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http://www.zentralblatt-math.org/zmath/en/advanced/?q=an:0922.34060
|
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Zbl 0922.34060
Philos, Ch.G.; Sficas, Y.G.
An oscillation criterion for first-order linear delay differential equations.
(English)
[J] Can. Math. Bull. 41, No. 2, 207-213 (1998). ISSN 0008-4395; ISSN 1496-4287/e
The authors consider the oscillation behavior of the delay differential equation $$x'(t)+p(t)x (t-\tau(t))=0$$ with $p,\tau\in C([0, \infty), [0,\infty))$, $t-\tau(t)$ is increasing, and $\lim_{t\to\infty}(t-\tau (t))= \infty$. It is shown that the equation is oscillatory if $M+{L^2\over 2(1-L)}+ {L^2\over 2}\lambda_0 >1$ where $$L=\lim \inf_{t\to \infty} \int^t_{t-\tau (t)}p(s)ds, \quad M=\lim \sup_{t\to \infty} \int^t_{t-\tau (t)}p(s)ds,$$ and $\lambda_0$ is the smaller real root of the equation $\lambda= e^{L\lambda}$. This result allows $$\lim\inf_{t\to\infty} \int^t_{t-\tau (t)}p(s) ds\le 1/e.$$
[Bingtuan Li (Tempe)]
MSC 2000:
*34K11 Oscillation theory of functional-differential equations
34C15 Nonlinear oscillations of solutions of ODE
Keywords: oscillation; first-order linear delay differential equations
Highlights
Master Server
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2013-06-18 21:02:39
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https://ryancquan.com/blog/2014/09/26/denoising-a-png-with-knn-imputation/
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Given a PNG image with noise, where noise is defined as having an RGB value equal to [0, 0, 0], we can use k-nearest neighbors imputation to fill in the zeros using nearest neighbor averaging. For each row with a zero value, we find the k-nearest neighbors using a Euclidean metric, confined to columns for which that row is not zero. The accuracy of this image restoration technique will vary depending on how we set our k parameter.
Here, we attempt to denoise Lena.png. The target image can be downloaded here. See what happens when we vary the k parameter.
So which value of k do you think produces the best image? Try denoising the image yourself with the script provided below:
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2017-06-24 20:38:31
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https://zbmath.org/?q=an:0614.65089
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## The h, p and h-p versions of the finite element method in 1 dimension. II. The error analysis of the h- and h-p versions.(English)Zbl 0614.65089
This paper concentrates on h and h-p versions of FEM. Generally, the h- version of FEM has the degree of elements fixed and the convergence is achieved by the refinement of the mesh. The h-p version combines both approaches, the p-version and h-version.
The same model problem as in Part I [ibid. 49, 577-612 (1986; reviewed above)] is considered. The paper is mainly concerned with the relation between the relative error in the energy norm and the number of degrees of freedom. The authors show that the selection of the mesh and degree of elements is essential for the performance of the method. More exactly, the proper selection of the h-p version leads to the exponential rate of convergence while the h-version with improper mesh, gives very low algebraic rate when a singularity is present.
Reviewer: C.-I.Gheorghiu
### MSC:
65L10 Numerical solution of boundary value problems involving ordinary differential equations 65L60 Finite element, Rayleigh-Ritz, Galerkin and collocation methods for ordinary differential equations 65L50 Mesh generation, refinement, and adaptive methods for ordinary differential equations 34B05 Linear boundary value problems for ordinary differential equations
Zbl 0614.65088
Full Text:
### References:
[1] Babuška, I., Gui, W.: Theh, p andh-p Versions of the Finite Element Method for One Dimensional Problem. Part I: The Error Analysis of thep-Version. Numer. Math.49, 577–612 (1986) · Zbl 0614.65088 [2] DeVore, R., Scherer, K.: Variable knot variable degree spline approximation tox {$$\beta$$}. In: Quantitative Approximation. Proceedings, Bonn, pp. 101–131 (1979) [3] Scherer, K.: On optimal global error bounds obtained by scaled local error estimates. Numer. Math.36, 257–277 (1981) · Zbl 0495.65006
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2022-06-25 17:38:46
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https://physics.stackexchange.com/questions/193937/how-fast-does-gravitational-information-travel?noredirect=1
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# How fast does gravitational information travel? [duplicate]
Imagine two objects with equal mass in empty space attracting each other. One of these objects moves tangentially with a very high speed (lets say 0.9c). (p1 = (0, 0) p2 = (1, 0) v1 = (0, 0) v2 = (0, 0.9c)). Would the direction of the force, acting on the resting mass point directly towards the other mass, or to a point, where the moving mass was some time ago. Einstein says, that information cant travel faster than light, so whats about gravity? Does the gravitational pull really point towards the center of mass, even if it its moving?
