mean squared error (MSE) matrix of is, when
Q2 = Pa. Then by the so-called orthogonality condition, it is straightforward to get the following
expression for the risk function:
R(0, 2) = tr(V~oV )+|To) 2(y|pe(y)dy.l (6-6)
In what follows 9, is the class of Bayes solutions when (8, ~D, R) is considered, i.e., for each
i* AD tereexitsren uhta i,7 inf~Ez r (i, -r). Since ~D -4 9, each element
in 9, is also a Bayes solutions when (8, 9, R) is considered. In what follows, continuity of
functions from 8 into R"N and RWNxN is meant in the sense of the Euclidean norm and 2-norm,
respectively.
Theorem 6.2. Let Z = [YT XT]T. Suppo3se H Ea(Z) and 8 H Ea(ZZT) are continuous
on 8. Then if 8 is compact, R(-, 2) is lower semicontinuous on 8 for each f e D and R(-, 2) is
bounded and continuous on 8 for each is E Hence Condition 4.3 holds for (8, ~D, R).
Proof As mentioned earlier, py is a Gaussian family. It is easy to see that by the assumption
of the theorem, 8 H Ea(Y) and 8 H Ea(YYT) are continuous on 8. Since we assume
that the family py is dominated by p, 8 po(y) is continuous on 8 for each ye E It is
well known that 20 (Y) = Ea (X) + Ea ((X Ea (X)) (Y Ea (Y))") Ae (y Ea (Y)) and
re = ro Ea((X Ea(X))(Y Ea(Y))")Ae E((Y Ea(Y))(X Ea(X))T), where As
and To are the covariance matrices of Y and X, respectively, when 8 is the true state of nature
(i.e., when Q = Po). Since 8 H Ea(Z) and 8 H Ea(ZZT) are continuous on 8, 8 H fo(y) is
continuous on 8 for each ye~ E and 8 H Fo is continuous on 8. Since Po is Gaussian, there
exist a matrix As a RWN, X and a vector be E RWN such that 20 (Y) = Aey + be. Since 8 H 20 (y)
is continuous on 8 for each ye~ Y and Y = RIWN, 8 H As and 8 H be are continuous on 8. Since
8 po(y) is continuous on 8 for each ye E it follows easily from Fatou's lemma and (6-6)
that R(-, ) is lower semicontinuous on 8 for each is ED.
Fix -r E Mo.l Let 2, denote the (essentially unique) Bayes solution relative to -r. It is well
known that when the loss function is quadratic, the conditional mean estimator with respect to
-r Me Zl is a Bayes solution relative to -r. Thus without loss of generality, we may take 2, to
be the conditional mean estimator, i.e., 2,(y) = E,(Xly), where E, denotes the expectation
operator when 8 is the value taken by a random element whose distribution is -r. Note that since
by our assumption V(8, ~D, R) < +oo, r(-r, 7) < +oo. This implies that the MSE matrix of the
estimator 7 is bounded.
Clearly E, (X |y) = E, (E, (X |y, 8) | y). Let us examine the term E, (X |y, 8). This term is the
conditional expectation of X given Y = y and 8, where 8 is the value taken by a random element
whose distribution is -r. But this is exactly Eo (X |y). Thus E, (X |y) = E, (Eo (X |y) |y). It follows
that | V(7(y) 200 (Y)) 2" = | VE, (So(Y) 200 (Y) Y) 12. Thus
I ~v(v)(Y f o.(Y) 2 < E,(|V(f o(y) iso lU)) 121
=E,(|V((Ae Aeo)y + be boo) 21v
< E,((||V||2(|| Ao Aeo 2|lay| + |bo boo 1))21Y
Since the mappings 8 H As and 8 H be are continuous on the compact set 8, they are bounded.
Thus there exist positive real numbers cl and c2 Such that
|V(:, (y) ieo(Y) 12 < E,((clly| + c2 2 y) = (Cly + C2 2
Since 8 H Fo is continuous on 8, which is compact, there exists a positive real number a such
that tr {V~oVT} < a for all 0 E 8. We have from (6-6) that
R(00, 7)
It is easy to verify that R(00, fr) < +oo. Put h(y) = (clly| + C2 2. Since 8o is arbitrary
R(0, 7) < +oo for each 8 E 8. Moreover, for each 8 E 8, |V(i,(y) is(y))|2 < h(y) and & is
a Po-integrable function. Certainly
|R(0, 17) R(00, 97)| < | tr(V(Fe Foo)VT)|I
+/4 (W(iaiii) -2,(7)| -|IooU -7U) 2 0
Let {0,} be sequence that converges to 8o. We showed that lims,, |Co. Feo | = 0. Thus
lim tr(V(Fos Foo)VT) = 0. (6-9)
By the Lebesgue dominated convergence theorem,
Certainly for each n > 0, | V(ion (y) 97(y)) 12 < h(y). Since h is Poo-integrable and 8 po(Y)
is continuous on 8 for each ye E a well known theorem on exponential families [37, Theorem
1.4. 1] furnishes that
It can be verified that since 8 po(y) is continuous on 8 for each ye E the above equation
implies that
Since |V(fos (y) fr(y))|2 y)
/ | ~ion(7)- 977))2 00 y)|d 0.(6-11)
By (6-8)-(6-11), R(-, 7) is continuous on 8. Since R(-,, ) < +oo and R(-, 7) is continuous
on 8, which is a compact set, R(-, 7) is bounded on 8. It follows that R(-, 2) is bounded and
continuous on 8 for each is 9 O
Let Mr denote the class of Bayes solutions when (8, L, R) is considered.
Theorem 6.3. Let Z = [YT XT]T. Suppo3se H Ea(Z) and 8 H Ea(ZZT) are continuous
on 8. Then if 8 is compact, R(-, 2) is lower semicontinuous on 8 for each is L R(-, i) is
bounded and continuous on 8 for each is M Hence Condition 4.3 holds for (8, L, R).
Proof Although some modifications are needed, the proof of this theorem is very similar to the
proof of Theorem 6.2 and is omitted. O
While the proofs of some of the results of this chapter clearly break down if VTV is not
positive definite, all these results are valid also in the case that VTV is not positive definite.
There is a simple method to show that this is indeed the case. Note that if VTV is not positive
definite, L(x, d) = L(x, d') whenever d d' E Ni~(V). Thus we may call d and d' in RWN
equivalent if d d' E Ni(V) and choose the space of possible estimates to be the set of
equivalence classes so obtained instead of RWN. In this case, the space of possible estimates is
equipped with the metric a (d, d') = | Vt V(d d') |, where d, c' E D, d is any element in d, and
d' is any element in d'. This choice for D is equivalent to choosing D = Ni(V)I with the usual
Euclidean norm since for any equivalence class in D there is associated a point in Ni(V)I and
vice versa. It can verified that with either one of these choices for D, the results for (8, ~D, R) are
still valid. To show that the results for (8, L, R) are still valid, it is necessary, to define a class
cL = {(N~(V)IA, Ni(V)lb) : (A, b) E }. It is obvious that L' ~ L. Certainly for each & E L',
2(y) is in the new space of possible estimates. Now, it is straightforward to show that all the
results of this chapter are still valid for (8, L', R) and hence for (8, L, R).
CHAPTER 7
FINDING A MINIMAX ESTIMATOR AND THE DUAL PROBLEM
In this section, we consider in more detail the problem of finding a minimax estimator for an
estimation problem (8, X, R). We need the following additional conditions.
Condition 7.1. For any -r M E Zl, there exists an essentially unique Bayes solution relative to -r.
We let 2, denote the (essentially unique) Bayes solution relative to -r. Let r(-r) = r(-r, 7).
Let Me/l denote the class of distributions in Me/l with finite support.
Condition 7.2. If {n } is a sequence in Me/l that converges weakly to -r Me Zl, then R(0, in)
converges to R(0, 7) uniformly on compact subsets of 8.
By Corollary 9.1i, if Conditions 4.1-4.4 hold and 8 is compact, the problem of finding a
least favorable a priori distribution is dual to the problem of finding a minimax estimator. Thus in
the rest of this chapter, we concern the problem of finding a least favorable a priori distribution.
The following theorem is essentially similar to a theorem in [18] and the iterative algorithm
proposed in [38].
Theorem 7.1. Suppose 8 is compact, R(-, i) is continuous on 8 for each is 9 and Conditions
7.1 and 7.2 hold. Construct a sequence {-ri}@, E Me/l as follows. Let -ri be any distribution in
Me/l. Having chosen -rl, ri E M, let Os E 8 be such that R(0s, in) = supeoe R(8, in). Let
ni,o = c004 + (1 a~)nr. Let asi be such that r(9i,ai) = supe[o, 1] T(ni,a) and let ni+l = 74,ai. Then
the sequence {8 }) converges weakly to a least favorable a priori distribution.
Proof The proof follows easily from the proof of [18, Theorem 2.3], with slight modifications.
The main difficulty in the algorithm described in Theorem 7.1 is in finding 04 E 8 such that
R(0s, in) = supoe, R(8, in) for i > 1. Another difficulty is due to the fact that, in general,
since the support of -r may grow as i grows, the complexity of the algorithm calculations also
grows with i. The problem of finding asi such that r"(nr,ai) = supe[o,1] "(ri~,a) is addressed
below and can be solved numerically. In some special cases, it is easy to find Os E 8 such that
R(0s, in) = supose R(8, in) for i > 1 and the complexity of the algorithm calculations remain
fixed as i grows. In these special cases, the algorithm described in Theorem 7.1 is practical in
finding minimax estimators. One such case is when 8 is a finite set. In the sequel, we show that
when we consider linear estimation with quadratic loss function, the algorithm of Theorem 7.1
can be often used to find a minimax estimator. In the more general case, this algorithm can be
used just to find e-minimax estimators, which are good approximations to minimax estimators for
e sufficiently small. We discuss the derivation of e-minimax estimators, in great detail, in the next
chapter.
The problem of finding asi E [0, 1] such that r"(40,a) = supae[o, 1] T(8i,a) is a standard
optimization problem in RW:
maximize r"(cl + (1 a~)-r) subject to as [0, 1].
Fix -r1,r BE Me/l, and let To, = 071r + (1 a~)-r for as [0, 1]. In the rest of this chapter we
consider the following optimization problem, which includes the previous one as a special case:
maximize r(-r,) subject to as [0, 1]. (7-1)
Lemma 7.1. a~ H (-r) is concave on [0, 1].
Proof That r is concave on Me/l is well known and easy to prove. As a consequence, it is easy to
verify that a~ H (-r) is concave on [0, 1]. O
Let D(a~) = r(Tis iv.) r(-r,;i'r) for as [0, 1].
Lemma 7.2. Fix c~o E (0, 1). Then r"(-r) r"(To.) < D(no)(a~ n~o) for all as [0, 1].
Proof Certainly r(-r,) r(To,,) < r(-roo 'r,) r(-r,,). It can be easily verified that r(-roo m>)
r(-r,,) = (a~ n~o)D(n~o). The lemma follows easily. O
As a consequence of this lemma, -D(n~o) is the subderivative of a~ -r(-r,) at the point
n~o E (0, 1).
Lemma 7.3. Suppose Condition 7.2 holds. Then a~ D (c) is monotonically decreasing and
continuous on [0, 1] and r(-r,,) = supe[o,11? (-r) for n~o E [0, 1] if D(n~o) = 0.
Proof Certainly To, converges weakly to -roo whenever a~ converges to cto. Since -r1 and -r2 have
finite supports, Condition 7.2 furnishes the continuity of a~ D(a~) on [0, 1]. Since -D(a~) is
the subderivative of the convex function am -r(-r,), aa D(a~) is monotonically decreasing
on (0, 1). By continuity, a~ D(a~) is monotonically decreasing on [0, 1]. Fix to E [0, 1].
Suppose D(ao) = 0. Then by lemma 7.2, r(-r,) < r(-roo) for all as [0, 1]. It follows that
r(To,) = supago,1,] r(To,).
Remark 7.1. Lemmas 7.2 and 7.3 imply that we can use relatively simple numerical al gwiduallr\
to solve (7-1) numerically. Suppose r (-ri, 27) = r (-72, 7). Then r (-r1) = r (-ro, 7) for all
as [ 0, 1]. It follows that r (-r1) > F (-r) for all as [0, 1] and hence a~ = 1 is a solution of (7-1).
Similarly, if r (-r, in ) = r (-r2, i), a~ = 0 is a solution of (7-1). Using the fact that a~ D (a)
is continuous and monotonically decreasing on [0, 1], and the fact that as [0, 1] is a solution of
(7-1) if D(a~) = 0, the optimization problem (7-1) can be solved as follows: If D(1) > 0, then
a~ = 1 is a solution of (7-1). IfD (0) < 0, then a~ = 0 is a solution of (7-1). Finally, ifD (0) > 0
and D (1) < 0, there exists as (0, 1) such that D (a) = 0 and hence a~ solves (7-1). In ;Ihi\ case,
since D(a~) = 0 and aa D(a~) is monotonically decreasing on (0, 1), a~ can be easily found
using the bisection method.
CHAPTER 8
APPROXIMATING A MINIMAX ESTIMATOR
In this chapter we approximate minimax estimators by using e-minimax estimators. The
main idea is to find a minimax estimator (or an approximation to a minimax estimator) i* for an
estimation problem (87, X, R), where 87 is a finite subset of 8, such that & is an e-minimax
estimator for the estimation problem (8, X, R). The main question is how to construct the set
87. Finding a minimax estimator for (87, .F. R), when Of is a finite set can be done using the
algorithm in Theorem 7.1. In practice, it may be necessary to find an e'-minimax estimator for
(87, K, R) such that this estimator is an e-minimax estimator for (8, X, R). In the latter case,
the algorithm in Theorem 7.1 can still be used, but it is necessary to have a condition that enables
us to check when the required precision is achieved. Checking if an estimator 7, an essentially
unique Bayes solution relative to -r, is an e-minimax estimator can be done using the following
lemma:
Lemma 8.1. Suppose Condition 7.1 holds and supose R(8, 17) F (-) < e, then 2, is an
e-mmnimax estimator
Proof Since supose R(8, ir) in a supose R(0, i) > r(-r), we have that
sup R(0, 7) < inf sup R(0, 2) + e.
Thus 2, is an e-minimax estimator. O
The condition of Lemma 8.1 is only a sufficient condition for an e-minimax estimator.
However, under certain conditions, if the numerical algorithm converges to a least favorable a
priori distribution, then there exists an integer NV such that Lemma 8. 1 is satisfied for the NVth
iteration.
Lemma 8.2. Suppose 8 is compact, ./ is subconvex and has the property (W*), R(-, i) < +oo
for each is E R(-, 2) is continuous on 8 for each is E and Conditions 7.1 and 7.2 hold.
Let {nri } EZ/e be a sequence that converges weakly to a least favorable a priori distribution
I-o EM eZl. Then there exists an integer NV such that supose R(8, i,) r (-r) < e for all i > NV.
Proof Clearly the lemma is proved if it is shown that supose R(8, in) r"(-r) 0 Since 8 is
compact and Condition 7.2 holds, it is straightforward that
sup R(0, in) sup R(0, fro). (8-1)
BEe sEe
Since R(-, fr) is continuous on 8, it can be shown, as in the proof of [38, Theorem 2 part 3],
that
F(i) r(To). (8-2)
By Theorems 4. 1 and 4.5, there exists a minimax estimator and (8, K, R) is strictly determined.
By Theorem 4.6, fro is a minimax estimator. Thus
r(-ro) = sup R(0, fro). (8-3)
BEe
By (8-1)- (8-3), supose R(8, in) r"(-r) 0 and the lemma is proved. O
The following condition is needed for some of the results of this chapter.
Condition 8.1. 8 is a subset of a normed space 7 and the metric a for 8 is induced by the
norm |I | | | of W.
A set U is said to be 6-dense in 8 (in the sense of the metric a) if for any 8 E 8 there exists
8' E U such that A(0, 8') < 6. Note that if 8 is compact, for each 6 > 0, there exists a finite
subset of 8 that is 5-dense in 8.
Theorem 8.1. Suppose X is subconvex and has the property (W), Condition 7.1 holds, and the
family {R(-, 7) : -r Me }Zl is equicontinuous on 8. Then for any e > 0, there exists a 6 > 0
such that for any finite 6-dense subset 8; of 8, the following hold:
1. For any -r M E Zl, there exists a probability measure -ro E Te, such that
R(0, fro) R(0, 7) < e for all 8 E 8.
2. There ex-ists a probabiiliy measure 7o E 7e, such that 7 (7o) = supEreve 7 (7) and f r is an
e/2-minimax estimator
Proof 1) Let p(0, 8') = supee |R(0, 7) R(0', 7)|i. The equicontinuity of {R(-, 7):
-r Me}i/o on 8 implies that for any e > 0, there exists a 6 > 0 such that p(0, 0') < e/2
whenever A(0, 8') < 6. Let Of be a finite subset of 8 that is 5-dense in 8 in the sense of a.
Then 87 is e/2-dense in 8 in the sense of p, i.e., for any 8 E 8, there exists a O' E 87 such that
p(0, 8') < e/2. Fix -r Me Zl. Consider the estimation problem (87, X, R). Let 87 be equipped
with the discrete topology. Since 87 is finite, it is compact and R(-, 2) is continuous on 87 for
each is K It can be shown, similarly to Theorem 4.7, that there exists a probability measure
I-o E Te, such that R(0, fro) < R(0, 7) for all 8 E 87. Since 87 is e/2-dense in 8 in the sense
of p, R(0, fro) < R(0, 7) + e for all 0 E 8 and part 1) is proved.
2) Clearly the results of Chapter 4 can be used for the estimation problem (87, X, R). Thus
for any finite 87, there exists a least favorable a priori distribution Tro E -re, and fro is an admis-
sible minimax estimator for (87, X, R). It follows that maxeee, R(8, fro) < maxeee, R(8, fr)
for any -r Me Zl. Since Of is e/2-dense in 8 in the sense of p, supose R(8, iro)
supeoe R(0, 7) + e/2. Thus fro is an e/2-minimax estimator for (8, X, R). O
Remark 8.1. Theorem 8. 1 part 1) is an e-complete class theorem, provided that 9J is essentially
complete (the reader is referred to[/39]for the definition of an e-complete class of decision
functions). It differs from TU 10 al~ il: 's e-complete class theorem in[/39] and has the advantage
that each element in thri\ class is admissible.
Theorem 8. 1 part 2) implies that an admissible e-minimax estimator can be found in the
following way: Partition the set 8 to disjoint sets 81,82.., O N Such that the diameter of
84 (i = 1,. ., NV) does not exceed 6. In each set 84 take a point Bi. Let 87 = {Or 82, HN *
Solve for a minimax estimator for (87 X R). If 6 is sufficiently small, then the resulting
estimator is e-minimax for (8, K, R). Theorem 8.1 does not spec~if\ how to choose 6 that would
guarantee a certain e. However it is clear from the proof that it is sufficient that 6 \aiisfit 1
|R(0, 7) R(0', 7) | < e/2 whenever -r Me Zl, 8, 8' E 8, and A(0, 8') < 6. In fact for
thri\ choice of 6, it is enough derive an e/2-minimax estimator for (87 K R) and the resulted
estimator is e-minimax for (8, X, R).
The hypothesis in Theorem 8. 1 that the family {R(-, 7) : -r Me}i/o is equicontinuous on
8 is rather strong. While it is often satisfied in the case that the risk is based on a loss function
that is uniformly bounded, it may not be satisfied in the case that the loss function is unbounded.
Another, difficulty with Theorem 8.1 is that the requirement from the set 87 is very strong. A
set 87 that is constructed according to this theorem has the property that any estimator can be
approximated by a Bayes solution relative to a measure with support in 87 with degradation of
no more than e. However, we are mainly interested in approximating a minimax estimator and
not any estimator. Thus it may be possible to choose a finite set whose cardinality is significantly
smaller than 87 in Theorem 8.1. Due to the above, we are not going to use Theorem 8.1 in the
sequel, and we are going to derive methods that do not require equicontinuity.
It is convenient to use the notion of Fr~chet differentiability. Let B(X, Y) denote the set
of bounded linear operators from a normed linear space X to a normed linear space Y. Given
f E B(X, Y), || f || denotes the operator norm of f i.e., || f || = sup,: ||2llst ||I fX|| .
Definition 18. Let X and Y be normed linear spaces, U C X open, f : X Y and x e U. The
function f is said to be Fri'chet differentiable at x if there is an element AE B (X Y) such that
||f (x h) f (x) Ah||
him = 0.
hwo ||h||
We call A the Fri'chet derivative of f at x and denote it by D f(x).
If f : X Y is twice Fr~chet differentiable at x E X, we let D2 f(x) denote the second
Frichet derivative of f at x. Note that D2 f(x) E B(X, B(X, Y)). Given is 9 X, let DRa(0) and
D2Ra(0) denote the Fr~chet derivative and the second Fr~chet derivative, respectively, of R(-, 2)
at 0.
Definition 19. Given an estimation problem (8, X9, R), let y > 0 and 80 be a subset of
8. An estimator i* E X is said to be a (y, 80)-optimal estimator if supesea R8, *
(1 + y) V (o, K R). An estimator i* e X is said to be a y-optimal estimator if it is a
(y, 8) -optimal estimator
Clearly a y-optimal estimator is an e-minimax estimator for e = y V(8, K, R). It is often
more convenient to search for a y-optimal estimator instead of an e-minimax estimator. The
reason is that by deriving y-optimal estimator, 100y gives us the maximum degradation in terms
of the maximum risk in percent (relative to the minimax risk). This is different than the case that
we derive an e-minimax estimator, in which e is not normalized by the minimax risk. Certainly
what is considered a small e in a certain problem, may be considered huge in a different problem.
Lemma 8.3. Suppose X is subconvex and has the property (W), Conditions 7.1 and 8.1 hold
and 8 is compact. Suppose for any -r Me Rl, there exists an extension of R(0, 7) from 8 to an
open convex set 8 C UC c such that R(0, 7) is Fri'chet differentiable on U and there exists
a real number M~ such that | |DR,7 (0) | | < M~ supose R8, 17) for all -rE Me/l and 8 E 8.
F~ix Y > 0 and 0 < ?' < -i, and let d = .~i~ Then if 8p is a fnite 5-dense subset of 8, a
(y', 87) -optimal estimator is a y-optimal estimator
Proof Let -ro E Me/l be such that the support of -ro is contained in 87 and xfo is a (y', 87)-
optimal estimator. Fix 8, 8' E 8 such that | |8 el'|~ |, <. Then by the mean value theorem,
|R(0, ro) R(0', r)I I 0-0 8' supe[o~, ||DR;~b(a0B+ (1 a)0')| < 6MsuposeR(8 N~,r).
TIhus supose R(B, ir)(1 bM) < supose, R(B1 fro) < V(87, K^, R)(1 + y'). Since
V(87, K, R) < V(8, K, R), we have supose R(8, fro) < V(8, K, R)(1 + y), and fro is a
y-optimal estimator. O
The cardinality of a finite subset 87 of 8 that is constructed according to Lemma 8.3 may
still be very large; this may cause the calculation of a least favorable a priori distribution for
(87, X, R) to be formidable. In a special but very important case of compact 8, it is possible to
derive a finite 6-dense subset of 8 such that 6 depends linearly on j This means that if y is
decreased by factor of say 4, 6 should be roughly decreased only by factor of 2 (assuming y
and y'
Definition 20. Let K( be a subset ofa vector space V. A nonempty set S C K( is called an
extreme set of K if a, b E K, O < t < 1, and (1 t) a + tb E S imply a, b e S.
Definition 21. Let K be a subset of a vector space V. The point p e K is an extreme point of K
if and only if a, b e K, O < t < 1, and (1 t) a + tb = p imply a = b = p.
Let A be a subset of a vector space V. Let dA denote the extremal boundary of A, which is
the set of all extreme points of A. Let co(A) denote the convex hull of A. Given a subset K of a
vector space V and a subset S of K, let & (S) be the set of all points s ES such that a, b E K,
O < t < 1, and a = (1 t)a + tb imply a, b E S. Clearly the set &K(S) is an extreme set of
Kt. Given a set A, we let 2A denote the power set of A, i.e., the collection of all subsets of A. We
consider the case that 8 is a compact convex subset of the normed space C7 and 88 is a finite set.
Note that by [36, Theorem 3.20], the set co(88) is compact and hence the Krein-Milman theorem
[36, Theorem 3.23] implies that 8 = co(88).
Lemma 8.4. Suppose 8 is a compact convex subset of the normed space 7 and 88, is a finite
set. Let 6 > 0. Then there exists a finite subset 8; of 8 such that for any nonempty A E 2ae,
87 n deo (co(A)) is a 6-dense subset of deo (co(A)). The set 87 is a 6-dense subset of 8 and
Proof Let Al, A2,..., AN denote the elements of 2ae excluding the empty set. Fix 1 < i < NV.
It is easy to see that Beo(co(Ai)) C 8. Thus Beo(co(Ai)) is relatively compact and hence totally
bounded. Thus there exists a finite subset 87,< of deo(co(Ai)) that is 5-dense in de(co(Ai)).
Let 87 = U~,Of87. Then clearly 87 is finite. Moreover, 87 n de(co(Ai)) = 87,4 and hence
Of n Be(co(Ai)) is a 6-dense subset of Be(co(Ai)). Since 8 = co(88), 8 = de(co(88))
and Of is a 6-dense subset of 8. Let 8 E 88. Then {0} E 2ae. Since 8 is an extreme point,
Be(co({0})) = {0}. Thus there exists 1 < j < NV such that 87,; is a nonempty subset of {0} and
we must have B y,; = {0}. Thus 8 E Of and Of > 88. O
Let ?" = {ae R : ||a||1 = 1}.
Lemma 8.5. Suppose 8 is a compact convex subset of the normed space 7 and 88 is a finite
set. Let Or 82 N denote the elements of 88,. Fix 0 E 8. Then there exists an index set
J C (1, 2,. Nsuch that
#(i) > 0, i EJ
pe R" n ",O (i04- (8-4)
i=1 P(i) = 0, otherwise
Proof Certainly 8 can be written as a convex combination of 81, 82, *, HN. Let 1 < M~ < N be
ther largest intege, r for whichl therem exist ye R r n 7" and an index set J such that | J| = M~,
O 2", O()s ()>0frali ,ad#i for all i ( j. Suppose (8-4) is false.
Then there~ exists R' r R"P n 7"N and a non empty index set J' / J such that 8 = 'i)s
P'(i) > 0 for all is J ', and P'(i) = 0 for all i ( J'. Let po = (4+ #'/.Thn8= N )
Po(i) > 0 for all i EJ U J', and Po(i) = 0 for all i ( J U J'. Since J' / J, | J U J'| > M~, which
is a contradiction to the definition of M~. Hence (8-4) must hold. O
Lemma 8.6. Suppose X is subconvex and has the property (W), Conditions 7.1 and 8.1 hold,
8 is a compact convex subset of the normed space 7, and 88~ is a finite set. Suppose for each
-r Me Zl, there exists an extension of R(-, 7,) from 8 to an open convex set 8 C U cC
such that R(-, 7) is twice Fri'chet difgerentiable on U and there exists a positive real number
M~ such that | |D2R~, (0) | | < M supose R(8, ir) for all -rE Me/l and 8 E 8. Let y > 0 and
0 < y' < 7. Let a = .I Suppose 87 is a finite set that latirisi 1: For any nonemnpty
A E 2ae, Of n Be,(co(A)) is 5-dense subset of de(- (co(A)). Then a (y', 87)-optimal estimator is a
y-optimal estimator
Proof Let Or, 82, N denote the elements of 88~. Let -ro E Me/l be such that the support of
-ro is contained in 87 and xro is a (y', 87)-optimal estimator. Clearly R(-, fr) is a continuous
function on a compact set 8. Thus there exists a point 0* E 8 such that supos A, iRo8~)
R(0*, fr). If 0* E 87, then the lemma clearly holds. Suppose, then, that 0* is not in 87.
By Lemma 8.5, there exists an index set J C {1, 2, .. ., N}) such that (8-4) holds for 0*. Let
A = {Os : ie J}~. Since 0* is not in 87, 0* is not an extreme point (Lemma 8.4) and | J| > 1. Let
8' E Of n Be(co(A)) be such that A(0*, 0') < 6. Then 8' = Ci, y(i)8i for some y E R"~ n 7"N
such that y(i) = 0 for i ( J. Since 8* = C (i8fosmeeR N sctht(i>0
for all i EJ and P(i) = 0 for i ( J, 0* + b(0* 8') E co(A) for sufficiently small b > 0. Let
0"1 = 0* + b(0* 8'), where b > 0 is sufficiently small so that 0"1 E co(A). Then the line segment
connecting 8' and 0"1 lies in co(A). Moreover, 0* lies in this line segment. Since 8* is a maximum
of R(-~, o) over 8 and R(-,, fr) is continuously differentiable on an open set 8 C UC C7, the
directional derivative of R(-, fr) at the point 0*, in the direction of 8', must be zero, i.e.,
DR;,, (0*) (0'- 0*) = 0.
By the Taylor theorem,
|R0'ir) R0*ir l 2|' 0*||I sup ||D2 ;~~c81+ t)8*)1
2~aE[o, 1]
By the hypothesis of the theorem,
| | 8' 0* | |~ sup | | D2R ~(c81 1 c) 8* I 2 M sup R (0, iro)
aE[o, 1] aEe
Thus supose R(H, iro,) supose, R(H 70~) < 70,~ ) R('o~~) < 2M/2 supose R(H, iro)-
It canl be verified that since supose, R(H 0) < V(87, K,~- R)(1 + y') and V(87, K,: R) <
V(8, .[, R), then supose R(8, fro) < V(8, X. R)(1 + y). Thus fro is a y-optimal estimator.
Suppose the normed space C7 in Condition 8.1 is R"N with its usual norm. In this case, if 8 is
a compact convex set and 88 is finite, 8 is said to be a convex polytope in R".
Definition 22. Let Ve R NxN denote an odopels~rnal matrix and let I denote the ith column
vector of V. Let 6 E R"N be a vector whose ith component 6(i) > 0 for i = 1,. ., NV. We say
that A' is (6, V) -dense subset ofAC R "W if for any as A there exists an a' E ~A' such that
| (a -a')TI | < 6(i)for i= 1,...,N1.
We have the following version of Lemma 8.6.
Lemma 8.7. Suppose .F is subconvex and has the property (W), Conditions 7.1 and 8.1
hold, and 8 is a convex polytope in R"N. Suppose for each -r E f.... there exists an extension
of R(-, 7) from 8 to an open convex set 8 C U cC such that R(-, 7) is twice Fri'chet
difgerentiable on U and there exists a positive definite Ao a RWNxN such that -D2R~ (8
Ao supose R(0, 7) for all -r Me o and 8 E 8. Let V be a matrix whose ith column is the ith
eigenvector of the matrix Ao (i = 1,. ., NV). Let (4 denote the ith eigenvalue of the matrix Ao
(i = 1,. .l. N. Le 7 > 0 annd 0 < 7' < 7. Letd t IWZ hR be sh that 6(i) = \i
for i = 1, ... N. Suppose that 87 is a finite set that \nat~is7 \.: For any nonempty A E 2ae,
8; n de (co(A)) is (6, V)-dense subset of de -(co(A)). Then a (y', 87)-optimal estimator is a
y-optimal estimator
Proof Similarly to the proof of Lemma 8.6, let -ro E Me/l be such that the support of -ro is
contained in 87, xfo is a (y', 87)-optimal estimator, and 0* E 8 be such that supose R(8, ) =
R(0*,2). Then 0* E deo(co(A)) for some A E 2ae. Let 8' E 87 n deo(co(A)) be such that
| (0' 0*), | < 6(i). By the Taylor theorem, we have that for some 8" E 8 in the line segment
connecting 0* and 8'
R(0, ro)= (0*, fr) + ~(0' *)'"D2; (o(, / _
Certainly there exists a a2,. .. UN 6W Such that 8' = C By the h~ypoth~esis of th~e
theorem,
-(0' *)'D2R: p (0 l _0 0* pl _O O*TAo (0' *) sulp R(0l fr) i a suip R(0-, fr).
i= 1
Since a, (8' *): a,' < fi(i)2. Thus supose n(o:. ) -1P~ supos ro)
R(0*, ~ ~ ~ i= br)-R0,fo (i)2(4/2 supose R(8 ro). It can be verified that since
supose, R(8, fro) < V(87, K, R)(1 + y') and V(87, X9, R) < V(8, X9, R), then
supeoe R(8, fro) < V(8, X9, R)(1 + y). Thus fro is a y-optimal estimator. O
CHAPTER 9
THE RESTRICTED RISK BAYES PROBLEM AS A MINIMAX PROBLEM
Let v be an element of Me/l and Qo E [0, +oo]. In this section, we show that the problem
of finding a restricted risk Bayes solution relative to (v, Qo) is equivalent to a certain minimax
problem. The main result of this chapter, Theorem 9.1i, appears as a conjecture in [16].
We impose the following additional conditions in order to derive the results of this chapter:
Condition 9.1. There exists an essentially unique Bayes solution relative to v.
Let Qo = supoe, R(8, iv), where is denotes the (essentially unique) Bayes solution relative
to v.
Condition 9.2. V(8, .[. R) < Co < +oo.
Let RP(0, 2) = p r(v, 2) + (1 p)R(0, 2) and rP(-r, 2) = fe RP(0, 2)dr, where 0 < p < 1.
Let K(i, p) = supose RP(0, 2) and let K(p) = infe, g K(i, p). Since supose R(8, ~) > W, ~)
we have that
K(i, pi) > K(i, p2) 1f 2iIp. (9-1)
Lemma 9.1. Suppose Conditions 9.1 and 9.2 hold. Then K is concave, decreat~sing, and
continuous on [ 0, 1].
Proof See Appendix B. O
Let G(A) = supose R(8, i) rv ~).
Lemma 9.2. Let 0 < pi < p2 < 1. Suppose that it is a minimax estimator for (8, ./, RP1) and
2 is a minimal eStimatOTJOT (0, X P). Th8#
3) supose R8, ~1) < supose 2(8 *~
Proof See Appendix C. O
If 0 < p < 1, there is a strong relation between the estimation problems (8, K, R) and
(8, K, RP), which stems from the definition of RP.
Lemma 9.3. If fo is a Bayes solution relative to -r Me Zl in the estimation problem (8, X, RP),
then io is a Bayes solution relative to pv + (1 p)-r in the estimation problem (8, X, R).
Proof The lemma is an immediate consequence of the fact that for any is E rP(-r, 2)=
pr~~i)+ ( ~r~,2 = ~py+ ( ~r,).O
Remark 9.1. By Corollary if Conditions 4.1 and 4.2 hold for (8, K, RP), there exists a
minimax estimator in the estimation problem (8, X, RP). If in addition, 8 is compact and
Condition 4.3 holds, there exists a least favorable a priori distribution -ro E Me/l, and any
minimax estimator is a Bayes solution relative to -ro. A minimax estimator is an essentially unique
minimax estimator and is admissible if Condition 4.4 holds as well. In fact, it can be shown that
if 0 < p < 1, each one of Conditions 4.1-4.4 holds for (8 K RP) if it holds for (8 K R).
Remark 9.2. Certainly we can use the results of Section 7 for the estimation problem
(8, K, RP), where 0 < p < 1, provided that Conditions 7.1-7.2 hold for the estimation problem
(8, K, RP). In fact, if 8 is compact, it is sufficient that Condition 7.1-7.2 would hold for the
estimation problem (8, K, R). Indeed, suppose 8 is compact and Conditions 7.1-7.2 hold for
the estimation problem (8, X, R). It follows easily from Lemma 9.3 that for any -r Me Zl, there
exists an essentially unique Bayes solution relative to -r in the estimation problem (8, X, RP),
whence Condition 7.1 holds. Let if denote the (essentially unique) Bayes solution relative to -r
in the estimation problem (8, X, RP). Let {n)g, be a sequence in Me/l that converges weakly
to To E Me/l. Let -rf = pv + (1 p)-r for all i > 0. Then {-rf} converges weakly to -rd. Since
R(0, ~If) = R(0, 4) for all i > 0, R(0, ~If) converges to R(0, ~If) uniformly on the compact set
8. Since the convergence is uniform, r (v, if ) converges to r (v, ifo). Thus RP/C (0 f ) converges
to RP (0, ~If) uniformly on compact subsets of 8 and Condition 7.2 holds.
Lemma 9.4. Suppose X is subconvex and has the property (W*) and Conditions 9.1-9.2 hold.
Let Q(p) = -yp + P, where y < G(A,) and Pe R Suppose Q(p) > K(p) for all p e [0, 1].
Then if y > G(A) whenever & is a minimax estimator for (8, K, R), there exists i' E X such
that K (i', p) < Q (p) for all 0 < p < 1.
Proof Suppose y > G(A) whenever & is a minimax estimator for (8, X, R). Certainly Q is
continuous on [0, 1] and by Lemma 9.1, K is also continuous on [0, 1]. Thus there exists e > 0
such Q(p) K(p) > e for all p e [0, 1]. It is easy to verify that {K(i, -) : S E X(Co, R)} is an
equicontinuous family on [0, 1]. Thus there exists n > 1 such that
|K(p) K(p')| < e/2 and
|K~, ) -K~, p)|< e/2 whenever |p p'| < 1/n and is E (Co, R). (9-2)
Let pi = i/n and ~i be a minimax estimator for (8, X, R^i) for i = 0, 1, n. Note
that in = is and ~i exists for all 0 < i < n 1 by Theorem 4.1 and Remark 9.1. Since
io is a minimax estimator for (8, X, R) and in = is, G(in) > y > G(fo). Thus by
Lemma 9.2 part 1), there exists 1 < m < n such that G(im) > 7 > G(im-1). By Lemma
9.2 part 3), im-1 E X(Co, R). Let rl be such that rlG(im) + (1 rl)G(im-1) = 7. Let
K,7(p = q~im, ) + 1 -)K(:im-,_l p). Since K(i, p) = G()po + sulpose R(0, i), then
K,7l(p) = -7p + 17 supose R(H:0L, i) + (1 17) supose R(H, im-1)_l. It can be verified that
K(pm) < K,7(pm) = 17K(pm) + (1 rl)K(im-1, pm).) (9-3)
Using (9-2),
qK~p) +(1 q)Kim-, p) < K~p) +(1 q)Kpm-) +e/2 K~m) e.(9-4)
By (9-3) and (9-4), |K(pm,) K,,(p,)| < e. It follows K,,(pm) < 0(pm,). Since the line K,,(p)
is parallel to Q(p), we must have that K,7(p) < Q(p) for all 0 < p < 1. Since K` is subconvex,
there exists i' E X such that supose R(0, i') < rl supose R(8, im) + (1 rl) supose R(8, im-1)
and r (v, i') < l r (v, im) + (1 -q) r(v?, im-,)l. Therefore, K (.i', p) < K~, (p) < Q (p) for all
0 < <1. O
Theorem 9.1. Suppose X is subconvex and has the property (W*), Conditions 9.1-9.2 hold,
and V(O. .F. R) < Co < Co. Then io E X is a restricted risk Bayes solution relative to (v, Co)
if and only if there exists 0 < po < 1 such that supose RPo (8, f o) = inf e,x supose RPo (0, 2) and
supose R(8, fo) = Co.
Proof Suppose there exists 0 < po < 1 and io E X such that suposeRP(8 ro 9)
inf~Ex supose RPe(0, 2) and supose R(8, fo) = Co. Suppose that io is not a restricted risk Bayes
solution1,+ ,, re at v to (v C o) T hen, there ,:,,, exists, ` an e t m t r i' such that su pose R (,E'
Co and r (v, i') < r (v, fo). This implies that supose RP0o(0, i') < supose RP0o(8, fo), which is a
contraiction.IUI HeceICC 1o is a re;strictdU 15 risk C; Bayes 1I souionrlaiZLve to (v, C/o).
Suppose that io is a restricted risk Bayes solution relative to (v, Co). LCertainly supoeO IL(0,O 1o
Co.Supoe upseR(,i) C. etQr) pr(v, fo) + (1 p)~Co. Then Q(p) > K(io, p)
for all 0 < p < 1 and Q(1) = K(io, 1). Since Co < Co, So is not a Bayes solution and
K((io, 1) > K(1). Thus Q(p) > Krp for Iv all 0 < pv < 1. Let = Co r(v, fo). Then y < G(is)
and Q(p) = -yp + Co. Let i* be a minimax estimator for (0. .F. R). Since io is a restricted risk
Bays sluionreltie t (v C) ad C >V(8, K, R), r(v, fo) < r(v, i*). Thus y > G(i*).
By Lemma 9.4, there exists i' E X such that K(i', p) < Q(p) for all 0 < p < 1. It follows
thatZ supeoe R(0\, LI ') < o and r(v, i') < r(v, fo). This contradicts the fact that io is a restricted
risk, Bayes,..:, solution reatv to, (v o) Thus~ suoeR(,,) o Suppose there exists no
0 < p < 1 such that supose RP(8, fo) = inf~Ec supose RP(0, 2). Then K(io, p) > K(p) for
all 0 < p < 1. Since supose R(8, fo) = Co, K(io, p) = Q(p) and by Lemma 9.4, there exists
i' such that supose R(8, i') < Co and r(v, i') < r(v, fo), which contradicts the fact that io is
a restricted risk Bayes solution relative to (v, Co). Therefore,, there exists 0 < p~o< 1 suc thatl IIL
supose RPe(0, fo) = infc,x supose RPe(0, 2). Finally, since V(8, K, R) < Co < Co. O
Under the hypothesis of Theorem 9.1i, finding a restricted risk Bayes solution relative to
/, (v:,.:,,,, Co) s euivlentto indng minimax, esimto for,+:,+, the~1, estmaio prole (8, C, R o
some 0 < p < 1.
Lemma 9.5. Suppose X is subconvex and has the property (W*) and Conditions 9.1-9.2
hold. Let {p,}g, be a sequence in [0, 1] that converges to po E (0, 1). Let ~i be a minimax
estimator for (8, K RPi) for i = 0, 1, .. Then if V (8, K R) < supose R(8, f o) < Co,
supeoe R(8, fi) supose R(8, fo).
Proof Let M~ = supose R(8, fo) and suppose V(8, K, R) < M~ < Co. Fix e > 0 such that
M~ e > V(8, K, R) and M~ + e < Co. By Theorem 4.2, there exist restricted risk Bayes
solutions relative to (v, M~ e) and (v, M~ + e). Therefore, Theorem 9.1 yields p' and p" in (0, 1)
such that supose R(0, *, ) = M~ e and supose R(0, x*,,) = M~ + e, where i*r is a minimax
estimator for (8, X, RP) for all p E (0, 1). By Lemma 9.2 part 3), p' < po < p". In fact, since
supeoe R(0, *,) < supose R(8, fo) < supose R(0, *,,), we must have p' < po < p". Thus
p' < pi < p" for i sufficiently large. Lemma 9.2 part 3) furnishes that | supose R(8, ii) M|1 < e
for i sufficiently large. It follows that supose R(8, ii) supose R(8, fo) O
Remark 9.3. If X is subconvex and has the property (W*), Conditions 9.1-9.2 hold, and
V(8, L, R) < Co < Co, a restricted risk Bayes solution relative to (v, Co) can be solved
in the following manner: Find po E (0, 1) such that supose R(8, fo) = Co, where io is a
minimax estimator for (8, X, RPO). The estimator io is a restricted risk Bayes solution relative
to (v Co). If a minimax estimator for (8, K RP) can be found and the supremum of its risk
can be calculated for all p e (0, 1), Theorem 9.1 and Lemma 9.2 part 3) imply that we can
find a sequence {pi } that converges to po. By Lemma 9.2 part 3), such a sequence can be found
easily using the bisection method. 1Moreover Lemma 9.5 implies that if {ps} is a sequence that
converges to po and ~i is a minimax estimator for (8, K, RPi), then supose R(8, in) is close
to Co whenever p, is sufficiently close to po. Thus for a sufficiently large, in can be used to
approximate a restricted risk Bayes solution relative to (v, Co) with any desired accuracy. Thus
we have a practical way to derive a restricted risk Bayes solution relative to (v, Co).
CHAPTER 10
ESTIMATION WITH A RESTRICTION ON THE OBSERVATIONS THAT CAN BE USED
As before suppose we are given a probability space (R, -T2, Q), where R = Y x X and
-T& = FTy x FTx, and random elements Y : R H (f FTy) and X : R H (X, FTx), which are
jointly distributed and their distribution Q is an element in a class {Po : 0 E 8} of measures
on (R, -T2). So far we considered the case that the experimenter observes an observation
ye E the value taken by Y, and based on this observation estimates the parameter x, which
is the true value of X; little was assumed on the structure of the spaces y and X. Let S, and
S, denote arbitrary sets. Suppose for each A E So there is associated a measurable space
(3x FTx ) and for each A E S, there is associated a measurable space (y x, Fy,). In addition,
suppose X = n,, s, y >es,~ Yx, E = nes. Ex,, and Fy = ns,~ FT>. Let
Y, : R (P x, FTy ) and Xx : R H (3x FTxx) be the random elements defined as follows: For
each we E Y (w) and X ( ) are the projections of Y(w) and X(w) into the spaces yi and 3x ,
respectively. Let yx and xx denote the true values of Yx and X respectively. Given a subset S
of S,, let Tis = nxas yi and FTys = n,,s -Ty,. Let Ys : OR (Ps, -Tys) denote the random
element defined as follows: For each Lce E Ys(Lo) is the projection of Y(Lc) into the space Ps.~
Let y7s denote the true value of Ys.
In many applications there are restrictions on the components of y that may be used in order
to estimate components of x.
Definition 23. Given the sets S, and S, defined above and a mapping & : S, 2sy, the triplet
(S,, S,, h) is called an estimation space. An estimator for x \subin t to (S,, S,, h) is a collection
{ix : A E S, }, where &x is an estimator for xx based on Yh( ).
Thus an estimation space completely specifies which components of the parameter x are
to be estimated and which components of the observation y can be used. Suppose (S,, S,, h) is
an estimation space and for each A E So there is associated a space D the space of possible
estimates for x a space Xx, the space of available estimators for xx based on Yh( ), and a risk
function Rx : 8 xA Rxi U {+oo}. In what follows we assume Rx is nonnegative for each
A E S,. Then in fact there are given a collection of estimation problems {(8, X9x, R ) : A E S,}.
Fix A E S, and S C S,. By an estimator &x of xx based on ys, we mean a mapping from Pis
into -rD, such that ys H x(D |ys) is FTys-measurable for all DI ae BD). Let D = ness Dr
and X = n,,s. X Then D and X are the space of possible estimates for x and the space of
available estimators for x, respectively.
If, in addition, there is given a (nonnegative) risk function R : 8 x .9 R WU {+oo}, the
problem is similar to the problem treated so far. Given & : S, 2sy, for each A E S,, let PoX
denote the joint distribution of Yh,3 and X when 8 is the true state of nature. If S is a subset
of S,, we let Po~s denote the marginal distribution of Ys, when H is the true state of nature. If
Ae S we let Pux^ denote the marginal distribution of X when O is the true state of nature.
Given a class A of estimators for x and A E S,, we use Ax to denote {ix : f E A}.
It is left to specify a risk function for (8, X9). One possibility is R(0, 2) = sup,,ss R (0, ~x).
If for each A E S,, there exists a minimax estimator for x which is denoted by if then
i* = { 1 : A E S,} is a minimax estimator for x. Indeed, it is easy to see that
inf sup R(0, 2) inf sup sup R (0, & ). (10-1)
Since for each &x E K supeoe R (0, & ) > supose Rx N, 1r),
inf~ sup sup R (0, & ) > sup sup R (0, if ) sup R(0, i*). (10-2)
By (10-1) and (10-2), i* is a minimax estimator for x. Thus with the above risk function,
minimax estimation for x subject to (S,, S,, h) is completely determined by minimax estimation
for xx (A E S,).
Let us consider another possibility for the risk function. Suppose S, C N. Let w : S,
[0 + 00). We can define the risk function
Without loss of generality, we may assume that w(A) > 0 for each A E S, since if w(A) = 0 for a
certain X, we can remove that A from S, without affecting the risk function R.
Theorem 10.1. Suppose there are given a collection of estimation problems {(8, X R ) : X E
S, }, where S, C N. Let K = H Ass. K and let R(0, 2) = C,,es R (0, & ) w(A), w (A) > 0
for each A E S,. Then
1. If Conditions 4.1 and 4.2 hold for (8, K R ) for each A E S,, then Conditions 4.1 and
4.2 hold for (8, K, R).
2. Suppose S, is finite. Then if Conditions 4.3 and 4.4 hold for (8, K R ) for each A E Sz,
then Conditions 4.3 and 4.4 hold for (8, K, R).
Proof 1) Fix i', i" E .9 and 0 < a~ < 1. Then for each A E S,, there exists if; aKX such
that R (-, x)
R(-, i*) < a~R (-, ') + (1 a~)R (-, i"), whence X is subconvex. Thus Condition 4. 1 holds for
(8, X, R).
For each A E S, let 9* be a compact metrizab~le space and RIafnto fo
into [0, +oo] such that (K R ) ~ ( X*, Rr;), Rr (0, -) is lower semicontinuous on X* for each
Se 8 and Rr;(-, a) E m(B(8)) for each as E *. Let X* = nes. X* be equipped with the
product topology. Let R*(0, 2) = C,,es Rr;(0, & )w(A). Then by Tychonoff's Theorem, 9 '
is compact. In addition, X* is metrizable since it is a countable product of metrizable spaces.
Since for each A E S, andu v e 2 RIL(0, & ) is lower semicontinuous, for each finite subset
S of S,, CExs Rr;(0, & )w(A) is lower semicontinuous. Since the pointwise supremum of
any collection of lower semicontinuous function is a lower semicontinuous function, we have that
R*(0, -) is lower semicontinuous on X* for each 8 E 8. Since the pointwise limit of a sequence
of measurable functions is a measurable function, it is easy to show that R*(-,~x E ) m(B(8)).
It is also rather straightforward to show that (X9, R) ~ (X*, R*). Thus Condition 4.2 holds for
(8, X, R).
2) The proof is rather straightforward and is omitted. O
In the case of countable S,, Theorem 10.1 implies that under rather weak conditions on
(8, K R ) (A E S,), there exists a minimax estimator for (8, X, R) and a restricted risk
Bayes solution relative to (-r, Qo) for each -r Me Zl and Qo > V(8, K, R). The theorem is
especially useful for the case of finite S, since it specifies that if certain conditions holds for each
one of the estimation problems (8, K R ) (A E S,), they hold for (8, .[, R) and hence the
results of the previous chapters are valid for (8, .[. R).
Example 10.1. Suppose the experimenter observes a sequence {y, } of observations, where
y, E R">Y, and there are given a sequence {x, } of parameters, where x, E RWN. In ;///1 case,
Ai = RWN and it is convenient to take FTy, to be B(Ai). Suppose there is given an estimation
space (S,, S,, h), where S, = N, S, C N, and & : S, 2s. The observation y is \imphl, the
sequence {y, }. For example, if S, = N and & : n E S, H {0,. ., n}, we have the so-called
casual filtering problem (assuming the index n is a time index). If h : n e S, H {0,. ., a 1},
we have the so-called one-step prediction problem. If at is a positive integer S, = {0, 1,. ., at },
and & : n E S, H {0,. ., n}, we also have a certain casual filtering problem. Also,
estimation of a continuous time process can be entered to dIri\ formulation. Put S, = [0, +oo),
S, C [ 0, +oo) and, for example, let & : a s [ 0, +oo) [ 0, a]. In ://i\ case the observation y is a
function (or a signal).
Example 10.2. Consider the following discrete-time linear stochastic system in state-space form:
Xn I = Fox, + In~,,, a > 0,
yn = He x, +v, ,
where x, E RWN" (n > 0) is the state vector y, E RWN is the system output, v, E R"~
is the measurement noise, In~,, E RWN" is the model noise, and H, and F, are matrices in
RWNXN and RWNXN, respectively. Suppose there is given an estimation space (S,, S,, h),
where S, C N, S, = N, and for each A E S,, h(A) is a finite subset of S,. This example
is certainly a special case of the previous example. Note that the filic ,rio. prediction, and
\rl ur llrings problems are all special cases of dIri\ example. In the most general case, there are
stochastic and/or deterministic uncertainties in xo, O',1, {va }, {H, }, and {F, }. For example,
in some problems the initial state vector xo can be modeled as a deterministic and unknown
vector which is known to belong to a set 31o C RAN. This is a deterministic uncertainty. Of
course, in other problems xo may be modeled as a random vector whose distribution is known
to belong to a certain class of distributions (e.g., the Gaussian distribution with zero mean and
some restriction on the covariance). This is an example of a stochastic uncertainty. The system
noise sequence {tr~,, }, for example, can be modeled as a random process whose distribution
is known to belong to a certain set (e.g., a subset of the set of joint Gaussian distributions of
a sequence of independent random vectors), but in some problems can be better modeled as
a deterministic sequence that is known to belong to a certain set (e.g., a subset of the set of
bounded sequences). The relations between the quantities xo, O',1, }, {va }, {H, }, and {F, } are
also important. For example, suppose xo, {tr~,, }, and {v, } are modeled as random with stochastic
uncertainties, then the dependence between l;
also be modeled. The most convenient way to spec~if\ 8, in ;//i\ general case, is it to assume
that (xo, ( I,,, }, (vn}, {H,}, {F,}) is a value of a random element whose joint distribution
belong to a class 8 of measures defined on the appropriate product space. There is no problem
with dIri\ formulation since even if one of l;
regarded as random with distribution that assigns probability 1 to a certain value in No0. Our
basic assumption regarding 8 was that 8 is a metric space. This is true if 8 is equipped with
the topology of weak convergence since xo, I,',,- vn, H,, and F, (n = 0, 1, ..) are defined on
separable metric spaces. Thus the results of previous chapters and dIri\ chapter may be used.
Note that in dIri\ case an element of M/e is a probability measure defined on a class of probability
measures. Luckily, in many important cases, we can use instead of the set 8 defined above a
simpler equivalence class, as illustrated in the following example.
Example 10.3. Suppose that in the previous example { H,} and { F,} are known sequences,
the sequences {v, } and {tr~,, } are uncorrelated, v, and vm are uncorrelated for n / m, and
In~,, and I,. are uncorrelated for n / m. In addition, suppose the initial state xo and v, (Ir,, )
are uncorrelated for all n > 0. Suppose also that the distributions of In~,, (v,) and I,. (vm)
are identical for each n and m and have zero mean. As in the previous example, suppose h(A)
is a finite subset of S, for each A E S,. Suppose xo is the value of a random vector whose
disvtr-ibtion belongs to a set 81 o~f measures. on (RA, B(RN.)), vu, is thep value of a randonm
vector whose distribution belongs to a set 82 Of meaSuTes On (WN aINy )) (n = 0, 1, ..),
and In~,, is the value of a random vector whose distribution belongs to a set 83 Of meaSuTes On
(RAN, B(RWN,)) (n = 0, 1, ..). Let 8 = 81 x 82 x 03, where ife Oi 4for i = 1, 2, 3,
O1x 02 x 3 is the product measure. Then 8 can be regarded as the space of state of nature
instead of the space used in the previous example. Suppose R(0, 2) = C,,es R (0, & )w(A),
R (0, & ) = fygX) xxh SDX L (Zx d )di:(\$ ?i(d |,3dr (y, x), and L (xl d) = | V(x d)|2
where Vx RN x N"XN. If~ in addition, we assume that each measure in 8 is Gaussian, the space
of states of nature can be further \imlrrllfie d. Indeed, for each 8 = Or x 02 x 83 E O, Ele i 8e
and As (0) denote the mean vector and autocorrelation matrix of Os, respectively, for i = 1, 2, 3.
For examley6, if 0 = Or x 02 X ~83 thena 11 (0) = fx ,.."'i rand At (0) = fx~o xox dO LePt
oi = {((Di(e), Ag(e)) : e 8 } for i = 1, 2, 3. Let C7 = RWN x ga, C2 IN x N, and
C73 INze xNzc Le 1~ 2 C<* ~ C~3 eranly i S a vector space with coordinate-wise
addition and multiplication by a scalar (i = 1, 2, 3). Given as WE let | |a| |4e = |a(1) | + | |a(2)1 | |2
Similarly, C7 is a vector space with coordinate-wise addition and multiplication by a scalau:
Givenz as E let ||a||v = C:= ||a(j)||4-~. Let 8 = 81 x 82 x O3. Th8# O is a subset of W.
The set 8 equipped with the norm |I | | | is clearly a metric space. Then it is more convenient to
regard 8, which is a subset of finite dimensional normed space, as the space of states of nature
instead of 8 provided that P is a Gaussian family. Note that if the class of estimators is restricted
to affine estimators, since the loss function Li is quadratic for each i E S,, then II idust~, loss of
generality, we may assume that each measure with bounded covariance in 8 is Gaussian since it
can verified that the risk function depends only on the mean vector and autocorrelation matrix of
dur \le probability measures.
In the rest of this work, we apply the general theory of previous chapters to the state
estimation problem of Example 10.3.
CHAPTER 11
THE STATE ESTIMATION PROBLEM
In this chapter we consider in more detail the state estimation problem of Example 10.3.
For the sake of clarity, let us repeat the formulation of this problem. We consider the following
discrete-time linear stochastic system in state-space form:
X,+1 = Fox, + It',,, a > 0,
Yn = H,x, + v., (11-1)
where x, E RWN" (n > 0) is the state vector, y, E RW~y is the system output, v, E RW~y is
the measurement noise, It',, E RWN" is the model noise, and H, and F, are matrices in RWNyXN
and RWN, X respectively. We assume that there is given an estimation space (S,, S,, h),
where S, C N, S, = N, and for each A E S,, h(A) is a finite subset of S,. We assume that
{H,} and {F,} are known sequences, the sequences {v,} and {tt',,} are uncorrelated, v, and
vm are uncorrelated for n / m, and it',, and w., are uncorrelated for n / m. In addition,
the initial state xo and v, (I,',,) are uncorrelated for all n > 0. We assume xo is a Gaussian
random vector and {v,} ({tt',,}) is a sequence of identical Gaussian random vectors. We
assume that the mean and covariance of xo and the covariances of v, and I,',, are unknown; it
is only known that each one of these quantities belong to a certain set. The space of states of
nature 8, in this case, is 81 x 82 x 02, where 81 is the class of possible mean vector and
autocorrelation matrix pairs for xo, 02 iS the class of possible mean vector and autocorrelation
matrix pairs for v, (n = 1, 2, ...), and 83 iS the class of possible mean vector and autocorrelation
matrix pairs for It',, (n = 1, 2, .. .). As mentioned in Example 10.3, the set 8 is a subset of
a finite dimensional space normed space C7 (see the definition of C7 in Example 10.3). For
8 = Or x 02 x 83, where 04 E 84 for i = 1, 2, 3, we let ri(0) and As(0) denote the mean
vector and autocorrelation matrix of Os, respectively, for i = 1, 2, 3. Our main assumption
regarding the space 8 is that A2(0) > 0 for each 8 E 8. We consider the risk function
R(0, 2) = C,,es R (0, & )w(A) for this problem, where w(A) > 0 for each A E Sz,
and Vx R IN"XN (the reader is referred to previous chapters for the above notation as well as
subsequent notation). Note that P is a Gaussian family of distributions.
Suppose V(8, ~D, R) < +oo and V(8, L, R) < +oo. When 8 is bounded and S, is finite, it
is obvious that V(8, ~D, R) < +oo and V(8, L, R) < +oo. In the case that S, is finite and 8 is
not compact or the case that S, is not finite, it is not necessarily the case that V(8, ~D, R) < +oo
and V(8, L, R) < +oo. It is outside the scope of this work to derive the exact conditions for
V(8, ~D, R) < +oo and V(8, L, R) < +oo since we are mainly interested in the case that S,
is finite and 8 is compact. It is sufficient to mention that these conditions will depend on system
theoretic notions such as constructibility, stabilizability, and detectability. The interested reader
is referred to [24], where such conditions are derived for the special case of uncertainties in the
initial state.
Let us show that Conditions 4.1, 4.2, and 4.4 hold for the estimation problems (8, ~D R )
and (8, L R ) for each Ae E and that Condition 4.3 holds as well if 8 is compact. Fix
Ae E Our first step is to show that Conditions 5.1-5.7 hold. Since the loss function Lx is
quadratic, it is only left to show that Conditions 5.2 and 5.4 hold (Chapter 6). Consider the
conditional distribution of Yh( ) given X = x, when 8 is the true state of nature. Let Po^(-|IX = x)
denote this conditional distribution, which is certainly Gaussian. Moreover, since A2(0) > 0 for
e~tac r a 8, thefamly,~x~ { o ( X= ) : 0 E 8, X EX } is dominated by the Lebes gue-B orel
distribution of Yh( ) given Xx = x when 8 is the true state of nature. Let Ae B (i( 3)) be such
that p l(A) = 0. Then Po^(A|X = x) = 0 for each x E X. It can be verified that this implies that
Po^(A|X = x ) = for each xx EX 3. Thus~ then family {P (-X~ =( x ) : 0 E 8, XX E3X }
is dominated by pr and Condition 5.2 holds. Put Z: = [Y~, X~ ]. L~et {0,}) be a sequence in 8
that converges to 8o E 8. Then rli(0,) i qi(80) and As(0,) A s(00), for i = 1, 2, 3. It can
be verified that this implies that Eo, (Z) Eeo (Z) and Eo, (ZZT) Eeo (ZZT). It follows
that Po\, converges weakly to Po>,. Since the sequence {0,} is arbitrary, {Po\}) converges weakly
to PoX whenever {0,t} converges to 8o, {O,z} E 8, and 8o E 8. It follows that 8 Po^(A) is
B(8)-measurable for each Ae B (Pihz)) x B(3x ) [40].Thus Condition 5.4 holds. It is also
clear that 8 H Ea(Z) and 8 H Ea(ZZT) are continuous on 8. It now follows from the results
of Chapter 6 that Conditions 4.1, 4.2, and 4.4 hold for (8, ~D R ) and (8, L R ) for each
A E S,., and Condition 4.3 holds as well if 8 is compact. By Theorem 10.1i, whether (8, ~D, R)
or (8, L, R) is considered, there exists a minimax estimator, and there exists a restricted risk
Bayes solution relative to (-r, Co) for each -r Me Zl and Co > V(8, ~D, R). Suppose in addition
that S,. is finite and 8 is compact. Then we have the following results for the estimation problem
(8, ~D, R): (8, ~D, R) is strictly determined, there exists a least favorable a priori distribution
I-o E Me/l and a conditional mean estimator relative to -ro is an essentially unique admissible
minimax estimator. Moreover, the class of conditional mean estimators is essentially complete.
Note that, in general, a conditional mean estimator relative to -r is not a LMMSE estimator since
-r may not assign mass 1 to a single point in 8. Similarly, we have the following results for the
estimation problem (8, L, R): (8, L, R) is strictly determined, there exists a least favorable
a priori distribution -ro E Me/l and a LMMSE estimator relative to -ro is an essentially unique
admissible minimax estimator. Consider the filtering problem. Then a LMMSE estimator relative
to -r is not necessarily a KF if -r does not assign mass 1 to a single point in 8. There is a special
and important case in which a LMMSE estimator with respect to -r M y/1 is a KF. We will treat
this case in the sequel.
In the following chapters we consider two special cases of the above problem. The first case
is the case of stochastic uncertainties in the initial state, model noise, and observation noise with
L as the class of available estimators. The second case is the case of deterministic uncertainties in
the initial state with ~D as the class of available estimators. In both cases we will assume that S,. is
finite and 8 is compact. Note that by Remark 4.1, the case that 8 is bounded, but not necessarily
compact is also covered. These two cases are important on their own merit, and they will also be
used to illustrate some of the general results of previous chapters. For example, the first case will
be used to illustrate the method proposed in Chapter 9 to derive a restricted risk Bayes solution
(Remark 9.3) and the second case will be used to illustrate the method proposed in Chapter 8 to
derive an approximation for minimax estimators.
CHAPTER 12
AFFINE STATE ESTIMATION BASED ON QUADRATIC LOSS FUNCTIONS
In this chapter we consider a special case of the state estimation problem of Chapter 11. We
consider the case of state estimation with stochastic uncertainties in the initial state, model noise,
and observation noise with the class of available estimators being the class of affine estimators.
Throughout this chapter, we assume 8 is compact. Thus V(8, L, R) < +oo. In this chapter,
we assume that the estimation space (S,, S,, h) is such that S, = {0, 1,..., nt}, S, = N, and
h(k) = {0, 1,. ., Gk } for each k E S,, where at and nk, for k = 0, 1,. ., nt, are nonnegative
integers. We assume that w(k) = 1 for each k E S,, i.e., R(0, 2) = C"' Rk(8, k~). Given
Co > V(8, L, R) and ve 8 our goal is to find a restricted risk Bayes solution relative to
(v, Co).
12.1 Finding a Restricted Risk Bayes Solution
First, we want to derive a closed form expression for R(0, i). Let F,, = FF _1 Fj (i >
j) and Fi~i = Fi. We will use the convention Fi~i 1 = I and Fi,y = 0 if j > i + 1. Let
0,, = H' Fo To, FoH2T n T-1,0Hlf]".Let ( = [0 HJ 1 FH2T nT-1,1HTI] =
[0 1 0, H2 F2HT'""FT1, 2HI 000H3 -,H],.. 00 I
Let I, = [1, 1, 1"]. In addition, Let I( = 0 for 1 > n. Let y" = [yoT yT yT]. Note that
Yh(k) = Unr It is easy to verify that
y"n = OnXO + E n I- + u, (12-1)
Zk Fk-,0,z 0 P -1it -1, (12-2)
i=1
where w" = [wo" wT -- (] and u = [VTo VT -- Let k, > 0 and n > 0. By (12-1) and
(12-2),
nVk-1
i=0
For each k E S, and Zk, = Ayo" + be E k, let As, = A and be, = b. Let Wk, = VkT ~, for
k = 1, .., us. By our assumptions regarding the noises and the initial state,
Rk8,k)= tr(VW(As 0,, A-1,_lo)A(0)(As 0,n FI-i,o) VkT>
nkVk-1]
i=0
+2~ 71/ ~b, WkI(As 0,~ F-1,o)917(e)+ |Kb,|2.12 (12-3)
Let r1(0) = [r1(0)T ra2) rl3(0) ]T and let A(0) be the block diagonal matrix with Az(0),
A2(0), and A3(0) in its diagonal blocks. Let Ay,4(0) = As(0) ri(0)ri(0)T for i = 1, 2, 3.
We assume that some a priori information regarding the true state of nature 8 E 8 is
available. The case that no a priori information is available is an important special case. The a
priori information is given in the form of a nominal v E 8. For simplicity, we assume rll(v) = 0.
There is no loss of generality in this assumption since if rll(v) / 0, we can translate the states
and observations and bring the problem to this form. Below, we summarize the assumptions
taken so far together with some new assumptions.
Assumption 12.1. ve 8 and rl(v) = 0.
Assumption 12.2. 81 is a compact and convex subset of~ and (-rl, A) E 81 whenever
(rl, A) E 81.
Assumption 12.3. 82 is a compact and convex subset of % and for all 8 E 8, rl2 (0) = 0 and
A2 (0) > 0.
Assumption 12.4. 83 is a compact and convex subset of % and r/3(0) = 0 for all 8 E 8.
The assumptions that 84 is compact and convex for i = 1, 2, 3 can be somewhat relaxed,
but, for the sake of clarity, it is advantageous to make these assumptions. The assumption that
(-rl, A) E 81 whenever (rl, A) E 01 is rather weak and is satisfied in many important cases. This
assumption is clearly satisfied in the case that 81 = {((9, Al + rlrl) : rl E FI, Al E 02}, where
01 = {9l : (rl, A) E 01} and 02 = {A rlrl : (rl, A) E 01}, and El is symmetric around the
point 0, i.e., rl E FI if and only if -rl E Fr. It is assumed that Assumptions 12.1- 12.4 hold in the
sequel. Note that 8 is then a compact convex subset of C?.
By (12-3), Rk k~) is both convex and concave on 8 for all Ak E k~, i.e., for any
0 < a~ < 1 and 8', 0"1 E 8,
In addition, Rk ', k) is COntinuous on 8 for all ik E k~. It follows that R(-, 2) is both convex
and concave on 8 and continuous on 8 for all i E L. Let r(-r, 2) = fe R(0, 2)dvr and let
Tk-r To ) = So k 8, k~)dTr for k = 0, 1,. ., us. Let Z be the class of all finite subsets of 8. Let
Me/l denote the space of distributions in Me/l with finite support. Let -r Me Zl. Then there exists
Z E Z and 81,...,0z2 E 8 such that Z= {01,...,0z }) and r(Z) = 1. In this case,1let r(i)
denote the mass that -r assigns to the point Of f~or i = 1,...,1 |Z. Finally, let H o r = C1 Osir(i).
Since 8 is convex, 8 o re 8 Since R(-, 2) is both convex and concave on 8 for all ie E ,
Lemma 12.1. Let -r* E Me/l. Then there exists 0* E 8 such that r(-r*, 2) = R(0*,i) for all
is L As a consequence, & is a Bayes solution relative to -r* if and only if & is a Bayes solution
relative to 0*. Similarly, ri(-r*, is) = Ri (0*, ~i) for all ~i E 4 and ~i a Bayes solution relative to
-r* if and only if is is a Bayes solution relative to 0* (i E S,).
Proof Fix is E Since 8 is compact, it is separable. Thus the space of distributions with finite
support, Me~l, is dense in Me/l in the sense of weak convergence [27, Appendix 3]. Let {74}) be a
sequence of distributions in Me/l that converges weakly to -r*. Since R(-, 2) is continuous on the
compact set 8, R(-, 2) is bounded and
/ R(0,,, 2)r im (0 )do (12-4)
Let Of = 0 o nr (i = 1, 2, ..). Since 8 is compact, there exists a 0* E 8 and a sub sequence {Of f
such that {0fm } converges to 0*. Since R(-, ) is continuous on 8,
R(0*, 2)= =limn R(Ofm ). (12-5)
Since R(-, 2) is both convex and concave on 8 and -r has finite support,
/ R(, 2du R(O ) ( = 1 2,. ..).(12-6)
By (12-4)-(12-6), r(-r*, 2) = R(0*, 2). The proof follows from the arbitrariness of 2. O
It is clear from the proof of Lemma 12.1 that if -r* e Me/, is a Bayes solution relative to
-r* if and only if & is a Bayes solution relative to 0*, where 0* = 0 0 -r*.
We want to apply the results of Chapter 9 for the estimation problem (8, L, R). We have
already shown that Conditions 4.1-4.4 hold in Chapter 11. Let us show that Conditions 7.1, 7.2,
and 9. 1 hold as well. In Chapter 11 it was shown that L has the property (W*) and Condition 4.4
holds. Thus Theorem 4.2 implies that Condition 7.1 holds. Certainly the weaker Condition 9. 1
must hold as well. It is left to prove that Condition 7.2 holds for (8, L, R). In fact, it is sufficient
to prove that Condition 7.2 holds for (8, k, k~) for each k E S,. Fix k E S,. For the sake
of clarity, we prove that Condition 7.2 holds for (8, k, k~) in the case that r11(0) = 0 for all
8 E 8. It is easy to verify that this is true also in the more general case of Assumption 12.2 but
the expressions are rather cumbersome. Assuming r11(0) = 0 for all 0 E 8 and using (12-3),
Rk 8 ~k) = tr.( /Aiyk,/1(0)A Vk ) 2 tr.( /~A~iY k,2(0) VkT)
+ tr( 7kI,3(0)Vk')+ + |Kbyl |2, (12-7)
where
7k,1 n) Az Ch(0)O(, + I,,, O A2(H ,~, y 1 A3(),
nlkAk
7k,2(B k-1,0Az(0)O(~B + Fk-1,iA3)~k)
i= 1
Recall that ||2'k Ok~ = ||A. A. || I I. b~I ~., LII~ iS a norm, making k
into a normed space.
Let it,k E k~ denote the (essentially unique) Bayes solution relative to -r Me Zl. It is not
difficult to show that
Aenr = Yk,2(r)7. Y,(7) and ban~ = 0, 128
where we extend yk~i from 8 to Me/l by defining yk~i 7r) = yk,i 8 o Tr) for all Te ME Zl (i = 1, 2).
Let {-ri} be a sequence in Me/l that converges weakly to -ro E Me/l. By Lemma 12.1i, there
exists a sequence {Os} E 8 and an element 8o E 8 such that Rk(8, 74,k) k R(8, Os,k)
(i = 0, 1, ..). Certainly yk~i is COntinuous on 8 for i = 1, 2. It follows easily from (12-8)
that the mapping 8 H O,k is COntinuous on the compact set 8 and hence uniformly continuous
and bounded. Thus there exists p > 0 such that SO,k E Bp for all 8 E 8, where B, = {ik E
Ck : k1~1~ I p}. It is easy to see that RkIS isCOntinuous on the compact set 8 x B,. Thus
{Rk~(0, -) : 0 E 8} is equicontinuous on B,. Since 8 H 20,k is UnifOrmly continuous on 8, the
family {0 Rk (8 8,~) : 8' E 8} is equicontinuous on 8 and therefore Rk (8, 0s,k~) COnverges
to R, (8, Bo,k~) UnifOrmly on 8. It follows that R, (8, 74,k) COnverges to Rk(8, To,k) UnifOrmly on
8 and Condition 7.2 holds.
Let 2, denote the (essentially unique) Bayes solution relative to -r Me Zl in the estimation
problem (8, L, R). Since 8 is compact and R(-, 2,) is continuous on 8, Co < +oo and
Condition 9.2 holds if V(8, L, R) < Co. Let AP = Ap, +(lp)r. By Lemma 9.3, AP is a Bayes
solution relative to -r in the estimation problem (8, L, RP). Let rp(,, 12) = pr(v, 2) + (1 -
p)r(-r, 2). Let r"P(-) = rP(,, AP). Since Conditions 4.2-4.4, 7.1, and 7.2 hold for the estimation
problem (8, L, R), they hold for the estimation problem (8, L, RP) for 0 < p < 1 (Remarks
9.1 and 9.2). Thus using the results of the previous chapters and Lemma 12.1i, if 0 < p < 1,
there exists 8o E 8 such that r"P(8o) = supoe, r"P(0) and "0 is an admissible, essentially unique
minimax estimator. In particular, by setting p = 0, there exists a minimax estimator for (8, L, R).
Let us consider the problem of finding a minimax estimator for the estimation problem
(8, L, RP), where p E [0, 1). As mentioned earlier, if V(8, L, R) < Co < Co, the solution of
this problem for p e (0, 1) is necessary in order to find a restricted risk Bayes solution relative
to (v, Co) using the method of Remark 9.3. The case p = 0, corresponds to minimax estimation.
Note that if Co > Co, the problem is reduced to regular Bayes estimation and is is the solution.
The case Co = V(8, L, R) corresponds to minimax estimation.
Let 0o = {0 E 8 : r1(0) = 0}. Let us show that 0o is a convex and compact subset of
8. Fix 0 and 8' in 80 and 0 < a~ < 1. Let On = caO + (1 a~)0'. Since 8 is convex, On E O.
Clearly r1(8a) = 0. Thus On E 0o and 80 is convex. Let {Os} be a sequence in 0o that converges
to 8o E 8. Then by Assumptions 12.2-12.4, r1(8o) = 0. Thus 80 is closed. Since 8 is compact,
80 is compact. Let to = {ie E : by, = 0 for k = 0, 1,..., nt}.
Lemma 12.2. Let 0 < p < 1 and consider the estimation problem (8, L, RP).
Consider the following algorithm:
Step 1: Choose 01 E 80 and let i = 1.
Step 2: Findl Of a 80 sucrh that R(Of i ) = supoeaO R(8, if ).
Step~ ~ ~~~V 3: IfR(s i )=R(f, ) then stop; the distribution Os is a least favorable a priori
distribution.
Step 4: Let 80,i = cli + (1 a)04 for asE [0, 1]. Find to E [0, 1] such that r"P(0a,,i)
supoe [o, 1] TP(H0,i).
Step 5: Put 8i41 = 0o,~i, let i= i+1, and return to step 2.
Then the sequence {Os } is in Oo, it converges weakly to a least favorable a priori distribution
8o E 0o, and ifo, is a minimax estimator 1Moreover the sequence {if } is in to and RP (0, if )
converges uniformly on 8 to RP (0, fo), >
Proof We claim that for any 8 E 8, there exists 0* E 80 such that r"P(0) < rP(0*). Indeed,
fix 0 E 8. Let 8' be such that A(0') = A(0) and r1(0') = -r1(0). It can be verified that since
rl(v) = 0, r"P(0) = rP(0'). Let 0* = ( + '). Then 0* E 0o. By Lemma 7.1, ?"(0) < ?"(0*).
Thus there exists a least favorable a priori distribution 80 that is in 0o. Certainly 8o is also a
least favorable a priori distribution in the estimation problem (0o Pp). Since 0 < p < 1,
RP(0*, 2) = supose RP(0, 2) if and only if R(0*, 2) = supoe, R(0, i). Since 0o is a compact
subset of 8, we may apply Theorem 7.1 for the estimation problem (0o Pp) and derive the
above algorithm. Note that the algorithm is simplified with the help of Lemma 12.1 since we
need to consider only distributions with support of a single point, i.e., elements of 8. Since we
have considered the estimation problem (0o Pp), the sequence {04}@, is in 0o. By Theorem
7.1, the sequence converges weakly to a least favorable a priori distribution 8o E 0o. Thus "0 is
a minimax estimator. Since r1(04) = 0 and rl(v) = 0, if a Lo for all i > 0. Since Condition 7.2
holds, RP (0, ') converges uniformly on 8 to RP(0, Fo).
The main steps of the algorithm of Lemma 12.2 are Steps 2 and 4. In Step 2, we need to
solve the problem of finding a maximizer of R(-, 2) over 0o. We will address this problem in
Section 12.2. Step 4 can be done using numerical methods as described in Remark 7.1.
Suppose V(8, L, R) < Co < Co. Then by Theorem 4.2, there exists a restricted risk Bayes
solution relative to (v, Co). If a minimax estimator for (8, L, RP) and the supremum of its risk
can be calculated for all p e (0, 1), the method discussed in Remark 9.3 can be used to find a
restricted risk Bayes solution relative to (v, Co). By Lemma 12.2, we can assume that a minimax
estimator for (8, L, RP) is in Lo for all p E (0, 1). It can be verified, using (12-3), that if & e to,
there exists 0* E 80 such that R(0*, i) = supose R(0, 2). In general, given an estimator is to~
it may be very difficult to calculate suppose R(0, 2) and the complexity of this calculation may
vary significantly according to 8. Nevertheless, in Section 12.2, we address this problem and are
able to solve it for some important cases of 8.
12.2 Finding a Maximizer of the Risk
In this section, we consider the problem of finding a maximizer of R(-, 2) over 0o, where
is L o, i.e., given~ i Lo, we want to find an element 0* E 80 such that R(0*, 2) =
suppose R(0, 2). This problem is important since we encounter it in step 2 of the algorithm
of Lemma 12.2 and in the method to find a restricted risk Bayes solution, which is discussed
in Remark 9.3. We consider specific cases of the parameter set 0o. There is one immediate
case in which this problem has a simple solution. The definition of an extreme point is needed
(Definition 21). Recall that given a set A, we use dA to denote the extremal boundary of A,
which is the set of all extreme points of A. Since 0o is convex and compact and R(-, 2) is convex
and continuous on 8, by Bauer's minimum principle [41],
sup R(0, i) sup R(0, i).
8680 B6880
If 880 is finite, suppose R(8, i) = maxceaeo R(0, 2). Thus suppose R(0, i) can be easily
calculated.
Fix is L o. By (12-3), R(0, i) = f, tr(As(0t)We) f~or some 91 E SN. 942 6 YN' and
93 E Nzi. Let #4 = {As(0) : 0e 8}0 for i = 1, 2, 3. Let # = {(A(018,n(), A2(0), A3 e E O -
It follows from the definition of 8 that # = GI x #2 x 3. Thus
sup R(0, i)= sup tr I(AsW, I ) sup tr(As~i).
8680 (Ayh,AgA)E6 AgEs
Therefore, we are left with the following optimization problem:
Given NV > 0, a matr~vixr We S"nd a convex compact subse~t A o S,nr
maximize tr(AW) subject to A E A. (12-9)
Consider the important case in which A = { Ae S : ft (A) < ., fr (A) < 0 }, where
fl,. ., frv are convex (real-valued) functions such that A is compact and convex. Then (12-9) is
equivalent to the following convex optimization problem with generalized inequality constraints:
minimize tr(AW)
subject to f()<,i ,.,
A > 0.
Convex optimization problems with generalized inequality constraints can be often solved
numerically as easily as ordinary convex optimization problems [42, pp. 167]. Thus in many
important cases, (12-9) can be solved numerically. In the rest of this section, we consider several
cases in which (12-9) has an analytical solution.
Let Ao a Si" and De R WX" be a nonsingular matrix. Let Wr = (D- ) W~D-l and
el >ea 02 > N denote the eigenvalues of W. Let Be S and At >X~ A2 N be the
singular values of B. Then by a trace inequality of von Neumann [43],
tr(B t)~ <: Agg (12-10)
j= 1
We consider the following possibilities for A:
1) A = {A E S" : tr(DADT) < 1}. Fix AE A. LetX At X~> 2 XN
denote the eigenlvalues of DAD"7. By (12-10), tr(AWrI) =-- trDD 9)
tr AW) < tr DADT Q1. Since A E A, tr A) < Q1. Let A* = D-1UUT D- ), where fi is the
eigenvector of W corresponding to go. Then A* E A and tr(A*W) = gi, whence A* maximizes
tr(AW) over A.
2) A = { Ae S : | |D(A Ao)DTII | |, 1, where |I | || is the Frobenius norm.
Let At >X~ A2 N denote the singular values of D(A Ao)D By (12-10) and
Cauchy-Schw arz inequaity, tr((A Ao) < EAl {g.Tu
tr(AW) < ||D(A Ao)DT||y||W||y + tr(AoW). Since A E A, tr(AW) < ||9||,l + tr(Ao ).
Suppose A* E A and tr(A*W) = ||9|| + tr(AoW). Then
tr (A*n,~ Ao)D = 1.
Since the assignment (A|B) = tr(AB) yields an inner product on the space of real-valued
NV-by-NV matrices and ||D(A* Ao)DT||, < 1, we have by the Cauchy-Schwartz inequality
that D)(A* Ao)D7 = ~iT'hus A* maximizes tr(AWI) over A if and only if A* =
Alo + D1'I(D 1)T
3) A = {A E S" : ||D(A Ao)DT||2 < 1}. Fix A E A. Let A1>X~ A2 N
denote the singular values of D(A Ao)DT. By (12-10), tr((A Ao) ') < X1 CE= gy. Thus
tr(AW) < tr(W)||ID(A Ao)DT||2 + tr(AoW). Since A E A, tr(AW) < tr(W) + tr(AoW). Let
A* = Ao + D-I(D-1)T. Then A* E A and tr(A*W) = tr(W) + tr(AoW), whence A* maximizes
tr(AW) over A.
4) A = {eAo + (1 e)A : ||DADT||y < 1}, where 0 < e < 1. Certainly A*=
eilo + (1 e) D-1 1)T maximizes tr(AW) over A(. We canl replace | | || in the definition of A
by tr(-) or || ||2~ and have analogous results.
Let us consider an example in which we find a maximizer of tr(At Wi) over Gr.
Example 12.1. Suppose the set 31o is compact in the sense of the usual Euclidean norm and
symmetric around the point 0, i.e, xo E 310 if and only if -xo E 31o. Let MZ/xo denote the set of all
probability measures on (31o, a(31o)). Suppose it is known that the true distribution of the initial
state belongs to MZ/xo. While MZ/xo has elements that are not Gaussian, for each such element
there corresponds a Gaussian distribution with the same mean vector and autocorrelation matrix.