• Gravitational effects propagate at the speed of light. – Ryan Unger Jul 13 '15 at 23:33
• @0celo7 Are you sure you want to major in nuclear engineering? From what I've seen, you have the aptitude to make some real contributions to more fundamental science and I question whether a mind like yours would be happy in an engineering discipline. – Selene Routley Jul 13 '15 at 23:44
• possible duplicate of The speed of gravity? – CoilKid Jul 13 '15 at 23:46
• – Kyle Oman Jul 14 '15 at 1:00
Gravity, like all cause-effect relationships, propagates at a maximum speed of $c$; indeed from the Einstein field equations small amplitude (linear limit) gravitational waves travel at precisely $c$.
A good idea for what is going on comes from an approximation of General Relativity called Gravitoelectromagnetism. This makes an approximate analogy between gravity and electrodynamics as the approximation has the same form as Maxwell's Equations. In electrodynamics, charges influence one another through retarded potentials as described by the Liénard-Weichert potentials and Feynman's delayed force formula; see the Physics SE Question "Do electrostatic fields really obey “action at a distance”? for more details. The force on a charge by another is roughly that calculated from the position of the latter at a time $d/c$ before the present, where $d$ is the distance separating the charges.
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2021-07-28 01:20:52
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https://forum.uipath.com/t/robot-activity-descompress-bz2-problem/462113
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# Robot activity descompress .bz2 problem
Hey guys!
I have a folder with different files to unzip, including some with .bz2 format and others with .zip format. For files in .zip format I have no problem decompressing them with UiPath activities, but the .bz2 format is being impossible for me to unzip, any suggestions? I would like it to be through an activity, but I have not found any that decompresses this format.
Regards!
If you have a separate third-party tool (for example https://www.7-zip.org/ ) that can decompress .bz2 files, you may create a UI automation workflow that can right-click on the respective file and access the right menu context by clicking on it and then choosing the Extract option.
Example that you need to automate:
You may try also the command line approach for the unattended automation:
7z x archive.bz2
More details:
Example (Inside the text.log.bz2, I have text.log file):
I am adding the environment System Variable Path for the 7z.exe file location as below:
From Start, I am accessing the Edit system environment variables
Edit the System variables → Path → New → C:\Program Files\7-Zip (path of the 7z.exe file)
Run the below command in a cmd.exe
7z x "C:\Marian\DownloadPackages\text.log.bz2"
Explanation: extracts all files from the archive text.log.bz2 to the current directory.
Results of the extracted activity:
If you need to extract all the files from the archive text.log.bz2 to a specific folder (for example to C:\Marian\DownloadPackages\ExtractedFiles), you may use the below command:
7z x "C:\Marian\DownloadPackages\text.log.bz2" -o"C:\Marian\DownloadPackages\ExtractedFiles"
Example:
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2022-08-09 07:24:25
|
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http://www.oranlooney.com/tags/math/
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## Craps Variants
Craps is a suprisingly fair game. I remember calculating the probability of winning craps for the first time in an undergraduate discrete math class: I went back through my calculations several times, certain there was a mistake somewhere. How could it be closer than $\frac{1}{36}$? (Spoiler Warning If you haven’t calculated these odds for yourself then you may want to do so before reading further. I’m about to spoil it for you rather thoroughly in the name of exploring a more general case.
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2018-11-17 10:42:18
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http://mathoverflow.net/revisions/21062/list
|
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
2) explains in an intuitive way why if $L$ is a finite separable extension of $K$, the map $(x,y) \mapsto Tr(xy)$, where $x,y$ are in $L$, is a non degenerated bilinear form on L?
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2013-06-20 00:47:24
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https://www.physicsforums.com/threads/f-i-pinch-hold-a-book-with-6n-on-either-side.691596/
|
F I pinch hold a book, with 6N on either side
1. May 13, 2013
dgamma3
If I pinch hold a book, with 6N on either side why is there only one force, of 6 newton on the free body diagram?
for example, the book has this free body diagram. http://i.imgur.com/pwtFZB6.png
there are two forces applied to each side of the book. so why is it, than in the free body diagram, there is only 1 force (of 6N). shouldn't you have two forces pointing away from one another, one with 6N (pointing right) and the other with -6N (pointing left), which would cancel out?
Edit: this is the explanation in the book.
Consider the free-body diagram below. The force of the fingers on the book is the reaction
force to the normal force of the book on the fingers, so is exactly equal and opposite the normal force on the fingers.
thanks
daniel
2. May 14, 2013
haruspex
I would guess that the F in the right hand picture should be an n.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook
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Draft saved Draft deleted
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2017-10-23 11:00:45
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https://tex.stackexchange.com/questions/213523/remove-appendix-tables-and-figures-from-list-of-figure-tables
|
# Remove appendix tables and figures from list of figure/tables
I'm writing a thesis and I have over hundred figures and tables in the appendix. But I don't want them to appear in List of figures / List of tables. - But I want the appendices and sections to appear in the table of content.
\documentclass[a4paper,12pt,icelandic]{report}
\usepackage[toc,page]{appendix}
\begin{document}
\tableofcontents
\listoffigures
\addcontentsline{toc}{chapter}{List of Figures}
\listoftables
\addcontentsline{toc}{chapter}{List of tables}
\chapter{Chapter 1}
\section{Section 1, chapter 1}
\section{Section 2, chapter 1}
\appendix
\addcontentsline{toc}{chapter}{Appendices}
\chapter{First appendix}
\section{Section 1}
\begin{figure} % DON'T WANT THIS TO APPEAR IN LIST OF FIGURES
\centering
\includegraphics[width=1\linewidth]{figure.png}
\caption{Caption of figure}
\label{Label of figure}
\end{figure}
\begin{table} % DON'T WANT THIS TO APPEAR IN LIST OF TABLES
\begin{tabular}{|c|c|}
\hline One & Two \\
\hline Three & Four \\
\hline
\end{tabular}
\caption{Caption of table}
\label{Label of table}
\end{table}
\end{document}
## 2 Answers
Right before the appendix, I redefine \addcontentsline to intercept {figure} and {table} calls and excise them, as follows (requires ifthen package):
\let\svaddcontentsline\addcontentsline
\renewcommand\addcontentsline[3]{%
\ifthenelse{\equal{#1}{lof}}{}%
{\ifthenelse{\equal{#1}{lot}}{}{\svaddcontentsline{#1}{#2}{#3}}}}
Here is my MWE (note: I added a body figure and table to demonstrate ability to discern report body from appendix)
\documentclass[a4paper,12pt,icelandic]{report}
\usepackage[toc,page]{appendix}
\usepackage{graphicx,ifthen}
\begin{document}
\tableofcontents
\listoffigures
\addcontentsline{toc}{chapter}{List of Figures}
\listoftables
\addcontentsline{toc}{chapter}{List of tables}
\chapter{Chapter 1}
\section{Section 1, chapter 1}
\begin{figure} % DO WANT THIS TO APPEAR IN LIST OF FIGURES
\centering
\includegraphics[width=1\linewidth]{example-image}
\caption{Body figure}
\end{figure}
\begin{table} % DO WANT THIS TO APPEAR IN LIST OF TABLES
\begin{tabular}{|c|c|}
\hline One & Two \\
\hline Three & Four \\
\hline
\end{tabular}
\caption{Body table}
\end{table}
\section{Section 2, chapter 1}
\let\svaddcontentsline\addcontentsline
\renewcommand\addcontentsline[3]{%
\ifthenelse{\equal{#1}{lof}}{}%
{\ifthenelse{\equal{#1}{lot}}{}{\svaddcontentsline{#1}{#2}{#3}}}}
\appendix
\addcontentsline{toc}{chapter}{Appendices}
\chapter{First appendix}
\section{Section 1}
\begin{figure} % DON'T WANT THIS TO APPEAR IN LIST OF FIGURES
\centering
\includegraphics[width=1\linewidth]{example-image}
\caption{Caption of figure}
\label{Label of figure}
\end{figure}
\begin{table} % DON'T WANT THIS TO APPEAR IN LIST OF TABLES
\begin{tabular}{|c|c|}
\hline One & Two \\
\hline Three & Four \\
\hline
\end{tabular}
\caption{Caption of table}
\label{Label of table}
\end{table}
\end{document}
ADDENDUM
A version of the redefinition that does not require the ifthen package:
\let\svaddcontentsline\addcontentsline
\renewcommand\addcontentsline[3]{%
\edef\qtest{#1}%
\def\qmatch{lof}%
\ifx\qmatch\qtest\else%
\def\qmatch{lot}%
\ifx\qmatch\qtest\else%
\svaddcontentsline{#1}{#2}{#3}%
\fi\fi%
}
• Thanks. This works just as I wanted to. Thank you very much. :) – Bjartmar Nov 23 '14 at 3:33
Load caption package and use \caption*:
\documentclass[a4paper,12pt,icelandic]{report}
\usepackage[toc,page]{appendix}
\usepackage{caption}
\usepackage[demo]{graphicx}
\begin{document}
\tableofcontents
\listoffigures
\addcontentsline{toc}{chapter}{List of Figures}
\listoftables
\addcontentsline{toc}{chapter}{List of tables}
\chapter{Chapter 1}
\section{Section 1, chapter 1}
\section{Section 2, chapter 1}
\appendix
\addcontentsline{toc}{chapter}{Appendices}
\chapter{First appendix}
\section{Section 1}
\begin{figure} % DON'T WANT THIS TO APPEAR IN LIST OF FIGURES
\centering
\includegraphics[width=1\linewidth]{figure.png}
\caption*{Caption of figure}
\label{Label of figure}
\end{figure}
\begin{table} % DON'T WANT THIS TO APPEAR IN LIST OF TABLES
\begin{tabular}{|c|c|}
\hline One & Two \\
\hline Three & Four \\
\hline
\end{tabular}
\caption*{Caption of table}
\label{Label of table}
\end{table}
\end{document}
• Thanks for the answer. I'm going to use Steven's method. To use your method I would have to open over 100 files to add the star. – Bjartmar Nov 23 '14 at 3:31
• Thank you for the solution! It solved my problem immediately without adding many new rules in premable! : ) – Vivian Jan 8 '17 at 20:48
• Minor improvement suggestion: use \captionsetup{list=no} at the start of wherever you need to disable listing, such as the start of your appendices, instead of \caption*. The former is meant for skipping the LoT and LoF, while the latter will (by default) eat up any numbering you may have, e.g., Table E.4 for the fourth table in appendix E. – Christopher Corley Jul 5 '17 at 2:03