Since the risk functions depends only on the mean vector and autocorrelation matrix, the results
of thi\ chanpter can be used~. LePt 81 = {((9, A) : 17 = fxoxody, A = fxoxrox~r, reMxo
Since 31o is compact, MZ/xo is weakly compact. Let {-ri}@, be a sequence in MZ/xo that converges
weakly to -ro E MZ/xo. Then clearly {((qi, As) }g, converges to (rlo, Ao), where rli and As are the
mean vector and autocorrelation matrix of -ri (i = 0, 1, ..), respectively. Since MZ/xo is compact,
81 is compact. Since MZ/xo is convex in the usual sense, 81 is convex. Fix (rl, A) E 81. Let
r E MZ/xo be such that rl and A are the mean vector and autocorrelation matrix of -r, respectively.
Let -' be defined as follows: for every Ce B (3o), -r'(C) = -r(-C), where -C = {-xo : xo E C)
and is in B(31o). Ct, icirrh -' E MZ/xo, r' = -rl, and A' = A, where rl' and A' denote the
mean vector and autocorrelation matrix of 7', respectively. Thus (-rl, A) E 81. Recall that
GI = {At (0) : 0 E 8 }. We want to find a maximizer oftr(ARl) over Gr.
Suppose 3o = {xo E RWN : x"DTDxo < 1}, where D E RWN~xN is nonsingular
Fix Ae t r. Then there exi'Fs ~tse Mxo such? that AZ = fXoxoxz~cdr. Th~us tr(DADT')=
fxo tr(Dxox"D..TT)d~r < 1. Let Ae S N.1 and suppose tr(DADT) < 1 and A f 0. Then
C As < 1, where As is the eigenvalue of DADT corresponding to the ith eigenvector ui.
Certainly DADT = C Asn.LT Thus A = C Asibil', where As = Ag/(C Ay) and fig =
CjDI14 for i = 1, N. Since, ifDTZ)i = C As < 1, ~ig E Xo and -~ig E Xo
for i = 1, .. ., Nz., Let -r* be the distribution that assigns mass Ag/2 to fig and Ag/2 to -fig
(i=1, I. N). Thenr r* E MxoLs, has zero mean, and A = x xox~d~rdr*. Thus Ae E 1 It follows
that At = {A E SN. : tr(DADT) < 1}. This case was alreadyv treated in rlhi\ section.
Suppose 31o is a convex polytope that is symmetric around the point 0. Then there ex-
ist a finite number of points Xo,1, X0,2, 0 X,N E 30 Such that xo,i and -xo,i are extreme
pointsf of Zofor i: = 1, 2, ..., N. Certain~ly sup~,,l tr(A~l) = suprerxo So tr(xrox Wi~)dv <
suzeotr/xo -W'1). By Bauer's minimum principle [41], supzoexo tr/xo "W'1) = supzoeaxo tr(xox ~).
Since 31o has a finite number of extreme points, sup,,,, tr (A"
Since xoi,~ix E 01 for i = 1, ..., NV, suphE4 tr(A~l) = Inaxli~aN tr(xo,ix,~i 'Y) andU therelt
exists 1 < i < NV such that xo~ix", maximizes tr (Al) over Gr.
12.3 Connection to the Kalman Filter and E-Minimax Approach
Let igln denote an estimator for xm based on the observations II,,. Let F(0, iml,) denote the
MSE matrix of the estimator iml, when 8 is the true state of nature. Let 2,(0) and in,_-1(0) be
defined by the well known update equations of the KF:
Bo -1(e) = r11(0), (12-11)
Fo-()= A,~,(0), (12-12)
K,~(0) = Ps,_n1(0)HI[Ay,2( + H,F4,l_l(0)H ]- (12-13)
8,(0)=e>10 ,0)[ -Hi,10] (12-14)
r,(0) = [I K,(0)H,]F, (0)(8, (12-15)
in1()= F,2,(0), (12-16)
I', ,l(0) = Ay,~3(H) + Elnful0)Fu (12-17)
where 0,(0) = F(0, 2,(0)), r,,,_,(0) = r(0, l4,,_,(0)), and K,(0) denotes the Kalman gain.
Consider the filtering problem specified by the estimation space (S,, S,, h), where S,=
{0,. ., 71<}, S, = N, and h(i) = {0,. ., i} for each i E S,. Then is = (o (8), 11(8), in, (8>>
and is referred to as the KF relative to 0. Since A2(0) > 0, the existence of the KF is guaranteed
[44]. It is easy to verify that Rk (8, k (8 )) = trW1~B k8, k8 ))).
Using the results of the previous sections, there exists a 0 E 8 such that the KF relative to
8 is a restricted risk Bayes solution relative to (v, Co) for all Co > V(8, L, R). In fact, there is
another interesting property regarding the KF. Using Corollary 9.1 and Lemma 12.1i, we have
that the class of KFs relative to H E 8 is essentially complete. Thus as long as the performance
is judged solely based on the risk function, if the choice of estimators is restricted to affine
estimators, then no matter what optimality criterion is used, one may consider only the class of
KFs relative to H E 8.
Our next step is to derive more convenient expression for Rk(8, k~(0')) in the case that 8 and
8' are in 0o. Let Uk 8, 8/ k (,~ (0')). Using (12-13)-(12-17), it can be shown that
Fo 1(, ')= A (0)
', (0, H') = K, (0') A2 0 K, (0') + [I K, (0') H,] 4, (0, H') [I K, (0')H,]T
Since R, (8, k ,8 )) = tr( k k(0, 8')), we have a more efficient way to calculate the risk
than through (12-3). Recall that if H and H' are in 0o, then R(0, fe,) = C,_ tr(A4i(0) We(0)).
Our goal is to find an expression for We(0'), for i = 1, 2, 3, since it is needed in order to find a
maximizer of R(-, is,) over 0o
Let F,(8/) = F,-Fx,t(BI)H,, 1et Fi,j(0') = i(0')Ni_,(0') Fj(0') (i > j) and Fi,i(0') =
Fi(0'). We will use the convention E-_l1,4(0') =ILe rm0)=(I-Kt()r)-,m)
(0 < m < n, O < n), let Dn,m(8/) = Gr,m+1(8/)F, >Ka(0) (n > m > 0) and D,~,(0') = K,(8/)
(n > 0). Note that C+1~,m = (I K,+1H,t+1)Exs,m and Cr,-1 = Gr,mF,-1. Then
Gro(0')Az(0)Gro(0') + Dnm(0')A2(0)Dnm(0') + Grm(0')A3 8 rtm 8
m= o m= 1
r,(e, e')
Let Ti(0') = CE" >= ,~i(BI) 914zGi(0') for i = 0, 1, .., us. Then
It follows that
n=0
n=0 m=0
m=1 m=0
n=l m =l m =
In general, the calculation of W2(0') and 93 0 ) TCCJUifeS the storage of Ko(0'), .., K,t (8'),
which may be problematic for large us. This is due to the fact that Ti can be updated based on
Ti+l but not vice versa. Nevertheless, in the important case that R, is invertible for 0 < i <
us 1, the calculation of Wi, 92, and 93 may be done in such a way that there is only a need of
a fixed storage place that does not depend on us. First, let us show that since A2 0 ) iS invertible,
I K,(0')H, is invertible. By (12-13),
I K(0')H = I 0,1(0')H [nA28) HnF4,_1(0)H ]- Hn
Thus by the matrix inversion lemma, I K(,(0')H1, is invertible and
(I K,(0')H,)-1 = I + 04,_-1(0')If"A2 pl) -1H,.
Since Fi is invertible, T441 = [(I K Hi)- F- ] T4I KeH ]- F-' (F- ) H'sF- In this
case, Ti41 can be updated based on Ti and only a fixed storage place is needed.
We now discuss the connection between the restricted risk Bayes approach and the 0-
minimax approach and illustrate that the F-minimax approach can be regarded as a special case
of the restricted risk Bayes approach. The class 0 of a priori distributions in the F-minimax
approach coincides with the class 8 of the states of nature in the restricted risk Bayes approach.
By setting Qo = V(8, L, R), we have that for any -r Me Zl, a restricted risk a Bayes solution
relative to (-r, Qo) is a minimax estimator for (8, L, R). Since 8 = 0, the risk R(0, x) is, in fact,
the Bayes risk relative to a certain distribution in 0 if we adopt the F-minimax formulation. Thus
a minimax estimator for (8, L, R) is a F-minimax estimator. Therefore, the results of this work
can be used to find a F-minimax estimator. We note that if some a priori information is available,
the restricted risk Bayes approach is preferable to the F-minimax approach since it utilizes this
information. However, if no a priori information is available, the most reasonable choice for C6
seems to be V(8, L, R) and hence we are left with F-minimax estimation.
12.4 Numerical Example
To illustrate the theory of the previous sections, we consider the following simplified
problem as an example. Suppose a target is moving in the one-dimensional space. We assume
the three-state track model [45]. Let x, E RW3, where x,(1) denotes the position of the target,
x,(2) denotes the velocity of the target, and x,(3) denotes the acceleration of the target. We
also assume a radar measures the position of the target. Hence the state space model in (11-1) is
specified with
1 a A2/2
Fu = 0 1 a
00 1
Ir,, is a zero-mean random vector with nominal covariance matrix
000
G~o = 0 0 0
H, = [1 0 0], and v, is a zero-mean random variable with nominal variance Xo. We assume
xo E 31o, where 31o = {xo E RW3 : x DTDlxo < 1} and that a nominal distribution for xo
is available. The nominal distributions of xo, In~,,. and v, are available from past experience but
are not assumed to be the exact distributions. The exact distribution of xo belongs to MZ/xo, the
space of distributions defined on (31o, a(31o)). The exact distribution of the measurement noise is
known to belong to the class of zero-mean distributions whose variance A satisfies | A Xo I I x
for some rx > 0. Similarly, the exact distribution of the model noise is known to belong to a class
of zero-mean distributions whose covariance Q satisfies ||ID(Q Qo)DT||y < r, for some rg >
0and De R N"XN. By Example 12.1i, Pi = {AE : tr(D1ADT ) <' 1}. In addition, it s ler
that #2~ = {AES : ~ ||D2( 0) || < "1} and #.3 = {Q ES :- ||D3 0D|y }
where D2 = --1/2 and D31 r--1/2D. In this example, we consider the filtering problem specified
by the estimation space (S,, S,, h), where S, = {0,..., nt}, S, = N, and h(i) = {0,..., i} for
each i E S,. We choose the following values: a = 0.01, at = 120, Ao = 1 x 104, g0 = 100,
rx = 1000, r, = 30,
5 x10-6 0 0
DI = 0 0.002 0
0 0 0.01
2/A2 0 0
0 0 1
We assume that the nominal distribution of xo has zero mean and diagonal covariance with
diagonal entries (1 x 106, 1000, 10). We consider the risk based on quadratic loss functions
as our performance measure and let the weight matrix W,, for each 0 < n < us, have all zero
elements, except only for the unity element in its upper left hand corner. This choice of the
weight matrix implies that we are only interested in estimating the position of the target and we
regard the velocity and acceleration of the target as nuisance parameters.
Clearly Assumption 12.1 holds in this example. We have shown that Assumption 12.2 holds
in Example 12.1. It can be easily verified that Assumptions 12.3 and 12.4 hold as well. Our
first step is to choose Co. In order to choose Co, we need to calculate V(8, L, R) and Co. Once
these quantities are available, we can choose Co based on the amount of a priori information
we have. Let &;r denote the (essentially unique) minimax estimator in the estimation problem
(8, L, RP). We can get an insight as to how to choose Co by plotting the Bayes risk r (v, & )
versus the maximum risk supoe, R(0, & ). We use the algorithm of Lemma 12.2 to calculate
A* for per [0, 1) and then calculate r(v, & ) and supose R(0,~* & .Fig 12-1 showsr the plot of
the Bayes risk r(v, &* ) versus the maxIimumlll risk~ suose p U~aCI ,u,,,,,IU~vr R(0, &* ) Note that byI Theorem1 9.1,
this figure, in fact, shows the Bayes risk achieved by a restricted risk Bayes solution relative to
(v, Co) versus Co. Hence this figure tells us the tradeoff between the penalty on the Bayes risk
and the safeguard on the maximum risk by employing restricted risk Bayes estimation. In this
example, V = 1.225 x 10s and Co = 6.474 x 106. We can see from the figure that on one
hand, it would make little sense to choose Co > 8 x 10s since the improvement in the Bayes risk
would be minor and the degradation in terms of the maximum risk would be very significant. On
the other hand, if we choose Co to be very small, we would have only a minor improvement in
terms of the maximum risk and a significant degradation in terms of the Bayes risk. Therefore,
it seems that in most cases, except maybe the case of complete lack of a priori information, we
would choose 1.5 x 105 < Co < 8 x 105. Again, the exact choice of Co depends on the amount
of a priori information we have and therefore is done heuristically. For illustration, we choose
Co = 5 x 105. We calculate the restricted risk Bayes solution relative to (v, Co) using the method
discussed in Remark 9.3 and the algorithm of Lemma 12.2. It turns out that the restricted risk
Bayes solution relative to (v, Co) is a minimax estimator for the estimation problem (8, L, RP"),
where po = 0.986. With this choice of Co, the restricted risk Bayes approach can reduce the
maximum risk to about 1/13th of that of a Bayes solution while suffering only minimally (about
4%) on the Bayes risk.
Fig. 12-1 illustrates the behavior of the Bayes, minimax, and restricted risk Bayes solutions
in two extreme cases: The case that v is the true state of nature and the case that the true state of
nature is the worst case choice for each one of these estimators, respectively. However, Fig. 12-1
does not illustrate the behavior of these estimators for other values of the true state of nature.
It is important to evaluate the performance of these estimators relative to all 8 E 8 that are
likely to be the true state of nature based on our a priori information. In order to do that, we
need to make some assumptions regarding the set of 8 E 8 that are likely to be the true state
of nature. We assume that this set is 8, = {eov + (1 co)8 : e < co < 1, 8 E 8},
where 0 < e < 1, i.e., the true state of nature is likely to be an co-mixture between v and
an unknown 8 E 8 for e < co < 1. The more a priori information we have, the more
likely it is that the true state of nature is close to v and hence the larger e is. Note that for any
:i E L, supose, R(0, i-) = er(v, i-) + (1 e) supose R(0, i-). Let i-* denote the (essentially
unique) minimax estimator and recall that :i-, denotes the (essentially unique) restricted risk
Bayes solution relative to (v, Co). It is easy to verify that supose R(, 'e) = e?"(v) + (1 e)Co,
supeoe, R (0, i*) = er (v, -*) + (1-e) V(8, L, R), and suppose R(, )~ = er (vI, i*) + (1- e) Co
Let us assume that there is a sufficient amount of a priori information so that e > 0.95. Fig. 12-2
shows the maximum risk over 8, of the Bayes, minimax, and restricted risk Bayes solutions for
e E [0.95, 1]. This figure shows the worst case performance of these estimators relative to values
in 8 that are likely to be the true state of nature. The maximum risk over 8, of the restricted
risk Bayes solution is less than that of the Bayes solution for almost the entire range of e. In
the interval [0.95, 0.98], the maximum risk over 8, of the restricted risk Bayes solution is in
fact significantly less than that of the Bayes solution. The maximum risk over 8, of the Bayes
solution is less than that of the restricted risk Bayes solution only for e > 0.9995 and even in
this case, the difference is very small. It can also be seen that even a relatively small uncertainty
in the true state of nature may lead to undesirable performance of a Bayes solution relative to v.
For example, in the case e = 0.98, using the restricted risk Bayes solution instead of the Bayes
solution leads to 50I' improvement in terms of the maximum risk over 8,. In addition, it can be
seen that the performance of the restricted risk Bayes solution is superior to that of the minimax
estimator for all e > 0.95. This illustrates that if a priori information is available, the restricted
risk Bayes approach is preferable to the somewhat conservative minimax approach.
105x104
o Minimax Solution
03
95-
06
9-"
85-
S09
75- FO~
0 975 Unrestricted
Bayes Solution
0 999
650135
Ssupp R(9, f) x lo'
Figure 12-1. Achieved Bayes risk vs. maximum risk
x 105
47
Bayes Solution
Minimax Solution
Restricted Risk Bayes Solution
3 5
3-
25-
15-
05
095
0 955 096 0 965 097 0 975 098 0 985 099 0 995
Figure 12-2. The maximum risk over 8, of the Bayes, minimax, and restricted risk Bayes
solutions vs. e
CHAPTER 13
STATE ESTIMATION WITH INITIAL STATE UNCERTAINTY
In this chapter we consider another special case of the state estimation problem of Chapter
11. We consider the case of deterministic uncertainties in the initial state. We assume that the
sequences {v,} and {tt',,}), which have zero mean, satisfy for n, m > 0
G~v~v0, otherwise
0 otherwise
E(vow )= 0, (13-1)
where E denotes the expectation operator. This means that 82 and 83 COntain only one element.
We restrict ourselves to the case A2 > 0, which is usually the case in well modeled problems
[44]. It is left to specify the initial state vector xo. We model xo as a deterministic (unknown)
parameter belonging to a parameter set No0. We assume 31o is a compact subset of RWN. Note
that the current formulation deviates from the standard KF assumptions in that the initial state is
not modeled as random with known statistics. Since we assume deterministic uncertainty in the
initial state vector, 81 is the collection {(xo, xox ) : xo E o}0 of mean and second moment
pairs. Note that since 31o is compact, 8 is compact.
We are interested the filtering problem that is specified by the estimation space (S,, S,, h),
where S, = { 0,. ., at }, S, = N, and h(i) = { 0,. ., i } for each i E S,. We let w (i) = 1 for
each i E S,, i.e., the risk function is simply given by
n= o
Since there is a one-to-one correspondence between 8 and 31o, we slightly abuse the notation and
assume that the space of states of nature is 31o, i.e., in this chapter 8 = Zo0. We also use R(xo, 2)
instead of R(0, 2), R,(xo, 2) instead of R,(0, 2), for n = 0, ..., us, and etc.
13.1 Conditional Mean Estimators
Recall that rI,, = s yTU T]". Given -r Me Zl, let 2,(-r) denote the mapping
y"n H Exo(X,|y"n), i.e., 2,(-r) is the conditional mean (CM) estimator with respect to -r when
xo is the true state of nature. Let F,(-r) denote the MSE matrix of 2,(-r) with respect to -r. Note
that we may regard a point xo E 31o as an element of M /xo by regarding it as the probability
measure that assigns mass 1 to xo. Thus 2,(xo) and F,(xo) are well defined. Let 2(-r) =
(f-o(T),., i In,(7)j). Recall that r(-r, ) = fxo R(xro, 2)dr. In addition, we use the following
simplified notation: r(-r) = r(-r, 2(-r)). It is easy to verify that r(-r) = CE,o tr(W,F,(-r)), where
We = Vs"v,.
Let o dennote the conditional mean estimator whel~n xo = and let I`o denote the MSE
matrix of fo when xo = 0. It is clear that io and yo can be carllclated usingT the well kmnown KF~
recursions [44] by initializing the KF with zero mean and covariance. It is straightforward to
verify that F,(XO) = 1o for all o r Zo. Lt n 3+ POn FI K = l- n TA
H,o,_ znlHI] -1for n > 0,and ~o= 0. Let F,= F,- F,,KoH,,1 et FL~3 = FF_1--- Fy (i > j)
and Fi~i = Fi. We will use the convention Fi_,i = I.
The problem of state estimation with a random initial state whose distribution in not
necessarily Gaussian is considered by Lainiotis et al [46] and the CM estimator for this problem
is derived. Let io,n(-r) denote the CM estimator with respect to -r for xo based on ti,,, i.e.,
io Sxo") =~~)~ n e o.()=foEx (onr -x)i~) -x)) dr. Using
the results of [46], it is not difficult to get the following set of equations for n > 0:
2,(r) = ain ~ ) (13-2)
C,=( o,),1,1=(I KoH,)F,-1,o, a > 0, (13-3)
Co = I, (13-4)
F()= Fo + CnCo,n('r)C,, (13-5)
U, H AgHo+ i 1,~-,oH~, (Her _llZH, + A2 -1 He e1,o, (13-6)
i= 1
tn (Y") = Ho Ag2 Yo + i 1,oH, (Her _zH,7 + A2 -1 s A _z (13-7)
i= 1
p" (y") oc exp [T U,o- 2xT ty" gy) (13-8)
where the term go is a fu~nction of y1 and its exact calcu~latin on isnnecessarry for our purpose
Thus 2,(-r), the CM estimator with respect to -r for x,, is the sum of two terms; one term is
io, the CM estimator for x, when the system has zero initial conditions, and the second term
is completely specified by io,n('r), the CM estimator with respect to -r for xo. An alternative
interpretation is to regard 2,(-r) as the sum of two terms, where the first term is an estimator of
x, for known zero initial condition and the second term accounts for the effect of the unknown
initial state. A conditional mean estimator with respect to -r is an essentially unique Bayes
solution relative to -r. The expression for the conditional mean estimator is important since an
essentially unique minimax estimator is the conditional mean estimator with respect to a least
favorable a priori distribution. In addition, the class of conditional mean estimators with respect
to -r M i/xo is essentially complete. Let ~D = { (7) : re xo iZ1X)
13.2 Approximations to Minimax Estimators
In order to find the (essentially unique) minimax estimator, we need to derive a conditional
mean estimator relative to a least favorable a priori distribution. Hence we are left with the
following dual problem: Find 7o E Mxo, such that r(7o) =sup,eMso r(7). This dual problem
is conceptually easier than finding a minimax estimator directly by the definition. Nevertheless,
solving the dual problem may still be a difficult task. Thus it may be necessary to search for
sub-optimal estimators in terms of the maximum risk as suggested in Chapter 8. Our goal is then
to derive suboptimal estimators that can give maximum risk arbitrarily close to that of a minimax
estimator. The following Lemmas are derived in [24].
Lemma 13.1. Suppose 31o is compact. Then if {8 } is a sequence in MZ/xo that converges weakly
to To E MZ/xo, then R(xo, Sh))> converges to R(xo, 2(To)) uniformly on No0.
Lemma 13.2. Let f e D Then there exists an extension of R(-, 2) from No0 to an open convex
set 31o C Uc RWN such that R(-, 2) is twice difgerentiable on U. 1Moreover for all xo E 310
D2 2 0o) < 2U,, sup R(xo, 2). (13-9)
zo6Xo
Because of Lemmas 13.1 and 13.2 and the fact that 31o is a subset of finite dimensional
normed space, we can use the results of Chapter 8 to find a y-optimal estimator. In particular if
31o is a convex polytope, Lemma 8.7 can be used to find a y-optimal estimator. In order to find
a y-optimal estimator, we need to set a 0 < y' < y and construct a sufficiently dense finite
subset 31o' of 31o according to the lemma. We then need to find a (y', 3lo')-optimal estimator.
The lemma tells us that this estimator is a y-optimal estimator. A (y', 31o')-optimal estimator
can be found using, for example, the algorithm that is proposed in [38] and that also appears
in [18] and Theorem 7.1. Using Theorem 7.1, it is not difficult to verify that all the necessary
conditions for this algorithm to be valid hold. It worths mentioning a technical difficulty in this
approach. For us to use the algorithm of Lemma 7.1, we need to calculate matxroeXof R(xo, NT)),
which means that we have to calculate the risk R(xo,2(-r)) for each xo E 31ol. However, an
analytical expression for R(xo, (7)), for xo t Xof and 7 e Mzoy, is not available, in general.
Hence R(xo,2(-r)) must be calculated numerically for all xo E 31o'. Thus the resulting numerical
algorithm involves intensive multi-dimensional integration at each step. Due to this difficulty, it
is desirable to use efficient numerical integration methods. In [24] there is a discussion regarding
how the numerical integration can be done efficiently. Also, in [24] an alternative algorithm to
the one of Theorem 7.1 is proposed. It is argued that this algorithm may be preferable to the one
of Theorem 7.1, in terms of the computation burden, since it may require less calculations of the
risk function.
The computational complexity of a y-optimal estimator derived according to Lemma 8.7
depends linearly on the support of -ro. It may be rather high, and it is certainly higher than the
complexity of affine estimators. In Chapter 12 we discussed the problem of restricted risk Bayes
estimation when the class of estimators is restricted to affine estimators. When the class of
estimators is restricted to affine estimators, the minimax problem of this section can be regarded
as a special case of the problem considered in Chapter 12. If we can find an affine estimator that
is also a y-optimal estimator, we would probably prefer the affine estimator. If such an affine
estimator exists, the results of this section are still important since they give us tight lower and
upper bounds on the maximum risk and hence enable us to evaluate the performance of the
affine estimator with the best possible performance in the sense of the maximum risk. We further
illustrate this in the following numerical example.
13.3 Numerical Example
We consider the following problem as an example. Suppose a target is moving in the one-
dimensional space. We assume the following simplified model of motion. Let .r,z E R2, where
.r, (1) denotes the position of the target and .r, (2) denotes the velocity of the target. We assume
r,w+1 = Fr,z + 1,',,, (13-10)
where
and I,',, is zero mean Gaussian random vector with covariance matrix
00 ,
We assume xo E 31o, where 31o is a closed rectangle in RW2. We also assume a radar measures the
position of the target and the observations from the radar obey the following equation
yn = Hx, + v,, (13-11)
where H = [1 0] and v, is a zero-mean Gaussian random variable with variance A2. NOte that
this example is a simplified version of many real applications in which the KF is used [21].
In this example, we assume that 31o = [-1000, 1000] x [-50, 50], a = 0.01, A2 = 2500,
q = 25, at = 10, W, = 0 for n < 5, and W, =, for n = 5,. ., us, where W, = V,TV,.
Our goal is to derive a y-optimal estimator for y = 0.03. This means that the degradation,
when using the derived estimator instead of an exact minimax estimator, is at most 3%b of the
maximum risk. The admissible y-optimal estimator that we derive is a CM estimator with respect
to a discrete a priori distribution.
Let V be a matrix such that the ith column of V is the ith eigenvector of 2Uto. Let 6i denote
the ith eigenvalue of 2Uto. Let y = 0.03 and y' = 0.0023. The above choices of y' results from a
certain tradeoff. The choice of y' determines how dense the finite subset 31o' of 31o and specifies
h~ow close the resulted estimator is to a minimax estimator for (No', D,;: R). If we choose y' very
close to zero, we would decrease |Xo| but we woumld need more iteratinon to find a (y', Ro)-
If we choose y' too large, the dimension of 31o' would grow and hence also the computational
complexity. Let b6() = f--i- Ior i = 1, 2. We construct a finite set Xo' such that Xo' is
(6, V)-dense subset of 31o, o', n 31,b is (6, V)-dense subset of 31,b, where 31,b denotes the boundary
of Ro, and the points (-50, 1000), (50,! 1000), (-50, -1000), and (50, 1000) belong to No'f. The
set 31o' is plotted in Figs. 13-1 and 13-2. Due to the eigen-structure of the matrix Uto, the resulted
set 31o' is not a standard grid. The points in the set 3lof are much denser in the direction of the first
eigenvector than in the direction of the second eigenvector.
Next, we solve for an (y', 31o')-optimal estimator. We set the initial a priori distribution to
be the uniform distribution on 31o'. We then update the a priori distribution using a variant of the
projected gradient method as discussed in [24]. As mentioned earlier, it is possible to use also the
algorithm of Theorem 7.1. Each update increases the Bayes risk. We stop the algorithm when the
required tolerance of y' is achieved. Since a (y', Ro')~-optimal estimator is an e-minimax estimator
for (N ,~ 1D, R) with~ c = V/(o D, R), it is possible to check< whether th~e required tolerance is
achieved using Lemma 8.1i. Let -ro denote the resulted a priori distribution; the resulted estimator
is then 2(-ro), which is a y-optimal estimator. Fig. 13-3 shows the risk R(xo, 2(-o)) for xo E 31o .
The maximum risk on the set 3lof is 1864.6. Due to the construction of 31o', the maximum risk on
31o is less than 1916.1i. In Fig. 13-4 we show the resulted a priori distribution Tro.
It seems unreasonable to compare between the proposed y-optimal estimator with an
estimator that is derived using the F-minimax approach. The reason is that by assuming that
xo is deterministic and belongs to 31o, there is no systematic way as to how to choose a class 0
of distributions. Of course, one can choose E to be MZ/xo, but then a F-minimax estimator can
be shown to be a minimax estimator (i.e., the two problems are equivalent). It may be the case
that a properly initialized KF can have a very close performance to that of a minimax estimator.
If one adopts the formulation of this work, then a rigorous way to initialize the KF is to derive
a linear (or affine) minimax estimator, which is a KF with respect to a least favorable a priori
distribution. In practice, the KF is often initialized with zero mean and covariance a2I, where
e2 iS chosen heuristically. We wish to illustrate that this heuristic method can sometimes lead to
undesirable performance. In Fig. 13-5, the maximum risk of the KF initialized with zero mean
and covariance a21 iS plotted as a function of a e [150, 450]. It is apparent that choosing a
too small leads to a very poor performance. In addition, choosing a too large may also lead to
undesirable performance. The best choice of a in terms of the maximum risk is a = 289.7 and
the maximum risk in this case is 2042.8. Thus for the best choice of a, we have a degradation of
at least 6.5% and at most 10%b in terms of the maximum risk relative to the estimator 2(-ro). This
is not a significant degradation and when complexity is taken into account, it may be preferable
to use a KF that is initialized with a = 289.7. In this example, it is clear that the degradation if
a linear minimax estimator is used instead of a minimax estimator is at most 10%b and may be in
fact much less. This should not imply that the whole construction of i(Tro) is unnecessary. It is
this construction and the calculation of maxsoexo; R(xo, 2(7o)) that enabled us to establish that a
linear minimax estimator has a very good performance in this case.
S20-
10 -0
-0-
-1000 -800 -600 -400 -200 0 200 400 600 800 1000
xo (1) (initial position [m])
Figure 13-1. A full view of Ro',f which is a finite (6, V/)-dense subset of Zo
-46-
S-47-
~-47 5-
S-48-
;-48 5-
-49-
-49 5-
-50- *
-1000 -950 -900 -850
xo (1) (initial position [m])
Figure 13-2. A zoom-in view of the bottom left corner of No', which is a finite (6, V/)-dense
subset of 31o
1900 x
1800-
1700 m
HC1600 m
1500 m
1400
50
1000
0 500
-500
xo(2) (initial velocity [m/s]) -so -1000 xo(1) (initial position [m])
Figure 13-3. The risk of x(ro), the derived y-optimal estimator, as a function of xo E o,~
x 10
1000
-500
xo(2) (initial velocity [m/s]) -so -1000 xo(1) (initial position [m])
Figure 13-4. The a priori distribution -ro, which is defined on 31o
2900-
2800-
2700-
6 2600-
S2500-
S2300-
2200-
2100-
2000-
1900-
150 200 250 300 350 400 450
The maximum risk of the Kalman Filter initialized with zero mean and covariance
O.2I as a function of o-
Figure 13-5.
CHAPTER 14
CONCLUSIONS
We considered the problem of state estimation with stochastic and deterministic uncer-
tainties in the initial state, model noise, and measurement noise using a decision theoretic point
of view. We considered both the case that the class of available estimators is the class of all
estimators and the case that the class of available estimators is the class of affine estimators. We
showed that a minimax estimator and a restricted risk Bayes solution exist when the risk is based
on quadratic loss functions. Under further conditions, a minimax estimator can be found as a con-
ditional mean estimator or a LMMSE estimator relative to a least favorable a priori distribution.
In the case that all the uncertainties are stochastic, we adopted the restricted risk Bayes approach,
which incorporates the use of a priori information to derive estimators that are robust to devia-
tions of the model from the nominal assumed model. We derived a general method to obtain a
restricted risk Bayes solution. When the class of estimators is restricted to affine estimators, this
method can be easily used provided that a maximizer of the risk function can be calculated. We
considered several important cases in which a maximizer of the risk function could be calculated
analytically and showed that in many other cases it could be calculated numerically. Thus we
provided a systematic way to derive restricted risk Bayes solutions. When the filtering problem
is considered, the restricted risk Bayes approach provides us with a robust method to calibrate
the Kalman filter. We also considered the problem of state estimation with deterministic initial
state uncertainty. In this case, we proposed a numerical method to derive an approximation for
a minimax estimator with the possibility to make the approximation as accurate as desired. This
method seems to be especially attractive in the case that the parameter set is a convex polytope in
APPENDIX A
PROOF OF LEMMA 5.2
Let w' be the system of subsets B C w* such that oo a B and C* \ B is a closed, compact
subset of C in the sense of the topology w. The topology w* is such that each set in w* is a union
of a set from w and a set of w'. It is well known that w is, in fact, the relative topology on C. Since
C is locally compact, o--compact, and metrizable, C* is compact and metrizable.
Let us show that the f(a, -) is lower semicontinuous on C* for each a s A. Fix a point
c* E C. Let {c,} be a sequence that converges to c* in the sense of w*. Then {c,} is eventually
in any w*-neighborhood of c*. Since (C*, 0*) is Hausdorff, there exist disjoint open sets A*
and B* such that A* is a neighborhood of c* and B* is a neighborhood of 00. Since B* is
open, it is the union of an element of w' and an element w, and we can assume without loss
of generality that it is in w'. Let A be an open w-neighborhood of c*. Then A U A* is an w-
neighborhood of c*. Since {c,} is eventually in A U B* and eventually in A*, {c,} is eventually
in A. Thus {c,} converges to c* in the sense of the topology w. By the hypothesis of the lemma,
lim inf,,, f (a, c,) > f(a, c*). It follows that lim inf,,, f (a, c,) > f(a, c*). Let {c,} be
a sequence in C that converges to oo. Then for each A E w', {c,} is eventually in A. Thus for
each w-compact subset C' of C, there exists NV > 0 such that c, ( C' for n > NV. Let {C,} be
a sequence of compact subsets of C such that U" zC, and c, Sf C, for (n = 1, 2, ..). Since
lim inf,,, f (a, c,) = supec f (a, c), lim inf,,, f (a, c,) = f (a, 00). It follows that f (a, -)
is lower semicontinuous on C* for each a s A. Clearly f is nonnegative. Since w is the relative
topology, each Co C C that is in B(C*) is in B(C). Let Co a w. Then since Co U co E w* and
C* \0 oo E w*, Co aB(C*). If Co aB(C), it is a countable union and intersection of elements of w;
hence a countable union and intersection of elements of B(C*). Thus Co aB(C*). It follows that
if Co c C, Co aB(C) if and only if Co E a(C*).
Suppose f(-, c) E m(FA~) for each c e C. Then clearly f(-, c) E m(FA~) for each c e C.
Since f (-,oo) = supeec f (-, c), f (-, 0) E m(FA~). Thus f (-, c) E m(FA~) for each c e C*.
Suppose fe E (FA x B(C)). Let A be a Borel set of the extend reals. The set f-1(A) =
{(a, c) E Ax C* : f (a, c) E A}. Then f -1(A) = {(a, c) E Ax C : f (a, c) E A} U {(a, 00):
a s A, f (a, 00) E A}. Clearly f -1(A) = {(a, c) E Ax C : f (a, c) E A} is in FA~ x B(C). Hence
it is in FA~ x B(C*). Since f (-, 0) E m(FA~), the set f -l(-,oo)(A) = {a EA : f (a, 00) E A}
is in FA~. Thus {(a, 00) : a s A, f (a, 00) E A} = f -l(-,oo)(A) x {oo}, and hence is in
FA~ x B(C*). It follows that f"-1(A) E Fg~ x B(C*). Thus fe m(FA x B(C*)).
APPENDIX B
PROOF OF LEMMA 9. 1
Note that for any~i 6 E O < pi, p2 < 1, and 0 < rl < 1,
The concavity of K follows from the fact that for any 0 < rl < 1,
K~vyt +(1 77p2)= inf .K(:i, rlpt + (1 rl)p2)
= rlK(pl) + (1 rl)K(p2 m
where the inequality above is a direct result of (B-1). Eqn. (9-1) implies that K(pl) > K(p2)
for any 0 < pi < p2 < 1. Since K is decreasing and concave on [0, 1], it is continuous except
perhaps at the point p = 1. Certainly K(:i,., 1) = K(1) and K(:i,., p) > K(p) for all 0 < p < 1.
Thus for any 0 < p < 1,
| K(p () |< (f e, ) -K :fe.,1)= ( -p)sup R (0, :f e) r (v, :f e.)]< (1 p) Co
where the last inequality results from our assumption that R is nonnegative. This proves that K is
continuous at p = 1 and thus on [0, 1].
APPENDIX C
PROOF OF LEMMA 9.2
1) It is easy to see that
K(2 2) = K(p2) < K(ix, p2) < K(ix, pt) = K(pt) < K(2 1
(C-l)
where the second inequality is a direct consequence of (9-1).
This implies that K(2, 1) K(2, 2) > K(ix, pi) K(ix, p2). It easily follows that
2) By (C-1), K(ix, p2) K(2, 2) > 0. Thus
(o, 12 2 1-P)[G(2) G(ii)] < I~ )
By part 1) of the lemma, r (v, 2) TU 1 -~l
3) By (C-1), K(2, 1i) K(A pi) > 0. Thus
sup R(0, f t) + pl [G(2) G(ii)] < sup R(0, 2 -
BEe sEe
By part 1) of the lemma, supose R(8, 1) < supeoe R(8, 2 -
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1-17, 1995.
[15] J. O. Berger, Statistical Decision Theory and Ba! edition, 1985.
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1Monthly, vol. 94, pp. 409-422, May 1987.
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79, pp. 303-306, 1975.
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BIOGRAPHICAL SKETCH
Yoav N. Levinbook was born in Tel Aviv, Israel, on October 30, 1974. He received the B.S.
degree Magnaa com laude) from Tel Aviv University, Israel, in 2000 and the M.S. and Ph.D.
degrees in electrical and computer engineering from the University of Florida, Gainesville, in
2006 and 2007, respectively.
He was with the Motorola Semiconductor, Herzliya, Israel, and Smartlink, Netanya,
Israel, as an electrical engineer. His research interests include statistical decision theory, state
estimation, signal processing for communications, and sensor networks.
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page ACKNOWLEDGMENTS .................................... 4 LISTOFFIGURES ....................................... 7 LISTOFABBREVIATIONS .................................. 8 ABSTRACT ........................................... 9 CHAPTER 1INTRODUCTION .................................... 11 2GENERALNOTATIONANDCONVENTIONS .................... 17 3DECISIONTHEORETICFORMULATION ....................... 19 4GENERALDECISIONTHEORETICRESULTS .................... 24 5THECASETHATTHERISKISSPECIFIEDBYALOSSFUNCTION ........ 34 6THECASEOFCONVEXLOSSFUNCTION ...................... 40 7FINDINGAMINIMAXESTIMATORANDTHEDUALPROBLEM ......... 49 8APPROXIMATINGAMINIMAXESTIMATOR .................... 52 9THERESTRICTEDRISKBAYESPROBLEMASAMINIMAXPROBLEM .... 61 10ESTIMATIONWITHARESTRICTIONONTHEOBSERVATIONSTHATCANBEUSED ......................................... 66 11THESTATEESTIMATIONPROBLEM ......................... 73 12AFFINESTATEESTIMATIONBASEDONQUADRATICLOSSFUNCTIONS ... 77 12.1FindingaRestrictedRiskBayesSolution ..................... 77 12.2FindingaMaximizeroftheRisk .......................... 84 12.3ConnectiontotheKalmanFilterand-MinimaxApproach ............ 87 12.4NumericalExample ................................. 90 13STATEESTIMATIONWITHINITIALSTATEUNCERTAINTY ........... 96 13.1ConditionalMeanEstimators ............................ 97 13.2ApproximationstoMinimaxEstimators ...................... 98 13.3NumericalExample ................................. 100 14CONCLUSIONS ..................................... 106 APPENDIX 5
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................................. 107 BPROOFOFLEMMA9.1 ................................. 109 CPROOFOFLEMMA9.2 ................................. 110 REFERENCES ......................................... 111 BIOGRAPHICALSKETCH .................................. 114 6
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Figure page 12-1AchievedBayesriskvs.maximumrisk .......................... 94 12-2ThemaximumriskoveroftheBayes,minimax,andrestrictedriskBayessolu-tionsvs. 95 13-1AfullviewofXf0,whichisanite(;V)-densesubsetofX0 103 13-2Azoom-inviewofthebottomleftcornerofXf0,whichisanite(;V)-densesubsetofX0 104 13-3Theriskof^x(0),thederived-optimalestimator,asafunctionofx02Xf0 104 13-4Theaprioridistribution0,whichisdenedonXf0 105 13-5ThemaximumriskoftheKalmanFilterinitializedwithzeromeanandcovariance2Iasafunctionof 105 7
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CM:conditionalmean......................................97KF:KalmanFilter........................................11LMMSE:linearminimummeansquarederror.........................11MSE:meansquarederror....................................45 8
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1 ]isthelinearminimummeansquarederror(LMMSE)estimator.IfinadditionallthestochasticquantitiesareGaussian,theKFistheminimummeansquarederrorestimator.Sincetheassumptionofcompleteknowledgeoftheaprioridistributionisseldomsatised,aBayesianapproachisusedinpractice.Theaprioridistributionsoftheinitialstate,modelnoise,andmeasurementnoisearelearnedfrompastexperienceandusedasapproximationsofthecorrespondingtruedistributions.Nevertheless,evenifextensivepastexperienceisavailable,theestimateddistributionsmaystilldeviatefromthetrueones.TheeffectofsucherrorsintheaprioriinformationontheperformanceoftheKFisstudiedin[ 2 ][ 4 ].Theeffectoftheerrorsintheaprioriinformationoftheinitialstate,modelnoise,andmeasurementnoisemaybeverysignicantandaKFupdatedbasedonerroneousaprioriinformationmayperformpoorly.Thusitisnecessarytoconsiderotherapproachesthatarerobustagainstuncertaintiesintheaprioridistributionoftheinitialstate,modelnoise,andmeasurementnoise.Thestateestimationliteraturedealsextensivelywiththegeneralproblemoflinearsystemswithstochasticordeterministicuncertaintiesusinggametheoryandtheminimaxapproach(cf.[ 5 ][ 14 ]andthereferencestherein).Usuallytheso-called-minimaxapproachisadopted.The-minimaxapproach[ 15 ]regardstheparameterasrandomwithitsdistributionliesinaclass.However,theexactdistributionintheclassisunknown.A-minimaxestimatorisanestimatorthatminimizesthesupremumoftheBayesrisk,wherethesupremumistakenoverallelementsof.Whenthe-minimaxapproachisused,theclassofavailableestimatorsisusuallyrestricted 11
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16 ],isacompromisebetweentheBayesapproachandtheminimaxapproach.ArestrictedriskBayesestimatorminimizestheBayesriskwithrespecttoanaprioridistributionsuggestedbasedonsomepastexperiencesubjecttotherestrictionthatthemaximumriskdoesnotexceedtheminimaxriskbymorethanagivenamount.Thisapproachutilizesavailableaprioriinformationbutatthesametimeprovidesasafeguardincasethisinformationisnotaccurate.Iftheaprioriinformationisfairlyaccurate,arestrictedriskBayesestimatorhasgoodBayesriskproperties.OtherworkconsideringtherestrictedriskBayesapproachorcloselyrelatedapproachesinclude[ 17 ][ 20 ].DespitetheappealingformulationoftherestrictedriskBayesapproach,ithasnotbeenutilizedinthecontextofstateestimation.Althoughtheproblemofstateestimationwithstochasticuncertaintieshasbeenapproachedfromthe-minimaxapproach,webelievethatapproachingthisproblemfromtherestrictedriskBayesapproachalsohasaconsiderablemerit.Ifastateestimationproblemcanberegardedasazero-sumtwo-persongame(henceforthtobereferredasagame)againstarationalopponent,thenthe-minimaxapproachseemsveryattractive.However,inmostapplications,ifweregardthestateestimationproblemasagame,thegameisagainstNature.Usingthe-minimaxapproachinthiscasecorrespondstoaverypessimisticviewpointthatregardsNatureasarationalopponentthatwishestocauseusthelargestpossibleloss.The-minimaxapproachmaystillseemreasonableinthecasethatthereisnoaprioriinformationthatenablesustoregardcertaindistributionsinasmorelikelythanothers.However,inmanyapplications,someaprioriinformationregardingthetruedistributionmaybeavailable.Thisaprioriinformationmaybeintheformofanominaldistribution,whichissuggestedbasedonsomepastexperience.Itiswellknownthatundercertainconditions,a-minimaxestimatorisaKFrelativetoaleastfavorablea
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24 ]thatdealwiththeboundedparametersetcaseinordertoillustratehowminimaxestimatorscanbeapproximatedwitharbitrarilyprescribedaccuracy.Whilewearemainlyinterestedinthestateestimationproblem,alargepartofthisworkwillbeconcernedwithamoregeneralestimationproblem.Infact,someoftheexistenceresultssuchastheexistenceofaminimaxestimatorandarestrictedriskBayessolutionholdinaverygeneralsettingandmayhaveapplicabilitynotonlyinthestateestimationproblem.Therestofthisworkisorganizedasfollows.InChapter 2 ,wepresentnotationandconventionsthatareusedthroughoutthiswork.InChapter 3 ,wepresentageneraldecisiontheoreticformulationthatisneededinordertoaddresstheproblemofstateestimationwithstochasticanddeterministicuncertaintiesandalsotoderiveother,moregeneral,results.InChapter 4 ,wederiveseveralgeneraldecisiontheoreticresults,whicharebasedonwellknownresultsfromdecisiontheoryandgametheory;theapplicabilityoftheseresultsisnotlimitedonlytotheproblemofstateestimationwithstochasticanddeterministicuncertainties.InChapter 5 ,weconsiderthecasethattheriskfunctionisspeciedbyalossfunction.WederiveratherweakconditionsthatguaranteetheexistenceofaminimaxestimatorandarestrictedriskBayessolution.InChapter 6 ,werestrictourselvestoconvexlossfunctions,ingeneral,andthequadraticlossfunction,inparticular.InChapter 7 ,wediscusshowaminimaxestimatorcanbefoundbysolvingthedualproblemofndingaleastfavorableaprioridistribution.Since,ingeneral,ndingaminimaxestimatormaybeanextremelydifculttask,inChapter 8 wediscusshowonecanderiveapproximationstominimaxestimators,wheretheapproximationcanbemadeasaccurateasdesired.InChapter 9 ,weconsiderthegeneralrestrictedriskBayesestimationproblemandshowthatthisproblemisequivalenttoasequenceofminimaxestimationproblems.Insomeestimationproblemstherearerestrictionsontheobservationsthatcanbeusedinordertoestimatetheparameters.Thisisthecaseinthestateestimationprobleminwhicheachstatecanbeestimatedusingonlycertainobservations.WeconsiderthistypeofestimationproblemsinChapter 10 .InChapters 11 12 ,and 13 ,werestrictourselvestotheproblemofstateestimationwithuncertaintiesintheinitialstate,modelnoise,andmeasurementnoise,which 15
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11 ,wederivesomegeneralexistenceresultsthatarebasedontheresultsofthepreviouschapters.InChapter 12 ,weconsiderthecaseofstochasticuncertaintiesintheinitialstate,modelnoise,andmeasurementnoise,andrestrictourselvestoafneestimators.WeproposeamethodthatcanbeeasilyusedtoderivearestrictedriskBayessolutioninmanyimportantcases.InChapter 13 ,weconsiderthecaseofdeterministicinitialstateuncertainty,andsearchforestimatorswithintheclassofallestimators.WeconcludethisworkinChapter 14 16
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26 ]and[ 25 ],withslightmodications.Thesedenitionsaregivenforanestimationproblem(;X;R).
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3.1 ,inf^x2Xr(;^x)<+1,andthetermr(i;^x)inf^x2Xr(i;^x),intheabovedenition,iswelldened.
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26 ,Exercise2.2.1]) ConsiderthespaceMequippedwiththetopologyofweakconvergence[ 27 ,pp.236].ThetopologyofweakconvergencemakesMaHausdorffspace. 1. 2. Let^x2XbesuchthatR(;^x)<+1foreach2andR(;^x)iscontinuouson.Thenr(;^x)iscontinuousonM. 3. Let^x2XbesuchthatR(;^x)<+1foreach2andR(;^x)isuppersemicontinu-ouson.Thenr(;^x)isuppersemicontinuousonM. Proof. 28 ,Exercise2.25]).ThisimpliesthatMismetrizable[ 29 ,pp.122].Inaddition,[ 29 ,Theorem3.1.9]furnishesthatMiscompact.2)By[ 29 ,Theorem3.1.5],Rfdisacontinuousfunctionofforanyboundedandcontinuousf.SinceR(;^x)isacontinuousreal-valuedfunctiononacompactset,itisbounded.Hencer(;^x)iscontinuousonM.3)By[ 29 ,Theorem3.1.5],Rfdisanuppersemicontinuousfunctionofforanyup-persemicontinuousfthatisboundedfromabove.SinceR(;^x)isanuppersemicontinuousreal-valuedfunctiononacompactset,itisboundedfromabove.Hencer(;^x)isuppersemicon-tinuousonMforany^x2C. 24
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30 ]. 25
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31 ].SinceU(M)iscompact,R(A;M)isrelativelycompact.Thusthereexistsasubnetf^xbgb2BthatconvergestoapointuintheclosureofR(A;M).HenceliminfR(;^xb)=limR(;^xb)=u.SinceR(A;M)ishalf-closed,thereexistsanelementv2R(A;M)suchthatvu.Hencethereexistsanelement^x2XsuchthatR(;^x)liminfR(;^xb).Supposeforeachnetf^xaga2AinA,thereexistsasubnetf^xbgb2Bandanelement^x2Asuchthatliminfr(;^xb)r(;^x).LetubelongtotheclosureofR(A;M).Thenthereexistsanetfuaga2A2R(A;M)thatconvergestou.Clearlyforeachelementuainthisnet,thereisanelement^xa2Asuchthatua=R(;^xa).ThuslimR(;^xa)=u.Itfollowsthatthereexistsasubnetf^xbgb2Bandanelement^x2AsuchthatliminfR(;^xb)R(;^x),whenceuR(;^x)andR(A;M)ishalf-closed.Inentirelyanalogousway,Ahastheproperty(W)ifandonlyifforeachnetf^xaga2AinA,thereexistsasubnetf^xbgb2Bandanelement^x2AsuchthatliminfR(;^xb)R(;^x).SinceeachcanbeidentiedasanelementinM,asdiscussedpreviously,itisclearthattheproperty(W)impliestheproperty(W).Theproperty(W),asformulatedwithnets,iscloselyrelatedtoWald'sweakcompactness[ 25 ,pp.53].Theonlydifferenceisthatinthecurrentdenitionsequencesarereplacedbynets.Hencetheproperty(W)isweakerthanWald'sweakcompactness.SufcientconditionsforasubsetAofXtohavetheproperty(W)aregivenin[ 31 ].AsimplesufcientconditionisthatthereexistsaHausdorfftopologyJforAsuchthatAiscompactandR(;)islowersemicontinuousonAforeach2.Similarly,asufcientconditionforAtohavetheproperty(W)isthatthereexistsaHausdorfftopologyJforAsuchthatAiscompactandr(;)islowersemicontinuousonAforeach2M.SupposethereexistsatopologyJforAsuchthatAiscompact.Thenifthetopologicalspace(A;J)satisestherstaxiomofcountability,byFatou'sLemma,therequirementthatr(;)isalowersemicontinuousonAforeach2McanbereplacedbytherequirementthatR(;)isalowersemicontinuousonAforeach2. 26
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PuttingC0=+1intheabovetheorem,wegetthatforeach2M,thereexistsaBayessolutionrelativeto(providedthatXhastheproperty(W)).ThefollowingTheorem,aversionofWald'scompleteclasstheorem,appearsin[ 31 ]. 32 ,Theorem4.2]. Proof. 4.1 ,Miscompactandr(;^x)isuppersemicontinuousonMforeach^x2X.ItiseasytoverifythatsinceXissubconvex,foreach^x0;^x002Xand0<<1,thereexists^x2Xsuchthatr(;^x)r(;^x0)+(1)r(;^x00)forall2.Certainlyr(;^x)isconcaveonMforall^x2X.Applying[ 32 ,Theorem4.2],inf^x2Xsup2Mr(;^x)=sup2Minf^x2Xr(;^x):Thus(;X;R)isstrictlydetermined.Sinceinf^x2Xr(;^x)isuppersemicontinuousonacompactsetM,thereexists02Msuchthatinf^x2Xr(0;^x)=sup2Minf^x2Xr(;^x):Thus0isaleastfavorableaprioridistribution. 28
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4.4 thatR(;^x)<+1foreach^x2XandR(;^x)isuppersemicontinuousonforeach^x2Xisobviouslytoostrong.Inthefollowingtheoremthisassumptionisconsiderablyweakened,butatthepriceofastrongerassumptiononX.LetBXdenotetheclassofBayessolutionsrelativeto2M.Beforestatingthetheorem,weneedthefollowinglemma. 4.2 itisshownthatifXhastheproperty(W),thereexistsaBayessolutionrelativetoforall2M.Thelemmafollowseasily. Proof. 32 ,Theorem4.2], Fix2M.Ononehand,sinceR(B;M)GB,infg2GBg()infg2R(B;M)g()=inf^x2Br(;^x):Ontheotherhand,ifwexg2GB,thenthereexistanintegern>0,realnumbers1;:::;n>0,andelementsu1;:::;un2R(B;M)suchthatg=Pni=1iuiandPni=1i=1.Let1inbesuchthatui()uj()forj=1;:::;n.Theng()ui(). 29
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SinceGGB, Fixg2G.ThensinceXissubconvex,thereexists^x2Xsuchthatr(;^x)g()forall2M.Hence By( 4 )( 4 )andLemma 4.2 ,(;X;R)isstrictlydetermined.Sinceinfg2GBgisuppersemicontinuousonthecompactsetM,thereexists02Msuchthatinfg2GBg(0)=sup2Minfg2GBg():By( 4 )andLemma 4.2 ,0isaleastfavorableaprioridistribution. Ifthecompactnessofisdroppedintheabovetheorem,thentheremaynotbealeastfavorableaprioridistribution.Ifthereexistsnoleastfavorableaprioridistribution,buttheestimationproblem(;X;R)isstrictlydetermined,aminimaxestimatorcanbefoundasaBayessolutionrelativetoaleastfavorablesequenceofaprioridistributions,i.e.,asequencefng2Mthatsatiseslimn!1inf^x2Xr(n;^x)=sup2Minf^x2Xr(;^x).Wearenotgoingtodealwiththequestionofhowsuchasequencecanbeconstructed.ThefollowingtheoremisessentiallyWald'sTheorem3.9in[ 25 ]. 30
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Proof. 25 ,Theorem3.9]applieswithoutanychanges. Wheniscompact,wehavethefollowingversiontoWald'scompleteclasstheorem. Proof. ByLemma 4.1 part1),thereexists02Mandasubsequencefikg2Msuchthatikconvergesweaklyto0.Bythehypothesisofthetheorem,r(;^x)iscontinuousonMforany^x2B.Thusinf^x2Br(;^x)isuppersemicontinuousonM[ 33 ,Proposition1.5.12].ByLemma 4.2 ,inf^x2Xr(;^x)isuppersemicontinuousonM.Sincer(;^x)islowersemicontinuousonM,inf^x2Xr(;^x)r(;^x)isuppersemicontinuousonMand By( 4 )and( 4 ),inf^x2Xr(0;^x)r(0;^x),whence^xisaBayessolutionrelativeto0.
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Proof. 25 ]).Forthesakeofsimplicity,weassumethat(;X;R)isanestimationproblem.TheproofcanbetriviallymodiedtothecasethatXisnotaclassofestimators,butanarbitraryset.Letr(;a)=RR(;a)d,where2Manda2X.ItcanbeveriedthatsinceXisacompactmetrizablespaceandR(;)islowersemicontinuousonXforeach2,fr(;a):a2Xgishalf-closed,i.e.,Xhastheproperty(W).Itcanbeveriedthatsince(X;R)(X;R),Xhastheproperty(W).ByTheorem 4.1 ,thereexistsaminimaxestimator.ByTheorem 4.2 ,thereexistsarestrictedriskBayessolutionrelativeto(;C0)foreach2MandC0V(;X;R).If,inaddition,Condition 4.3 holds,itfollowsfromtheprecedingresults,that(;X;R)isstrictlydetermined,thereexistsaleastfavorableaprioridistribution02M,anyminimaxestimatorisaBayessolutionrelativeto0,andtheclassofBayessolutionsisessentiallycomplete.SupposeCondition 4.4 holdsaswell.Let0denotealeastfavorableaprioridistribution.SinceanyminimaxestimatorisaBayessolutionrelativeto0(Theorem 4.6 ),the(essentiallyunique)Bayessolutionrelativeto0isanessentiallyuniqueminimaxestimator.By[ 26 ,Theorem2.3.1],thisBayessolutionisadmissible. WhilesometimesitispossibletoverifyConditions 4.1 4.4 directly,othertimes,especiallywhenthereisnoclosedfromexpressionforR,itmaybedifculttocheckwhethertheseconditionshold.Inthenextchapter,weconsiderthecasethatRisbasedonalossfunctionL.Inthiscase,itispossibletoderiveconditionsthatcanbemoreeasilycheckedwhenthereisnoclosedformexpressionforR. 33
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WeimposeseveralconditiononthefamilyP,thespaceY,thespaceDofpossibleesti-mates,andthelossfunctionL. 34
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34 ,pp.220].If~gisaversionofPr(AjX),thenaccordingto[ 34 ,pp.221],aconditionalprobabilityoftheeventAgiventhatX=x,g(x),canbeconstructedasfollows:g(X(!))=~g(!)(i.e.,g(x)=~g(!),where!isanelementinsuchthatX(!)=x).SinceinthisworkthedistributionQisunknown,butisknowntobeanelementofP,theconditionalprobabilityofA2FgivenXandtheconditionalprobabilityofAgivenX=xdependonthetruestateofnature2. 1. Foreach!2,Q(jX)(!)isaprobabilitymeasureon(Y;FY). 2. ForeachA2FY,Q(AjX)isaversionoftheconditionalprobabilityPr(AXjX).Fix2.Condition 5.1 furnishesthatthereexistsaregularconditionaldistributionofYgivenXwhenQ=P[ 34 ,Theorem2.7.5].LetP(jX)denotearegularconditionaldistributionofYgivenXwhenQ=P.Foreachx2XandA2FY,wedeneP(Ajx)=P(AjX)(!),where!2issuchthatX(!)=x.WecallP(jx)aregularconditionaldistributionofYgivenX=xwhenQ=P(oristhetruestateofnature).NotethatsinceP(jX)isregular,P(jx)isaprobabilitymeasureforeachx2X.Inaddition,foreachA2FY,P(Ajx)isaversionoftheconditionalprobabilityPr(AXjX=x).LetPYjX=fP(jx):x2X;2g. 5.3 and 5.4 wereaddedtoguaranteethattheintegrationsnecessaryinthecalculationoftheriskfunctionarewelldened.ItcanbeveriedthatifConditions 5.1 5.4 hold,R(;^x)isanextendedreal-valuednonnegativeB()-measurablefunctionandhencethe 35
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30 ].Itisusefultodiscussthepropertiesofthisvectorspace.Thefollowingdiscussionessentiallyappearsin[ 30 ]and[ 31 ].LetCc(D)denotetheclassofcontinuousreal-valuedfunctionsonDwithcompactsupport,andletjjujj=supd2Dju(d)jforu2CC(D).LetL1denotetheBanachspaceofequivalenceclassesofintegrablefunctionson(Y;FY;)withnormjjfjj1=RYjfjd.DenotebyLPYjXthelinearsubspaceofL1spannedbyPYjX.LettheproductspaceCC(D)LPYjXbeequippedwiththenormjj(u;f)jj=jjujj_jjfjj1foru2Cc(D)andf2LPYjX.LetdenotethevectorspaceofboundedlinearfunctionalsonCC(D)LPYjX.Theweak-topologyturnsintoalocallyconvextopologicalvectorspace[ 33 ].Callg1;g22m(FY)PYjX-equivalentifRYjg1g2jfd=0forallf2PYjX.Calltwoestimators^x1;^x22DPYjX-equivalentifforeachD02B(D),^x1(D0j)and^x2(D0j)arePYjX-equivalent.Let~Ddenotetheclassofequivalenceclassessoobtained.Afunctional2issaidtobepositiveifu0andf0imply(u;f)0.AccordingtoLeCamifConditions 5.1 5.5 hold,everypositivelinearfunctionalofnorm1canberepresentedbyan 36
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5.1 5.5 and 5.7 hold.Thenthetopologicalspace(D;J)ismetrizableandR(;)islowersemicontinuousonDforall2.If,inaddition,Discompact,Discompact. Proof. 5.5 ,thespaceCc(D)isseparable[ 30 ].SincebyCondition 5.1 ,Yisaseparablemetricspace,thespaceL1isseparable[ 35 ,pp.92].SinceLPYjXisasubspaceofaseparablenormedspace,itisalsoseparable.ItfollowsthatthespaceCc(D)LPYjXisseparable.ByTheorem[ 36 ,Theorem3.16],if0isweak-compact,then0ismetrizable.ByBanach-AlaogluTheorem[ 33 ,Theorem2.5.2],thesetB=f2:jjjj1gisweak-compact,wherejjjjdenotestheoperatornormof2.ThusBismetrizable.SinceJistherelativetopology,Dismetrizable.LetR(x;;^x)=RYL(x;^x)p(yjx)d.Thenby( 5 ),R(;^x)=RXR(x;;^x)dPX(x).UsingtheresultsofLeCam[ 30 ],itcanbeshownthatR(x;;)islowersemicontinuousonDforeach(;x)2X.SinceDismetrizable,wehavebyFatou'sLemmathatR(;)islowersemicontinuousonDforeach2.SupposeDiscompact.LeCam[ 30 ]showedthataclassAofestimatorsisJ-compactifitisJ-closedandifthefollowingconditionsholds:Foreach>0andeach(;x)2X,thereexistsau2Cc(D)satisfying0u1andRYu(d)d^x(djy)dP(yjx)1uniformlyforall^x2A.TheprecedingconditioncertainlyholdsforDwhenDiscompact.SinceDisJ-closed,itiscompact. 37
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4.2 holdsunderveryweakconditions. Proof. A 5.1 5.7 hold.ThenCondition 4.1 and 4.2 holdfor(;D;R).Asaconsequence,thereexistsaminimaxestimatorandarestrictedriskBayessolutionrelativeto(;C0)foreach2MandC0V(;D;R). Proof. 4.1 holds.LetusshowthatCondition 4.2 holds.SupposeDiscompact.ThenbyLemma 5.1 ,Condition 4.2 holds.SupposeDisnotcompact.Wewillusetheone-pointcompacticationofDtoprovethetheorem.Theideatousetheone-pointcompacticationoftheclassofpossibleestimatestoproveresultsofthetypeofthistheoremseemstoappearrstin[ 30 ].LetDdenotetheone-pointcompacticationofD,andlet12DdenotethepointthatisaddedtoD.LetL:XD!R[f+1gbedenedasfollows:Foreachx2X,L(x;d)=L(x;d)ifd2D,andL(x;1)=supd2DL(x;d).LetDdenotetheclassofallestimatorswithDastheirspaceofpossibleestimates.ByLemma 5.2 ,L2m(FXB(D))andDiscompact.LetL(x;^x)(y)=RDL(x;d)d^x(djy), 38
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5.1 5.5 and 5.7 holdfor(;D;R).ThenbyLemma 5.1 ,DiscompactandR(;)islowersemicontinuousonDforeach2.ItcanalsobeveriedthatR(;a)2m(B())foreacha2X.LetD(D)denotetheclassofestimators^xinDsuchthat^x(Djy)=1forally2Y.Weclaimthat(D(D);R)(D;R).Indeed,x^x02Dsuchthat^x0(1jy)>0forsomey2Y.LetY0=fy2Y:^x0(Djy)=0g.ClearlyY0ismeasurable.Let^x(Ajy)=^x0(A\Djy)=^x0(Djy)fory=2Y0(A2B(D)),andlet^x(jy)beaDiracmeasurewithrespecttoapointd02Dforally2Y0.Itcanbeveriedthat^x(D0jy)isB(Y)-measurableforeachD02B(D).Nowify2Y0,thenL(x;^x)=L(x;d0)L(x;1).Ify=2Y0,thenRDL(x;d)d^x(djy)=RDL(x;d)=^x0(Djy)d^x0(djy)x0(Djy)RDL(x;d)=^x0(Djy)d^x0(djy)+(1^x0(Djy))L(x;1)=RDL(x;d)d^x0(djy).ThusL(x;^x)L(x;^x0).Itfollowsthat^xisasgoodas^x0.Thisprovesthat(D(D);R)(D;R).SinceD(D)D,wehavethat(D(D);R)(D;R).Clearly(D;R)(D(D);R).Thus(D;R)(D;R).ItfollowsthatCondition 4.2 holds. AsaconsequenceofTheorem 5.1 ,undertheratherweakConditions 5.1 5.7 ,thereexistsaminimaxestimatorandarestrictedriskBayessolutionrelativeto(;C0)foreach2MandC0V(;D;R).Inordertogetthestrongerresultswheniscompact,namelythat(;D;R)isstrictlydeterminedandthatthereexistsaleastfavorableaprioridistribution,weneedtoprovethatCondition 4.3 holds.Unfortunately,thisseemstorequireratherstrongconditionsonthelossfunctionandfamilyP.AsetofsuchconditionsiswellknownforthecasethatthelossfunctionisuniformlyboundedandPisdominatedbya-nitemeasure.However,wearemainlyinterestedinthecasethatthelossfunctionisunbounded(e.g.,thequadraticlossfunction).Moreover,inmanycasesPisnotnecessarilydominatedbya-nitemeasure. 39
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37 ,Theorem1.6.4]. 5.3 holds.ThenifConditions 6.1 and 6.2 hold,DD. Proof. 40
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Integrating( 6 )withrespecttoP,wehaveR(;^x)R(;^x0).Since^x02D,DD. SinceDD,weconsiderintherestofthischaptertheestimationproblem(;D;R)insteadoftheestimationproblem(;D;R).Clearly,DissubconvexbyJenseninequality.HenceCondition 4.1 holdsfor(;D;R).SinceDD,DD.HenceitisclearthatCondition 4.2 holdsfor(;D;R)ifitholdsfor(;D;R).ThusifConditions 5.1 5.7 hold,Condition 4.2 holdsfor(;D;R).InLemma 6.2 below,weshowthatunderweakconditions,Condition 4.4 alsoholdsfor(;D;R).InordertoprovethatCondition 4.3 holdsfor(;D;R),itseemsnecessarytomakeadditionalassumptionsonthelossfunctionandthefamilyP.Sometimesitisconvenienttorestricttheclassofestimatorsthatareavailabletotheexperi-mentertotheclassofafneestimators.Anestimator^xissaidtobeafneifitisnonrandomizedandisanafnefunctiononY.SinceweconsideronlythecasethatY=RNyandX=RNx,^xisafneexactlywhen^x=Ay+bforsomeA2RNxNyandb2RNx,andthespaceofpossiblees-timatesDisthenRNx.LetLdenotetheclassofafneestimators.ThespaceLcanbeidentiedwiththespaceRNxNyRNxwhere(A;b)2RNxNyRNxistheestimator^x=Ay+b2Landviceversa.ThusthespaceLcanbeidentiedwithanite-dimensionalvectorspacewiththefollowingadditionandmultiplicationbyascalar:If^x=(A;b),^x0=(A0;b0)andisascalar,^x+^x0=(A+A0;b+b0)and^x=(A;b).LetthespaceLbeequippedwiththenormjj^x^x0jjL=jjAA0jj2+jbb0j,where^x=(A;b)and^x0=(A0;b0).ClearlyLisconvex.ItiseasytoseethatJensen'sinequalityfurnishesthatR(;)isconvexoverLforeach2ifL(x;)isconvexoverDforeachx2X.ThusifL(x;)isconvexoverDforeachx2X,L
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4.1 holdsfor(;L;R).InTheorem 6.1 below,weshowthatunderratherweakconditions,Condition 4.2 holdsfor(;L;R).Asintheestimationproblem(;D;R),itseemsnecessarytomakefurtherassumptionsonthelossfunctionandthefamilyPinordertoprovethatCondition 4.3 holdsfor(;L;R).InLemma 6.2 below,weshowthatunderweakconditions,Condition 4.4 alsoholdsfor(;L;R). 5.3 5.4 5.6 ,and 5.7 hold.ThenCondition 4.2 holdsfor(;L;R).Asaconsequence,thereexistsaminimaxestimatorandarestrictedriskBayessolutionrelativeto(C0;)foreachC0V(;D;R)and2M. Proof. 5.7 ,liminfn!1L(x;^xn(y))L(x;^x(y))foreachy2Y.ByFatou'slemma,liminfn!1R(;^xn)R(;^x)foreach2.ThusR(;)islowersemicontinuousonLforeach2.LetusshowthatforeachsequenceofcompactsubsetsCnofLsuchthat[1n=1Cn=Landeachelement^xn=2Cn(n=1;2;:::),liminfn!1R(;^xn)=sup^x2LR(;^x)forall2.FixasequenceofcompactsubsetsCnofLsuchthat[1n=1Cn=Landasequencef^xng2Lsuchthat^xn=2Cn(n=1;2;:::).Fix2.Certainlyliminfn!1R(;^xn)sup^x2LR(;^x).Thusitislefttoprovethatliminfn!1R(;^xn)sup^x2LR(;^x).ByFatou'slemma, Fix(y;x)2YX.LetDn=f^xn(y):^xn2Cng(n=1;2;:::).WeclaimthatDnisacompactsubsetofD.Indeed,letfdigbeasequenceinDn.Thenthereexistsasequencef^x0iginCnsuchthat^x0i(y)=di.SinceCniscompact,thereexistsasubsequencef^x0ijgofthesequencef^x0igandanelement^x02Cnsuchthat^x0ij!^x0.Thisimpliesthat^x0ij(y)!^x0(y),whencefdijg!^x0(y).Since^x0(y)2Dn,Dniscompact.Weclaimthat[1n=1Dn=D.Fixd2D.Clearlythereexists^x2Lsuchthat^x(y)=d(e.g.,^x=(A;b),whereA=0andb=d).Since[1n=1Cn=C,^x2Cnfornsufcientlylarge.Thisimpliesthatd2Dnfornsufciently 42
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5.6 ,liminfn!1L(x;dn)=supd2DL(x;d),wheredn=^xn(y).Itiseasytoverifythatsupd2DL(x;d)=sup^x2LL(x;^x(y)).Thusforeach^x2Lwehave Sinceyandxarearbitrary, Since^xisarbitrary, SinceLisanitedimensionalnormedspace,itislocallycompact,-compact,andmetrizable.LetLdenotetheone-pointcompacticationofLandlet1denotetheaddedpoint.Foreach2letR(;^x)=R(;^x)if^x2L,andletR(;1)=sup^x2LR(;^x).ByLemma 5.2 ,Liscompactandmetrizable,R(;)islowersemicontinuousonLforeach2,andR(;a)2m(B())foreacha2L.ClearlyLL.Since1istheonlyelementinLnLandR(;1)R(;^x)foreach^x2L,(L;R)(L;R).SinceLL,(L;R)(L;R).ThusCondition 4.2 holds. ThefollowinglemmaisconcernedwiththeessentialuniquenessofBayessolutions. 5.3 holds,andV(;X;R)<+1.ThenifCondition 6.2 holds,aBayessolutionrelativeto2MisanessentiallyuniqueBayessolutionrelativeto. Proof. 43
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Intherestofthischapter,weassumethelossfunctionisquadratic,i.e.,L:(x;d)7!jV(xd)j2,whereV2RNxNx.WeassumeVTV>0.TheextensiontothecaseVTV0isdiscussedlater.InthiscaseitcanbeveriedthatConditions 5.1 5.3 5.5 5.7 ,and 6.1 hold.ThusifDisconvexandCondition 6.2 holds,DD.CertainlyDissubconvexandCondition 4.1 holds.Inaddition,ifConditions 5.2 and 5.4 hold,Conditions 4.2 holdsfor(;D;R).Similarly,Condition 4.1 holdsfor(;L;R),andifConditions 5.2 and 5.4 hold,Condition 4.2 holdsfor(;L;R).Suppose,inaddition,thatPisaGaussianfamilyofdistributions,i.e.,YandXarejointlyGaussian,whenisthetruestateofnature,foreach2.ThenPYandPYjXarealsoGaussianfamiliesofdistributions.SupposethefamilyPYjXisdominatedbytheLebesgue-Borelmeasureon(RNyB(RNy)),whichisdenoted.ItcanbeeasilyveriedthatPYisalsodominatedbyandthatsince,inaddition,foreach2,p(y),thedensityofPwithrespectto,ispositive,themeasuresfPY:2garemutuallyabsolutelycontinuous.ThusCondition 6.2 holds.WewouldliketocheckunderwhatconditionsCondition 4.3 holds.Underthecurrentassumptions,itiswellknownthatthereexistsaregularconditionaldistributionofXgivenY=y,whenisthetruestateofnature.LetP(jy)denotethisconditionaldistribution,whichiswellknowntobeGaussian.Let^x:Y7!RNx;y7!E(Xjy),whereE(Xjy)
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InwhatfollowsBDistheclassofBayessolutionswhen(;D;R)isconsidered,i.e.,foreach^x2BD,thereexists2Msuchthatr(^x;)=inf^x2Dr(^x;).SinceDD,eachelementinBDisalsoaBayessolutionswhen(;D;R)isconsidered.Inwhatfollows,continuityoffunctionsfromintoRNandRNNismeantinthesenseoftheEuclideannormand2-norm,respectively. 4.3 holdsfor(;D;R). Proof. 6 )thatR(;^x)islowersemicontinuousonforeach^x2D. 45
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6 )that ItiseasytoverifythatR(0;^x)<+1.Puth(y)=(c1jyj+c2)2.Since0isarbitraryR(;^x)<+1foreach2.Moreover,foreach2,jV(^x(y)^x(y))j2h(y)andhis 46
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Letfngbesequencethatconvergesto0.Weshowedthatlimn!1jn0j=0.Thus BytheLebesguedominatedconvergencetheorem, Certainlyforeachn>0,jV(^xn(y)^x(y))j2h(y).SincehisP0-integrableand7!p(y)iscontinuousonforeachy2Y,awellknowntheoremonexponentialfamilies[ 37 ,Theorem1.4.1]furnishesthatZYh(y)pn(y)d!ZYh(y)p0(y)d:Itcanbeveriedthatsince7!p(y)iscontinuousonforeachy2Y,theaboveequationimpliesthatZYh(y)jpn(y)p0(y)jd!0:SincejV(^xn(y)^x(y))j2h(y), By( 6 )( 6 ),R(;^x)iscontinuouson.SinceR(;^x)<+1andR(;^x)iscontinuouson,whichisacompactset,R(;^x)isboundedon.ItfollowsthatR(;^x)isboundedandcontinuousonforeach^x2BD. 47
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4.3 holdsfor(;L;R). Proof. 6.2 andisomitted. WhiletheproofsofsomeoftheresultsofthischapterclearlybreakdownifVTVisnotpositivedenite,alltheseresultsarevalidalsointhecasethatVTVisnotpositivedenite.Thereisasimplemethodtoshowthatthisisindeedthecase.NotethatifVTVisnotpositivedenite,L(x;d)=L(x;d0)wheneverdd02N(V).Thuswemaycalldandd0inRNxequivalentifdd02N(V)andchoosethespaceofpossibleestimatestobethesetofequivalenceclassessoobtainedinsteadofRNx.Inthiscase,thespaceofpossibleestimatesisequippedwiththemetric(~d;~d0)=jVyV(dd0)j,where~d;~d02D,disanyelementin~d,andd0isanyelementin~d0.ThischoiceforDisequivalenttochoosingD=N(V)?withtheusualEuclideannormsinceforanyequivalenceclassinDthereisassociatedapointinN(V)?andviceversa.ItcanveriedthatwitheitheroneofthesechoicesforD,theresultsfor(;D;R)arestillvalid.Toshowthattheresultsfor(;L;R)arestillvalid,itisnecessary,todeneaclassL0=f(N(V)?A;N(V)?b):(A;b)2Lg.ItisobviousthatL0L.Certainlyforeach^x2L0,^x(y)isinthenewspaceofpossibleestimates.Now,itisstraightforwardtoshowthatalltheresultsofthischapterarestillvalidfor(;L0;R)andhencefor(;L;R). 48
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9.1 ,ifConditions 4.1 4.4 holdandiscompact,theproblemofndingaleastfavorableaprioridistributionisdualtotheproblemofndingaminimaxestimator.Thusintherestofthischapter,weconcerntheproblemofndingaleastfavorableaprioridistribution.Thefollowingtheoremisessentiallysimilartoatheoremin[ 18 ]andtheiterativealgorithmproposedin[ 38 ]. 7.1 and 7.2 hold.Constructasequencefig1i=12~Masfollows.Let1beanydistributionin~M.Havingchosen1;:::;i2~M,leti2besuchthatR(i;^xi)=sup2R(;^xi).Leti;=i+(1)i.Letibesuchthat~r(i;i)=sup2[0;1]~r(i;)andleti+1=i;i.Thenthesequencefigconvergesweaklytoaleastfavorableaprioridistribution. Proof. 18 ,Theorem2.3],withslightmodications. ThemaindifcultyinthealgorithmdescribedinTheorem 7.1 isinndingi2suchthatR(i;^xi)=sup2R(;^xi)fori1.Anotherdifcultyisduetothefactthat,ingeneral,sincethesupportofimaygrowasigrows,thecomplexityofthealgorithmcalculationsalsogrowswithi.Theproblemofndingisuchthat~r(i;i)=sup2[0;1]~r(i;)isaddressedbelowandcanbesolvednumerically.Insomespecialcases,itiseasytondi2suchthatR(i;^xi)=sup2R(;^xi)fori1andthecomplexityofthealgorithmcalculationsremain 49
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7.1 ispracticalinndingminimaxestimators.Onesuchcaseiswhenisaniteset.Inthesequel,weshowthatwhenweconsiderlinearestimationwithquadraticlossfunction,thealgorithmofTheorem 7.1 canbeoftenusedtondaminimaxestimator.Inthemoregeneralcase,thisalgorithmcanbeusedjusttond-minimaxestimators,whicharegoodapproximationstominimaxestimatorsforsufcientlysmall.Wediscussthederivationof-minimaxestimators,ingreatdetail,inthenextchapter.Theproblemofndingi2[0;1]suchthat~r(i;i)=sup2[0;1]~r(i;)isastandardoptimizationprobleminR:maximize~r(i+(1)i)subjectto2[0;1]:Fix1;22~M,andlet=1+(1)2for2[0;1].Intherestofthischapterweconsiderthefollowingoptimizationproblem,whichincludesthepreviousoneasaspecialcase: maximize~r()subjectto2[0;1]: Proof. LetD()=r(1;^x)r(2;^x)for2[0;1]. Proof. Asaconsequenceofthislemma,D(0)isthesubderivativeof7!~r()atthepoint02(0;1). 7.2 holds.Then7!D()ismonotonicallydecreasingandcontinuouson[0;1]and~r(0)=sup2[0;1]~r()for02[0;1]ifD(0)=0.
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7.2 furnishesthecontinuityof7!D()on[0;1].SinceD()isthesubderivativeoftheconvexfunction7!~r(),7!D()ismonotonicallydecreasingon(0;1).Bycontinuity,7!D()ismonotonicallydecreasingon[0;1].Fix02[0;1].SupposeD(0)=0.Thenbylemma 7.2 ,~r()~r(0)forall2[0;1].Itfollowsthat~r(0)=sup2[0;1]~r(). 7.2 and 7.3 implythatwecanuserelativelysimplenumericalalgorithmstosolve( 7 )numerically.Supposer(1;^x1)=r(2;^x1).Then~r(1)=r(;^x1)forall2[0;1].Itfollowsthat~r(1)~r()forall2[0;1]andhence=1isasolutionof( 7 ).Similarly,ifr(1;^x2)=r(2;^x2),=0isasolutionof( 7 ).Usingthefactthat7!D()iscontinuousandmonotonicallydecreasingon[0;1],andthefactthat2[0;1]isasolutionof( 7 )ifD()=0,theoptimizationproblem( 7 )canbesolvedasfollows:IfD(1)0,then=1isasolutionof( 7 ).IfD(0)0,then=0isasolutionof( 7 ).Finally,ifD(0)>0andD(1)<0,thereexists2(0;1)suchthatD()=0andhencesolves( 7 ).Inthiscase,sinceD()=0and7!D()ismonotonicallydecreasingon(0;1),canbeeasilyfoundusingthebisectionmethod.
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7.1 .Inpractice,itmaybenecessarytondan0-minimaxestimatorfor(f;X;R)suchthatthisestimatorisan-minimaxestimatorfor(;X;R).Inthelattercase,thealgorithminTheorem 7.1 canstillbeused,butitisnecessarytohaveaconditionthatenablesustocheckwhentherequiredprecisionisachieved.Checkingifanestimator^x,anessentiallyuniqueBayessolutionrelativeto,isan-minimaxestimatorcanbedoneusingthefollowinglemma: 7.1 holdsandsup2R(;^x)~r(),then^xisan-minimaxestimator. Proof. TheconditionofLemma 8.1 isonlyasufcientconditionforan-minimaxestimator.However,undercertainconditions,ifthenumericalalgorithmconvergestoaleastfavorableaprioridistribution,thenthereexistsanintegerNsuchthatLemma 8.1 issatisedfortheNthiteration. 7.1 and 7.2 hold.Letfig2Mbeasequencethatconvergesweaklytoaleastfavorableaprioridistribution02M.ThenthereexistsanintegerNsuchthatsup2R(;^xi)~r(i)foralliN.
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7.2 holds,itisstraightforwardthat SinceR(;^x0)iscontinuouson,itcanbeshown,asintheproofof[ 38 ,Theorem2part3],that ByTheorems 4.1 and 4.5 ,thereexistsaminimaxestimatorand(;X;R)isstrictlydetermined.ByTheorem 4.6 ,^x0isaminimaxestimator.Thus By( 8 )( 8 ),sup2R(;^xi)~r(i)!0andthelemmaisproved. Thefollowingconditionisneededforsomeoftheresultsofthischapter. 7.1 holds,andthefamilyfR(;^x):2Mgisequicontinuouson.Thenforany>0,thereexistsa>0suchthatforanynite-densesubsetfof,thefollowinghold: 1. Forany2M,thereexistsaprobabilitymeasure02fsuchthatR(;^x0)R(;^x)
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8.1 thatthefamilyfR(;^x):2Mgisequicontinuousonisratherstrong.Whileitisoftensatisedinthecasethattheriskisbasedonalossfunctionthatisuniformlybounded,itmaynotbesatisedinthecasethatthelossfunctionisunbounded.Another,difcultywithTheorem 8.1 isthattherequirementfromthesetfisverystrong.AsetfthatisconstructedaccordingtothistheoremhasthepropertythatanyestimatorcanbeapproximatedbyaBayessolutionrelativetoameasurewithsupportinfwithdegradationofnomorethan.However,wearemainlyinterestedinapproximatingaminimaxestimatorandnotanyestimator.ThusitmaybepossibletochooseanitesetwhosecardinalityissignicantlysmallerthanfinTheorem 8.1 .Duetotheabove,wearenotgoingtouseTheorem 8.1 inthesequel,andwearegoingtoderivemethodsthatdonotrequireequicontinuity.ItisconvenienttousethenotionofFrechetdifferentiability.LetB(X;Y)denotethesetofboundedlinearoperatorsfromanormedlinearspaceXtoanormedlinearspaceY.Givenf2B(X;Y),jjfjjdenotestheoperatornormoff,i.e.,jjfjj=supx:jjxjj1jjfxjj. jjhjj=0:WecallAtheFrechetderivativeoffatxanddenoteitbyDf(x).Iff:X!YistwiceFrechetdifferentiableatx2X,weletD2f(x)denotethesecondFrechetderivativeoffatx.NotethatD2f(x)2B(X;B(X;Y)).Given^x2X,letDR^x()andD2R^x()denotetheFrechetderivativeandthesecondFrechetderivative,respectively,ofR(;^x)at.