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2020-02-25 03:49:44
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http://www.gradesaver.com/textbooks/math/algebra/intermediate-algebra-6th-edition/chapter-3-section-3-1-graphing-equations-exercise-set-page-126/67
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## Intermediate Algebra (6th Edition)
Published by Pearson
# Chapter 3 - Section 3.1 - Graphing Equations - Exercise Set: 67
#### Answer
(-10,-30) is in the third quadrant
#### Work Step by Step
We are given that the location of the ordered pair is (-10,-30). Therefore, the point has a negative x-coordinate and a negative y-coordinate, which corresponds to the third quadrant. Because $|−10|\lt|-30|$, we know that the point is farther away from the x-axis than it is from the y-axis.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2018-04-23 04:17:28
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http://www.dailykos.com/story/2009/12/09/812363/-Probability-for-Kossacks-the-fallacy-of-the-inverse
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In the blogosphere, we often see arguments like this:
The odds of [some event] happening by random chance are incredibly low. Therefore [event] was rigged or influenced to happen that way.
I've seen this argument applied to variations in voting patterns, timelines of terror attacks, the timing of bad news, and coincidences in general. People are prone to suspect an evil actor behind unwanted events, and a mathematical argument seems to confirm our suspicions. It's a very compelling argument.
It is also bogus, as you can observe any time you play cards. Shuffle a deck of cards and look at the outcome: the probability of that shuffle happening by random chance is 1 in 80658175170943878571660636856403766975289505440883277824000000000000. Obviously this was rigged by a conspiracy while you weren't looking!
This is a common logical fallacy known as the confusion of the inverse. I explain below.
The fallacy centers around the ambiguity of this question:
What is the probability that event X happened by chance?
This question has two confusable meanings:
1. Under random chance, what is the probability of X happening?
2. If X happens, what is the probability that random chance is the cause?
To write this conditionally, question (1) asks for Pr[X|Chance] and question (2) asks for Pr[Chance|X]. In probability, a "|" means "given that"; or to put it another way, everything to the left of the bar is the thing you are wondering about, and everything to the right of the bar is stuff you either know, or assume to be true.
These expressions are not the same: they are inverses of one another, and have very different meanings and often very different values. The conspiratorial blogger should be making a decision based on (2), but almost always people mistakenly compute (1), yielding an impressively tiny number that doesn't really mean anything, but which can sway an audience of laypeople.
Look everyone! Little tiny numbers!
But this fallacy is bigger than simply using the wrong formula. It also employs pseudoscience and argument from incredulity. First the incredulity: you're supposed to believe that an event couldn't have happened because its probability is so impressively tiny. What is missing is the context: incredibly tiny is in fact perfectly normal.
In reality, most anything that ever happens by chance has probability well nigh 0. You spill some salt, and the scattering of grains takes on one of inexhaustibly many different outcomes, each with infinitesimal probability. (1) is small! And yet that incredibly improbable outcome did happen by chance, without any reason to suspect the unseen aid of space lasers.
Even more mind-numbing is the fact that it will never happen again: any individual card shuffle is so unlikely that, once seen, you can be guaranteed that it will never be seen again, for the rest of the lifetime of the universe, assuming the shuffling is fair. There is something weirdly unintuitive about observing an event happen right in front of you, and immediately declaring that it can never happen.
This can be counter-intuitive because when we hear "probability 0," or even "odds of 1 in a million," we think "this can't have happened." But that's not what probability means. A low probability does not mean an event can't have happened; it does mean that if you predict that specific event to happen, in advance, then you are not going to be right.
If that confuses you, just remember the old joke about the farmer who would shoot the side of his barn, and then paint a bullseye wherever he hit. The probability of landing on the bullseye is only low if you declare the bullseye in advance. That's what the number Pr[X] ultimately describes.
Now, multiply by the probability of aliens
Okay, the second fallacy behind the confusion of the inverse: usually, the probability Pr[Chance|X] can't even be computed, because it doesn't have a well-defined value. So making a mathematical argument is pseudo-scientific to begin with.
Often we can compute formula (1) (this is why so many people mistakenly use that value,) but to get formula (2) you need to know numbers that you can't possibly know. We can see this using Bayes Rule: the expressions (1) and (2) are related as follows:
Pr[Chance|X] = Pr[X|Chance](Pr[Chance]/Pr[X])
The right side contains parts that often have no meaningful value. Pr[X] is the probability of X happening by any cause, from coincidence to conspiracies to space lasers. You don't know that number. And Pr[Chance] is the overall probability of no conspiracy, no space lasers. If an election is going to happen tomorrow, what is the probability that it will be in some way rigged? 1? 0.1? 0.001? How do you know?
How would we even get an estimate of that number? By examining previous elections? That would only tell us the odds of election-rigging happening and being caught. And what elections do we count? All the races in that same district? There aren't enough elections, and election personnel and machines change too rapidly, to get any useful estimate of the odds of a conspiracy. It's like trying to tell if a coin is fair by observing three coin flips.
It is possible to compute this type of number in very carefully designed experimental circumstances, which is what scientists do. You compare results of drug XYZZY versus the results of a placebo, and you design the experiment so that you know the precise numbers of each case. Likewise, in engineering, we can use these formulas because we know all the probabilities of everything---having built everything. If you want to decide if a received signal represents a 0 or a 1, you know Pr[0] and Pr[1], because you built the transmitter. But none of this applies if you try to analyze, after the fact, an event that happened in the wild.
There's one more mistake behind the fallacy of the inverse: the idea that we can draw a box around an event and determine the odds. The odds of what? Where do you draw the box? How much detail do you include?
For example: I roll six dice and get 3, 4, 1, 5, 6 ,2. What are the odds? One in 46656? Well, maybe if you only look at the numbers on top. Suppose you consider the exact positions where the dice fell, or their orientation---what are the odds of them landing like that? Much smaller, certainly.
When people examine a real-world event and try to compute the odds of it happening, they have to choose what detail to include and what not to include. This event resolution can make the probability much smaller, much larger, and that much more meaningless as a number.
So now that you know how it's done...kids...don't do it
I hope this diary will innoculate your brain against a common mathematical misconception. If ever you see someone arguing over whether the Governor's memo was intentional and they bust out the odds, remember that those odds are often meaningless, and any argument based on them is largely pseudo-quantitative.
I guess the moral of this diary is this: mathematics is a tool for evil, and if any of us try to convince you of something using mathematics, you should assume the opposite is true.
Ha ha, just kidding, the preceding sentence is false. Happy Wednesday.
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Comment Preferences
• Tip Jar(19+ / 0-)
Linking to a news article is journalism in the same sense that putting a Big Mac on a paper plate is cooking.
• I love this diary. Thank you!(3+ / 0-)
Recommended by:
Caj, gooners, Jane Lew
disclaimer: I oppose the escalation and any contrary discussion of said escalation is just that.
[ Parent ]
• you have indeed brought...(0+ / 0-)
...the quality shit.
Prison rape is not funny.
[ Parent ]
• Cool(1+ / 0-)
Recommended by:
Caj
but I think you might want to repost those equations as images.
• I figured, screw it(3+ / 0-)
Recommended by:
pontechango, G2geek, XNeeOhCon
I was advised to use MathML, but it didn't work. Also it was horrible and painful; I will never do that again.
Especially considering that probability is one of the few branches of math where most expressions can be written in plain old Roman letters, with brackets and a vertical pipe.
I keep a running list of things that worked perfectly well before the tech sector tried to "improve" it by intruducing hilarious complexity and hundreds of new failure modes. Like VHS movies, overhead projectors, telephones, and analog television. I guess I'll add typesetting: $$e^{i\pi}=-1$$ was far too complicated, and clearly they had to replace that with 20 lines of angle brackets that won't display.
[ Parent ]
• I don't understand why HTML doesn't just accept..(1+ / 0-)
Recommended by:
Caj
...simple TeX expressions using some simple HTML tag.
How hard can it be to render them properly?
illegal, n. A term used by descendents of European immigrants to refer to descendants of Indigenous Americans
[ Parent ]
• telephones and stolen elections. (0+ / 0-)
Telephones:
I'm a telephone switching systems engineer and I agree that the network has been broken. Not long ago we had hardwired residential phones that always worked, and digital PBX office phones that always worked. Today we have sexy cellphones that take pictures and send telegrams (text messages) but sound like dog doo and interfere with conversation. And we have IP-based office phones that cost twice as much to install but are less functional and require frequent service to keep 'em running. Bah humbug to all of that.