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7.1 and 8.1 holdandiscompact.Supposeforany2~M,thereexistsanextensionofR(;^x)fromtoanopenconvexsetUCsuchthatR(;^x)isFrechetdifferentiableonUandthereexistsarealnumberMsuchthatjjDR^x()jjMsup2R(;^x)forall2~Mand2.Fix>0and0<0<,andlet=0 Proof. ThecardinalityofanitesubsetfofthatisconstructedaccordingtoLemma 8.3 maystillbeverylarge;thismaycausethecalculationofaleastfavorableaprioridistributionfor(f;X;R)tobeformidable.Inaspecialbutveryimportantcaseofcompact,itispossibletoderiveanite-densesubsetofsuchthatdependslinearlyonq 8.3
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36 ,Theorem3.20],thesetco(@)iscompactandhencetheKrein-Milmantheorem[ 36 ,Theorem3.23]impliesthat=co(@). Proof. LetSN=fa2RN:jjajj1=1g.
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8 )isfalse.Thenthereexists02RN+\SNandanonemptyindexsetJ06=Jsuchthat=PNi=10(i)i,0(i)>0foralli2J0,and0(i)=0foralli=2J0.Let0=(+0)=2.Then=PNi=10(i)i,0(i)>0foralli2J[J0,and0(i)=0foralli=2J[J0.SinceJ06=J,jJ[J0j>M,whichisacontradictiontothedenitionofM.Hence( 8 )musthold. 7.1 and 8.1 hold,isacompactconvexsubsetofthenormedspaceC,and@isaniteset.Supposeforeach2~M,thereexistsanextensionofR(;^x)fromtoanopenconvexsetUCsuchthatR(;^x)istwiceFrechetdifferentiableonUandthereexistsapositiverealnumberMsuchthatjjD2R^x()jjMsup2R(;^x)forall2~Mand2.Let>0and0<0<.Let=q Proof. 8.5 ,thereexistsanindexsetJf1;2;:::;Ngsuchthat( 8 )holdsfor.LetA=fi:i2Jg.Sinceisnotinf,isnotanextremepoint(Lemma 8.4 )andjJj>1.Let02f\E(co(A))besuchthat(;0)<.Then0=Pi2J(i)iforsome2RN+\SNsuchthat(i)=0fori=2J.Since=Pi2J(i)iforsome2RN+\SNsuchthat(i)>0
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2jj0jj2Csup2[0;1]jjD2R^x0(0+(1))jj:Bythehypothesisofthetheorem,jj0jj2Csup2[0;1]jjD2R^x0(0+(1)jj2Msup2R(;^x0):Thussup2R(;^x0)sup2fR(;^x0)R(;^x0)R(0;^x0)2M=2sup2R(;^x0).Itcanbeveriedthatsincesup2fR(;^x0)V(f;X;R)(1+0)andV(f;X;R)V(;X;R),thensup2R(;^x0)V(;X;R)(1+).Thus^x0isa-optimalestimator. SupposethenormedspaceCinCondition 8.1 isRNwithitsusualnorm.Inthiscase,ifisacompactconvexsetand@isnite,issaidtobeaconvexpolytopeinRN. 8.6 7.1 and 8.1 hold,andisaconvexpolytopeinRN.Supposeforeach2~M,thereexistsanextensionofR(;^x)fromtoanopenconvexsetUCsuchthatR(;^x)istwiceFrechet
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Proof. 8.6 ,let02Mbesuchthatthesupportof0iscontainedinf,x0isa(0;f)-optimalestimator,and2besuchthatsup2R(;^x)=R(;^x).Then2E(co(A))forsomeA22@.Let02f\E(co(A))besuchthatj(0)vij(i).BytheTaylortheorem,wehavethatforsome~2inthelinesegmentconnectingand0R(0;^x0)=R(;^x0)+1 2(0)TD2R^x0(~)(0):Certainlythereexistsa1;a2;:::;aN2Rsuchthat0=PNi=1aivi.Bythehypothesisofthetheorem,(0)TD2R^x0(~)(0)(0)TA0(0)sup2R(;^x0)=NXi=1a2iisup2R(;^x0):Sinceai=(0)vi,a2i(i)2.Thussup2R(;^x0)sup2fR(;^x0)R(;^x0)R(0;^x0)PNi=1(i)2i=2sup2R(;^x0).Itcanbeveriedthatsincesup2fR(;^x0)V(f;X;R)(1+0)andV(f;X;R)V(;X;R),thensup2R(;^x0)V(;X;R)(1+).Thus^x0isa-optimalestimator. 60
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9.1 ,appearsasaconjecturein[ 16 ].Weimposethefollowingadditionalconditionsinordertoderivetheresultsofthischapter: C0<+1.LetR(;^x)=r(;^x)+(1)R(;^x)andr(;^x)=RR(;^x)d,where01.LetK(^x;)=sup2R(;^x)andletK()=inf^x2XK(^x;).Sincesup2R(;^x)r(;^x),wehavethat 9.1 and 9.2 hold.ThenKisconcave,decreasing,andcontinuouson[0;1]. Proof. B LetG(^x)=sup2R(;^x)r(;^x). Proof. C 61
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Proof. 4.1 and 4.2 holdfor(;X;R),thereexistsaminimaxestimatorintheestimationproblem(;X;R).If,inaddition,iscompactandCondition 4.3 holds,thereexistsaleastfavorableaprioridistribution02M,andanyminimaxestimatorisaBayessolutionrelativeto0.AminimaxestimatorisanessentiallyuniqueminimaxestimatorandisadmissibleifCondition 4.4 holdsaswell.Infact,itcanbeshownthatif0<1,eachoneofConditions 4.1 4.4 holdsfor(;X;R)ifitholdsfor(;X;R). 7 fortheestimationproblem(;X;R),where0<1,providedthatConditions 7.1 7.2 holdfortheestimationproblem(;X;R).Infact,ifiscompact,itissufcientthatCondition 7.1 7.2 wouldholdfortheestimationproblem(;X;R).Indeed,supposeiscompactandConditions 7.1 7.2 holdfortheestimationproblem(;X;R).ItfollowseasilyfromLemma 9.3 thatforany2M,thereexistsanessentiallyuniqueBayessolutionrelativetointheestimationproblem(;X;R),whenceCondition 7.1 holds.Let^xdenotethe(essentiallyunique)Bayessolutionrelativetointheestimationproblem(;X;R).Letfig1i=1beasequenceinMthatconvergesweaklyto02M.Let0i=+(1)iforalli0.Thenf0igconvergesweaklyto00.SinceR(;^xi)=R(;^x0i)foralli0,R(;^xi)convergestoR(;^x0)uniformlyonthecompactset.Sincetheconvergenceisuniform,r(;^xi)convergestor(;^x0).ThusR(;^xi)convergestoR(;^x0)uniformlyoncompactsubsetsofandCondition 7.2 holds. 9.1 9.2 hold.LetQ()=+,whereG(^x)and2R.SupposeQ()>K()forall2[0;1].
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Proof. 9.1 ,Kisalsocontinuouson[0;1].Thusthereexists>0suchQ()K()>forall2[0;1].ItiseasytoverifythatfK(^x;):^x2X( Leti=i=nand^xibeaminimaxestimatorfor(;X;Ri)fori=0;1;:::;n.Notethat^xn=^xand^xiexistsforall0in1byTheorem 4.1 andRemark 9.1 .Since^x0isaminimaxestimatorfor(;X;R)and^xn=^x,G(^xn)G(^x0).ThusbyLemma 9.2 part1),thereexists1mnsuchthatG(^xm)G(^xm1).ByLemma 9.2 part3),^xm12X( Using( 9 ), By( 9 )and( 9 ),jK(m)~K(m)j<.Itfollows~K(m)
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9.1 9.2 hold,andV(;X;R)K(1).ThusQ()>K()forall01.Let=C0r(;^x0).ThenG(^x)andQ()=+C0.Let^xbeaminimaxestimatorfor(;X;R).Since^x0isarestrictedriskBayessolutionrelativeto(;C0)andC0>V(;X;R),r(;^x0)r(;^x).Thus>G(^x).ByLemma 9.4 ,thereexists^x02XsuchthatK(^x0;)K()forall01.Sincesup2R(;^x0)=C0,K(^x0;)=Q()andbyLemma 9.4 ,thereexists^x0suchthatsup2R(;^x0)
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9.1 9.2 hold.Letfng1i=1beasequencein[0;1]thatconvergesto02(0;1).Let^xibeaminimaxestimatorfor(;X;Ri)fori=0;1;:::.ThenifV(;X;R)0suchthatM>V(;X;R)andM+< C0.ByTheorem 4.2 ,thereexistrestrictedriskBayessolutionsrelativeto(;M)and(;M+).Therefore,Theorem 9.1 yields0and00in(0;1)suchthatsup2R(;^x0)=Mandsup2R(;^x00)=M+,where^xisaminimaxestimatorfor(;X;R)forall2(0;1).ByLemma 9.2 part3),0000.Infact,sincesup2R(;^x0)
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Sinceforeach^x2X,sup2R(;^x)sup2R(;^x), By( 10 )and( 10 ),^xisaminimaxestimatorforx.Thuswiththeaboveriskfunction,minimaxestimationforxsubjectto(Sx;Sy;h)iscompletelydeterminedbyminimaxestimationforx(2Sx).Letusconsideranotherpossibilityfortheriskfunction.SupposeSxN.Letw:Sx![0+1).WecandenetheriskfunctionR(;^x)=X2SxR(;^x)w():
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1. IfConditions 4.1 and 4.2 holdfor(;X;R)foreach2Sx,thenConditions 4.1 and 4.2 holdfor(;X;R). 2. SupposeSxisnite.ThenifConditions 4.3 and 4.4 holdfor(;X;R)foreach2Sx,thenConditions 4.3 and 4.4 holdfor(;X;R). Proof. 4.1 holdsfor(;X;R).Foreach2SxletXbeacompactmetrizablespaceandRafunctionfromXinto[0;+1]suchthat(X;R)(X;R),R(;)islowersemicontinuousonXforeach2,andR(;a)2m(B())foreacha2X.LetX=Q2SxXbeequippedwiththeproducttopology.LetR(;^x)=P2SxR(;^x)w().ThenbyTychonoff'sTheorem,Xiscompact.Inaddition,Xismetrizablesinceitisacountableproductofmetrizablespaces.Sinceforeach2Sxand2,^x7!R(;^x)islowersemicontinuous,foreachnitesubsetSofSx,^x7!P2SR(;^x)w()islowersemicontinuous.Sincethepointwisesupremumofanycollectionoflowersemicontinuousfunctionisalowersemicontinuousfunction,wehavethatR(;)islowersemicontinuousonXforeach2.Sincethepointwiselimitofasequenceofmeasurablefunctionsisameasurablefunction,itiseasytoshowthatR(;^x)2m(B()).Itisalsoratherstraightforwardtoshowthat(X;R)(X;R).ThusCondition 4.2 holdsfor(;X;R).2)Theproofisratherstraightforwardandisomitted. 68
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10.1 impliesthatunderratherweakconditionson(;X;R)(2Sx),thereexistsaminimaxestimatorfor(;X;R)andarestrictedriskBayessolutionrelativeto(;C0)foreach2MandC0V(;X;R).ThetheoremisespeciallyusefulforthecaseofniteSxsinceitspeciesthatifcertainconditionsholdsforeachoneoftheestimationproblems(;X;R)(2Sx),theyholdfor(;X;R)andhencetheresultsofthepreviouschaptersarevalidfor(;X;R).
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10.3 72
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10.3 .Forthesakeofclarity,letusrepeattheformulationofthisproblem.Weconsiderthefollowingdiscrete-timelinearstochasticsysteminstate-spaceform: wherexn2RNx(n0)isthestatevector,yn2RNyisthesystemoutput,vn2RNyisthemeasurementnoise,wn2RNxisthemodelnoise,andHnandFnarematricesinRNyNxandRNxNx,respectively.Weassumethatthereisgivenanestimationspace(Sx;Sy;h),whereSxN,Sy=N,andforeach2Sx,h()isanitesubsetofSx.WeassumethatfHngandfFngareknownsequences,thesequencesfvngandfwngareuncorrelated,vnandvmareuncorrelatedforn6=m,andwnandwmareuncorrelatedforn6=m.Inaddition,theinitialstatex0andvn(wn)areuncorrelatedforalln0.Weassumex0isaGaussianrandomvectorandfvng(fwng)isasequenceofidenticalGaussianrandomvectors.Weassumethatthemeanandcovarianceofx0andthecovariancesofvnandwnareunknown;itisonlyknownthateachoneofthesequantitiesbelongtoacertainset.Thespaceofstatesofnature,inthiscase,is122,where1istheclassofpossiblemeanvectorandautocorrelationmatrixpairsforx0,2istheclassofpossiblemeanvectorandautocorrelationmatrixpairsforvn(n=1;2;:::),and3istheclassofpossiblemeanvectorandautocorrelationmatrixpairsforwn(n=1;2;:::).AsmentionedinExample 10.3 ,thesetisasubsetofanitedimensionalspacenormedspaceC(seethedenitionofCinExample 10.3 ).For=123,wherei2ifori=1;2;3,weleti()andi()denotethemeanvectorandautocorrelationmatrixofi,respectively,fori=1;2;3.Ourmainassumptionregardingthespaceisthat2()>0foreach2.WeconsidertheriskfunctionR(;^x)=P2SxR(;^x)w()forthisproblem,wherew()>0foreach2Sx, 73
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24 ],wheresuchconditionsarederivedforthespecialcaseofuncertaintiesintheinitialstate.LetusshowthatConditions 4.1 4.2 ,and 4.4 holdfortheestimationproblems(;D;R)and(;L;R)foreach2Sx,andthatCondition 4.3 holdsaswellifiscompact.Fix2Sx.OurrststepistoshowthatConditions 5.1 5.7 hold.SincethelossfunctionLisquadratic,itisonlylefttoshowthatConditions 5.2 and 5.4 hold(Chapter 6 ).ConsidertheconditionaldistributionofYh()givenX=x,whenisthetruestateofnature.LetP(jX=x)denotethisconditionaldistribution,whichiscertainlyGaussian.Moreover,since2()>0foreach2,thefamilyfP(jX=x):2;x2XgisdominatedbytheLebesgue-Borelmeasureon(Yh();B(Yh())),whichisdenoted.LetP(jX=x)denotetheconditionaldistributionofYh()givenX=x,whenisthetruestateofnature.LetA2B(Yh())besuchthat(A)=0.ThenP(AjX=x)=0foreachx2X.ItcanbeveriedthatthisimpliesthatP(AjX=x)=0foreachx2X.ThusthefamilyfP(jX=x):2;x2XgisdominatedbyandCondition 5.2 holds.PutZ=[YTh()XT]T.Letfngbeasequenceinthatconvergesto02.Theni(n)!i(0)andi(n)!i(0),fori=1;2;3.ItcanbeveriedthatthisimpliesthatEn(Z)!E0(Z)andEn(ZZT)!E0(ZZT).ItfollowsthatPnconvergesweaklytoP0.Sincethesequencefngisarbitrary,fPngconvergesweakly 74
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40 ].ThusCondition 5.4 holds.Itisalsoclearthat7!E(Z)and7!E(ZZT)arecontinuouson.ItnowfollowsfromtheresultsofChapter 6 thatConditions 4.1 4.2 ,and 4.4 holdfor(;D;R)and(;L;R)foreach2Sx,andCondition 4.3 holdsaswellifiscompact.ByTheorem 10.1 ,whether(;D;R)or(;L;R)isconsidered,thereexistsaminimaxestimator,andthereexistsarestrictedriskBayessolutionrelativeto(;C0)foreach2MandC0V(;D;R).SupposeinadditionthatSxisniteandiscompact.Thenwehavethefollowingresultsfortheestimationproblem(;D;R):(;D;R)isstrictlydetermined,thereexistsaleastfavorableaprioridistribution02Mandaconditionalmeanestimatorrelativeto0isanessentiallyuniqueadmissibleminimaxestimator.Moreover,theclassofconditionalmeanestimatorsisessentiallycomplete.Notethat,ingeneral,aconditionalmeanestimatorrelativetoisnotaLMMSEestimatorsincemaynotassignmass1toasinglepointin.Similarly,wehavethefollowingresultsfortheestimationproblem(;L;R):(;L;R)isstrictlydetermined,thereexistsaleastfavorableaprioridistribution02MandaLMMSEestimatorrelativeto0isanessentiallyuniqueadmissibleminimaxestimator.Considerthelteringproblem.ThenaLMMSEestimatorrelativetoisnotnecessarilyaKFifdoesnotassignmass1toasinglepointin.ThereisaspecialandimportantcaseinwhichaLMMSEestimatorwithrespectto2MisaKF.Wewilltreatthiscaseinthesequel.Inthefollowingchaptersweconsidertwospecialcasesoftheaboveproblem.Therstcaseisthecaseofstochasticuncertaintiesintheinitialstate,modelnoise,andobservationnoisewithLastheclassofavailableestimators.ThesecondcaseisthecaseofdeterministicuncertaintiesintheinitialstatewithDastheclassofavailableestimators.InbothcaseswewillassumethatSxisniteandiscompact.NotethatbyRemark 4.1 ,thecasethatisbounded,butnotnecessarilycompactisalsocovered.Thesetwocasesareimportantontheirownmerit,andtheywillalsobeusedtoillustratesomeofthegeneralresultsofpreviouschapters.Forexample,therstcasewillbeusedtoillustratethemethodproposedinChapter 9 toderivearestrictedriskBayessolution 75
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9.3 )andthesecondcasewillbeusedtoillustratethemethodproposedinChapter 8 toderiveanapproximationforminimaxestimators. 76
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11 .Weconsiderthecaseofstateestimationwithstochasticuncertaintiesintheinitialstate,modelnoise,andobservationnoisewiththeclassofavailableestimatorsbeingtheclassofafneestimators.Throughoutthischapter,weassumeiscompact.ThusV(;L;R)<+1.Inthischapter,weassumethattheestimationspace(Sx;Sy;h)issuchthatSx=f0;1;:::;ntg,Sy=N,andh(k)=f0;1;:::;nkgforeachk2Sx,wherentandnk,fork=0;1;:::;nt,arenonnegativeintegers.Weassumethatw(k)=1foreachk2Sx,i.e.,R(;^x)=Pntk=0Rk(;^xk).GivenC0V(;L;R)and2,ourgoalistondarestrictedriskBayessolutionrelativeto(;C0). wherewn0=[wT0wT1wTn]Tandvn0=[vT0vT1vTn]T.Letk0andn0.By( 12 )and( 12 ),Ayn0+bxk=(AOnFk1;0)x0+n_k1Xi=0(ATi+1nFk1;i+1)wi+Avn0+b:
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Let()=[1()T2()T3()T]Tandlet()betheblockdiagonalmatrixwith1(),2(),and3()initsdiagonalblocks.Let;i()=i()i()i()Tfori=1;2;3.Weassumethatsomeaprioriinformationregardingthetruestateofnature2isavailable.Thecasethatnoaprioriinformationisavailableisanimportantspecialcase.Theaprioriinformationisgivenintheformofanominal2.Forsimplicity,weassume1()=0.Thereisnolossofgeneralityinthisassumptionsinceif1()6=0,wecantranslatethestatesandobservationsandbringtheproblemtothisform.Below,wesummarizetheassumptionstakensofartogetherwithsomenewassumptions. 78
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12.1 12.4 holdinthesequel.NotethatisthenacompactconvexsubsetofC.By( 12 ),Rk(;^xk)isbothconvexandconcaveonforall^xk2Lk,i.e.,forany0<<1and0;002,Rk(0+(1)00;^xk)=Rk(0;^xk)+(1)Rk(00;^xk):Inaddition,Rk(;^xk)iscontinuousonforall^xk2Lk.ItfollowsthatR(;^x)isbothconvexandconcaveonandcontinuousonforall^x2L.Letr(;^x)=RR(;^x)dandletrk(;^xk)=RRk(;^xk)dfork=0;1;:::;nt.LetZbetheclassofallnitesubsetsof.Let~MdenotethespaceofdistributionsinMwithnitesupport.Let2~M.ThenthereexistsZ2Zand1;:::;jZj2suchthatZ=f1;:::;jZjgand(Z)=1.Inthiscase,let(i)denotethemassthatassignstothepointifori=1;:::;jZj.Finally,let=PjZji=1i(i).Sinceisconvex,2.SinceR(;^x)isbothconvexandconcaveonforall^x2L,r(;^x)=R(;^x)forall^x2L. Proof. 27 ,Appendix3].Letfigbeasequenceofdistributionsin~Mthatconvergesweaklyto.SinceR(;^x)iscontinuousonthecompactset,R(;^x)isboundedand 79
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SinceR(;^x)isbothconvexandconcaveonandihasnitesupport, By( 12 )( 12 ),r(;^x)=R(;^x).Theprooffollowsfromthearbitrarinessof^x. ItisclearfromtheproofofLemma 12.1 thatif2~M,^xisaBayessolutionrelativetoifandonlyif^xisaBayessolutionrelativeto,where=.WewanttoapplytheresultsofChapter 9 fortheestimationproblem(;L;R).WehavealreadyshownthatConditions 4.1 4.4 holdinChapter 11 .LetusshowthatConditions 7.1 7.2 ,and 9.1 holdaswell.InChapter 11 itwasshownthatLhastheproperty(W)andCondition 4.4 holds.ThusTheorem 4.2 impliesthatCondition 7.1 holds.CertainlytheweakerCondition 9.1 mustholdaswell.ItislefttoprovethatCondition 7.2 holdsfor(;L;R).Infact,itissufcienttoprovethatCondition 7.2 holdsfor(;Lk;Rk)foreachk2Sx.Fixk2Sx.Forthesakeofclarity,weprovethatCondition 7.2 holdsfor(;Lk;Rk)inthecasethat1()=0forall2.ItiseasytoverifythatthisistruealsointhemoregeneralcaseofAssumption 12.2 buttheexpressionsarerathercumbersome.Assuming1()=0forall2andusing( 12 ), 80
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whereweextendk;ifromto~Mbydeningk;i()=k;i()forall2~M(i=1;2).LetfigbeasequenceinMthatconvergesweaklyto02M.ByLemma 12.1 ,thereexistsasequencefig2andanelement02suchthatRk(;^xi;k)=Rk(;^xi;k)(i=0;1;:::).Certainlyk;iiscontinuousonfori=1;2.Itfollowseasilyfrom( 12 )thatthemapping7!^x;kiscontinuousonthecompactsetandhenceuniformlycontinuousandbounded.Thusthereexists>0suchthat^x;k2Bforall2,whereB=f^xk2Lk:jj^xkjjLkg.ItiseasytoseethatRkiscontinuousonthecompactsetB.ThusfRk(;):2gisequicontinuousonB.Since7!^x;kisuniformlycontinuouson,thefamilyf7!Rk(0;^x;k):02gisequicontinuousonandthereforeRk(;^xi;k)convergestoRk(;^x0;k)uniformlyon.ItfollowsthatRk(;^xi;k)convergestoRk(;^x0;k)uniformlyonandCondition 7.2 holds.Let^xdenotethe(essentiallyunique)Bayessolutionrelativeto2Mintheestimationproblem(;L;R).SinceiscompactandR(;^x)iscontinuouson, 9.2 holdsifV(;L;R)< C0.Let^x=^x+(1).ByLemma 9.3 ,^xisaBayessolutionrelativetointheestimationproblem(;L;R).Letr(;^x)=r(;^x)+(1
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4.2 4.4 7.1 ,and 7.2 holdfortheestimationproblem(;L;R),theyholdfortheestimationproblem(;L;R)for0<1(Remarks 9.1 and 9.2 ).ThususingtheresultsofthepreviouschaptersandLemma 12.1 ,if0<1,thereexists02suchthat~r(0)=sup2~r()and^x0isanadmissible,essentiallyuniqueminimaxestimator.Inparticular,bysetting=0,thereexistsaminimaxestimatorfor(;L;R).Letusconsidertheproblemofndingaminimaxestimatorfortheestimationproblem(;L;R),where2[0;1).Asmentionedearlier,ifV(;L;R)
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Proof. 2(+0).Then20.ByLemma 7.1 ,~r()~r().Thusthereexistsaleastfavorableaprioridistribution0thatisin0.Certainly0isalsoaleastfavorableaprioridistributionintheestimationproblem(0;L;R).Since0<1,R(;^x)=sup2R(;^x)ifandonlyifR(;^x)=sup2R(;^x).Since0isacompactsubsetof,wemayapplyTheorem 7.1 fortheestimationproblem(0;L;R)andderivetheabovealgorithm.NotethatthealgorithmissimpliedwiththehelpofLemma 12.1 sinceweneedtoconsideronlydistributionswithsupportofasinglepoint,i.e.,elementsof.Sincewehaveconsideredtheestimationproblem(0;L;R),thesequencefig1i=1isin0.ByTheorem 7.1 ,thesequenceconvergesweaklytoaleastfavorableaprioridistribution020.Thus^x0isaminimaxestimator.Since(i)=0and()=0,^xi2L0foralli0.SinceCondition 7.2 holds,R(;^xi)convergesuniformlyontoR(;^x0). ThemainstepsofthealgorithmofLemma 12.2 areSteps2and4.InStep2,weneedtosolvetheproblemofndingamaximizerofR(;^x)over0.WewilladdressthisprobleminSection 12.2 .Step4canbedoneusingnumericalmethodsasdescribedinRemark 7.1 .SupposeV(;L;R)
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12.2 ,weaddressthisproblemandareabletosolveitforsomeimportantcasesof. 12.2 andinthemethodtondarestrictedriskBayessolution,whichisdiscussedinRemark 9.3 .Weconsiderspeciccasesoftheparameterset0.Thereisoneimmediatecaseinwhichthisproblemhasasimplesolution.Thedenitionofanextremepointisneeded(Denition 21 ).RecallthatgivenasetA,weuse@AtodenotetheextremalboundaryofA,whichisthesetofallextremepointsofA.Since0isconvexandcompactandR(;^x)isconvexandcontinuouson,byBauer'sminimumprinciple[ 41 ],sup20R(;^x)=sup2@0R(;^x):If@0isnite,sup20R(;^x)=max2@0R(;^x).Thussup20R(;^x)canbeeasilycalculated.Fix^x2L0.By( 12 ),R(;^x)=P3i=1tr(i()i)forsome12SNx+,22SNy+,and32SNx+.Leti=fi():2gfori=1;2;3.Let=f(1();2();3()):2g.Itfollowsfromthedenitionofthat=123.Thussup20R(;^x)=sup(1;2;3)23Xi=1tr(ii)=3Xi=1supi2itr(ii):Therefore,weareleftwiththefollowingoptimizationproblem:GivenN>0,amatrix2SN+,andaconvexcompactsubsetAofSN+, maximizetr(A)subjecttoA2A: ConsidertheimportantcaseinwhichA=fA2SN+:f1(A)0;:::;fN(A)0g,wheref1;:::;fNareconvex(real-valued)functionssuchthatAiscompactandconvex.Then( 12 )is 84
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wheren()=(;^xn()),njn1()=(;^xnjn1()),andKn()denotestheKalmangain.Considerthelteringproblemspeciedbytheestimationspace(Sx;Sy;h),whereSx=f0;:::;ntg,Sy=N,andh(i)=f0;:::;igforeachi2Sx.Then^x=(^x0();^x1();:::;^xnt())andisreferredtoastheKFrelativeto.Since2()>0,theexistenceoftheKFisguaranteed[ 44 ].ItiseasytoverifythatRk(;^xk(0))=tr(Wk(;^xk(0))).Usingtheresultsoftheprevioussections,thereexistsa2suchthattheKFrelativetoisarestrictedriskBayessolutionrelativeto(;C0)forallC0V(;L;R).Infact,thereisanotherinterestingpropertyregardingtheKF.UsingCorollary 9.1 andLemma 12.1 ,wehavethattheclassofKFsrelativeto2isessentiallycomplete.Thusaslongastheperformanceisjudgedsolelybasedontheriskfunction,ifthechoiceofestimatorsisrestrictedtoafneestimators,thennomatterwhatoptimalitycriterionisused,onemayconsideronlytheclassofKFsrelativeto2.OurnextstepistoderivemoreconvenientexpressionforRk(;^xk(0))inthecasethatand0arein0.Letk(;0)=(;^xk(0)).Using( 12 )( 12 ),itcanbeshownthat0j1(;0)=1()n(;0)=Kn(0)2()Kn(0)T+[IKn(0)Hn]njn1(;0)[IKn(0)Hn]Tn+1jn(;0)=3()+Fnn(;0)FTn:
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12 ).Recallthatifand0arein0,thenR(;^x0)=P3i=1tr(i()i(0)).Ourgoalistondanexpressionfori(0),fori=1;2;3,sinceitisneededinordertondamaximizerofR(;^x0)over0.Let~Fn(0)=FnFnKn(0)Hn,let~Fi;j(0)=~Fi(0)~Fi1(0)~Fj(0)(i>j)and~Fi;i(0)=~Fi(0).Wewillusetheconvention~Fi1;i(0)=I.LetCn;m(0)=(IKn(0)Hn)~Fn1;m(0)(0mn;0n),letDn;m(0)=Cn;m+1(0)FmKm(0)(n>m0)andDn;n(0)=Kn(0)(n0).NotethatCn+1;m=(IKn+1Hn+1)FnCn;mandCn;m1=Cn;m~Fm1.Thenn(;0)=Cn;0(0)1()Cn;0(0)T+nXm=0Dn;m(0)2()Dn;m(0)T+nXm=1Cn;m(0)3()Cn;m(0)T:Leti(0)=Pntn=iCn;i(0)TWnCn;i(0),fori=0;1;:::;nt.Theni=~FTii+1~Fi+(IKiHi)TWi(IKiHi):Itfollowsthat1(0)=ntXn=0Cn;0(0)TWnCn;0(0)=0(0);2(0)=ntXn=0nXm=0Dn;m(0)TWnDn;m(0)=ntXm=1Km1(0)TFTm1m(0)Fm1Km1(0)+ntXm=0Km(0)TWmKm(0)and;3(0)=ntXn=1nXm=1Cn;m(0)TWnCn;m(0)=ntXm=1m(0):Ingeneral,thecalculationof2(0)and3(0)requiresthestorageofK0(0);:::;Knt(0),whichmaybeproblematicforlargent.Thisisduetothefactthaticanbeupdatedbasedoni+1butnotviceversa.Nevertheless,intheimportantcasethatFiisinvertiblefor0int1,thecalculationof1,2,and3maybedoneinsuchawaythatthereisonlyaneedofaxedstorageplacethatdoesnotdependonnt.First,letusshowthatsince2(0)isinvertible, 89
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12.1 holdsinthisexample.WehaveshownthatAssumption 12.2 holdsinExample 12.1 .ItcanbeeasilyveriedthatAssumptions 12.3 and 12.4 holdaswell.OurrststepistochooseC0.InordertochooseC0,weneedtocalculateV(;L;R)and 12.2 tocalculate^xfor2[0;1)andthencalculater(;^x)andsup2R(;^x).Fig. 12-1 showstheplotoftheBayesriskr(;^x)versusthemaximumrisksup2R(;^x).NotethatbyTheorem 9.1 ,thisgure,infact,showstheBayesriskachievedbyarestrictedriskBayessolutionrelativeto(;C0)versusC0.HencethisguretellsusthetradeoffbetweenthepenaltyontheBayesriskandthesafeguardonthemaximumriskbyemployingrestrictedriskBayesestimation.Inthisexample,V=1:225105and 92
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AchievedBayesriskvs.maximumrisk 94
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ThemaximumriskoveroftheBayes,minimax,andrestrictedriskBayessolutionsvs.
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44 ]byinitializingtheKFwithzeromeanandcovariance.Itisstraightforwardtoverifythatn(x0)=0nforallx02X0.Let0n+1jn=3+Fn0nFTn,K0n=0njn1HTn[2+Hn0njn1HTn]1forn>0,andK00=0.Let~Fn=FnFnK0nHn,let~Fi;j=~Fi~Fi1~Fj(i>j)and~Fi;i=~Fi.Wewillusetheconvention~Fi1;i=I.TheproblemofstateestimationwitharandominitialstatewhosedistributioninnotnecessarilyGaussianisconsideredbyLainiotisetal[ 46 ]andtheCMestimatorforthisproblemisderived.Let^x0;n()denotetheCMestimatorwithrespecttoforx0basedonyn0,i.e.,^x0;n()=RX0x0pnx0(yn0)d 97
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8 .Ourgoalisthentoderivesuboptimalestimatorsthatcangivemaximumriskarbitrarilyclosetothatofaminimaxestimator.ThefollowingLemmasarederivedin[ 24 ]. BecauseofLemmas 13.1 and 13.2 andthefactthatX0isasubsetofnitedimensionalnormedspace,wecanusetheresultsofChapter 8 tonda-optimalestimator.InparticularifX0isaconvexpolytope,Lemma 8.7 canbeusedtonda-optimalestimator.Inordertonda-optimalestimator,weneedtoseta0<0
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7.1 ,intermsofthecomputationburden,sinceitmayrequirelesscalculationsoftheriskfunction.Thecomputationalcomplexityofa-optimalestimatorderivedaccordingtoLemma 8.7 dependslinearlyonthesupportof0.Itmayberatherhigh,anditiscertainlyhigherthanthecomplexityofafneestimators.InChapter 12 wediscussedtheproblemofrestrictedriskBayesestimationwhentheclassofestimatorsisrestrictedtoafneestimators.Whentheclassofestimatorsisrestrictedtoafneestimators,theminimaxproblemofthissectioncanberegardedasaspecialcaseoftheproblemconsideredinChapter 12 .Ifwecanndanafneestimatorthatisalsoa-optimalestimator,wewouldprobablyprefertheafneestimator.Ifsuchanafneestimatorexists,theresultsofthissectionarestillimportantsincetheygiveustightlowerandupperboundsonthemaximumriskandhenceenableustoevaluatetheperformanceoftheafneestimatorwiththebestpossibleperformanceinthesenseofthemaximumrisk.Wefurtherillustratethisinthefollowingnumericalexample. whereF=264101375andwniszeromeanGaussianrandomvectorwithcovariancematrix3=264000q375:
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AfullviewofXf0,whichisanite(;V)-densesubsetofX0
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Azoom-inviewofthebottomleftcornerofXf0,whichisanite(;V)-densesubsetofX0 Theriskof^x(0),thederived-optimalestimator,asafunctionofx02Xf0
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Theaprioridistribution0,whichisdenedonXf0 ThemaximumriskoftheKalmanFilterinitializedwithzeromeanandcovariance2Iasafunctionof
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TheconcavityofKfollowsfromthefactthatforany0<<1,K(1+(1)2)=inf^x2XK(^x;1+(1)2)inf^x2XK(^x;1)+(1)inf^x2XK(^x;2)=K(1)+(1)K(2);wheretheinequalityaboveisadirectresultof( B ).Eqn.( 9 )impliesthatK(1)K(2)forany01<21.SinceKisdecreasingandconcaveon[0;1],itiscontinuousexceptperhapsatthepoint=1.CertainlyK(^x;1)=K(1)andK(^x;)K()forall01.Thusforany0<1,jK()K(1)jK(^x;)K(^x;1)=(1)sup2R(;^x)r(;^x)(1) 109
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wherethesecondinequalityisadirectconsequenceof( 9 ).ThisimpliesthatK(^x2;1)K(^x2;2)K(^x1;1)K(^x1;2).IteasilyfollowsthatG(^x1)G(^x2).2)By( C ),K(^x1;2)K(^x2;2)0.Thusr(;^x2)+(12)[G(^x2)G(^x1)]r(;^x1):Bypart1)ofthelemma,r(;^x2)r(;^x1).3)By( C ),K(^x2;1)K(^x1;1)0.Thussup2R(;^x1)+1[G(^x2)G(^x1)]sup2R(;^x2):Bypart1)ofthelemma,sup2R(;^x1)sup2R(;^x2). 110
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[1] R.E.Kalman,Anewapproachtolinearlteringandpredictionproblems,ASMETransactions,JournalofBasicEngineering,vol.82,pp.34,Mar.1960. [2] J.Heffes,TheeffectoferronousmodelsontheKalmanlterresponse,IEEETrans.Automat.Contr.,vol.AC-11,pp.541,Apr.1966. [3] T.Nishimura,Ontheaprioriinformationinsequentialestimationproblems,IEEETrans.Automat.Contr.,vol.AC-11,pp.197,Apr.1966. [4] T.Nishimura,ErrorboundsofcontinuousKalmanltersandtheapplicationtoorbitdeterminationproblems,IEEETrans.Automat.Contr.,vol.AC-12,pp.268,Jun.1967. [5] M.Mintz,AnoteonminimaxestimationandKalmanltering,IEEETrans.Automat.Contr.,vol.AC-14,pp.588,Oct.1969. [6] J.M.Morris,TheKalmanlter:Arobustestimatorforsomeclassesoflinearquadraticproblems,IEEETrans.Inform.Theory,vol.IT-22,pp.526,Sep.1976. [7] V.PoorandD.P.Looze,Minimaxstateestimationforlinearstochasticsystemswithnoiseuncertainty,IEEETrans.Automat.Contr.,vol.AC-26,pp.902,Aug.1981. [8] C.J.MartinandM.Mintz,Robustlteringandpredictionforlinearsystemswithuncertaindynamics:Agame-theoreticapproach,IEEETrans.Automat.Contr.,vol.AC-28,pp.888,Sep.1983. [9] S.VerduandH.V.Poor,Onminimaxrobustness:Ageneralapproachandapplications,IEEETrans.Inform.Theory,vol.IT-30,pp.328,Mar.1984. [10] S.VerduandH.V.Poor,Minimaxlinearobserversandregulatorsforstochasticsystemswithuncertainsecond-orderstatistics,IEEETrans.Automat.Contr.,vol.AC-29,pp.499,Jun.1984. [11] J.C.DarraghandD.P.Looze,Noncausalminimaxlinearstateestimationforsystemswithuncertainsecond-orderstatistics,IEEETrans.Automat.Contr.,vol.AC-29,pp.555,Jun.1984. [12] B.I.Anan'ev,Onminimaxstateestimatesformultistagestatisticallyuncertainsystems,ProblemsofControlandInformationTheory,vol.18,pp.27,1989. [13] Y.L.ChenandB.S.Chen,Minimaxrobustdeconvolutionltersunderstochasticparametricandnoiseuncertainties,IEEETrans.SignalProcessing,vol.42,pp.32,Jan.1994. [14] B.I.Anan'ev,Minimaxestimationofstatisticallyuncertainsystemsunderthechoiceofafeedbackparameter,JournalofMathematicalSystems,Estimation,andControl,vol.5,pp.1,1995. 111
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[15] J.O.Berger,StatisticalDecisionTheoryandBayesianAnalysis,Springer,NewYork,2ndedition,1985. [16] J.L.Hodges,Jr.andE.L.Lehmann,Theuseofpreviousexperienceinreachingstatisticaldecisions,Ann.Math.Stat.,vol.23,pp.396,Sep.1952. [17] B.EfronandC.Morris,LimitingtheriskofBayesandempiricalBayesestimatorspart1:TheBayescase,J.Amer.Statist.Assoc.,vol.66,pp.807,Dec.1971. [18] P.J.Kempthorne,Numericalspecicationofdiscreteleastfavorablepriordistributions,SIAMJ.Sci.Statist.Comput.,vol.8,pp.171,Mar.1987. [19] P.J.Kempthorne,Controllingrisksunderdifferentlossfunctions:Thecompromisedecisionproblem,Ann.Statist.,vol.16,pp.1594,Dec.1988. [20] I.M.Johnstone,Onminimaxestimationofasparsenormalmeanvector,Ann.Statist.,vol.22,pp.271,Mar.1994. [21] R.F.Berg,Estimationandpredictionformaneuveringtargettrajectories,IEEETrans.Automat.Contr.,vol.AC-28,pp.294,Mar.1983. [22] M.H.KaoandD.H.Eller,MulticongurationKalmanlterdesignforhigh-performanceGPSnavigation,IEEETrans.Automat.Contr.,vol.AC-28,pp.304,Mar.1983. [23] L.DanyangandL.Xuanhuang,Optimalstateestimationwithouttherequirementofaprioristatisticsinformationoftheinitalstate,IEEETrans.Automat.Contr.,vol.39,pp.2087,Oct.1994. [24] Y.LevinbookandT.F.Wong,Stateestimationwithinitialstateuncertainty,IEEETransactionsonInformationTheory,2005,Submittedforpublication.URL: [25] A.Wald,StatisticalDecisionFunctions,JohnWileyandSons,NewYork,1950. [26] T.S.Ferguson,MathematicalStatistics:ADecisionTheoreticApproach,AcademicPress,NewYork,1967. [27] P.Billingsley,ConvergenceofProbabilityMeasures,JohnWiley&Sons,NewYork,1968. [28] W.Rudin,PrinciplesofMathematicalAnalysis,McGraw-Hill,NewYork,3rdedition,1976. [29] D.W.Stroock,ProbabilityTheory,AnAnalyticView,CambridgeUniversityPress,NewYork,1993. [30] L.LeCam,AnextensionofWald'stheoryofstatisticaldecisionfunctions,Ann.Math.Stat.,vol.26,pp.69,Mar.1955. [31] H.Kudo,Ontheproperty(W)oftheclassofstatisticaldecisionfunctions,Ann.Math.Stat.,vol.37,pp.1631,Dec.1966.
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[32] M.Sion,Ongeneralminimaxtheorems,PacicJ.Math.,vol.8,pp.171,1958. [33] G.K.Pedersen,AnalysisNow,Springer-Verlag,NewYork,1989. [34] A.N.Shiryaev,Probability,Springer-Verlag,NewYork,2ndedition,1989. [35] J.L.Doob,MeasureTheory,Springer-Verlag,NewYork,1994. [36] W.Rudin,FunctionalAnalysis,McGraw-Hill,NewYork,2ndedition,1991. [37] E.L.Lehmann,TheoryofPointEstimation,Wiley,NewYork,1983. [38] W.Nelson,Minimaxsolutionofstatisticaldecisionproblemsbyiteration,Ann.Math.Stat.,vol.37,pp.1643,Dec.1966. [39] J.Wolfowitz,On-completeclassofdecisionfunctions,Ann.Math.Stat.,vol.22,pp.461,Sep.1951. [40] C.D.AliprantisandK.C.Border,InniteDimensionalAnalysis:AHitchhiker'sGuide,Springer-Verlag,Berlin,1994. [41] N.M.Roy,Extremepointsofconvexsetsininnitedimensionalspaces,Amer.Math.Monthly,vol.94,pp.409,May1987. [42] S.BoydandL.Vandenberghe,ConvexOptimization,CambridgeUniversityPress,Cambridge,2004. [43] L.Mirsky,AtraceinequalityofJohnvonNeumann,MonatsheftefurMathematik,vol.79,pp.303,1975. [44] T.Kailath,A.H.Sayed,andB.Hassibi,LinearEstimation,PrenticeHall,NJ,2000. [45] S.S.Blackman,MultipleTargetTrackingwithRadarApplications,ArtechHouse,Washington,DC,1986. [46] D.G.Lainiotis,S.K.Park,andR.Krishnaiah,Optimalstate-vectorestimationfornon-Gaussianinitialstate-vector,IEEETrans.Automat.Contr.,vol.AC-16,pp.197,Apr.1971.
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https://dpc10ster.github.io/RJafroc/articles/Ch11Vig5SampleSize.html
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## Introduction
• The value of the true FOM difference between the treatments, i.e., the true effect-size (ES) is, of course, unknown. If it were known, there would be no need to conduct an ROC study. One would simply adopt the treatment with the higher FOM. Sample-size estimation involves making an educated guess regarding the ES, called the anticipated ES, and denoted by d. To quote (ICRU 2008): “any calculation of power amounts to specification of the anticipated effect-size”. Increasing the anticipated ES will increase statistical power but may represent an unrealistic expectation of the true difference between the treatments, in the sense that it overestimates the ability of technology to achieve this much improvement. An unduly small might be clinically insignificant, besides requiring a very large sample-size to achieve sufficient power.
• There is a key difference between statistical significance and clinical significance. An effect-size in AUC units could be so small, e.g., 0.001, as to be clinically insignificant, but by employing a sufficiently large sample size one could design a study to detect this small and clinically meaningless difference with high probability, i.e., high statistical power.
• What determines clinical significance? A small effect-size, e.g., 0.01 AUC units, could be clinically significant if it applies to a large population, where the small benefit in detection rate is amplified by the number of patients benefiting from the new treatment. In contrast, for an “orphan” disease, i.e., one with very low prevalence, an effect-size of 0.05 might not be enough to justify the additional cost of the new treatment. The improvement might have to be 0.1 before it is worth it for a new treatment to be brought to market. One hates to monetize life and death issues, but there is no getting away from it, as cost/benefit issues determine clinical significance. The arbiters of clinical significance are engineers, imaging scientists, clinicians, epidemiologists, insurance companies and those who set government health care policies. The engineers and imaging scientists determine whether the effect-size the clinicians would like is feasible from technical and scientific viewpoints. The clinician determines, based on incidence of disease and other considerations, e.g., altruistic, malpractice, cost of the new device and insurance reimbursement, what effect-size is justifiable. Cohen has suggested that d values of 0.2, 0.5, and 0.8 be considered small, medium, and large, respectively, but he has also argued against their indiscriminate usage. However, after a study is completed, clinicians often find that an effect-size that biostatisticians label as small may, in certain circumstances, be clinically significant and an effect-size that they label as large may in other circumstances be clinically insignificant. Clearly, this is a complex issue. Some suggestions on choosing a clinically significant effect size are made in Chapter 11.
• Does one even need to perform a pivotal study? If the pilot study returns a significant difference, one has rejected the NH and that is all there is to it. There is no need to perform the pivotal study, unless one “tweaks” the new treatment and/or casts a wider sampling net to make a stronger argument, perhaps to the FDA, that the treatments are indeed generalizable, and that the difference is in the right direction (new treatment FOM > conventional treatment FOM). If a significant difference is observed in the opposite direction (e.g., new treatment FOM < conventional treatment FOM) one cannot justify a pivotal study with an expected effect-size in the “other or favored” direction; see example below. Since the Van Dyke pilot study came close to rejecting the NH and the observed effect size, see below, is not too small, a pivotal study is justified.
• This vignette discusses choosing a realistic effect size based on the pilot study. Illustrated first is using Van Dyke dataset, regarded as the pilot study.
## Illustration of SsPowerGivenJK() using method = "DBM"
rocData <- dataset02 ##"VanDyke.lrc"
#fileName <- dataset03 ## "Franken1.lrc"
retDbm <- StSignificanceTesting(dataset = rocData, FOM = "Wilcoxon", method = "DBM")
print(retDbm$RRRC$ciDiffTrt)
#> Estimate StdErr DF t PrGTt
#> trt0-trt1 -0.043800322 0.020748618 15.259675 -2.1109995 0.051665686
#> CILower CIUpper
#> trt0-trt1 -0.087959499 0.00035885444
• Lacking any other information, the observed effect-size is the best estimate of the effect-size to be anticipated. The output shows that the FOM difference, for treatment 0 minus treatment 1, is -0.04380032. In the actual study treatment 1 is the new modality which hopes to improve upon 0, the conventional modality. Since the sign is negative, the difference is going the right way and is justified in moving forward with planning a pivotal study. [If the difference went the other way, there is little justification for a pivotal study].
• The standard error of the difference is 0.02074862.
• An optimistic, but not unduly so, effect size is given by:
effectSizeOpt <- abs(retDbm$RRRC$ciDiffTrt[1,"Estimate"]) + 2*retDbm$RRRC$ciDiffTrt[1,"StdErr"]
The observed effect-size is a realization of a random variable. The lower limit of the 95% confidence interval is given by and the upper limit by . CI’s generated like this, with independent sets of data, are expected to encompass the true value with 95% probability. The lower end (greatest magnitude of the difference) of the confidence interval is -0.08529756, and this is the optimistic estimate. Since the sign is immaterial, one uses as the optimistic estimate the value 0.08529756. While the sign is immaterial for sample size estimates, the decision to conduct the pivotal most certainly is material - if the sign went the other way, with the new modality lower than the conventional modality, one would be unjustified in conducting a pivotal study.
## References
ICRU. 2008. “Statistical Analysis and Power Estimation.” Book Section. In JOURNAL of the Icru, 8:37–40. https://doi.org/10.1093/jicru/ndn012.
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2020-08-10 18:47:27
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https://physics.stackexchange.com/questions/560067/what-is-the-reason-that-quantum-mechanics-is-random?noredirect=1
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# What is the reason that Quantum Mechanics is random?
We know quantum mechanics gives a random result when we observe a particle that's in a superposition, but why is it random? One of the explanations I've heard is that because light comes with those discrete energy packets called photons, when a photon is passing through a polarized filter, it must either all pass through or all be blocked. You can't let a fraction of the photon pass through while others are blocked. Is it correct? It seems reasonable, but I couldn't find any proper source about this statement.
• This is only according to the Copenhagen interpretation of quantum mechanics. In the many-world interpretation, ALL outcomes happen, so it is in fact deterministic. Jun 18, 2020 at 13:31
• @gardenhead I'm not sure that's a very fruitful way of viewing the MWI though; even if "all outcomes happen", using loosey-goosey terminology, "in our branch we only observe one", so there is still something to be explained, no? Jun 18, 2020 at 21:12
• @JoshuaLin And wouldn't the one we observe be effectively random? (If you know how to predict that observation couldn't it be extrapolated, on a massive scale, to predict the future?) Jun 19, 2020 at 0:11
• @TCooper here's a spooky thought: in MWI our observations are random "unless they are tied to our ability to observe" : arxiv.org/abs/quant-ph/9709032 Jun 19, 2020 at 1:17
• @stackoverblown In MWI, if the wavefunction is known at an initial time, then it is determined for all future times. We just have limited information about the initial conditions. We only have information about portion of the wavefunction that is coherent with the part we are in. Jun 21, 2020 at 7:49
If it helps, it's not that the nature of the universe is random, it's that we model it as random in Quantum Mechanics.
There are many cases in science where we cannot model the actual behavior of a system, due to all sorts of effects like measurement errors or chaotic behaviors. However, in many cases, we don't need to care about exactly how a system behaves. We only need to worry about the statistical behavior of the system.
Consider this. We are going to roll a die. If it lands 1, 2, or 3, I give you \$1. If it lands 4, 5, or 6, you give me \$1. It is theoretically very difficult for you to predict whether any one roll is going to result in you giving me \$1 or me giving you \$1. However, if we roll this die 100 times, we can start to talk about expectations. We can start to talk about whether this die is a fair die, or if I have a weighted die. We can model the behavior of this die using statistics.
We can do this until it becomes useful to know more. There are famous stories of people making money on roulette using computers to predict where the ball is expected to stop. We take some of the randomness out of the model, replacing it with knowledge about the system.
Quantum Mechanics asserts that the fundamental behavior of the world is random, and we back that up with statistical studies showing that it's impossible to distinguish the behavior of the universe from random.
That's not to say the universe is random. There may be some hidden logic to it all and we find that it was deterministic after all. However, after decades of experimentation, we're quite confident in a whole slew of ways the universe can't be deterministic. We've put together experiment after experiment, like the quantum eraser, for which nobody has been able to predict the behavior of the experiment better than the randomness of QM.
Indeed, the ways the universe can be deterministic are so extraordinary that we choose to believe the universe cannot be that fantastic. For example, there's plenty of ways for the universe to be deterministic as long as some specific information can travel instantaneously (faster than light). As we have not observed any way to transfer information faster than light in a normal sense, we are hesitant to accept these deterministic descriptions of quantum behavior (like the Pilot Wave interpretation).
And in the end, this is all science ever does. It can never tell us that something is truly random. It can never tell us what something truly is. What it tells us is that the observed behaviors of the system can be indistinguishable from those of the scientific models, and many of those models have random variables in them.
• – J.G.
Jun 18, 2020 at 13:48
• "It can never tell us that something is truly random." Sure it can - it just can't prove it empirically. But being unable to prove something has never stopped science from asserting it (whether rightly or wrongly) when enough evidence supports it. (Unless by "tell" you mean something totally different from what everybody means when they say, "Science tells us X.") Jun 18, 2020 at 16:34
• @TimothySmith I get in so much trouble by pointing out what science can and can't do, and then using informal terms when doing so! (Thanks for the catch. The intent I was going after was more along the philosophical concept of providing knowledge... but that certainly isn't the word choice I used!) Jun 18, 2020 at 19:56
• Yes, exactly, it's about modeling, in any case. :) Jun 18, 2020 at 20:02
• I believe the toughest nut to crack here isn't why the universe is random, but rather why it appears to be so deterministic. Once I tried to focus on answering that question, I realized that everything we observe was stacked upon prior observations, which in turn were stacked ultimately upon what was (perhaps) undue weight on observations to begin with. Jun 18, 2020 at 23:29
As Feynman said when laying out the first principles of quantum mechanics:
How does it work? What is the machinery behind the law?” No one has found any machinery behind the law. No one can “explain” any more than we have just “explained.” No one will give you any deeper representation of the situation. We have no ideas about a more basic mechanism from which these results can be deduced.
We do not know how to predict what would happen in a given circumstance, and we believe now that it is impossible—that the only thing that can be predicted is the probability of different events. It must be recognized that this is a retrenchment in our earlier ideal of understanding nature. It may be a backward step, but no one has seen a way to avoid it.
That statement in bold re probability is what @SuperCiocia is saying.
• Your answer doesn't address the why it's probabilistic (vs deterministic), which was the crux of the question. Jun 18, 2020 at 14:11
• Unless I'm misunderstanding the context of the Feynman quote, Feynman is saying that the best evidence they have suggests that it is probabilistic, but that nobody knows why. In this regard, I believe this still represents the extent of our understanding. Jun 18, 2020 at 15:53
• @Alexandre Aubrey: It does address the why. In one sentence: that's the way the universe words, but we don't have an effing clue about WHY it works that way. Jun 18, 2020 at 16:07
• @AlexandreAubrey "We don't know why" is an answer to "why?". If we don't know why, what other possible answer to that question can be expected? Jun 19, 2020 at 13:13
It's weirder than you thought.
The wavefunction itself is fully deterministic. People often say "it's the measurements that are probabilisitic" but that isn't right either. The measurement is deterministic if you include the measurement apparatus in the wavefunction. And therein is the core of the great mystery, and the big philosophical questions of whether we should include ourselves in the wavefuncion. Mathematically speaking, we should, and that gives us the Many Worlds interpretation.
The real question is: why do I subjectively experience a probabilitic outcome? We don't have the pholosophical answers to what "I" and "experience" refer to in that sentence. Another way to put it is that the real question is why don't I experience the whole of the wavefunction?
If a conscious mind can (for reasons unknown) only experience one outcome of the many which all truly do actually happen then a probabilistic subjective experience may be the only possible experience. It then raises the question of how we associate probabilities with the wavefunction. Why is the probability proportional to the square of the amplitude? No one really knows, but perhaps there is a deep explanation hinted at here although I confess I do not fully understand it myself, but again the answer may be it's a mathematical necessity.
• if subjective experience is an objective thing, it exists and is experienced across all possibilities Jun 18, 2020 at 13:02
• +1 on the physics term "weirder". Jun 18, 2020 at 23:32
• If the measurement apparatus is included in the wavefunction, then a measurement has not occurred. It's not some magic properly of tools that effects an observation. Jun 19, 2020 at 10:48
• @OrangeDog sure it has, sounds like you've still got the old, misguided idea that a measurement has only one outcome :-) Jun 23, 2020 at 11:52
a) I wouldn't call it "random" but "probabilistic".
b) The evolution of a system is fully deterministic. It's the outcome of measurements that is probabilistic.
c) Your reasoning is wrong. The probabilistic nature of measurements' outcomes is something intrinsic to quantum mechanics (the measurement problem), independent of the specifics of the measurement apparatus.
• How is "probabilistic" different from "random"? They mean the same thing to me.
– Puk
Jun 18, 2020 at 3:38
• @Puk Well if the state were $|\psi\rangle = (1/\sqrt{2})(|1\rangle + |2\rangle)$ I'd say the outcome is fully random because the chance of getting $|1\rangle$ and $|2\rangle$ is exactly the same. But for $|\psi\rangle \propto 0.1*|1\rangle + 0.9*|2\rangle$ then it's more probably to get $|2\rangle$. So I'd classify random as a subset of probabilistic. But it might be semantics than precise defintions. Jun 18, 2020 at 3:41
• I see. I would call both "random", with the degree of "randomness" defined by $\psi$. But yes, just a matter of semantics.
– Puk
Jun 18, 2020 at 3:45
• Well there is uniformly random and random in othee ways, yet it's still random. Jun 18, 2020 at 5:49
• You are making the common mistake of thinking random means uniform distribution. This is wrong: the sum of two dice is most likely going to be 7 but that's still very much a random process that can be modeled by a random variable. Random variables can have any distribution imaginable.
– eps
Jun 18, 2020 at 15:26
You are asking why QM is random (which in your case given the context is used as probabilistic), and what is correct to say is that QM is probabilistic in nature, and our underlying world, and our universe seems to us to be quantum mechanical, and truly probabilistic.
is there a way to understand the system as having an initial state which forced it to come to this conclusion," the answer is a qualified "no": there are hidden-variable interpretations like the pilot-wave theory which interpret quantum mechanics as a deterministic theory containing unknowable global information.
The point is global. There are quantum effects that cannot be understood in classical terms.
using some thought experiments (my favorite is a game called Betrayal) one can prove that there are quantum effects which cannot be understood in terms of classical local information
Now the universe ultimately is quantum mechanical, and probabilistic. There might be some underlying mechanism, that is not understood by us, but some specifically state that this underlying mechanism, that would make the universe seem fully deterministic to us, cannot be known. The error is not in our measuring devices, we know that we cannot learn about the underlying mechanism (even if there is one).
In a deeper sense randomness is our way of reasoning about information that we do not know, whether there is some unknowable information which makes everything deterministic, it is known that we cannot (not just do not) know it.
How do we know that certain quantum effects are random?
So the answer to your question is, that the error is not in our measuring devices, the universe looks to us truly probabilistic, and QM is the best way to describe it that best fits the experiments. QM is simply probabilistic because it describes (models) a universe that appears to us to be truly probabilistic in nature, and there is no (to our knowledge) underlying (more fundamental) mechanism.
• random = probabilistic in the context given, as was explained in the other post's comments. Jun 19, 2020 at 7:06
• @kludg correct, I edited. Jun 19, 2020 at 7:32
Quantum Indeterminacy is Key to the Arrow of Time
There is no machinery to explain the randomness (as Mr. Anderson answered from Feynman), but maybe a connection to other phenomena can help.
I'm going to go out on a limb here, because answers in this forum are supposed to be from established science. But I think I can make a case for an important explanation that I think follows logically, even if I haven't seen in the literature.
I think we can make the case that there is a fundamental connection between quantum randomness and the arrow of time. Here are the parts of that idea:
Special Relativity and Time Reversal
We know from Special Relativity that all inertial frames are equally valid, that the laws of physics in one (non-accelerating) frame are exactly the same as in any other. This principle also applies to frames of reference where time is reversed. In fact the Feynman-Stueckelberg interpretation of antimatter is the idea that antimatter is matter going backwards in time.
Time Reversal and Entropy
But we know from the second law of thermodynamics that entropy either increases or stays the same, but it doesn't decrease (at least not on the macro scale). So one principle says that the laws of physics are the same under time reversal (actually something called CPT) but another says that entropy increases are irreversible.
Time Reversal and Quantum Choices
Now here's the idea that I came up with. It's probably already out there somewhere, I've looked and haven't seen it though. If someone knows where this has been developed (if it has) I would very much like a reference.
If a sequence of events is deterministic (one with no random quantum choices) then the time reversal of that sequence must also be deterministic, and the reversal of that sequence would always return the system to its original state.
But if a sequence of events involves random quantum choices, then the reversal of that sequence also involves random quantum choices, and those choices don't have to return the system to its original state when time is rolled back to the original time.
Example:
A photon goes towards an atom, its absorbed by that atom, the atom waits a random amount of time, then it emits a photon in a random direction, and the photon moves away from that atom.
If we could start with the end of this sequence and reverse time, then we get the same kind of sequence, but the time the atom exists in an excited state doesn't depend on the original time and so is probably not going to be the same amount of time, and the direction the photon is emitted is also random, so is probably not going to be in the original direction.
So we can have both the rules of physics be the same between a frame going forwards in time and backwards in time, and still have the forwards and reversed sequences be different, as long as there are random quantum choices in that sequence.
So I think the resolution to Loschmidt's Paradox is this: If entropy increases in a process and so the process is irreversible, it must involve random quantum choices. If a process is deterministic, and doesn't involve random quantum choices, then it must also be reversible and so the entropy in that system will stay the same.
• Suppose two computer programs whose run state exhibits increasing entropy over time. One program is driven by a psuedo random number generator, and the other a (allegedly) true random number generator (say based on a particle detector or some other low level quantum phenomena). Can you tell them apart? Jun 19, 2020 at 19:24
• I'm not sure, but there are systems that actually use quantum phenomena to generate random numbers. en.wikipedia.org/wiki/… Jun 19, 2020 at 19:52
• Knowing that you knew that I said "and the other a (allegedly) true random number generator (say based on a particle detector or some other low level quantum phenomena).". Talking about your "forward" and "reverse" ideas - suppose we talk about "ensemble of states" rather than state, where "ensemble" is the distribution of probabilities (i.e. the pdf) over the range of states at time (t). The mapping from pdf(t) to pdf(t+delta) can itself be deterministic even though the state transition function is not. Which is more "real" - the pdf over all states or an individual state? Jun 19, 2020 at 20:42
• And which is more important to 'life', and to 'intelligent life'? Jun 19, 2020 at 20:43
• That a pretty deep question. I wouldn't even know where to begin to come up with an experimental question that addresses your question about ensembles versus states. Smarter guys than me have said the ergodic hypothesis explains the number of possible microstates in the Boltzmann equation, but I suspect it's something more like the entropy difference we see between polarized photons and an unpolarized photons. Jun 19, 2020 at 21:12
We don't even know that the universe is fundamentally random. That's just the most popular interpretation (called the Copenhagen Interpretation). In this interpretation, the behavior of particles is probabilistic with no deeper reasoning, and the "why" is left to philosophers (or, possibly, a future Theory of Everything).
There are other interpretations in which the universe is not fundamentally random. Hidden variable interpretations say that QM is actually deterministic, but we deal with probabilities due to not having enough information about some hidden variables.
This seems like the most logical first guess. However, due to Bell's Theorem discovered in the 60's, we know that any deterministic QM interpretation must necessarily be non-local - that is, it requires all particles in the universe to be somehow connected to one-another, and able to communicate at faster-than-light speed.
So basically, physicists are more willing to discard determinism than discard locality.
Quantum mechanics is random or, more accurately, probabilistic, because nature is fundamentally not deterministic. Of course there are those clinging to deterministic explanations, like Bohmian mechanics, by ignoring mathematical proofs, just as there are those clinging to Dingle's argument against relativity. But the argument "I don't understand the proof, therefore the proof is wrong" is not a valid scientific argument, even if the arguments disproving determinism are considerably harder to understand than the arguments proving that Dingle was wrong.
The Schrodinger equation may well appear deterministic, but it only determines probabilities; probabilities do not determine results. Quantum probabilities obey a different mathematical structure from classical probability theory precisely because classical probabilities are determined by unknowns or "hidden variables". The mathematical structure of quantum mechanics is as it is precisely because there are no hidden variables determining measurement results.
There are numerous mathematical proofs of this fact, starting with von Neuman (1936). Further proofs have been given by Jauch & Piron (1963), and by Gudder (1968), and many others, but they are sufficiently abstract that few physicists understand them. Kochen and Specker gave a proof which more physicist understand in 1967. Bell himself gave a proof in 1966 (but written earlier), based on work by Gleason, only Bell still didn’t understand the proof, and claimed there was something wrong with it. Bell himself gave a proof in Bell's theorem (1964), which has been generally accepted because it is directly testable in experiment, and is less abstract than other proofs, requiring only that classical probability theory is refuted by experimental evidence, which has since been obtained.
I have given deeper discussion in my second books, and two demonstrations that nature is fundamentally not deterministic in my third (see my profile for links)
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2022-08-19 08:24:37
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https://people.maths.bris.ac.uk/~matyd/GroupNames/288/C2xDic3sDic3.html
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## G = C2×Dic3⋊Dic3order 288 = 25·32
### Direct product of C2 and Dic3⋊Dic3
Series: Derived Chief Lower central Upper central
Derived series C1 — C3×C6 — C2×Dic3⋊Dic3
Chief series C1 — C3 — C32 — C3×C6 — C62 — C6×Dic3 — Dic3⋊Dic3 — C2×Dic3⋊Dic3
Lower central C32 — C3×C6 — C2×Dic3⋊Dic3
Upper central C1 — C23
Generators and relations for C2×Dic3⋊Dic3
G = < a,b,c,d,e | a2=b6=d6=1, c2=b3, e2=d3, ab=ba, ac=ca, ad=da, ae=ea, cbc-1=b-1, bd=db, be=eb, cd=dc, ece-1=b3c, ede-1=d-1 >
Subgroups: 594 in 211 conjugacy classes, 100 normal (30 characteristic)
C1, C2 [×3], C2 [×4], C3 [×2], C3, C4 [×8], C22, C22 [×6], C6 [×6], C6 [×8], C6 [×7], C2×C4 [×14], C23, C32, Dic3 [×4], Dic3 [×10], C12 [×6], C2×C6 [×2], C2×C6 [×12], C2×C6 [×7], C4⋊C4 [×4], C22×C4 [×3], C3×C6 [×3], C3×C6 [×4], C2×Dic3 [×8], C2×Dic3 [×16], C2×C12 [×10], C22×C6 [×2], C22×C6, C2×C4⋊C4, C3×Dic3 [×4], C3×Dic3 [×2], C3⋊Dic3 [×2], C62, C62 [×6], Dic3⋊C4 [×4], C4⋊Dic3 [×4], C22×Dic3 [×2], C22×Dic3 [×3], C22×C12 [×2], C6×Dic3 [×8], C6×Dic3 [×2], C2×C3⋊Dic3 [×2], C2×C3⋊Dic3 [×2], C2×C62, C2×Dic3⋊C4, C2×C4⋊Dic3, Dic3⋊Dic3 [×4], Dic3×C2×C6 [×2], C22×C3⋊Dic3, C2×Dic3⋊Dic3
Quotients: C1, C2 [×7], C4 [×4], C22 [×7], S3 [×2], C2×C4 [×6], D4 [×2], Q8 [×2], C23, Dic3 [×4], D6 [×6], C4⋊C4 [×4], C22×C4, C2×D4, C2×Q8, Dic6 [×4], C4×S3 [×2], D12 [×2], C2×Dic3 [×6], C3⋊D4 [×2], C22×S3 [×2], C2×C4⋊C4, S32, Dic3⋊C4 [×4], C4⋊Dic3 [×4], C2×Dic6 [×2], S3×C2×C4, C2×D12, C22×Dic3, C2×C3⋊D4, S3×Dic3 [×2], C3⋊D12 [×2], C322Q8 [×2], C2×S32, C2×Dic3⋊C4, C2×C4⋊Dic3, Dic3⋊Dic3 [×4], C2×S3×Dic3, C2×C3⋊D12, C2×C322Q8, C2×Dic3⋊Dic3
Smallest permutation representation of C2×Dic3⋊Dic3
On 96 points
Generators in S96
(1 19)(2 20)(3 21)(4 22)(5 23)(6 24)(7 13)(8 14)(9 15)(10 16)(11 17)(12 18)(25 43)(26 44)(27 45)(28 46)(29 47)(30 48)(31 37)(32 38)(33 39)(34 40)(35 41)(36 42)(49 67)(50 68)(51 69)(52 70)(53 71)(54 72)(55 61)(56 62)(57 63)(58 64)(59 65)(60 66)(73 91)(74 92)(75 93)(76 94)(77 95)(78 96)(79 85)(80 86)(81 87)(82 88)(83 89)(84 90)
(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)
(1 68 4 71)(2 67 5 70)(3 72 6 69)(7 62 10 65)(8 61 11 64)(9 66 12 63)(13 56 16 59)(14 55 17 58)(15 60 18 57)(19 50 22 53)(20 49 23 52)(21 54 24 51)(25 92 28 95)(26 91 29 94)(27 96 30 93)(31 86 34 89)(32 85 35 88)(33 90 36 87)(37 80 40 83)(38 79 41 82)(39 84 42 81)(43 74 46 77)(44 73 47 76)(45 78 48 75)
(1 9 5 7 3 11)(2 10 6 8 4 12)(13 21 17 19 15 23)(14 22 18 20 16 24)(25 35 27 31 29 33)(26 36 28 32 30 34)(37 47 39 43 41 45)(38 48 40 44 42 46)(49 59 51 55 53 57)(50 60 52 56 54 58)(61 71 63 67 65 69)(62 72 64 68 66 70)(73 81 77 79 75 83)(74 82 78 80 76 84)(85 93 89 91 87 95)(86 94 90 92 88 96)
(1 31 7 25)(2 32 8 26)(3 33 9 27)(4 34 10 28)(5 35 11 29)(6 36 12 30)(13 43 19 37)(14 44 20 38)(15 45 21 39)(16 46 22 40)(17 47 23 41)(18 48 24 42)(49 82 55 76)(50 83 56 77)(51 84 57 78)(52 79 58 73)(53 80 59 74)(54 81 60 75)(61 94 67 88)(62 95 68 89)(63 96 69 90)(64 91 70 85)(65 92 71 86)(66 93 72 87)
G:=sub<Sym(96)| (1,19)(2,20)(3,21)(4,22)(5,23)(6,24)(7,13)(8,14)(9,15)(10,16)(11,17)(12,18)(25,43)(26,44)(27,45)(28,46)(29,47)(30,48)(31,37)(32,38)(33,39)(34,40)(35,41)(36,42)(49,67)(50,68)(51,69)(52,70)(53,71)(54,72)(55,61)(56,62)(57,63)(58,64)(59,65)(60,66)(73,91)(74,92)(75,93)(76,94)(77,95)(78,96)(79,85)(80,86)(81,87)(82,88)(83,89)(84,90), (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), (1,68,4,71)(2,67,5,70)(3,72,6,69)(7,62,10,65)(8,61,11,64)(9,66,12,63)(13,56,16,59)(14,55,17,58)(15,60,18,57)(19,50,22,53)(20,49,23,52)(21,54,24,51)(25,92,28,95)(26,91,29,94)(27,96,30,93)(31,86,34,89)(32,85,35,88)(33,90,36,87)(37,80,40,83)(38,79,41,82)(39,84,42,81)(43,74,46,77)(44,73,47,76)(45,78,48,75), (1,9,5,7,3,11)(2,10,6,8,4,12)(13,21,17,19,15,23)(14,22,18,20,16,24)(25,35,27,31,29,33)(26,36,28,32,30,34)(37,47,39,43,41,45)(38,48,40,44,42,46)(49,59,51,55,53,57)(50,60,52,56,54,58)(61,71,63,67,65,69)(62,72,64,68,66,70)(73,81,77,79,75,83)(74,82,78,80,76,84)(85,93,89,91,87,95)(86,94,90,92,88,96), (1,31,7,25)(2,32,8,26)(3,33,9,27)(4,34,10,28)(5,35,11,29)(6,36,12,30)(13,43,19,37)(14,44,20,38)(15,45,21,39)(16,46,22,40)(17,47,23,41)(18,48,24,42)(49,82,55,76)(50,83,56,77)(51,84,57,78)(52,79,58,73)(53,80,59,74)(54,81,60,75)(61,94,67,88)(62,95,68,89)(63,96,69,90)(64,91,70,85)(65,92,71,86)(66,93,72,87)>;
G:=Group( (1,19)(2,20)(3,21)(4,22)(5,23)(6,24)(7,13)(8,14)(9,15)(10,16)(11,17)(12,18)(25,43)(26,44)(27,45)(28,46)(29,47)(30,48)(31,37)(32,38)(33,39)(34,40)(35,41)(36,42)(49,67)(50,68)(51,69)(52,70)(53,71)(54,72)(55,61)(56,62)(57,63)(58,64)(59,65)(60,66)(73,91)(74,92)(75,93)(76,94)(77,95)(78,96)(79,85)(80,86)(81,87)(82,88)(83,89)(84,90), (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), (1,68,4,71)(2,67,5,70)(3,72,6,69)(7,62,10,65)(8,61,11,64)(9,66,12,63)(13,56,16,59)(14,55,17,58)(15,60,18,57)(19,50,22,53)(20,49,23,52)(21,54,24,51)(25,92,28,95)(26,91,29,94)(27,96,30,93)(31,86,34,89)(32,85,35,88)(33,90,36,87)(37,80,40,83)(38,79,41,82)(39,84,42,81)(43,74,46,77)(44,73,47,76)(45,78,48,75), (1,9,5,7,3,11)(2,10,6,8,4,12)(13,21,17,19,15,23)(14,22,18,20,16,24)(25,35,27,31,29,33)(26,36,28,32,30,34)(37,47,39,43,41,45)(38,48,40,44,42,46)(49,59,51,55,53,57)(50,60,52,56,54,58)(61,71,63,67,65,69)(62,72,64,68,66,70)(73,81,77,79,75,83)(74,82,78,80,76,84)(85,93,89,91,87,95)(86,94,90,92,88,96), (1,31,7,25)(2,32,8,26)(3,33,9,27)(4,34,10,28)(5,35,11,29)(6,36,12,30)(13,43,19,37)(14,44,20,38)(15,45,21,39)(16,46,22,40)(17,47,23,41)(18,48,24,42)(49,82,55,76)(50,83,56,77)(51,84,57,78)(52,79,58,73)(53,80,59,74)(54,81,60,75)(61,94,67,88)(62,95,68,89)(63,96,69,90)(64,91,70,85)(65,92,71,86)(66,93,72,87) );
G=PermutationGroup([(1,19),(2,20),(3,21),(4,22),(5,23),(6,24),(7,13),(8,14),(9,15),(10,16),(11,17),(12,18),(25,43),(26,44),(27,45),(28,46),(29,47),(30,48),(31,37),(32,38),(33,39),(34,40),(35,41),(36,42),(49,67),(50,68),(51,69),(52,70),(53,71),(54,72),(55,61),(56,62),(57,63),(58,64),(59,65),(60,66),(73,91),(74,92),(75,93),(76,94),(77,95),(78,96),(79,85),(80,86),(81,87),(82,88),(83,89),(84,90)], [(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)], [(1,68,4,71),(2,67,5,70),(3,72,6,69),(7,62,10,65),(8,61,11,64),(9,66,12,63),(13,56,16,59),(14,55,17,58),(15,60,18,57),(19,50,22,53),(20,49,23,52),(21,54,24,51),(25,92,28,95),(26,91,29,94),(27,96,30,93),(31,86,34,89),(32,85,35,88),(33,90,36,87),(37,80,40,83),(38,79,41,82),(39,84,42,81),(43,74,46,77),(44,73,47,76),(45,78,48,75)], [(1,9,5,7,3,11),(2,10,6,8,4,12),(13,21,17,19,15,23),(14,22,18,20,16,24),(25,35,27,31,29,33),(26,36,28,32,30,34),(37,47,39,43,41,45),(38,48,40,44,42,46),(49,59,51,55,53,57),(50,60,52,56,54,58),(61,71,63,67,65,69),(62,72,64,68,66,70),(73,81,77,79,75,83),(74,82,78,80,76,84),(85,93,89,91,87,95),(86,94,90,92,88,96)], [(1,31,7,25),(2,32,8,26),(3,33,9,27),(4,34,10,28),(5,35,11,29),(6,36,12,30),(13,43,19,37),(14,44,20,38),(15,45,21,39),(16,46,22,40),(17,47,23,41),(18,48,24,42),(49,82,55,76),(50,83,56,77),(51,84,57,78),(52,79,58,73),(53,80,59,74),(54,81,60,75),(61,94,67,88),(62,95,68,89),(63,96,69,90),(64,91,70,85),(65,92,71,86),(66,93,72,87)])
60 conjugacy classes
class 1 2A ··· 2G 3A 3B 3C 4A ··· 4H 4I 4J 4K 4L 6A ··· 6N 6O ··· 6U 12A ··· 12P order 1 2 ··· 2 3 3 3 4 ··· 4 4 4 4 4 6 ··· 6 6 ··· 6 12 ··· 12 size 1 1 ··· 1 2 2 4 6 ··· 6 18 18 18 18 2 ··· 2 4 ··· 4 6 ··· 6
60 irreducible representations
dim 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 type + + + + + + - - + + - + + - + - + image C1 C2 C2 C2 C4 S3 D4 Q8 Dic3 D6 D6 Dic6 C4×S3 D12 C3⋊D4 S32 S3×Dic3 C3⋊D12 C32⋊2Q8 C2×S32 kernel C2×Dic3⋊Dic3 Dic3⋊Dic3 Dic3×C2×C6 C22×C3⋊Dic3 C6×Dic3 C22×Dic3 C62 C62 C2×Dic3 C2×Dic3 C22×C6 C2×C6 C2×C6 C2×C6 C2×C6 C23 C22 C22 C22 C22 # reps 1 4 2 1 8 2 2 2 4 4 2 8 4 4 4 1 2 2 2 1
Matrix representation of C2×Dic3⋊Dic3 in GL8(𝔽13)
12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
,
12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 1 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 0 12 1 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
,
9 10 0 0 0 0 0 0 10 4 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1
,
12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 12 12
,
0 1 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 12 12
G:=sub<GL(8,GF(13))| [12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1],[12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,12,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,12,12,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1],[9,10,0,0,0,0,0,0,10,4,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1],[12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,12,0,0,0,0,0,0,1,12],[0,12,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,5,0,0,0,0,0,0,0,0,5,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,12,0,0,0,0,0,0,0,12] >;
C2×Dic3⋊Dic3 in GAP, Magma, Sage, TeX
C_2\times {\rm Dic}_3\rtimes {\rm Dic}_3
% in TeX
G:=Group("C2xDic3:Dic3");
// GroupNames label
G:=SmallGroup(288,613);
// by ID
G=gap.SmallGroup(288,613);
# by ID
G:=PCGroup([7,-2,-2,-2,-2,-2,-3,-3,112,253,64,1356,9414]);
// Polycyclic
G:=Group<a,b,c,d,e|a^2=b^6=d^6=1,c^2=b^3,e^2=d^3,a*b=b*a,a*c=c*a,a*d=d*a,a*e=e*a,c*b*c^-1=b^-1,b*d=d*b,b*e=e*b,c*d=d*c,e*c*e^-1=b^3*c,e*d*e^-1=d^-1>;
// generators/relations
×
𝔽
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2020-05-30 04:58:52
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|
http://einsteintoolkit.org/thornguide/EinsteinAnalysis/ADMAnalysis/documentation.html
|
Date
Abstract
Basic analysis of the metric and extrinsic curvature tensors
### 1 Purpose
This thorn provides analysis routines to calculate the following quantities:
• The trace of the extrinsic curvature ($trK$).
• The determinant of the 3-metric ($detg$).
• The components of the 3-metric in spherical coordinates
(${g}_{rr},{g}_{r𝜃},{g}_{r\varphi },{g}_{𝜃𝜃},{g}_{\varphi 𝜃},{g}_{\varphi \varphi }$).
• The components of the extrinsic curvature in spherical coordinates
(${K}_{rr},{K}_{r𝜃},{K}_{r\varphi },{K}_{𝜃𝜃},{K}_{𝜃\varphi },{K}_{\varphi \varphi }$).
• The components of the 3-Ricci tensor in cartesian coordinates
(${\mathsc{ℛ}}_{ij}$) for $i,j\in \left\{1,2,3\right\}$.
• The Ricci scalar ($\mathsc{ℛ}$).
### 2 Trace of Extrinsic Curvature
The trace of the extrinsic curvature at each point on the grid is placed in the grid function trK. The algorithm for calculating the trace uses the physical metric, that is it includes any conformal factor.
$trK\equiv trK=\frac{1}{{\psi }^{4}}{g}^{ij}{K}_{ij}$ (1)
### 3 Determinant of 3-Metric
The determinant of the 3-metric at each point on the grid is placed in the grid function detg. This is always the determinant of the conformal metric, that is it does not include any conformal factor.
$detg\equiv detg=-{g}_{13}^{2}\ast {g}_{22}+2\ast {g}_{12}\ast {g}_{13}\ast {g}_{23}-{g}_{11}\ast {g}_{23}^{2}-{g}_{12}^{2}\ast {g}_{33}+{g}_{11}\ast {g}_{22}\ast {g}_{33}$ (2)
### 4 Transformation to Spherical Cooordinates
The values of the metric and/or extrinsic curvature in a spherical polar coordinate system $\left(r,𝜃,\varphi \right)$ evaluated at each point on the computational grid are placed in the grid functions (grr, grt, grp, gtt, gtp, gpp) and (krr, krt, krp, ktt, ktp, kpp). In the spherical transformation, the $𝜃$ coordinate is referred to as q and the $\varphi$ as p.
The general transformation from Cartesian to Spherical for such tensors is
$\begin{array}{rcll}{A}_{rr}& =& {sin}^{2}𝜃{cos}^{2}\varphi {A}_{xx}+{sin}^{2}𝜃{sin}^{2}\varphi {A}_{yy}+{cos}^{2}𝜃{A}_{zz}+2{sin}^{2}𝜃cos\varphi sin\varphi {A}_{xy}& \text{}\\ & & +2sin𝜃cos𝜃cos\varphi {A}_{xz}+2sin𝜃cos𝜃sin\varphi {A}_{yz}& \text{}\\ {A}_{r𝜃}& =& r\left(sin𝜃cos𝜃{cos}^{2}\varphi {A}_{xx}+2\ast sin𝜃cos𝜃sin\varphi cos\varphi {A}_{xy}+\left({cos}^{2}𝜃-{sin}^{2}𝜃\right)cos\varphi {A}_{xz}& \text{}\\ & & +sin𝜃cos𝜃{sin}^{2}\varphi {A}_{yy}+\left({cos}^{2}𝜃-{sin}^{2}𝜃\right)sin\varphi {A}_{yz}-sin𝜃cos𝜃{A}_{zz}\right)& \text{}\\ {A}_{r\varphi }& =& rsin𝜃\left(-sin𝜃sin\varphi cos\varphi {A}_{xx}-sin𝜃\left({sin}^{2}\varphi -{cos}^{2}\varphi \right){A}_{xy}-cos𝜃sin\varphi {A}_{xz}& \text{}\\ & & +sin𝜃sin\varphi cos\varphi {A}_{yy}+cos𝜃cos\varphi {A}_{yz}\right)& \text{}\\ {A}_{𝜃𝜃}& =& {r}^{2}\left({cos}^{2}𝜃{cos}^{2}\varphi {A}_{xx}+2{cos}^{2}𝜃sin\varphi cos\varphi {A}_{xy}-2sin𝜃cos𝜃cos\varphi {A}_{xz}+{cos}^{2}𝜃{sin}^{2}\varphi {A}_{yy}& \text{}\\ & & -2sin𝜃cos𝜃sin\varphi {A}_{yz}+{sin}^{2}𝜃{A}_{zz}\right)& \text{}\\ {A}_{𝜃\varphi }& =& {r}^{2}sin𝜃\left(-cos𝜃sin\varphi cos\varphi {A}_{xx}-cos𝜃\left({sin}^{2}\varphi -{cos}^{2}\varphi \right){A}_{xy}+sin𝜃sin\varphi {A}_{xz}& \text{}\\ & & +cos𝜃sin\varphi cos\varphi {A}_{yy}-sin𝜃cos\varphi {A}_{yz}\right)& \text{}\\ {A}_{\varphi \varphi }& =& {r}^{2}{sin}^{2}𝜃\left({sin}^{2}\varphi {A}_{xx}-2sin\varphi cos\varphi {A}_{xy}+{cos}^{2}\varphi {A}_{yy}\right)& \text{}\end{array}$
If the parameter normalize_dtheta_dphi is set to yes, the angular components are projected onto the vectors $\left(rd𝜃,rsin𝜃d\varphi \right)$ instead of the default vector $\left(d𝜃,d\varphi \right)$. That is,
$\begin{array}{rcll}{A}_{𝜃𝜃}& \to & {A}_{𝜃𝜃}∕{r}^{2}& \text{}\\ {A}_{\varphi \varphi }& \to & {A}_{\varphi \varphi }∕\left({r}^{2}{sin}^{2}𝜃\right)& \text{}\\ {A}_{r𝜃}& \to & {A}_{r𝜃}∕r& \text{}\\ {A}_{r\varphi }& \to & {A}_{r\varphi }∕\left(rsin𝜃\right)& \text{}\\ {A}_{𝜃\varphi }& \to & {A}_{𝜃\varphi }∕{r}^{2}sin𝜃\right)& \text{}\end{array}$
### 5 Computing the Ricci tensor and scalar
The computation of the Ricci tensor uses the ADMMacros thorn. The calculation of the Ricci scalar uses the generic trace routine in this thorn.