Hint: When buying an office phone system, use digital handsets within each building, use IP-telephony for home office extensions and to connect between different branches of the company if any. Lower cost, more reliable outcomes.
Stolen elections:
The 2004 election outcomes were analyzed by independent statisticians in regard to one variable: the reported instances where a voter found that Diebold electronic voting machines registered a vote for the opposite presidential candidate. In other words, you touched the screen for candidate A, and the machine showed a vote for candidate B, and the error persisted despite trying to re-enter the correct choice, to the point where you called an election official into the voting booth to try to fix it.
What they found was that the reported vote-flips favored Bush over Kerry almost without exception (many reported flips toward Bush, almost no reported flips toward Kerry), to a degree that was millions to one against chance. Conclusion: either a systemic flaw in the machines, or deliberate tampering, was responsible for those outcomes.
This was not a case of the kinds of logical fallacies you cite. There were two possible values for the variable: flip toward Bush and flip toward Kerry. The null hypothesis was clear to begin with: that vote flips would be evenly distributed between candidates (equal probability of an outcome in either direction, like a coin flip). The operationalization used information from county election officials rather than from partisan sources, so it should have picked up both types of flips (toward Bush and toward Kerry) equally well.
Note also, if you try to offset for tabulated vote totals, then you end up even further from the null hypothesis: if for example 55% of the voters in a district voted for Bush, then you would expect 55% of the reported flips to be complaints from Bush voters that the machines flipped their votes from Bush to Kerry, rather than 50%. Sticking with 50% effectively gives a higher weighting to each instance of a Bush-to-Kerry flip.
These results were not anecdotal, and they stood up to peer review.
So at least for the 2004 Presidential race, we can say with reasonable certainty that the outcome was affected by a systemic flaw in the voting machines.
This does not give us "proof" that the voting machines were rigged, since the systemic flaw could have been caused by factors other than hacking. However, in combination with circumstantial evidence such as the behavior of Diebold employees in making "last minute rush updates" to the machines in the field, and the partisan statements from the president of Diebold that he would "deliver" the election for Bush, we can reasonably say that the 2004 presidential election was stolen.
• One nitpick(0+ / 0-)
The null hypothesis was clear to begin with: that vote flips would be evenly distributed between candidates (equal probability of an outcome in either direction, like a coin flip).
This would not be the null hypothesis for a capacitive touch-screen.
If a bunch of touch screens are defective or miscalibrated and the reading sticks to one spot, you would not expect the spots to be uniformly distributed over the screen, with equal likelihood of being stuck on a candidate at the top of the screen and a candidate further down. I would not be surprised at all if the stuck areas tended to be at the edges rather than the middle, for example.
But the fact that these reported errors persisted despite multiple attempts strongly suggest that they were not intentionally programmed misbehavior, unless the programmer's intention was to get caught rather than affect the election's outcome.
A programmer could simply have the machine record the wrong vote without making anything unusual on screen. Even if they were forced to display Bush on screen to record a Bush vote, it would not make sense to continually flip the vote after multiple attempts to select Kerry.
That doesn't rule out intentional malfeasance completely; a broken machine could be the result of voter vandalism.
[ Parent ]
• is this a new series?(0+ / 0-)
disclaimer: I oppose the escalation and any contrary discussion of said escalation is just that.
• Very nice synopsis.(2+ / 0-)
Recommended by:
Caj, gooners
But the odds of any person who already regularly employs this fallacy actually reading it, understanding it, and changing = 52357390589253 to 1
;-P
I wish I would have added this to this diary
This also reminds me of the idiotic argument some IDers use against scientific theories of the development of life on Earth (one example was in Ben Stein's terrible movie "Expelled").
"The first reaction of a progressive should be not to look at who is the target of hate, but to reject the hate first." -RandomActsOfReason
• Also, source code for this diary(1+ / 0-)
Recommended by:
social democrat
How do you get those big numbers? On a Mac or any Unix-like operating system, open a terminal and use the bc command.
Actually, it's a bit of a pain to use it directly; I invoke it from Tcl. To get the numbers for this diary I opened a terminal, typed tclsh, and entered the following at the % prompt:
% proc math e { join [join [eval exec echo \"$e\" | bc]] "" } % math 1+1 2 % math 6^6 46656 % % for {set f [set i 1]} {$i<=52} {incr i} {set f [math $i*$f]} <br>% set f
80658175170943878571660636856403766975289505440883277824000000000000
• Er...(1+ / 0-)
Recommended by:
social democrat
That <br> should be a new line.
I better not typeset anything else today; maybe on my next attempt I'll break an ankle.
[ Parent ]
• for windows users...(0+ / 0-)
...the equivalent in PowerShell would be:
C:>PowerShell
PS C:> 1+1
2
PS C:> [Math]::Pow(6,6)
46656
PS C:>$f = 1; 1..52 |% {$f *= $_ };$f
8.06581751709439E+67
Prison rape is not funny.
[ Parent ]
• Dueling banjos(0+ / 0-)
That's nice, concise syntax. Let me add that to the Tcl interpreter:
% proc know what {
if ![info complete $what] {error "incomplete command(s)$what"}
proc unknown args $what\n[info body unknown] } % know {if {![catch {math$args} res]} {return $res}} [testing...] % 1+1 2 % 2^60 1152921504606846976 [seems to work. Now the 1..52 part:] % know { if {[regexp {(.*)\.\.(.*)} [lindex$args 0] - a b]} {
set c [lindex $args 1] upvar 1 ; for {set$a} {$<=$b} {incr _} {uplevel 1 $c} <br> return } } % set f 1; 1..52 {set f [math$f*$_]} % puts$f
80658175170943878571660636856403766975289505440883277824000000000000
Truly, Tcl is the Play-Doh of languages.
[ Parent ]
• Again with the weirdness(0+ / 0-)
That <br> should be a newline, and there should be two underscores after the upvar 1.
I guess it's a good sign if writing the code is easier than posting it.
For more info on Tcl, check out the awesome Wikibook --- from which I got the know procedure.
[ Parent ]
• to avoid the <br> weirdness...(0+ / 0-)
...use <tt> rather than <blockquote> (or perhaps <tt> within <blockquote>); adding <br> for line breaks won't be necessary.
As far as the ability to screw with the interpreter and so "import" new syntax, I gotta say that's pretty frickin' cool. Not sure how often I'd use it, but then I have been know to overload an operator from time to time just to get cleaner, more intuitive syntax.
For instance, PowerShell would really benefit from having the exponentiation operator ^. I'm not sure I know how to get there from here, though.
Prison rape is not funny.
[ Parent ]
• see my posting "telephones and stolen elections."(0+ / 0-)
2004 was stolen and that's not CT, it's a logical conclusion based on findings of independent statisticians analyzing reported vote-flips on Diebold voting machines. Details in my posting.
And FYI, I routinely fight CT where I see it, most obviously re. 9/11.
• Once is happenstance,(0+ / 0-)
twice is coincidence.
Three times is enemy action.
"If all else fails... immortality can always be assured by spectacular error."
• Quote of the week.(2+ / 0-)
Recommended by:
Caj, G2geek
There is something weirdly unintuitive about observing an event happen right in front of you, and immediately declaring that it can never happen.
illegal, n. A term used by descendents of European immigrants to refer to descendants of Indigenous Americans
• Heraclitus, CA 500 B.C.(1+ / 0-)
Recommended by:
Caj
"Fascism should more properly be called corporatism because it is the merger of state and corporate power." -- Benito Mussolini
[ Parent ]
• What are the odds???(2+ / 0-)
Recommended by:
Caj, G2geek
I tell my friends that if you're dealt five cards, the odds are 100% that you'll get five cards. And the odds against getting those particular cards are approximately 3 million to 1. So the odds of getting a royal flush (in a particular suit) in five cards is 3 million to one, but the odds of getting 2 diamond, 5 heart, 6 heart, J spade, and Q diamond are also 3 million to one.
But nobody ever looks at their hand and says, "Wow! What are the odds of getting 2, 5, 6, J, Q?"
Listen, strange women lying in ponds distributing swords is no basis for a system of government.
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2015-05-27 06:35:29
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https://au.mathematicstip.com/9427-46-linear-functions-calculations-mathematics.html
|
4.6: Linear Functions - Calculations - Mathematics
As you hop into a taxicab in Las Vegas, the meter will immediately read $3.50; this is the “drop” charge made when the taximeter is activated. This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2017. Using descriptive variables, we choose (m) for miles and (C) for Cost in dollars as a function of miles: (C(m)). We know for certain that (C(0)=3.50), since the$3.50 drop charge is assessed regardless of how many miles are driven. Since $2.67 is added for each mile driven, then [C(1)=3.50+2.67=6.17 onumber ] If we then drove a second mile, another$2.67 would be added to the cost:
[C(2)=3.50+2.67+2.67=3.50+2.67(2)=8.84 onumber ]
If we drove a third mile, another $2.67 would be added to the cost: [C(3)=3.50+2.67+2.67+2.67=3.50+2.67(3)=11.51 onumber ] From this we might observe the pattern, and conclude that if (m) miles are driven, (C(m)=3.50+2.67m) because we start with a$3.50 drop fee and then for each mile increase we add $2.67. It is good to verify that the units make sense in this equation. The$3.50 drop charge is measured in dollars; the $2.67 charge is measured in dollars per mile. [C(m)=3.50 ext{ dollars}+left(2.67dfrac{ ext{dollars}}{ ext{mile}} ight)left(m; ext{ miles} ight) onumber ] When dollars per mile are multiplied by a number of miles, the result is a number of dollars, matching the units on the 3.50, and matching the desired units for the C function. Notice this equation (C(m)=3.50+2.67m) consisted of two quantities. The first is the fixed$3.50 charge which does not change based on the value of the input. The second is the $2.67 dollars per mile value, which is a rate of change. In the equation, this rate of change is multiplied by the input value. Looking at this same problem in table format we can also see the cost changes by$2.67 for every 1 mile increase.