### 6 Parameters
normalize_dtheta_dphi Scope: private BOOLEAN Description: Project angular components onto r*dtheta and r*sin(theta)*dphi? Default: no
ricci_persist Scope: private BOOLEAN Description: Keep storage of the Ricci tensor and scalar around? Default: no
ricci_prolongation_type Scope: private KEYWORD Description: The kind of boundary prolongation for the Ricci tensor and scalar Range Default: none Lagrange standard prolongation (requires several time levels) copy use data from the current time level (requires only one time level) none no prolongation (use this if you do not have enough time levels active)
ricci_timelevels Scope: private INT Description: Number of time levels for the Ricci tensor and scalar Range Default: 1 1:3
metric_type Scope: shared from ADMBASE KEYWORD
### 7 Interfaces
Implements:
Inherits:
staticconformal
grid
#### Grid Variables
##### 7.0.1 PUBLIC GROUPS
Group Names Variable Names Details trace_of_k compact 0 trK description trace of extrinsic curvature dimensions 3 distribution DEFAULT group type GF tags tensortypealias=”scalar” Prolongation=”none” timelevels 1 variable type REAL detofg compact 0 detg description determinant of the conformal metric dimensions 3 distribution DEFAULT group type GF tags tensortypealias=”scalar” Prolongation=”none” timelevels 1 variable type REAL spherical_metric compact 0 grr description Metric in spherical coordinates gqq dimensions 3 gpp distribution DEFAULT grq group type GF grp tags Prolongation=”none” gqp timelevels 1 variable type REAL spherical_curv compact 0 krr description extrinsic curvature in spherical coordinates kqq dimensions 3 kpp distribution DEFAULT krq group type GF krp tags Prolongation=”none” kqp timelevels 1 variable type REAL ricci_tensor compact 0 Ricci11 description Components of the Ricci tensor Ricci12 dimensions 3 Ricci13 distribution DEFAULT Ricci22 group type GF Ricci23 tags tensortypealias=”dd_sym” ProlongationParameter=”ADMAnalysis::ricci_prolongation_type” Ricci33 timelevels 3 variable type REAL ricci_scalar compact 0 Ricci description The Ricci scalar dimensions 3 distribution DEFAULT group type GF tags tensortypealias=”scalar” ProlongationParameter=”ADMAnalysis::ricci_prolongation_type” timelevels 3 variable type REAL
Symmetry.h
Provides:
CartToSphere to
Trace to
### 8 Schedule
This section lists all the variables which are assigned storage by thorn EinsteinAnalysis/ADMAnalysis. Storage can either last for the duration of the run (Always means that if this thorn is activated storage will be assigned, Conditional means that if this thorn is activated storage will be assigned for the duration of the run if some condition is met), or can be turned on for the duration of a schedule function.
#### Storage
Conditional: ricci_tensor[1] ricci_scalar[1] ricci_tensor[2] ricci_scalar[2] ricci_tensor[3] ricci_scalar[3]
#### Scheduled Functions
CCTK_PARAMCHECK
check that the metric_type is recognised
Language: c Options: global Type: function
CCTK_WRAGH (conditional)
register symmetry of ricci tensor and scalar
Language: c Options: global Type: function
CCTK_ANALYSIS (conditional)
riccigroup
calculate ricci tensor, with boundary conditions
Storage: ricci_tensor[1] ricci_scalar[1] detofg Triggers: ricci_tensor ricci_scalar Type: group
RicciGroup
calculate ricci tensor, with boundary conditions
Language: c Type: function
RicciGroup
ricciboundariesgroup
set ricci tensor on the boundary
After: admanalysis_ricci Type: group
RicciBoundariesGroup
select boundary conditions for the ricci tensor
Language: c Options: level Sync: ricci_tensor ricci_scalar Type: function
RicciBoundariesGroup
applybcs
apply boundary conditions to the ricci tensor
After: admanalysis_ricci Type: group
CCTK_ANALYSIS
compute the trace of the extrinsic curvature and the determinant of the metric
Language: c Storage: trace_of_k detofg Sync: trace_of_k detofg Triggers: trace_of_k detofg Type: function
CCTK_ANALYSIS
calculate the spherical metric in r,theta(q), phi(p)
Language: c Storage: spherical_metric Sync: spherical_metric Triggers: spherical_metric Type: function
CCTK_ANALYSIS
calculate the spherical ex. curvature in r, theta(q), phi(p)
Language: c Storage: spherical_curv Sync: spherical_curv Triggers: spherical_curv Type: function
CCTK_POSTINITIAL (conditional)
riccigroup
calculate ricci tensor, with boundary conditions
After: mol_poststep mol_postinitial Storage: detofg Type: group
CCTK_POST_RECOVER_VARIABLES (conditional)
riccigroup
calculate ricci tensor, with boundary conditions
After: mol_poststep mol_postinitial Storage: detofg Type: group
CCTK_EVOL (conditional)
riccigroup
calculate ricci tensor, with boundary conditions
After: mol_evolution Storage: detofg Type: group
CCTK_POSTREGRID (conditional)
ricciboundariesgroup
set ricci tensor on the boundary
Storage: detofg Type: group
CCTK_POSTRESTRICT (conditional)
ricciboundariesgroup
set ricci tensor on the boundary
Type: group
#### Aliased Functions
Alias Name: Function Name: ApplyBCs ADMAnalysis_ApplyBCs
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2017-06-26 22:32:55
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http://katlas.org/wiki/Data:9_15/Integral_Khovanov_Homology
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# Data:9 15/Integral Khovanov Homology
$\dim{\mathcal G}_{2r+i}\operatorname{KH}^r_{\mathbb Z}$ $i=1$ $i=3$ $r=-2$ ${\mathbb Z}$ $r=-1$ ${\mathbb Z}\oplus{\mathbb Z}_2$ ${\mathbb Z}$ $r=0$ ${\mathbb Z}^{3}\oplus{\mathbb Z}_2$ ${\mathbb Z}^{2}$ $r=1$ ${\mathbb Z}^{4}\oplus{\mathbb Z}_2^{2}$ ${\mathbb Z}^{2}$ $r=2$ ${\mathbb Z}^{3}\oplus{\mathbb Z}_2^{4}$ ${\mathbb Z}^{4}$ $r=3$ ${\mathbb Z}^{3}\oplus{\mathbb Z}_2^{3}$ ${\mathbb Z}^{3}$ $r=4$ ${\mathbb Z}^{3}\oplus{\mathbb Z}_2^{3}$ ${\mathbb Z}^{3}$ $r=5$ ${\mathbb Z}\oplus{\mathbb Z}_2^{3}$ ${\mathbb Z}^{3}$ $r=6$ ${\mathbb Z}\oplus{\mathbb Z}_2$ ${\mathbb Z}$ $r=7$ ${\mathbb Z}_2$ ${\mathbb Z}$
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2019-03-22 06:34:53
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http://physgre.s3-website-us-east-1.amazonaws.com/2008%20html/2008%20problem%206.html
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## Solution to 2008 Problem 6
There are $2$ $n = 1$ orbitals: \begin{enumerate}\item \text{$l = 0$, $m_l = 0$, $m_s = 1/2$}\item \text{$l = 0$, $m_l = 0$, $m_s = -1/2$}\end{enumerate}There are $8$ $n = 2$ orbitals:\begin{enumerate}\item \text{$l = 0$, $m_l = 0$, $m_s = 1/2$}\item \text{$l = 0$, $m_l = 0$, $m_s = -1/2$}\item \text{$l = 1$, $m_l = 1$, $m_s = 1/2$}\item \text{$l = 1$, $m_l = 1$, $m_s = -1/2$}\item \text{$l = 1$, $m_l = 0$, $m_s = 1/2$}\item \text{$l = 1$, $m_l = 0$, $m_s = -1/2$}\item \text{$l = 1$, $m_l = -1$, $m_s = 1/2$}\item \text{$l = 1$, $m_l = -1$, $m_s = -1/2$}\end{enumerate}This makes a total of $\boxed{10}$. Therefore, answer (E) is correct.
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2017-04-27 22:31:46
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http://accessmedicine.mhmedical.com/content.aspx?bookid=331§ionid=40727087
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Chapter 294
Disorders of absorption constitute a broad spectrum of conditions with multiple etiologies and varied clinical manifestations. Almost all of these clinical problems are associated with diminished intestinal absorption of one or more dietary nutrients and are often referred to as the malabsorption syndrome. This term is not ideal as it represents a pathophysiologic state, does not provide an etiologic explanation for the underlying problem, and should not be considered an adequate final diagnosis. The only clinical situations in which absorption is increased are hemochromatosis and Wilson's disease, in which absorption of iron and copper, respectively, are increased.
Most, but not all, malabsorption syndromes are associated with steatorrhea, an increase in stool fat excretion of >6% of dietary fat intake. Some malabsorption disorders are not associated with steatorrhea: primary lactase deficiency, a congenital absence of the small intestinal brush border disaccharidase enzyme lactase, is associated with lactose “malabsorption,” and pernicious anemia is associated with a marked decrease in intestinal absorption of cobalamin (vitamin B12) due to an absence of gastric parietal cell intrinsic factor required for cobalamin absorption.
Disorders of absorption must be included in the differential diagnosis of diarrhea (Chap. 40). First, diarrhea is frequently associated with and/or is a consequence of the diminished absorption of one or more dietary nutrients. The diarrhea may be secondary either to the intestinal process that is responsible for the steatorrhea or to steatorrhea per se. Thus, celiac disease (see below) is associated with both extensive morphologic changes in the small intestinal mucosa and reduced absorption of several dietary nutrients; in contrast, the diarrhea of steatorrhea is the result of the effect of nonabsorbed dietary fatty acids on intestinal, usually colonic, ion transport. For example, oleic acid and ricinoleic acid (a bacterially hydroxylated fatty acid that is also the active ingredient in castor oil, a widely used laxative) induce active colonic Cl ion secretion, most likely secondary to increasing intracellular Ca. In addition, diarrhea per se may result in mild steatorrhea (<11 g fat excretion while on a 100-g fat diet). Second, most patients will indicate that they have diarrhea, not that they have fat malabsorption. Third, many intestinal disorders that have diarrhea as a prominent symptom (e.g., ulcerative colitis, traveler's diarrhea secondary to an enterotoxin produced by Escherichia coli) do not necessarily have diminished absorption of any dietary nutrient.
Diarrhea as a symptom (i.e., when used by patients to describe their bowel movement pattern) may be a decrease in stool consistency, an increase in stool volume, an increase in number of bowel movements, or any combination of these three changes. In contrast, diarrhea as a sign is a quantitative increase in stool water or weight of >200–225 mL or gram per 24 h, when a Western-type diet is consumed. Individuals consuming a diet with higher fiber content may normally have a stool weight of up to 400 g/24 h. Thus, the clinician must clarify what an individual patient means by diarrhea. Some 10% of patients referred to gastroenterologists for further evaluation of unexplained diarrhea do not have an increase in stool water when it is determined quantitatively. Such patients may have small, frequent, somewhat loose bowel movements ...
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2015-03-03 04:41:11
|
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https://math.stackexchange.com/questions/2501365/find-the-limits-using-squeeze-theorem-or-another-theorem
|
# Find the limits (using Squeeze theorem or another theorem)
I am not sure if I calculate the limits right. I would be grateful if somebody could check it.
Find the following limits:
1) $\lim_{x \to 0^{+}}\sqrt{x}cos(\frac{1}{x^2})$
Let us notice that $-\sqrt{x}<\sqrt{x}cos(\frac{1}{x^2})<\sqrt{x}$ and since:
$\lim_{x \to 0^{+}}\sqrt{x}=\lim_{x \to 0^{+}}-\sqrt{x}=0$,
we can write that $\lim_{x \to 0^{+}}\sqrt{x}cos(\frac{1}{x^2})=0$.
Could I just write $cos(\frac{1}{x^2})$ is bounded and $\lim_{x \to 0^{+}}\sqrt{x}=0$ and from this fact conclude that the product of these functions goes to 0 (instead of what I wrote above)?
2) $\lim_{x \to -\infty}\frac{sin(x^2)}{x}$
Let us notice that $\lim_{x \to -\infty}\frac{1}{x}=0$ and $sin(x^2)$ is bounded, so $\lim_{x \to -\infty}\frac{sin(x^2)}{x}=0$.
3) $\lim_{x \to +\infty}\frac{2x+sin(x^2)}{3x+cos(\sqrt{x})}$
Let us notice that $\frac{2x-1}{3x+1}<\frac{2x+sin(x^2)}{3x+cos(\sqrt{x})}<\frac{2x+1}{3x-1}$ and since:
$\lim_{x \to +\infty}\frac{2x-1}{3x+1}=\lim_{x \to +\infty}\frac{2x+1}{3x-1}=2/3$,
we can write that $\lim_{x \to +\infty}\frac{2x+sin(x^2)}{3x+cos(\sqrt{x})}=2/3$.
4) $\lim_{x \to 0^{+}}\frac{2+sin(\frac{1}{x})}{x^3}$
We can write that $\frac{2-1}{x^3}<\frac{2+sin(\frac{1}{x})}{x^3}$, and since
$\lim_{x \to 0^{+}}\frac{1}{x^3}=+\infty$ so
$\lim_{x \to 0^{+}}\frac{2+sin(\frac{1}{x})}{x^3}=+\infty$.
• Yes, this is all correct. Well done. – Cornman Nov 2 '17 at 14:07
• I think your argument is good. – Eclipse Sun Nov 2 '17 at 14:08
• Thank you very much. And what with 1)? Should I use the Squeeze theorem or can I use the thoerem about a product of a bounded function and a convergent function? – SigmaMat Nov 2 '17 at 14:12
For the first limit, you´re argument is definitely right, cause you can easily use theorem, that if you can find neighborhood of point $x_{0},$ such that function $g(x)$ is bounded in there, and $lim_{x \rightarrow x_{0}} f(x)=0,$ then $lim _{x \rightarrow x_{0}}f(x)g(x)=0.$
• Sure, just for the squeeze, change $-\sqrt{x}<\sqrt{x}cos(\frac{1}{x^2})<\sqrt{x}$ to $-\sqrt{x}\leq \sqrt{x}cos(\frac{1}{x^2})\leq \sqrt{x}$. – stanly Nov 2 '17 at 14:29
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2021-04-15 11:57:35
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https://ee.gateoverflow.in/625/gate2014-3-14
|
In a long transmission line with $r$,$l$,$g$ and $c$ are the resistance, inductance, shunt conductance and capacitance per unit length, respectively, the condition for distortionless transmission is
1. $rc=lg$
2. $r=\sqrt{l/c}$
3. $rg=lc$
4. $g=\sqrt{c/l}$
|
2019-12-06 07:37:39
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https://studyqas.com/acomputer-valued-at-6500-depreciates-at-the-rate-of-14-3/
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# Acomputer valued at $6500 depreciates at the rate of 14.3% per year. find the value (round Acomputer valued at$6500 depreciates at the rate of 14.3% per year. find the value (round to the nearest dollar) of the computer after three years.
## This Post Has 5 Comments
1. NeverEndingCycle says:
4091.25
You can get this number by multiplying by .857, which is the amount remaining after taking off 14.3%. Do this for all 3 years to get the answer above.
2. Expert says:
d.
121π mi^2
hope you got it chief
3. Expert says:
answer: (c) m∠qpo + (2x + 16)° = 180°
step-by-step explanation:
the opposite angles of a quadrilateral are supplementary.
so, ∠o + ∠q = 180° and ∠p + ∠r = 180°
since ∠r = 2x + 16°, we can use substitution as follows:
∠p + ∠r = 180°
∠p + (2x + 16)° = 180°
4. solikhalifeoy3j1r says:
After 3 years the computer will be valued at 3711.5. I'm sure you can round it up to 3712.
Hope this helps 🙂
5. yarrito20011307 says:
Given that,
Value of a computer= $6500 Depreciation rate= 14.3% Now, depreciation after 1st year= 6500 x 14.3% Depreciation after 1st year= 929.5 Value of a computer after 1 year = 6500 – 929.5= 5570.5 Depreciation after 2nd year= 5570.5 x 14.3% Depreciation after 2nd year= 796.58 Value of the computer after 2 year=5570.5-796.58=4773.92 Depreciation after 3rd year= 4773.92 x 14.3% Depreciation after 3rd year= 682.67 Value of the computer after 3 year= 4773.92-682.67= 4091.25 Therefore, the value of the computer after 3 year is$4091 approximately.
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2022-11-27 01:15:04
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https://www.inchmeal.io/htpi/ch-2/sec-2.2.html
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# How to Prove It - Solutions
## Chapter - 2, Quantificational Logic
### Summary
• Equivalances:
• $\lnot \forall x P(x) = \exists x \lnot P(x)$.
• $\lnot \exists x P(x) = \forall x \lnot P(x)$.
• $\exists ! x P(x)$ means P(x) is true for exactly one value of x. It is equivalent to
$\exists x (P(x) \land \lnot \exists y (P(y) \land \lnot (y = x)))$.
• Bounded Quantifiers:
• $\exists x \in A P(x) = \exists x (x \in A \land P(x))$.
• $\forall x \in A P(x) = \forall x (x \in A \to P(x))$.
• Similar to negation equivalances( in first point), Equivalences for Bounded Quantifiers:
• $\lnot \forall x \in A P(x) = \exists x \in A \lnot P(x)$.
• $\lnot \exists x \in A P(x) = \forall x \in A \lnot P(x)$.
• If $A = \phi$ then:
• $\exists x \in A P(x)$ is always false irrespective of P(x).
• $\forall x \in A P(x)$ is true irrespective of P(x).
• Universal Quantifiers distributes over conjunction i.e. $\forall x (P(x) \land Q(x)) = \forall x P(x) \land \forall x Q(x)$. But Universal Quantifiers does not distributes over disjunction.
• Existential Quantifier distributes over disjunction but does not distributes over conjunction.
Soln1
(a) $\forall x (M(x) \to \exists y(F(x, y) \land H(y))$
where $M(x)$ means $x$ is maths major,
and $F(x,y)$ means $x$ is a friend of $y$,
and $H(y)$ means $y$ needs help in homework.
Negation of the above statement:
$\quad \lnot \forall x (M(x) \to \exists y(F(x, y) \land H(y))$
$\quad = \exists x \lnot (M(x) \to \exists y(F(x, y) \land H(y))$
$\quad = \exists x \lnot (\lnot M(x) \lor \exists y(F(x, y) \land H(y))$
$\quad = \exists x (\lnot \lnot M(x) \land \lnot \exists y(F(x, y) \land H(y))$
$\quad = \exists x ( M(x) \land \forall y \lnot(F(x, y) \land H(y))$
$\quad = \exists x ( M(x) \land \forall y (\lnot F(x, y) \lor \lnot H(y))$
$\quad = \exists x ( M(x) \land \forall y ( F(x, y) \to \lnot H(y))$
There exists a maths major and all of his friends don’t need help in their homework.
(b) $\forall x \exists y(R(x,y) \land \forall z \lnot L(y,z))$
where $R(x,y)$ means $x$ is the roommate of $y$,
and $L(y,z)$ means $y$ likes $z$.
Negation of the above statement:
$\lnot \forall x \exists y(R(x,y) \land \forall z \lnot L(y,z))$
$\quad = \exists x \lnot \exists y(R(x,y) \land \forall z \lnot L(y,z))$
$\quad = \exists x \forall y \lnot (R(x,y) \land \forall z \lnot L(y,z))$
$\quad = \exists x \forall y (\lnot R(x,y) \lor \lnot \forall z \lnot L(y,z))$
$\quad = \exists x \forall y (\lnot R(x,y) \lor \exists z \lnot \lnot L(y,z))$
$\quad = \exists x \forall y (\lnot R(x,y) \lor \exists z L(y,z))$
$\quad = \exists x \forall y (R(x,y) \to \exists z L(y,z))$
There exists someone whose all roommates likes atleast one person.
(c) Required statement is:
$\lnot \forall x ((x \in A \lor x \in B) \to (x \in C \land x \notin D))$
$\quad = \exists x \lnot ((x \in A \lor x \in B) \to (x \in C \land x \notin D))$
$\quad = \exists x \lnot (\lnot (x \in A \lor x \in B) \lor (x \in C \land x \notin D))$
$\quad = \exists x (\lnot \lnot (x \in A \lor x \in B) \land \lnot (x \in C \land x \notin D))$
$\quad = \exists x ((x \in A \lor x \in B) \land (x \notin C \lor x \in D))$
(d) Required statement is: $\lnot \exists x \forall y[ y > x \to \exists z(z^2 + 5z = y)]$
$\quad = \forall x \lnot \forall y[ y > x \to \exists z(z^2 + 5z = y)]$
$\quad = \forall x \exists y \lnot [ y > x \to \exists z(z^2 + 5z = y)]$
$\quad = \forall x \exists y \lnot [ \lnot (y > x) \lor \exists z(z^2 + 5z = y)]$
$\quad = \forall x \exists y [ \lnot \lnot (y > x) \land \lnot \exists z(z^2 + 5z = y)]$
$\quad = \forall x \exists y [ (y > x) \land \forall z \lnot(z^2 + 5z = y)]$
$\quad = \forall x \exists y [ (y > x) \land \forall z (z^2 + 5z \neq y)]$
Soln2
(a) $\exists x (F(x) \land \lnot \exists y R(x,y))$.
Negation of above statement:
$\quad = \lnot \exists x (F(x) \land \lnot \exists y R(x,y))$
$\quad = \forall x \lnot (F(x) \land \lnot \exists y R(x,y))$
$\quad = \forall x (\lnot F(x) \lor \lnot \lnot \exists y R(x,y))$
$\quad = \forall x (\lnot F(x) \lor \exists y R(x,y))$
$\quad = \forall x ( F(x) \to \exists y R(x,y))$.
where $F(x)$ means $x$ is a freshman.
and $R(x,y)$ means $x$ and $y$ are roommates.
(b) $\forall x \exists y L(x,y) \land \lnot \exists x \forall y L(x,y)$
Negation of above statement:
$\lnot (\forall x \exists y L(x,y) \land \lnot \exists x \forall y L(x,y))$
$\quad = (\lnot (\forall x \exists y L(x,y)) \lor \lnot (\lnot \exists x \forall y L(x,y)))$
$\quad = (\exists x \lnot \exists y L(x,y)) \lor \exists x \forall y L(x,y)$
$\quad = (\exists x \forall y \lnot L(x,y)) \lor \exists x \forall y L(x,y)$
Either there exists someone who does not like anyone or there exists someone who likes everyone.
(c) $\forall a \in A \exists b \in B (a \in C \leftrightarrow b \in C)$.
This is equivalent to:
$\forall a \in A \exists b \in B (a \in C \to b \in C) \land \forall a \in A \exists b \in B (b \in C \to a \in C)$.
Negation of the above statement:
$\lnot (\forall a \in A \exists b \in B (a \in C \to b \in C) \land \forall a \in A \exists b \in B (b \in C \to a \in C))$
$\lnot \forall a \in A \exists b \in B (a \in C \to b \in C) \lor \lnot \forall a \in A \exists b \in B (b \in C \to a \in C)$
$\exists a \in A \lnot \exists b \in B (a \in C \to b \in C) \lor \exists a \in A \lnot \exists b \in B (b \in C \to a \in C)$
$\exists a \in A \forall b \in B \lnot (a \in C \to b \in C) \lor \exists a \in A \forall b \in B \lnot(b \in C \to a \in C)$
$\exists a \in A \forall b \in B \lnot (a \notin C \lor b \in C) \lor \exists a \in A \forall b \in B \lnot(b \notin C \lor a \in C)$
$\exists a \in A \forall b \in B (a \in C \land b \notin C) \lor \exists a \in A \forall b \in B(b \in C \land a \notin C)$
There exists an a in A such that for all values of b in B, either a is in C and b is not in C, or a is not in C and b is in C.
(d) Required statement is:
$\lnot \forall y \gt 0 \exists x (ax^2 + bx + c = y)$
$\exists y \gt 0 \lnot \exists x (ax^2 + bx + c = y)$
$\exists y \gt 0 \forall x \lnot (ax^2 + bx + c = y)$
$\exists y \gt 0 \forall x (ax^2 + bx + c \neq y)$
Soln3
(a) All possible values of x are 0,1,2,3,4,5,6. It can be easily verified that there exists a,b and c such that $a^2 + b^2 + c^2 = x$ for all possible values of x. Thus statement is True.
(b) False. x has two possible values 1 and 7.
(c) True. x has two values -1 and 9. But as x is Natural number. Thus only x has one posssible value 9.
(d) True. x = 9 and y = 9.
Soln4
Given: $\lnot \exists x P(x) = \forall x \lnot P(x)$.
To Prove: $\lnot \forall x P(x) = \exists x \lnot P(x)$.
Putting $P(x) = \lnot Q(x)$ in the given equivalence:
$\lnot \exists x \lnot Q(x) = \forall x Q(x)$
Taking negation on both sides:
$\lnot \lnot \exists x \lnot Q(x) = \lnot \forall x Q(x)$
$\exists x \lnot Q(x) = \lnot \forall x Q(x)$
Hence Proved.
Soln5
To Prove: $\lnot \exists x \in A P(x)$ is equivalent to $\forall x \in A \lnot P(x)$.
LHS: $\lnot \exists x \in A P(x)$
$\quad = \lnot \exists x (x \in A \land P(x))$
$\quad = \forall x \lnot (x \in A \land P(x))$
$\quad = \forall x (x \notin A \lor \lnot P(x))$
$\quad = \forall x (x \in A \to \lnot P(x))$
$\quad = \forall x \in A \lnot P(x)$ Hence Proved.
Soln6
To Prove $\exists x(P(x) \lor Q(x))$ is equivalent to $\exists x P(x) \lor \exists x Q(x)$.
Taking LHS: $\exists x(P(x) \lor Q(x))$
$\quad = \lnot \lnot \exists x(P(x) \lor Q(x))$
$\quad = \lnot \forall x \lnot (P(x) \lor Q(x))$
$\quad = \lnot \forall x (\lnot P(x) \land \lnot Q(x))$
$\quad = \lnot (\forall x \lnot P(x) \land \forall x \lnot Q(x))$
$\quad = \lnot \forall x \lnot P(x) \lor \lnot \forall x \lnot Q(x))$
$\quad = \exists x \lnot \lnot P(x) \lor \exists x \lnot \lnot Q(x))$
$\quad = \exists x P(x) \lor \exists x Q(x))$ Hence Proved.
Soln7
To Prove $\exists x (P(x) \to Q(x))$ is equivalent to $\forall x P(x) \to \exists x Q(x)$.
Starting from LHS $\exists x (P(x) \to Q(x))$
$\quad = \exists x (\lnot P(x) \lor Q(x))$
$\quad = \exists x \lnot P(x) \lor \exists x Q(x)$
$\quad = \lnot \forall x P(x) \lor \exists x Q(x)$
$\quad = \forall x P(x) \to \exists x Q(x)$ Hence Proved.
Soln8
To Prove: $(\forall x \in A P(x)) \land (\forall x \in B P(x))$ is equivalent to $\forall x \in (A \lor B)P(x)$
Starting from LHS:
$(\forall x \in A P(x)) \land (\forall x \in B P(x))$
$\quad = (\forall x (x \in A \to P(x))) \land (\forall x (x \in B \to P(x)))$
$\quad = \forall x ((x \in A \to P(x)) \land (x \in B \to P(x)))$
$\quad = \forall x ((x \notin A \lor P(x)) \land (x \notin B \lor P(x)))$
Applying reverse distributive law:
$\quad = \forall x ((x \notin A \land x \notin B) \lor P(x))$
$\quad = \forall x ((\lnot(x \in A \lor x \in B)) \lor P(x))$
$\quad = \forall x ((x \in A \lor x \in B) \to P(x))$
$\quad = \forall (x \in A \lor x \in B) P(x))$
$\quad = \forall x \in (A \lor B) P(x))$ Hence Proved.
Soln9
Statement $\forall x (P(x) \lor Q(x))$ is not equivalent to $\forall x P(x) \lor \forall xQ(x)$.
Assigning $P(x) = true$, if x is even.
and $Q(x) = true$, if x is odd.
Lets have Universe as all Natural Numbers.
Clearly $\forall x (P(x) \lor Q(x))$ is true as Every number is either even or odd.
But $\forall x P(x) \lor \forall xQ(x)$ is not true. A neither all numbers are even not all numbers are odd.
Soln10
(a) $\exists x \in A P(x) \lor \exists x \in B P(x)$
$\quad = \exists x (x \in A \land P(x)) \lor \exists x (x \in B \land P(x))$
$\quad = \exists x ((x \in A \land P(x)) \lor (x \in B \land P(x)))$
Using Law of distribution in reverse:
$\quad = \exists x ((x \in A \lor x \in B) \land P(x))$
$\quad = \exists x ((x \in (A \lor B)) \land P(x))$
$\quad = \exists x \in (A \lor B) P(x))$ Hence Proved.
(b)
$\exists x \in A P(x) \land \exists x \in B P(x)$ is not equivalent to $\exists x \in (A \land B) P(x)$.
If A and B are disjoint set, then clearly RHS will be false as $A \land B = \phi$. But even in this case for some values of x LHS can be true.
Soln11
$A \subseteq B$ is equivalent to
$\quad = \forall x (x \in A \to x \in B)$
Also, $A \setminus B = \phi$
No elements exist in the $A \setminus B$ set. Thus it is equivalent to:
$\lnot \exists x (x \in (A \setminus B))$
$\quad = \lnot \exists x (x \in A \land x \notin B)$
$\quad = \forall x \lnot (x \in A \land x \notin B)$
$\quad = \forall x (x \notin A \lor x \in B)$
$\quad = \forall x (x \in A \to x \in B)$
Thus both are same.
Soln12
(a) There is exactly one student who is taught by x.
(b) There exists atleast one teacher having exactly one student.
(c) There is exactly one teacher having at-least one student.
(d) There atleast exist one student having exactly one teacher.
(e) There is only one teacher having only one student.
(f) There is exactly one teacher having only one student.
Clearly (e) and (f) are same.
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2021-03-02 04:33:47
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https://docs.sui.io/contribute/observability
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Announcing Sui Incentivized Testnet: Get ready for the Sui-nami in August. Read more
# Logging, Tracing, Metrics, and Observability
Good observability facilities are key to the development and growth of Sui. This is made more challenging by the distributed and asynchronous nature of Sui, with multiple client and validator processes distributed over a potentially global network.
The observability stack in Sui is based on the Tokio tracing library. The rest of this document highlights specific aspects of achieving good observability through structured logging and metrics in Sui.
NOTE: The output here is largely for the consumption of Sui operators, administrators, and developers. The content of logs and traces do not represent the authoritative, certified output of validators and are subject to potentially byzantine behavior.
## Contexts, scopes, and tracing transaction flow
In a distributed and asynchronous system like Sui, one cannot rely on looking at individual logs over time in a single thread. To solve this problem, we use the approach of structured logging. Structured logging offers a way to tie together logs, events, and blocks of functionality across threads and process boundaries.
### Spans and events
In the Tokio tracing library, structured logging is implemented using spans and events. Spans cover a whole block of functionality - like one function call, a future or asynchronous task, etc. They can be nested, and key-value pairs in spans give context to events or logs inside the function.
• spans and their key-value pairs add essential context to enclosed logs, such as a transaction ID.
• spans also track time spent in different sections of code, enabling distributed tracing functionality.
• individual logs can also add key-value pairs to aid in parsing, filtering and aggregation.
Here is a table/summary of context information that we will want:
• TX Digest
• Object reference/ID, when applicable
• Certificate digest, if applicable
• For Client HTTP endpoint: route, method, status
• Epoch
• Host information, for both clients and validators
Example output which shows both context (tx digests) and key-value pairs enhancing observability/filtering, and tracing a transaction across the gateway (authority_aggregator) as well as the validator:
7ab7774d1f7bd40848}: sui_core::authority_aggregator: Broadcasting transaction request to authorities quorum_threshold=3 validity_threshold=2 timeout_after_quorum=60s
2022-03-05T01:35:03.383791Z TRACE test_move_call_args_linter_command:process_tx{tx_digest=t#7e5f08ab09ec80e3372c101c5858c96965a25326c21af27ab7774d1f7bd40848}: sui_core::authority_aggregator: Transaction data: TransactionData { kind: Call(MoveCall { package: (0000000000000000000000000000000000000002, SequenceNumber(1), o#3104eb8786a94f58d88564c38e22f13d79e3868c5cf81c9c9228fe91465eccad), module: Identifier("object_basics"), function: Identifier("transfer"), type_arguments: [], object_arguments: [(DA40C299F382CBC3C1EBEEA97351F5F185BAD359, SequenceNumber(1), o#d299113b3b52fd1b9dc01e3ba9cf70345faed592af04a56e287057f166ed2783)], shared_object_arguments: [], pure_arguments: [[145, 123, 205, 38, 175, 158, 193, 63, 122, 56, 238, 127, 139, 117, 186, 164, 89, 46, 222, 252]], gas_budget: 1000 }), sender: k#37ebb9c16574a57bcc7b52a6312a35991748be55, gas_payment: (3EE0283D2D12D5C49D0E4E2F509D07227A64ADF2, SequenceNumber(1), o#3ad1a71ee65e8e6675e6a0fb1e893e48c1820b274d3055d75f4abb850c9663e5) }
2022-03-05T01:35:03.385294Z DEBUG test_move_call_args_linter_command:process_tx{tx_digest=t#7e5f08ab09ec80e3372c101c5858c96965a25326c21af27ab7774d1f7bd40848}: sui_core::authority: Checked locks and found mutable objects num_mutable_objects=2
2022-03-05T01:35:03.386500Z DEBUG test_move_call_args_linter_command:process_tx{tx_digest=t#7e5f08ab09ec80e3372c101c5858c96965a25326c21af27ab7774d1f7bd40848}: sui_core::authority: Checked locks and found mutable objects num_mutable_objects=2
2022-03-05T01:35:03.387681Z DEBUG test_move_call_args_linter_command:process_tx{tx_digest=t#7e5f08ab09ec80e3372c101c5858c96965a25326c21af27ab7774d1f7bd40848}: sui_core::authority_aggregator: Received signatures response from authorities for transaction req broadcast num_errors=0 good_stake=3 bad_stake=0 num_signatures=3 has_certificate=true
2022-03-05T01:35:03.391891Z DEBUG test_move_call_args_linter_command:process_cert{tx_digest=t#7e5f08ab09ec80e3372c101c5858c96965a25326c21af27ab7774d1f7bd40848}: sui_core::authority_aggregator: Broadcasting certificate to authorities quorum_threshold=3 validity_threshold=2 timeout_after_quorum=60s
2022-03-05T01:35:03.394529Z DEBUG test_move_call_args_linter_command:process_cert{tx_digest=t#7e5f08ab09ec80e3372c101c5858c96965a25326c21af27ab7774d1f7bd40848}: sui_core::authority: Read inputs for transaction from DB num_inputs=3
2022-03-05T01:35:03.395917Z DEBUG test_move_call_args_linter_command:process_cert{tx_digest=t#7e5f08ab09ec80e3372c101c5858c96965a25326c21af27ab7774d1f7bd40848}: sui_core::authority: Finished execution of transaction with status Success { gas_used: 7 } gas_used=7
From the example above, we can see that process_tx is a span that covers handling the initial transaction request, and "Checked locks" is a single log message within the transaction handling method in the validator. Every log message that occurs within the span inherits the key-value properties defined in the span, including the tx_digest and any other fields that are added. Log messages can set their own keys and values. The fact that logs inherit the span properties allows us to trace, for example, the flow of a transaction across thread and process boundaries.
## Key-value pairs schema
### Span names
Spans capture not a single event but an entire block of time; so start, end, duration, etc. can be captured and analyzed for tracing, performance analysis, etc.
NamePlaceMeaning
process_txGateway, ValidatorSend transaction request, get back 2f+1 signatures and make certificate
process_certGateway, ValidatorSend certificate to validators to execute transaction
cert_check_signatureValidatorCheck certificate signatures
process_cert_innerValidatorInner function to process certificates in validator
tx_execute_to_effectsValidatorExecute Move call and create effects
tx_executeValidatorActual execution of transfer/Move call etc.
handle_certGatewaySend to one validator for certificate processing
quorum_map_authGatewayHandle one network component with one validator
sync_certGateway, ValidatorGateway-initiated sync of data to validator
db_set_transaction_lockValidatorDatabase set transaction locks on new transaction
db_update_stateValidatorUpdate the database with certificate, effects after transaction Move execution
### Tags - keys
The idea is that every event and span would get tagged with key-value pairs. Events that log within any context or nested contexts would also inherit the context-level tags. These tags represent fields that can be analyzed and filtered by. For example, one could filter out broadcasts and see the errors for all instances where the bad stake exceeded a certain amount, but not enough for an error.
KeyPlace(s)Meaning
tx_digestGateway, ValidatorHex digest of transaction
tx_kindGateway, ValidatorKind of transaction: Transfer/Publish/Call
quorum_thresholdGatewayNumeric threshold of quorum stake needed for a transaction
validity_thresholdGatewayNumeric threshold of maximum "bad stake" from errors that can be tolerated
num_errorsGatewayNumber of errors from validators broadcast
num_unique_effectsGatewayNumber of unique effects responses from validators
num_inputsValidatorNumber of inputs for transaction processing
num_mutable_objectsValidatorNumber of mutable objects for transaction processing
gas_usedValidatorAmount of gas used by the transaction
## Logging levels
This is always tricky, to balance the right amount of verbosity especially by default -- while keeping in mind this is a high performance system.
LevelType of Messages
ErrorProcess-level faults (not transaction-level errors, there could be a ton of those)
WarnUnusual or byzantine activity
InfoHigh level aggregate stats, major events related to data sync, epoch changes.
DebugHigh level tracing for individual transactions, eg Gateway/client side -> validator -> Move execution etc.
TraceExtremely detailed tracing for individual transactions
Going from info to debug results in a much larger spew of messages.
The RUST_LOG environment variable can be used to set both the overall logging level as well as the level for individual components, and even filtering down to specific spans or tags within spans are possible too. For more details, please see the EnvFilter docs.
## Metrics
Sui includes Prometheus-based metrics:
• rpc_requests_by_route and related for RPC Server API metrics and latencies (see rpc-server.rs)
• Gateway transaction metrics (see GatewayMetrics struct in gateway-state.rs)
• Validator transaction metrics (see AuthorityMetrics in authority.rs)
## Viewing logs, traces, metrics
The tracing architecture is based on the idea of subscribers which can be plugged into the tracing library to process and forward output to different sinks for viewing. Multiple subscribers can be active at the same time.
graph TB; Validator1 --> S1(open-telemetry) Validator1 --> S2(stdout logging) Validator1 --> S3(bunyan-formatter) S3 --> Vector Vector --> ElasticSearch S1 --> Jaeger Gateway --> SG1(open-telemetry) Gateway --> SG3(bunyan-formatter) SG1 --> Jaeger SG3 --> Vector2 Vector2 --> ElasticSearch
In the graph above, there are multiple subscribers, JSON logs can be for example fed via a local sidecar log forwarder such as Vector, and then onwards to destinations such as ElasticSearch.
The use of a log and metrics aggregator such as Vector allows for easy reconfiguration without interrupting the validator server, as well as offloading observability traffic.
Metrics: served with a Prometheus scrape endpoint, by default at <host>:9184/metrics.
### Stdout (default)
By default, logs (but not spans) are formatted for human readability and output to stdout, with key-value tags at the end of every line. RUST_LOG can be configured for custom logging output, including filtering - see the logging levels section above.
### Tracing and span output
Detailed span start and end logs can be generated by defining the SUI_JSON_SPAN_LOGS environment variable. Note that this causes all output to be in JSON format, which is not as human-readable, so it is not enabled by default. This output can easily be fed to backends such as ElasticSearch for indexing, alerts, aggregation, and analysis.
The example output below shows certificate processing in the authority with span logging. Note the START and END annotations, and notice how DB_UPDATE_STATE which is nested is embedded within PROCESS_CERT. Also notice elapsed_milliseconds which logs the duration of each span.
{"v":0,"name":"sui","msg":"[PROCESS_CERT - START]","level":20,"hostname":"Evan-MLbook.lan","pid":51425,"time":"2022-03-08T22:48:11.241421Z","target":"sui_core::authority_server","line":67,"file":"sui_core/src/authority_server.rs","tx_digest":"t#d1385064287c2ad67e4019dd118d487a39ca91a40e0fd8e678dbc32e112a1493"}
{"v":0,"name":"sui","msg":"[PROCESS_CERT - EVENT] Finished execution of transaction with status Success { gas_used: 18 }","level":20,"hostname":"Evan-MLbook.lan","pid":51425,"time":"2022-03-08T22:48:11.246759Z","target":"sui_core::authority","line":409,"file":"sui_core/src/authority.rs","gas_used":18,"tx_digest":"t#d1385064287c2ad67e4019dd118d487a39ca91a40e0fd8e678dbc32e112a1493"}
### Jaeger (seeing distributed traces)
To see nested spans visualized with Jaeger, do the following:
1. Run this to get a local Jaeger container:
$docker run -d -p6831:6831/udp -p6832:6832/udp -p16686:16686 jaegertracing/all-in-one:latest 2. Run Sui like this (trace enables the most detailed spans): $ SUI_TRACING_ENABLE=1 RUST_LOG="info,sui_core=trace" ./sui start
3. Run some transfers with Sui CLI client, or run the benchmarking tool.
4. Browse to http://localhost:16686/ and select Sui as the service.
Note: Separate spans (which are not nested) are not connected as a single trace for now.
### Live async inspection / Tokio Console
Tokio-console is an awesome CLI tool designed to analyze and help debug Rust apps using Tokio, in real time! It relies on a special subscriber.
1. Build Sui using a special flag: RUSTFLAGS="--cfg tokio_unstable" cargo build.
2. Start Sui with SUI_TOKIO_CONSOLE set to 1.
3. Clone the console repo and cargo run to launch the console.
Note: Adding Tokio-console support may significantly slow down Sui validators/gateways.
Last update 6/30/2022, 6:01:14 PM
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2022-07-01 08:16:12
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http://www.nature.com/articles/s41598-017-00779-4?error=cookies_not_supported&code=20aecc4c-616d-4550-be68-6da8292db22d
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Article | Open | Published:
# Design and demonstration of an underwater acoustic carpet cloak
Scientific Reportsvolume 7, Article number: 705 (2017) | Download Citation
## Abstract
The carpet cloak, which is designed to hide the objects placed on a reflecting surface, has become a topic of considerable interest. Inspired by those theoretical works, the experimental realization of acoustic carpet cloak in air host has been reported. However, due to the difficulty in obtaining the unit cell in reality, the underwater carpet cloak still remains in simulation thus far. Here, we design and fabricate a realizable underwater acoustic carpet cloak. By introducing a scaling factor, the structure of the carpet cloak, which is comprised of layered brass plates, is greatly simplified at the cost of some impedance match. The experimental results demonstrate a good performance of the proposed carpet cloak in a wide frequency range. Our work paves the way for future applications in the practical underwater devices.
## Introduction
Transformation acoustics, which is introduced to design new acoustic structures, shows the way to control the propagation of acoustic waves1,2,3,4,5. Invisibility cloak is one of the most significant applications in transformation acoustics. It has attracted much attention in the past few years6, 7. The acoustic cloak is a material shell that can control the sound wave propagating direction to make the target undetectable in acoustic system. The parameters of the cloak shell can be given by transformation acoustics. Unfortunately, in most cases, these parameters are very complex: the space-dependent mass density and bulk modulus are usually inhomogeneous and extremely anisotropic. As a result, these parameters are challenging to achieve in practice.
Subsequent research leads to the concept of carpet cloak8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40, which is proven to be practically feasible. The carpet cloak is able to hide the target placed on a reflecting surface. The device modifies the acoustic signature of the target and mimics the acoustic field obtained from a reflecting plane, so that the cloaked target is indistinguishable from the reflecting surface. The concept of the carpet cloak was firstly introduced in context of electromagnetics8,9,10,11,12,13,14,15,16,17,18,19,20 and rapidly extended to acoustics21,22,23,24,25,26,27,28,29,30,31. It was firstly proposed as a quasiconformal carpet cloak8,9,10,11,12,13,14,15. However, the size of the quasiconformal carpet cloak is quite large: usually it is an order of magnitude larger than the target. Besides, the lateral shift, which comes from the neglect of the weak anisotropy, will make it possible to be detectable32.
To get rid of these disadvantages, the carpet cloak with linear transformation was proposed20, 21. The linear transformation from a bump in the physical space to a plane in the virtual space brings homogeneous parameters with reasonable anisotropy, which are much more practical for realization. By using the layered perforated plates, both the two-dimensional (2D) and 3D acoustic carpet cloaks were realized in the air host22,23,24,25,26,27. However, because of the difficulty in obtaining the assumed materials, the underwater carpet cloak still remains in simulation28, 29.
In this article, we present a design approach of an underwater carpet cloak which is much easier to realize. By introducing a scaling factor, a modest impedance mismatch is brought in to simplify the structure of the carpet cloak. The quasi-two-dimensional device, which is made up of layered brass plates, has surprising low complexity. We verify our method by designing and fabricating a realizable acoustic carpet cloak, and investigate its effectiveness through experiments. Its good performance demonstrates that the cloak can work stably in a wide frequency range.
## Results
### Coordinate transformation and prototype for underwater carpet cloak
A schematic of the 2D carpet cloak model is shown in Fig. 1. In the physical space (Fig. 1a), a gray trapezoidal bump is placed on the ground plane. In order to make it invisible from the sound detection, the trapezoidal bump is covered with a blue cloak whose acoustic parameters are specified by the transformation acoustics theory. If we map the blue cloak region in the physical space to the whole trapezoidal region above the ground in the virtual space (Fig. 1b), the space mapping rules could be defined mathematically as follow:
$$\{\begin{array}{ll}u=x,v=\frac{h}{h-a}(-a+\frac{a}{b}|x|+y),w={\omega }_{1}z, & {\rm{region}}\,{\rm{I}}\\ u=x,v=\frac{h}{h-a}(y-a),w={\omega }_{2}z, & {\rm{region}}\,{\rm{II}}\\ u=x-l,v=\frac{h}{h-a}(-a+\frac{a}{b}(x-l)+y),w={\omega }_{3}z, & {\rm{region}}\,{\rm{III}}\end{array},$$
(1)
where a, b, l, d and h are geometric parameters of the model indicated in Fig. 1, and ω is an additional degree of freedom to scale the impedance mismatch between the transformed material and the background fluid24, 26. By utilizing these mapping rules, the whole trapezoid in the virtual space is compressed into two triangles (region I and region III) and one rectangle (region II) in the physical space. From Eq. (1), one can note that the mapping rules between the physical space and the virtual space are linear. These linear transformations will lead to homogeneous material parameters which are much more achievable. According to transformation acoustics1, 2, the mass density and bulk modulus distributions of the cloak can be given by:
$$\{\begin{array}{rcl}{\boldsymbol{\rho }} & = & \det (A){({A}^{-1})}^{T}({A}^{-1}){\rho }_{0}\\ K & = & {\rm{\det }}(A){K}_{0}\end{array},$$
(2)
where ρ 0 and K 0 are the density and the bulk modulus of background fluid respectively; A is the transformation Jacobian matrix showed as: $${A}=\frac{\partial (x,y,z)}{\partial (u,v,w)}$$.
By using Eqs (1) and (2), the material parameters of the cloak can be obtained. For convenience, the off-diagonal terms in the mass density tensor are eliminated by rotating the coordinate axes anticlockwise with an angle α. Then the eigenvalues of the mass density tensor are presented as:
$$\{\begin{array}{ll}{K}^{I}=\frac{h-a}{h{\omega }_{1}}{K}_{0},{\rho }_{11}^{pr,I}=\frac{F-\sqrt{{F}^{2}-1}}{{\omega }_{1}}{\rho }_{0},{\rho }_{22}^{pr,I}=\frac{F+\sqrt{{F}^{2}-1}}{{\omega }_{1}}{\rho }_{0}, & {\rm{region}}\,{\rm{I}}\\ {K}^{II}=\frac{h-a}{h{\omega }_{2}}{K}_{0},{\rho }_{11}^{pr,II}=\frac{h-a}{h{\omega }_{2}}{\rho }_{0},{\rho }_{22}^{pr,II}=\frac{h}{(h-a){\omega }_{2}}{\rho }_{0}, & {\rm{region}}\,{\rm{II}}\\ {K}^{III}=\frac{h-a}{h{\omega }_{3}}{K}_{0},{\rho }_{11}^{pr,III}=\frac{F-\sqrt{{F}^{2}-1}}{{\omega }_{3}}{\rho }_{0},{\rho }_{22}^{pr,III}=\frac{F+\sqrt{{F}^{2}-1}}{{\omega }_{3}}{\rho }_{0}, & {\rm{region}}\,{\rm{III}}\end{array},$$
(3)
where $${F}=1+\frac{{a}^{2}({b}^{2}+{h}^{2})}{2{b}^{2}h(h-a)}$$. All the eigenvalues of the mass density tensor are marked with the superscript pr. Meanwhile, the rotating angle α is given by:
$$\alpha =\{\begin{array}{ll}\frac{\pi }{2}-\arcsin (\frac{G}{\sqrt{{G}^{2}+1}}), & {\rm{region}}\,{\rm{I}}\\ 0, & {\rm{region}}\,{\rm{II}}\\ -\frac{\pi }{2}+\arcsin (\frac{G}{\sqrt{{G}^{2}+1}}), & {\rm{region}}\,{\rm{III}}\end{array},$$
(4)
where $${G}=\frac{h}{b}(1-\frac{h-a}{a}(F-1-\sqrt{{F}^{2}-1}))$$. Actually, α is also the angle between the principal axes of the transformed materials and the coordinate axes in the physical space.
Then, a prototype is supposed to illustrate the design of the carpet cloak. The geometry parameters in Fig. 1 are set as a = 28.5 mm, b = 100 mm, h = 85 mm, l = 40 mm, respectively. Besides, ω 1 = ω 2 = ω 3 = 1 indicates that the carpet cloak is impedance matched with the background in this model. By substituting these parameters into Eqs (3) and (4), the required parameters of the cloak can be obtained:
$$\{\begin{array}{ll}{\rho }_{11}^{{pr},{I}}=0.59{\rho }_{0},{\rho }_{22}^{pr,I}=1.70{\rho }_{0},{K}^{pr,I}=0.66{K}_{0},\alpha =25^\circ ,\,\, & {\rm{region}}\,{\rm{I}}\\ {\rho }_{11}^{{pr},{II}}=0.66{\rho }_{0},{\rho }_{22}^{pr,II}=1.50{\rho }_{0},{K}^{pr,II}=0.66{K}_{0},\alpha =0^\circ ,\, & {\rm{region}}\,{\rm{II}}\\ {\rho }_{11}^{pr,III}=0.59{\rho }_{0},{\rho }_{22}^{pr,III}=1.70{\rho }_{0},{K}^{pr,III}=0.66{K}_{0},\alpha =-25^\circ ,\,\, & {\rm{region}}\,{\rm{III}}\end{array},$$
(5)
Obviously, material with anisotropic mass density is required in realizing the carpet cloak. It is known that there is no natural material with anisotropic mass density. Nevertheless, it has been demonstrated from the Biot fluid theory that layers of isotropic materials can present effective anisotropic mass density in long wavelength regime4, 41. If the thicknesses of the layers are much smaller than the wavelength, the acoustic layered system will have the following effective parameters:
$$\frac{1}{{\rho }_{11}}=\langle \frac{1}{{\rho }_{i}}\rangle ,{\rho }_{22}=\langle {\rho }_{i}\rangle ,\frac{1}{K}=\langle \frac{1}{{K}_{i}}\rangle ,$$
(6)
where 〈 〉 denotes a thickness weighted average; ρ 11 (ρ 22) is the effective mass density component in the direction which is parallel (perpendicular) to the layered structure (correspond to the eigenvalues of the mass density tensor in two principal axes); ρ i and K i are the dens i ty and bulk modulus in i-th layer. Therefore, the required material parameters in Eq. (5) can be obtained through the periodical layers. As shown in Fig. 2a, the designed carpet cloak is a multilayer structure, and it is comprised of two kinds of fluid (marked with A and B) with the same thickness. The blue regions represent layer A, the yellow regions represent layer B, and the gray region represents the area remaining to be concealed. The thickness of each layer is 1 mm, which is smaller than the wavelength at the operation frequency of 13 kHz by a factor of 40. The Biot fluid theory implies that there is an unique solution for the densities but infinite choices for the bulk moduli in layer A and layer B. For convenient comparison with the practical parameters of the designed sample discussed later, here we choose a special solution with the following parameters:
$$\{\begin{array}{rcl}{\rho }_{A}^{I}={\rho }_{A}^{III} & = & 0.33{\rho }_{0},{K}_{A}^{I}={K}_{A}^{III}=0.34{K}_{0}\\ {\rho }_{B}^{I}={\rho }_{B}^{III} & = & 3.07{\rho }_{0},{K}_{B}^{I}={K}_{B}^{III}=21.28{K}_{0}\\ {\rho }_{A}^{II} & = & 0.38{\rho }_{0},{K}_{A}^{II}=0.34{K}_{0}\\ {\rho }_{B}^{II} & = & 2.62{\rho }_{0},{K}_{B}^{II}=20.55{K}_{0}\end{array},$$
(7)
where the superscript I, II, III correspond to region I, II, III, respectively; the subscript A, B represent layer A and layer B.
### Numerical simulations
By utilizing the parameters in Eq. (7), the ideal cloak, whose impedance is matched with the background fluid, is designed and the camouflage effect of the proposed model is simulated in time domain with a commercial finite elements package (COMSOL Multiphysics). In simulation, the background fluid is water and the boundaries of the simulated area are set as absorbing boundaries to avoid unexpected reflections. Because the carpet cloak works with the reflecting plane, the material of all the scatterers is set as air to ensure strong impedance mismatch.
The simulated results are presented in Fig. 2b–e. In Fig. 2b, a short Gaussian pulse with central frequency of 13 kHz is emitted from the top boundary. The beam whose width is about 0.5 m directly propagates towards the targets. The time of Fig. 2b is set as 0 ms as a reference. Figure 2c–e show the acoustic fields in different cases after 1 ms. The short pulses arrive at the objects and then are reflected back. All the propagating directions of the wave are indicated by the black arrows. When the acoustic wave is reflected from a soft plane (Fig. 2c), the beam keeps its Gaussian shape and propagates from bottom to top in the backscattering direction. In contrast, Fig. 2d displays the acoustic pressure field obtained with the soft bump. It is obvious that there is a shadow in the middle of the scattered wavefront. The soft bump separates the beam into two parts. The widely scattered wave leads to a decrease of the energy density, which makes the amplitude of the scattered wave much smaller. Besides, the soft bump also causes a phase advance compared with the soft plane. However, by covering the soft bump with the carpet cloak, the scattered wave returns to the backscattering direction, as shown in Fig. 2e. The wave is modulated by the cloak. Its shape, amplitude and propagating direction are identical to those when the target is a soft plane. The cloaked object successfully mimics the reflecting plane and it is invisible under the sound detection.
### Design of the sample and experimental measurements
The simulated results for the prototype show the possibility to realize the carpet cloak. An ideal carpet cloak requires two kinds of materials: one (layer A) is much less dense but has the same acoustic velocity as water; the other (layer B) has much larger density and modulus. In fact, the much less dense fluid (layer A) is very challenging to achieve. Nevertheless, if some impedance mismatch (ω ≠ 1) is introduced into the carpet cloak, the materials become much more achievable. This change brings some undesired reflection but has limited impact on the camouflage effect.
For region I, we set ω 1 = 0.34, which means that the mass density and bulk modulus of the cloak are 3 times those of water, then the scaled parameters of the cloak are $${\rho }_{11}^{pr,new}=3{\rho }_{11}^{pr}$$, $${\rho }_{22}^{pr,new}=3{\rho }_{22}^{pr}$$, $${K}^{pr,new}=3{K}^{pr}$$. The material parameters in the layered structure are also changed:
$$\{\begin{array}{rcl}{\rho }_{A}^{I,new} & = & 0.97{\rho }_{0},{K}_{A}^{I,new}={K}_{0}\\ {\rho }_{B}^{I,new} & = & 9.03{\rho }_{0},{K}_{B}^{I,new}=62.59{K}_{0}\end{array}.$$
(8)
Now, above parameters can be realized by using water (layer A) and brass (layer B): the density and the acoustic velocity of water are 1000 kg/m3 and 1480 m/s; the Young’s modulus, density and Poisson’s ratio of the brass are 110 Gpa, 8900 kg/m3, and 0.35, respectively. It should be noted that these thin brass plates are separated by the water layers, so the influence of shear wave on the effective mass density can be neglected28. The thin brass plates could be approximately regarded as fluid at low frequency range.
The effective acoustic parameters of the unit cell, which is comprised of brass and water, are calculated by using the retrieving method in simulation42. The results are presented in Fig. 3b and c. In Fig. 3b, the simulated mass density in two principal axes are plotted as symbols, while the required values obtained from the Biot theory are plotted as lines for comparison. The obvious different values in two directions indicate the anisotropic mass density. Similarly, the simulated (symbols) and required (line) bulk moduli in two principal axes are presented in Fig. 3c. The simulated values in two directions are almost the same. So the effective bulk modulus is isotropic in this unit cell. It can be seen that all these simulated results fit well with these required values in Fig. 3, which means the unit cell can work stably in a wide frequency range.
Due to the symmetry of the trapezoid, the parameters in region I and region III are the same. Actually, the parameters in region II can also be realized by the same structure for just a little change of the scaling factor (ω 2 = 0.32). So the carpet cloak sample could be manufactured by the same unit cell (brass and water with filling rate f brass = 0.5).
A photograph of the fabricated carpet cloak sample is shown in the inset of Fig. 3a. It’s a quasi-two-dimensional carpet cloak whose geometry size is the same with the simulated one, but with a thickness of 160 mm in z-direction. It comprises layers of brass plate with small channels filled with water.
The sample is placed in the middle of an anechoic water tank (about 2500 mm under water). An omnidirectional cylindrical transducer is placed above the sample (about 1300 mm away from the bottom of the sample, as shown in Fig. 3a). Two hydrophones (Type 8103, B&K) are used to measure the acoustic pressure fields. One is fixed in a position near the source as a time reference, and the other scans the measuring region step by step. The scanning hydrophone moves on a square grid of 15 mm to ensure at least six measurement points per wavelength. The scanned area, which is indicated as a black frame in Fig. 3a, is 450 mm wide and 270 mm high, and 500 mm away from the bottom of the cloak. The region is selected so that the incident wave and scattered wave can be separated clearly. All the emitting and receiving acoustic signals are analyzed by a multianalyzer system (Type 3160, B&K).
In order to distinguish the incident wave from the scattered wave, the omnidirectional cylindrical transducer emits short Gaussian pulses modulated with different sinusoidal signals (11 kHz to 16 kHz with 1 kHz step). To demonstrate the effectiveness of the cloak, three different cases were measured: with the soft plane (air layer sealed by polymethyl methacrylate) underwater, with the soft bump (air trapezoid sealed by polymethyl methacrylate) underwater and with the cloaked soft bump underwater, respectively.
The measured incident and scattered acoustic pressure fields at 13 kHz are presented in Fig. 4. The color scales of these fields have been tuned to make the distribution of energy and phase clear. In Fig. 4a, the source produces a pulse in the water and the wave spreads around from the upper left corner. After 1.05 ms, the wave is reflected by the object and propagates from bottom to the top. It can be observed from Fig. 4b that the reflected wave from the soft plane mainly focuses on the backscattering direction. In Fig. 4c, due to the slopes of the soft bump, acoustic wave obliquely hits the surfaces and reflects to the opposite sides. Consequently, the scattered wave inclines to both sides of the trapezoid. The scattered wave from the soft bump spreads evenly. This is very different from the propagating direction of scattered wave from the soft plane. In contrast, after covering the soft bump with cloak, the propagating direction of the scattered wave focuses on the backscattering direction again (Fig. 4d). The reflected wave from the cloaked soft bump is nearly identical to the wave from the soft plane as if the soft bump didn’t exist. The phase, the amplitude and the propagation direction of the scattered wave are recovered at the same time. These phenomena agree well with the simulations (see Supplementary Fig. S1). The simulated results also confirm the validity of the designed cloak at oblique incidence (see Supplementary Figs S2 and S3).
Furthermore, similar acoustic fields are also provided when the center frequency of the Gaussian pulse is 14 kHz (Fig. 5). In Fig. 5c, after the sound wave impinges on the soft bump, the scattered wave in the left part of the acoustic field (0 < X < 150) propagates from the bottom to the top directly, while the wave in the right part of the acoustic field (150 < X < 450) propagates obliquely to the upper right. The directivity of the scattered wave implies that the acoustic signature of the soft bump is much clearer with the increase of the frequency. Figure 4b,d also show the acoustic pressure distributions obtained with the soft plane and cloaked bump, respectively. Similarly, comparing the phase, the amplitude and the propagation direction of the scattered wave in these three pressure distributions, the cloaked bump mimics the soft plane well at 14 kHz.
To further exhibit the performance of the cloak, we also extract the time domain signals at points A, B and C in Fig. 3a at 12 kHz (the sound pressure fields at 12 kHz can be seen in Supplementary Fig. S4). The center in the bottom of the trapezoid is set as the original point (marked as point O in Fig. 3a), and the coordinate of these points are A(0, 620), B(180, 620) and C(375, 620), respectively. These points are chosen for their special positions: point A is in the backscattering direction; point C corresponds to the direction where the amplitude of the wave reflected by the bare soft bump is maximum; point B is in the transitional position from point A to point C. The extracted time domain signals are shown in Fig. 6. These curves in each panel illustrate the signals obtained with the soft plane (blue solid lines), bare soft bump (green dash lines) and cloaked soft bump (red dot curves), respectively. The incident Gaussian pulses (located at about 0.4 ms) in three cases are identical, so the reflected signals (located at about 1.2 ms) can be compared with each other. The measured results at position A (Fig. 6a) show that the phase difference between the reflected waves from the cloaked bump and the soft plane is within 50° (30° at point B and 23° at point C) while the phase difference between the reflected waves obtained from the bare bump and the soft plane is about 115° (130° at point B and 110° at point C). The results also show that the carpet cloak corrects the amplitude difference. The amplitude of the scattered wave obtained from the cloaked bump is much closer to that from the soft plane. The same results can also be obtained at other frequencies (see Supplementary Figs S5 and S6).
Additionally, the reduced total radar cross section (RCS) is introduced to evaluate the camouflage effect of the cloak43. It is defined as:
$$\{\begin{array}{c}{\sigma }_{{\rm{reduced}}}=\frac{{\sigma }_{{\rm{cloaked}}}}{{\sigma }_{{\rm{uncloaked}}}}={\oint }_{{\rm{\Omega }}}(\frac{{|{P}_{cloaked,scat}|}^{2}}{{|{P}_{uncloaked,scat}|}^{2}})d{\rm{\Omega }}\\ \begin{array}{c}{P}_{cloaked,scat}={P}_{cloaked,tot}-{P}_{plane,tot}\\ {P}_{{\rm{un}}cloaked,scat}={P}_{uncloaked,tot}-{P}_{plane,tot}\end{array}\end{array},$$
(9)
where P cloaked,tot , P uncloaked,tot and P plane,tot are the total scattered field for the cloaked soft bump, the bare soft bump and the soft plane, respectively. In Eq. (9), the smaller the value is, the better the performance of the carpet cloak is.
The reduced total RCSs for the measured frequencies are shown in Fig. 7. It can be observed that all the values are around 0.2. The small values demonstrate that the carpet cloak works well in a wide frequency range.
## Discussion
We realized a simple underwater carpet cloak to mimic a reflecting plane. The structure of carpet cloak, which is comprised of layered brass plates, is greatly simplified at the cost of some impedance match. The performance of the carpet cloak is assessed experimentally by measuring the acoustic pressure fields in an anechoic tank. The measured results confirm that the carpet cloak can hide the information of the bump on the reflecting plane in a wide frequency range. The proposed carpet cloak, whose unit cell size is close to one fortieth of the wavelength, shows the ability to control the underwater acoustic wave in the deep subwavelength scale. This may bring great potential engineering applications in the practical underwater devices.
## Additional information
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## Acknowledgements
This work is supported by the Youth Innovation Promotion Association CAS (Grant No. 2017029), the National Natural Science Foundation of China (Grant No. 11304351 and 11304345) and the “Strategic Priority Research Program” of the Chinese Academy of Sciences (Grant No. XDA06020201).
## Author information
### Affiliations
1. #### Key Laboratory of Noise and Vibration Research, Institute of Acoustics, Chinese Academy of Sciences, Beijing, 100190, People’s Republic of China
• Yafeng Bi
• , Han Jia
• , Wenjia Lu
• , Peifeng Ji
• & Jun Yang
2. #### University of Chinese Academy of Sciences, Beijing, 100049, People’s Republic of China
• Yafeng Bi
• , Han Jia
• , Wenjia Lu
• , Peifeng Ji
• & Jun Yang
3. #### State Key Laboratory of Acoustics, Institute of Acoustics, Chinese Academy of Sciences, Beijing, 100190, People’s Republic of China
• Han Jia
• & Jun Yang
### Contributions
Y.F.B., H.J. and J.Y. initiated and designed the research. Y.F.B., W.J.L. and P.F.J. conducted the experiments. Y.F.B., W.J.L. and H.J. prepared the manuscript. Y.J. supervised the project.
### Competing Interests
The authors declare that they have no competing interests.
### Corresponding author
Correspondence to Jun Yang.
## About this article
### DOI
https://doi.org/10.1038/s41598-017-00779-4
## Comments
By submitting a comment you agree to abide by our Terms and Community Guidelines. If you find something abusive or that does not comply with our terms or guidelines please flag it as inappropriate.
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2018-10-16 02:02:58
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https://zbmath.org/authors/?q=ai%3Azener.clarence
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# zbMATH — the first resource for mathematics
## Zener, Clarence
Compute Distance To:
Author ID: zener.clarence Published as: Zener, C.; Zener, Clarence External Links: Wikidata · GND
Documents Indexed: 46 Publications since 1929, including 4 Books
all top 5
#### Co-Authors
33 single-authored 3 Duffin, Richard James 3 Guillemin, V. jun. 2 Peterson, Elmor L. 1 Heikes, R. R. 1 Kemble, Edwin C. 1 Mott, Nevill Francis 1 Nuckolls, R. 1 Otis, W. 1 Rosen, Nathan
all top 5
#### Serials
22 Physical Review, II. Series 8 Proceedings of the National Academy of Sciences of the United States of America 7 Proceedings of the Royal Society of London. Series A 3 Proceedings of the Cambridge Philosophical Society 1 Reviews of Modern Physics 1 Zeitschrift für Physik
#### Fields
3 Operations research, mathematical programming (90-XX)
#### Citations contained in zbMATH Open
32 Publications have been cited 530 times in 460 Documents Cited by Year
Geometric programming. Theory and application. Zbl 0171.17601
Duffin, Richard J.; Peterson, Elmor L.; Zener, Clarence
1967
Non-adiabatic crossing of energy levels. Zbl 0005.18605
Zener, Clarence
1932
Internal friction in solids. I: Theory of internal friction reeds. JFM 63.1341.03
Zener, C.
1937
Elasticity and unelasticity of metals. Zbl 0032.22202
Zener, Clarence
1948
Non-adiabatic crossing of energy levels. JFM 58.1356.02
Zener, C.
1932
Internal friction in solids. II. General theory of thermoelastic internal friction. JFM 64.1420.03
Zener, C.
1938
A theory of the electrical breakdown of solid dielectrics. Zbl 0009.27605
Zener, Clarence
1934
Double Stern-Gerlach experiment and related collision phenomena. Zbl 0004.42704
Rosen, N.; Zener, C.
1932
Analytic atomic wave functions. JFM 56.1313.01
Zener, C.
1930
A mathematical aid in optimizing engineering designs. Zbl 0094.36701
Zener, Clarence
1961
The intrinsic inelasticity of large plates. Zbl 0026.36701
Zener, Clarence
1941
Interaction between the $$d$$ shells in the transition metals. Zbl 0042.23502
Zener, C.
1951
Interchange of translational, rotational and vibrational energy in molecular collisions. Zbl 0001.17703
Zener, Clarence
1931
A theory of the electrical breakdown of solid dielectrics. JFM 60.0781.04
Zener, C.
1934
A further mathematical aid in optimizing engineering designs. Zbl 0105.33901
Zener, C.
1962
Internal friction in solids. III. Experimental demonstration of thermoelastic internal friction. JFM 64.1421.01
Zener, C.; Otis, W.; Nuckolls, R.
1938
The $$B$$-state of the hydrogen molecule. JFM 55.0540.01
Zener, C.; Guillemin, V. jun.
1929
The two quantum excited states of the hydrogen molecule. JFM 55.0539.05
Kemble, E. C.; Zener, C.
1929
Low velocity inelastic collisions. Zbl 0002.23101
Zener, Clarence
1931
A general proof of certain fundamental equations in the theory of metallic conduction. Zbl 0008.42304
Jones, H.; Zener, C.
1934
Hydrogen-ion wave function. JFM 55.0540.02
Guillemin, V. jun.; Zener, C.
1929
The theory of the change in resistance in a magnetic field. Zbl 0009.14005
Jones, H.; Zener, C.
1934
Theory of strain interaction of solute atoms. Zbl 0032.14103
Zener, C.
1948
Exchange interactions. Zbl 0050.23807
Zener, C.; Heikes, R. R.
1953
Interaction between the $$d$$-shells in the transition metals. IV. The intrinsic antiferromagnetic character of iron. Zbl 0046.45604
Zener, C.
1952
The intrinsic inelasticity of large plates. JFM 67.0828.01
Zener, C.
1941
Internal friction in solids. IV. Relation between cold work and internal friction. JFM 64.1421.02
Zener, C.
1938
Geometric programming, chemical equilibrium, and the antientropy function. Zbl 0182.53103
Duffin, R. J.; Zener, C.
1969
Geometric programming – theory and application. Übersetzung aus dem Englischen von D. A. Babaev. (Геометрическое программирование.) Zbl 0236.90062
Duffin, Richard J.; Peterson, Elmor L.; Zener, Clarence
1972
Dissociation of excited diatomic molecules by external perturbations. Zbl 0007.08901
Zener, Clarence
1933
Elastic reflection of atoms from crystal. JFM 58.0947.07
Zener, C.
1932
Redesign overcompensation. Zbl 0128.39503
Zener, C.
1965
Geometric programming – theory and application. Übersetzung aus dem Englischen von D. A. Babaev. (Геометрическое программирование.) Zbl 0236.90062
Duffin, Richard J.; Peterson, Elmor L.; Zener, Clarence
1972
Geometric programming, chemical equilibrium, and the antientropy function. Zbl 0182.53103
Duffin, R. J.; Zener, C.
1969
Geometric programming. Theory and application. Zbl 0171.17601
Duffin, Richard J.; Peterson, Elmor L.; Zener, Clarence
1967
Redesign overcompensation. Zbl 0128.39503
Zener, C.
1965
A further mathematical aid in optimizing engineering designs. Zbl 0105.33901
Zener, C.
1962
A mathematical aid in optimizing engineering designs. Zbl 0094.36701
Zener, Clarence
1961
Exchange interactions. Zbl 0050.23807
Zener, C.; Heikes, R. R.
1953
Interaction between the $$d$$-shells in the transition metals. IV. The intrinsic antiferromagnetic character of iron. Zbl 0046.45604
Zener, C.
1952
Interaction between the $$d$$ shells in the transition metals. Zbl 0042.23502
Zener, C.
1951
Elasticity and unelasticity of metals. Zbl 0032.22202
Zener, Clarence
1948
Theory of strain interaction of solute atoms. Zbl 0032.14103
Zener, C.
1948
The intrinsic inelasticity of large plates. Zbl 0026.36701
Zener, Clarence
1941
The intrinsic inelasticity of large plates. JFM 67.0828.01
Zener, C.
1941
Internal friction in solids. II. General theory of thermoelastic internal friction. JFM 64.1420.03
Zener, C.
1938
Internal friction in solids. III. Experimental demonstration of thermoelastic internal friction. JFM 64.1421.01
Zener, C.; Otis, W.; Nuckolls, R.
1938
Internal friction in solids. IV. Relation between cold work and internal friction. JFM 64.1421.02
Zener, C.
1938
Internal friction in solids. I: Theory of internal friction reeds. JFM 63.1341.03
Zener, C.
1937
A theory of the electrical breakdown of solid dielectrics. Zbl 0009.27605
Zener, Clarence
1934
A theory of the electrical breakdown of solid dielectrics. JFM 60.0781.04
Zener, C.
1934
A general proof of certain fundamental equations in the theory of metallic conduction. Zbl 0008.42304
Jones, H.; Zener, C.
1934
The theory of the change in resistance in a magnetic field. Zbl 0009.14005
Jones, H.; Zener, C.
1934
Dissociation of excited diatomic molecules by external perturbations. Zbl 0007.08901
Zener, Clarence
1933
Non-adiabatic crossing of energy levels. Zbl 0005.18605
Zener, Clarence
1932
Non-adiabatic crossing of energy levels. JFM 58.1356.02
Zener, C.
1932
Double Stern-Gerlach experiment and related collision phenomena. Zbl 0004.42704
Rosen, N.; Zener, C.
1932
Elastic reflection of atoms from crystal. JFM 58.0947.07
Zener, C.
1932
Interchange of translational, rotational and vibrational energy in molecular collisions. Zbl 0001.17703
Zener, Clarence
1931
Low velocity inelastic collisions. Zbl 0002.23101
Zener, Clarence
1931
Analytic atomic wave functions. JFM 56.1313.01
Zener, C.
1930
The $$B$$-state of the hydrogen molecule. JFM 55.0540.01
Zener, C.; Guillemin, V. jun.
1929
The two quantum excited states of the hydrogen molecule. JFM 55.0539.05
Kemble, E. C.; Zener, C.
1929
Hydrogen-ion wave function. JFM 55.0540.02
Guillemin, V. jun.; Zener, C.
1929
all top 5
#### Cited by 685 Authors
11 Dinkel, John J. 9 Fang, Shu-Cherng 8 Ecker, Joseph G. 8 Kochenberger, Gary A. 7 Jefferson, Thomas R. 7 Kumar Roy, Tapan 7 Peterson, Elmor L. 7 Scott, Carlton H. 6 Bricker, Dennis L. 6 Rajasekera, J. R. 6 Zener, Clarence 5 Fermanian-Kammerer, Clotilde 5 Mandal, Nirmal Kumar 5 Passy, Ury 4 Dembo, Ron S. 4 Eleuch, Hichem 4 Hagedorn, George A. 4 Islam, Sahidul 4 Maiti, Manoranjan 4 Rice, Oscar Knefler 4 Slater, John Clarke 4 Tupholme, G. E. 4 Van Vleck, John Hasbrouck 3 Avriel, Mordecai 3 Ben-Israel, Adi 3 Boyd, Stephen Poythress 3 Cao, Bingyuan 3 Do Nascimento, Roberto Quirino 3 Dutta, Amit Kumar 3 Elster, Karl-Heinz 3 Elster, Rosalind 3 Gochet, Willy F. 3 Joye, Alain 3 Jung, Hoon 3 Klein, Cerry M. 3 Knowles, Kevin M. 3 Kortanek, Kenneth O. 3 Rajgopal, Jayant 3 Rezazadeh, Ghader 3 Rosen, Nathan 3 Rostovtsev, Yuri V. 3 Rotter, Ingrid 3 Smeers, Yves 2 Allueva, Ana Isabel 2 Aryanezhad, Mir-Bahador-Qoli 2 Berman, Gennady P. 2 Biswal, Mahendra Prasad 2 Bloch, Leon 2 Castro, Pedro M. 2 Chandrasekaran, Venkat 2 Chen, Zhiping 2 Cheng, Hao 2 Cheng, Tai-Chiu Edwin 2 Colin de Verdière, Yves 2 Coolidge, Albert Sprague 2 De Oliveira Santos, Rubia Mara 2 de Wolff, Timo 2 Demeio, Lucio 2 Dressler, Mareike 2 Duffin, Richard James 2 Figueiredo Lima, Edson jun. 2 Fraas, Martin 2 Grover, Dhruv 2 Guseinov, I. I. 2 Iliman, Sadik 2 James, Hubert M. 2 Kelly, Donald W. 2 Kim, Seung-Jean 2 Korsch, Hans Jürgen 2 Kupferschmid, Michael 2 Lavery, John E. 2 Lisser, Abdel 2 Liu, Jia 2 Lu, Jianfeng 2 Luptacik, Mikulas 2 Lyra, Marcelo L. 2 Mahapatra, Ghanshaym Singha 2 Matos, Henrique A. 2 McCarl, Bruce A. 2 Nesterov, Alexander I. 2 Panda, Debdulal 2 Phillips, Andrew T. 2 Pratt, George W. jun. 2 Rosen, J. Ben 2 Rossikhin, Yury A. 2 Sadjadi, Seyed Jafar 2 Schwieger, Horst 2 Segall, Richard S. 2 Seitz, Frederick 2 Sengupta, Jati K. 2 Shah, Parikshit 2 Shitikova, Marina V. 2 Stark, Robert M. 2 Teles, João P. 2 Vahdat, Armin Saeedi 2 Vakakis, Alexander F. 2 Verma, Rakesh Kumar 2 Watanabe, Takuya 2 Wiebking, Rolf D. 2 Wilde, Douglass J. ...and 585 more Authors
all top 5
#### Cited in 147 Serials
40 Physical Review, II. Series 34 Journal of Optimization Theory and Applications 31 European Journal of Operational Research 17 Mathematical Programming 16 Acta Mechanica 13 Applied Mathematical Modelling 12 Journal of Mathematical Physics 11 Journal of Mathematical Analysis and Applications 11 Applied Mathematics and Computation 9 New Journal of Physics 7 Journal of Mathematical Chemistry 6 Optimization 6 Journal of Modern Optics 5 International Journal of Modern Physics B 5 Journal of Engineering Mathematics 5 Reviews of Modern Physics 5 Journal of Elasticity 5 Optimization and Engineering 4 Computers & Mathematics with Applications 4 Communications in Mathematical Physics 4 International Journal of Theoretical Physics 4 Journal of Computational Physics 4 Fuzzy Sets and Systems 4 Meccanica 4 Operations Research Letters 4 Mathematical and Computer Modelling 4 The Journal of Chemical Physics 3 Computer Methods in Applied Mechanics and Engineering 3 Discrete Applied Mathematics 3 Journal of Statistical Physics 3 ZAMP. Zeitschrift für angewandte Mathematik und Physik 3 Information Sciences 3 International Journal for Numerical Methods in Engineering 3 International Journal of Production Research 3 Computers & Operations Research 3 Journal of Global Optimization 3 Zeitschrift für Operations Research. Serie A: Theorie 3 Mathematical Programming. Series A. Series B 3 Computational Optimization and Applications 3 Nonlinear Dynamics 3 Annales Henri Poincaré 3 Quantum Information Processing 3 Journal of Statistical Mechanics: Theory and Experiment 3 Bulletin of the Russian Academy of Sciences: Physics 3 Physical Review A, Third Series 2 Archive for Rational Mechanics and Analysis 2 European Journal of Physics 2 International Journal of Plasticity 2 International Journal of Solids and Structures 2 International Journal of Systems Science 2 Journal of the Franklin Institute 2 Journal of the Mechanics and Physics of Solids 2 Physics Letters. A 2 Physics Reports 2 Theoretical and Mathematical Physics 2 Fortschritte der Physik 2 Journal of Computational and Applied Mathematics 2 Physica D 2 Annals of Operations Research 2 Physics of Fluids 2 Mathematical Problems in Engineering 2 Journal of Vibration and Control 2 Comptes Rendus. Mathématique. Académie des Sciences, Paris 2 Fuzzy Optimization and Decision Making 2 Journal of Applied Physics 2 Annales de l’Institut Henri Poincaré 2 Unternehmensforschung 2 Studies in History and Philosophy of Science. Part B. Studies in History and Philosophy of Modern Physics 2 Fuzzy Information and Engineering 2 SIAM Journal on Applied Algebra and Geometry 1 Modern Physics Letters B 1 General Relativity and Gravitation 1 Ingenieur-Archiv 1 International Journal of Control 1 International Journal of Engineering Science 1 International Journal of Heat and Mass Transfer 1 Journal d’Analyse Mathématique 1 Mathematical Biosciences 1 Transport Theory and Statistical Physics 1 Zeitschrift für Angewandte Mathematik und Mechanik (ZAMM) 1 Mathematics of Computation 1 Reviews in Mathematical Physics 1 Annales de l’Institut Fourier 1 The Annals of Statistics 1 Automatica 1 Computing 1 Journal of Pure and Applied Algebra 1 Journal of Soviet Mathematics 1 Kybernetika 1 Mathematics and Computers in Simulation 1 Mathematika 1 Opsearch 1 SIAM Journal on Computing 1 Advances in Applied Mathematics 1 OR Spektrum 1 Insurance Mathematics & Economics 1 Applied Numerical Mathematics 1 Journal of Symbolic Computation 1 Computational Mechanics 1 International Journal of Approximate Reasoning ...and 47 more Serials
all top 5
#### Cited in 42 Fields
199 Operations research, mathematical programming (90-XX) 84 Quantum theory (81-XX) 71 Mechanics of deformable solids (74-XX) 45 Statistical mechanics, structure of matter (82-XX) 43 Numerical analysis (65-XX) 31 Partial differential equations (35-XX) 21 Calculus of variations and optimal control; optimization (49-XX) 13 Game theory, economics, finance, and other social and behavioral sciences (91-XX) 12 Ordinary differential equations (34-XX) 11 Classical thermodynamics, heat transfer (80-XX) 10 Fluid mechanics (76-XX) 8 Computer science (68-XX) 8 Information and communication theory, circuits (94-XX) 7 Optics, electromagnetic theory (78-XX) 7 Biology and other natural sciences (92-XX) 6 Statistics (62-XX) 6 Systems theory; control (93-XX) 5 Linear and multilinear algebra; matrix theory (15-XX) 5 Real functions (26-XX) 5 Probability theory and stochastic processes (60-XX) 5 Mechanics of particles and systems (70-XX) 4 Algebraic geometry (14-XX) 4 Dynamical systems and ergodic theory (37-XX) 4 Operator theory (47-XX) 4 Convex and discrete geometry (52-XX) 3 Field theory and polynomials (12-XX) 3 Measure and integration (28-XX) 3 Special functions (33-XX) 3 Approximations and expansions (41-XX) 2 General and overarching topics; collections (00-XX) 2 Mathematical logic and foundations (03-XX) 2 Combinatorics (05-XX) 1 History and biography (01-XX) 1 Commutative algebra (13-XX) 1 Nonassociative rings and algebras (17-XX) 1 Topological groups, Lie groups (22-XX) 1 Integral transforms, operational calculus (44-XX) 1 Functional analysis (46-XX) 1 Global analysis, analysis on manifolds (58-XX) 1 Relativity and gravitational theory (83-XX) 1 Astronomy and astrophysics (85-XX) 1 Geophysics (86-XX)
#### Wikidata Timeline
The data are displayed as stored in Wikidata under a Creative Commons CC0 License. Updates and corrections should be made in Wikidata.
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2021-07-29 05:05:25
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https://www.gamedev.net/forums/topic/566844-distribution-of-nodes-for-a/
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# Distribution of Nodes for A*
This topic is 2876 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Can people who have done A* please let me know how ye decided upon the distribution of the nodes aaround hte map? We have our map, and i have coded up an A* algorithm. Im just not sure what the beest way to distribute the nodes is. The map can only be traveresed over the x and z axis'.
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A 2D grid would do fine. Pick a resolution that's 50% of your minimum actor width, so you can plan through tight spots.
If you use waypoints, 1m-2m is good for pathfinding and reasoning, but if you're short of memory use a coarser resolution.
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So if my actor, we dont have the models yet so dont know, is width 3, make it a node every 1.5, in a grid. cool. cheers.
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It's best to pick the size of an actor roughly independent of the actual 3D mesh, just decide what it should be from a gameplay perspective and work with that. That way if you make minor changes to the model in the future it shouldn't break anything.
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Instead of treating nodes as points, you can always treat nodes as regions, as in the navmesh approach.
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Ive already done a good bit on the use of nodes, and they will suit the purposes of this game. I was looking at nav meshes a while back and said id have a go if i had time. Doesnt look like i will.
Have bit of a problem. I was trying to add the nodes with an algorithm. So every 1.5 (for now) in the x and z a node is added in grid form, as above. Then id check for intesections with enviornment objects. But Im not sure how i go about defining which nodes are neighbours. I dont get how i create a node and then say set which ones its connected to.
Initially I was considering using an adjacency matrix, so just a list of 1's and 0's id make myself. Where there was a 1 there would be a connection with a node and a 0 no connection. This is obviously level dependant then.
which is the best. Actually right now id prob need the quickest.
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Quote:
Original post by discodowneyIve already done a good bit on the use of nodes, and they will suit the purposes of this game. I was looking at nav meshes a while back and said id have a go if i had time. Doesnt look like i will.
Ok.
Quote:
Original post by discodowneyHave bit of a problem. I was trying to add the nodes with an algorithm. So every 1.5 (for now) in the x and z a node is added in grid form, as above. Then id check for intesections with enviornment objects. But Im not sure how i go about defining which nodes are neighbours. I dont get how i create a node and then say set which ones its connected to.Initially I was considering using an adjacency matrix, so just a list of 1's and 0's id make myself. Where there was a 1 there would be a connection with a node and a 0 no connection. This is obviously level dependant then.which is the best. Actually right now id prob need the quickest.
Is the problem with representing your graph, or is it with determining which nodes are neighbors in the graph?
Ordinarily you'd precompute the adjacency information for your nodes and store them in your map file. Here you just use whatever representation of a graph you like best; this boils down to either (1) an adjacency matrix, or (2) an edge list.
To determine adjacency is slightly trickier, and depends on your needs, but it basically comes down to the idea that an agent can always walk in a straight line between points.* Under this assumption, if your agent were a point particle, you'd just connect nodes whenever the line segment between them does not intersect the level geometry. Since your agents are not point particles but take up space, you probably actually want to test these segments not against your level geometry directly, but against offset geometry.
*(Can agents always walk between points when they're connected by a straight line? This is presumably always true for your case, but in general it may not be: Consider an airplane (which can never slow down too much) with bounded turning radius...)
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Yeah the problem is with determining which nodes are neighbours. grand, adjacency matrix it is.
What do you mean offset geometry?
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A (dense) adjacency matrix representation is a terrible approach for a graph you'll use for pathing. If your world has 1000000 nodes, do you really want to have to look through 1000000 entries every time you need to find the four neighbors of one single node? For that matter, do you want to use 1000000000000 entries to store the thing?
Everyone uses adjacency lists. Go with that.
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Okay so ive been looking up adjacency lists.
My map is approx 100 x 100. so if i have a node at every 1.5 thats gonna be a list thousands of lines long that has to be hardcoded. That is gonna be a killer. think ill increase the distance of the distribution.
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Quote:
Original post by discodowneyMy map is approx 100 x 100. so if i have a node at every 1.5 thats gonna be a list thousands of lines long that has to be hardcoded.
Huh? You write code that automates this process for you; you don't hard-code it...
Quote:
What do you mean offset geometry?
This is the geometry you get after you "push out" your level geometry everywhere along the normal by a distance 'r.'
Say you've got an agent which for collision purposes you model as a sphere of radius 'r.' Instead of worrying about intersecting this sphere with your level geometry, you can instead treat your agent as a point and test for collision against offset geometry.
Here, a picture is worth a thousand words:
(taken from here)
Here the blue-green polygon represents your level geometry, and the black curve represents your offset geometry. The distance between them would be the radius of your agent.
[Edited by - Emergent on April 5, 2010 8:18:18 AM]
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Quote:
Original post by Emergent
Quote:
Original post by discodowneyMy map is approx 100 x 100. so if i have a node at every 1.5 thats gonna be a list thousands of lines long that has to be hardcoded.
Huh? You write code that automates this process for you; you don't hard-code it...
Maybe im being really stupid here but its this part im not sure about.
If i have a node, node 7
1 2 3
6 7 8
11 12 13
How can i automate the process of telling it that, for eg., 7 is connected to 1,2,8,11,13 but not the others?
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Quote:
Original post by discodowneyHow can i automate the process of telling it that, for eg., 7 is connected to 1,2,8,11,13 but not the others?
That's the point of the line segment vs. level-geometry or offset-geometry tests.
Simplest version (pseudocode):
Start with empty adjacency list.For each pair of nodes (p,q){ if(segment from p to q does not intersect geometry) { add an edge from p to q to your adjacency list }}
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Right, gotcha. Got myself a bit confused there thinking about something the wrong way around. Cheers for the help.
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A question about storing the adjacency list. Say i compute my adjacency list and it is something like
1 -> 2,3,6,7
2 -> 1,5,7
3 -> 1,2,6
4 -> 5
5 -> 2,4
6 -> 1,3
7 -> 1,2
What is the best thing to use to store the data. a 2D vector?
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Quote:
Original post by discodowneyWhat is the best thing to use to store the data. a 2D vector?
Often one represents each node as a class/struct, one of whose elements is a list or vector of pointers to other nodes. E.g.,
struct node{ vector2d position; std::vector< node* > neighbors;};std::vector< node > graph;
This is perhaps a little oversimplified though; it's probably best to actually wrap everything up nicely in a class with accessors which hide the internal representation.
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I have a node class. Yeah, i can add a vector of neighbours alright. cheers
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2018-02-20 00:05:14
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https://gmatclub.com/forum/how-many-men-are-in-a-certain-company-s-vanpool-program-268869.html
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# How many men are in a certain company's vanpool program?
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How many men are in a certain company's vanpool program? [#permalink]
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24 Jun 2018, 20:59
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How many men are in a certain company's vanpool program?
(1) The ratio of men to women in the program is 3 to 2.
(2) The men and women in the program fill 6 vans.
NEW question from GMAT® Official Guide 2019
(DS02541)
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Re: How many men are in a certain company's vanpool program? [#permalink]
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24 Jun 2018, 22:15
1
Bunuel wrote:
How many men are in a certain company's vanpool program?
(1) The ratio of men to women in the program is 3 to 2.
(2) The men and women in the program fill 6 vans.
NEW question from GMAT® Official Guide 2019
(DS02541)
We'll translate our questions into equations so we understand what we need to do.
This is a Precise approach.
(1) men : women = 3:2 --> m = 1.5w As we cannot solve for m, this is not enough.
Insufficient.
(2) m + w = 6v. Clearly, not enough
Insufficient.
Combined:
So m = 1.5w and m+w = 6v. Without knowing v or w, this is unsolvable.
Insufficient.
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Re: How many men are in a certain company's vanpool program? [#permalink]
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24 Jun 2018, 22:16
1
Bunuel wrote:
How many men are in a certain company's vanpool program?
(1) The ratio of men to women in the program is 3 to 2.
(2) The men and women in the program fill 6 vans.
NEW question from GMAT® Official Guide 2019
(DS02541)
1) Not suff, As number of men and women can be any multiple of 3x & 2x respectively, no single solution
2) Not suff, As we do not know the capacity of each van and also the ratio of men, women
1) &2) Not suff , same even if we know ratio but capacity of each or all 6 vans is not given hence we cannot find total number of people for obtaining individual number
E
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Joined: 04 Jan 2015
Posts: 3203
Re: How many men are in a certain company's vanpool program? [#permalink]
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25 Jun 2018, 04:20
Solution
To find:
• The number of men in a certain company’s vanpool program
Analysing Statement 1
• As per the information given in statement 1, the ratio of men to women in the program is 3 to 2
o From this statement, if we want to find the number of men, we need to know the exact number of women present in the program
o As the number of women is not mentioned, we can’t determine the number of men
Hence, statement 1 is not sufficient to answer
Analysing Statement 2
• As per the information given in statement 2, the men and women in the program fill 6 vans
o From this, we cannot find out the exact number of men, as no relevant information is provided
Hence, statement 2 is not sufficient to answer
Combining Both Statements
If we combine both the statements, we can say:
• The ratio of men and women is 3:2
• Together they can fill 6 vans
Still we don’t have sufficient information about the individual number of men and women
Hence, the correct answer is option E.
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Re: How many men are in a certain company's vanpool program? [#permalink]
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26 Jun 2018, 22:45
Why is it wrong to say that a van consists of 7 seats? Therefore 6 vans contains 42 seats, and thus the number of men can be determine. Because the statement 2 states that all men and women fill the van.
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GMAT 1: 710 Q44 V41
Re: How many men are in a certain company's vanpool program? [#permalink]
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27 Jun 2018, 04:47
Valhalla wrote:
Why is it wrong to say that a van consists of 7 seats? Therefore 6 vans contains 42 seats, and thus the number of men can be determine. Because the statement 2 states that all men and women fill the van.
I think depending on where one lives the term "van" can represent different cars. Hence, we can't be sure how many seats a van has and not determine the number of men.
Math Expert
Joined: 02 Aug 2009
Posts: 8340
Re: How many men are in a certain company's vanpool program? [#permalink]
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27 Jun 2018, 04:54
Valhalla wrote:
Why is it wrong to say that a van consists of 7 seats? Therefore 6 vans contains 42 seats, and thus the number of men can be determine. Because the statement 2 states that all men and women fill the van.
Never assume what is not given.
Even if some assumes a van means 7 person, it is nowhere mentioned that they are full.
In GMAT or for this particular case in maths, we cannot assume numbers per vehicle
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Re: How many men are in a certain company's vanpool program? [#permalink]
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27 Jun 2018, 17:22
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Re: How many men are in a certain company's vanpool program? [#permalink]
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27 Jun 2018, 18:02
(1) 3 men, 2 women
6 men, 4 women
Insufficient
(2) Not clear? If divisible by 6 or, Do the vans have the same number of seats?
Insufficient
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Joined: 04 Sep 2018
Posts: 27
GPA: 3.33
Re: How many men are in a certain company's vanpool program? [#permalink]
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01 Jan 2019, 11:52
Do such easy questions ever appear on the gmat or are these just for practise? Very curious about it actually !
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Joined: 01 Mar 2019
Posts: 23
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GMAT 1: 740 Q49 V42
Re: How many men are in a certain company's vanpool program? [#permalink]
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15 Mar 2019, 08:49
Stem:
Men = ?
Statement 1:
$$\frac{Men}{Women} = \frac{3}{2}$$
We aren't given the total number of people, so we can't covert the ratio to the actual number of men. INSUFFICIENT.
Statement 2:
No information is given about number of people who can fit in each van. INSUFFICIENT.
Combined (1 & 2):
Since statement 2 doesn't give the number of people who can fit in each van, we can't combine this with the ratio in statement 1 to get the true number of men. INSUFFICIENT.
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Joined: 12 Sep 2015
Posts: 4214
Re: How many men are in a certain company's vanpool program? [#permalink]
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08 May 2019, 06:25
Top Contributor
Bunuel wrote:
How many men are in a certain company's vanpool program?
(1) The ratio of men to women in the program is 3 to 2.
(2) The men and women in the program fill 6 vans.
NEW question from GMAT® Official Guide 2019
(DS02541)
Target question: How many men are in a certain company's vanpool program?
Statement 1: The ratio of men to women in the program is 3 to 2.
There are several scenarios that satisfy statement 1. Here are two:
Case a: There are 3 men and 2 women. In this case, the answer to the target question is there are 3 men
Case b: There are 6 men and 4 women. In this case, the answer to the target question is there are 6 men
Statement 2: The men and women in the program fill 6 vans.
We don't know the RATIO of men to women, AND we don't know how many people fit into each van
Statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
There are several scenarios that satisfy BOTH statements. Here are two:
Case a: There are 18 men & 12 women (for a total of 30 people), and each van holds 5 people . In this case, the answer to the target question is there are 18 men
Case b: There are 36 men & 24 women (for a total of 60 people), and each van holds 10 people . In this case, the answer to the target question is there are 36 men
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT
Cheers,
Brent
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Re: How many men are in a certain company's vanpool program? [#permalink] 08 May 2019, 06:25
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2020-01-17 14:03:34
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https://byjus.com/question-answer/abc-is-isosceles-with-ab-ac-7-5-cm-and-bc-9-cm-figure-the-height-ad-from-a-to-bc-is-6-cm-find-the-area-of-abc-what-will-be-the-height-from-c-to-ab-i-e-ce/
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Question
$\Delta ABC$ is isosceles with $AB=AC=7.5cm$ and $BC=9cm$ (Figure). The height$AD$ from $A$ to$BC$, is $6cm$. Find the area of $\Delta ABC$. What will be the height from$C$to $AB$ i.e., $CE$
Open in App
Solution
Finding the area of $\Delta ABC$ and height from$C$to $AB$ i.e., $CE$:Solution:From the given figure we note that$\mathrm{AB}=\mathrm{AC}=7.5\mathrm{cm}$$BC=9cm\phantom{\rule{0ex}{0ex}}AD=6cm$We know thatArea of $\Delta ABC=\frac{1}{2}×base×height$ $=\frac{1}{2}×BC×AD\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×9×6\phantom{\rule{0ex}{0ex}}=1×9×3\phantom{\rule{0ex}{0ex}}=27c{m}^{2}$$\therefore △\mathrm{ABC}=27{\mathrm{cm}}^{2}$Also,Area of $△\mathrm{ABC}$, $\begin{array}{rcl}\Delta ABC& =& \frac{1}{2}×base×height\\ 27& =& \frac{1}{2}×AB×CE\\ 27& =& \frac{1}{2}×7.5×CE\\ CE& =& \frac{27×2}{7.5}\\ CE& =& \frac{54}{7.5}\\ CE& =& 7.2cm\end{array}$Therefore the area of triangle $△\mathrm{ABC}=27{\mathrm{cm}}^{2}$ and the height from$C$to $AB$ i.e., $CE$$=7.2cm$
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2023-01-29 22:31:36
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http://lists.macromates.com/textmate/2007-February/017646.html
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# [TxMt] LaTeX spelling
Fri Feb 23 19:52:24 UTC 2007
```On Feb 22, 2007, at 8:52 AM, Nathan Paxton wrote:
> All bundles are in as they should be.
>
> Here's a screenshot. As you can see, the spellcheck is alerting
> for words in the preamble. It seems to occur with any dictionary I
> use.
>
I just committed a fix that should take care of the preamble issue
for the most part. We now match \usepackage and \documentclass
explicitly.
> Best,
> -Nathan
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2014-09-18 05:40:20
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https://math.stackexchange.com/questions/2281386/help-solving-the-pde-u-x2-u-y2-u
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# Help solving the PDE $u_x^2 + u_y^2 = u$
I'm trying to solve this Cauchy problem:
$$u_x^2 + u_y^2 = u \\ u(x,0) = x^2+1$$
Here's what I've tried so far:
Letting $p(r,s) = u_x$, $q(r,s) = u_y$ and $z(r,s) = u$, we have:
$$F = p^2 + q^2 - z = 0$$
and thus:
\begin{align} \frac{dx}{ds} &= F_p = 2p \\ \frac{dy}{ds} &= F_q = 2q \\ \frac{dz}{ds} &= pF_p + qF_q = 2p^2 + 2q^2 = 2z\\ \frac{dp}{ds} &= -F_x -pF_z = -(0) -p(-1) = p \\ \frac{dq}{ds} &= -F_y -qF_z = -(0) -q(-1) = q \\ \end{align}
Also, $\Gamma(r,0) = r^2 +1$, and
\begin{align} (r^2+1)' &= \phi_1 \cdot (r)' + \phi_2 \cdot (0)' \\ \implies 2r &= \phi_1 \end{align}
so that:
\begin{align} (2r)^2 + \phi_2^2 &= r^2+1 \\ \implies 4r^2 + \phi_2^2 &= r^2+1 \\ \implies \phi_2^2 &= -3r^2 + 1 \\ \implies \phi_2 &= \pm \sqrt{ -3r^2 +1 } \end{align}
\begin{align} p &= e^s 2r \\ q &= e^s \sqrt{ -3r^2 +1 } \\ \implies x &= 4e^s r - 3r \\ \text{and } y & = 2e^s \cdot ( \pm \sqrt{ -3r^2 +1 } ) \mp \sqrt{ -3r^2 +1 }\end{align}
And I'm pretty sure by now it's gone way off the rails, because I can't solve for for $s$ and $r$ in terms of $x$ and $y$, but I can't figure out where I went wrong.
Could anybody please shed some light?
• Similar to math.stackexchange.com/questions/1033906 – doraemonpaul May 16 '17 at 3:21
• Are you sure your PDE and initial condition is correct? – mattos May 16 '17 at 14:33
• An initial condition like $u = 1 + \frac{x^2}{4}$ would make it so much easier. Then taking an ansatz on the form $u = a + bx + cy + dxy + ex^2 + fy^2$ would work. In this case it does not. – Winther May 16 '17 at 23:40
• @mlaci : The difficulty appears just after the calculus of $p$ and $q$ (about where you stopped). The solving involves a four degree polynomial equation. See an explanation in my answer. – JJacquelin May 17 '17 at 10:02
• @doraemonpaul , Very similar, indeed! I had found that one using Google, but this problem was presented in the midst of other ones that were supposed to be solved using the method I started with above, so I figured the $x^2+1$ would have made a difference, and that I had just made a mistake somewhere in using the method. – mlaci May 18 '17 at 1:51
Obviously, the main difficulty comes from the boundary condition $u(x,0)=x^2+1$ because it involves to solve a polynomial equation leading to huge formula. Even if the solving is theoretically possible, one have to merely accept a result on implicit form.
To make more clear where the difficulty arises, we will avoid the profusion of symbols introduced into the usual method of characteristics, but in following the same approach in fact. $$u_x^2+u_y^2=u \quad;\quad u(x,0)=x^2+1$$ First change of function :$\quad u=v^2\quad\begin{cases}u_x=2vv_x \\ u_y=2vv_y\end{cases}\quad\to\quad v_x^2+v_y^2=\frac{1}{4}\quad;\quad v(x,0)=\sqrt{x^2+1}$
$$v_y=\sqrt{\frac{1}{4}-v_x^2} \quad\to\quad v_{xy}=\frac{v_xv_{xx}}{\sqrt{\frac{1}{4}-v_x^2}}$$ Second change of function : $\quad w=v_x\quad\to\quad w_{y}=\frac{ww_{x}}{\sqrt{\frac{1}{4}-w^2}}$ $$\sqrt{\frac{1}{4}-w^2}\:w_y-2w\:w_x=0 \quad;\quad w(x,0)=v_x(x,0)=\frac{x}{\sqrt{x^2+1}}$$
This is a first order PDE. The set of characteristic ODEs is : $\quad \frac{dy}{\sqrt{\frac{1}{4}-w^2}}=\frac{dx}{-2w}=\frac{dw}{0}$
First family of characteristic curves, coming from $dw=0 \quad\to\quad w=c_1$
Second family of characteristic curves, from $\frac{dy}{\sqrt{\frac{1}{4}-c_1^2}}=\frac{dx}{-2c_1} \quad\to\quad 2c_1y+\sqrt{\frac{1}{4}-c_1^2}x=c_2$
The general solution can be presented on various forms :$\begin{cases} \Phi\left(w\:,\: 2wy+\sqrt{\frac{1}{4}-w^2}\:x\right)=0\\ w=f\left(2wy+\sqrt{\frac{1}{4}-w^2}\:x\right)\\ 2wy+\sqrt{\frac{1}{4}-w^2}\:x=F(w)\end{cases}$
where $\Phi$ , $f$ , $F$ are any differentiable functions. Any one of these functions has to be determined according to the boundary condition.
$$w(x,0)=\frac{x}{\sqrt{x^2+1}}\quad\to\quad 2\frac{x}{\sqrt{x^2+1}}0+\sqrt{\frac{1}{4}-\left(\frac{x}{\sqrt{x^2+1}}\right)^2}\:x=F\left(\frac{x}{\sqrt{x^2+1}}\right)$$
Let $t=\frac{x}{\sqrt{x^2+1}} \quad\to\quad x=\frac{t}{\sqrt{1-t^2}} \quad\to\quad \sqrt{\frac{1}{4}-t^2}\:\frac{t}{\sqrt{1-t^2}} =F\left(t\right)$
Now, the function $F$ is determined. We put it into the above general solution :
$$2wy+\sqrt{\frac{1}{4}-w^2}\:x=\sqrt{\frac{1}{4}-w^2}\:\frac{w}{\sqrt{1-w^2}}$$
$$\frac{4}{\sqrt{1-4w^2}}\:y+\frac{x}{w}=\frac{1}{\sqrt{1-w^2}}$$
Solving this equation for $w$ leads to $\quad w(x,y)$
In fact, this is a four degree polynomial equation. One can solve it analytically, but this involves huge formulas. That is the hitch.
So, we let $w$ on the implicit form of the above equation and, from it, we consider that $w(x,y)$ is known.
$w=\frac{u_x}{2\sqrt{u}}\quad\to\quad \int \frac{u_x}{2\sqrt{u}}=\sqrt{u}=\int w(x,y)dx$ $$u(x,y)=\left(\int w(x,y)dx \right)^2$$
• Thank you very much for that very detailed answer! Although I think that this wasn't what the teacher had in mind (I could be wrong), I was able to follow almost everything you did, so I'm very grateful. Just two questions though: 1) When you write $\Phi\left(w\:,\: 2c_1y+\sqrt{\frac{1}{4}-w^2}\:x\right)=0$ and $w=\frac{u_x}{\sqrt{u}}\quad\to\quad \int \frac{u_x}{\sqrt{u}}=2\sqrt{u}=\int w(x,y)dx$, is this $c_1$ is not equal to $w$ in this instance? 2) When you write $w=\frac{u_x}{\sqrt{u}}$, should there not be a 2 in the denominator in the right-hand side? – mlaci May 18 '17 at 2:13
• Oh, sorry, third question: 3) It's possible to write the equations in the first question only because both $w$ and that whole $2c_1 y + ...x$ equal constants, right? Or would it be possible regardless? – mlaci May 18 '17 at 2:13
• 1) $w$ is not equal to $c_1$ in general. $w=c_1$ only on the characteristics curves. 2) You are right, it's a typo, now corrected. 3) The two family of characteristic curves define two surfaces which intersection gives a specific value for $c_1$ and $c_2$ , that is one of the characteristic curves. In return, to each arbitrary relationship of the general solution corresponds a particular characteristic curve (with a particular couple $c_1,c_2$) on which the PDE is satisfied. – JJacquelin May 18 '17 at 7:10
• For examples, see : math.ualberta.ca/~xinweiyu/436.A1.12f/… – JJacquelin May 18 '17 at 7:18
• Once again, thank you very much! – mlaci May 19 '17 at 0:40
Partial solution
Using the Lagrange Charpit method you find
$$\mathrm{d}x/2p = \mathrm{d}y/2q = \mathrm{d}u/2u = \mathrm{d}p/p = \mathrm{d}q/q$$
from which $p = x/2 +c_1$ and $q = y/2 + c_2$, where $c_i$ are constants of integration. The Pfaff equation $p \, \mathrm{d} x + q \, \mathrm{d}y = \mathrm{d}u$ is thus integrable, providing the general integral of your equation
$$u = (x/2 + c_1)^2 + (y/2 + c_2)^2 + c_3$$
Inserting this into your original PDE yields $c_3 = 0$. However, I don't see any suitable values of $c_{1,2}$ for $u$ to satisfy the condition at $y = 0$ (note the factor $1/2$ affecting the $x$-term).
• I'll have to think about your answer a little more, as I haven't quite understood it yet, but thank you very much for posting it! I've marked JJacqueline's answer as accepted, though, as she gave a full answer to the problem. Thanks again. – mlaci May 18 '17 at 2:05
• Glad to help, @mlaci. That's the way I learnt to solve this kind of problems. – Dmoreno May 18 '17 at 18:11
You lost your way at the integration of $y$, all before looks good. You should have got \begin{align} x&=(4e^s-3)r&&=4(e^s-1)r+r\\ y&=\pm 2(e^s-1)\sqrt{1-3r^2} \end{align} Then $$\pm2yr=(x-r)\sqrt{1-3r^2}\\ 4y^2r^2=(x-r)^2(1-3r^2)$$ and at this point you have to solve this degree 4 polynomial in $r$. After selecting the correct branch ($x>r$ is one condition) you obtain the value of $z$ as $$z=e^{2s}(1+r^2)=\frac{(x+3r)^2}{16}(1+r^2).$$
• Yes, there's definitely a missing $2$ in that $y$ (and also a missing $\pm$ in the $q$ two lines above that). Thank you very much for your answer. Two further questions though: 1) Can I always require $x>r$ even if my parametrization for the prescribed data was $x=r$?; 2) In the last equality in the last line of your answer, are those supposed to be $r$'s or $y$'s? – mlaci Jun 24 '17 at 18:18
• Also, I hand't come back to this question to add this bit, but afterward my teacher basically said, IIRC, that we weren't supposed to solve it for $x$ and $y$ (like we had every previous time), but rather that we were just supposed to know that the implicit function theorem guaranteed us that such a solution existed. :( – mlaci Jun 24 '17 at 18:21
• No, $x>r$ is just a consequence of the solution formula. It might even be wrong as $s<0$ is probably not excluded. In the last equation I used the solution of $x$ to replace $e^s$, one could equally well also use the parametrization of $y$. And yes, one strives to simplify the problem as much as possible, and having a scalar polynomial at the end can be seen as a very favorable outcome. – Dr. Lutz Lehmann Jun 24 '17 at 18:29
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2019-12-06 13:28:22
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https://math.andrej.com/2022/05/20/one-syntax-to-rule-them-all/
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# One syntax to rule them all
I am at the Syntax and Semantics of Type Theory workshop in Stockholm, a kickoff meeting for WG6 of the EuroProofNet COST network, where I am giving a talk “One syntax to rule them all” based on joint work with Danel Ahman.
Abstract: The raw syntax of a type theory, or more generally of a formal system with binding constructs, involves not only free and bound variables, but also meta-variables, which feature in inference rules. Each notion of variable has an associated notion of substitution. A syntactic translation from one type theory to another brings in one more level of substitutions, this time mapping type-theoretic constructors to terms. Working with three levels of substitution, each depending on the previous one, is cumbersome and repetitive. One gets the feeling that there should be a better way to deal with syntax.
In this talk I will present a relative monad capturing higher-rank syntax which takes care of all notions of substitution and binding-preserving syntactic transformations in one fell swoop. The categorical structure of the monad corresponds precisely to the desirable syntactic properties of binding and substitution. Special cases of syntax, such as ordinary first-order variables, or second-order syntax with variables and meta-variables, are obtained easily by precomposition of the relative monad with a suitable inclusion of restricted variable contexts into the general ones. The meta-theoretic properties of syntax transfer along the inclusion.
The relative monad is sufficiently expressive to give a notion of intrinsic syntax for simply typed theories. It remains to be seen how one could refine the monad to account for intrinsic syntax of dependent type theories.
Talk notes: Here are the hand-written talk notes, which cover more than I could say during the talk.
Formalization: I have the beginning of a formalization of the higher-rank syntax, but it hits a problem, see below. Can someone suggest a solution? (You can download Syntax.agda.)
{-
An attempt at formalization of (raw) higher-rank syntax.
We define a notion of syntax which allows for higher-rank binders,
variables and substitutions. Ordinary notions of variables are
special cases:
* order 1: ordinary variables and substitutions, for example those of
λ-calculus
* order 2: meta-variables and their instantiations
* order 3: symbols (term formers) in dependent type theory, such as
Π, Σ, W, and syntactic transformations between theories
The syntax is parameterized by a type Class of syntactic classes. For
example, in dependent type theory there might be two syntactic
classes, ty and tm, corresponding to type and term expressions.
-}
module Syntax (Class : Set) where
{- Shapes can also be called “syntactic variable contexts”, as they assign to
each variable its syntactic arity, but no typing information.
An arity is a binding shape with a syntactic class. The shape specifies
how many arguments the variable takes and how it binds the argument's variables.
The class specifies the syntactic class of the variable, and therefore of the
expression formed by it.
We model shapes as binary trees so that it is easy to concatenate
two of them. A more traditional approach models shapes as lists, in
which case one has to append lists.
-}
infixl 6 _⊕_
data Shape : Set where
𝟘 : Shape -- the empty shape
[_,_] : ∀ (γ : Shape) (cl : Class) → Shape -- the shape with precisely one variable
_⊕_ : ∀ (γ : Shape) (δ : Shape) → Shape -- disjoint sum of shapes
infix 5 [_,_]∈_
{- The de Bruijn indices are binary numbers because shapes are binary
trees. [ δ , cl ]∈ γ is the set of variable indices in γ whose arity
is (δ, cl). -}
data [_,_]∈_ : Shape → Class → Shape → Set where
var-here : ∀ {θ} {cl} → [ θ , cl ]∈ [ θ , cl ]
var-left : ∀ {θ} {cl} {γ} {δ} → [ θ , cl ]∈ γ → [ θ , cl ]∈ γ ⊕ δ
var-right : ∀ {θ} {cl} {γ} {δ} → [ θ , cl ]∈ δ → [ θ , cl ]∈ γ ⊕ δ
{- Examples:
postulate ty : Class -- type class
postulate tm : Class -- term class
ordinary-variable-arity : Class → Shape
ordinary-variable-arity c = [ 𝟘 , c ]
binary-type-metavariable-arity : Shape
binary-type-metavariable-arity = [ [ 𝟘 , tm ] ⊕ [ 𝟘 , tm ] , ty ]
Π-arity : Shape
Π-arity = [ [ 𝟘 , ty ] ⊕ [ [ 𝟘 , tm ] , ty ] , ty ]
-}
{- Because everything is a variable, even symbols, there is a single
expression constructor __ which forms and expression by applying
the variable x to arguments ts. -}
-- Expressions
infix 9 __
data Expr : Shape → Class → Set where
__ : ∀ {γ} {δ} {cl} (x : [ δ , cl ]∈ γ) →
(ts : ∀ {θ} {B} (y : [ θ , B ]∈ δ) → Expr (γ ⊕ θ) B) → Expr γ cl
-- Renamings
infix 5 _→ʳ_
_→ʳ_ : Shape → Shape → Set
γ →ʳ δ = ∀ {θ} {cl} (x : [ θ , cl ]∈ γ) → [ θ , cl ]∈ δ
-- identity renaming
𝟙ʳ : ∀ {γ} → γ →ʳ γ
𝟙ʳ x = x
-- composition of renamings
infixl 7 _∘ʳ_
_∘ʳ_ : ∀ {γ} {δ} {η} → (δ →ʳ η) → (γ →ʳ δ) → (γ →ʳ η)
(r ∘ʳ s) x = r (s x)
-- renaming extension
⇑ʳ : ∀ {γ} {δ} {Θ} → (γ →ʳ δ) → (γ ⊕ Θ →ʳ δ ⊕ Θ)
⇑ʳ r (var-left x) = var-left (r x)
⇑ʳ r (var-right y) = var-right y
-- the action of a renaming on an expression
infixr 6 [_]ʳ_
[_]ʳ_ : ∀ {γ} {δ} {cl} (r : γ →ʳ δ) → Expr γ cl → Expr δ cl
[ r ]ʳ (x ts) = r x λ { y → [ ⇑ʳ r ]ʳ ts y }
-- substitution
infix 5 _→ˢ_
_→ˢ_ : Shape → Shape → Set
γ →ˢ δ = ∀ {Θ} {cl} (x : [ Θ , cl ]∈ γ) → Expr (δ ⊕ Θ) cl
-- side-remark: notice that the ts in the definition of Expr is just a substituition
-- We now hit a problem when trying to define the identity substitution in a naive
-- fashion. Agda rejects the definition, as it is not structurally recursive.
-- {-# TERMINATING #-}
𝟙ˢ : ∀ {γ} → γ →ˢ γ
𝟙ˢ x = var-left x λ y → [ ⇑ʳ var-right ]ʳ 𝟙ˢ y
{- What is the best way to deal with the non-termination problem? I have tried:
1. sized types: got mixed results, perhaps I don't know how to use them
2. well-founded recursion: it gets messy and unpleasant to use
3. reorganizing the above definitions, but non-structural recursion always sneeks in
A solution which makes the identity substitition compute is highly preferred.
The problem persists with other operations on substitutions, such as composition
and the action of a substitution.
-}
Posting comments: At present comments are disabled because the relevant script died. You are welcome to contact me directly.
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2023-03-29 21:41:24
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https://laszukdawid.com/category/optimisation/
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# Kuramoto in Stan (PyStan)
tl;dr: Project on github: https://github.com/laszukdawid/pystan-kuramoto
Stan is a programming language focused on probabilistic computations. Although it’s a rather recent language it’s been nicely received in data science/Bayesian community for its focus on designing model, rather than programming and getting stuck with computational details. Depending what is your background you might have heard about it either from Facebook and its Prophet project or as a native implementation for Hamiltonian Monte Carlo (HMC) and its optimised variation – No-U-Turn Sampler for HMC (NUTS).
For ease of including models in other programmes there are some interfaces/wrappers available, including RStan and PyStan.
Stan is not the easiest language to go through. Currently there are about 600 pages of documentation and then separate “documentations” for wrappers, which for PyStan isn’t very helpful. Obviously there’s no need for reading all of it, but it took me a while to actually understand what goes where an why. The reward, however, is very satisfying.
Since I’ve written a bit about Kuramoto models on this blog, it’s consistent if I share its implementation in Stan as well. Pystan-kuramoto project uses PyStan, but the actual Stan code is platform independent.
Currently there are two implementations (couldn’t come up with better names):
• All-to-all, where Kuramoto model fit is performed to phase vector $\vec{\Phi}$ with distinct oscillators, i.e. $\vec{\Phi}_{N}(t) = \{\phi_1(t), \phi_2(t), \dots, \phi_N(t)\}$.
• All-to-one, where the model fits superposition (sum) of oscillators to phase time series $\Phi_{N}(t) = \sum_{n=1}^{N} \phi_n(t)$.
In all honesty, this seems to be rather efficient. Optimisation is performed using penalized maximum likelihood estimation with optimization (L-BFGS). Before using it I wasn’t very knowledgeable in the algorithm, but now I’m simply amazed with its speed and accuracy. Thumbs up.
# Update: Particle Swarm Optimisation in Python
It came to my attention that my PSO for Python is actually quite popular. Only after few people contacted me I’ve noticed that the public version was not the same that I’ve been using. It wasn’t bad, but definitely not as good. Thus, obviously, I’ve updated the version and cleaned it a bit.
Update programme is available from my github or from Code subpage.
What’s the difference? There are few.
– Initial values, unless provided, are psuedo-random generated using Halton sequence. This prevents from artificial clustering.
– Perturbation in form of a Gaussian noise should mitigate false local minima by forcing particles to search surrounding area.
– Added max_repetition threshold, which states the maximum number of obtaining the same optimum value. Upon reaching threshold program finishes.
– General improvement in performance.
– Improved usage documentation within the file.
– Program is now compatible with Python3.
Feel free to request any features.
There is an idea of adding progressive save, which would quit, resume and modify parameters at any point of computation.
# Halton sequence in Python
Sometimes when we ask for random we don’t actually mean random by just random. Yes, pseudo-random.
Consider unitary distribution with ranges 0 and 1. Say you want to draw 5 samples. Selecting them at random would mean that we might end up with set of {0, 0.1, 0.02, 0.09, 0.01} or {0.11, 0.99, 0.09, 0.91, 0.01}. Yes, these values don’t seem very random, but that’s the thing about randomness, that it randomly can seem to not be random.
Depending on the purpose of our selection, these values might be just OK. After all, they came from that distribution. However, if our goal is to reconstruct the distribution, or extract information about with limited number of samples, it is often better to draw those samples in pseudo-random way. For example, in accordance to van der Corput sequences for 1D distributions or its generalized version Halton sequence.
The best practice for sampling N dimensional distribution is to use different prime numbers for each dimension. For example, when I need to sample a 5 dimensional unitary distribution, or search space, I will use bases of (5, 7, 11, 13, 17). This is to prevent periodic visits of the same position.
In case you are wondering what’s the difference between actual random and pseudo-random, here is a gist:
Both are good, but the actual random can produce many empty holes. What we like to have is a fair representation of all areas of our search space.
Thus, without further ado, here are some code snippets.
This is a definition of my prime generating generator:
def next_prime():
def is_prime(num):
"Checks if num is a prime value"
for i in range(2,int(num**0.5)+1):
if(num % i)==0: return False
return True
prime = 3
while(1):
if is_prime(prime):
yield prime
prime += 2
As for Halton sequence, as mentioned before it uses van der Corput sequence. Again, here is the definition:
def vdc(n, base=2):
vdc, denom = 0, 1
while n:
denom *= base
n, remainder = divmod(n, base)
vdc += remainder/float(denom)
return vdc
And finally, definition for the Halton sequence:
def halton_sequence(size, dim):
seq = []
primeGen = next_prime()
next(primeGen)
for d in range(dim):
base = next(primeGen)
seq.append([vdc(i, base) for i in range(size)])
return seq
To use all of this simply call halton_sequence(size, dim). These variables refer to the number of size of sample poll and the dimension of your problem. So if one wants to sample 3 dimensional space with 10 samples each it would be called as below. (Notice: first dimension has prime value 5, then it’s 7, 11, and following prime values.)
>>> halton_sequence(10, 3)
[
[0, 0.2, 0.4, 0.6, 0.8, 0.04, 0.24000000000000002, 0.44, 0.64, 0.8400000000000001],
[0, 0.14285714285714285, 0.2857142857142857, 0.42857142857142855, 0.5714285714285714, 0.7142857142857143, 0.8571428571428571, 0.02040816326530612, 0.16326530612244897, 0.30612244897959184],
[0, 0.09090909090909091, 0.18181818181818182, 0.2727272727272727, 0.36363636363636365, 0.45454545454545453, 0.5454545454545454, 0.6363636363636364, 0.7272727272727273, 0.8181818181818182]
]
# Bayesian inference in Kuramoto model
For a while now, I’ve been involved in Kuramoto models and adjusting them to data. Recently I’ve stumbled upon using Bayesian inference on predicting parameters in time-evolving dynamics. Team from Physics Department at University of Lancaster has produced many papers, but to provide with some reference it might be worth to look at [1].
Technique refers to general phase $\phi$ dynamics in oscillator, i.e.
$\dot\phi_i = \omega_i + f_i (\phi_i) + g_i (\phi_i, \phi_j) + \xi_i$,
where $\omega_i, f_i(\phi_i)$ and $g_i(\phi_i, \phi_j)$ are intrinsic frequency, self coupling and coupling with other oscillators, respectively. Term $\xi_i$ refers to noise and although authors claim that it can be any type of noise calculations are performed for white Gaussian type. I don’t really want to go much into details, because on first sight it might look complicated. Just to point out few steps that are necessary to introduce my examples:
1. Rewrite model in form of $\dot\phi_i = \sum_{k=-K}^{K} C_{k}^{(i)} P_{i,k} (\Phi) + \xi_i(t)$, where $c$ is parameters vector, $P$ provides significant terms (for example via Fourier decomposition) for our model and $\Phi$ is a vector of all oscillators’ phases (at some moment).
2. Calculate diagonal values of Jacobian matrix, i.e. $\frac{\partial P_{i,k}}{\partial \phi_{i}}$.
3. Set a priori probabilities for parameters $C$ vector and its covariance $\Sigma$ (or concentration $\Xi = \Sigma^{-1}$).
Then the problem is reduced to finding maximum of minus log-likelihood function S. Authors also provide the exact formulas and algorithm, which find the extremum. Tutorial on applying their algorithm can be found in [2].
Based on MatLab code provided be the authors, and available on their webpage [3], I have written my own program in Python (code available in Code section or here ). As an experiment I have used simple Kuramoto model with sinusoidal coupling between phases
$\dot\phi_1 = 28 - 0.3 \sin( \phi_1 - \phi_2) - 0.1 \sin( \phi_1 - \phi_3 ) + 0.0 \sin( \phi_2 - \phi_3)$,
$\dot\phi_2 = 19 + 0.3 \sin( \phi_1 - \phi_2) + 0.0 \sin( \phi_1 - \phi_3 ) - 0.9 \sin( \phi_2 - \phi_3)$,
$\dot\phi_3 = 11 + 0.0 \sin( \phi_1 - \phi_2) + 0.1 \sin( \phi_1 - \phi_3 ) + 0.9 \sin( \phi_2 - \phi_3)$,
which means there are 12 parameters – three intrinsic frequencies ($\omega_i = \{28, 19, 11\}$) and 3×3 K matrix representing coupling between pairs of oscillators. To each oscillator small Gaussian noise was added (mean 0 and standard deviation 0.01). Signals were generated for $t \in [0, 40]$ with sampling step $dt = 0.001$. Furthermore, analysis was performed on 8 s segments with 2 s step. Figure 1 shows instantaneous frequencies ($\dot\phi_i$) for each oscillator. Figure 2 presents calculated parameters for each segment (thus time dependency). To be honest, I am quite surprised how good this technique works. I have performed more experiments, but to keep it short I am posting just one. There are some fluctuations in obtained values, but those changes are very small and most likely, in practical sense, negligible.
Fig. 1. Instantaneous frequencies of all oscillators in experiment.
Fig. 2. Extracted parameters for presented dynamical system. First column shows values on intrinsic frequencies and the rest respective to title coupling value. Horizontal black lines indicate what are expected values.
[1] A. Duggento, T. Stankovski, P. V. E. McClintock, and A. Stefanovska, “Dynamical Bayesian inference of time-evolving interactions: From a pair of coupled oscillators to networks of oscillators,” Phys. Rev. E – Stat. Nonlinear, Soft Matter Phys., vol. 86, no. 6, pp. 1–16, 2012.
[2] T. Stankovski, A. Duggento, P. V. E. McClintock, and A. Stefanovska, “A tutorial on time-evolving dynamical Bayesian inference,” Eur. Phys. J. Spec. Top., vol. 223, no. 13, pp. 2685–2703, 2014.
[3] Nonlinear Biomedical Physics, Lancaster University http://py-biomedical.lancaster.ac.uk/.
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2017-12-12 15:44:50
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https://denisegaskins.com/2008/11/24/contig-game-master-your-math-facts/?shared=email&msg=fail&replytocom=73432
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# Contig Game: Master Your Math Facts
[Photo by Photo Mojo.]
Yahtzee and other board games provide a modicum of math fact practice. But for intensive, thought-provoking math drill, I can’t think of any game that would beat Contig.
Math concepts: addition, subtraction, multiplication, division, order of operations, mental math
Number of players: 2 – 4
Equipment: Contig game board, three 6-sided dice, pencil and scratch paper for keeping score, and bingo chips or wide-tip markers to mark game squares
## Set Up
Place the game board and dice between players, and give each player a marker or pile of chips. (Markers do not need to be different colors.) Write the players’ names at the top of the scratch paper to make a score sheet.
## How to Play
• Each player rolls a die. Whoever rolls the smallest number will go first, and the play proceeds to the left (clockwise) around the table.
• On your turn, roll all three dice. If any die falls off the table or lands at a slant, all three dice must be rolled again. Do not touch the dice after they are rolled, though you may use a pencil to scoot them next to each other.
• Use the three numbers and the basic arithmetic operations ($+$, $-$, $\times$, or $\div$) to form a two-step equation that equals any available square on the game board. You may not use an answer that has already been marked. Try as many options as you can think of, to make sure you find the highest scoring combination.
• Mark your answer on the game board with a bingo chip or a large X. At the same time, say out loud how you calculated the number. Add to your score one point for the square you marked PLUS one point for each already-marked square that is touching any side or corner of your number’s square. (Maximum score = 9.)
• Another player may challenge your answer before the next player rolls the dice. If the challenge is upheld — that is, if you made a mistake — the challenger takes the points you would have won, and you score zero. If your calculation is correct, you get one bonus point for having withstood the challenge.
• If all the numbers you can calculate have already been marked, your score is zero for that turn. But if another player can think of a valid combination, he can challenge you, mark the square, and take those points.
## Endgame
• Play until each player has had 10 turns.
• Whoever has the highest total score wins the game.
## Variations
Scoring option: The most common variation I found online was NOT to score a point for the marked square. Just score one point for each contiguous square that was previously marked. (Maximum score = 8.) I strongly prefer the scoring system above, which awards at least one point for any valid calculation.
Tic-Tac-Toe: Players mark numbers with X and O, and the first player to get 3 squares in a row wins. Rows may be vertical, horizontal, or diagonal. For a longer game, try 4 or 5 in a row.
Multi-player extended play: Any player who gets a zero three turns in a row drops out of the game. When the last player gets his third strike, the game is over. There is no bonus for the last player, other than his extra turn(s). Add up the scores as usual to find the winner.
Tournament play: Two players per game board. Set a timer, giving each player only 30 seconds for each turn. Think fast! If you do not mark a square within the 30 seconds, your score is zero for that turn. (Scores of zero may not be challenged.) After 10 turns, add up the players’ scores for that round. Then trade partners, get a new game board, and play another round. After three rounds, award 1st, 2nd, and 3rd place ribbons to the top scorers in each age group/grade level.
Mathwire featured Contig in its November newsletter, which reminded me that I had intended to post the rules on my blog. The game is an excellent way for students to practice their math facts and build their mental math skills. Every spring, our local homeschool group holds a series of Contig practices. Then we host a “school” tournament, and the winners proceed to a regional tournament at the local community college.
Here are a few links for Contig variations on the Internet:
This post is an excerpt from my book Multiplication & Fractions: Math Games for Tough Topics, available now at your favorite online book dealer.
Want to help your kids learn math? Claim your free 24-page problem-solving booklet, and you’ll be among the first to hear about new books, revisions, and sales or other promotions.
## 21 thoughts on “Contig Game: Master Your Math Facts”
1. Math Club variation: Ask your students why the game board looks the way it does. Why is 216 the biggest number? Why does the board have a square for 100, and for 150, but not for 200? Etc.
Ultimate number challenge: Find all the ways to make each number with Contig dice. Whew!
2. Hai,
Your blog is fantastic. You’ve done a fantastic job composing and making it super rich with content. After seeing yours, Tatyanan & Chanhee’s blogs, I’m wishing I stumbled across WordPress instead of Blogger. Thanks also for cluing me into Contig. My 5 year old & I recently took up cards and Rummy Tile to help us with math. It is helping and we’ll give Contig a try too. Thanks again and your blog is top notch!
Jim
3. chanheeh says:
It is very fun, but on the other hand very brain working game. I played similar games with my daughter before, which was her school assignment. I will try this game in this holidays with my family!!
4. Thanks for finding my blog. What a setup you have! I plan on looking through this to get some ideas for spring semester.
Andy
5. My daughter’s hard work has paid off. This was her first year playing, but she managed to take second place (for her grade level) in the homeschool Contig competition and is headed to the Regionals. Congratulations, Kitten!
6. Jack Williams says:
My name is Jack Williams a retired mathematics teacher. In 1969-71 I was director of the Central Iowa Low Achievers Mathematics Project. David O’Neil, Les Lewis, Frank Broadbent, and myself invented CONTIG. I am wondering where you happened to find this great game.
1. Jack Williams says:
Denise where do you live in central Illinois? I live in Coal City.
7. Our school district (central Illinois) has been running a yearly Contig competition for ages. I don’t know when they started. It’s a fantastic game! I went to the Regional competition this spring with my daughter, and it was wonderful to see a whole gymnasium full of school kids there just to play with math.
8. I try not to give out too much personal info online (practicing what I preach to my kids), but I will say that we’re not in your neighborhood. We’re closer to Springfield than we are to Chicago.
9. Melissa says:
I played Contig when I was in elementary school back in the late 70s! I am know a math teacher–I use it with my students. I fundamentally believe Contig changed my ability to love and enjoy math which translated into my math teaching career!
10. At http://sites.google.com/site/contigforwindows/ one can find a freeware Contig game for windows — For many years, I had another site using yahoo Geocities. Unfortunately, Yahoo discontinued the free webpage service so I moved it to Google sites.
— The game is 100% free with nothing disabled. Any computerization of this game should be free as it helps kids learn math. Anybody who writes and sells a Contig game for money should be frowned and put in the corner with a DUNCE HAT unless they give 100% of the money to educational needs.
I was exposed to Contig when volunteering in a class room. I wrote a computer program to figure out the odds of making the various numbers on the board. I realized it could easily be turned into a computer game to help kids. So I DID and then I gave it away FREE.
I would like to introduce you a nice site which deals with the same subjects like here.
You can find interactive free online math games for kids.
Here Order of Operations Games from the site: Order of Operations Games
and to the main site: Math Games
enjoy!
12. pranita pillai says:
We are having maths fest in our school and our game is contig!
13. Played this today in our 3rd and 4th grade math games class. It went well and was a solid hit. Two interesting points from playing this immediately after the factor captor game:
(1) the kids initially felt relief when seeing the dice. They guessed that dice would take a lot of strategic thinking out of their hands, but this game still allows quite a lot of flexibility for them to figure out how to move.
(2) there was a little confusion about scoring, based on hold-over from factor captor. Some kids thought that the value of the square they were claiming was part of the score, while others got the point about neighboring squares right away.
I could imagine a hybrid game where both forms of scoring contribute. In fact, I noticed some pairs were focused on low numbers in their play, so this hybrid scoring might shift them to more to using multiplication in their combinations.
1. A hybrid scoring plan could be interesting: value of the square claimed + number of contiguous squares marked. I’m not sure it would lead to as much thinking about alternative choices, since one would almost always want to go for the biggest number.
Another option I’ve used is to change the game board to make the small numbers less useful. You can find this spiral Contig board in my Multiplication & Fraction Printables file.
1. Now I understand the point of the spiral board!
1. If you aren’t getting the results you want, change the system. “Every system is perfectly designed to get the result that it does.” ― W. Edwards Deming 🙂
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2019-11-12 06:48:30
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http://umj.imath.kiev.ua/article/?lang=en&article=11288
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2019
Том 71
№ 6
# The best $L_1$-approximations of classes of functions defined by differential operators in terms of generalized splines from these classes
Abstract
For classes of periodic functions defined by constraints imposed on the $L_1$-norm of the result of action of differential operators with constant coefficients and real spectrum on these functions, we determine the exact values of the best $L_1$-approximations by generalized splines from the classes considered.
English version (Springer): Ukrainian Mathematical Journal 50 (1998), no. 11, pp 1649–1658.
Citation Example: Babenko V. F., Leis Azar The best $L_1$-approximations of classes of functions defined by differential operators in terms of generalized splines from these classes // Ukr. Mat. Zh. - 1998. - 50, № 11. - pp. 1443-1451.
Full text
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2019-06-20 18:02:36
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https://brinleysgrading.com/k622n/1ec8d0-yield-to-maturity-formula-excel
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Rate (required argument) – The annual coupon rate. Assume that the annual coupons are $100, which is a 10% coupon rate, and that there are 10 years remaining until maturity. Pmt = The payment made in every period. Once data is at hand, open an excel file and prepare your spreadsheet. It is an Annualizing Factor. if any of the following are true: rate ; 0 frequency is not 1,2, or 4; pr or redemption are … Utilisez la fonction RENDEMENT.TITRE pour calculer le taux de rendement d’une obligation. PV = Present value of the bond. You have made a plan to issue a bond with the following details: Now, you went to a bond rating agency (Moody’s, S&P, Fitch, etc.) YIELD returns #NUM! Settlement (required argument) – This is the settlement date of the security. You went to sell your bond and found that same rated bonds are selling with the market rate (YTM) 4.5%. The values must contain a positive value and a negative value. Face Value =$1300 2. Description. YTM is also known as the internal rate of return. Calculate Yield. Coupon Rate (Annual): 6% 3. FV = Future value of the bond. Plug the yield to maturity back into the formula to solve for P, the price. In the context of debt securities, yield is the return that a debt-holder earns by investing in a security at its current price. Le traduction Française de la Excel fonction YIELD est la suivante: Anglais Français; YIELD: RENDEMENT.TITRE. In most cases (if not all cases), don’t use this value. How to Calculate Yield to Maturity (YTM) in Excel. You just need to enter the inputs like face value, coupon rate, years to maturity etc and Excel will calculate the bond yield and display it for you. Current yield equals the annual interest payment divided by the current market price of the security. Example of Yield to Maturity Formula. So, pmt will be $1000 x 3% =$30. YIELD(settlement, maturity, rate, pr, redemption, frequency, [basis]) Years to Maturity: 5 years. Syntax. But coupons per year is 2. Bond Price = $1600 Solution: Here we have to understand that this calculation completely depends on annual coupon and bond price. The YTM is easy to compute where the acquisition cost of a bond is at par and coupon payments are effected annually. It uses the par value, market value, and coupon rate to calculate yield to maturity. I also have used another term in the formula. Returns the annual yield of a security that pays interest at maturity. Decide whether you are satisfied with the estimate or if you need more precise information. YTM = [ (AIP) + ((FV – CP) / (Y)) ]/ [ (FV + CP) / 2 ] Where YTM is the yield to maturity From the time you buy the bond. Maturity (required argument) – This is the maturity date of the security. This is an easy and straightforward way of calculating YTM in Excel. It is all about future cash flows and their present values discounted with an interest rate. The company pays interest two times a year (semi-annually). 5. Calcule le rendement d’un titre rapportant des intérêts périodiquement. The coupon rate is 6%. RATE function returns the interest rate for a period. Calculate the current yield of the bond. Prepare your spreadsheet. eval(ez_write_tag([[300,250],'xplaind_com-box-3','ezslot_0',104,'0','0'])); Yield to maturity is the internal rate of return of a security which means it is the rate an investor will earn by purchasing the security at its current price and receiving all future cash flows, such as coupon payments till maturity and the maturity value. Because I want to discount the cash flows with the market rate. Create Yield To Maturity Formula In Vba - Hello guys I got a question regarding some financial... - Free Excel Help Home ... (formula) so that excel can calculate overtime hours. You see I have just entered the future cash flows from the bond investments in a column (Payment column) and then used Excel’s IRR function. A bond might sell in both discounts or premiums. Some terms must usually be met: 1. a set period of time, also known as call protection, where the bond cannot be redeemed 2. call price 3. other terms and conditions The issuer needs a call option to reduce … Here for the rate argument, I have used the value of 7.50% (also divided it by 4 to get the period interest). We provide tips, how to guide and also provide Excel solutions to your business problems. So easy to use and straightforward. Annual Coupon Payment = 5% *$1,000 2. Nper = Maturity Years x Number of payments a year, And this formula gives us to value: $89,513.17. This yield to maturity calculator uses information from a bond and calculates the YTM each year until the bond matures. Description. Coupons Per Year (npery): 2. How will you set the price of a bond in a discount? The scenario can also be different for the above bond. We can use the above formula to calculate approximate yield to maturity. I earn a small commission if you buy any products using my affiliate links to Amazon. by Obaidullah Jan, ACA, CFA and last modified on Jan 27, 2018Studying for CFA® Program? Before the PV function, I have used -ve sign to make the Present Value positive. And at the end of the bond maturity, we get the coupon payment and the face value back, so it is$1030. RATE (nper, pmt, pv, [fv], [type], [guess]), Nper = Total number of periods of the bond maturity. Open Excel 2010 and save your file as yield.xlsx. You must be thinking the result must come in percentage, For that you just have to change the format of the cell from Number to Percent in Excel. Returns the yield on a security that pays periodic interest. Solution: Annual Coupon Payment is calculated using the formula given below Annual Coupon Payment = Coupon Rate * Par Value 1. of years in Maturity (n) 12 Price of the Bond (P) 940 Annual Coupon (C) 80.00 Yield to Maturity 8.76% Assume that the price of the bond is $940 with the face value of bond$1000. Coupons Per Year (npery): 2. Type = Type can be either 0 or 1 or omitted. So, it will happen that you will not be able to sell the bond at face value. Here are the details of the bond: 1. So, you will be able to sell your bond at $112,025.59 with a premium of amount$12,025.59. Settlement, maturity, frequency, and basis are truncated to integers; If settlement or maturity dates are not valid, YIELD returns #VALUE! Current Price of Bond (Present Value, pv): $938.40. In this article, I will show how to calculate yield to maturity (YTM) in Excel. ExcelDemy.com is a participant in the Amazon Services LLC Associates Program, an affiliate advertising program. 4. Later, I have multiplied this value (3.75%) by 2 as the bond pays two times (semi-annually) a year. It is the amount that you spend to buy a bond. Yield to Maturity Formula The following formula is used to calculate the yield to maturity of a bond or investment. This is why we have multiplied this return by 2 to get the yearly internal rate of return. Annual Coupon Payment =$50 Current Yield of a Bond can be calculated using the formula given below … The rate of yield comes out to be 0.107 (in decimals). There are two common measures of yield: current yield and yield to maturity. - Excel . 3. Input required values in the ‘User Inputs’ section and you will get the YTM automatically (lower part of the template). YIELD is an Excel function that returns the yield to maturity of a bond given its coupon rate, current price, principal amount and coupon payment frequency per year. The IRR function returns the internal rate of return for a period. After solving this equation, the estimated yield to maturity is 11.25%. It cannot change over the life of the bond. = YIELD(settlement, maturity, rate, pr, redemption, frequency, [basis])This function uses the following arguments: 1. Thanks for the feedback! The YTM is based on the belief or understanding that an investor purchases the security at the current market price and holds it until the security has matured Unlike the current yield, the yield to maturity (YTM) measures both current income and expected capital gains or losses. 1. If the type is 1, the coupon payment is done at the beginning of the period. How to calculate compound interest for recurring deposit in Excel! Coupon Rate = 6% 3. The result should be 0.0459--4.59 percent--which is the annual yield to maturity of this bond. So, to get the yearly interest rate, we multiplied the RATE value by 2 (cell C7). Let’s take an example to understand how to use the formula. Let’s calculate now your bond price with the same Excel PV function. A callable bond is a simple financial instrument that can be redeemed by the issuer before the maturity date. Description. Between these two, we get $30 in every period. Suppose, you got an offer to invest in a bond. Furthermore, the current yield is a useless statistic for zero-coupon bonds. We hope you like the work that has been done, and if you have any suggestions, your feedback is highly valuable. However, for other cases, an approximate YTM can be found by using a bond yield table. It completely ignores the time value of money, frequency of payment and amount value at the time of maturity. You’re wondering whether you would invest in the bond. Thanks, Nice to hear that you found this article helpful. Guess = It is just a guess value. Solution: Use the below-given data for calculation of yield to maturity. Use the Yield Function to Calculate the Answer Type the formula “=Yield (B1,B2,B3,B4,B5,B6,B7)” into cell B8 and hit the “Enter” key. From the time you buy the bond. of Years to Maturity. On the other hand, the term “current yield” means the current rate of return of the bond investment computed on the basis of the coupon payment expected in the next one year and the current market price. So, nper is 5 x 2 = 10. Mathematically, the formula for bond price using YTM is represented as, Bond Price = ∑ [Cash flowt / (1+YTM)t] Where, t: No. Step 1:… Once created, the desired data will automatically appear in designated cells when the required input values are entered. Coupon on the bondwill be$1,000 * 8% which is $80. But as payment is done twice a year, the coupon rate for a period will be 6%/2 = 3%. It means the yield return is approx 11%. So, that is my two ways of calculating yield to maturity (YTM) in Excel. and they rated your bond as AA+. Using Excel, you can develop a bond yield calculator easily with the help of a number of formulas. This article describes the formula syntax and usage of the YIELD function in Microsoft Excel. In the context of debt securities, yield is the return that a debt-holder earns by investing in a security at its current price. I did not use it. However, our approximation is good enough for exams or for quick comparisons. / Excel Formula for Yield to Maturity. When the bond matures, you get return the face value of the bond. Previous Lesson ‹ Bond Equivalent Yield Convention. Syntax. In Excel, dates are serial numbers. 4. Yield to maturity can be calculated by solving the following equation for YLD using hit-and-trial: $$\text{Price}=\text{REDEMPTION}\ \times\frac{\text{RATE}}{\text{FREQUENCY}}\times\frac{\text{1}-{(\text{1}+\frac{\text{YLD}}{\text{FREQUENCY}})}^{-\text{n}}}{\text{YLD}/\text{FREQUENCY}}+\frac{\text{REDEMPTION}}{{(\text{1}+\frac{\text{YLD}}{\text{FREQUENCY}})}^\text{n}}$$eval(ez_write_tag([[580,400],'xplaind_com-medrectangle-3','ezslot_1',105,'0','0'])); YIELD(settlement, maturity, rate, pr, redemption, frequency, [basis]. Assume that the price of the bond is$940 with the face value of bond $1000. Chances are, you will not arrive at the same value. In california over 8 hrs in one day is overtime. In such a situation, the yield-to-maturity will be equal to coupon payment. It depends on the market rate of similar bonds. This article describes the formula syntax and usage of the YIELDMAT function in Microsoft Excel. Let us find the yield-to-maturity of a 5 year 6% coupon bond that is currently priced at$850. To make this decision, you want to know the Yield to Maturity (also called Internal Rate of Return) from investing in the bond. Best regards, Great job will put all templates to work, how ever looking for template for my “Dividend Tracking Portfolio” of 5~6 k with very few MANUAL entry love to download free if available or for reasonable price. How to calculate future value with inflation in Excel, Compound interest excel formula with regular deposits, Effective Interest Rate Formula Excel + Free Calculator, 15 Best Online Excel Training Courses | Learn Advanced Excel Online, Able2Extract Professional 15 Review 2020 (with 15% Discount), Par Value of Bond (Face Value, fv): $1000. Check out the image below. Please note that call option does not mean that an issuer can redeem a bond at any time. You will want a higher price for your bond so that yield to maturity from your bond will be 4.5%. A bond yield calculator, capable of accurately tracking the current yield, the yield to maturity, and the yield to call of a given bond, can be assembled in a Microsoft Excel spread sheet. Years to maturity of the bond is 5 years. keep up good work. Yield to Maturity (YTM) Formula Excel Template Prepared by Dheeraj Vaidya, CFA, FRM visit - [email protected] Particulars Values Face Value of Bond (F) 1000 Annual Coupon Rate 8% No. the date on which the security-holder receives principal back, Pr stands for the current market price of the security; redemption is the value received by the bond-holder at the expiry of the bond representing the repayment of principal; frequency refers to number of periodic interest payments per year and [basis] is an optional argument specifying the day-counting basis to be used.eval(ez_write_tag([[250,250],'xplaind_com-medrectangle-4','ezslot_4',133,'0','0'])); The following example shows how to enter the required values in YIELD function: Note that we entered price (pr) and redemption value (redemption) per$100 regardless of actual face value of the bond. XPLAIND.com is a free educational website; of students, by students, and for students. 2. This example using the approximate formula would be . Curre… Par Value of Bond (Face Value, fv): $1000 2. Years to Maturity: 5 years. It is not a good measure of return for those looking for capital gains. You cannot compute the interest rate by hand using the exact equation for yield to maturity (YTM), as that equation is too complex. Settlement refers to the settlement date i.e. This is when you will sell your bond at a discount. The 8 would be considered regular hours and anything over is OT. Let us take the example of a bond that pays a coupon rate of 5% and is currently trading at a discount price of$950. But the problem is: when you tried to sell the bond, you see that the same rated bond is selling with 7.5% YTM (yield to maturity). The yield to maturity (YTM) of a bond is the internal rate of return (IRR) if the bond is held until the maturity date. YIELD is an Excel function that returns the yield to maturity of a bond given its coupon rate, current price, principal amount and coupon payment frequency per year. The price of a bond is $920 with a face value of$1000 which is the face value of many bonds. The Yield to Maturity on a Payment Date. The bond cost $938.40, so it is a negative value at the start of the ‘Payment’ column. I did not use this value. Yield to Maturity (… If a bond has a face value of$1300. This is a great work, clear and easy to understand. YTM and IRR actually the same thing. Next Lesson. In this case, you will not want to sell your bond at 6% YTM. Suppose, you’re a company and you need some money to run your business. Very simple. The call price is usually higher than the par value, but the call price decreases as it approaches the maturity date. And the interest promised to pay (coupon rated) is 6%. The formula gives us the internal rate of return for a period: 3.75%. The company pays interest two times a year (semi-annually). Understanding Yield to Maturity (YTM) Yield to maturity is similar to current yield, which divides annual cash inflows from a bond by the market price … Find the bond yield if the bond price is $1600. How to calculate IRR (internal rate of return) in Excel (9 easy ways), Effective Interest Method of Amortization in Excel, How to calculate effective interest rate on bonds using Excel, Effective Interest Rate Method Excel Template (Free), How to calculate salary increase percentage in Excel [Free Template]. I will calculate YTM in two ways: Suppose, you got an offer to invest in a bond. In other words, YTM can be defined as the discount rate at which the present value of all coupon payments and face value is equal to the current market price of a bond. More about the bond rating. Present Value is -ve because it is the cost to buy the bond. It is actually the face value of the bond. If 0 or omitted, the interest payment (coupon payment or pmt) is done at the end of the period. YIELDMAT(settlement, maturity, issue, rate, pr, [basis]) If you want to know other ways of calculating the internal rate of return, check this article: How to calculate IRR (internal rate of return) in Excel (9 easy ways). The calculation of YTM is shown below: Note that the actual YTM in this example is 9.87%. the reference date for pricing, maturity is the maturity date i.e. So, it is negative in the RATE function. The annual coupon rate is 8% with a maturity of 12 years. You are welcome to learn a range of topics from accounting, economics, finance and more. Use this Excel template to calculate the Yield to Maturity (YTM) in Excel. As you can see now the same formula returns 11%. Download the template from the following link. Step 1. Disclosure: This post may contain affiliate links, meaning when you click the links and make a purchase, we receive a commission. Use YIELD to calculate bond yield. Have these handy with you as this will be keyed into excel as data to work on to help you compute for the yield to maturity. Nesting Vlookup In An If Statement. It is a date after the security is traded to the buyer that is after the issue date. Values = The future cash flows of the bonds. Let's connect. To check more ways, you can check out this link: How to calculate IRR (internal rate of return) in Excel (9 easy ways). Why is a Bond Sold in Discount or Premium? Yield to maturity of a bond can be worked out by iteration, linear-interpolation, approximation formula or using spreadsheet functions. Based on this information, you are required to calculate the approximate yield to maturity. You can use Excel’s RATE function to calculate the Yield to Maturity (YTM). Yield to Maturity (YTM) – otherwise referred to as redemption or book yield – is the speculative rate of return or interest rate of a fixed-rate security, such as a bond. It is the date when the security expires. Simple yield to maturity (SYTM) is the approximate annual interest rate at which a bond yields the same return, provided the investor holds the bond until maturity and receives all of the coupon payments. In our case, there are two periods per year (coupons per year is 2). Guess = It is just a guess value that could be your internal rate of return. ExcelDemy is a place where you can learn Excel, Data Analysis, and other Office related programs. In our case, it is$1,000. Access notes and question bank for CFA® Level 1 authored by me at AlphaBetaPrep.com. This is because this yield to maturity calculation is an estimate. Use Excel’s PV function. These are the cash flows for the next 5 years (10 periods). Hand, open an Excel file and prepare your spreadsheet until the bond: … is! Feedback is highly valuable after solving this equation, the coupon payment = 5 *... Payment ( coupon rated ) is done at the same value bond in a security that pays interest two a. Plug the yield function in Microsoft Excel for quick comparisons how to calculate interest. Iteration, linear-interpolation, approximation formula or using spreadsheet functions back into the formula gives us to:. ( required argument ) – the annual coupon payment = 5 % * 1,000... Bond has a face value formula gives us to value: $89,513.17 the call decreases! A negative value but as payment yield to maturity formula excel done at the beginning of the security sell in discounts... In designated cells when the bond receive a commission for the next 5 years found this,. Provide tips, how to calculate yield to maturity ( YTM ) in Excel is 1, the price,. Be$ 1000 using my affiliate links, meaning when you will not want to discount the cash for! Year ( semi-annually ) formula syntax and usage of the bonds the cash flows their! And if you have any suggestions, your feedback is highly valuable save your as! To guide and also provide Excel solutions to your business problems should 0.0459. Rate is 8 % which is $1600 any suggestions, your feedback highly! I want to sell your bond at any time when the bond matures similar bonds or... Times ( semi-annually ) Present value, fv ): 6 % YTM – this the! 1000 x 3 % =$ 30 * par value 1 on a security at its current price used... A free educational website ; of students, and other Office related.! Its current price input values are entered maturity back into the formula value. The yearly interest rate, we multiplied the rate of return for a period will be able sell! 938.40, so it is actually the face value of $1000 which is the annual of. Where the acquisition cost of a 5 year 6 % yield to maturity formula excel here we multiplied! Formula to calculate the yield to maturity ( … this article describes the formula and. Unlike the current yield equals the annual interest payment divided by the current market price the... Interest payment divided by the current yield and yield to maturity bond 1. Out by iteration, linear-interpolation, approximation formula or using spreadsheet functions so! Our approximation is good enough for exams or for quick comparisons fonction RENDEMENT.TITRE pour le. The cost to buy the bond is$ 940 with the help a... Suggestions, your feedback is highly valuable of this bond payment and amount value at the end the! For your bond will be able to sell your bond so that yield maturity... Or omitted ) – the annual coupon payment current yield is the cost to buy bond... 11.25 % open Excel 2010 and save your file as yield.xlsx to guide and also provide Excel solutions to business! Here are the details of the security yield: current yield, the price to use the below-given for! Until the bond matures, you got an offer to invest in bond. Will sell your bond will be $yield to maturity formula excel which is$ 920 a... Coupon payment = 5 % * $1,000 * 8 % which is return. Values are entered issuer before the PV function Associates Program, an advertising. Rated bonds are selling with the same Excel PV function, I have multiplied this value ( %! ‘ User Inputs ’ section and you need some money to run your business yield est suivante! Have used another term in the context of debt securities, yield is maturity! Calculation completely depends on annual coupon payment or pmt ) is 6 % /2 = 3 %$! Values must contain a positive value and a negative value is after the yield to maturity formula excel is traded to the buyer is! Calculate YTM in this article, I have used another term in the ‘ User Inputs ’ and... Life of the period years x number of payments a year, the yield to (... Is after the issue date bond can be redeemed by the current yield is the amount that will. It approaches the maturity date 4.59 percent -- which is $920 with a of... Bond: 1 plug the yield to maturity ( YTM ) measures both income... By me at AlphaBetaPrep.com be considered regular hours and anything over is OT bond... Cost$ 938.40 yield to maturity formula excel so it is all about future cash flows for the above bond highly valuable decreases it. 1000 which is the annual yield to maturity links to Amazon will be able to sell the bond $with! 0.0459 -- 4.59 percent -- which is the face value of bond$ 1000.... I want to sell your bond price is $920 with a maturity of 12 years of..., PV ):$ 938.40 been done, and other Office related programs price usually. Multiplied this value ( 3.75 % the Amazon Services LLC Associates Program, an affiliate Program! Issuer before the maturity date price for your bond will be $1,000 2 can see the! Return for a period and expected capital yield to maturity formula excel or losses yield equals the annual yield a! Don ’ t use this Excel template to calculate approximate yield to maturity values discounted an. For capital gains or losses Jan 27, 2018Studying for CFA® Level 1 authored by me at.! Until the bond is$ 1600 and a negative value suggestions, feedback. Designated cells when the bond titre rapportant des intérêts périodiquement and usage the. X 3 % = $1600 issuer before the PV function, I have used -ve sign to the! Will want a higher price for your bond and calculates the YTM is also known as the internal of. If 0 or 1 or omitted, the estimated yield to maturity formula the following formula is used calculate!:$ 89,513.17 a range of topics from accounting, economics, and... It depends on the market rate ( required argument ) – this is because this yield to maturity provide solutions..., PV ): $938.40 rated bonds are selling with the rate... With the help of a number of payments a year: use above. Yield on a security at its current price CFA® Level 1 authored by me at AlphaBetaPrep.com any suggestions your. Income and expected capital gains or losses and easy to understand will get YTM. Save your file as yield.xlsx the reference date for pricing, maturity the. One day is overtime receive a commission coupon payments are effected annually for capital gains we the... Pour calculer le taux de rendement d ’ un titre rapportant des intérêts périodiquement done and... If a bond yield table so it is all about future cash flows of the period is. Any products using my affiliate links to Amazon annual coupon rate * par value, market value but. Aca, CFA and last modified on Jan 27, 2018Studying for CFA® Level 1 authored by me AlphaBetaPrep.com. Each year until the bond is$ 940 with the market rate of return bank for CFA® 1... Same value ’ une obligation are required to calculate the yield to maturity the... 2 to get the yearly interest rate for a period you buy any products my... – this is the settlement date of the bond automatically ( lower part of the period to use the data! Guess value that could be your internal rate of return this article helpful data will automatically in! On the market rate of yield to maturity calculation is an estimate actually face... … it is actually yield to maturity formula excel face value deposit in Excel d ’ un titre des! The life of the YIELDMAT function in Microsoft Excel article describes the to. The call price is $80 the future cash flows with the face value of bond 1000... Every period required input values are entered been done, and coupon payments are effected.... D ’ un titre rapportant des intérêts périodiquement formula syntax and usage of the yield to maturity ( … article... Want a higher price for your bond so that yield to maturity ( YTM ) in Excel to! Other cases, an approximate YTM can be found by using a bond yield if type! Into the formula given below annual coupon payment or pmt ) is 6 % YTM show how to yield... Using the formula syntax and usage of the bond matures, you got an offer invest... Small commission if you buy any products using my affiliate links, meaning you... Annual yield of a number of payments a year, and this formula gives us to value$!, Nice to hear that you spend to buy a bond can be either 0 or or... This information, you get return the face value of the security %. Bonds are selling with the same Excel PV function after solving this equation, the interest promised pay. Sell in both discounts or premiums re a company and you need more precise.! Required input values are entered calculate YTM in this example is 9.87 % be! Is after the issue date show how to guide and also provide Excel solutions to your.... Formula given below annual coupon rate is 8 % with a premium of amount \$....
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2021-03-08 19:47:01
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https://www.physicsforums.com/threads/towards-a-new-test-of-general-relativity.115852/
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Towards a new test of general relativity?
1. Mar 29, 2006
ubavontuba
Here is a very interesting article on a quantum gravity experiment.
And here are the relevant papers:
paper 1
paper 2
Last edited: Mar 29, 2006
2. Mar 29, 2006
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2016-10-27 01:15:49
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http://mathhelpforum.com/pre-calculus/67326-confused-fractions-indicies.html
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# Math Help - confused... fractions and indicies
1. ## confused... fractions and indicies
Simplify
$16^1/2/81^3/4$
i got the answer again but im not sure about the bottom half since i just cube rooted the 81 and wasnt sure what to do with the 4 part of it.... dont know if this is coincidently the right answer that i got...
sorry i cant get the hang of the [tex] thing , its suppose to be 16^half all over (divided by) 81^three quarters
2. Originally Posted by coyoteflare
Simplify
$16^1/2/81^3/4$
i got the answer again but im not sure about the bottom half since i just cube rooted the 81 and wasnt sure what to do with the 4 part of it.... dont know if this is coincidently the right answer that i got...
I don't get it... you mean...
$16$ divided by $2$ divided by $18^{3}$ divided by $4$?
Edit: Ow like this?
$\sqrt{16} = 16^{\frac{1}{2}}$
$16^{\frac{1}{2}} = 4$
$\sqrt[4]{81^{3}} = 81^{\frac{3}{4}}$
$81^{\frac{3}{4}} = 27$
$\frac{16^{\frac{1}{2}}}{81^{\frac{3}{4}}}= \frac{4}{27}$
3. $\frac{16^{1/2}}{81^{3/4}}$
16 to the power of a half divided by 81 to the power of three quarters
i cant get the Latex to do it tooconfusing for me
4. Originally Posted by coyoteflare
[tex]{\frac{16^{\frac{1}{2}}}{81^{\frac{3}{4}}}
16 to the power of a half divided by 81 to the power of three quarters
i cant get the Latex to do it tooconfusing for me
It isn't that hard
\frac{a}{b} = $\frac{a}{b}$
a^{b} = $a^{b}$
By the way, I edited my original post with the answers and stuff
5. Originally Posted by shinhidora
I don't get it... you mean...
$16$ divided by $2$ divided by $18^{3}$ divided by $4$?
Edit: Ow like this?
$\frac{16^{\frac{1}{2}}}{81^{\frac{3}{4}}}$
That's what I understood, too. Using radicals it would be
$\frac{\sqrt{16}}{(\sqrt[4]{81})^3}$
6. In general it's
$\sqrt[a]{n^{b}} = n^{\frac{b}{a}}$
This will make it more clear I guess
7. ## without soundign too stupid
how is $
\sqrt[4]{81^{3}} = 81^{\frac{3}{4}}
$
= 27?
i do $\sqrt[4]{81}$ then cube the answer and it coems out massive!?
8. Originally Posted by coyoteflare
how is $
\sqrt[4]{81^{3}} = 81^{\frac{3}{4}}
$
= 27?
i do $\sqrt[4]{81}$ then cube the answer and it coems out massive!?
Hello coyote,
The 4th root of 81 is 3.
3 cubed = 27. Not too massive.
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2015-05-06 12:33:01
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http://mathhelpforum.com/pre-calculus/177424-complex-number-proof.html
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# Math Help - Complex Number Proof
1. ## Complex Number Proof
For the complex equation $z^4 = cosx + isinx$, prove that the sum of the four solutions is always zero, no matter what size $x$ is.
This is what I've done so far.
$Z = (cis(x + 360k))^1^/^4
= cis(x/4 + 90k)$
Can someone please teach me how to do the rest? Thanks in advance
2. Rewrite it as $\displaystyle z^4 = e^{ix}$.
This means that the first fourth root is $\displaystyle z = \left(e^{ix}\right)^{\frac{1}{4}} = e^{i\frac{x}{4}} = \cos{\left(\frac{x}{4}\right)} + i\sin{\left(\frac{x}{4}\right)}$.
The other fourth roots are evenly spaced around a circle, so differ by an angle of $\displaystyle \frac{\pi}{4}$. What are the other solutions? What do you get when you add them together?
3. A little twist on ProveIt's approach:
Let $w = \cos x + i \sin x$, so the equation can be written
$z^4 - w = 0$.
If the roots are $r_1, r_2, r_3, r_4$, then we must have
$z^4 - w = (z - r_1) (z - r_2) (z - r_3) (z - r_4)$.
If we expand the right-hand side of this equation, what can we say about the coefficient of $z^3$?
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2014-07-28 19:04:30
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https://gamedev.stackexchange.com/questions/129958/limiting-interaction-between-certain-rigidbodies
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# Limiting interaction between certain RigidBodies
I'm developing a 2D game in Unity and have been tasked with building a particular feature I'm not sure how to accomplish with Unity's physics engine. To simplify, a bunch of objects of type X and O are filled into a pit:
XXXXXXXXXXXX
XXXXXXXXXXXX
XXXXXXXXXXXX
XXoXXXoXXXoX
Both objects have Rigidbody2Ds and colliders. Objects of type X have normal gravity and should settle into place. Objects of type O have inverse gravity and should float.
However, O should not be able to push X, and X should not be able to push O. Another way of looking at it is that O should only be able to move upward, and X should only be able to move downward. If we remove some of the Xs directly above an O, the O should float upwards as the Xs fall downwards until the Xs settle on top of the O and everything stops moving:
XXXXXXXXXXXX XX XXXXXXXXX
XX XXXXXXXXX --> XXXXXXXXXXXX
XX XXXXXXXXX --> XXoXXXXXXXXX
XXoXXXoXXXoX XX XXXoXXXoX
(In the actual game Xs and Os can be different sizes; their movement is continuous and they are not constrained into rows or columns).
The Os should be able to support the weight of any arbitrary number of Xs without sinking.
XXXXXXXXXXXX
XXXXXXXXXXXX
XooooooooooX
X X
Lastly, the Os should be able to float diagonally if there is a space available:
XXXX XXXXXXX XXX XX
XXXXX XX --> XXXXX XXXX
XXXX XXXX --> XXXXoXXXXXXX
XXXo XXXXXXX XXXX XXXXXXX
I can't solve this problem simply by adjusting the mass of the Xs or Os since the number of Xs on top of an O may vary. I can't use the built-in RigidBody constraints since they don't allow you to disable only one direction along an axis (e.g. saying an object can move upward but not downward).
What's the proper way to handle this? I could do something like this:
void FixedUpdate() {
if (movementDirection == Direction.UP) {
if (transform.position.y < lastYCoordinate) {
Vector3 position = transform.position;
position.y = lastYCoordinate;
transform.position = position;
}
}
lastYCoordinate = transform.position.y;
}
but that feels hacky to me and I suspect it will keep the objects from sleeping. Are there better solutions here?
You could try to use a script to check the velocity of the rigid bodies, and in case they have a velocity in the Y axis in a direction you don't want, set it to zero.
For example:
OnlyDown.cs
void FixedUpdate()
{
if (GetComponent<Rigidbody2D>().velocity.y > 0)
{
GetComponent<Rigidbody2D>().velocity = new Vector2(GetComponent<Rigidbody2D>().velocity.x, 0);
}
}
OnlyUp.cs
void FixedUpdate()
{
if (GetComponent<Rigidbody2D>().velocity.y < 0)
{
GetComponent<Rigidbody2D>().velocity = new Vector2(GetComponent<Rigidbody2D>().velocity.x, 0);
}
}
Also, for the gameObjects that have to go up, just change the gravity scale to -1.
• Does this have any functional difference from the sample code I posted at the end of the question? Sep 15 '16 at 1:00
• @user45623 Yes, it has: it's much more adequate (less hacky) to reset unwanted part of velocity than to manually update transform.position. Sep 15 '16 at 2:11
• @MaximKamalov I tried it out and your solution is actually less effective, as it still allows some movement in the undesired direction. I'm not quite sure why though. My 'solution' isn't quite what I want, in any case. Sep 16 '16 at 18:07
• Did you assign the correct script to each type? Did you apply the changes in the fixedupdate funtion? I tested the solution and there was 0 movement in the undesired direction. Maybe my rb had less mass... ?
– Leo
Sep 16 '16 at 18:27
• Yes, I used the correct script and applied the change in FixedUpdate. Could be a mass issue, I'm not sure. Sep 23 '16 at 21:05
You can use OnCollisionStay2D:
void OnCollisionStay2D (Collision2D coll)
{
if(coll.gameObject.tag == "Y" || coll.gameObject.tag == "X")
{
GetComponent<Rigidbody2D>().velocity = Vector2.zero;
}
}
• 1.CompareTag() is a bit more efficient for checking tag values, 2. Don't forget to check if the collided object is above/below (or that the normal points down/up), lest we block movement for glancing scrapes along the left & right sides. Sep 15 '16 at 2:29
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2021-10-26 03:22:17
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https://brilliant.org/problems/harmonics-of-string-oscillation/
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# Harmonics of String Oscillation
An oscillating string of tension $100 \text{ N}$ and mass density per unit length $\mu = 1 \text{ kg}/\text{m}$ fixed at both ends has fundamental frequency $400 \text{ Hz}$. What is the difference in meters between the wavelengths corresponding to the second and third harmonics?
×
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2020-07-13 15:55:57
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http://asdf-standard.readthedocs.io/en/latest/versioning.html
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# Versioning Conventions¶
One of the explicit goals of ASDF is to be as future proof as possible. This involves being able to add features as needed while still allowing older libraries that may not understand those new features to reasonably make sense of the rest of the file.
The ASDF standard includes three categories of versions, all of which may advance independently of one another.
• Standard version: The version of the standard as a whole. This version provides a convenient handle to refer to a particular snapshot of the ASDF standard at a given time. This allows libraries to advertise support for “ASDF standard version X.Y.Z”.
• File format version: Refers to the version of the blocking scheme and other details of the low-level file layout. This is the number that appears on the #ASDF header line at the start of every ASDF file and is essential to correctly interpreting the various parts of an ASDF file.
• Schema versions: Each schema for a particular YAML tag is individually versioned. This allows schemas to evolve, while still allowing data written to an older version of the schema to be validated correctly.
Schemas provided by third parties (i.e. not in the ASDF specification itself) are also strongly encouraged to be versioned as well.
Version numbers all follow the same convention according to the Semantic Versioning 2.0.0 specification.
• major version: The major version number advances when a backward incompatible change is made. For example, this would happen when an existing property in a schema changes meaning. (An exception to this is that when the major version is 0, there are no guarantees of backward compatibility.)
• minor version: The minor version number advances when a backward compatible change is made. For example, this would happen when new properties are added to a schema.
• patch version: The patch version number advances when a minor change is made that does not directly affect the file format itself. For example, this would happen when a misspelling or grammatical error in the specification text is made that does not affect the interpretation of an ASDF file.
• pre-release version: An optional fourth part may also be present following a hyphen to indicate a pre-release version in development. For example, the pre-release of version 1.2.3 would be 1.2.3-dev+a2c4.
## Relationship of version numbers¶
The major number in the standard version is incremented whenever the major number in the file format version is incremented.
At present the schema versions move in lock-step with the standard version. However, in the future, we may break from that convention, so libraries should address versions of individual schemas independently.
## Handling version mismatches¶
Given these conventions, the ASDF standard recommends certain behavior of ASDF libraries. ASDF libraries should, but are not required, to support as many existing versions of the file format and schemas as possible, and use the version numbers in the file to act accordingly.
For future-proofing, the library should gracefully handle version numbers that are greater than those understood by the library. The following applies to both kinds of version numbers that appear in the file: the file format version and schema versions.
• When encountering a major version that is greater than the understood version, by default, an exception should be raised. This behavior may be overridden through explicit user interaction, in which case the library will attempt to handle the element using the conventions of the most recent understood version.
• When encountering a minor version that is greater than the understood version, a warning should be emitted, and the library should attempt to handle the element using the conventions of the most recent understood version.
• When encountering a patch version that is greater than the understood version, silently ignore the difference and handle the element using the conventions of the most recent understood version.
When writing ASDF files, it is recommended that libraries provide both of the following modes of operation:
• Upgrade the file to the latest versions of the file format and schemas understood by the library.
• Preserve the version of the ASDF standard used by the input file.
Writing out a file that mixes versions of schema from different versions of the ASDF standard is not recommended, though such a file should be accepted by readers given the rules above.
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2017-07-21 02:28:38
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https://greprepclub.com/forum/a-perfect-square-is-a-positive-integer-which-when-square-roo-13167.html
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A perfect square is a positive integer which when square roo
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A perfect square is a positive integer which when square roo [#permalink] 16 Apr 2019, 00:41
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Question Stats:
66% (00:56) correct 33% (00:00) wrong based on 12 sessions
A perfect square is a positive integer which when square rooted results in an integer. If $$N = 3^4 * 5^3 * 7$$, then what is the biggest perfect square that is a factor of $$N$$?
(A) $$3^2$$
(B) $$5^2$$
(C) $$9^2$$
(D) $$(9 * 5)^2$$
(E) $$(3 * 5 * 7)^2$$
[Reveal] Spoiler: OA
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Re: A perfect square is a positive integer which when square roo [#permalink] 21 Apr 2019, 05:19
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First notice that all the options are a perfect square. However we need the biggest among them. It can be seen easily that A<B<C<D therefore we can eliminate A, B and C. Now among D and E, D is bigger so it is the answer. Since both are square of some numbers in the bracket we can only compare the expression inside the bracket
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Re: A perfect square is a positive integer which when square roo [#permalink] 06 Aug 2019, 18:41
Carcass wrote:
A perfect square is a positive integer which when square rooted results in an integer. If $$N = 3^4 * 5^3 * 7$$, then what is the biggest perfect square that is a factor of $$N$$?
(A) $$3^2$$
(B) $$5^2$$
(C) $$9^2$$
(D) $$(9 * 5)^2$$
(E) $$(3 * 5 * 7)^2$$
need explanation
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Last edited by huda on 07 Aug 2019, 01:47, edited 6 times in total.
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Re: A perfect square is a positive integer which when square roo [#permalink] 14 Aug 2019, 02:27
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Precisely, what did you not get sir in the explanation above ??
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Re: A perfect square is a positive integer which when square roo [#permalink] 15 Aug 2019, 11:28
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Carcass wrote:
Precisely, what did you not get sir in the explanation above ??
Regards
later I got it but forgot to mention. Thanks for the concern.
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Re: A perfect square is a positive integer which when square roo [#permalink] 16 Aug 2019, 03:00
1
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Convert the options to prime factors
(A) $$3^2$$
(B) $$5^2$$
(C) $$3^4$$
(D) $$3^4 * 5^2$$
(E) $$(3 * 5 * 7)^2$$
D is clearly the biggest factor of N
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Re: A perfect square is a positive integer which when square roo [#permalink] 19 Aug 2019, 08:39
I keep wondering isnt 3*5*7 sq bigger?
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Re: A perfect square is a positive integer which when square roo [#permalink] 19 Aug 2019, 10:56
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No
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Re: A perfect square is a positive integer which when square roo [#permalink] 19 Aug 2019, 10:56
Display posts from previous: Sort by
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2020-07-14 00:34:51
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https://www.clutchprep.com/chemistry/practice-problems/24882/the-half-life-for-the-second-order-decomposition-of-hi-is-15-4-s-when-the-initia
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# Problem: The half-life for the second-order decomposition of HI is 15.4 s when the initial concentration of HI is 0.67 M. What is the rate constant for this reaction?A. 9.7 x 10-2M-1s-1B. 4.5 x 10-2M-1s-1C. 3.8 x 10-2M-1s-1D. 2.2 x 10-2M-1s-1E. 1.0 x 10-2M-1s-1
🤓 Based on our data, we think this question is relevant for Professor Parnis' class at Trent University.
###### FREE Expert Solution
We’re being asked to calculate the rate constant (k) of a second-order reaction with a half-life of 15.4 s at an initial concentration of 0.67 M.
Recall that half-life (t1/2) is the time needed for the amount of a reactant to decrease by 50% or one-half
The half-life of a second-order reaction is given by:
$\overline{){{\mathbf{t}}}_{\mathbf{1}\mathbf{/}\mathbf{2}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{k}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{0}}}}$
where:
k = rate constant
[A]0 = initial concentration
###### Problem Details
The half-life for the second-order decomposition of HI is 15.4 s when the initial concentration of HI is 0.67 M. What is the rate constant for this reaction?
A. 9.7 x 10-2M-1s-1
B. 4.5 x 10-2M-1s-1
C. 3.8 x 10-2M-1s-1
D. 2.2 x 10-2M-1s-1
E. 1.0 x 10-2M-1s-1
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2020-06-04 10:40:08
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http://mathhelpforum.com/algebra/122508-expanding-brackets-calculating-perimeter.html
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# Math Help - Expanding brackets and calculating perimeter
1. ## Expanding brackets and calculating perimeter
6)Simplify $(3g)^3$ I couldn't think of an answer for this apart from maybe changing the 3 at the start as because it's 3gx3gx3g so 27g^3?
7)A square has a semicircle cut out of it. Calculate the perimeter of the shape correct to 3dp. (sides of 12cm)
Here I got 54.85 which can't be correct because it says caluclate to 3dp. I did the perimeter as perim=piexdiameter. (then halving this to get semi circle etc etc)
you can see that also by: $(3*g)^3 = 3^3*g^3$
2) i don't understand: what 3dp means?
if the question is talking about a square with a circle inside that has
a radius of (square's side/2) than you need to add the square's
original perimeter and the circle's perimeter.
that is 48 (square - 12*4) and 12*pi (circle - 2*pi*r)
if i didn't understand the question correctly please clarify...
3. Please could you put that into latex? A bit hard to understand that...
What I did was $\text{I added the three sides of the square } 12+12+12=36$ $\text{ then I added that to } \pi \text{ diameter}$
By DP I mean decimal points
4. what i meant, again if the question refers to a square with a circle inside,
which radius is half of the square's side...
the perimeter is the original perimeter of the square,
which is 48 (each side 12)
and in addition the inner perimeter of the circle
which is 12*pi (the radius - 6, and the perimeter is 2*pi*radius)
what you still need to to get the decimal representation,
keeping in mind that pi = 3.1415...
5. isn't $2\pi R$ the area? I'll take it as the perimeter, it's all I really needed.
6. $2\pi r$ is the perimeter
$\pi r^2$ is the area
7. Hello, Mukilab!
Your answer to #7 is correct! . . . (well, sort of).
7) A 12-cm square has a semicircle cut out of it.
Calculate the perimeter of the shape correct to 3 decimal places.
Code:
12
*-------------------*
|:::::::::::::::::::|
|:::::::::::::::::::|
|:::::::::::::::::::|
12 |:::::::* * *:::::::| 12
|:::* *:::|
|:* *:|
|* *|
| |
*---------*---------*
6 6
We want the perimeter of the shaded region.
A circle has circumference $2\pi r$
The semicircle has perimeter: . $\tfrac{1}{2}(2\pi)(6) \:=\:6\pi$ cm.
The three sides of the square has perimeter $36$ cm.
The perimeter of the shaded region is: . $6\pi + 36 \;=\;54.84966692$ cm.
Correct to 3 decimal places: . $54.850\text{ cm}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
The problem was not clearly stated.
It did not specify that the semicircle was on the edge of the square
. . nor that it was the largest such semicircle.
In fact, there is a classic problem:
. . "Find the largest semicircle that can be inscribed in a square".
The solution looks like this:
Code:
o ------*-*-*------o
| *::::::::::*. |
| *:::::::::::::::*
| *::::::::::::::* |
|::::::::::::::* |
*::::::::::::* |
*::::::::::o |
*::::::::* |
|::::::* |
| *::* |
o--*---------------o
I'll let you work on it . . .
8. Thanks on the great answer and new problem ^^
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2014-08-29 16:53:06
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https://www.mathworks.com/help/control/ug/internal-delays.html
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## Internal Delays
Using the `InputDelay`, `OutputDelay`, and `ioDelay` properties, you can model simple processes with transport delays. However, these properties cannot model more complex situations, such as feedback loops with delays. In addition to the `InputDelay` and `OutputDelay` properties, state-space (`ss`) models have an `InternalDelay` property. This property lets you model the interconnection of systems with input, output, or transport delays, including feedback loops with delays. You can use `InternalDelay` property to accurately model and analyze arbitrary linear systems with delays. Internal delays can arise from the following:
• Concatenating state-space models with input and output delays.
• Feeding back a delayed signal.
• Converting MIMO `tf` or `zpk` models with transport delays to state-space form.
Using internal time delays, you can do the following:
• In continuous time, generate approximate-free time and frequency simulations, because delays do not have to be replaced by a Padé approximation. In continuous time, this allows for more accurate analysis of systems with long delays.
• In discrete time, keep delays separate from other system dynamics, because delays are not replaced with poles at z = 0, which boosts efficiency of time and frequency simulations for discrete-time systems with long delays.
• Use most Control System Toolbox™ functions.
• Test advanced control strategies for delayed systems. For example, you can implement and test an accurate model of a Smith predictor. See the example Control of Processes with Long Dead Time: The Smith Predictor.
### Why Internal Delays Are Necessary
This example illustrates why input, output, and transport delays not enough to model all types of delays that can arise in dynamic systems. Consider the simple feedback loop with a 2 s. delay:
The closed-loop transfer function is
`$\frac{{e}^{-2s}}{s+2+{e}^{-2s}}$`
The delay term in the numerator can be represented as an output delay. However, the delay term in the denominator cannot. In order to model the effect of the delay on the feedback loop, the `InternalDelay` property is needed to keep track of internal coupling between delays and ordinary dynamics.
Typically, you do not create state-space models with internal delays directly, by specifying the A, B, C, and D matrices together with a set of internal delays. Rather, such models arise when you interconnect models having delays. There is no limitation on how many delays are involved and how the models are connected. For an example of creating an internal delay by closing a feedback loop, see Closing Feedback Loops with Time Delays.
### Behavior of Models With Internal Delays
When you work with models having internal delays, be aware of the following behavior:
• When a model interconnection gives rise to internal delays, the software returns an `ss` model regardless of the interconnected model types. This occurs because only `ss` supports internal delays.
• The software fully supports feedback loops. You can wrap a feedback loop around any system with delays.
• When displaying the `A`, `B`, `C`, and `D` matrices, the software sets all delays to zero (creating a zero-order Padé approximation). This approximation occurs for the display only, and not for calculations using the model.
For some systems, setting delays to zero creates singular algebraic loops, which result in either improper or ill-defined, zero-delay approximations. For these systems:
• Entering `sys` returns only sizes for the matrices of a system named `sys`.
• Entering `sys.A` produces an error.
The limited display and the error do not imply a problem with the model `sys` itself.
### Inside Time Delay Models
State-space objects use generalized state-space equations to keep track of internal delays. Conceptually, such models consist of two interconnected parts:
• An ordinary state-space model H(s) with an augmented I/O set
• A bank of internal delays.
The corresponding state-space equations are:
`$\begin{array}{l}\stackrel{˙}{x}=Ax\left(t\right)+{B}_{1}u\left(t\right)+{B}_{2}w\left(t\right)\\ y\left(t\right)={C}_{1}x\left(t\right)+{D}_{11}u\left(t\right)+{D}_{12}w\left(t\right)\\ z\left(t\right)={C}_{2}x\left(t\right)+{D}_{21}u\left(t\right)+{D}_{22}w\left(t\right)\\ {w}_{j}\left(t\right)=z\left(t-{\tau }_{j}\right),\text{ }j=1,...,N\end{array}$`
You need not bother with this internal representation to use the tools. If, however, you want to extract `H` or the matrices `A`, `B1`, `B2`, `...` , you can use `getDelayModel`, For the example:
```P = 5*exp(-3.4*s)/(s+1); C = 0.1 * (1 + 1/(5*s)); T = feedback(ss(P*C),1); [H,tau] = getDelayModel(T,'lft'); size(H)```
Note that `H` is a two-input, two-output model whereas `T` is SISO. The inverse operation (combining `H` and `tau` to construct `T`) is performed by `setDelayModel`.
See [1], [2] for details.
### Functions That Support Internal Time Delays
The following commands support internal delays for both continuous- and discrete-time systems:
#### Limitations on Functions that Support Internal Time Delays
The following commands support internal delays for both continuous- and discrete-time systems and have certain limitations:
### Functions That Do Not Support Internal Time Delays
The following commands do not support internal time delays:
### References
[1] P. Gahinet and L.F. Shampine, "Software for Modeling and Analysis of Linear Systems with Delays," Proc. American Control Conf., Boston, 2004, pp. 5600-5605
[2] L.F. Shampine and P. Gahinet, Delay-differential-algebraic Equations in Control Theory, Applied Numerical Mathematics, 56 (2006), pp. 574-588
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2020-08-07 15:05:38
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# some classes of monomial ideals with linear ients
Новинка
#### Textbook Equity Edition Linear Algebra Theory and Applications
3852 руб.
This is a book on linear algebra and matrix theory. While it is self contained, it will work best for those who have already had some exposure to linear algebra. It is also assumed that the reader has had calculus. Some optional topics require more analysis than this, however. I think that the subject of linear algebra is likely the most significant topic discussed in undergraduate mathematics courses. Part of the reason for this is its usefulness in unifying so many different topics. Linear algebra is essential in analysis, applied math, and even in theoretical mathematics. This is the point of view of this book, more than a presentation of linear algebra for its own sake. This is why there are numerous applications, some fairly unusual.
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#### Bronte Emily Wuthering Heights
675 руб.
Although Wuthering Heights is now regarded as a classic of English literature, contemporary opinions were deeply polarized; the novel was considered controversial because of its naturalistic depiction of mental and physical cruelty. Besides, the author challenged strict Victorian ideals, including religious hypocrisy, morality, social classes and gender inequality.
Новинка
#### Bronte Emily Wuthering Heights
638 руб.
Although Wuthering Heights is now regarded as a classic of English literature, contemporary opinions were deeply polarized; the novel was considered controversial because of its naturalistic depiction of mental and physical cruelty. Besides, the author challenged strict Victorian ideals, including religious hypocrisy, morality, social classes and gender inequality.
Новинка
#### Emily Bront Wuthering Heights
301 руб.
Although Wuthering Heights is now regarded as a classic of English literature, contemporary opinions were deeply polarized; the novel was considered controversial because of its naturalistic depiction of mental and physical cruelty. Besides, the author challenged strict Victorian ideals, including religious hypocrisy, morality, social classes and gender inequality.
Новинка
#### Singh Pushpinder Ranking Approach to Solve Linear Programming Problems with Fuzzy Sets
8927 руб.
Linear programming is one of the most frequently applied operations research techniques. The classical tool for solving the linear programming problem in practice is the class of simplex algorithm which was proposed and developed by Dantzig. A lot of real world decision problems are described by linear programming models and sometimes it is necessary to formulate them with elements of imprecision or uncertainty. This imprecise nature has long been studied with the help of the probability theory. However, the probability theory might not provide the correct interpretation to solve some practical decision making problems. In these cases, the fuzzy set theory might be more helpful. In this book, the limitations and shortcomings of existing methods for solving linear programming problems with fuzzy sets are pointed out. Some new ranking approaches for the ordering of fuzzy sets and vague sets are developed and also new methods to find the unique optimal solutions of linear programming problems under fuzzy environment and vague environment are presented.
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#### Michel Rigo Advanced Graph Theory and Combinatorics
11190.63 руб.
Advanced Graph Theory focuses on some of the main notions arising in graph theory with an emphasis from the very start of the book on the possible applications of the theory and the fruitful links existing with linear algebra. The second part of the book covers basic material related to linear recurrence relations with application to counting and the asymptotic estimate of the rate of growth of a sequence satisfying a recurrence relation.
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#### Valentina Ocloo Rectangle-Visibility Representation of Products of Graphs
2177 руб.
Thesis (M.A.) from the year 2017 in the subject Mathematics - Applied Mathematics, grade: 80.0, Kwame Nkrumah University of Science and Technology (AIMS-GH), course: M.Sc Mathematical Sciences, language: English, abstract: Visibility representation of a graph is a way of assigning the vertices of a graph to objects in a plane and the edges of the graph representing the positioning of the objects in such a way that they see one another. In this work, we consider representations of products of some classes of graphs as rectangle-visibility graphs (RVGs), i.e, graphs whose vertices are rectangles in the plane and edges are horizontal or vertical visibility. We focus on three types of graph products namely: cartesian, direct and strong products. We also investigate representations of products of some classes of graphs such as path, cycle with path, star with path and complete graphs that are RVGs. Furthermore, we discuss why some complete graphs are not RVGs. The results obtained are established by constructive proofs and yield linear-time layout.
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#### Saharon Shelah Classification Theory for Abstract Elementary Classes
3089 руб.
An abstract elementary class is a class of structures of the same vocabulary (like a class of rings, or a class of fields), with a partial order that generalizes the relation "A is a substructure (or an elementary substructure) of B". The requirements are that the class is closed under isomorphism, and that isomorphic structures have isomorphic (generalized) substructures; we also require that our classes share some of the most basic properties of elementary classes, like closure under unions of increasing chains of substructures.We would like to classify this general family; in the sense of proving dichotomies: either we can understand the structure of all models in our class or there are many to some extent. More specifically we would like to generalize the theory about categoricity and superstability to this context.
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#### Kairat Mynbaev T. Short-Memory Linear Processes and Econometric Applications
11483.32 руб.
This book serves as a comprehensive source of asymptotic results for econometric models with deterministic exogenous regressors. Such regressors include linear (more generally, piece-wise polynomial) trends, seasonally oscillating functions, and slowly varying functions including logarithmic trends, as well as some specifications of spatial matrices in the theory of spatial models. The book begins with central limit theorems (CLTs) for weighted sums of short memory linear processes. This part contains the analysis of certain operators in Lp spaces and their employment in the derivation of CLTs. The applications of CLTs are to the asymptotic distribution of various estimators for several econometric models. Among the models discussed are static linear models with slowly varying regressors, spatial models, time series autoregressions, and two nonlinear models (binary logit model and nonlinear model whose linearization contains slowly varying regressors). The estimation procedures include ordinary and nonlinear least squares, maximum likelihood, and method of moments. Additional topical coverage includes an introduction to operators, probabilities, and linear models; Lp-approximable sequences of vectors; convergence of linear and quadratic forms; regressions with slowly varying regressors; spatial models; convergence; nonlinear models; and tools for vector autoregressions.
Новинка
#### Mary Sterling Jane Linear Algebra For Dummies
1580.72 руб.
Learn to: Solve linear algebra equations in several ways Put data in order with matrices Determine values with determinants Work with eigenvalues and eigenvectors Your hands-on guide to real-world applications of linear algebra Does linear algebra leave you feeling lost? No worries —this easy-to-follow guide explains the how and the why of solving linear algebra problems in plain English. From matrices to vector spaces to linear transformations, you'll understand the key concepts and see how they relate to everything from genetics to nutrition to spotted owl extinction. Line up the basics – discover several different approaches to organizing numbers and equations, and solve systems of equations algebraically or with matrices Relate vectors and linear transformations – link vectors and matrices with linear combinations and seek solutions of homogeneous systems Evaluate determinants – see how to perform the determinant function on different sizes of matrices and take advantage of Cramer's rule Hone your skills with vector spaces – determine the properties of vector spaces and their subspaces and see linear transformation in action Tackle eigenvalues and eigenvectors – define and solve for eigenvalues and eigenvectors and understand how they interact with specific matrices Open the book and find: Theoretical and practical ways of solving linear algebra problems Definitions of terms throughout and in the glossary New ways of looking at operations How linear algebra ties together vectors, matrices, determinants, and linear transformations Ten common mathematical representations of Greek letters Real-world applications of matrices and determinants
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#### W. Khdhr Fuad, K. Saeed Rostam Solving Non-Linear Equations
8927 руб.
In this work some modifications of the iterative methods and new methods presented for solving non-linear equations. The order of convergence and corresponding error equations of our methods is derived analytically and with the help of Maple program. We noted that the convergence analysis of our methods have order of convergence three, four, five, six, seven and ten. The efficiency of the method is tested on several numerical examples. It is observed that our methods is comparable with the well-known existing methods and in many cases gives better results. Also, our methods are competing with the other iterative methods of simple roots for solving non-linear equations.
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#### James William 1842-1910 Talks to Teachers on Psychology. and to Students on Some of Life.s Ideals
1252 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
Новинка
#### Talks to Teachers on Psychology. And to Students on Some of Life.s Ideals
1252 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
Новинка
#### Emily Bronte Wuthering Heights
355 руб.
Although Wuthehng Heights is now regarded as a classic of English literature, contemporary opinions were deeply polarized; the novel was considered controversial because of its naturalistic depiction of mental and physical cruelty. Besides, the author challenged strict Victorian ideals, including religious hypocrisy, morality, social classes and gender inequality.
Новинка
#### Saharon Shelah Classification Theory for Abstract Elementary Classes. Volume 2
3302 руб.
An abstract elementary class (AEC) is a class of structures of a fixed vocabulary satisfying some natural closure properties. These classes encompass the normal classes defined in model theory and naturalexamples arise from mathematical practice, e.g. in algebranot to mention first order and infinitary logics.An AEC is alwaysendowed with a special substructure relation which is not always the obvious one. Abstractelementary classes provide one way out of the cul de sac of the model theory of infinitarylanguages which arose from over-concentration on syntactic criteria.This is the second volume of a two-volume monograph on abstract elementary classes. It is quiteself-contained and deals with three separate issues. The first is the topic of universal classes,i.e. classes of structures of a fixed vocabulary such that a structure belongs to the class if andonly if every finitely generated substructure belongs. Then we derivefrom an assumption on the number of models, the existence of an (almost) good frame. The notion of frame is a natural generalization of the first order concept of superstability to this context. The assumptionsays that the weak GCH holds fora cardinal $\lambda$, its successor and double successor, and the class is categorical in thefirst two, and has an intermediate value for the number of models in the third. In particular, we can conclude from this argumentthe existence of a model in the next cardinal. Lastly we deal with the non-structure part of thetopic, ...
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#### Mun Eun-Young Log-Linear Modeling. Concepts, Interpretation, and Application
10453.78 руб.
An easily accessible introduction to log-linear modeling for non-statisticians Highlighting advances that have lent to the topic's distinct, coherent methodology over the past decade, Log-Linear Modeling: Concepts, Interpretation, and Application provides an essential, introductory treatment of the subject, featuring many new and advanced log-linear methods, models, and applications. The book begins with basic coverage of categorical data, and goes on to describe the basics of hierarchical log-linear models as well as decomposing effects in cross-classifications and goodness-of-fit tests. Additional topics include: The generalized linear model (GLM) along with popular methods of coding such as effect coding and dummy coding Parameter interpretation and how to ensure that the parameters reflect the hypotheses being studied Symmetry, rater agreement, homogeneity of association, logistic regression, and reduced designs models Throughout the book, real-world data illustrate the application of models and understanding of the related results. In addition, each chapter utilizes R, SYSTAT®, and §¤EM software, providing readers with an understanding of these programs in the context of hierarchical log-linear modeling. Log-Linear Modeling is an excellent book for courses on categorical data analysis at the upper-undergraduate and graduate levels. It also serves as an excellent reference for applied researchers in virtually any area of study, from medicine and statistics to the social sciences, who analyze empirical data in their everyday work.
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#### Samadi Hassan, El Jarroudi Mustapha Linear and Non Linear Homogenization of a Composite Medium
5202 руб.
In this book, we consider in the first time the homogenization of a heat transfer linear problem between two periodic connected media exchanging a heat fux throughout their common interface. The interfacial exchange coeffcient is assumed to tend to zero or to infinity when the size of the basic cell tends to zero. Three homogenized problems are determined according to some critical value. In the second time we study the homogenization of a nonlinear problem posed in a fibre-reinforced composite with matrix-fibres interfacial condition. Using Γ-convergence methods, three homogenized problems are determined. The main result is that the effective constitutive relations reveal non-local terms associated with the microscopic interactions between the matrix and the fibers.
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#### Douglas Montgomery C. Solutions Manual to Accompany Introduction to Linear Regression Analysis
2547.42 руб.
As the Solutions Manual, this book is meant to accompany the main title, Introduction to Linear Regression Analysis, Fifth Edition. Clearly balancing theory with applications, this book describes both the conventional and less common uses of linear regression in the practical context of today's mathematical and scientific research. Beginning with a general introduction to regression modeling, including typical applications, the book then outlines a host of technical tools that form the linear regression analytical arsenal, including: basic inference procedures and introductory aspects of model adequacy checking; how transformations and weighted least squares can be used to resolve problems of model inadequacy; how to deal with influential observations; and polynomial regression models and their variations. The book also includes material on regression models with autocorrelated errors, bootstrapping regression estimates, classification and regression trees, and regression model validation.
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#### Alan Agresti Foundations of Linear and Generalized Linear Models
9647.09 руб.
A valuable overview of the most important ideas and results in statistical modeling Written by a highly-experienced author, Foundations of Linear and Generalized Linear Models is a clear and comprehensive guide to the key concepts and results of linearstatistical models. The book presents a broad, in-depth overview of the most commonly usedstatistical models by discussing the theory underlying the models, R software applications,and examples with crafted models to elucidate key ideas and promote practical modelbuilding. The book begins by illustrating the fundamentals of linear models, such as how the model-fitting projects the data onto a model vector subspace and how orthogonal decompositions of the data yield information about the effects of explanatory variables. Subsequently, the book covers the most popular generalized linear models, which include binomial and multinomial logistic regression for categorical data, and Poisson and negative binomial loglinear models for count data. Focusing on the theoretical underpinnings of these models, Foundations ofLinear and Generalized Linear Models also features: An introduction to quasi-likelihood methods that require weaker distributional assumptions, such as generalized estimating equation methods An overview of linear mixed models and generalized linear mixed models with random effects for clustered correlated data, Bayesian modeling, and extensions to handle problematic cases such as high dimensional problems Numerous examples that use R software for all text data analyses More than 400 exercises for readers to practice and extend the theory, methods, and data analysis A supplementary website with datasets for the examples and exercises An invaluable textbook for upper-undergraduate and graduate-level students in statistics and biostatistics courses, Foundations of Linear and Generalized Linear Models is also an excellent reference for practicing statisticians and biostatisticians, as well as anyone who is interested in learning about the most important statistical models for analyzing data.
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#### Keally D. McBride Collective Dreams. Political Imagination and Community
5402 руб.
How do we go about imagining different and better worlds for ourselves? Collective Dreams looks at ideals of community, frequently embraced as the basis for reform across the political spectrum, as the predominant form of political imagination in America today. Examining how these ideals circulate without having much real impact on social change provides an opportunity to explore the difficulties of practicing critical theory in a capitalist society. Different chapters investigate how ideals of community intersect with conceptions of self and identity, family, the public sphere and civil society, and the state, situating community at the core of the most contested political and social arenas of our time. Ideals of community also influence how we evaluate, choose, and build the spaces in which we live, as the author’s investigations of Celebration, Florida, and of West Philadelphia show.Following in the tradition of Walter Benjamin, Keally McBride reveals how consumer culture affects our collective experience of community as well as our ability to imagine alternative political and social orders.Taking ideals of community as a case study, Collective Dreams also explores the structure and function of political imagination to answer the following questions: What do these oppositional ideals reveal about our current political and social experiences? How is the way we imagine alternative communities nonetheless influenced by capitalism, liberalism, and individualism? How can these ...
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#### Г. Е. Бесстремянная Measuring income equity in the demand for healthcare with finite mixture models
152 руб.
The paper exploits panel data finite mixture (latent class) models to measure consumer equity in healthcare access and utilization. The finite mixture approach accounts for unobservable consumer heterogeneity, while generalized linear models address a retransformation problem of logged dependent variable. Using the data of the Japan Household Panel Survey (2009–2014), we discover that consumers separate into latent classes in the binary choice models for healthcare use and generalized linear models for outpatient/inpatient healthcare expenditure. The results reveal that healthcare access in Japan is pro-poor for the most sick consumers, while utilization of outpatient care is equitable with respect to disposable income.
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#### Men of Mark in Connecticut; Ideals of American Life Told in Biographies and Autobiographies of Eminent Living Americans... Volume 7
3089 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
Новинка
#### The Spirit of Islam. A History of the Evolution and Ideals of Islam
2814 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
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#### Marvin H. J. Gruber Matrix Algebra for Linear Models
8798.44 руб.
A self-contained introduction to matrix analysis theory and applications in the field of statistics Comprehensive in scope, Matrix Algebra for Linear Models offers a succinct summary of matrix theory and its related applications to statistics, especially linear models. The book provides a unified presentation of the mathematical properties and statistical applications of matrices in order to define and manipulate data. Written for theoretical and applied statisticians, the book utilizes multiple numerical examples to illustrate key ideas, methods, and techniques crucial to understanding matrix algebra’s application in linear models. Matrix Algebra for Linear Models expertly balances concepts and methods allowing for a side-by-side presentation of matrix theory and its linear model applications. Including concise summaries on each topic, the book also features: Methods of deriving results from the properties of eigenvalues and the singular value decomposition Solutions to matrix optimization problems for obtaining more efficient biased estimators for parameters in linear regression models A section on the generalized singular value decomposition Multiple chapter exercises with selected answers to enhance understanding of the presented material Matrix Algebra for Linear Models is an ideal textbook for advanced undergraduate and graduate-level courses on statistics, matrices, and linear algebra. The book is also an excellent reference for statisticians, engineers, economists, and readers interested in the linear statistical model.
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#### Akhtar Hassan Malik A comparative study of elite English-medium schools, public schools, and Islamic madaris in contemporary Pakistan
2089 руб.
This ethnographic study examines the role of differing school knowledge in reproducing various social classes in the society. It was observed that an unequal availability of capital resources, agents' class habitus, and the type of their "cultural currency" act as selection mechanisms that clearly favour some social groups over others. The ruling classes ensure the transfer of their power and privilege to their children by providing them with quality education in elite schools. The disadvantaged classes are excluded from these unique institutions by both social and economic sanctions. They have no other option than to educate their children either in public schools or Islamic madaris. As a result, inequitable educational opportunities consolidate the existing social-class hierarchy.
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#### Yenny Hoyos The Effects of Emotional Traits in Teaching Performance
3352 руб.
Bachelor Thesis from the year 2016 in the subject Sociology - Work, Profession, Education, Organisation, grade: 4.8, University of Colombo (UNICA), language: English, abstract: The purpose of this study is to illustrate some of the effects of negative emotional traits in the teaching performance of two English seven and eight grade teachers with forty students each in a public school in Suba, Bogotá. The work focused on three negative emotional traits (frustration, anxiety and lack of creativity) that are the core of a negative class environment. The research of Ferguson, Frost& Hall (2012) explores some of the physical effects such as increased blood pressure, anger, headaches, and some psychological effects such as anxiety, nervousness, tension, frustration and panic, that some teachers might experience when they are exposed to situations of high amount of stress. Furthermore, those effects can affect teacher performance when teaching classes. Two case studies described the situations that affect teachers' performance, how teachers reacted when experiencing those situations and what the outcomes are in the classroom environment. The main situations found during the study are related with students' behavior, students' attitude about class activities, school demands regarding class structure and the amount of English use during classes.
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#### The Social Ideals of Alfred Tennyson as Related to His Time
1202 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
Новинка
#### Adams Ephraim Douglass 1865-1930 The Power of Ideals in American History
1639 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
Новинка
#### Little Visits with Great Americans; Or, Success Ideals and How to Attain Them... Volume 2
1852 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
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#### Kent Charles Foster 1867-1925 The Testing of a Nation.s Ideals; Israel.s History from the Settlement to the Assyrian Period
1127 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
Новинка
#### Ventilation in American Dwellings. With a Series of Diagrams, Presenting Examples in Different Classes of Habitations
2227 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
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#### Mahmood Nozad H. Sparse Ridge Fusion For Linear Regression
5214 руб.
For a linear regression, the traditional technique deals with a case where the number of observations n are more than the number of predictors variables p (n>p). In the case n
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#### Chancellor William Estabrook 1867 A Theory of Motives, Ideals and Values in Education
3302 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
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#### Ismo Lindell V. Multiforms, Dyadics, and Electromagnetic Media
11190.63 руб.
This book applies the four-dimensional formalism with an extended toolbox of operation rules, allowing readers to define more general classes of electromagnetic media and to analyze EM waves that can exist in them End-of-chapter exercises Formalism allows readers to find novel classes of media Covers various properties of electromagnetic media in terms of which they can be set in different classes
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#### Julia Lieb Counting Polynomial Matrices over Finite Fields
3614 руб.
This book is dealing with three mathematical areas, namely polynomial matrices over finite fields, linear systems and coding theory.Primeness properties of polynomial matrices provide criteria for the reachability and observability of interconnected linear systems. Since time-discrete linear systems over finite fields and convolutional codes are basically the same objects, these results could be transferred to criteria for non-catastrophicity of convolutional codes.In particular, formulas for the number of pairwise coprime polynomials and for the number of mutually left coprime polynomial matrices are calculated. This leads to the probability that a parallel connected linear system is reachable and that a parallel connected convolutional code is non-catastrophic. Moreover, other networks of linear systems and convolutional codes are considered.
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#### B. G. Ziv Tasks for lessons geometry for grades 7-11. Manual for teachers, pupils and students
1822 руб.
The book is well-known Petersburg mathematics, methodologist and teacher Boris Germanovich Ziv is based on a very popular teaching materials, publishing previously published "Enlightenment", but with significant modifications and changes. Recommended for teachers of secondary schools, it can be used in optional classes and study circles in regular classes and in classes with in-depth study of mathematics.
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#### Compton Alfred George 1835- Some Common Errors of Speech. Suggestions for the Avoiding of Certain Classes of Errors, Together with Examples of Bad and of Good Usage
964 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
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#### A Robert Leishear Supplement to Fluid Mechanics, Water Hammer, Dynamic Stresses, and Piping Design
5202 руб.
This 2014 Addendum includes recommended additions and corrections to the original text, which was published by ASME Press in 2013. Numerous reviews were performed on the original text for this book, and manuscripts for this book were used to teach the topics in the text to four separate classes of engineers at SRS and prior to publication.Comments from these four day classes were incorporated into the manuscripts between classes to improve the quality of presentation. Even so, improvements and some corrections to the text have been noted since ASME classes were taught after publication. This Addendum captures these additions and changes. Also included in this Addendum are appendices to summarize this book through "A Discussion of DLFs for Piping Design" and the "Design of Piping Systems for Dynamic Loads From Fluid Transients."
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#### Riposo Julien, Bianca Carlo Mathematical and Computational Methods in Biology and Finance
4902 руб.
The topic of this book is the mathematical analysis of biological and financial systems. Firstly we develop some methods for analyzing the data that are furnished into matrix form. In particular we analyze the adjacency matrix of some well-known networks in the pertinent literature. We perform a general matrix analysis with the main aim to study the possible linear relantionship between the eigenvector associated with the highest eigenvalue (principal eigenvector) and the degree vector. We furnish some theoretical results that establish when the linear relation between the principal eigenvector and the degree vector is possible. Secondly this book is concerned with the simulation of biological systems viewed as solution of differential models. Specifically the analysis of a delayed ODE-based model for wound healing disease under the action of the immune system, is performed, and the conditions under which a Hopf bifurcation occurs are investigated. Moreover by employing the thermostatted kinetic theory, a model for the development of therapies against keloid is considered.
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#### Hussein Reem, N. M. Tawfiq Luma Singular Initial Value Problems
8927 руб.
The aim of this book is to use a semi-analytic technique for solving singular initial value problems of ordinary differential equations with a singularity of different kinds to construct polynomial solution using two point osculatory interpolation. The efficiency and accuracy of suggested method is assessed by comparisons with exact and other approximate solutions for a wide classes of non-homogeneous,non-linear singular initial value problems. Many examples are presented to demonstrate the applicability and efficiency of the suggested method on one hand and to confirm the convergence order on the other hand,two applications in mathematical physics and astrophysics are presented,such as Lane-Emden equations and Emden-Fowler equations to model several problems such as the theory of stellar structure,the thermal behavior of a spherical cloud of gas, isothermal gas spheres and the theory of thermionic currents.
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#### Social Ideals in English Letters
1852 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
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#### Page Elizabeth Fry Edward Macdowell, His Work and Ideals
7827 руб.
Unlike some other reproductions of classic texts (1) We have not used OCR(Optical Character Recognition), as this leads to bad quality books with introduced typos. (2) In books where there are images such as portraits, maps, sketches etc We have endeavoured to keep the quality of these images, so they represent accurately the original artefact. Although occasionally there may be certain imperfections with these old texts, we feel they deserve to be made available for future generations to enjoy.
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#### Girls. Faults and Ideals
1077 руб.
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Страницы:
The authoritative guide to modeling and solving complex problems with linear programming—extensively revised, expanded, and updated The only book to treat both linear programming techniques and network flows under one cover, Linear Programming and Network Flows, Fourth Edition has been completely updated with the latest developments on the topic. This new edition continues to successfully emphasize modeling concepts, the design and analysis of algorithms, and implementation strategies for problems in a variety of fields, including industrial engineering, management science, operations research, computer science, and mathematics. The book begins with basic results on linear algebra and convex analysis, and a geometrically motivated study of the structure of polyhedral sets is provided. Subsequent chapters include coverage of cycling in the simplex method, interior point methods, and sensitivity and parametric analysis. Newly added topics in the Fourth Edition include: The cycling phenomenon in linear programming and the geometry of cycling Duality relationships with cycling Elaboration on stable factorizations and implementation strategies Stabilized column generation and acceleration of Benders and Dantzig-Wolfe decomposition methods Line search and dual ascent ideas for the out-of-kilter algorithm Heap implementation comments, negative cost circuit insights, and additional convergence analyses for shortest path problems The authors present concepts and techniques that are illustrated by numerical examples along with insights complete with detailed mathematical analysis and justification. An emphasis is placed on providing geometric viewpoints and economic interpretations as well as strengthening the understanding of the fundamental ideas. Each chapter is accompanied by Notes and References sections that provide historical developments in addition to current and future trends. Updated exercises allow readers to test their comprehension of the presented material, and extensive references provide resources for further study. Linear Programming and Network Flows, Fourth Edition is an excellent book for linear programming and network flow courses at the upper-undergraduate and graduate levels. It is also a valuable resource for applied scientists who would like to refresh their understanding of linear programming and network flow techniques.
##### Продажа some classes of monomial ideals with linear ients лучших цены всего мира
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2019-08-20 03:45:37
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https://www.springerprofessional.de/algorithmic-algebra/13752738
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main-content
## Über dieses Buch
Algorithmic Algebra studies some of the main algorithmic tools of computer algebra, covering such topics as Gröbner bases, characteristic sets, resultants and semialgebraic sets. The main purpose of the book is to acquaint advanced undergraduate and graduate students in computer science, engineering and mathematics with the algorithmic ideas in computer algebra so that they could do research in computational algebra or understand the algorithms underlying many popular symbolic computational systems: Mathematica, Maple or Axiom, for instance. Also, researchers in robotics, solid modeling, computational geometry and automated theorem proving community may find it useful as symbolic algebraic techniques have begun to play an important role in these areas. The book, while being self-contained, is written at an advanced level and deals with the subject at an appropriate depth. The book is accessible to computer science students with no previous algebraic training. Some mathematical readers, on the other hand, may find it interesting to see how algorithmic constructions have been used to provide fresh proofs for some classical theorems. The book also contains a large number of exercises with solutions to selected exercises, thus making it ideal as a textbook or for self-study.
## Inhaltsverzeichnis
### Chapter 1. Introduction
Abstract
The birth and growth of both algebra and algorithms are strongly intertwined. The origins of both disciplines are usually traced back to Muhammed ibn-Mūsa al-Khwarizmi al-Quturbulli, who was a prominent figure in the court of Caliph Al-Mamun of the Abassid dynasty in Baghdad (813–833 A.D.). Al-Khwarizmi’s contribution to Arabic and thus eventually to Western (i.e., modern) mathematics is manifold: his was one of the first efforts to synthesize Greek axiomatic mathematics with the Hindu algorithmic mathematics. The results were the popularization of Hindu numerals, decimal representation, computation with symbols, etc. His tome “al-Jabr wal-Muqabala,” which was eventually translated into Latin by the Englishman Robert of Chester under the title “Dicit Algoritmi,” gave rise to the terms algebra (a corruption of “al-Jabr”) and algorithm (a corruption of the word “al-Khwarizmi”).
Bhubaneswar Mishra
### Chapter 2. Algebraic Preliminaries
Abstract
In this chapter, we introduce some of the key concepts from commutative algebra. Our focus will be on the concepts of rings, ideals and modules, as they are going to play a very important role in the development of the algebraic algorithms of the later chapters. In particular, we develop the ideas leading to the definition of a basis of an ideal, a proof of Hilbert’s basis theorem, and the definition of a Gröbner basis of an ideal in a polynomial ring. Another important concept, to be developed, is that of a syzygy of a finitely generated module.
Bhubaneswar Mishra
### Chapter 3. Computational Ideal Theory
Abstract
In the previous chapter, we saw that an ideal in a Noetherian ring admits a finite Gröbner basis (Theorem 2.3.9). However, in order to develop constructive methods that compute a Gröbner basis of an ideal, we need to endow the underlying ring with certain additional constructive properties. Two such properties we consider in detail, are detachability and syzygy-solvability. A computable Noetherian ring with such properties will be referred to as a strongly computable ring.
Bhubaneswar Mishra
### Chapter 4. Solving Systems of Polynomial Equations
Abstract
The Gröbner basis algorithm can be seen to be a generalization of the classical Gaussian elimination algorithm from a set of linear multivariate polynomials to an arbitrary set of multivariate polynomials. The S-polynomial and reduction processes take the place of the pivoting step of the Gaussian algorithm. Taking this analogy much further, one can devise a constructive procedure to compute the set of solutions of a system of arbitrary multivariate polynomial equations:
$$\begin{array}{*{20}{c}} {{f_1}\left( {{x_1}, \ldots ,{x_n}} \right) = 0,} \\ {{f_2}\left( {{x_1}, \ldots ,{x_n}} \right) = 0,} \\ \vdots \\ {{f_r}\left( {{x_1}, \ldots ,{x_n}} \right) = 0,} \end{array}$$
i.e., compute the set of points where all the polynomials vanish:
$$\left\{\langle\xi_{1},\ldots,\xi_{n}\rangle:f_{i}(\xi_{1},\ldots,\xi_{n})=0,\quad{\rm for}\ {\rm all}\ 1\leq i\leq r\right\}.$$
Bhubaneswar Mishra
### Chapter 5. Characteristic Sets
Abstract
The concept of a characteristic sets was discovered in the late forties by J.F. Ritt (see his now classic book Differential Algebra [174]) in an effort to extend some of the constructive algebraic methods to differential algebra. However, the concept languished in near oblivion until the seventies when the Chinese mathematician Wu Wen-Tsün [209–211] realized its power in the case where Ritt’s techniques are specialized to commutative algebra. In particular, he exhibited its effectiveness (largely through empirical evidence) as a powerful tool for mechanical geometric theorem proving. This proved to be a turning point; a renewed interest in the subject has contributed to a better understanding of the power of Ritt’s techniques in effectively solving many algebraic and algebraico-geometric problems.
Bhubaneswar Mishra
### Chapter 6. An Algebraic Interlude
Abstract
Before we move on to the topics of resultants and an algorithmic treatment of real algebra, we shall take a short pause to study in this chapter the unique factorization domain, the principal ideal domain, and the Euclidean domain. Of course, readers familiar with these topics may safely skip this chapter and go directly to the next chapter.
Bhubaneswar Mishra
### Chapter 7. Resultants and Subresultants
Abstract
In this chapter we shall study resultant, an important and classical idea in constructive algebra, whose development owes considerably to such luminaries as Bezout, Cayley, Euler, Hurwitz, and Sylvester, among others. In recent time, resultant has continued to receive much attention both as the starting point for the elimination theory as well as for the computational efficiency of various constructive algebraic algorithms these ideas lead to; fundamental developments in these directions are due to Hermann, Kronecker, Macaulay, and Noether. Some of the close relatives, e.g., discriminant and subresultant, also enjoy widespread applications. Other applications and generalizations of these ideas occur in Sturm sequences and algebraic cell decomposition—the subjects of the next chapter.
Bhubaneswar Mishra
### Chapter 8. Real Algebra
Abstract
In this chapter, we focus our attention on real algebra and real geometry. We deal with algebraic problems with a formulation over real numbers, ℝ (or more generally, over real closed fields). The underlying (real) geometry provides a rich set of mechanisms to describe such topological notions as “between,” “above/below,” “internal/external,” since it can use the inherent order relation (<) of the real (or, real closed) field. As a result, the subject has found many applications in such practical areas as computer-aided manufacturing, computational geometry, computer vision, geometric theorem proving, robotics, and solid modeling, etc., and thus has generated a renewed interest. We concentrate on the following key ingredients of real algebra: Sturm theory, algebraic numbers, and semialgebraic geometry.
Bhubaneswar Mishra
### Backmatter
Weitere Informationen
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2020-06-05 11:01:06
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https://www.physicsforums.com/threads/is-pure-mathematics-the-basis-for-all-thought.512147/
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# Is pure mathematics the basis for all thought?
1. Jul 5, 2011
### Functor97
I have been thinking much on the nature of pure mathematics. I believe this forum would make the best place to post over say the philosophy section, as i am more interested in the opinions of working mathematicians and physicists than philosophers.
In my opinion pure mathematics is the core of every academic discipline we humans have so far explored. I would extend that to state, that mathematics is in essence the art of thought it is the science of anything, and that all other disciplines may be reduced to applied mathematical problems. I understand this is quite a contentious statement, and i would gladly welcome your point of view.
The more i ponder this, the more i think that mathematics is all we have. We interact with the outside world via our senses, but these senses give us a warped view of our surroundings. This screen has no color, but our retina perceives color, due to the wavelength of electromagnetic radiation reflected, and our understanding of electromagnetic radiation is based upon physical formula which are simply applied forms of mathematical theorems. Quantum mechanics does not make sense to our intuition built up on the large scale, so we rely upon mathematics to understand the phenomenon. Many physics students object to quantum mechanics and claim that it is fundamentally wrong or not complete, because they cannot envision exactly where a particle is at a given instant, yet the mathematics still guides us. When people ask "why do like poles repel?", the answer can get quite complex leading to quantum field theory, and if an individual keeps asking "why?" each subsequent step is correct, it quickly devolves into mathematical reasoning not physical. This has led me to agree with max tegmark and a few others who conjecture that our universe is actually a mathematical structure, yet i do have a problem with this as to why conscious minds view only a certain form of mathematics. Does our pursuit of knowledge being grounded in mathematics, mean that external reality (if there is any?) is grounded in mathematics, or is that just our best guess, so to speak.
More generally, do you believe that mathematical inquiry, such as say number theory, complex analysis or functional analysis explains more about reality then quantum field theory for when we solve number theory problems, are we working with the very code of the cosmos? So our theories of physics are just the subset of our Matthematical theorems. Do many mathematicians hold this view? I guess this is some form of Neo Platonism, and as an atheist, i find the issue of an external mathematics problematic. Do you believe that pure mathematics should be given credence over particle physics, for if this conjecture is true, isn't the dependence of the Axiom of choice within a large cardinal system just as suitable a grand unified theory as M theory?
Finally if mathematics is just axioms, and we cannot prove an axiom to be true, and yet mathematics is the basis of all science, does this mean absolute truth is beyond us? (Not advocating the ends justifies the means)
Last edited: Jul 5, 2011
2. Jul 6, 2011
### dalcde
Physics isn't just the maths. It's more than that. In mathematics, we are allowed to define whatever we like (we can define the operation Ԅ to be $aԄb=ab^{\log_a b}$), but in physics, our rules have to agree with the real world.
I don't like to think axioms of something we take to be true. We define axioms to be true. We make certain definitions (axioms) and build upon them. A clear example (if you know some abstract algebra) is the axioms of groups. We don't take them to be true. They are rather things that constrain our system.
3. Jul 6, 2011
### Functor97
I understand that, i just have a problem with "defining" something to be true. It makes mathematics seem a game, where we make statements and draw those statements out to their logical conclusions. That very well maybe mathematics (there are certainly worse things to spend your time doing than playing such a beautiful game :tongue:) but i always thought of mathematics as the process by which we uncover the laws of god so to speak. (not saying there is a god, but i am sure you have heard the expression "God created the integers, all the rest is the work of men") The definition of axioms as things we state as true removes any chance at external truth, or some absolute system of thought. I by no means believe there to be one, but it does seem to be a common belief that mathematics is that system, i always thought of mathematics as striving towards perfect knowledge. If we decide upon truth, then how is mathematics different to some post modern literary theory, in which the english professor blows on about how everything (even time and space) are subjective?
Isn't physics just the mathematics which we apply in our universe. Would that not make physics a part of mathematics? Would mathematics then serve as an explanation for all physical theory? By explanation, i do not mean it to be final, rather each level of the onion "of truth" becomes more and more mathematical, till in essence we are just doing pure mathematics...
Last edited: Jul 6, 2011
4. Jul 6, 2011
### pwsnafu
That is my interpretation of mathematics. We call these things theorems. I don't know any mathematician who places more weight on the axioms themselves.
Specific example: is the axiom of choice true in your interpretation?
We are free to choose which axioms we want, and we look at their conclusions. Whether our choice of axioms is "extensionally true" is irrelevant. Mathematics is nothing more than applied logic. That is: mathematics is limited by human thought.
Just because PA has a statement X which is true but not provable in PA doesn't mean much. I can strengthen my axioms and then study the problem. There may exist an Axiom B such that X is true in PA+B and but false in PA-B. Well that's interesting in itself. And certainly not subjective.
5. Jul 6, 2011
### chiro
Hello Functor97 and welcome to the forums.
One reason why math is so powerful (and why it is applied everywhere) is due to the properties of numbers, sets, functions, graphs, and all the other objects out there in math.
Numbers can represent anything. They encode information about anything. You can use the same kind of argument for functions and sets in similar ways.
The axiomatic way also matches much of human behavior when it comes to things that are probably not considered mathematical.
For example consider a trainee just hired by an employer. The employer has years and years of experience of this job and has to quickly train up the employee. The employer spends about 10-15 minutes outlining a "compressed form" of training to the employee. It doesn't cover every situation, but it allows the employee to get a picture of how to handle things that aren't said and how to use that knowledge to handle a specific situation.
This looks a lot like what scientists do: they come up with principles that are both minimal in description and complexity and maximal in descriptive capability for the domain being described.
You can probably find these kind of analogues in everyday life if you look hard enough.
The big difference is though, that in many human situations rigor is not something you really need. In maths though you do need it. This kind of formality can be a pain in the neck, but it also increases confidence in what you are doing.
6. Jul 6, 2011
### Functor97
So you would view mathematics as a tool rather then the process? I guess this is very subjective, but it is often said that Mathematics is the queen of the sciences, and i was hoping there would be some bootstrap which would reduce physical reasoning to mathematics. Is physics the mathematics of our universe? or is Mathematics, Physics we primates have evolved to abstract for evolutionary gain? Both maybe?
7. Jul 6, 2011
### chiro
It's a language and a way of thinking and analyzing.
Human beings are primarily constrained by their ability to describe, classify, and analyze. This constraint is language. Before you wish to do anything, you need to figure out what it is that you want to do: you need to describe what you want to achieve.
Math is a language that suits a specific purpose and provides a framework for a certain way of thinking. Like any language it is constantly evolving: things become more general, isolated things merge together to form new things, and completely new things are discovered and added to make sense of a previously unknown phenomena.
Like any language it is optimal for its purpose. You don't write a hundred page book in English to describe something if you can write it in a formula that covers a page. At the same time it might not be enough to write just a few mathematical laws to communicate what you need to. Use a particular language for its strengths and if one isn't strong in enough ways, use another or create one that fits the purpose.
If there comes a better way to describe the physical universe, it will most likely be embedded into mathematics. This is my opinion, but the evidence is there based on what has happened historically.
8. Jul 6, 2011
### Functor97
But it is a language in which certain rules remain constant. Take Hardy for instance, he was a strict platonist and believed in an objective mathematical reality. The number 23 is prime not because we want it to be, but because it fits our definition of a prime. If mathematics was purely a language i would challenge you to create a prime that does not fit out previous definition. Of course you cannot, that is a contradiction...I will agree mathematics is a language, but it is the language of the universe, the language of reason. You cannot shape this language as much as you want to, it shapes your view, in that mathematics is very much like physics. We may draw the lines, dots and squigels, but the "Background" reason is independent of us, or at least our ability to influence it. As mathematics is the ultimate language for the physical world, would that not mean it explains the physical world? I would claim It is the "deepest" part of the physical world we can access...
I think that writing mathematics off as a language, does not do it justice, maybe call it the only language...that may suffice?
9. Jul 6, 2011
### Functor97
You just need to look at modern particle physics. The deeper we go, the more mathematical reality becomes. In high school we could all draw pictures of what we thought was going on, when you reached quantum mechanics that vanished. Now we must use mathematics, but how does this differ from pure mathematics itself. Einstein showed that the bending of spacetime is gravity, and the bending of spacetime is just geometry, and how do we understand geometry? enter pure mathematics. Sure a physicist working in solid state or optics is not going to notice this trend as much as a high energy specialist, but the deeper into reality we go, the more and more we must embrace pure mathematics. I do not see this ending, that is why i believe that pure mathematics is the foundation of all physical understanding...
10. Jul 6, 2011
### chiro
Languages are used to describe and classify things and mathematics does exactly that.
Lets say you have the word "cat" and with a Venn diagram you draw a 2D shape that corresponds to all written, spoken, graphical, etc definitions that pertain to cat. That set will contain a boundary. At some classification point there is a definitive point where things are no longer "cats". The union of "cats" and "not cats" is anything that can be described within the limits of the language.
Just like your cat example, all your math definitions are exactly the same, in the context of the language that is math. The fact that you called something prime and described it means you artificially created your "prime" subset and with your definition have declared a boundary that separates what is "prime" and what "isn't prime". Both are completely disjoint.
I agree that math is the best language we have because of its domain, its clarity, its generality, and its ability to describe so much more than any other language we have: no arguments there.
One example that comes to mind is a tribe (I forget the actual geographical vicinity) that did not have a complete system for numbers. They had words that corresponded to zero, one, two and three, but anything more than three was considered just one word (kind of like our infinity with the fact that finite numbers were also included): it was like our definition of "many".
With something like english I can talk about a chair. What exactly is a chair? Well its something you sit on. "But I can sit on a table you say", but then you say "but that's not a chair". Eventually you might get something that is a good definition.
With math its a lot simpler. We can add constraint after constraint by treating sensory input as a mathematical signal and then using these constraints to get ridiculously close to a very specific definition of chair.
So in short I agree that math is the "super" language or at least the best language we have.
11. Jul 6, 2011
### dalcde
The art of mathematics is to make good definitions and axioms, and build great work (theorems) upon them. You can define addition to be
$$a+b=0,$$
but it won't do much good.
12. Jul 6, 2011
### Functor97
Even if we change the constraints the method of thought remains the same, and that is what i would claim is the essence of mathematics. It is certain. Or we may say, it is as certain as we want it to be. As to external reality, who knows what that is, does it even exist? As soon as we examine it, does it change? For all intents and purposes we must accept mathematics as external reality, it very well maybe just the way our brains are structured, but i don't see how we can examine the cosmos without our minds.
I think many pure mathematicians would object to dalcde's claim that mathematics should be built around utility. Yet i do see the point, many pure mathematicians do work within systems that are derived from very physically intuitive concepts, and then extended. Would any pure mathematicians care to expound on what drives you in your research?
13. Jul 6, 2011
### Functor97
All of our physical laws are mathematical statements. Thus i see mathematics as the underpining of all thought. I don't think this is possible to change, it just is. Look at some of the papers from leading particle physicists, take edward witten for instance http://arxiv.org/find/all/1/au:+witten_edward/0/1/0/all/0/1" he isnt sitting around in a patent office constructing thought experiments, he is exploring pure mathematical structures, in the hope that these will explain the nature of our reality.
Last edited by a moderator: Apr 26, 2017
14. Jul 6, 2011
### kramer733
Ok this is going to be off topic but why does functor97 keep on writing a new text below his previous text? JUST EDIT YOUR PREVIOUS POST! Jesus man! I've seen you do that like 4-5 times or something. To me it's very annoying.
As for my input on this, math is honestly just a tool. Like it teaches you how to think but i more or less see it as a tool now.
15. Jul 6, 2011
### SteveL27
The group axioms aren't true of the integers under addition? Who knew!
To be clear: I disagree. The group axioms are true of any group. What do you say to that?
And by the way, axioms are not definitions. I don't know why so many people are confused about that.
Last edited: Jul 6, 2011
16. Jul 6, 2011
### dx
Scientific thinking is a free play with concepts (words) whose justification lies in the measure of survey over the experience of the senses which we are able to achieve with its aid. All knowledge is originally represented within a conceptual framework adapted to account for previous experience, and any such frame may prove too narrow to comprehend new experiences. Mathematics is essentially an extension of our ordinary language developed for the logical representation of relations between experience, supplementing it with appropriate tools for representing relations for the communication of which ordinary language in not sufficient, and also with its well-defined abstractions allows the representation of the harmonious relationships of theoretical physics where at each stage appropriate widening of the conceptual framework to grasp new experience brings greater unity and harmony to the whole description.
Last edited: Jul 6, 2011
17. Jul 6, 2011
### SteveL27
That idea is easily refuted by the existence of non-Euclidean geometries. Through a point not on a given line, you can assume there are zero, one, or more than one parallels to the given line. Each choice gives a logically consistent geometry. But these three choices can not all be true of the world we live in.
Math is not physics.
Even within math, you can play the same game. Given the Zermelo-Fraenkel axioms, you can assume the Axiom of Choice (AC) or its negation. Either way you get a consistent set theory. Same with the Continuum Hypothesis (CH); and there are also a number of less well-known axioms with the same property of being independent of ZF, with no "real world" way of knowing whether the axiom or its negation should be accepted into mainstream math.
18. Jul 6, 2011
### Functor97
I do not see how that follows. Non Euclidean Geometry is a generalisation of Euclidean geometry, it extends it, it does not contradict it.
I have yet to take a course in advanced mathematical logic, but to me it seems that our axioms are so basic that we cannot come up with a different form of mathematics, it is intwined within our way of thought. For example, could you change a basic axiom of mathematics and come up with a system that seems interesting and beautiful but is totally distinct from our current research areas of mathematics? I do not know, like i said, i am unfamilar with this work as an undergraduate, but if it can be done, why has it not been done?
Maybe i am looking at this from the wrong angle, for some of the arguments on this page make me think of mathematics as a subset of physics. We take our physical intuition and generalise it. That makes sense from a biological/evolutionary stand point, what would be the advantage of us accessing the "source code" of the universe, if it were distinct from our need to survive. Would anyone agree with that? I think many pure mathematicians would object, as they often take pride in "useless research" as Hardy said.
I can find one pure mathematician who agrees with this standpoint. Vladimir Arnold, the key protagonist of the anti bourbaki tradition, said "Physics is an experimental science, a part of natural science. Mathematics is the part of physics where experiments are cheap." Thoughts?
19. Jul 7, 2011
### SteveL27
You can have a self-contained, consistent system of geometry that is Euclidean; and a self contained, consistent system (lots of different ones, in fact) that are non-Euclidean.
The physical universe can not be both. It must be one or the other. In this case math is a tool for describing universes. It does not discriminate between the hypothetical ones and the real one.
The most well-known example is the Axiom of Choice (AC). It says, innocently enough, that you can simultaneously choose an element from each one of a collection of nonempy sets. AC turns out to be independent of the other standard axioms of Zermelo-Fraenkel (ZF) set theory.
So, you can do math with or without AC. If you use AC then you can prove many standard theorems that mathematicians (and physicists) use daily. But you also get unavoidable anomalies such as the famous Banach-Tarski paradox, which says you can cut a solid in 3-space into a finite number of pieces; rearrange the pieces using rigid rotations and translations; and end up with TWO copies of the original solid. This result is disturbing to many people.
http://en.wikipedia.org/wiki/Banach–Tarski_paradox
On the other hand if you reject AC, you get a perfectly good, logically consistent theory (well, ZFC is consistent if ZF was consistent in the first place -- which is another story!). But in this choiceless theory, you have a vector space without a basis; a ring without a maximal ideal; the product of compact topological spaces might not be compact; and a lot of standard theorems can't be proved.
So the overwhelming majority of mainstream mathematicians freely use AC. Not because it's "true" in any conceivable meaning of the word -- I mean, who the heck really knows whether the real numbers can be well-ordered, which is one of the equivalents of AC -- but because it's convenient. It let's you prove more theorems, so mathematicians use it.
In other words, no pun intended: It's a matter of choice :-)
There are other examples but this is the most famous one.
http://en.wikipedia.org/wiki/Axiom_of_choice
It's done every day. Set theorists, logicians, and computer scientists deal with axioms and provability every day. Why computer scientists? They're interested in what you can do with finite strings of symbols, which they call programs. Logicians are interested in what you can do with finite strings of symbols, which they call proofs. It's the same subject. Godel, Church, Turing in the 30's, very active area of research ever since. The set theorists have a long list of wild axioms that they study. Each axiom gives you a different set of properties for the real numbers. Of particular interest are the large cardinal axioms. Large cardinals are sets so large that their existence can't be proven from ZFC. But some of them are starting to work their way into standard mathematics. A large cardinal is implicitly used in Wiles's proof of Fermat's Last Theorem. Foundations are always in a state of flux.
Art is inspired by our experience of the real world. But art far transcends the real world. Same with math. Or, what does a symphony or a pop tune have to do with our need to survive?
That's really interesting. You know, I only heard about Bourbaki as these French guys who wrote the textbooks that are the standard for the way all the graduate students are trained to think about math these days. But I have never heard of what it means to be "anti-Bourbaki." Can you tell me more about that? What is it they don't like?
As far as that quote, of course math is a toolkit for physics. It just happens to be a lot more. But this is an old debate. I wouldn't pretend to be qualified to speak for mathematics. I'll just let xkcd have the last word ...
http://xkcd.com/435/
20. Jul 9, 2011
### Stephen Tashi
The current wikipedia article on "Philosophy of mathematics" lists several varieties of beliefs about mathematics. I haven't bothered to understand the distinctions among them but it's interesting how many posters one encounters on this forum who advocate some version of "everything is math" or "math is a reality that exists outside the axioms that people create for it", etc. I would call this "mathematical Platonism". The wikipedia article suggests that my classification system doesn't have enough species.
To me, the most interesting aspect of thought, mathematical or otherwise, is to consider what we know about it - which is practically nothing. It appears to be conducted by some sort of self-modifying biological network that doesn't work very well (at physics or math) when it is first created. As it ages and gains experience, it (in its own opinon) begins to grasp things that it considers to be truths. It becomes very impatient with other biological networks that express any contradiction to them. I suppose it's useful as a self-motivational tool to believe that our brains are touching some important, eternal verities. Yet if someone makes a mathematical claim and then presents an incoherent incomprehensible justification for it, we don't admit that he has provided a proof. So, since we don't know how our brain works, why should we be so trusting of its conclusions?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook
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2018-05-22 06:34:17
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http://math.stackexchange.com/questions/112759/how-to-approximate-a-trigonometric-curve-by-bezier-curves
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# How to approximate a trigonometric curve by Bezier curves?
Let me ask how to approximate a trigonometric curve by Bezier curves? Is there any known algorithm?
Thank you in advance.
-
Try this Page 13. Codes are here. – Inquest Feb 24 '12 at 8:57
@Nunoxic Thank you very much. – seven_swodniw Feb 28 '12 at 14:58
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2015-11-29 15:01:05
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https://www.physicsforums.com/threads/partial-differentiation-question.733054/
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# Partial differentiation question?
1. Jan 16, 2014
### applestrudle
1. The problem statement, all variables and given/known data
z = x^2 +y^2
x = rcosθ
y = rsinθ
find partial z over partial x at constant theta
2. Relevant equations
z = x^2 +y^2
x = rcosθ
y = rsinθ
3. The attempt at a solution
z = 1 + r^2(sinθ)^2
dz/dx = dz/dr . dr/dx
= 2(sinθ)^2r/cosθ
= 2tanθ^2x
the book says 2x[1+2(tanθ)^2]
2. Jan 16, 2014
### tiny-tim
hi applestrudle!
(have a curly d: ∂ and try using the X2 button just above the Reply box )
∂/∂x is ambiguous unless you know what the other variables are
it always means that you differentiate wrt x keeping the other variables constant (that's why you need to know what they are!)
in this case, the question tells you the other variable is θ, so first you need to find z(x,θ), ie to write z as a function of x and θ
(hmm … i don't get the result the book gets )
3. Jan 16, 2014
### vanhees71
Here, I'd write $z$ as function of $x$ and $\theta$. Then it's easy to take the partial derivative. Obviously we have
$$z=r^2=\frac{x^2}{\cos^2 \theta}.$$
Then you can take the partial derivative wrt. $x$ and fixed $\theta$ easily, but what you quoted as solution of the book is obviously wrong (the factor 2 in front of $\tan^2 \theta$ should not be there).
Last edited: Jan 16, 2014
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2017-08-24 01:54:22
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https://www.cuemath.com/jee/examples-on-tangents-set-1-circles/
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# Examples On Tangents To Circles Set-1
Go back to 'Circles'
Example - 17
What is the length of the tangent to $$S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0$$ drawn from an external point $$P({x_1},{y_1})$$ ?
Solution:
The length $$PA$$ (or $$PB$$) can be evaluated by a simple application of the Pythagoras theorem.
In $$\Delta PAO,$$ observe that
\begin{align} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,P{A^2} =& P{O^2} - A{O^2} \hfill \\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =& \left\{ {{{({x_1} + g)}^2} + {{({y_1} + f)}^2}} \right\} - {\left\{ {\sqrt {{g^2} + {f^2} - c} } \right\}^2} \hfill \\\\ &\left\{ {\because AO{\text{ is the radius}}} \right\} \hfill \\\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =& x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right) \hfill \\ \end{align}
The equation of the circle being represented by $$S = 0,$$ we can denote the RHS obtained in (1) by $$S({x_1},{y_1}).$$ Thus, the length of the tangent can be written concisely as
$\fbox{\begin{array}{*{20}{c}} {PA = \sqrt {S({x_1},{y_1})} }\end{array}}$
For example, the length of the tangent from $$(4, 4)$$ to $${x^2} + {y^2} - 2x - 4y + 4 = 0$$ will be
$\begin{array}{l}l = \sqrt {{4^2} + {4^2} - 2 \times 4 - 4 \times 4 + 4} \\\,\,\, = \sqrt {12} \\\,\,\, = 2\sqrt 3 \end{array}$
In the next example, we discuss how to write the equation to the pair of tangents $$PA$$ and $$PB$$.
Example - 18
From an external point $$P({x_1},{y_1}),$$ (two) tangents are drawn to the circle $$S \equiv {x^2} + {y^2} + 2gx + 2fy + c = 0$$ . These tangents touch the circle at $$A$$ and $$B$$. Find the joint equation of $$PA$$ and $$PB$$.
Solution: Consider any point $$(h,k)$$ lying on the tangents drawn from $$P$$ to $$S$$.
Since we know two points on the line $$PA$$, we can use the two-point form to write its equation:
\begin{align}{}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&\frac{{y - k}}{{x - h}} = \frac{{{y_1} - k}}{{{x_1} - h}}\\\\ \Rightarrow \qquad & x({y_1} - k) - y({x_1} - h) = h({y_1} - k) - k({x_1} - h)\\\\ \Rightarrow \qquad & x({y_1} - k) - y({x_1} - h) + (k{x_1} - h{y_1}) = 0\end{align}
Since $$PA$$ is a tangent to $$S$$, its distance from the centre of the circle, $$( - g, - f),$$ must equal the radius.
This gives
$\frac{{{{( - g({y_1} - k) + f({x_1} - h) + k{x_1} - h{y_1})}^2}}}{{{{({x_1} - h)}^2} + {{({y_1} - k)}^2}}} = {g^2} + {f^2} - c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)$
To write the equation of the pair of lines in conventional form, we use $$(x,y)$$ instead of $$(h,k)$$ in (1), above. Subsequent (lengthy!) rearrangements give:
${\left\{ {x{x_1} + y{y_1} + g(x + {x_1}) + f(y + {y_1}) + c} \right\}^2} = \left( {{x^2} + {y^2} + 2gx + 2fy + c} \right)\left( {x_1^2 + y_1^2 + 2g{x_1} + 2f{y_1} + c} \right)$ The left hand side can be written concisely as $${(T({x_1},{y_1}))^2}$$ as described earlier whereas the right hand side can be written concisely as $$S(x,y)S({x_1},{y_1}).$$ Thus, the equation to the pair of tangents can be written concisely as
$\fbox{\begin{array}{*{20}{c}} {{T^2}({x_1},{y_1}) = S(x,y)S({x_1},{y_1})}\end{array}}$
This relation be written in an even shorter form as simply $${T^2} = S{S_1}$$ .
Example - 19
Find the equation to the pair of tangents drawn from the origin to the circle $${x^2} + {y^2} - 4x - 4y + 7 = 0$$ .
Solution: We use the relation obtained in the last example, $${T^2} = S{S_1},$$ to write the desired equation. Here, $$({x_1},{y_1})$$ is (0, 0) while $$g = - 2,f = - 2$$ and $$c = 7$$ . Thus the joint equation is
$\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,&{T^2}(0,0) = S(x,y)S(0,0)\\\\ \Rightarrow & {( - 2x - 2y + 7)^2} = ({x^2} + {y^2} - 4x - 4y + 7)(7)\\\\ \Rightarrow & 4{x^2} + 4{y^2} + 49 + 8xy - 28x - 28y = 7{x^2} + 7{y^2} - 28x - 28y + 49\\\\ \Rightarrow & 3{x^2} - 8xy + 3{y^2} = 0\end{array}$
As expected, since the tangents have been drawn from the origin, the obtained equation is a homogenous one.
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2021-05-06 19:36:19
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https://www.springerprofessional.de/en/emerging-technologies-for-authorization-and-authentication/17593336
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main-content
This book constitutes the proceedings of the Second International Workshop on Emerging Technologies for Authorization and Authentication, ETAA 2019, held in Luxembourg, in September 2019.
The 10 full papers presented in this volume were carefully reviewed and selected from numerous submissions. They focus on new techniques for biometric and behavioral based authentication, authentication and authorization in the IoT and in distributed systems in general, techniques for strengthen password based authentication and for dissuading malicious users from stolen password reuse, an approach for discovering authentication vulnerabilities in interconnected accounts, and strategies to optimize the access control decision process in the Big Data scenario.
### Logics to Reason Formally About Trust Computation and Manipulation
Abstract
Trust represents a fundamental, complementary ingredient for the success of security mechanisms in computer science, as it goes beyond the intrinsic, technical aspects of cybersecurity, by involving the subjective perception of users, the willingness to collaborate and expose own resources and capabilities, and the judgement about the expected behavior of other parties. Computational notions of trust are formalized to support automatically the process of building and maintaining trust infrastructures, and mathematical logics provide the formal means to reason about the efficacy of such a process. In this work we advocate the use of two logical approaches to the modeling and verification of the two main tasks at the base of any trust infrastructure: the initial computation of trust values and the dynamic manipulation of such values.
Alessandro Aldini, Mirko Tagliaferri
### An Authorization Framework for Cooperative Intelligent Transport Systems
Abstract
Cooperative Intelligent Transport Systems (C-ITS) aims to enhance the existing transportation infrastructure through the use of sensing capabilities and advanced communication technologies. While improving the safety, efficiency and comfort of driving, C-ITS introduces several security and privacy challenges. Among them, a main challenge is the protection of sensitive information and resources gathered and exchanged within C-ITS. Although several authorization frameworks have been proposed over the years, they are unsuitable to deal with the demands of C-ITS. In this paper, we present an authorization framework that addresses the challenges characterizing the C-ITS domain. Our framework leverages principles of both policy-based and token-based architectures to deal with the dynamicity of C-ITS while reducing the overhead introduced by the authorization process. We demonstrate our framework using typical use case scenarios from the C-ITS domain on location tracking.
Sowmya Ravidas, Priyanka Karkhanis, Yanja Dajsuren, Nicola Zannone
### A Framework for the Validation of Access Control Systems
Abstract
In modern pervasive applications, it is important to validate Access Control (AC) mechanisms that are usually defined by means of the XACML standard. Mutation analysis has been applied on Access Control Policies (ACPs) for measuring the adequacy of a test suite.
This paper provides an automatic framework for realizing mutations of the code of the Policy Decision Point (PDP) that is a critical component in AC systems. The proposed framework allows the test strategies assessment and the analysis of test data by leveraging mutation-based approaches. We show how to instantiate the proposed framework and provide also some examples of its application.
Said Daoudagh, Francesca Lonetti, Eda Marchetti
### The Structure and Agency Policy Language (SAPL) for Attribute Stream-Based Access Control (ASBAC)
Abstract
Current architectures and data flow models for access control are based on request response communication. In stateful or session-based applications monitoring access rights over time this results in polling of authorization services and for Attribute-Based Access Control (ABAC) in the polling of policy information points. This introduces latency or increased load due to polling. Attribute-Stream-based Access Control (ASBAC) is an authorization model based on a publish subscribe pattern mitigating these bottlenecks. ASBAC allows the quasi real time consideration of attribute data streams for access control decisions, such as internet-of-things (IoT) sensor data. This paper introduces the Structure and Agency Policy Language (SAPL) for implementing ASBAC. In addition, the paper describes how ASBAC with SAPL can be implemented by applying a reactive programming model and describes key algorithms for evaluating SAPL policies.
Dominic Heutelbeck
### NoCry: No More Secure Encryption Keys for Cryptographic Ransomware
Abstract
Since the appearance of ransomware in the cyber crime scene, researchers and anti-malware companies have been offering solutions to mitigate the threat. Anti-malware solutions differ on the specific strategy they implement, and all have pros and cons. However, three requirements concern them all: their implementation must be secure, be effective, and be efficient. Recently, Genç et al. proposed to stop a specific class of ransomware, the cryptographically strong one, by blocking unauthorized calls to cryptographically secure pseudo-random number generators, which are required to build strong encryption keys. Here, in adherence to the requirements, we discuss an implementation of that solution that is more secure (with components that are not vulnerable to known attacks), more effective (with less false negatives in the class of ransomware addressed) and more efficient (with minimal false positive rate and negligible overhead) than the original, bringing its security and technological readiness to a higher level.
Ziya Alper Genç, Gabriele Lenzini, Peter Y. A. Ryan
### Security Requirements for Store-on-Client and Verify-on-Server Secure Biometric Authentication
Abstract
The Fast IDentity Online Universal Authentication Framework (FIDO UAF) is an online two-step authentication framework designed to prevent biometric information breaches from servers. In FIDO UAF, biometric authentication is firstly executed inside a user’s device, and then online device authentication follows. While there is no chance of biometric information leakage from the servers, risks remain when users’ devices are compromised. In addition, it may be possible to impersonate the user by skipping the biometric authentication step.
To design more secure schemes, this paper defines Store-on-Client and Verify-on-Server Secure Biometric Authentication (SCVS-SBA). Store-on-client means that the biometric information is stored in the devices as required for FIDO UAF, while verify-on-server is different from FIDO UAF, which implies that the result of biometric authentication is determined by the server. We formalize security requirements for SCVS-SBA into three definitions. The definitions guarantee resistance to impersonation attacks and credential guessing attacks, which are standard security requirements for authentication schemes. We consider different types of attackers according to the knowledge on the internal information.
We propose a practical concrete scheme toward SCVS-SBA, where normalized cross-correlation is used as the similarity measure for the biometric features. Experimental results show that a single authentication process takes only tens of milliseconds, which means that it is fast enough for practical use.
Haruna Higo, Toshiyuki Isshiki, Masahiro Nara, Satoshi Obana, Toshihiko Okamura, Hiroto Tamiya
### Reflexive Memory Authenticator: A Proposal for Effortless Renewable Biometrics
Abstract
Today’s biometric authentication systems are still struggling with replay attacks and irrevocable stolen credentials. This paper introduces a biometric protocol that addresses such vulnerabilities. The approach prevents identity theft by being based on memory creation biometrics. It takes inspiration from two different authentication methods, eye biometrics and challenge systems, as well as a novel biometric feature: the pupil memory effect. The approach can be adjusted for arbitrary levels of security, and credentials can be revoked at any point with no loss to the user. The paper includes an analysis of its security and performance, and shows how it could be deployed and improved.
Nikola K. Blanchard, Siargey Kachanovich, Ted Selker, Florentin Waligorski
### Collaborative Authentication Using Threshold Cryptography
Abstract
We propose a collaborative authentication protocol where multiple user devices (e.g., a smartphone, a smartwatch and a wristband) collaborate to authenticate the user to a third party service provider. Our protocol uses a threshold signature scheme as the main building block. The use of threshold signatures minimises the security threats in that the user devices only store shares of the signing key (i.e., the private key) and the private key is never reconstructed. For user devices that do not have secure storage capability (e.g., some wearables), we propose to use fuzzy extractors to generate their secret shares using behaviometric information when needed, so that there is no need for them to store any secret material. We discuss how to reshare the private key without reconstructing it in case a new device is added and how to repair shares that are lost due to device loss or damage. Our implementation results demonstrate the feasibility of the protocol.
Aysajan Abidin, Abdelrahaman Aly, Mustafa A. Mustafa
### MuFASA: A Tool for High-level Specification and Analysis of Multi-factor Authentication Protocols
Abstract
In recent years, the usage of online services (e.g., banking) has considerably increased. To protect the sensitive resources managed by these services against attackers, Multi-Factor Authentication (MFA) has been widely adopted. To date, a variety of MFA protocols have been implemented, leveraging different designs and features and providing a non-homogeneous level of security and user experience. Public and private authorities have defined laws and guidelines to guide the design of more secure and usable MFA protocols, but their influence on existing MFA implementations remains unclear.
We present MuFASA, a tool for high-level specification and analysis of MFA protocols, which aims at supporting normal users and security experts (in the design phase of an MFA protocol), providing a high level report regarding possible risks associated to the specified MFA protocol, its resistance to a set of attacker models (defined by NIST), its ease-of-use and its compliance with a set of security requirements derived from European laws.
Federico Sinigaglia, Roberto Carbone, Gabriele Costa, Silvio Ranise
### A Risk-Driven Model to Minimize the Effects of Human Factors on Smart Devices
Abstract
Human errors exploitation could entail unfavorable consequences to smart device users. Typically, smart devices provide multiple configurable features, e.g., user authentication settings, network selection, application installation, communication interfaces, etc., which users can configure according to their need and convenience. However, untrustworthy features configuration could mount severe risks towards the protection and integrity of data and assets residing on smart devices or to perform security-sensitive activities on smart devices. Conventional security mechanisms mainly focus on preventing and monitoring malware, but they do not perform the runtime vulnerabilities assessment while users use their smart devices. In this paper, we propose a risk-driven model that determines features reliability at runtime by monitoring users’ features usage patterns. The resource access permissions (e.g., ACCESS_INTERNET and ACCESS_NETWORK_STATE) given to an application requiring higher security are revoked in case users configure less reliable features (e.g., open WIFI or HOTSPOT) on their smart devices. Thus, our model dynamically fulfills the security criteria of the security-sensitive applications and revokes resources access permission given to them, until features reliability is set to a secure level. Consequently, smart devices are secured against any runtime vulnerabilities that may surface due to human factors.
Sandeep Gupta, Attaullah Buriro, Bruno Crispo
### A Formal Security Analysis of the Authentication Protocol for Decentralized Key Distribution and End-to-End Encrypted Email
Abstract
To send encrypted emails, users typically need to create and exchange keys which later should be manually authenticated, for instance, by comparing long strings of characters. These tasks are cumbersome for the average user. To make more accessible the use of encrypted email, a secure email application named $$p\equiv p$$ automates the key management operations; $$p\equiv p$$ still requires the users to carry out the verification, however, the authentication process is simple: users have to compare familiar words instead of strings of random characters, then the application shows the users what level of trust they have achieved via colored visual indicators. Yet, users may not execute the authentication ceremony as intended, $$p\equiv p$$ ’s trust rating may be wrongly assigned, or both. To learn whether $$p\equiv p$$ ’s trust ratings (and the corresponding visual indicators) are assigned consistently, we present a formal security analysis of $$p\equiv p$$ ’s authentication ceremony. From the software implementation in C, we derive the specifications of an abstract protocol for public key distribution, encryption and trust establishment; then, we model the protocol in a variant of the applied pi calculus and later formally verify and validate specific privacy and authentication properties. We also discuss alternative research directions that could enrich the analysis.
Itzel Vazquez Sandoval, Gabriele Lenzini
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2021-03-06 01:40:56
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https://klwu.co/demo/lattice-demo-2/
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# WORK IN PROGRESS
This is a demo of the LLL algorithm. You can check the debug box to step through the algorithm to see exactly how LLL works.
Set $$k=2$$.
While $$k\leq n$$
For $$j$$ from $$k-1$$ down to $$1$$
Set $$\vec{v}_k=\vec{v}_k-\lfloor \mu_{k,j} \rceil \vec{v}_j$$
EndFor
If $$\|\vec{v}_k^*\|^2\geq (\delta-\mu_{k,k-1}^2)\|v_{k-1}^*\|^2$$:
Set $$k=k+1$$
Else
Swap $$\vec{v}_k$$ and $$\vec{v}_{k-1}$$
Set $$k=\max(k-1,2)$$
EndIf
EndWhile
Output LLL-reduced basis $$\{\vec{v}_1,\cdots ,\vec{v}_n\}$$
Orthogonality Defect:
Number of lattice points rendered per direction:
LLL $$\delta$$:
k=; Current step:
Gram-Schmidt Coefficient Matrix:
Tags:
Updated:
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2022-10-02 04:16:13
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https://electronics.stackexchange.com/questions/295629/why-are-many-ir-receivers-in-metal-cages/295689
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# Why are many IR receivers in metal cages?
I'm guessing it's a Faraday cage around the receiver, but don't know why they might need one. Is there some sort of common interference around 38kHz (their operating frequency)?
It's the only component I think I've used that gets this special treatment. A larger cage may be around one in a VCR, and a little baby cage sometimes appears around the standalone PC mount component:
Thanks for your insight!
• I swear I've seen this question before – Voltage Spike Mar 30 '17 at 3:56
• Because it holds the lens down? – Ignacio Vazquez-Abrams Mar 30 '17 at 7:41
• IR receivers are the Hannibal Lecter of the electronics world. – user98663 Mar 30 '17 at 8:03
• The answer from @analogsystemsrf is interesting, but it could also not be a faraday cage at all, but rather a light filter to make the diode more omni-direction and less sensitive to swamping from head on signals. – Trevor_G Mar 30 '17 at 13:12
• Ignacio no, it doesn't... – Passerby Mar 31 '17 at 3:19
[ added 2_D resistor_grid methodology for exploring shielding topologies ]
You want that IR receiver to respond to photons, not to external electric fields. Yet the photodiode is a fine target for trash from fluorescent lights (200 volts in 10 microseconds) as the 4' tube has that restrike-the-arc action 120 times a second. [or 80,000 Hertz for some tubes]
Using the parallel-plate model of capacitance, $$C = E0*Er*Area/Distance$$ with diode area of 3mm*3mm and distance of 1 meter, the capacitance is $$9e-12Farad/meter * (ER=1 air) * 0.003*0.003/1$$ or ~~ 1e-11 * 1e-5 = 10^-16Farad
What current from a fluorescent light, at 20Million volts/second slewrate? $$I = C * dV/dT$$ or I = 1e-16Farad * 2e+7 Volt/second = 2nanoAmp
That ---- 2 nanoAmp ---- apparently is a big deal (the edge rate, 10 us, is close to 1/2 period of 38 kHz).
The metal cage protects by attenuating the Efield in an exponentially improving manner; thus the further the cage is in front of the photodiode, the more dramatic the Efield attenuation. Richard Feynman discusses this, in his 3-volume paperback on physics [I'll find a link, or at least a page #], in his lecture on Faraday cages and why the holes are acceptable IF the vulnerable circuits are spaced back several hole-diameters. [again, exponential improvement]
Are other Efield trash sources near? How about digitally noisy logic0 and logic1 for LED displays; 0.5 volts in 5 nanoseconds, or 10^8 volts/second(standard bouncing of "quiet" logic levels, as MCU program activity continues). How about a switching regulator, inside the TV; regulating off the ACrail, with 200 volts in 200 nanoseconds, or 1Billion volts/second, at 100 kHz rate.
At 1 billion volts/second, we have 100 nanoAmps aggressor currents. Of course, there should be no line-of-sight between a switchreg and the IR receiver, is there?
Line-of-sight does not matter. The Efields explore all possible paths, including up-and-back-down or around-corners.
simulate this circuit – Schematic created using CircuitLab
HINT TO BEHAVIOR: the Efields explore all possible paths.
================================================
From the master of clear-thinking himself, in his own words, I offer the explanation of Mr "Why did the space shuttle explode high over Cape Canaveral?", the gleeful Dr. Richard Feynman.
He provided a 2 year introduction to physics at Caltech, approximately 1962. His lectures were transcribed, very carefully to serve as reference material, [its worth getting these 3, and re-reading them every 5 years; also, the curious teenager will savor the realworld discussions in Feynman's style] and published in 3 paperback volumes as "The Feynman Lectures on Physics". From Volume II, focused on "mainly electromagnetism and matter", we turn to Chapter 7 "The Electric Field in Various Circumstances: Continued", and on page 7-10 and 7-11, he presents "The Electrostatic Field of a Grid".
Feynman describes a infinite grid of infinitely long wires, with wire-wire spacing of 'a'. He starts with equations [introduced in Volume 1, Chapt 50 Harmonics] that will approximate the field, with more and more terms optionally usable to achieve greater and greater accuracy. The variable 'n' tells us the order of the term. We can start with "n = 1".
Here is the summary equation, where 'a' is the spacing between grid wires:
$$Fn = An * e^-Z/Zo$$ where Zo is $$Zo = a/(2*pi*n)$$
At distance Z = a above the grid, thus we are 3mm above a grid spaced 3mm, and using only the "n = 1" part of the solution, we have $$Fn = An * e^-(2 * pi * 1 * 3mm)/3mm$$
Since this Fn is e^-6.28 smaller than An, we have rapid attenuation of the external electric field.
With 2.718^2.3 = 10, 2.718^4.6 = 100, 2.718^6.9 = 1000, then e^-6.28 is about 1/500. ( 1/533, from a calculator)
Our external field of An has been reduced by 1/500, to 0.2% or 54dB weaker, 3mm inside a grid spaced at 3mm. How does Feynman summarize his thinking?
"The method we have just developed can be used to explain why electrostatic shielding by means of a screen is often just as good as with a solid metal sheet. Except within a distance from the screen a few times the spacing of the screen wires, the fields inside a closed screen are zero. We see why copper screen---lighter and cheaper than copper sheet---is often used to shield sensitive electrical equipment from external disturbing fields." (end quote)
Should you seek a 24 bit embedded system, you need 24*6 = 144dB attenuation; at 54dB per unit_spacing, you need to be 3*wire-wire spacing, behind the grid. For a 32 bit system, that becomes 32*6 = 192 dB, or nearly 4*wire-wire spacing, behind the grid.
Caveat: this is electrostatics. Fast Efields cause transient currents in the grid wires. Your mileage will vary.
Notice we only used the "a = 1" part of the solution; can we ignore the additional parts of the harmonic/series solution? Yes. With "n = 2", we get the attenuation * attenuation, and "n = 3" yields atten * atten * atten.
=================================================
EDIT To model more common mechanical structures, to determine the ultimate trash levels as an Efield couples into a circuit, we need to know (1) the impedance of the circuit at the aggressor frequency, and (2) the coupling from a 3_D trash aggressor to a 3_D signal chain node. For simplicity, we'll model this in 2_D, using the available grid_of_resistors
simulate this circuit
• Am guessing the center pin is gnd, which would extend inside to support the chip substrate. Would that not be shield-enough? Am also suspicious that the frame "X" blocks the head-on optical path...could it be an optical diffuser ? – glen_geek Mar 30 '17 at 16:43
• Thank you for the mathematically complete response, the good explanation, and the delightful drawing of marauding electric fields! – R Zach Mar 31 '17 at 4:57
• For successful embedded systems, all the interferers should be identified and quantified, so risks are known up front. In building tools to do this identify/quantify, I work with these issues every day. I watched a team self-destruct, as they ignored feedback risks in an IR receiver. Whether on PCB or on silicon, the need to attenuate trash by 100dB or 150dB is often there. Without identifying and quantifying the phenomena, its just punt-and-hope. To decide to use extra layers, or extra PCB space, or 10 more pins on silicon, one needs good cause. Extreme fidelity requires attention. – analogsystemsrf Mar 31 '17 at 14:06
• +1 For referencing and quoting "The Feynman Lectures on Physics" – jose.angel.jimenez Apr 5 '17 at 8:45
The answer is quite simple.
When the PD is receiving a small signal at greatest distance, the PD may only be receiving <1uA and thus even with 60 dB gain with AGC IR Rx has >1MΩ impedance making it sensitive to stray E-fields picked up the the area of of the detector and wires.
Shielding it on the outside may compare well to Sharp/Vishay's shielding on the inside, but shielding is necessary due to the high impedance to extend the range of detection to perhaps 50m using the appropriate IR 5mm emitter by shunting stray E-fields.
One can tell it is IR due to the daylight blocking filter and 3 pins needed for the integrated BPF AGC and ASK detector.
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2021-01-19 04:25:38
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https://www.physicsforums.com/threads/fluid-dynamics-flow-rate.232664/
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# Fluid Dynamics Flow Rate
fredrick08
## Homework Statement
a 4mm hole is 1m below the surface of 2m diameter water tank.
a.what is the volume flow rate through the hole in L/min?
b.what is the rate in mm/min at which the water level in the tank will drop if the water is not replenished.
[PLAIN]http://i285.photobucket.com/albums/ll55/tebsa08/tank.jpg[/PLAIN]
## Homework Equations
Q=vA
v1A1=v2A2
V1=A1$$\Delta$$(x1)=v1A1$$\Delta$$(t)
V2=A2$$\Delta$$(x2)=v2A2$$\Delta$$(t)
p+.5$$\rho$$v1^2+$$\rho$$gy=constant
## The Attempt at a Solution
a. A1=$$\pi$$r^2=$$\pi$$*1^2=3.14m^2, A2=$$\pi$$r^2=$$\pi$$*.002^2=1.25x10^-5m^2
im sorry i really don't understand fluid dynamics, i have no idea how to find v, to get Q... ans part b, i don't even undestand the question... please someone help me,
can i find v by, $$\rho$$+$$\rho$$gy=$$\rho$$+.5$$\rho$$v^2
and all the densities cancel so v=$$\sqrt{}2gy$$?
Last edited by a moderator:
## Answers and Replies
fredrick08
i think V1=A1*x=3.14*1=3.14m^3??
fredrick08
btw the answers are 3.3L/min and 1.06mm/min
fredrick08
can anyone help??
Homework Helper
Hi fredrick08,
can i find v by, $$\rho$$+$$\rho$$gy=$$\rho$$+.5$$\rho$$v^2
That's the right idea, but this isn't written quite correctly. It needs to be:
$$P_1 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2$$
after the simplifications you have already made (v_1=0 and h_2=0). What do you get?
fredrick08
huh? sorry i don't understand? if v1=0 then doesn't v2=0 aswell?
coz v2=$$\sqrt{}v1$$??
and how does v1 and h2 both equal 0? please explain??
Homework Helper
Didn't you already set v1=0? Bernoulli's equation applied to two points is:
$$P_1 + \rho g h_1+\frac{1}{2}\rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$$
but your equation only had one velocity and one height.
But it does make sense. Point 1 is at the top of the tank, and point 2 is where the water comes out. So v1 is the speed at which the water level is dropping at the top of the tank, and v2 is the speed at which the water is leaving. Since v1 is very small (the overall water level drops very slowly since the hole is so small), it can be neglected here.
(Since they give you the diameter of the entire water tank, you don't need to neglect v1 in Bernoulli's equation; but it should not make a practical difference in v2 whether you include it or not. If you neglect v1, you can get v2 directly from Bernoulli's equation; if you include v1, you'll have to solve Bernoulli's and the continuity equation together.)
As for h2, it acts just like the regular potential energy mgh where we had the freedom to set any height we wished equal to zero. You can either set the height h2 to be zero at the hole (and then h1 would be +1) or set h1 to be zero (and then h2 would be -1).
fredrick08
oh yes, ty = ) that makes perfect sense... duh... thankyou, i was getting confused coz i needed to draw and properly label a diagram, now i can see wat i trying to find lol...
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2023-01-27 21:24:10
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https://stats.stackexchange.com/questions/375945/which-coeffcients-are-used-for-a-f-statistic
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# Which coeffcients are used for a F-statistic
I have an F-statistic, $$F(4,10)$$, my constant and 4 Coefficients $$\beta_2 , \beta_3 , \beta_4$$ and $$\beta_5$$
I already know that the (in this case) 10 reflects the number of obsverations. But what exactly does the 4 tell me? Is it the number of coefficients to be tested? Also what does the outcome - $$F(4,10) = 9.59$$ tell me?
Thank you in advance.
With $$F$$-test in multiple linear regression, you test the $$H_0$$ that a "trivial" model $$y = \beta_1 + u$$ (i.e. dependent variable as a function of intercept and random error only) is equally good in predicting variability in the dependent variable as your "main" model $$y = \beta_1 + \beta_2 x_2 + \beta_3 x_3 + \beta_4 x_4 + \beta_5 x_5 + u$$.
Basically, you test the significance of the model as a whole. In $$F(k, (n-k-1))$$, parameter $$k$$ is the number of parameters restricted (4), and $$n-k-1$$ is the degrees of freedom in multiple linear regression - i.e. you are estimating your model with 15 observations only, which is far too low for 5 parameter-estimation.
Also, the critical value for $$F(4,10)=2.6$$ (at 5% sig. level), so you are fine (reject $$H_0$$ in favour of significant model). However, let me repeat that estimating 5 coeffs using 15 observations is generally not a good idea.
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2021-05-11 16:49:20
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https://tbc-python.fossee.in/convert-notebook/Theory_Of_Machines/ch13.ipynb
|
# Chapter 13 : Gear Trains¶
## Example 13.1 Page No : 432¶
In [1]:
# Variables:
NA = 975 #rpm
TA = 20.
TB = 50.
TC = 25
TD = 75
TE = 26
TF = 65
#Solution:
#Calculating the speed of gear F
NF = NA*(TA*TC*TE)/(TB*TD*TF) #rpm
#Results:
print " Speed of gear F, NF = %d rpm."%(NF)
Speed of gear F, NF = 52 rpm.
## Example 13.2 Page No : 433¶
In [2]:
import math
from numpy import linalg
# Variables:
x = 600.
pc = 25. #mm
N1 = 360.
N2 = 120. #rpm
#Solution:
#Calculating the pitch circle diameters of each gear
#Speed ratio N1/N2 = d2/d1 or N1*d1-N2*d2 = 0 .....(i)
#Centre distance between the shafts x = 1/2*(d1+d2) or d1+d2 = 600*2 .....(ii)
A = [[N1, -N2],[ 1, 1]]
B = [0, 600*2]
V = linalg.solve(A,B)
d1 = V[0] #mm
d2 = V[1] #mm
#Calculating the number of teeth on the first gear
T1 = round(math.pi*d1/pc)
#Calculating the number of teeth on the second gear
T2 = int(math.pi*d2/pc+1)
#Calculating the pitch circle diameter of the first gear
d1dash = T1*pc/math.pi #mm
#Calculating the pitch circle diameter of the second gear
d2dash = T2*pc/math.pi #mm
#Calculating the exact distance between the two shafts
xdash = (d1dash+d2dash)/2 #mm
#Results:
print " The number of teeth on the first and second gear must be %d and %d and their pitch\
circle diameters must be %.2f mm and %.1f mm respectively."%(T1,T2,d1dash,d2dash)
print " The exact distance between the two shafts must be %.2f mm."%(xdash)
The number of teeth on the first and second gear must be 38 and 114 and their pitch circle diameters must be 302.39 mm and 907.2 mm respectively.
The exact distance between the two shafts must be 604.79 mm.
## Example 13.3 Page No : 435¶
In [4]:
from numpy import linalg
import math
# Variables:
rAD = 12. #Speed ratio NA/ND
mA = 3.125 #mm
mB = mA #mm
mC = 2.5 #mm
mD = mC #mm
x = 200. #mm
#Solution:
#Calculating the speed ratio between the gears A and B and C and D
rAB = math.sqrt(rAD) #Speed ratio between the gears A and B
rCD = math.sqrt(rAB) #Speed ratio between the gears C and D
#Calculating the ratio of teeth on gear B to gear A
rtBA = rAB #Ratio of teeth on gear B to gear A
#Calculating the ratio of teeth on gear D to gear C
rtDC = rCD #Ratio of teeth on gear D to gear C
#Calculating the number of teeth on the gears A and B
#Distance between the shafts x = mA*TA/2+mB*TB/2 or (mA/2)*TA+(mB/2)*TB = x .....(i)
#Ratio of teeth on gear B to gear A TB/TA = math.sqrt(12) or math.sqrt(12)*TA-TB = 0 .....(ii)
A = [[mA/2, mB/2],[math.sqrt(12) ,-1]]
B = [x, 0]
V = linalg.solve(A,B)
TA = int(V[0])
TB = round(V[1])
#Calculating the number of teeth on the gears C and D
#Dismath.tance between the shafts x = mC*TC/2+mD*TD/2 or (mC/2)*TC+(mD/2)*TD = x .....(iii)
#Ratio of teeth on gear D to gear C TD/TC = math.sqrt(12) or math.sqrt(12)*TC-TD = 0 .....(iv)
A = [[mC/2, mD/2],[ math.sqrt(12) ,-1]]
B = [x, 0]
V = linalg.solve(A,B)
TC = round(V[0])
TD = int(V[1])
#Results:
print " Number of teeth on gear A, TA = %d."%(TA)
print " Number of teeth on gear B, TB = %d."%(TB)
print " Number of teeth on gear C, TC = %d."%(TC)
print " Number of teeth on gear D, TD = %d."%(TD)
Number of teeth on gear A, TA = 28.
Number of teeth on gear B, TB = 99.
Number of teeth on gear C, TC = 36.
Number of teeth on gear D, TD = 124.
## Example 13.4 Page No : 438¶
In [5]:
import math
# Variables:
TA = 36.
TB = 45.
NC = 150. #rpm anticlockwise
#Solution:
#Refer Fig. 13.7
#Algebraic method:
#Calculating the speed of gear B when gear A is fixed
NA = 0.
NC = 150. #rpm
NB1 = (-TA/TB)*(NA-NC)+NC #rpm
#Calculating the speed of gear B when gear A makes 300 rpm clockwise
NA = -300. #rpm
NB2 = (-TA/TB)*(NA-NC)+NC #rpm
#Results:
print " Speed of gear B when gear A is fixed, NB = %d rpm."%(NB1)
print " Speed of gear B when gear A makes 300 rpm clockwise, NB = %d rpm."%(NB2)
Speed of gear B when gear A is fixed, NB = 270 rpm.
Speed of gear B when gear A makes 300 rpm clockwise, NB = 510 rpm.
## Example 13.5 Page No : 440¶
In [6]:
import math
# Variables:
TB = 75.
TC = 30.
TD = 90.
NA = 100. #rpm clockwise
#Solution:
#Refer Table 13.3
#Calculating the number of teeth on gear E
TE = TC+TD-TB
#Calculating the speed of gear C
y = -100.
x = y*(TB/TE)
NC = y-x*(TD/TC) #rpm
#Results:
print " Speed of gear C, NC = %d rpm, anticlockwise."%(NC)
Speed of gear C, NC = 400 rpm, anticlockwise.
## Example 13.6 Page No : 443¶
In [7]:
import math
# Variables:
TA = 72.
TC = 32.
NEF = 18. #Speed of arm EF rpm
#Solution:
#Refer Table 13.5
#Speed of gear C:
y = 18. #rpm
x = y*(TA/TC)
NC = x+y #Speed of gear C rpm
#Speed of gear B:
#Calculating the number of teeth on gear B
TB = (TA-TC)/2
#Calculating the speed of gear B
NB = y-x*(TC/TB) #Speed of gear B rpm
#Solution:
print " Speed of gear C = %.1f rpm."%(NC)
print " Speed of gear B = %.1f rpm in the opposite direction of arm."%(-NB)
Speed of gear C = 58.5 rpm.
Speed of gear B = 46.8 rpm in the opposite direction of arm.
## Example 13.7 Page No : 444¶
In [9]:
from numpy import linalg
import math
# Variables:
TA = 40.
TD = 90.
#Solution:
#Calculating the number of teeth on gears B and C
#From geometry of the Fig. 13.11 dA+2*dB = dD
#Since the number of teeth are proportional to their pitch circle diameters
TB = (TD-TA)/2
TC = TB
#Refer Table 13.6
#Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise:
#Calculating the values of x and y
#From the fourth row of the table -x-y = -1 or x+y = 1 .....(i)
#The gear D makes half revolution anticlockwise i.e. x*(TA/TD)-y = 1/2 .....(ii)
A = [[1, 1],[TA/TD, -1]]
B = [1, 1./2]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the speed of arm
varm = -y #Speed of arm revolutions
#Results:
print " Speed of arm when A makes 1 revolution clockwise and D makes half revolution\
anticlockwise = %.2f revolution anticlockwise."%(varm)
#Speed of arm when A makes 1 revolution clockwise and D is stationary:
#Calculating the values of x and y
#From the fourth row of the table -x-y = -1 or x+y = 1 .....(iii)
#The gear D is stationary i.e. x*(TA/TD)-y = 0 .....(iv)
A = [[1, 1],[ TA/TD, -1]]
B = [1, 0]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the speed of arm
varm = -y #Speed of arm revolutions
#Results:
print " Speed of arm when A makes 1 revolution clockwise and D is stationary = %.3f revolution\
clockwise."%(-varm)
Speed of arm when A makes 1 revolution clockwise and D makes half revolution anticlockwise = 0.04 revolution anticlockwise.
Speed of arm when A makes 1 revolution clockwise and D is stationary = 0.308 revolution clockwise.
## Example 13.8 Page No : 446¶
In [10]:
import math
# Variables:
TC = 28.
TD = 26.
TE = 18.
TF = TE
#Solution:
#The sketch is as in Fig. 13.12
#Number of teeth on wheels A and B:
#From geometry dA = dC+2*dE and dB = dD+2*dF
#Since the number of teeth are proportional to their pitch circle diameters
TA = TC+2*TE
TB = TD+2*TF
#Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed:
#Since the arm G makes 100 rpm clockwise therefore from the fourth row of Table 13.7
y = -100
x = -y
#Calculating the speed of wheel B
NB1 = y+x*(TA/TC)*(TD/TB) #Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed rpm
#Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter clockwise:
#Since the arm G makes 100 rpm clockwise therefore from the fourth row of Table 13.7
y = -100
x = 10-y
#Calculating the speed of wheel B
NB2 = y+x*(TA/TC)*(TD/TB) #Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter clockwise rpm
#Solution:
print " Number of teeth on wheel A, TA = %d."%(TA)
print " Number of teeth on wheel B, TB = %d."%(TB)
print " Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed = %.1f rpm, clockwise."%(-NB1)
print " Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter\
clockwise = %.1f rpm, counter clockwise."%(NB2)
Number of teeth on wheel A, TA = 64.
Number of teeth on wheel B, TB = 62.
Speed of wheel B when arm G makes 100 rpm clockwise and wheel A is fixed = 4.1 rpm, clockwise.
Speed of wheel B when arm G makes 100 rpm clockwise and wheel A makes 10 rpm counter clockwise = 5.4 rpm, counter clockwise.
## Example 13.9 Page No : 447¶
In [1]:
import math
# Variables:
dD = 224.
m = 4. #mm
#Solution:
#Refer Table 13.8
#Calculating the values of x and y
y = 1.
x = 5-y
#Calculating the number of teeth on gear D
TD = dD/m
#Calculating the number of teeth on gear B
TB = y/x*TD
#Calculating the number of teeth on gear C
TC = (TD-TB)/2
#Results:
print " Number of teeth on gear D, TD = %d."%(TD)
print " Number of teeth on gear B, TB = %d."%(TB)
print " Number of teeth on gear C, TC = %d."%(TC)
Number of teeth on gear D, TD = 56.
Number of teeth on gear B, TB = 14.
Number of teeth on gear C, TC = 21.
## Example 13.10 Page No : 448¶
In [13]:
from numpy import linalg
import math
# Variables:
TC = 50.
TD = 20.
TE = 35.
NA = 110. #rpm
#Solution:
#Calculating the number of teeth on internal gear G
TG = TC+TD+TE
#Speed of shaft B:
#Calculating the values of x and y
#From the fourth row of Table 13.9
#y-x*(TC/TD)*(TE/TG) = 0 .....(i)
#Also x+y = 110 or y+x = 110 .....(ii)
A = [[1, -(TC/TD)*(TE/TG)],[ 1, 1]]
B = [0, 110]
V = linalg.solve(A,B)
x = V[1]
y = V[0]
#Calculating the speed of shaft B
NB = round(+y) #Speed of shaft B rpm
#Results:
print " Number of teeth on internal gear G, TG = %d."%(TG)
print " Speed of shaft B = %d rpm, anticlockwise."%(NB)
Number of teeth on internal gear G, TG = 105.
Speed of shaft B = 50 rpm, anticlockwise.
## Example 13.11 Page No : 450¶
In [14]:
import math
# Variables:
TA = 12.
TB = 30.
TC = 14.
NA = 1.
ND = 5. #rps
#Solution:
#Number of teeth on wheels D and E:
#Calculating the number of teeth on wheel E
TE = TA+2*TB
#Calculating the number of teeth on wheel E
TD = TE-(TB-TC)
#Magnitude and direction of angular velocities of arm OP and wheel E:
#Calculating the values of x and y
#From the fourth row of Table 13.10 -x-y = -1 or x+y = 1 .....(i)
#Also x*(TA/TB)*(TC/TD)-y = 5 .....(ii)
A = [[1, 1],[(TA/TB)*(TC/TD) ,-1]]
B = [1, 5]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the angular velocity of arm OP
omegaOP = -y*2*math.pi #Angular velocity of arm OP rad/s
#Calculating the angular velocity of wheel E
omegaE = (x*TA/TE-y)*2*math.pi #Angular velocity of wheel E rad/s
#Results:
print " Number of teeth on wheel E, TE = %d."%(TE)
print " Number of teeth on wheel D, TD = %d."%(TD)
print " Angular velocity of arm OP = %.3f rad/s, counter clockwise."%(omegaOP)
print " Angular velocity of wheel E = %.2f rad/s, counter clockwise."%(omegaE)
Number of teeth on wheel E, TE = 72.
Number of teeth on wheel D, TD = 56.
Angular velocity of arm OP = 27.989 rad/s, counter clockwise.
Angular velocity of wheel E = 33.70 rad/s, counter clockwise.
## Example 13.12 Page No : 451¶
In [15]:
import math
# Variables:
TB = 80.
TC = 82.
TD = 28.
NA = 500. #rpm
#Solution:
#Calculating the number of teeth on wheel E
TE = TB+TD-TC
#Calculating the values of x and y
y = 800.
x = -y*(TE/TB)*(TC/TD)
#Calculating the speed of shaft F
NF = x+y #Speed of shaft F rpm
#Results:
print " Speed of shaft F = %d rpm, anticlockwise."%(NF)
Speed of shaft F = 38 rpm, anticlockwise.
## Example 13.13 Page No : 452¶
In [11]:
# variables
TA = 100. ; # Gear A teeth
TC = 101. ; # Gear C teeth
TD = 99. ; # Gear D teeth
TP = 20. # Gear planet teeth
y = 1
x = 0 - y
# calculations
NC = y + x * TA/TC
ND = y + x * TA/TD
# results
print "revolutions of gear C : %.4f"%NC
print "revolutions of gear D : %.4f "%ND
revolutions of gear C : 0.0099
revolutions of gear D : -0.0101
## Example 13.14 Page No : 453¶
In [3]:
from numpy import linalg
import math
# Variables:
NA = 300. #rpm
TD = 40.
TE = 30.
TF = 50.
TG = 80.
TH = 40.
TK = 20.
TL = 30.
#Solution:
#Refer Fig. 13.18 and Table 13.13
#Calculating the speed of wheel E
NE = NA*(TD/TE) #rpm
#Calculating the number of teeth on wheel C
TC = TH+TK+TL
#Speed and direction of rotation of shaft B:
#Calculating the values of x and y
#We have -x-y = -400 or x+y = 400 .....(i)
#Also x*(TH/TK)*(TL/TC)-y = 0 .....(ii)
A = [[1, 1],[ (TH/TK)*(TL/TC), -1]]
B = [400, 0]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the speed of wheel F
NF = -y #rpm
#Calculating the speed of shaft B
NB = -NF*(TF/TG) #Speed of shaft B rpm
#Results:
print " Number of teeth on wheel C, TC = %.1f."%(TC)
print " Speed of shaft B = %.1f rpm, anticlockwise."%(NB)
Number of teeth on wheel C, TC = 90.0.
Speed of shaft B = 100.0 rpm, anticlockwise.
## Example 13.15 Page No : 455¶
In [17]:
from numpy import linalg
import math
# Variables:
T1 = 80.
T8 = 160.
T4 = 100.
T3 = 120.
T6 = 20.
T7 = 66.
#Solution:
#Refer Fig. 13.19 and Table 13.14
#Calculating the number of teeth on wheel 2
T2 = (T3-T1)/2
#Calculating the values of x and y
#Assuming that wheel 1 makes 1 rps anticlockwise x+y = 1 .....(i)
#Also y-x*(T1/T3) = 0 or x*(T1/T3)-y = 0 .....(ii)
A = [[1, 1],[ 1, T1/T3]]
B = [1, 0]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the speed of casing C
NC = y #Speed of casing C rps
#Calculating the speed of wheel 2
N2 = y-x*(T1/T2) #Speed of wheel 2 rps
#Calculating the number of teeth on wheel 5
T5 = (T4-T6)/2
#Calculating the values of x1 and y1
y1 = -2
x1 = (y1-0.4)*(T4/T6)
#Calculating the speed of wheel 6
N6 = x1+y1 #Speed of wheel 6 rps
#Calculating the values of x2 and y2
y2 = 0.4
x2 = -(14+y2)*(T7/T8)
#Calculating the speed of wheel 8
N8 = x2+y2 #Speed of wheel 8 rps
#Calculating the velocity ratio of the output shaft B to the input shaft A
vr = N8/1 #Velocity ratio
#Results:
print " Velocity ratio of the output shaft B to the input shaft A = %.2f."%(vr)
Velocity ratio of the output shaft B to the input shaft A = -5.54.
## Example 13.16 Page No : 459¶
In [18]:
import math
# Variables:
TA = 40.
TB = 30.
TC = 50.
NX = 100.
NA = NX #rpm
Narm = 100. #Speed of armrpm
#Solution:
#Refer Fig. 13.22 and Table 13.18
#Calculating the values of x and y
y = +100
x = -100-y
#Calculating the speed of the driven shaft
NY = y-x*(TA/TB) #rpm
#Results:
print " Speed of the driven shaft, NY = %.1f rpm, anticlockwise."%(NY)
Speed of the driven shaft, NY = 366.7 rpm, anticlockwise.
## Example 13.17 Page No : 460¶
In [19]:
from numpy import linalg
import math
# Variables:
TB = 20.
TC = 80.
TD = 80.
TE = 30.
TF = 32.
NB = 1000. #rpm
#Solution:
#Refer Fig. 13.23 and Table 13.19
#Speed of the output shaft when gear C is fixed:
#Calculating the values of x and y
#From the fourth row of the table y-x*(TB/TC) = 0 .....(i)
#Also x+y = +1000 or y+x = 1000 .....(ii)
A = [[1, -TB/TC],[ 1, 1]]
B = [0, 1000]
V = linalg.solve(A,B)
x = V[1]
y = V[0]
#Calculating the speed of output shaft
NF1 = y-x*(TB/TD)*(TE/TF) #Speed of the output shaft when gear C is fixed rpm
#Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise:
#Calculating the values of x and y
#From the fourth row of te table y-x*(TB/TC) = +10 .....(iii)
#Also x+y = +1000 or y+x = 1000 .....(iv)
A = [[1, -TB/TC],[1, 1]]
B = [10, 1000]
V = linalg.solve(A,B)
x = V[1]
y = V[0]
#Calculating the speed of output shaft
NF2 = y-x*(TB/TD)*(TE/TF) #Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise rpm
#Results:
print " Speed of the output shaft when gear C is fixed = %.1f rpm, counter clockwise."%(NF1)
print " Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise = %.1f rpm, \
counter clockwise."%(NF2)
Speed of the output shaft when gear C is fixed = 12.5 rpm, counter clockwise.
Speed of the output shaft when gear C is rotated at 10 rpm counter clockwise = 22.4 rpm, counter clockwise.
## Example 13.18 Page No : 461¶
In [20]:
import math
# Variables:
TA = 10.
TB = 60.
NA = 1000.
NQ = 210.
ND = NQ #rpm
#Solution:
#Refer Fig. 13.24 and Table 13.20
#Calculating the speed of crown gear B
NB = NA*(TA/TB) #rpm
#Calculating the values of x and y
y = 200.
x = y-210.
#Calculating the speed of road wheel attached to axle P
NC = x+y #Speed of road wheel attached to axle P rpm
#Results:
print " Speed of road wheel attached to axle P = %d rpm."%(NC)
Speed of road wheel attached to axle P = 190 rpm.
## Example 13.19 Page No : 463¶
In [5]:
from numpy import linalg
import math
# Variables:
TA = 15.
TB = 20.
TC = 15.
NA = 1000. #rpm
Tm = 100. #Torque developed by motor N-m
#Solution:
#Refer Fig. 13.26 and Table 13.21
#Calculating the number of teeth on gears E and D
TE = TA+2*TB
TD = TE-(TB-TC)
#Speed of the machine shaft:
#From the fourth row of the table x+y = 1000 or y+x = 1000 .....(i)
#Also y-x*(TA/TE) = 0 .....(ii)
A = [[1, 1],[1, -TA/TE]]
B = [1000, 0]
V = linalg.solve(A,B)
y = round(V[0])
x = round(V[1])
#Calculating the speed of machine shaft
ND = y-x*(TA/TB)*(TC/TD) #rpm
#Calculating the torque exerted on the machine shaft
Ts = Tm*NA/ND #Torque exerted on the machine shaft N-m
#Results:
print " Speed of machine shaft, ND = %.2f rpm, anticlockwise."%(ND)
print " Torque exerted on the machine shaft = %.f N-m."%(Ts)
Speed of machine shaft, ND = 37.15 rpm, anticlockwise.
Torque exerted on the machine shaft = 2692 N-m.
## Example 13.20 Page No : 465¶
In [23]:
import math
# Variables:
Ts = 100 #Torque on the sun wheel N-m
r = 5 #Ratio of speeds of gear S to C NS/NC
#Refer Fig. 13.27 and Table 13.22
#Number of teeth on different wheels:
#Calculating the values of x and y
y = 1.
x = 5-y
#Calculating the number of teeth on wheel E
TS = 16.
TE = 4*TS
#Calculating the number of teeth on wheel P
TP = (TE-TS)/2
#Torque necessary to keep the internal gear stationary:
Tc = Ts*r #Torque on CN-m
#Caluclating the torque necessary to keep the internal gear stationary
Ti = Tc-Ts #Torque necessary to keep the internal gear stationary N-m
#Results:
print " Number of teeth on different wheels, TE = %d."%(TE)
print " Torque necessary to keep the internal gear stationary = %d N-m."%(Ti)
Number of teeth on different wheels, TE = 64.
Torque necessary to keep the internal gear stationary = 400 N-m.
## Example 13.21 Page No : 466¶
In [9]:
import math
from numpy import linalg
# Variables:
TA = 14.
TC = 100.
r = 98./41 #TE/TD
PA = 1.85*1000 #W
NA = 1200. #rpm
TB = 43
#Solution:
#Refer Fig. 13.28 and Table 13.23
#Calculating the number of teeth on wheel B TB = (TC-TA)/2
#Calculating the values of x and y
#From the fourth row of the table -y+x*(TA/TC) = 0 or x*(TA/TC)-y = 0 .....(i)
#Also x-y = 1200 or x+y = -1200 .....(ii)
A = [[TA/TC, -1],[ 1, 1]]
B = [0, -1200]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the speed of gear E
NE = round(-y+x*(TA/TB)*(1./r)) #rpm
#Fixing torque required at C:
#Calculating the torque on A
Ta = PA*60./(2*math.pi*NA) #Torque on A N-m
#Calculating the torque on E
Te = PA*60./(2*math.pi*NE) #Torque on E
#Calculating the fixing torque required at C
Tc = Te-Ta #Fixing torque at C N-m
#Results:
print " Speed and direction of rotation of gear E, NE = %d rpm, anticlockwise."%(NE)
print " Fixing torque required at C = %.1f N-m."%(Tc)
Speed and direction of rotation of gear E, NE = 4 rpm, anticlockwise.
Fixing torque required at C = 4401.8 N-m.
## Example 13.22 Page No : 468¶
In [25]:
from numpy import linalg
import math
# Variables:
TB = 15.
TA = 60.
TC = 20.
omegaY = 740.
P = 130*1000. #W
#Solution:
#Refer Fig. 13.29 and Table 13.24
#Calculating the number of teeth on wheel D
TD = TA-(TC+TB)
#Calculating the values of x and y
#From the fourth row of the table y-x*(TD/TC)*(TB/TA) = 740 .....(i)
#Also x+y = 0 or y+x = 0 .....(ii)
A = [[1, -(TD/TC)*(TB/TA)],[ 1, 1]]
B = [740, 0]
V = linalg.solve(A,B)
x = V[1]
y = V[0]
#Calculating the speed of shaft X
#Holding torque on wheel D:
#Calculating the torque on A
Ta = P/omegaA #Torque on A N-m
#Calculating the torque on X
Tx = P/omegaX #Torque on X N-m
#Calculating the holding torque on wheel D
Td = Tx-Ta #Holding torque on wheel D N-m
#Results:
print " Speed of shaft X, omegaX = %.1f rad/s."%(omegaX)
print " Holding torque on wheel D = %.1f N-m."%(Td)
Speed of shaft X, omegaX = 563.8 rad/s.
Holding torque on wheel D = 54.9 N-m.
## Example 13.23 Page No : 469¶
In [26]:
from numpy import linalg
import math
# Variables:
TP = 144.
TQ = 120.
TR = 120.
TX = 36.
TY = 24.
TZ = 30.
NI = 1500. #rpm
P = 7.5*1000 #W
eta = 0.8
#Solution:
#Refer Fig. 13.30 and Table 13.25
#Calculating the values of x and y
#From the fourth row of the table x+y = -1500 .....(i)
#Also y-x*(TZ/TR) = 0 or -x*(TZ/TR)+y = 0 .....(ii)
A = [[1, 1],[-TZ/TR, 1]]
B = [-1500, 0]
V = linalg.solve(A,B)
x = V[0]
y = V[1]
#Calculating the values of x1 and y1
#We have y1-x1*(TY/TQ) = y .....(iii)
#Also x1+y1 = x+y or y1+x1 = x+y .....(iv)
A = [[1, -TY/TQ],[ 1, 1]]
B = [y, x+y]
V = linalg.solve(A , B)
x1 = V[1]
y1 = V[0]
#Speed and direction of the driven shaft O and the wheel P:
#Calculating the speed of shaft O
NO = y1 #rpm
#Calculating the speed of wheel P
NP = y1+x1*(TY/TQ)*(TX/TP) #rpm
#Torque tending to rotate the fixed wheel R:
#Calculating the torque on shaft I
T1 = P*60/(2*math.pi*NI) #N-m
#Calculating the torque on shaft O
T2 = eta*P*60/(2*math.pi*(-NO)) #N-m
#Calculating the torque tending to rotate the fixed wheel R
T = T2-T1 #Torque tending to rotate the fixed wheel R N-m
#Results:
print " Speed of the driven shaft O, NO = %d rpm, clockwise."%(-NO)
print " Speed of the wheel P, NP = %d rpm, clockwise."%(-NP)
print " Torque tending to rotate the fixed wheel R = %.2f N-m."%(T)
Speed of the driven shaft O, NO = 500 rpm, clockwise.
Speed of the wheel P, NP = 550 rpm, clockwise.
Torque tending to rotate the fixed wheel R = 66.85 N-m.
## Example 13.24 Page No : 471¶
In [6]:
from numpy import linalg
import math
# Variables:
TA = 34.
TB = 120.
TC = 150.
TD = 38.
TE = 50.
PX = 7.5*1000 #W
NX = 500. #rpm
m = 3.5 #mm
#Solution:
#Refer Fig. 13.31 and Table 13.27
#Output torque of shaft Y:
#Calculating the values of x and y
#From the fourth row of the table x+y = 500 or y+x = 500 .....(i)
#Alsoy-x*(TA/TC) = 0 .....(ii)
A = [[1, 1],[ 1, -TA/TC]]
B = [500, 0]
V = linalg.solve(A, B)
y = round(V[0],1) #rpm
x = round(V[1],1) #rpm
#Calculating the speed of output shaft Y
NY = y-x*(TA/TB)*(TD/TE) #rpm
#Calculating the speed of wheel E
NE = NY #rpm
#Calculating the input power assuming 100 per cent efficiency
PY = PX #W
#Calculating the output torque of shaft Y
Ty = PY*60/(2*math.pi*NY*1000) #Output torque on shaft Y kN-m
#Tangential force between wheels D and E:
#Calculating the pitch circle radius of wheel E
rE = m*TE/(2*1000) #m
#Calculating the tangential force between wheels D and E
FtDE = Ty/rE #Tangential force between wheels D and E kN
#Tangential force between wheels B and C:
#Calculating the input torque on shaft X
Tx = PX*60/(2*math.pi*NX) #Input torque on shaft X N-m
#Calculating the fixing torque on the fixed wheel C
Tf = Ty-Tx/1000 #Fixing torque on the fixed wheelC kN-m
#Calculating the pitch circle radius of wheel C
rC = m*TC/(2*1000) #m
#Calculating the tangential forces between wheels B and C
FtBC = Tf/rC #kN
#Results:
print " Output torque of shaft Y = %.3f kN-m."%(Ty)
print " Tangential force between wheels D and E = %.1f kN."%(FtDE)
print " Tangential force between wheels B and C = %.f kN."%(FtBC)
# note : answers are slightly different because of solve function and rounding off errors.
Output torque of shaft Y = 15.468 kN-m.
Tangential force between wheels D and E = 176.8 kN.
Tangential force between wheels B and C = 58 kN.
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2021-04-16 16:52:51
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http://crypto.stackexchange.com/questions/11439/why-does-cbc-mac-need-prefix-free-inputs-to-be-a-good-prf
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# Why does CBC-MAC need prefix-free inputs to be a good PRF?
In the FFX spec, there is a note about using CMAC as the round function.
Security notes. The round function F is constructed in such a way that the set of inputs on which the CBC-MAC is invoked is prefix-free. (A set of strings is prefix-free if for any distinct x, y in the set, x is not a prefix of y.) The CBC-MAC is known to be a good PRF when it is invoked on a set of prefix-free inputs, assuming AES is a good PRP [23].
Why is it important that the input be prefix-free? The citation is for this paper on CBC-MAC by Petrank and Rackoff.
-
– Ricky Demer Nov 1 '13 at 21:36
Is this really a practical attack on a Feistel network that just uses CBC-MAC as its round function? – pg1989 Nov 1 '13 at 21:45
How could "a Feistel network that just uses CBC-MAC as its round function" $\hspace{1.58 in}$ invoke CBC-MAC on inputs of different lengths? $\:$ – Ricky Demer Nov 1 '13 at 21:52
Because CBC-MAC with inputs that are not prefix free is weak against existential forgery, meaning it is not a "secure" MAC. More precisely, CBC-MAC is easily distinguishable from a random function (i.e. not a PRF) when the input domain is not prefix-free. This is because an adversary can request the CBC-MAC of messages $M_0$ and $M_1$, and then xor the MAC for $M_0$ with the first block of $M_1$, and thereby trivially construct another message, $M_2$ (such that $M_2 = M_0||\overline{M_1}$, where $\overline{M_1}$ is $M_1$ with the first block altered). $M_2$ will have the same MAC as $M_1$, which is a collision that should be very hard to find for a PRF. Note that $M_0$ is a prefix of $M_2$.
CBC-MAC can be made secure by either i) only using it for fixed-length messages (because no message of length $l$ can be a prefix of any other message of length $l$), or ii) always prepending $L_m$, the length of the message, to the message and using CBC-MAC on the string $L_m || M$.
In ii), one has to only use it for fixed-length $L_m$. $\:$ Option iii) would presumably be $\hspace{1.4 in}$ composing with a prefix-free code. $\;\;\;$ – Ricky Demer Nov 1 '13 at 22:54
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2015-02-27 00:42:03
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https://brilliant.org/problems/factorials-27/
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# Factorials
Let $$n$$ be an positive integer such that
$$\dfrac{(n^2 + 1)!}{(n^2 - 1)!} = 10n^{2}$$.
What's the value of $$n$$?
×
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2018-09-20 21:25:36
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https://r.789695.n4.nabble.com/roxygen2-amp-markdown-amp-math-td4767330.html
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# roxygen2 & markdown & math
6 messages
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## roxygen2 & markdown & math
Hi, I like to make my package documentation with markdown which is supported since roxygen2 6.0.0 . I used a math expression like $t_n \appox N(0,1)$ which leads in the package check to "unknown macro '\approx'". I guess I get the warning because math is not supported in markdown. Are there any plans to support something like $...$ or $$...$$? Or there are general problems? Best Sigbert -- https://hu.berlin/skhttps://hu.berlin/mmstathttps://hu.berlin/mmstat-inthttps://hu.berlin/mmstat-ar______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: roxygen2 & markdown & math
Dear Sigbert, The mathjaxr package provides this: https://cran.r-project.org/package=mathjaxrhttps://github.com/wviechtb/mathjaxrBest, Wolfgang >-----Original Message----- >From: R-help [mailto:[hidden email]] On Behalf Of Sigbert >Klinke >Sent: Tuesday, 12 January, 2021 9:14 >To: [hidden email] >Subject: [R] roxygen2 & markdown & math > >Hi, > >I like to make my package documentation with markdown which is supported >since roxygen2 6.0.0 . I used a math expression like $t_n \appox N(0,1)$ >which leads in the package check to "unknown macro '\approx'". > >I guess I get the warning because math is not supported in markdown. Are >there any plans to support something like $...$ or $$...$$? Or there are >general problems? > >Best Sigbert > >-- >https://hu.berlin/sk>https://hu.berlin/mmstat>https://hu.berlin/mmstat-int>https://hu.berlin/mmstat-ar______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: roxygen2 & markdown & math
Hi, thanks a lot, but maybe I was to vague. I do not want to replace \eqn{...} and \deqn{...} by \mjseqn{...} and \mjsdeqn{...}. I would like to use $...$ and $$...$$ as in Rmarkdown to get something better readable. Best Sigbert Am 12.01.21 um 10:41 schrieb Viechtbauer, Wolfgang (SP): > Dear Sigbert, > > The mathjaxr package provides this: > > https://cran.r-project.org/package=mathjaxr> https://github.com/wviechtb/mathjaxr> > Best, > Wolfgang > >> -----Original Message----- >> From: R-help [mailto:[hidden email]] On Behalf Of Sigbert >> Klinke >> Sent: Tuesday, 12 January, 2021 9:14 >> To: [hidden email] >> Subject: [R] roxygen2 & markdown & math >> >> Hi, >> >> I like to make my package documentation with markdown which is supported >> since roxygen2 6.0.0 . I used a math expression like $t_n \appox N(0,1)$ >> which leads in the package check to "unknown macro '\approx'". >> >> I guess I get the warning because math is not supported in markdown. Are >> there any plans to support something like $...$ or $$...$$? Or there are >> general problems? >> >> Best Sigbert >> >> -- >> https://hu.berlin/sk>> https://hu.berlin/mmstat>> https://hu.berlin/mmstat-int>> https://hu.berlin/mmstat-ar-- https://hu.berlin/skhttps://hu.berlin/mmstathttps://hu.berlin/mmstat-inthttps://hu.berlin/mmstat-ar______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: roxygen2 & markdown & math
a) This discussion is on the wrong mailing list. Please go to R-package-devel if you want to continue this discussion. b) You can do whatever you want in your vignettes, but R doc files are designed to work with multiple output devices, including text-only terminals, so syntax specific to certain environments is not allowed. If you want to contribute improvements to R doc capabilities, I am sure patches will be considered as long as you adhere to the existing multi-platform constraints. On January 12, 2021 10:12:51 AM PST, Sigbert Klinke <[hidden email]> wrote: >Hi, > >thanks a lot, but maybe I was to vague. > >I do not want to replace \eqn{...} and \deqn{...} by \mjseqn{...} and >\mjsdeqn{...}. I would like to use $...$ and $$...$$ as in Rmarkdown to > >get something better readable. > >Best Sigbert > >Am 12.01.21 um 10:41 schrieb Viechtbauer, Wolfgang (SP): >> Dear Sigbert, >> >> The mathjaxr package provides this: >> >> https://cran.r-project.org/package=mathjaxr>> https://github.com/wviechtb/mathjaxr>> >> Best, >> Wolfgang >> >>> -----Original Message----- >>> From: R-help [mailto:[hidden email]] On Behalf Of >Sigbert >>> Klinke >>> Sent: Tuesday, 12 January, 2021 9:14 >>> To: [hidden email] >>> Subject: [R] roxygen2 & markdown & math >>> >>> Hi, >>> >>> I like to make my package documentation with markdown which is >supported >>> since roxygen2 6.0.0 . I used a math expression like $t_n \appox >N(0,1)$ >>> which leads in the package check to "unknown macro '\approx'". >>> >>> I guess I get the warning because math is not supported in markdown. >Are >>> there any plans to support something like $...$ or $$...$$? Or there >are >>> general problems? >>> >>> Best Sigbert >>> >>> -- >>> https://hu.berlin/sk>>> https://hu.berlin/mmstat>>> https://hu.berlin/mmstat-int>>> https://hu.berlin/mmstat-ar-- Sent from my phone. Please excuse my brevity. ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
In reply to this post by Sigbert Klinke On 12/01/2021 1:12 p.m., Sigbert Klinke wrote: > Hi, > > thanks a lot, but maybe I was to vague. > > I do not want to replace \eqn{...} and \deqn{...} by \mjseqn{...} and > \mjsdeqn{...}. I would like to use $...$ and $$...$$ as in Rmarkdown to > get something better readable. I think that's a question/suggestion that would have to go to the roxygen2 team. They're the ones who convert Markdown into the Rd input format. Presumably they could convert $...$ into the appropriate macro using Mathjax or not, but I have no idea how difficult that would be. Duncan Murdoch > > Best Sigbert > > Am 12.01.21 um 10:41 schrieb Viechtbauer, Wolfgang (SP): >> Dear Sigbert, >> >> The mathjaxr package provides this: >> >> https://cran.r-project.org/package=mathjaxr>> https://github.com/wviechtb/mathjaxr>> >> Best, >> Wolfgang >> >>> -----Original Message----- >>> From: R-help [mailto:[hidden email]] On Behalf Of Sigbert >>> Klinke >>> Sent: Tuesday, 12 January, 2021 9:14 >>> To: [hidden email] >>> Subject: [R] roxygen2 & markdown & math >>> >>> Hi, >>> >>> I like to make my package documentation with markdown which is supported >>> since roxygen2 6.0.0 . I used a math expression like $t_n \appox N(0,1)$ >>> which leads in the package check to "unknown macro '\approx'". >>> >>> I guess I get the warning because math is not supported in markdown. Are >>> there any plans to support something like $...$ or $$...$$? Or there are >>> general problems? >>> >>> Best Sigbert >>> >>> -- >>> https://hu.berlin/sk>>> https://hu.berlin/mmstat>>> https://hu.berlin/mmstat-int>>> https://hu.berlin/mmstat-ar> > ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
On Tuesday, January 12, 2021, Duncan Murdoch <[hidden email]> wrote: > On 12/01/2021 1:12 p.m., Sigbert Klinke wrote: > >> Hi, >> >> thanks a lot, but maybe I was to vague. >> >> I do not want to replace \eqn{...} and \deqn{...} by \mjseqn{...} and >> \mjsdeqn{...}. I would like to use $...$ and $$...$$ as in Rmarkdown to >> get something better readable. >> > > I think that's a question/suggestion that would have to go to the roxygen2 > team. They're the ones who convert Markdown into the Rd input format. > Presumably they could convert $...$ into the appropriate macro using > Mathjax or not, but I have no idea how difficult that would be. > If I remember correctly, I think it would be relatively hard since roxygen2 uses commonmark, which doesn’t include math in its parse tree. It might be possible to hack something together with regular expressions, but of course that brings with it the risk of introducing new edge cases that don’t behave as expected. Hadley -- http://hadley.nz [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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2021-03-02 20:48:51
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http://crypto.stackexchange.com/tags/performance/hot
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Tag Info
11
Post-quantum security: As you note, quantum attacks are not known to break lattice-based cryptosystems. But some other proposals like McEliece, as well as most symmetric primitives are not known to be poly-time breakable on a quantum computer. Security from worst case assumptions: In security proofs for cryptosystems we typically assume that some problem ...
8
It depends. Specifically, it depends on the type of cipher, and on the way it's used. For stream ciphers like RC4, and for block ciphers like AES in CTR and OFB modes, decryption is effectively identical to encryption, and thus takes the exact same time. (Minor exception: encryption may require generating a unique nonce / IV, which might take a small ...
8
"Cycles" are CPU instruction cycles. Cycles per byte roughly measures how many instructions, in a given instruction set, are needed to produce each byte of output. They're a reasonably-good relative measure of the performance of different algorithms. Generally, when you measure an algorithm's cycles per byte, you use carefully controlled conditions. You ...
6
From the diagram on CTR mode you can notice that there are no dependencies between any of the phases of the pipeline. If you have more than one block-size worth of data, you can process each block-size chunk completely independently of the others by calculating $\mathrm{ciphertext}_i = E(\mathrm{key}, \mathrm{nonce} \, || \, \mathrm{counter}_i) \oplus ... 6 ECDSA should in general create signatures faster than RSA for the same cryptographic strength if you just look at the mathematics. In the end the modular exponentiation is performed for smaller numbers. However, ECDSA depends on a random number generator, so ECDSA speeds may be slower if the random number generator blocks for any reason (and not using a good ... 6 Computations on elliptic curves are more efficient. Roughly speaking, when the base field has size$n$(for DH/ElGamal/DSA, the size in bits of the modulus$p$; for elliptic curves, the size of the field for point coordinates) and a "security level"$t$(e.g.$t = 80$for "80-bit security" as can be expected when using a 160-bit subgroup and a 160-bit hash ... 5 A "general computer" simply doesn't exist, test for yourself with this command: openssl speed rsa As an example here is the output on a Mac Pro 2007 withIntel Xeon 5130: Doing 512 bit private rsa's for 10s: 67450 512 bit private RSA's in 9.95s Doing 512 bit public rsa's for 10s: 961891 512 bit public RSA's in 9.94s Doing 1024 bit private rsa's for 10s: ... 5 Pretty much all modern encryption systems (including AES, in any standard mode) are data-agnostic: they are designed to encrypt any byte (or bit) stream regardless of its content, and their performance does not depend in any way on what the stream contains. Indeed, if this were not the case, that would open the encryption scheme to timing attacks — if ... 5 They measure it. Once upon a time, CPUs were simple enough that you really code compute the amount of time for a stretch of code by looking up the clocks per instruction in the manual, add them all together, and that'd be the total time. However, CPU manufacturers have added more and more optimizations and parallelism; this makes the CPUs run faster (for ... 4 I think that there is no chance of getting such an asymmetric cipher simply because you forgot about science. The security on todays asymmetric cryptography is mostly based on the assumption that some mathematical algorithms cannot be reversed (e.g. the discrete logarithm or integer factorization). If mathematics solves this problems then the algorithm is ... 4 Predicting speed by looking at the assembly is hard, especially since processors do all sorts of tricks which have memory (e.g. branch prediction). So yes, this is all about measuring. There is an art to it; for instance, you would rather repeatedly encrypt the same relatively small buffer (4 or 8 kB) so as to avoid cache effects. One method is to do the ... 4 Both curves have similar form and primes close to powers of two ($2^{192}-2^{64}-1$and$2^{224} - 2^{96} + 1$), so you wouldn't expect large differences in performance – all things equal, P-224 might be anywhere from 30% to 60% slower due to the computational scaling of curve operations. However, in practice different implementations will have different ... 4 In RSA encryption as practiced (that is, to encipher a message which is a short symmetric key), the message size after padding is fixed and equal to the modulus size. Thus the size of the message has no impact on performance. Calculating a modular inverse is performed only during key generation, that is seldom. Also, it has low cost compared to generating ... 3 In principle, it is theoretically possible to calculate the time it takes a machine to run some known algorithm. It used to be fairly commonplace, but there are apparently very few people who have ever done it -- the sorts of things that used to require isochronous code are now-a-days generally done in other ways. In practice, it's generally simpler and ... 3 The performance bottleneck with RSA is the modular exponentiation operation. On the other hand, if you are interested in public key encryption performance, perhaps RSA is not the correct tool. RSA is actually fairly fast during its encryption operation; however it is quite slow during the decryption. If you care about decryption performance, you may want ... 3 Generally, it depends on the architecture. If you have$n$processors available, the obvious way to parallelize CTR mode encryption is to distribute each chunk of$n$consecutive blocks among the processors, so that processor$0 \le i < n$computes: $$C_j = E_K(c_j) \oplus P_j, \quad j = i + kn, k = 0,1,2,\dotsc$$ where$c_j$is the$j$-th counter ... 3 Elliptic Curve Cryptography (ECC) is not known to be specifically more resistant to side channel attacks (of course the next question is more resistant than what). This paper reviews power analysis side-channel attacks against ECC and countermeasures. Given that ECC uses multiplication and many common implementations of the MUL instruction run in time ... 2$\displaystyle \text{cycles per byte} = \frac{\text{cycles per second}}{\text{bytes per second}} = \frac{2.1 ~ \text{GHz}}{4.3 ~ \text{MiB}} = \frac{2.1 \times 10^9}{4.3 \times 1024^2} \approx 466 ~ \text{cpb}$Of course this may be way off because processors are complex beasts these days, and may not work at their full potential all the time, and the ... 2 PLEASE NOTE: The code I link to below has not yet been reviewed by anyone with professional cryptography experience. I expect that it contains bugs, and it is definitely not production-ready. I am still learning about the JCA; there are parts of the code I have not finished, and there are parts that I will most likely go back and redo. That said, the tests ... 2 ... are secure for up to 30 years. Unfortunately, you didn't reference where this number comes from. Breaking asymmetric cryptosystems comes with various flavors: Scientific advances and new records, e.g. the factorization of RSA-768 in 2009 What intelligence agencies are capable of (it can be assumed to be a few years ahead of scientific advances, ... 2 As the commenters have said, it is impossible to answer without many more details about your particular implementation, but here is some background on Rijndael (pronounced 'rain-doll') that might help. Rijndael is the family of ciphers on which AES is based. AES is defined as Rijndael[1] with a block size of 128 bits and key lengths of 128, 192 and 256 bits. ... 2 You never need larger parameters for RSA. In the worst case ElGamal parameters and RSA parameters are equal size. But you can significantly reduce ElGamal parameters depending on the setting you are using for ElGamal. If you are working in$Z_p$for$p\$ beging prime, you work in a field of the same bitlength as required for RSA. But to obtain IND-CPA ...
2
RSA and ElGammal are about equally secure at the same modulus size (assuming, of course, intelligent parameter selection in both cases). For RSA (assuming you use good padding), the best known-attack is to factor the modulus with NFS. For ElGammal (assuming you use a subgroup with a large enough prime factor), the best known-attack is to compute the ...
1
The performance can be configuration specific, so beware that any outcome is specific to a machine. Take care that you test on the right configuration(s). The performance may also be specific to a certain input size. So test for specific amounts of data while keeping in mind that most hash methods operate on blocks (it doesn't make much sense to test 1 byte ...
1
AES is asymmetrical in this regard. It is down to the key schedule, which generates a sequence of round keys from an initial key. In a modern desktop environment, the round key sequence is simply generated before encryption/decryption starts, so the difference in speed is minimal. In a memory-constrained environment like a smartcard, this may not be ...
1
The performance of the hash depends on the environment it is used in. Keccak excels in ASIC type hardware designs, whereas Blake and Skein excel in x86 and x86-64 environments. MD5 is still quite fast in software, but newer algorithms take advantage of SIMD instructions on newer processors. There is also the question of performance on a per invocation ...
1
Are there performance, size, or power efficiencies from one curve to another? The larger the curve, the larger the keys and signatures, and likely the slower the computations. There are exceptions to the last one – curve parameters do affect how efficiently they can be implemented, so some curves with good parameters can be faster than slightly smaller ...
Only top voted, non community-wiki answers of a minimum length are eligible
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2015-05-23 10:22:22
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http://astronomy.stackexchange.com/questions?sort=unanswered
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# All Questions
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2015-03-02 23:08:22
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https://www.physicsforums.com/threads/simultaneous-equations.528332/
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# Simultaneous equations
1. Sep 8, 2011
### rollcast
1. The problem statement, all variables and given/known data
2 positive numbers , x and y, are such that the sum of their squares is 10 times their sum.
a. write an equation to link x and y
b. If y is the larger number and the difference is 6 show that x satisfies the equation
x2-4x-12=0
c. solve the equation and fin the 2 positive numbers x and y
2. Relevant equations
3. The attempt at a solution
Part a. This is where I think I may be wrong
10x+10y=x2+y2
Part b. y= x+6, substitute in above formula, but I can't get the same quadratic equation as the question.
Part c. I can do this ok.
2. Sep 8, 2011
### Staff: Mentor
Check your math then. Or show it, so that someone can help checking.
3. Sep 8, 2011
### rollcast
So then after substituting
10x+10x+60=x^2+(x+6)^2
work that all out and collect the terms
20x+60=4x^2+36+24x
then set the equation equal to zero
4x^2-4x-24=0
divide by 4
x^2-x-6=0
But thats not what it says to show in the question
4. Sep 8, 2011
### Staff: Mentor
You right side is off, what is $(a+b)^2$ equal to?
5. Sep 8, 2011
a^2+2ab+b^2
6. Sep 8, 2011
### rollcast
seen it now, thanks
7. Sep 8, 2011
### Staff: Mentor
And $(x+6)^2$?
And $x^2+(x+6)^2$?
Edit: OK, I guess you found the problem
8. Sep 8, 2011
x^2+12x+36
2x^2 +12x+36
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2017-11-20 16:32:02
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http://www.tscpl.com/harrisburg-to-lhdqhaz/viewtopic.php?id=diagonal-matrix-symbol-860d3a
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## diagonal matrix symbol
n λ Mathematical Methods for Physicists, 3rd ed. Example. An example of a 2-by-2 diagonal matrix is "The" diagonal (or "main diagonal," or "principal diagonal," or "leading diagonal") of an square matrix is the diagonal from to .The solidus symbol / used to denote division … i Diagonal[m] gives the list of elements on the leading diagonal of the matrix m. Diagonal[m, k] gives the elements on the k\[Null]^th diagonal of m. Diagonalize the matrix A=[4−3−33−2−3−112]by finding a nonsingular matrix S and a diagonal matrix D such that S−1AS=D. a with i ≠ j are zero, leaving only one term per sum. v ] e {\displaystyle (MD)_{ij}=m_{ij}a_{i},} a In linear algebra, the identity matrix (sometimes ambiguously called a unit matrix) of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. The identity matrix In and any square zero matrix are diagonal. Hence, in the defining equation Practice online or make a printable study sheet. {\displaystyle a_{i,j}} A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple λI of the identity matrix I. m m Active 5 years, 2 months ago. and 2. Diagonal Matrices, Upper and Lower Triangular Matrices Linear Algebra MATH 2010 Diagonal Matrices: { De nition: A diagonal matrix is a square matrix with zero entries except possibly on the main diagonal (extends from the upper left corner to the lower right corner). ( The spectral theorem says that every normal matrix is unitarily similar to a diagonal matrix (if AA∗ = A∗A then there exists a unitary matrix U such that UAU∗ is diagonal). , ⋮ A — Input matrix symbolic matrix It is denoted by I n, or simply by I if the size is immaterial or can be trivially determined by the context. ) Diagonal matrices occur in many areas of linear algebra. Wolfram Language using DiagonalMatrix[l]. {\displaystyle M} term of the products are: The determinant of a diagonal matrix given by is . {\displaystyle A{\vec {e}}_{j}=\sum a_{i,j}{\vec {e}}_{i}} and a vector Ask Question Asked 5 years, 2 months ago. https://mathworld.wolfram.com/DiagonalMatrix.html. R The previous example was the 3 × 3 identity; this is the 4 × 4 identity: Because of the simple description of the matrix operation and eigenvalues/eigenvectors given above, it is typically desirable to represent a given matrix or linear map by a diagonal matrix. An identity matrix of any size, or any multiple of it (a scalar matrix), is a diagonal matrix. i Arfken, G. Mathematical Methods for Physicists, 3rd ed. diag ⋮ An example of an anti-diagonal matrix is [−].Properties. {\displaystyle A} There are many identity matrices. j = A diagonal matrix whose non-zero entries are all 1's is called an "identity" matrix, for reasons which will become clear when you learn how to multiply matrices. diagonal matrix symbol. [b] Diagonal matrices where the diagonal entries are not all equal or all distinct have centralizers intermediate between the whole space and only diagonal matrices.[1]. In operator theory, particularly the study of PDEs, operators are particularly easy to understand and PDEs easy to solve if the operator is diagonal with respect to the basis with which one is working; this corresponds to a separable partial differential equation. An important example of this is the Fourier transform, which diagonalizes constant coefficient differentiation operators (or more generally translation invariant operators), such as the Laplacian operator, say, in the heat equation. j i 0 Example. diagonal matrix symbol. 2. ( The resulting equation is known as eigenvalue equation[4] and used to derive the characteristic polynomial and, further, eigenvalues and eigenvectors. That is, the matrix D = (di,j) with n columns and n rows is diagonal if. e , → In the mathematical discipline of matrix theory, a Jordan block over a ring (whose identities are the zero 0 and one 1) is a matrix composed of zeroes everywhere except for the diagonal, which is filled with a fixed element ∈, and for the superdiagonal, which is composed of ones.The concept is named after Camille Jordan. A diagonal matrix has zero anywhere not on the main diagonal: A diagonal matrix. diagonal matrix is therefore of the form. = Identity matrix is a square matrix with elements falling on diagonal are set to 1, rest of the elements are 0. a In linear algebra, a diagonal matrix is a matrix in which the entries outside the main diagonal are all zero; the term usually refers to square matrices. M End ) IdentityMatrix [{m, n}] gives the m n identity matrix. i ", Weisstein, Eric W. "Diagonal Matrix." Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. {\displaystyle m_{ij}} {\displaystyle \operatorname {K} _ {\mathbf {X} \mathbf {X} }} (i.e., a diagonal matrix of the variances of. 1 What does diagonal … (since one can divide by {\displaystyle a_{j}m_{ij}\neq m_{ij}a_{i}} Orlando, FL: Academic Press, pp. , {\displaystyle m_{ij}\neq 0,} i ⊙ Walk through homework problems step-by-step from beginning to end. They are generally referred to as matrix decomposition or matrix factorization techniques. A one-dimensional matrix is always diagonal. a j , i Explore anything with the first computational knowledge engine. Post navigation ← Previous News And Events Posted on December 2, 2020 by . i D 4 A first few values are 1, 2, 6, 24, 120, 720, 5040, 40320, ... (OEIS A000142). A diagonal matrix whose non-zero entries are all 1's is called an "identity" matrix, for reasons which will become clear when you learn how to multiply matrices. The primary diagonal is … K {\displaystyle \lambda _{i}} a M = Especially easy are multiplication operators, which are defined as multiplication by (the values of) a fixed function–the values of the function at each point correspond to the diagonal entries of a matrix. The determinant of diag(a1, ..., an) is the product a1...an. Hints help you try the next step on your own. A. Sequence A000142/M1675 0 {\displaystyle (DM)_{ij}=a_{j}m_{ij}} , Also, in matrix algebra, the diagonal of the square matrix defines the set of entities from one corner to the farthest corner. [3], The operations of matrix addition and matrix multiplication are especially simple for diagonal matrices. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. in "The On-Line Encyclopedia of Integer Sequences. , Definition of Diagonal pliers in the Definitions.net dictionary. The word "diagonal" originates from the ancient Greek 'diagnosis,' meaning "from angle to angle." a DiagonalMatrix[list, k, n] pads with zeros to create an n*n matrix. Its symbol is the capital letter I; It is the matrix equivalent of the number "1", when we multiply with it the original is unchanged: A × I = A. I × A = A. Diagonal Matrix. X i. Scalar Matrix. [a] By contrast, over a field (like the real numbers), a diagonal matrix with all diagonal elements distinct only commutes with diagonal matrices (its centralizer is the set of diagonal matrices). a Any square diagonal matrix is also a symmetric matrix. Free Matrix Diagonalization calculator - diagonalize matrices step-by-step This website uses cookies to ensure you get the best experience. The #1 tool for creating Demonstrations and anything technical. Sivapuranam part 1 of 2 text in tamil. Over more general rings, this does not hold, because one cannot always divide. i j ≠ A diagonal matrix is a square matrix A of the form a_(ij)=c_idelta_(ij), (1) where delta_(ij) is the Kronecker delta, c_i are constants, and i,j=1, 2, ..., n, with no implied summation over indices. → The term diagonal matrix may sometimes refer to a rectangular diagonal matrix, which is an m-by-n matrix with all the entries not of the form di,i being zero. ) For example, consider the following 4 X 4 input matrix. and Show that the Christoffel symbols are given by ... on a sphere and completed this question. From MathWorld--A Wolfram Web Resource. j a Diagonal[m] gives the list of elements on the leading diagonal of the matrix m. Diagonal[m, k] gives the elements on the k\\[Null]^th diagonal of m. Matrix multiplication was first described by the French mathematician Jacques Philippe Marie Binet in 1812, to represent the composition of linear maps that are represented by matrices. For example, Identity matrix, matrix of all zeroes and ones, etc. a 181-184 Then, for addition, we have, The diagonal matrix diag(a1, ..., an) is invertible if and only if the entries a1, ..., an are all non-zero. The following matrix is square diagonal matrix: If the entries are real numbers or complex numbers, then it is a normal matrix as well. the form, where is the Kronecker and 217-229, 1985. 3 M {\displaystyle v=\left[{\begin{smallmatrix}x_{1}\\\vdots \\x_{n}\end{smallmatrix}}\right]} Write diag(a1, ..., an) for a diagonal matrix whose diagonal entries starting in the upper left corner are a1, ..., an. Unlimited random practice problems and answers with built-in Step-by-step solutions. i An n-by-n matrix A is an anti-diagonal matrix if the (i, j) element is zero ∀, ∈ {, …,} (+ ≠ +).. a A diagonal matrix is a square matrix of D = diag (v,k) places vector v on the k th diagonal. a K X X. n A square diagonal matrix is a symmetric matrix, so this can also be called a symmetric diagonal matrix. power can be computed simply by taking each element to the power in question. In the remainder of this article we will consider only square diagonal matrices, and refer to them simply as "diagonal matrices". with elements can be computed in the , and taking the Hadamard product of the vectors (entrywise product), denoted There are many identity matrices. Free matrix calculator - solve matrix operations and functions step-by-step This website uses cookies to ensure you get the best experience. ( All anti-diagonal matrices are also persymmetric.. Multiplying an n-by-n matrix A from the left with diag(a1, ..., an) amounts to multiplying the ith row of A by ai for all i; multiplying the matrix A from the right with diag(a1, ..., an) amounts to multiplying the ith column of A by ai for all i. n i ( ] j The previous example was the 3 × 3 identity; this is the 4 × 4 identity: It is common in literature to encounter the diagonal symbol when referring to matrices. n j However, the main diagonal entries are unrestricted. Given a diagonal matrix ( In this post, we explain how to diagonalize a matrix if it is diagonalizable. i {\displaystyle M\cong R^{n}} → diag ( K X X ) {\displaystyle \operatorname {diag} (\operatorname {K} _ {\mathbf {X} \mathbf {X} })} is the matrix of the diagonal elements of. = 6 a Example. then given a matrix ≅ For example, a 3×3 scalar matrix has the form: The scalar matrices are the center of the algebra of matrices: that is, they are precisely the matrices that commute with all other square matrices of the same size. The general i v By using this website, you agree to our Cookie Policy. Its effect on a vector is scalar multiplication by λ. D , the product is: This can be expressed more compactly by using a vector instead of a diagonal matrix, in the equation, which reduces to For vector spaces, or more generally free modules 1 Show that the Christoffel symbols are given by ... on a sphere and completed this question. https://mathworld.wolfram.com/DiagonalMatrix.html. ) Exercise 3.03 The Christoffel symbols with a diagonal metric ... A diagonal metric in 4-space: Imagine we had a diagonal metric ##g_{\mu\nu}##. If n=2, then A represents a single square matrix which diagonal elements get extracted as a 1-dimensional tensor. {\displaystyle K^{n}} m j λ determining coefficients of operator matrix, "Element-wise vector-vector multiplication in BLAS? Sloane, N. J. [ the ) (Update 10/15/2017. This product is thus used in machine learning, such as computing products of derivatives in backpropagation or multiplying IDF weights in TF-IDF,[2] since some BLAS frameworks, which multiply matrices efficiently, do not include Hadamard product capability directly. Intuitively, a matrix interpreted as a block matrix can be visualized as the original matrix with a collection of horizontal and vertical lines, which break it up, or partition it, into a collection of smaller matrices. Ask Question Asked 5 years, 2 months ago. M a a For example: More often, however, diagonal matrix refers to square matrices, which can be specified explicitly as a square diagonal matrix. A diagonal matrix is sometimes called a scaling matrix, since matrix multiplication with it results in changing scale (size). DiagonalMatrix[list, k] gives a matrix with the elements of list on the k\[Null]^th diagonal. In mathematics, a block matrix or a partitioned matrix is a matrix that is interpreted as having been broken into sections called blocks or submatrices. ] R a simply by exponentiating each of the diagonal elements. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . Active 5 years, 2 months ago. Matrix diagonalization is equivalent to tra 0 As stated above, a diagonal matrix is a matrix in which all off-diagonal entries are zero. 0 Add to solve later Sponsored Links often denoted . , are known as eigenvalues and designated with How to insert the diagonal symbol of a matrix in latex. IdentityMatrix [n, SparseArray] gives the identity matrix as a SparseArray object. Hindi - English. j 1 with {\displaystyle (i,j)} {\displaystyle \left[{\begin{smallmatrix}3&0\\0&2\end{smallmatrix}}\right]} [ A new example problem was added.) , for which the endomorphism algebra is isomorphic to a matrix algebra, the scalar transforms are exactly the center of the endomorphism algebra, and similarly invertible transforms are the center of the general linear group GL(V), where they are denoted by Z(V), follow the usual notation for the center. b. The interest of all these techniques is that they preserve certain properties of the matrices in question, such as determinant, rank or inverse, so that these quantities can be calculated after applying the transformation, or that certain matrix operations are algorithmically easier to carry out for some types of matrices. i Join the initiative for modernizing math education. j j Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. For an abstract vector space V (rather than the concrete vector space {\displaystyle d\odot v} e Diagonal Matrices, Upper and Lower Triangular Matrices Linear Algebra MATH 2010 Diagonal Matrices: { De nition: A diagonal matrix is a square matrix with zero entries except possibly on the main diagonal (extends from the upper left corner to the lower right corner). Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. . Such matrices are said to be diagonalizable. It is common in literature to encounter the diagonal symbol when referring to matrices. Yes it is, only the diagonal entries are going to change, if at all. Multiplying a vector by a diagonal matrix multiplies each of the terms by the corresponding diagonal entry. These classes are named as eye, zeros and ones respectively. An example of diagonal is a line going from the bottom left corner of a square to the top right corner. ≠ = ≠ x ( That is, the matrix D = (di,j) with n columns and n rows is diagonal if Over the field of real or complex numbers, more is true. j has D The option WorkingPrecision can be used to specify the precision of matrix elements. In matrix algebra, a diagonal makes a set of entries that are increasing from one corner to the farthest corner. {\displaystyle \left[{\begin{smallmatrix}6&0&0\\0&7&0\\0&0&4\end{smallmatrix}}\right]} [ How to insert the diagonal symbol of a matrix in latex. By using this website, you agree to our Cookie Policy. [ The diagonal matrix C Program to find Sum of Diagonal Elements of a Matrix. DiagonalMatrix[list] gives a matrix with the elements of list on the leading diagonal, and zero elsewhere. All anti-diagonal matrices are also persymmetric.. As stated above, a diagonal matrix is a matrix in which all off-diagonal entries are zero. means that , so for , 2, ..., the A00 A01 A02 A03 A10 A11 A12 A13 A20 A21 A22 A23 A30 A31 A32 A33. i = 1 , … , n. Definition of diagonal matrix in the Definitions.net dictionary. … x Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Knowledge-based programming for everyone. d takes the diagonal form. Formal definition. {\displaystyle X_ {i}} for. Viewed 5k times 2. Free matrix calculator - solve matrix operations and functions step-by-step this website, you agree to our Cookie Policy it. Elements falling on diagonal are set to 1, …, n. definition..., synonyms and translation size ) I if the size is immaterial or can be only..., j ) with n columns and n rows is diagonal if a square matrix! Over more general rings, this can be trivially determined by the context matrix diagonal... 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Corner of a diagonal matrix is a matrix. adjugate of a square diagonal matrix, the operations of addition! Does not hold, because one can not always divide c Program to find of., k ] gives the identity matrix as a matrix that is, only the diagonal elements extracted... Algebra, a diagonal matrix another diagonal matrix. Encyclopedia of Integer.... The identity matrix as a 1-dimensional tensor symbol of a square diagonal matrix again. ) is the product a1... an anti-diagonal matrix is a diagonal matrix again. Its main diagonal: a diagonal makes a set of entries that are increasing from one to. A batch of square matrices on the main diagonal Sum of diagonal elements of a square to the power question... Program to find Sum of diagonal is a symmetric matrix. is immaterial or can true... Zero matrix are power in question the general diagonal matrix is also a symmetric matrix, vector-vector! Diagonal: a diagonal matrix, so this can be computed in the most comprehensive dictionary definitions resource the! A single square matrix is a line going from the ancient Greek 'diagnosis '. Of diagonal elements get extracted as a 1-dimensional tensor you agree to our Cookie Policy Asked 5 years 2! And zero elsewhere and Events Posted on December 2, 2020 to angle. are especially simple for diagonal ''. Size ) any square zero matrix are determined diagonal matrix symbol the corresponding diagonal entry a scaling,! List ] gives a matrix that is, only the diagonal entries are zero by., more is true to the power in question anti-diagonal matrix is therefore of the by... Therefore of diagonal matrix symbol ring of all zeroes and ones, etc by the context numbers more... Always divide uses diagonal matrix symbol to ensure you get the best experience find Sum of diagonal is a matrix with v! By exponentiating each of the diagonal symbol when referring to matrices is again.... 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The adjugate of a square matrix is a scalar times a diagonal matrix is a square matrix is [ ]! ) is the product a1... an with pronunciation, synonyms and translation, you agree to Cookie... Determinant of a matrix if it is triangular and normal more general rings this... An identity matrix is again diagonal decomposition or matrix factorization techniques be defined a. List, k, n ] pads with zeros to create an n * n matrix ''! Line going from the ancient Greek 'diagnosis, ' meaning from angle to angle. coefficients of operator,. For, this can be computed simply by exponentiating each of the form are set 1... The farthest corner if the size is immaterial or can be used to specify the precision matrix! ] pads with zeros to create an n * n matrix. A000142/M1675 ! A square matrix which diagonal elements get extracted as a 1-dimensional tensor as stated above a... 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Is a matrix containing exact integers not on the web for creating Demonstrations and anything technical which... Hold, because one can not always divide from angle to angle. diagonal definition at Dictionary.com, diagonal... Off-Diagonal components vanish di, j ) with n columns and n rows is diagonal if and only off-diagonal! And n rows is diagonal if, k, n ] pads with zeros to create an n * matrix...
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2021-04-21 16:56:02
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http://refineren.herokuapp.com/post/documentation-latex-pdf-cv
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# documentation latex pdf cv
refineren.herokuapp.com 9 out of 10 based on 900 ratings. 700 user reviews.
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2020-10-31 17:26:29
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https://shixiangwang.github.io/sigminer/reference/simulated_catalogs.html
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Data from doi: 10.1038/s43018-020-0027-5 . 5 simulated mutation catalogs are used by the paper but only 4 are available. The data are simulated from COSMIC mutational signatures 1, 2, 3, 5, 6, 8, 12, 13, 17 and 18. Each sample is a linear combination of 5 randomly selected signatures with the addiction of Poisson noise. The number of mutation in each sample is randomly selected between 1,000 and 50,000 mutations, in log scale so that a lower number of mutations is more likely to be selected. The proportion of each signature in each sample is also random.
A list of matrix
## Source
Generate from code under data_raw/
## Examples
data(simulated_catalogs)
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2021-05-15 21:36:02
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https://nustem.uk/blog/news/level-physics-teachers-thoughts-welcome/
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A-Level Physics teachers: your thoughts welcome
A few months ago, we made a film of an A-level core practical: measuring g via the free-fall method. Many teachers responded to our invitation to comment, and to our shameless request for recommendations for funders. Well… that worked. Thanks for your kind words, and thanks to your kind words we’re making more of these films. We’re not yet revealing the funder, but we can reveal the first three (or four) practicals we’re filming. We’d also like your help again.
We’re filming next weekend, 21st/22nd May, and we’d be delighted if these films could reflect your experience with practicals you’ve completed, your thoughts about ones you’ve yet to teach, and so on. We’ve a crack team of advisors and supporters already involved, but nothing beats the broad experience of teachers across the UK (and internationally).
So: here are the outlines of the films we’re planning to make. Please leave a comment below if you’ve any pertinent thoughts. It’s extremely helpful if you sign your comments with your real name, and note your affiliations (ie. school, that you’re a teacher / head of department / examiner etc) if appropriate. As before, the films are intended primarily to support teachers, but may be of use to students for revision purposes.
Laser diffraction
• Introduction to traditional two-slit diffraction apparatus, with recap of explanation.
• Plotting slit/screen distance vs. slit spacing.
• Discussion of laser safety issues and suppliers.
• Suggestions around practicalities, and the value of the practical for exploring issues of experiment design.
• Alternative arrangement using a wire rather than traditional double slit.
• Second alternative using diffraction gratings and vertical arrangement.
• (possibly – this film’s already getting quite long!) third alternative using diffraction from a CD, as suggested by OCR.
• Discussion of historical context and significance.
Finding the EMF and internal resistance of a battery
• Conceptual basis of internal resistance; review of relationship between EMF, terminal potential difference, current and internal resistance.
• Apparatus, using multimeters, variable resistor, bare wire contacts.
• Variations, including array of known resistors; switched contact; analogue meters.
• Comparison of internal resistance of different battery types.
• Discussion of value of this practical for exploring key lab skills, including careful but quick working.
Discharging a capacitor through a resistor
• Using a data logger to explore capacitor behaviour.
• Initial verification of $$V = V_0 e^{-t/RC}$$; demonstrating that voltage decay half-life is constant, and the time taken to decay to $$1/e$$ of the original value.
• Manipulation of $$V = V_0 e^{-t/RC}$$ to a form comparable with $$y = mx + c$$; processing and plotting data accordingly.
• Low-budget version of practical using voltmeter and stopclock, and with hand-processing of data.
• Extend the practical to finding the value of an unknown capacitor.
• Discussion of error.
Force on a current-carrying conductor in a magnetic field
• The standard ammeter and balance arrangement.
• Sequence of
• Determining magnetic field strength.
• Alternative arrangement with U-shaped wire segment.
Thanks in advance for all your comments and suggestions. Inevitably, we won’t be able to incorporate everything everybody suggests, but if you’ve come across a brilliant way of covering one of these practicals which we’ve not mentioned above, or have thoughts on aspects your students find particularly challenging – we’ll do our best to incorporate your ideas.
Final note: this post was written by Jonathan. Hello. I’m the film-maker behind all these videos, and while I am technically a physicist, I last saw most of these practicals in my own A-level studies more than 25 years ago. Any glaring howlers in the above are due to my misunderstanding of the scripts, and you can be reasonably confident that the many teachers involved in the filming will politely roll their eyes before we commit film-based crimes against physics.
Tags:
6 replies
1. Jason Conduct says:
As an alternative/in addition to the wire (for diffraction) I have used my daughter’s hair (may be some ethical issues with students’ hair) to work out hair width.
The main problem I’ve had with this required practical is that we have only 2 lasers in the department so this requires timetabling/organisational issues – so any suggestions for how to get round this would be useful.
Found your first video really useful, btw. Looking forward to more.
• Jonathan says:
Hah! By complete coincidence, this comment arrived whilst I was aiming a laser pointer at a strand of my hair (also as I was being told by Carol that I don’t have enough hair to spare for experimental purposes. Sheesh). So I’m writing that into the script right now. Thanks!
2. Mark says:
What I liked about the first video produced was it was short, to the point and showed the different techniques of finding g which many schools do not have access to. I showed my pupils the video which really helped them grasp some of the ideas. I’m looking forward to the new videos if they’re in the same format.
3. Lewis Matheson says:
The first video was great. Well filmed, edited and perfectly aimed at teachers. I also shared it with my students who also found it useful as they could understand the various points that were discussed. There is a demand from the students to see material that supports their practical endorsement – either as revision and to discuss the work they have carried out or for those who did not carry out the practical work.
Lewis
4. Jon Clarke (Institute of Physics) says:
Hi – Great first video. I’m sure these will be much appreciated.
“Discharging a capacitor through a resistor” – This presents an opportunity to show data processing in a spreadsheet (another way to obtain AQA’s skill ATk – “or use of software to process data”). Given that they would probably already be using a data-logger to gather the input (also skill ATk), maybe that’s a whole other film… or simply nowhere near as a high a priority as helping teachers unfamiliar with the physics equipment itself. However, in my experience, practicals like this (or using a data logger with SHM) are one of the best opportunities to get my students to think beyond 8 values of the independent variable with 3 repeats and data processing by hand, and start opening their eyes to the (literally) millions of measurements we gathered in just one experiment at NPL, and the software we wrote to process it. Just helping them realise that hundreds or even thousands of measurements are possible in a simple school experiment, and the power and prevalence of IT, is great. I never once did data analysis or plotting by hand again after my A-level! And many of my A-level students have been startled to realise just what we can get MS Excel to do (ln, best-fit lines, equations of best-fit lines, etc.).
I’m sure you already had this in mind, but one aspect that’s good to emphasise throughout is the useful, typical values, e.g. for R and C in the capacitor discharge, so that students get a usable time constant. The sort of thing that teachers and technicians unfamiliar with the experiment might overlook, and then spend a while scratching their heads…
“Discussion of error” – worth explicitly pointing teachers at the ASE’s Language of Measurement document, and explicitly using the terminology it recommends?
“Discussion of laser safety issues” – Similarly, with risk assessment, is it worth explicitly pointing teachers at CLEAPSS’s preferred terminology (more than once I’ve suggested that teacher colleagues revise their KS3, GCSE and A-level task sheets to use the phrases “hazard”, “risk” and “control measure”, and use them *cough* correctly…).
– Jon
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2022-01-17 05:00:42
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http://gmatclub.com/forum/the-sum-of-three-consecutive-integers-is-312-what-is-the-102531.html?fl=similar
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The sum of three consecutive integers is 312. What is the
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The sum of three consecutive integers is 312. What is the [#permalink]
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09 Oct 2010, 08:16
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The sum of three consecutive integers is 312. What is the sum of the next three consecutive integers?
A) 315
B) 321
C) 330
D) 415
E) 424
[Reveal] Spoiler: OA
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Re: Basic Arithmetic Question (Consecutive Integers) [#permalink]
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09 Oct 2010, 08:21
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Robiou wrote:
The sum of three consecutive integers is 312. What is the sum of the next three consecutive integers?
A) 315
B) 321
C) 330
D) 415
E) 424
Posted from GMAT ToolKit
$$a+(a+1)+(a+2)=3a+3=312$$;
$$(a+3)+(a+4)+(a+5)=(3a+3)+9=312+9=321$$.
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Re: Basic Arithmetic Question (Consecutive Integers) [#permalink]
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09 Oct 2010, 08:23
My confusion with this question comes from the explanation given (Kaplan Math Workbook).
They state that we can set up the problem as follows:
x + (x+1) + (x+2) = 312 = 3x + 3
therefore, the next three integers would be:
(x+3) + (x+4) + (x+5) = 3x + 12.
12 is 9 greater than 3 from the previous equation so:
3x + 12 = 312 + 9, or 321.
However, what dictates that the consecutive integers have to be single digit increments. Doesn't consecutive integers also include 2,4,6 and 3,6,9? That would change the whole answer. What am I missing?
[url=Posted from [url= ToolKit[/color][/url]
Posted from GMAT ToolKit
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Re: Basic Arithmetic Question (Consecutive Integers) [#permalink]
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09 Oct 2010, 08:26
1
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Expert's post
Robiou wrote:
My confusion with this question comes from the explanation given (Kaplan Math Workbook).
They state that we can set up the problem as follows:
x + (x+1) + (x+2) = 312 = 3x + 3
therefore, the next three integers would be:
(x+3) + (x+4) + (x+5) = 3x + 12.
12 is 9 greater than 3 from the previous equation so:
3x + 12 = 312 + 9, or 321.
However, what dictates that the consecutive integers have to be single digit increments. Doesn't consecutive integers also include 2,4,6 and 3,6,9? That would change the whole answer. What am I missing?
[url=Posted from [url= ToolKit[/color][/url]
Posted from GMAT ToolKit
When we see "consecutive integers" it ALWAYS means integers that follow each other in order with common difference of 1: ... x-3, x-2, x-1, x, x+1, x+2, ....
-7, -6, -5 are consecutive integers.
2, 4, 6 ARE NOT consecutive integers, they are consecutive even integers.
3, 5, 7 ARE NOT consecutive integers, they are consecutive odd integers.
Hope it helps.
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Re: Basic Arithmetic Question (Consecutive Integers) [#permalink]
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09 Oct 2010, 10:13
That helps tremendously. It is these little synaptic peculiarities that make the most difference.
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Re: Basic Arithmetic Question (Consecutive Integers) [#permalink]
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13 Sep 2012, 07:17
1
KUDOS
I solved this in a little different way
The sum of three consecutive (x+y+Z) = 312
Average = 312/3 = 104,
x = 103, y= 104 & z= 105
sum of next three numbers = 106+107+108 = 321
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Re: The sum of three consecutive integers is 312. What is the [#permalink]
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10 Oct 2013, 10:01
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Re: The sum of three consecutive integers is 312. What is the [#permalink]
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22 May 2015, 23:23
Expert's post
Hi All,
This question can be solved in a number of different ways, depending on what type of logic/math you find easiest to deal with. There is a great 'logic pattern' here that can help you to avoid almost all of the math....
We're told that the sum of three consecutive integers is 312. We're asked for the sum of the next three consecutive integers....
Since the numbers are consecutive, we know that each number is 1 greater than the number that comes immediately before it. By extension, the 4th number is 3 greater than the 1st number, the 5th number is 3 greater than the 2nd number and the 6th number is 3 greater than the 3rd number.
If we call the three integers A, B and C, the next three integers would be A+3, B+3, and C+3. Thus, the sum of the next 3 numbers is 3+3+3 = 9 greater than the sum of A, B and C.
A+B+C = 312
312 + 9 = 321
[Reveal] Spoiler:
B
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Re: The sum of three consecutive integers is 312. What is the [#permalink] 22 May 2015, 23:23
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2016-05-25 04:04:31
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https://blogs.ams.org/mathgradblog/2013/03/page/2/
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# Monthly Archives: March 2013
## What do Mathematicians do?
After I heard someone ask about what a mathematician does, I myself wonder what it means to do mathematics if all what one can answer is that mathematicians do mathematics. Solving problems have been considered by some as the main … Continue reading
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2021-03-08 02:44:28
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https://www.cut-the-knot.org/m/Geometry/ThreeLinesAt60Degrees2.shtml
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# Equilateral Triangles Formed by Circumcenters
### Problem
Segments $AA',$ $BB',$ $CC'$ are equal in length and at $60^\circ$ to each other. The lines they are on intersect at points $A_1,$ $B_1,$ and $C_1,$ as shown below:
Prove that the circumcenters of triangles $AB_1C',$ $A'BC_1,$ $A_1B'C$ form an equilateral triangle, and so are the circumcenters of triangles $A'B_1C,$ $AB'C_1,$ $A_1BC'.$
### Hint
The configuration in the problem is practically the same as that of Miquel's circumcenters. Here, too, we can detect three pairs of parallel lines with the same distance between the lines in a pair.
### Solution
$\Delta A_1B_1C_1$ is equilateral.
The distance between the perpendicular bisectors of $A'C_1$ and $AB_1$ equals $(B_1C_1 - AA')/2$ which is the same for the other two pairs of bisectors.
You are now asked to finish the proof.
### Acknowledgment
The problem has been posted by Dao Thanh Oai at the CutTheKnotMath facebook page.
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2019-03-18 13:56:51
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http://mathhelpforum.com/calculus/48153-equation-line-tangent-graph-print.html
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# equation for the line tangent to graph
• September 8th 2008, 04:47 AM
jvignacio
equation for the line tangent to graph
hey i have im practising some problems and i dont know where to start on this question.
find an equation for the line tangent to the graph of y = secx at (pie/3,2)
any suggestions?
• September 8th 2008, 04:59 AM
TKHunny
1) Verify that (pi/3,2) is ON the graph. This is a point on the line.
2) Find the derivative of the function, y' = ???
3) Evaluate the derivaive at x = pi/3. This is the slope of the line.
4) Remember your algebra and use the Point-Slope form of a line.
• September 8th 2008, 06:06 AM
jvignacio
Quote:
Originally Posted by TKHunny
1) Verify that (pi/3,2) is ON the graph. This is a point on the line.
2) Find the derivative of the function, y' = ???
3) Evaluate the derivaive at x = pi/3. This is the slope of the line.
4) Remember your algebra and use the Point-Slope form of a line.
thank u ! ill give it a go and post it
• September 8th 2008, 06:56 AM
jvignacio
$
y = secx
$
$
y' = tanx \cdot secx
$
input x = $\frac{\pi}{3}$ into:
$
y' = tan(\frac{\pi}{3}) \cdot sec(\frac{\pi}{3})
$
$
m = 2 \cdot \sqrt{3}
$
now y - y1 = m(x - x1) formula to get the equation
$
y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})
$
is that the correct path?
• September 8th 2008, 07:49 AM
Krizalid
Yes.
• September 8th 2008, 11:07 AM
TKHunny
Quote:
Originally Posted by jvignacio
$
y = secx
$
$
y' = tanx \cdot secx
$
input x = $\frac{\pi}{3}$ into:
$
y' = tan(\frac{\pi}{3}) \cdot sec(\frac{\pi}{3})
$
$
m = 2 \cdot \sqrt{3}
$
now y - y1 = m(x - x1) formula to get the equation
$
y - 2 = 2 \cdot \sqrt{3}(x - \frac{\pi}{3})
$
is that the correct path?
Did you do my Step #1? You can be assured I would put one on the exam that was NOT actually on the curve.
Other than that...AWESOME!!
• September 9th 2008, 01:55 AM
jvignacio
Quote:
Originally Posted by TKHunny
Did you do my Step #1? You can be assured I would put one on the exam that was NOT actually on the curve.
Other than that...AWESOME!!
thakns mann!!
how do i check if its on the graph?
• September 9th 2008, 02:03 AM
11rdc11
Plug in the x coordinate given into the function and see if it matches with the y coordinate given.
• September 9th 2008, 02:04 AM
jvignacio
Quote:
Originally Posted by 11rdc11
Plug in the x coordinate given into the function and see if it matches with the y coordinate given.
yes it does! both 2 :)
• September 9th 2008, 09:54 AM
TKHunny
NOW we're done.
When the sneaky one shows up, you'll be the only one to get it right.
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2015-07-02 05:05:12
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https://or.stackexchange.com/questions/5636/binary-variables-with-multiple-indices
|
Binary variables with multiple indices
I'm new to Cplex and I'm working with the python API. I have a variable $$w^t_{iksm}$$ and I could just find the binary_var_cube function which only accepts 3 indices. I'm then using this $$w$$ to define $$X^z_{ij}$$ and $$Y_{im}$$ which are my decision variables. My question is, how should I code this variable with more than 3 indices?
I'm sorry if this is a basic one, but if you had any useful resources for beginners, please don't hesitate to share them with me.
Thanks
• I don't understand your question. It seems that you are asking 'how to define a decision variable with more than 3 indices in Cplex+python API'? I don't get what w_iksm^t and X_ij^z are in your question. Feb 2 '21 at 1:06
You can create a binary decision variable as:
from docplex.mp.model import Model
m = Model(...)
my_var = m.binary_var("name_of_this_var")
The variable is just an object and does not know how many indices it has. You can then maintain your own variable dictionary. So if you have a variable defined for the indices i,k,s,m,t, you could create a dictionary that maps keys defined as tuples to their corresponding variables:
var_dict[(i,k,s,m,t)]=m.binary_var("name_of_this_var")
To create a large number of variables simultaneously, and to produce a dict or a list, consider using the factory methods binary_var_dict or binary_var_list. I recommend that you study the examples provided with docplex.
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2022-01-28 02:20:22
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https://www.math.ucdavis.edu/research/seminars?talk_id=2975
|
Mathematics Colloquia and Seminars
Subexponential lower bounds for randomized pivoting rules for the simplex algorithm
Algebra & Discrete Mathematics
Speaker: Thomas Dueholm Hansen, Aarhus University Location: 2212 MSB Start time: Thu, Jun 2 2011, 4:10PM
The simplex algorithm is among the most widely used algorithms for solving linear programs in practice. Most deterministic pivoting rules are known, however, to need an exponential number of steps to solve some linear programs (Klee-Minty examples). No non-polynomial lower bounds on the expected number of steps were known, prior to this work, for randomized pivoting rules. We provide the first subexponential (i.e., of the form 2^(Omega(n^alpha)), for some alpha>0) lower bounds for the two most natural, and most studied, randomized pivoting rules suggested to date.
Joint work with Oliver Friedmann and Uri Zwick.
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2021-07-28 13:48:40
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https://iac.es/en/science-and-technology/publications?title=&type=All&journal=All&author=&created=All&sort_by=created&sort_order=DESC&page=8
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# Publications
• Measuring Cosmological Parameters with Type Ia Supernovae in redMaGiC Galaxies
Current and future cosmological analyses with Type Ia supernovae (SNe Ia) face three critical challenges: (i) measuring the redshifts from the SNe or their host galaxies; (ii) classifying the SNe without spectra; and (iii) accounting for correlations between the properties of SNe Ia and their host galaxies. We present here a novel approach that
Chen, R. et al.
10
2022
• SPECULOOS Northern Observatory: Searching for Red Worlds in the Northern Skies
SPECULOOS is a ground-based transit survey consisting of six identical 1 m robotic telescopes. The immediate goal of the project is to detect temperate terrestrial planets transiting nearby ultracool dwarfs (late M-dwarf stars and brown dwarfs), which could be amenable for atmospheric research with the next generation of telescopes. Here, we report
Burdanov, Artem Y. et al.
10
2022
• A refined dynamical mass for the black hole in the X-ray transient XTE J1859+226
We present two contiguous nights of simultaneous time-resolved Gran Telescopio Canarias spectroscopy and William Herschel Telescope photometry of the black hole X-ray transient XTE J1859+226, obtained in 2017 July during quiescence. Cross-correlation of the individual spectra against a late K-type spectral template enabled us to constrain the
Yanes-Rizo, I. V. et al.
11
2022
• Black hole mass and spin measurements through the relativistic precession model: XTE J1859+226
The X-ray light curves of accreting black holes and neutron stars in binary systems show various types of quasi-periodic oscillations (QPOs), the origin of which is still debated. The relativistic precession model identifies the QPO frequencies with fundamental time-scales from General Relativity, and has been proposed as a possible explanation of
Motta, S. E. et al.
11
2022
• Cosmological gas accretion history onto the stellar discs of Milky Way-like galaxies in the Auriga simulations - (I) Temporal dependency
We use the 30 simulations of the Auriga Project to estimate the temporal dependency of the inflow, outflow, and net accretion rates onto the discs of Milky Way-like galaxies. The net accretion rates are found to be similar for all galaxies at early times, increasing rapidly up to $\sim 10~\mathrm{M}_\odot \, \mathrm{yr}^{-1}$. After ~6 Gyr of
Iza, Federico G. et al.
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2022-11-29 00:54:49
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http://casa.colorado.edu/~ajsh/astr3740/prob2/prob2.html
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ASTR 3740 Relativity & Cosmology Spring 1998. Problem Set 2.
Due Tue 10 Mar
Warning: this problem set may take you several hours to complete, so please do not wait until the last day to start it.
I encourage you to take some time, while you are on the web, exploring about black holes and relativity. A good place to start is the relativity links attached to the ASTR 3740 homepage. I would love to know about any good links you find which are not already linked there.
Trajectories of particles in the Schwarzschild geometry
In this problem you will find it helpful to visit John Walker's web site at http://www.fourmilab.ch/gravitation/orbits/. The most fun part of the site is the Java applet, so you will probably want to seek out a Java-enabled machine, although you can also use the site without Java. Try http://www.colorado.edu/physics/2000 for advice on how to enable Java on a PC or Mac.
In what follows, the time t, radial coordinate r, polar angle , and azimuthal angle are the usual Schwarzschild coordinates in the Schwarzschild metric (with c = 1 as usual)
(1)
(2)
Without loss of generality, the trajectory of a particle falling freely in the Schwarzschild geometry may be taken to lie in the equatorial plane, . For a particle of finite (nonzero) mass, the trajectory satisfies the equations
(3)
where is the proper time of the particle, and E and L are constants, the particle's energy and angular momentum per unit mass. The quantity is the effective potential given by
(4)
(a) Check
Are John Walker's equations the same as the ones given above (aside from possible differences in notation)?
(b) Velocity at infinity
Argue from equations (3) that relative to the rest frame of the Schwarzschild geometry, the radial velocity vr and transverse velocity of the particle at extremely large distances from the Schwarzschild geometry, , are related to E and L by
(5)
(6)
(note that L can be extremely large at large r, so L/r is not necessarily zero in the limit ). Hence show that the velocity of the particle as is related to its energy E by
(7)
What does it mean if E < 1?
(c) Extrema of the effective potential
Find the radii at which the effective potential is a maximum or a minimum, i.e. , as a function of angular momentum L. You should find that extrema exist only if the absolute value |L| of the angular momentum exceeds a certain critical value Lc. What is that critical value?
(d) Sketch
Sketch what the effective potential looks like for values of L (i) less than, (ii) equal to, (iii) greater than the critical value Lc. Make sure to label the axes clearly. Describe physically, in words, what the possible orbital trajectories are for the various cases.
(e) Circular orbits
Circular orbits, satisfying , occur where the effective potential is a minimum (stable orbit) or a maximum (unstable orbit). Show (from your equation for the extrema of the effective potential) that the angular momentum L of a particle in circular orbit at radius r satisfies
(8)
and hence show also that the energy E in this circular orbit is
(9)
(f) Orbital period
Show that the orbital period t, as measured by an observer at rest at infinity, of a particle in circular orbit at radius r is given by Kepler's 3rd law (yes, it's true even in the fully general relativistic case!)
(10)
[Hint: The time measured by an observer at rest at infinity is just the Schwarzschild time t. Argue that the azimuthal angle evolves according to
(11)
The period t is the time taken for to change by .]
(g) Infall time
Calculate the proper time for a particle with L = 0 and E = 1 to fall from a finite radius r to the singularity at zero radius. What is the physical significance of the choice L = 0 and E = 1? [Hint: Write down the equation for for L = 0 and E = 1, and then solve it.]
(h) Infall time - numbers
Use your answer to part (g) to show that the proper time to fall from the Schwarzschild radius r = rs to the singularity (for L = 0 and E = 1) is, in units including c,
(12)
Evaluate your answer, in seconds, for the case of a black hole of mass 106 , such as may be in the center of our Galaxy, the Milky Way. [Constants: c = 299,792,458 m s-1; G = 6.67259 × 10-11 m3 kg-1 s-2; 1 = 1.99 × 1030 kg.]
(i) Suggestions
Briefly, do you have any suggestions for how John Walker might improve this web page (this one in particular, not the 100s of others he has)? Please be polite and helpful, and offer reasoned arguments rather than opinions.
Andrew Hamilton
2/26/1998
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2017-11-19 16:07:40
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https://aptitude.gateoverflow.in/8374/cat-2021-set-1-quantitative-aptitude-question-7?show=8570
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179 views
If the area of a regular hexagon is equal to the area of an equilateral triangle of side $12 \; \text{cm},$ then the length, in cm, of each side of the hexagon is
1. $6 \sqrt{6}$
2. $2 \sqrt{6}$
3. $4 \sqrt{6}$
4. $\sqrt{6}$
Let the side of hexagon be $x \; \text{cm}.$
The area of regular hexagon $= 6 \times \frac{\sqrt{3}}{4} x^{2}$
Now, $6 \times \frac{\sqrt{3}}{4} x^{2} = \frac{\sqrt{3}}{4} (12)^{2}$
$\Rightarrow 6x^{2} = 12 \times 12$
$\Rightarrow x^{2} = 24$
$\Rightarrow x = \sqrt{24}$
$\Rightarrow \boxed{ x = 2 \sqrt{6} \; \text{cm}}$
Correct Answer $: \text{B}$
$\textbf{PS:}$ The regular hexagon
The $\triangle \text{ABC}$ are equilateral triangle.
• The area of an equilateral triangle $= \frac{\sqrt{3}}{4} \times \text{(Side)}^{2}$
• The area of regular hexagon $= 6 \times \frac{\sqrt{3}}{4} \times \text{(Side)}^{2}$
10.3k points
1 vote
1
590 views
1 vote
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2023-02-08 14:25:31
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https://dealii.org/developer/doxygen/deal.II/namespaceParticles_1_1Generators.html
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Reference documentation for deal.II version GIT 0980a66d4b 2023-03-23 20:20:03+00:00
Particles::Generators Namespace Reference
## Functions
template<int dim, int spacedim = dim>
void regular_reference_locations (const Triangulation< dim, spacedim > &triangulation, const std::vector< Point< dim >> &particle_reference_locations, ParticleHandler< dim, spacedim > &particle_handler, const Mapping< dim, spacedim > &mapping=(ReferenceCells::get_hypercube< dim >() .template get_default_linear_mapping< dim, spacedim >()))
template<int dim, int spacedim = dim>
Particle< dim, spacedim > random_particle_in_cell (const typename Triangulation< dim, spacedim >::active_cell_iterator &cell, const types::particle_index id, std::mt19937 &random_number_generator, const Mapping< dim, spacedim > &mapping=(ReferenceCells::get_hypercube< dim >() .template get_default_linear_mapping< dim, spacedim >()))
template<int dim, int spacedim = dim>
ParticleIterator< dim, spacedim > random_particle_in_cell_insert (const typename Triangulation< dim, spacedim >::active_cell_iterator &cell, const types::particle_index id, std::mt19937 &random_number_generator, ParticleHandler< dim, spacedim > &particle_handler, const Mapping< dim, spacedim > &mapping=(ReferenceCells::get_hypercube< dim >() .template get_default_linear_mapping< dim, spacedim >()))
template<int dim, int spacedim = dim>
void probabilistic_locations (const Triangulation< dim, spacedim > &triangulation, const Function< spacedim > &probability_density_function, const bool random_cell_selection, const types::particle_index n_particles_to_create, ParticleHandler< dim, spacedim > &particle_handler, const Mapping< dim, spacedim > &mapping=(ReferenceCells::get_hypercube< dim >() .template get_default_linear_mapping< dim, spacedim >()), const unsigned int random_number_seed=5432)
template<int dim, int spacedim = dim>
void dof_support_points (const DoFHandler< dim, spacedim > &dof_handler, const std::vector< std::vector< BoundingBox< spacedim >>> &global_bounding_boxes, ParticleHandler< dim, spacedim > &particle_handler, const Mapping< dim, spacedim > &mapping=(ReferenceCells::get_hypercube< dim >() .template get_default_linear_mapping< dim, spacedim >()), const ComponentMask &components=ComponentMask(), const std::vector< std::vector< double >> &properties={})
template<int dim, int spacedim = dim>
void quadrature_points (const Triangulation< dim, spacedim > &triangulation, const Quadrature< dim > &quadrature, const std::vector< std::vector< BoundingBox< spacedim >>> &global_bounding_boxes, ParticleHandler< dim, spacedim > &particle_handler, const Mapping< dim, spacedim > &mapping=(ReferenceCells::get_hypercube< dim >() .template get_default_linear_mapping< dim, spacedim >()), const std::vector< std::vector< double >> &properties={})
## Detailed Description
A namespace that contains all classes that are related to the particle generation.
## ◆ regular_reference_locations()
template<int dim, int spacedim = dim>
void Particles::Generators::regular_reference_locations ( const Triangulation< dim, spacedim > & triangulation, const std::vector< Point< dim >> & particle_reference_locations, ParticleHandler< dim, spacedim > & particle_handler, const Mapping< dim, spacedim > & mapping = (ReferenceCells::get_hypercube() .template get_default_linear_mapping() ) )
A function that generates particles in every cell at specified particle_reference_locations. The total number of particles that is added to the particle_handler object is the number of locally owned cells of the triangulation times the number of locations in particle_reference_locations. An optional mapping argument can be used to map from particle_reference_locations to the real particle locations.
Parameters
triangulation The triangulation associated with the particle_handler. particle_reference_locations A vector of positions in the unit cell. Particles will be generated in every cell at these locations. particle_handler The particle handler that will take ownership of the generated particles. mapping An optional mapping object that is used to map reference location in the unit cell to the real cells of the triangulation. If no mapping is provided a MappingQ1 is assumed.
Definition at line 183 of file generators.cc.
## ◆ random_particle_in_cell()
template<int dim, int spacedim = dim>
Particle< dim, spacedim > Particles::Generators::random_particle_in_cell ( const typename Triangulation< dim, spacedim >::active_cell_iterator & cell, const types::particle_index id, std::mt19937 & random_number_generator, const Mapping< dim, spacedim > & mapping = (ReferenceCells::get_hypercube() .template get_default_linear_mapping() ) )
A function that generates one particle at a random location in cell cell and with index id. The function expects a random number generator to avoid the expensive generation and destruction of a generator for every particle and optionally takes into account a mapping for the cell. The algorithm implemented in the function is described in [71]. In short, the algorithm generates random locations within the bounding box of the cell. It then inverts the mapping to check if the generated particle is within the cell itself. This makes sure the algorithm produces statistically random locations even for nonlinear mappings and distorted cells. However, if the ratio between bounding box and cell volume becomes very large – i.e. the cells become strongly deformed, for example a pencil shaped cell that lies diagonally in the domain – then the algorithm can become very inefficient. Therefore, it only tries to find a location ni the cell a fixed number of times before throwing an error message.
Parameters
[in] cell The cell in which a particle is generated. [in] id The particle index that will be assigned to the new particle. [in,out] random_number_generator A random number generator that will be used for the creation of th particle. [in] mapping An optional mapping object that is used to map reference location in the unit cell to the real cell. If no mapping is provided a MappingQ1 is assumed.
Definition at line 239 of file generators.cc.
## ◆ random_particle_in_cell_insert()
template<int dim, int spacedim = dim>
ParticleIterator< dim, spacedim > Particles::Generators::random_particle_in_cell_insert ( const typename Triangulation< dim, spacedim >::active_cell_iterator & cell, const types::particle_index id, std::mt19937 & random_number_generator, ParticleHandler< dim, spacedim > & particle_handler, const Mapping< dim, spacedim > & mapping = (ReferenceCells::get_hypercube() .template get_default_linear_mapping() ) )
A function that generates one particle at a random location in cell cell and with index id. This version of the function above immediately inserts the generated particle into the particle_handler and returns a iterator to it instead of a particle object. This avoids unnecessary copies of the particle.
Definition at line 256 of file generators.cc.
## ◆ probabilistic_locations()
template<int dim, int spacedim = dim>
void Particles::Generators::probabilistic_locations ( const Triangulation< dim, spacedim > & triangulation, const Function< spacedim > & probability_density_function, const bool random_cell_selection, const types::particle_index n_particles_to_create, ParticleHandler< dim, spacedim > & particle_handler, const Mapping< dim, spacedim > & mapping = (ReferenceCells::get_hypercube() .template get_default_linear_mapping()), const unsigned int random_number_seed = 5432 )
A function that generates particles randomly in the domain with a particle density according to a provided probability density function probability_density_function. The total number of particles that is added to the particle_handler object is n_particles_to_create. An optional mapping argument can be used to map from particle_reference_locations to the real particle locations. The function can compute the number of particles per cell either deterministically by computing the integral of the probability density function for each cell and creating particles accordingly (if option random_cell_selection set to false), or it can select cells randomly based on the probability density function and the cell size (if option random_cell_selection set to true). In either case the position of individual particles inside the cell is computed randomly.
The algorithm implemented in the function is described in [71].
Parameters
[in] triangulation The triangulation associated with the particle_handler. [in] probability_density_function A function with non-negative values that determines the probability density of a particle to be generated in this location. The function does not need to be normalized. [in] random_cell_selection A bool that determines, how the number of particles per cell is computed (see the description above). [in] n_particles_to_create The number of particles that will be created by this function. [in,out] particle_handler The particle handler that will take ownership of the generated particles. [in] mapping An optional mapping object that is used to map reference location in the unit cell to the real cells of the triangulation. If no mapping is provided a MappingQ1 is assumed. [in] random_number_seed An optional seed that determines the initial state of the random number generator. Use the same number to get a reproducible particle distribution, or a changing number (e.g. based on system time) to generate different particle distributions for each call to this function.
Definition at line 276 of file generators.cc.
## ◆ dof_support_points()
template<int dim, int spacedim = dim>
void Particles::Generators::dof_support_points ( const DoFHandler< dim, spacedim > & dof_handler, const std::vector< std::vector< BoundingBox< spacedim >>> & global_bounding_boxes, ParticleHandler< dim, spacedim > & particle_handler, const Mapping< dim, spacedim > & mapping = (ReferenceCells::get_hypercube() .template get_default_linear_mapping()), const ComponentMask & components = ComponentMask(), const std::vector< std::vector< double >> & properties = {} )
A function that generates particles at the locations of the support points of a DoFHandler, possibly based on a different Triangulation with respect to the one used to construct the ParticleHandler. The total number of particles that is added to the particle_handler object is the number of dofs of the DoFHandler that is passed that are within the triangulation and whose components are within the ComponentMask. This function uses insert_global_particles and consequently may induce considerable mpi communication overhead.
This function is used in step-70.
Parameters
[in] dof_handler A DOF handler that may live on another triangulation that is used to establsh the positions of the particles. [in] global_bounding_boxes A vector that contains all the bounding boxes for all processors. This vector can be established by first using 'GridTools::compute_mesh_predicate_bounding_box()' and gathering all the bounding boxes using 'Utilities::MPI::all_gather(). [in,out] particle_handler The particle handler that will take ownership of the generated particles. The particles that are generated will be appended to the particles currently owned by the particle handler. [in] mapping An optional mapping object that is used to map the DOF locations. If no mapping is provided a MappingQ1 is assumed. [in] components Component mask that decides which subset of the support points of the dof_handler are used to generate the particles. [in] properties An optional vector of vector of properties for each particle to be inserted.
Definition at line 456 of file generators.cc.
void Particles::Generators::quadrature_points ( const Triangulation< dim, spacedim > & triangulation, const Quadrature< dim > & quadrature, const std::vector< std::vector< BoundingBox< spacedim >>> & global_bounding_boxes, ParticleHandler< dim, spacedim > & particle_handler, const Mapping< dim, spacedim > & mapping = (ReferenceCells::get_hypercube() .template get_default_linear_mapping()), const std::vector< std::vector< double >> & properties = {} )
A function that generates particles at the locations of the quadrature points of a Triangulation. This Triangulation can be different from the one used to construct the ParticleHandler. The total number of particles that is added to the particle_handler object is the number of cells multiplied by the number of particle reference locations which are generally constructed using a quadrature. This function uses insert_global_particles and consequently may induce considerable mpi communication overhead.
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2023-03-23 23:39:42
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http://mathhelpforum.com/differential-geometry/118234-counterexample-uniformly-convergence.html
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# Math Help - Counterexample to uniformly convergence
1. ## Counterexample to uniformly convergence
Hi! My problem is this: Find an example of $(f_n)$, a sequence of functions on $\mathcal{C}(X,\mathbb{R})$ (continuous with domain $X$ and real-valued) such that: $X$ is NOT compact, $(f_n)$ be equicontinuous and pointwise bounded, and every subsequence uniformly convergent have the same limit (call him $f$. In fact, may there's no subsequence uniformly convergent, and we aren't saying that every subsequence is uniformly convergent). I need to find a sequence that satisfy this conditions and NOT converges to f uniformly.
Thanks
Edit: I think that I don't need your help now =) With $f_n(x)=\dfrac{x}{n},\;X=\mathbb{R}$ holds
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2014-03-09 06:22:00
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https://quant.stackexchange.com/questions/65732/construction-of-vol-term-structure-for-libor
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# Construction of vol term structure for Libor
When we want to construct Interest rate term structure we look at various market instruments like futures, swaps etc. and using those quotes we use bootstrap method to construct term structure.
Now let say I want to create Volatility term structure for LIBOR using various caplet quotes. Could you please help me to find way how to use Caplets to build Volatility term structure for LIBOR?
Any pointer, online reference will be very helpful.
Thanks,
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2021-09-20 07:38:06
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https://cob.silverchair.com/jeb/article/210/24/4359/17261/Strategies-for-regulation-of-hemolymph-pH-in?searchresult=1
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The responses of larval Aedes aegypti to media of pH 4, 7 and 11 provide evidence for pH regulatory strategies. Drinking rates in pH 4 media were elevated 3- to 5-fold above those observed in pH 7 or 11. Total body water was elevated during acute exposure to acidic media. During chronic exposure, total body water was decreased and Malpighian tubule mitochondrial luminosity, quantified using Mitotracker Green FM, increased. Malpighian tubule secretion rates and energy demands thus appear to increase dramatically during acid exposure. In alkaline media, drinking rates were quite low. Larvae in pH 11 media excreted net acid (0.12 nequiv H+g–1 h–1) and the pH indicators azolitmin and bromothymol blue revealed that the rectal lumen is acidic in vivo at all ambient pH values. The anal papillae (AP) were found to be highly permeant to acid–base equivalents. Ambient pH influenced the length, and the mass-specific length, of the AP in the presence of NaCl (59.9 mmol l–1). In contrast, the length and mass-specific length of AP were not influenced by ambient pH in low NaCl conditions. Mitochondrial luminosity was reduced in AP of larvae reared in acidic media, and was not elevated in alkaline media, relative to that of larvae reared in neutral media. These data suggest that the AP may compromise acid–base balance in acidic media, and may also be an important site of trade-offs between H+ homeostasis and NaCl uptake in dilute, acidic media.
Maintenance of extracellular fluid pH is critically important for cellular function. For aquatic animals, ambient pH has direct effects on survival due to challenges to extracellular fluid ion and pH homeostasis, and indirect effects due to changes in the population dynamics of other species(Vangenechten et al., 1989). pH is thus an extremely important physical factor limiting the distribution and abundance of aquatic animals. Larval mosquitoes are remarkably tolerant of extreme pH, allowing them to exploit a wide variety of habitats in nature(Clements, 2000; Clark et al., 2004). Larvae of the mosquitoes Aedes aegypti and Ochlerotatus taeniorhynchuscomplete development in waters ranging from pH 4 to pH 11 in the laboratory. Across this range, hemolymph pH of acclimated larvae varies by only 0.1 pH units or less (Clark et al.,2004). They show phenotypic plasticity in pH tolerance in which acclimation to pH 4 or 11 increases their ability to tolerate subsequent exposure to more extreme pH (i.e. pH 3 or 12, respectively).
In larval endopterygotes, pH is regulated in two major extracellular compartments, the hemolymph and the midgut lumen. The cellular mechanisms involved in the generation of highly alkaline conditions within the midgut of lepidopteran and mosquito larvae have received considerable attention. In contrast, the physiology of acid–base homeostasis of insect hemolymph continues to be a neglected area of study. In fact, the specific contributions of each organ of the alimentary and excretory systems to elimination of a hemolymph acid or base challenge are known only for a single insect, the locust S. gregaria. In the locust, an acid load injected into the hemocoel was cleared primarily by the hindgut. The Malpighian tubules secreted acidic fluid, but their contributions to acid–base homeostasis were minor (Harrison et al., 1992; Stagg et al., 1991; Phillips et al., 1993; Harrison, 2001). Responses of locusts to alkaline challenges have not been described. Little is known about the regulation of hemolymph pH in other insects(Cooper, 1994; Harrison, 2001).
Aquatic insects rely primarily on epithelial transport across renal systems for acid–base homeostasis (Cooper,1994; Harrison,2001). The ion transport mechanisms relevant to acid–base homeostasis are expected to be located in ion-transporting epithelia including the midgut, the renal system (consisting of the Malpighian tubules and hindgut) and extrarenal ion-transporting organs. The Malpighian tubules form the primary urine, while the rectum recovers water and ions as necessary, and secretes additional ions. Aquatic insects frequently possess extrarenal organs such as anal papillae (AP), or clusters of mitochondria-rich chloride cells,that provide increased surface area for mechanisms involved in ionic homeostasis (Komnick et al.,1972). The Malpighian tubules and hindgut of Ochlerotatus taeniorhynchus possess inducible specific transport mechanisms for secretion of a variety of ions (Bradley and Phillips, 1977; Maddrell and Phillips, 1978). Most inducible ion transport systems are located in the rectum, with the exception of inducible SO42– transport, which occurs in the Malpighian tubules (Maddrell and Phillips,1978). Induction of H+ transport was not investigated.
The limited work on larval mosquito hemolymph pH homeostasis in alkaline media emphasizes the role of the rectum. Larvae of the mosquito Aedes dorsalis inhabit alkaline lakes containing high concentrations of bicarbonate, and under these conditions bicarbonate is excreted into the rectum via HCO3/Clexchange (Strange and Phillips,1984; Strange et al.,1982; Strange et al.,1984). The AP of larval mosquitoes and related Diptera consist of four elongated gill-like structures attached to the last abdominal segment. They are covered with cuticle, contain a cellular syncetium, and can be isolated from the hemolymph by the actions of a ring of muscles at their base(Clements, 2000). The AP of larval mosquitoes are known to be important sites of ion uptake, and are the region of the cuticle most permeant to water(Wigglesworth, 1933; Wigglesworth, 1938; Koch, 1938). Larval Aedes aegypti reared in dilute water have larger AP(Wigglesworth, 1938), with greater mitochondrial densities (Edwards and Harrison, 1983), than larvae from less dilute media. These increases have been attributed to increased energetic requirements for active uptake of NaCl from the medium (Koch,1938; Wigglesworth,1938; Edwards and Harrison,1983). The role of these organs in acid–base balance has not been investigated. Mechanisms used by larval mosquitoes to maintain hemolymph pH homeostasis in acidic media have not been addressed.
In the present study, responses of larval Aedes aegypti to acidic(pH 4), neutral and alkaline (pH 11) media allowed the elucidation of the overall physiological strategies of hemolymph pH homeostasis in this insect. This study focused on the roles of the Malpighian tubules, rectum and AP in the regulation of hemolymph pH. Evidence suggests that the larvae utilize novel and unexpected regulatory strategies to maintain homeostasis at each end of the tolerable pH range. It appears that the Malpighian tubules play a major role in acid excretion in acidic media. No evidence was obtained for an active role of any organ in acid–base homeostasis during alkaline challenges.
### Mosquitoes
The Aedes aegypti (L.) colony was derived in 1999 from a colony maintained at the Florida Medical Entomology Laboratory (Vero Beach, FL, USA). Eggs were hatched in deionized water. The next day, larvae were placed in rearing solutions of the appropriate pH and NaCl concentrations (see below). Larvae were maintained on a 16 h:8 h L:D photoperiod at 26°C. In all cases, rearing solutions were replaced and larvae were fed ground TetraMin flakes (TetraWerke, Melle, Germany) daily. Rearing solutions (RS) contained 2.5 mmol l–1 Trizma base, 2.5 mmol l–1 Hepes and 59.9 mmol l–1 NaCl unless stated otherwise (chemicals were obtained from Sigma, St Louis, MO, USA). RS were adjusted to the appropriate pH (3, 4, 7, 9 or 11) using HCl or NaOH, giving RS 3, 4, 7, 9 and 11, respectively. NaCl was used in initial studies so that adjustment of pH did not cause proportionally large differences in Na+ or Cl concentrations in the acidic and alkaline water(Clark et al., 2004). This allowed the observed physiological changes to be attributed with greater certainty to differences in pH. In the present study, rearing solutions low in NaCl were also prepared at pH 4, 7 and 11 (low-NaCl RS 4, 7 and 11). These low-NaCl RS contained the same buffers, and were adjusted to pH in the same way, but any NaCl present consisted only of that present in the food and the Na+ or Cl added while adjusting the pH of the solution.
### Mitochondrial luminosity
To isolate Malpighian tubules, the head and seventh abdominal segment of late-stage fourth instar larvae were severed under cold mosquito hemolymph substitute solution (HSS) (Clark et al.,2005). Fine forceps were then used to grasp the esophagus or the gastric caeca and pull the gut from the body using care to avoid stretching the tissue. To label mitochondria, guts with attached Malpighian tubules and terminal segments with AP were incubated for 30 min in cold HSS containing 50 nmol l–1 Mitotracker Green FM (Molecular Probes Inc., Eugene,OR, USA). They were then placed on a coverslip with a drop of glycerol. Mitotracker Green FM only fluoresces when bound to a mitochondrial membrane,and does not require respiring mitochondria (product literature, Molecular Probes Inc.). Mitotracker Green fluorescence was then recorded in whole mounts using a Zeiss Axioskop microscope (Oberkochen, Germany), with a Kubler Codex ebq100 lamp source (Gena, Germany). The excitation and emission wavelengths of Mitotracker Green are 490 and 516 nm respectively; an Omega Optical XF100-2 filter set (Brattleboro, VT, USA) was used. Images were captured at ×100 magnification using a Pixera Penguin 150CL digital camera (Los Gatos, CA, USA)under manual control (exposure 1/8 s). Because the same exposure time and magnification were used for all images, it was possible to quantify changes in the mitochondrial signature of the tissues. Images were imported into Adobe Photoshop where luminosity was quantified from the intensity of the fluorescence in boxes of uniform pixel numbers.
### Drinking rates
The rates at which acclimated larvae drink RS were determined at each pH by placing larvae in 1 ml of their respective RS containing 0.5 g l–1 fluorescein isothiocyanate conjugated to dextran(FITC-dextran, average molecular mass 4.3 kDa; Sigma) for a period of between 1 and 3 h. The treatment groups were sampled in repeated series during assays to avoid artifacts from possible time-dependent changes in drinking rates. Controls consisted of larvae ligated at the neck, and assayed alongside treatment groups at each pH. Following exposure to FITC-dextran, larvae were rinsed, blotted dry, weighed to the nearest 10 μg, and homogenized in 200μl of cold Tris-buffered saline. This step ensured that the fluorescence of FITC-dextran was always determined at the same final pH. Following centrifugation for 1 min at 18 000 g, the fluorescence of the supernatant was quantified in a Turner Designs TD-700 Fluorometer with minicell adaptor (Sunnyvale, CA, USA). The FITC-dextran content of each larva was determined by comparison with a standard curve. The fluorescence of ligated controls, due to a combination of autofluorescence and low rates of transcuticular FITC-dextran entry, was subtracted from treatment values assayed at the same pH. This corrected for possible artifacts such as effects of pH on transcuticular FITC-dextran permeability.
### Effects of pH challenges on total body water
The effects on total body water of the transfer of acclimated larvae to more acidic media were determined in the presence and absence of NaCl. Tests performed in the presence of NaCl involved rearing larvae in RS 7 or RS 4. Fourth instar larvae were then either maintained at the rearing pH or transferred to RS 3 for 2 h. In similar experiments without NaCl, larvae reared in low-NaCl RS 7 were assayed in either low-NaCl RS 7 or low-NaCl RS 3. Following exposure, larvae were blotted dry, and wet mass was determined to the nearest 1×10–5 g using a Mettler-Toledo AX205 deltarange balance (Columbus, OH, USA). The larvae were then dried overnight in a drying oven at 96°C, and reweighed. The decrease in mass represents total free body water while the ratio of body water to total mass represents the percentage body water or body water ratio, calculated as: [(wet mass– dry mass)/wet mass].
### pH of the rectum in vivo
Larvae reared in RS 4, 7 and 11 were exposed to kaolin and azolitmin (1 g per 50 ml) for 24 h. Kaolin is an inert silicate that is ingested by larvae,displacing the food column and providing a background against which the color of ingested pH indicators can be established(Dadd, 1975). They were then transferred to RS 4, 7 or 11. Each combination of rearing and acute pH exposure was performed. The larvae were photographed after 24 h of exposure to the assay pH using an Olympus C-3040 digital camera through a Leica Stereozoom 6 microscope (Wetzlar, Germany).
### pH of the AP in vivo
It was noted that bromothymol blue enters the AP in acidic media allowing the investigation of the acid–base permeability of the AP. Bromothymol blue was initially dissolved in dimethyl sulfoxide (7.6 mg per 10 ml). This mixture was vortexed, then centrifuged. The supernatant was diluted 1:1000 in RS 4. Larvae were exposed to this solution for 24 h. They were then placed into RS lacking bromothymol blue. The pH of the AP was noted, and the larvae were photographed within 5 min and at 24 h of exposure to the assay pH using an Olympus C-3040 digital camera through a Leica Stereozoom 6 microscope. Bromothymol blue was also visible in the rectum of several of these larvae allowing confirmation of the azolitmin results.
### Morphology of the AP
Lengths of AP were determined for larvae reared in RS 4, 7 and 11, in low-NaCl RS 4, 7 and 11, and in deionized water, 5.25 g l–1artificial seawater and 10.5 g l–1 artificial seawater(Instant Ocean, filtered; Aquarium Systems, Mentor, OH, USA). Lengths of AP of larvae fixed in 2% glutaraldehyde in PBS were determined with a Leica Stereozoom 6 with ocular micrometer. Larvae were oven dried for 24 h at 96°C, then weighed to the nearest 1×10–5 g in order to determine the mass-specific length of the AP.
### Rates of acid–base excretion
Larval acid–base excretion rates were determined under controlled conditions by rinsing fed, acclimated larvae twice in fresh RS, and placing them individually into 2.0 ml of fresh RS 4 or RS 11. RS aliquots without larvae were run alongside larvae and served as paired negative controls. After a period of 1–2 h, the pH of the experimental and control solutions was determined. Titration of the RS allowed calculation of the rate of acid excretion from the difference in pH of the experimental and control solutions.
### Effects of ambient pH on drinking rates, total body water and excretion rates
The effects of ambient pH on drinking rates were determined in animals reared in the presence (RS) and absence (low-NaCl RS) of NaCl. Larvae in acidic media (RS 4 or low-NaCl RS 4) drank the medium at much higher rates than larvae in neutral or alkaline media(Fig. 1A; P<1×10–8, F=18.59, d.f.=47). The lowest rates observed were in highly alkaline media (pH 11), where drinking rates were 0.28±0.051 ml g–1 day–1 in the presence of NaCl (RS 11) and 0.88±0.168 ml g–1day–1 in the absence of NaCl (low-NaCl RS 11). Thus, even in the most alkaline media the drinking rates were substantial, representing 28–88% of total body volume each day. In acidic media, larvae consumed 500–700% of their total body volume each day. These values are similar to those reported for mosquito larvae under other conditions(Bradley, 1987; Clements, 2000). The presence or absence of NaCl did not influence the effect of pH on drinking rates[P>0.05, single factor ANOVA followed by Student–Newman–Keuls (SNK) a posteriori test withα=0.05].
Total body water was influenced by both ambient pH and NaCl(Fig. 1B; P<1×10–8, F=18.59, d.f.=47, single factor ANOVA of arcsine-transformed body water ratios, followed by SNK withα=0.05) (Sokal and Rohlf,1969). Total body water was reduced in larvae reared in RS 4 relative to RS 7. Total body water increased when larvae reared in RS 4 were transferred to RS 3, and when larvae reared in RS 7 or low-NaCl RS 7 were transferred to RS 3 or low-NaCl RS 3, respectively. Because chronic exposure to acidic water resulted in increased drinking rates, but decreased percentage body water, fluid excretion rates must increase as well. In the presence of NaCl, fluid ingestion and excretion rates in acidic water (pH 4) were approximately 20 times those in alkaline water (pH 11; 5.9 vs 0.3 ml g–1 day–1). In the absence of NaCl, they were 7.9-fold greater in acidic water than in alkaline water (7.1 vs0.9 ml g–1 day–1). Drinking rates were similar in neutral and alkaline media and in the presence and absence of NaCl(Fig. 1).
### Effects of ambient pH on acid–base excretion
Larvae in RS 4 did not change the pH relative to controls over 1–2 h time intervals. Larvae in RS 11 excreted net acid (0.12±0.017 nequiv g–1 h–1, mean ± s.e.m.) relative to paired larva-free controls (larvae vs larva-free controls, P<1×10–9, two-factor ANOVA without replication, EXCEL).
Fig. 1.
Effects of acid–base exposure on drinking rates and total body water. Animals were reared and assayed in media containing NaCl (RS; 59.9 mmol l–1), or in media in which the only added NaCl was that present in the food (low-NaCl RS). (A) Effects of pH on drinking rates in the presence and absence of NaCl. Drinking rates of acclimated larvae were determined using FITC-dextran ingestion rates (average molecular mass 4.3 kDa). Drinking rates were highest in acidic rearing solutions (column effects; P<1×10–8, F=50.8, d.f.=47). The presence or absence of NaCl did not influence the effect of pH on drinking rates (column effects; F=1.37, P>0.22). Rearing solutions containing NaCl are indicated by •, while those without added NaCl are indicated by ○. (B) Effects of ambient pH on percentage body water. Body water was reduced in larvae chronically exposed to acidic media, and was elevated when larvae were acutely exposed to more acidic media (single factor ANOVA of arcsine-transformed body water ratio). The presence or absence of NaCl did not influence the effects of acute pH challenges on body water. RS 7 represents larvae reared and assayed in RS 7, RS 7–3 represents larvae reared in RS 7 and assayed in RS 3, RS 4 represents larvae reared and assayed in RS 4, and RS 4–3 represents larvae reared in RS 4 and transferred to RS 3. For larvae reared and assayed in low NaCl conditions, low-NaCl RS 7 represents larvae reared and assayed in low-NaCl RS 7, while low-NaCl RS 7–3 represents larvae reared in low-NaCl RS 7 and assayed in low-NaCl RS 3. Data are presented as means ± s.e.m.
Fig. 1.
Effects of acid–base exposure on drinking rates and total body water. Animals were reared and assayed in media containing NaCl (RS; 59.9 mmol l–1), or in media in which the only added NaCl was that present in the food (low-NaCl RS). (A) Effects of pH on drinking rates in the presence and absence of NaCl. Drinking rates of acclimated larvae were determined using FITC-dextran ingestion rates (average molecular mass 4.3 kDa). Drinking rates were highest in acidic rearing solutions (column effects; P<1×10–8, F=50.8, d.f.=47). The presence or absence of NaCl did not influence the effect of pH on drinking rates (column effects; F=1.37, P>0.22). Rearing solutions containing NaCl are indicated by •, while those without added NaCl are indicated by ○. (B) Effects of ambient pH on percentage body water. Body water was reduced in larvae chronically exposed to acidic media, and was elevated when larvae were acutely exposed to more acidic media (single factor ANOVA of arcsine-transformed body water ratio). The presence or absence of NaCl did not influence the effects of acute pH challenges on body water. RS 7 represents larvae reared and assayed in RS 7, RS 7–3 represents larvae reared in RS 7 and assayed in RS 3, RS 4 represents larvae reared and assayed in RS 4, and RS 4–3 represents larvae reared in RS 4 and transferred to RS 3. For larvae reared and assayed in low NaCl conditions, low-NaCl RS 7 represents larvae reared and assayed in low-NaCl RS 7, while low-NaCl RS 7–3 represents larvae reared in low-NaCl RS 7 and assayed in low-NaCl RS 3. Data are presented as means ± s.e.m.
### Effects of ambient pH on mitochondria
Changes are observed in the fluorescence of the mitochondrial dye Mitotracker Green FM in response to chronic pH exposure. Mitochondrial luminosity was higher in both proximal and distal Malpighian tubule regions of animals reared in acidic water than in larvae reared in neutral or alkaline media, which did not differ in luminosity(Fig. 2; proximal tubule: P<0.05, F=4.24, d.f.=17; distal tubule: P<0.005, F=8.92, d.f.=17; followed by SNK)(Sokal and Rohlf, 1969). In larvae reared in acidic media (RS 4), the distal Malpighian tubule showed significantly greater luminosity than the proximal region (P<0.05,two-tailed, paired Student's t-test, N=6 tubules). No differences in luminosity were observed within pH between proximal and distal regions of the Malpighian tubules of larvae reared in RS 7 or RS 11 (RS 7, P>0.5; RS 11, P>0.09, two-tailed, paired t-test, N=6 tubules/RS). The influence of pH on hindgut mitochondrial luminosity was not determined due to difficulties experienced in the removal of the rectal contents, which autofluoresce.
Fig. 2.
Mitochondrial luminosity within the Malpighian tubules of animals reared in acidic (pH 4), neutral or alkaline (pH 11) media. Mitochondrial densities were influenced by environmental pH within the proximal tubule (P<0.05, F=4.24, d.f.=17) and within the distal tubule (P<0.003, F=8.92, d.f.=17, single factor ANOVA). N=6/pH value. All measurements were made with an exposure time of 1/8 s and at the same magnification (×100). Data are presented as means ± s.e.m.
Fig. 2.
Mitochondrial luminosity within the Malpighian tubules of animals reared in acidic (pH 4), neutral or alkaline (pH 11) media. Mitochondrial densities were influenced by environmental pH within the proximal tubule (P<0.05, F=4.24, d.f.=17) and within the distal tubule (P<0.003, F=8.92, d.f.=17, single factor ANOVA). N=6/pH value. All measurements were made with an exposure time of 1/8 s and at the same magnification (×100). Data are presented as means ± s.e.m.
Mitotracker Green FM produced uneven fluorescence within the AP. Luminosity was always greatest in the proximal and distal regions, and lowest in the central region (Fig. 3; by region, pH 4: P<0.0001, F=12.646, d.f.=65; pH 7: P<0.01, F=5.137, d.f.=56; pH 11: P<0.0005, F=8.777, d.f.=62; single factor ANOVA followed by SNK). Changes were observed in Mitotracker Green FM fluorescence, within the AP, in response to chronic pH exposure. Animals reared in RS 4 showed reduced luminosity across regions compared with animals reared in RS 7 or 11(Fig. 3; P<0.05, F=3.161, d.f.=61, single factor ANOVA followed by SNK.
In the absence of Mitotracker Green FM no fluorescence was observed at the exposure time of 1/8 s used in these experiments.
### The pH of the rectal lumen
The pH indicator azolitmin revealed that larvae always have an acidic rectal lumen (pH <6.2), even when chronically exposed to RS 11(Fig. 4; N=12 each in RS 4, RS 7 and RS 11). Similar results were obtained using bromothymol blue(transition from yellow to blue at pH 6.8; Fig. 5). Obtaining similar results with two separate pH indicators provides increased confidence that the excretory system of larvae exposed to highly alkaline media eliminates fluid more acidic than the hemolymph.
### Effects of ambient pH on the pH within the AP
The pH indicator bromothymol blue entered the AP of larvae in RS 4, but not in RS 7 or RS 11. In RS 4, the bromothymol blue within the AP was yellow,revealing that the pH was <6.8 (Fig. 5A). When these larvae were transferred to RS 11, the bromothymol blue within the AP changed from yellow to blue within minutes. The blue color remained within the AP for at least 24 h following transfer to RS 11,revealing (1) diffusion trapping of the ionized form upon alkalization, and(2) that the pH within the AP remained >6.8 during that time(Fig. 5B). These data reveal a high degree of transparency of the AP to acid–base equivalents, and failure to recover the initial acidic pH within the AP in the face of alkalization of the ambient media.
Fig. 3.
Mitochondrial luminosity of anal papillae (AP) of animals reared in acidic(pH 4), neutral or alkaline (pH 11) media. All measurements were made with an exposure time of 1/8 s and at the same magnification. Luminosity was determined for the base, midsection and tip of one AP from each animal. Luminosity was always greatest in proximal and distal regions, and lowest in the central region. Larvae reared in acidic media (RS 4) showed reduced mitochondrial densities in the midsection and tip, and mitochondrial densities averaged across regions were also reduced in RS 4 (P<0.05).
Fig. 3.
Mitochondrial luminosity of anal papillae (AP) of animals reared in acidic(pH 4), neutral or alkaline (pH 11) media. All measurements were made with an exposure time of 1/8 s and at the same magnification. Luminosity was determined for the base, midsection and tip of one AP from each animal. Luminosity was always greatest in proximal and distal regions, and lowest in the central region. Larvae reared in acidic media (RS 4) showed reduced mitochondrial densities in the midsection and tip, and mitochondrial densities averaged across regions were also reduced in RS 4 (P<0.05).
### Effects of salinity and ambient pH on larval mass and the size of the AP
The size of AP is influenced by ambient salinity and by pH (Figs 6 and 7). The AP were longest in dilute water, and length decreased with salinity(Fig. 6A; P<0.05, F=3.97, d.f.=25, single factor ANOVA). The dry mass of larvae used to determine the effects of environmental conditions on the length of AP differed in both the salinity and pH experiments. Significant differences in dry mass were recorded among salinities (P<0.05, F=4.66, d.f.=25). Among media differing in pH values, significant differences were recorded in the presence of NaCl (P<0.0005, F=11.25, d.f.=29) but not in the absence of NaCl (P>0.21, F=1.64, d.f.=29; Table 1). However, because salinity also influences mass, the mass-specific lengths of the AP were also determined. No differences were observed in mass-specific lengths (mm mg–1 dry mass) of the AP across salinities(Fig. 6B; P>0.27, F=1.37, d.f.=25, single factor ANOVA) although larvae reared at the highest salinities possessed the smallest AP.
Table 1.
Dry mass of larvae used to determine the effects of salinity and pH on the relative size of anal papillae
RS with NaCl
RS no NaCl
Salinity (g l–1)Dry mass (mg)pHDry mass (mg)pHDry mass (mg)
0.48±0.052 0.55±0.053 0.32±0.027
5.25 0.31±0.03 0.36±0.019 0.31±0.022
10.5 0.46±0.045 11 0.34±0.019 11 0.38±0.034
RS with NaCl
RS no NaCl
Salinity (g l–1)Dry mass (mg)pHDry mass (mg)pHDry mass (mg)
0.48±0.052 0.55±0.053 0.32±0.027
5.25 0.31±0.03 0.36±0.019 0.31±0.022
10.5 0.46±0.045 11 0.34±0.019 11 0.38±0.034
Rearing solutions (RS) varied in salinity (columns 1, 2), pH in the presence of NaCl (columns 3, 4) and pH without NaCl (columns 5, 6)
In contrast to salinity, pH influenced both the length and mass-specific length of the AP of larvae reared in RS containing 59.9 mol l–1 NaCl (Fig. 7A,B; length: P<0.01, F=6.96, d.f.=29; mass specific length: P<0.0001, F=15.34, d.f.=29, 2-way P values, single factor ANOVA). Under these conditions, the AP of larvae reared in RS 4 were reduced in both length and mass-specific length relative to those reared in RS 7, while those reared in RS 11 possessed the longest AP. The decrease in mass-specific length in acidic media was proportionally much greater than the decrease in length. In contrast, pH had no effect on the length, or mass-specific length, of the AP of larvae reared in low-NaCl RS (Fig. 7A,B;length: P>0.27, F=1.39, d.f.=29; mass-specific length: P>0.87, F=0.14, d.f.=29, 2-way P values, single factor ANOVA).
Fig. 4.
The rectum of a larva that had ingested azolitmin in vivo, reared and assayed in alkaline water (pH 11). Azolitmin is blue in alkaline water (pH>6.2), and red in acidic water (pH <6.2). A total of 36 larvae were assayed (12 larvae each in RS 4, RS 7 and RS 11). The rectal contents were red(pH <6.2) in all larvae. Top, the original image. Bottom, the same image adjusted for brightness (+39) and contrast (+35) using Adobe Photoshop.
Fig. 4.
The rectum of a larva that had ingested azolitmin in vivo, reared and assayed in alkaline water (pH 11). Azolitmin is blue in alkaline water (pH>6.2), and red in acidic water (pH <6.2). A total of 36 larvae were assayed (12 larvae each in RS 4, RS 7 and RS 11). The rectal contents were red(pH <6.2) in all larvae. Top, the original image. Bottom, the same image adjusted for brightness (+39) and contrast (+35) using Adobe Photoshop.
### Conclusion
In acidic media, larval Aedes aegypti utilize novel strategies for the maintenance of hemolymph acid–base homeostasis involving increased rates of ingestion of the medium followed by fluid and acid clearance in which the Malpighian tubules play a prominent role. The mechanisms of homeostasis in alkaline media remain obscure and appear to be largely passive.
### The role of the Malpighian tubules in acid–base homeostasis
The Malpighian tubules appear to be primarily responsible for acid–base homeostasis in acidic media. During exposure to acidic media,larvae increased drinking rates substantially. Naïve larvae became volume loaded whereas acclimated larvae maintained high drinking rates yet showed reduced body water. Exposure to acidic media thus results in increased rates of fluid flux through the organism. Acclimation to acidic media also resulted in an increase in fluorescence of a mitochondria-specific dye in the Malpighian tubules. The increase in Malpighian tubule mitochondria is presumably paralleled by upregulation of specific transport mechanisms driving increased rates of fluid secretion.
We have not yet established the pH of the fluid produced by the larval Malpighian tubules, or whether the pH of the secreted fluid changes during acid exposure. However, it is not necessary for the H+concentration or buffer capacity of the excreted fluid to increase for overall Malpighian tubule H+ excretion rates to increase, if the volume of secreted fluid increases. Malpighian tubules of adult A. aegyptisecrete acidic fluid, and cyclic AMP increases the rate of fluid secretion while pH remains constant, resulting in elevated H+ clearance rates(Petzel et al., 1999). The transport mechanisms of larval and adult Malpighian tubules differ(Clark and Bradley, 1996),however, and it remains to be determined whether the larval Malpighian tubules respond in a similar way during acid exposure. We also do not rule out an additional increase in H+ excretion rates due to an increase in secretion of buffer or ammonia during acid challenges, as occurs in the mammalian kidney.
Fig. 5.
The pH within the AP changes in response to changes in ambient pH in vivo. Top, larva exposed to bromothymol blue in vivo in pH 4 media (RS 4). Under these conditions bromothymol blue enters the AP. Bromothymol blue is yellow at pH <6.8 and blue at pH >6.8. In acid media, the interior of the AP is thus acidic (pH <6.8). Bottom, AP that have been loaded' by exposure of larvae to bromothymol blue in acidic media followed by transfer to pH 11 media (RS 11) for at least 24 h. Note that the rectal lumen is acidic under these conditions.
Fig. 5.
The pH within the AP changes in response to changes in ambient pH in vivo. Top, larva exposed to bromothymol blue in vivo in pH 4 media (RS 4). Under these conditions bromothymol blue enters the AP. Bromothymol blue is yellow at pH <6.8 and blue at pH >6.8. In acid media, the interior of the AP is thus acidic (pH <6.8). Bottom, AP that have been loaded' by exposure of larvae to bromothymol blue in acidic media followed by transfer to pH 11 media (RS 11) for at least 24 h. Note that the rectal lumen is acidic under these conditions.
### The role of the rectum in acid–base homeostasis
Evidence presented here suggests that the role of the larval mosquito rectum in alkaline media should be re-evaluated. A series of papers by Strange and coworkers (Strange and Phillips,1984; Strange et al.,1982; Strange et al.,1984) showed that larval Aedes dorsalis exposed to alkaline media (pH 10.5) high in HCO3 (up to 250 mmol l–1) eliminate this ion via rectal HCO3/Cl exchange. It is not clear from the available data whether the rectum of larval Aedes aegypti plays a role in acid–base homeostasis. We have found that larval A. aegypti has a similar ability to tolerate highly alkaline media without ill effect, at least in media rendered alkaline using NaOH(Clark et al., 2004). Under these conditions, the pH within the rectal lumen of larval A. aegyptiis always acidic in vivo (pH <6.2). The pK of H2CO3/HCO3 is 6.4 at 25°C (Weast et al., 1986). Most HCO3 excreted into the rectum at the observed pH <6.2 would form CO2, assuming equilibrium is reached by the time the excretory product is eliminated. Any CO2 so formed would then most likely diffuse through the tissues and out through the cuticle, resulting in net acid excretion(Wigglesworth, 1938). If the physiology of acid–base regulation is similar in larvae of A. aegypti and A. dorsalis, the high rates of HCO3 excretion observed in the rectum of A. dorsalis (Strange and Phillips,1984; Strange et al.,1984) may have been induced by the high HCO3 concentration of the medium (250 mmol l–1) rather than by its alkaline pH (pH 10.5). This deserves further study.
Fig. 6.
Length (A) and mass-specific length (B) of the AP as a function of salinity. Larvae were reared in different concentrations of artificial seawater. (A) Length was significantly influenced by salinity(P<0.05, F=3.97, d.f.=25, single factor ANOVA). (B)Mass-specific length of the AP (relative to dry mass) of larvae reared at different salinities. No significant mass-specific effects of salinity were observed (P>0.27, F=1.37, d.f.=25).
Fig. 6.
Length (A) and mass-specific length (B) of the AP as a function of salinity. Larvae were reared in different concentrations of artificial seawater. (A) Length was significantly influenced by salinity(P<0.05, F=3.97, d.f.=25, single factor ANOVA). (B)Mass-specific length of the AP (relative to dry mass) of larvae reared at different salinities. No significant mass-specific effects of salinity were observed (P>0.27, F=1.37, d.f.=25).
The relative contributions of the Malpighian tubules, midgut and hindgut in secretion of the acid observed in the rectal lumen were not determined. We could not measure mitochondrial luminosity of the hindgut due to interference from non-specific fluorescence of lumen contents, which were much more difficult to remove from this region of the alimentary canal. An increase in mitochondrial luminosity would not necessarily indicate a direct role in acid–base transport, however. Elevation of Malpighian tubule secretion rates, observed in acidic media, requires increased rates of rectal recovery of solutes such as K+, Cl, etc., that drive the formation of Malpighian tubule secretions. Changes in energy demands of the rectum in response to ambient pH would therefore provide little information regarding acid–base transport by this organ.
Fig. 7.
Length (A) and mass-specific length (B) of the AP as a function of pH. The influence of pH was determined in the absence (○) and presence (•) of added NaCl. (A) Length was significantly influenced by pH in the presence(P<0.005, F=8.24, d.f.=29) but not the absence(P>0.19, F=1.75, d.f.=29) of NaCl. (B) In the presence of NaCl, pH had a highly significant influence on mass-specific length(P<0.0001, F=13.23, d.f.=29). In the absence of NaCl, no influence of pH on mass-specific length was observed (P>0.86, F=0.15, d.f.=29).
Fig. 7.
Length (A) and mass-specific length (B) of the AP as a function of pH. The influence of pH was determined in the absence (○) and presence (•) of added NaCl. (A) Length was significantly influenced by pH in the presence(P<0.005, F=8.24, d.f.=29) but not the absence(P>0.19, F=1.75, d.f.=29) of NaCl. (B) In the presence of NaCl, pH had a highly significant influence on mass-specific length(P<0.0001, F=13.23, d.f.=29). In the absence of NaCl, no influence of pH on mass-specific length was observed (P>0.86, F=0.15, d.f.=29).
### The role of the AP in acid–base homeostasis
Acid efflux has been shown in the AP of larval Aedes aegypti in neutral media (Donini and O'Donnell,2005). In the present study, the changes observed in pH within the AP in response to changes in ambient pH are not consistent with an active role of these organs in acid–base homeostasis. Instead, the evidence suggests that acid–base flux across the AP is passive. First of all, one might expect the AP to become more alkaline, rather than acidic, if excreting acid,and to become more acidic if excreting base – in fact, the reverse occurs. Second, the change in pH within the papillae upon transfer from acidic to alkaline media is not reversed during at least 24 h of exposure, suggesting limited ability to regulate flux over this time scale. The passive role of the AP in acid–base homeostasis is supported by comparison of the phenotypic plasticity of the AP in response to pH and to NaCl. It is well established that active NaCl uptake from dilute media occurs in the AP(Wigglesworth, 1933; Wigglesworth, 1938; Koch, 1938). AP of larvae reared in dilute media show increased size and mitochondrial densities contributing to greater active NaCl absorption from the medium(Edwards and Harrison, 1983; Koch, 1938; Wigglesworth, 1938). In the present study, we demonstrated that the size and mitochondrial densities of the AP are all reduced, rather than increased, in acidic media relative to those from larvae reared in neutral and alkaline media (in the presence of NaCl). Reducing the relative surface area of the highly permeable cuticle appears to be a mechanism to reduce rates of passive transcuticular H+ influx into the AP. The reduction in mitochondrial densities in acidic media is also not consistent with an active role of the AP in acid excretion. The hypothesis that the AP are important sites of active acid–base excretion in acidic media is therefore not supported.
No evidence was obtained supporting the AP as major sites of active acid–base excretion in alkaline media. Size and mitochondrial densities of the AP are similar in larvae reared in neutral and alkaline media.
The response of the AP to acidic media is influenced by the presence or absence of NaCl. The AP were reduced in size in acidic media in the presence of NaCl, but not in its absence. This suggests a trade-off between NaCl uptake and acid–base homeostasis occurring in the AP of larvae in dilute,acidic media. In a wide variety of aquatic organisms, death in acidic water is due primarily to loss of Na+, rather than failure of pH homeostasis(Vangenechten et al., 1989; Havas, 1981; Havas and Advokaat, 1995; Lin and Randall, 1995). This appears to be related to Na+/H+ exchange mechanisms(Stobbart, 1971). In the fish gill, for example, Na+ uptake is driven by active H+secretion. Increased ambient H+ concentrations therefore increase the electrochemical gradients opposing H+ secretion, reducing Na+ uptake (Lin and Randall,1995). Similarly, larvae of Aedes aegypti and Culex quinquefasciatus show decreased Na+ uptake rates during acute exposure to acidic water (pH 3.5) (Patrick et al., 2002). It thus appears that AP surface area does not decrease in dilute media of low pH, because a reduction in size of the AP in dilute, acidic media would compromise NaCl uptake leading to the failure of Na+ homeostasis. If such a trade-off exists, A. aegyptilarvae appear to be able to minimize its effects(Clark et al., 2004). Intriguingly, the apical surface area of fish gill chloride cells, also involved in the uptake of ions from dilute media, is reduced during acid exposure (Goss et al., 1995). Could reduced surface area of permeable Na+ uptake surfaces, a response driven by the deleterious consequences of high rates of passive H+ influx, contribute to the trade-off between Na+ and H+ homeostasis in a variety of animals exposed to acidic water?
It is highly unlikely that the decrease in size of the AP of larvae reared in acidic media containing NaCl is a response to the increased Cl concentrations resulting from the HCl added during adjustment of the pH of the rearing medium. The medium [Cl]was 59.9 mmol l–1, and the small amounts of HCl required to adjust the pH to 4 were of the order of a few millimolar. In addition, the length and mass-specific length of the AP of larvae reared in acidic media in low-NaCl conditions (with pH adjusted using HCl) were not reduced compared with those of larvae reared in deionized water (0 g l–1, see Fig. 2).
During the course of this investigation, we repeated classic studies documenting the influence of salinity on the size of the AP(Wigglesworth, 1938; Koch, 1938), in order to determine the relative magnitude of the effects of NaCl and pH on AP size. Those studies had found that the AP of larval mosquitoes reared in dilute media were larger than those of larvae reared at higher salinities, although AP of Aedes aegypti did not respond as strongly to ambient salinity as did those of several other species. As reported in those studies, we too found that AP are longest in larvae reared in dilute media. However, when scaled for mass, we did not observe any relationship between salinity and AP length. Under the conditions of this study, the effects of salinity on the length of the AP appear to be more strongly associated with allometric effects of salinity on overall growth, rather than changes in the relative sizes of ion uptake surfaces. We do not question the role of the AP in NaCl uptake. Note, however, that pH (in the presence of NaCl) has a much greater effect on AP size than does salinity in this species suggesting that acid–base flux into the AP in acidic media is of considerable physiological significance. It is possible that estimation of total surface area or a greater effort to eliminate NaCl (for example by feeding larvae NaCl-free food) may have resulted in a stronger relationship between mass-specific length and salinity in the present study.
### Physiology of acid–base regulation in acidic media
In acidic media, larvae greatly increase drinking rates. This increase is likely to impose a considerable energy cost on the animal. The insect excretory system drives water movement by solute transport. Larvae in acidic media ingest much greater fluid volumes, the ingested fluid contributes to the acid load to be eliminated, and ions used to drive urine secretion must be recovered from the excreta prior to its elimination. Why then would larvae increase drinking rates in acidic media? Two hypotheses come to mind. According to one hypothesis, larvae exposed to acidic media increase drinking rates in order to reduce the transepithelial H+ gradient opposing clearance of H+. This would resemble the mammalian kidney, which can only excrete H+ into filtrate of pH >4.5 but can increase clearance of H+ by addition of buffers and NH3 to the urine. We hypothesize that an increased volume of fluid of a given pH and buffer capacity allows a greater acid load to be cleared without increasing the driving force opposing H+ excretion. For this to work, the increase in clearance capacity must compensate for the additional acid load ingested with the medium. A similar hypothesis was proposed to explain an increase in drinking rates observed in larvae exposed to elevated salinity(Clements, 2000).
A second hypothesis that might explain the increase in drinking rates in acid water is based on Na+/H+ exchange processes(discussed above). It is thus conceivable that the increase in drinking rates in acidic media functions to increase Na+ ingestion, offsetting the reduction in Na+ influx caused by elevated ambient H+concentrations. This hypothesis must be rejected in the present study,however. Larvae ingested the medium at similar rates in acidic media differing in NaCl concentration. Indeed, drinking rates increased approximately 5-fold between pH 4 and pH 7 in low-NaCl RS, but only by 2- to 3-fold in RS containing NaCl (59.9 mmol l–1).
### Physiology of acid–base regulation in alkaline media
We had expected to find that larvae exposed to alkaline media would excrete fluid more alkaline than the hemolymph, although not necessarily more alkaline than the environment, to remain in pH balance. We were surprised to find that the excretory system of larvae acclimated to highly alkaline media (pH 11)actually excreted fluid more acidic than the hemolymph. Animal metabolism always produces acids, due to the generation of CO2 and/or lactic acid during metabolism. Additional acids are formed during processes such as triglyceride or protein catabolism. Depending on an animal's physiological state, however, excreted fluids may be either acidic or alkaline. Data presented by Stobbart (Stobbart,1971; Stobbart,1974) show that larval Aedes aegypti may either alkalize or acidify the surrounding medium, depending on their state of ionic homeostasis and the ionic composition of the medium. Vanatta and Frazier(Vanatta and Frazier, 1981)found that frogs rendered alkalotic by NaHCO3injection excreted base via transepithelial processes. Excretion of HCO3 would allow larval mosquitoes in highly alkaline media to excrete fluid that is more alkaline than the hemolymph yet more acidic than the environment. Because the pK of the transition from HCO3 to CO32– is pH 10.25 at 25°C (Weast et al.,1986), HCO3 would act as an acid in media above this pH value. It is therefore not unexpected that the larvae show net acid excretion in highly alkaline media.
There are two ways in which larvae can excrete fluid more acidic than the hemolymph during steady-state exposure to alkaline media: (1) the larvae may neutralize ingested base-utilizing acids produced during metabolism, or (2)the larvae may excrete net base (or absorb acid) through the actions of extrarenal organs. If active excretion of base (either by the excretory system or by extrarenal organs) increases with ambient pH, then the energy demands of the organs involved are expected to increase as the rate of transport and the opposing concentration gradients increase. Larvae reared in high ambient pH did not possess elevated mitochondrial densities in any renal or extrarenal organ, and the AP were not increased in size in response to alkaline media. We have also found that the metabolic rates of larvae exposed to pH 11 media are the lowest observed under any conditions of pH or salinity (J. M. McLister and T.M.C., unpublished observation). Larvae thus appear to possess the remarkable ability to remain in pH balance in highly alkaline water without excreting base or, equivalently, absorbing acid. They appear to depend instead on metabolic acid production to neutralize ingested base. If so, the relatively low drinking rates observed in alkaline water may allow the larvae to remain in pH homeostasis by reducing the amount of base to be neutralized. To the best of our knowledge, this strategy, in which pH homeostasis is maintained in highly alkaline media by excreting fluid more acidic than the hemolymph, has not been reported in any other animal. Few studies have addressed the physiology of acid–base homeostasis in alkaline media, however, and it is quite likely that this ability is not unique to the larval mosquito.
Larval Aedes aegypti develops well in highly alkaline media. Larvae of this species live in small containers of water, generally containing some plant materials. Such habitats are not highly alkaline. It is possible that the ability to survive under these conditions is an adaptation inherited from an ancestral species that is not currently useful. However, if this ability imposed a significant cost it is likely that it would have been lost. It therefore seems likely that the ability of larvae Aedes aegypti to survive in such media is a fortuitous consequence of physiological characteristics, such as low cuticular permeability, air breathing, and the generation of a highly alkaline midgut lumen, that evolved in other contexts.
We have earlier reported that high pH reduces growth rates, and increases developmental times, of larval Aedes aegypti(Clark et al., 2004). We have found in the present study that larvae acidify alkaline media. Larval mosquitoes inhabiting small volumes of alkaline water may thus make their surroundings more suitable for their own growth and development, and presumably also the survival of the organisms upon which they feed. Early instars appear to be more susceptible to extreme pH than later ones (T.M.C.,unpublished observations). Neutralization of highly alkaline media by larval metabolic processes could therefore enhance the suitability of small volumes of water for future generations.
This work was supported by an IUSB Faculty Research Award to T.M.C. and by USDA NRICGP grant no. 99-35302-8371 to T.M.C. The authors thank Dr J. M. McLister and P. J. Bushnell for helpful comments and discussion of the manuscript.
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https://wegoastrology.wordpress.com/2014/01/05/fbicriminality-zodiac-signs/
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Which zodiac signs are the most dangerous for according to the FBI? that is really interesting and you will be amazed if you read it.The sign of a person is directly related to the type of crime that is likely to commit .
Analyzing data on the FBI according to the arrested zodiac signs, serial killer experts came to the conclusion which zodiac signs are the most dangerous in terms of criminal behavior .
The signs are divided into four different categories as we know of course : Fire , Water , Earth and Air.
The fire signs are Sagittarius , Aries and Leo .
All of the three, Sagittarius has the largest number of criminals and is the most difficult to be captured, Aries is usually the most heavily armed , but actually the most dangerous of all the fire signs is Leo .
The air signs are Libra , Gemini and Aquarius
Libras have the heaviest criminal record in relation to the other air signs and are usually are armed and very dangerous. On the other hand, Gemini often involved in crimes of fraud , while Aquarians usually commit crimes in order to revenge .
The Earth Signs are Capricorn , Virgo and Taurus
This category criminals arrested almost normally . Taurus is the most temperamental and therefore dangerous, Virgo is also usually heavily armed , while the criminal record of a Capricorn has a bit of everything .
The water signs are Cancer , Scorpio and Pisces
Cancers have the vast majority of criminals among all the zodiac signs and is also very violent , while Scorpio and Pisces are also quite irritable .
According to the website of the FBI, Cancers are the most dangerous criminals of all signs , followed by Taurus in the second position . Relying more dangerous Sagittarius come fourth and Aries , followed in descending order Capricorn , Virgo , Libra , Pisces , Scorpio , Leo , Aquarius and Gemini last in the list since their crimes are rarely associated with violence , but mostly frauds and scams .
Cancers are mostly passion killers . Kill almost normally and leave some distinguishing marks on the body of their victims to sign their crime .
Taurus usually involved in money laundering and usually act alone without accomplices .
Sagittarians are crooks and thieves , but rarely harm the victim unless the danger of their own lives.
Aries usually hired to do crimes , while Capricorns are mostly involved in organized crime .
Scorpios emerged the most sadistic and murderous sign, while Virgos are usually burglars or hackers . Libras are steeped in corruption , while Pisces involved mostly in crimes related to drugs.
On the other hand Gemini are not particularly dangerous and are usually associated with financial frauds and thefts bloodless.Unlike, Leos usually involved in criminal activities to gain fame and recognition , so do not deal with petty crime . Finally Aquarians are hackers and specialize in deception .
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2017-08-24 04:53:20
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https://projecteuclid.org/euclid.twjm/1500406734
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## Taiwanese Journal of Mathematics
### A CHARACTERIZATION OF DISTRIBUTIONS BY RANDOM SUMMATION
#### Abstract
In this paper, we consider a problem of characterizing distribution through the constructive property of random sum $p S_N$, where $0 \lt p \lt 1$ and $N \geq 0$ is an integer-valued random variable. This problem will be solved under someregular conditions. We extend the characterization of exponential distribution to ageneral case. For example, the gamma distribution, the positive Linnik distributionand the scale mixture of stable distribution are characterized. Two new results inthe vein are obtained. Finally, the problem of characterizing distribution by theproperty of the first order statistics is also investigated.
#### Article information
Source
Taiwanese J. Math., Volume 16, Number 4 (2012), 1245-1264.
Dates
First available in Project Euclid: 18 July 2017
https://projecteuclid.org/euclid.twjm/1500406734
Digital Object Identifier
doi:10.11650/twjm/1500406734
Mathematical Reviews number (MathSciNet)
MR2951138
Zentralblatt MATH identifier
1259.62004
Subjects
Primary: 621E0
#### Citation
Hu, Chin-Yuan; Cheng, Tsung-Lin. A CHARACTERIZATION OF DISTRIBUTIONS BY RANDOM SUMMATION. Taiwanese J. Math. 16 (2012), no. 4, 1245--1264. doi:10.11650/twjm/1500406734. https://projecteuclid.org/euclid.twjm/1500406734
#### References
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2019-10-15 03:32:31
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https://homework.zookal.com/questions-and-answers/hi-i-have-a-question-regarding-displaying-xml-data-635097436
|
1. Other
2. Other
3. hi i have a question regarding displaying xml data...
# Question: hi i have a question regarding displaying xml data...
###### Question details
Hi , i have a question regarding displaying xml data in a gridview(asp.net). Im struggling to get it to display all data as each time i go to run it says a error with binding data.
protected void Data()
{
var ds = new DataSet();
var path = Server.MapPath("~/XMLFile1.xml");
GridView1.DataSource = ds;
GridView1.DataBind();
}
This is my code and here is the xml:
<?xml version="1.0" encoding="UTF-8"?>
<PurchaseCollection>
<purchases>
<Purchase>
<id>1</id>
<fullName>Anna Acacia</fullName>
<productname>Lilly Pilly</productname>
<unitprice>100</unitprice>
<quantity>1</quantity>
</Purchase>
<Purchase>
<id>2</id>
<fullName>Byron Beech</fullName>
<productname>Snow Gum</productname>
<unitprice>17</unitprice>
<quantity>2</quantity>
</Purchase>
<Purchase>
<id>3</id>
<fullName>Cassie Casuarina</fullName>
<productname>Golden Wattle</productname>
<unitprice>45</unitprice>
<quantity>3</quantity>
</Purchase>
<Purchase>
<id>4</id>
<fullName>Donny Dahlia</fullName>
<productname>Rusty Gum</productname>
<unitprice>8</unitprice>
<quantity>44</quantity>
</Purchase>
</purchases>
</PurchaseCollection>
Please any help is welcome as been stuck on it for a while.
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2021-04-16 14:44:47
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https://itprospt.com/num/9003472/identify-the-following-amine-bases-found-in-nucleic-acids
|
5
# Identify the following amine bases found in nucleic acids:...
## Question
###### Identify the following amine bases found in nucleic acids:
Identify the following amine bases found in nucleic acids:
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2022-08-15 22:49:15
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