(m) 0 1 2 3 (C(m)) 3.5 6.17 8.84 11.51
It is important here to note that in this equation, the rate of change is constant; over any interval, the rate of change is the same.
Graphing this equation, (C(m)=3.50+2.67m) we see the shape is a line, which is how these functions get their name: linear functions.
When the number of miles is zero the cost is $3.50, giving the point (0, 3.50) on the graph. This is the vertical or (C(m)) intercept. The graph is increasing in a straight line from left to right because for each mile the cost goes up by$2.67; this rate remains consistent.
In this example, you have seen the taxicab cost modeled in words, an equation, a table and in graphical form. Whenever possible, ensure that you can link these four representations together to continually build your skills. It is important to note that you will not always be able to find all 4 representations for a problem and so being able to work with all 4 forms is very important.
Definition: Linear Function
A linear function is a function whose graph produces a line. Linear functions can always be written in the form
(f(x)=b+mx) or (f(x)=mx+b); they’re equivalent
where
• (b) is the initial or starting value of the function (when input, x = 0), and
• (m) is the constant rate of change of the function
Many people like to write linear functions in the form (f(x)=b+mx) because it corresponds to the way we tend to speak: “The output starts at (b) and increases at a rate of (m).”
For this reason alone we will use the (f(x)=b+mx) form for many of the examples, but remember they are equivalent and can be written correctly both ways.
Definition: slope and increasing/decreasing
(m) is the constant rate of change of the function (also called slope). Slope The slope determines if the function is an increasing function or a decreasing function.
(f(x)=b+mx) is an increasing function if (m>0)
(f(x)=b+mx) is a decreasing function if (m<0)
If (m=0), the rate of change zero, and the function (f(x)=b+0x=b) is just a horizontal line passing through the point (0, (b)), neither increasing nor decreasing.
Example (PageIndex{1})
Marcus currently owns 200 songs in his iTunes collection. Every month, he adds 15 new songs. Write a formula for the number of songs, (N), in his iTunes collection as a function of the number of months, (m). How many songs will he own in a year?
Solution
The initial value for this function is 200, since he currently owns 200 songs, so (N(0)=200). The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. With this information, we can write the formula:
[N(m)=200+15m onumber ]
(N(m)) is an increasing linear function. With this formula we can predict how many songs he will have in 1 year (12 months):
[N(12)=200+15(12)=200+180=380 onumber ] Marcus will have 380 songs in 12 months.
Exercise (PageIndex{1})
If you earn $30,000 per year and you spend$29,000 per year write an equation for the amount of money you save after y years, if you start with nothing.
“The most important thing, spend less than you earn!(http://www.thesimpledollar.com/2009/...than-you-earn/)
(S(y)=30,000y - 29,000y = 1000y) $1000 is saved each year. Definition: Calculating rate of change Given two values for the input,(x_{1} { m ; and; }x_{2}), and two corresponding values for the output, (y_{1} { m ; and; }y_{2}), or a set of points, ((x_{1} { m ,; ; }y_{1} )) and((x_{2} { m ,; ; }y_{2} )), if we wish to find a linear function that contains both points we can calculate the rate of change, m: [m=dfrac{ m change; in; output}{ m change; in; input} =dfrac{Delta y}{Delta x} =dfrac{y_{2} -y_{1} }{x_{2} -x_{1} }] Rate of change of a linear function is also called the slope of the line. Note in function notation,(y_{1} =f(x_{1} )) and (y_{2} =f(x_{2} )), so we could equivalently write [m=dfrac{fleft(x_{2} ight)-fleft(x_{1} ight)}{x_{2} -x_{1} }] Example (PageIndex{2}) The population of a city increased from 23,400 to 27,800 between 2002 and 2006. Find the rate of change of the population during this time span. Solution The rate of change will relate the change in population to the change in time. The population increased by (27800-23400=4400)people over the 4 year time interval. To find the rate of change, the number of people per year the population changed by: [dfrac{4400 ext{ people}}{4 ext{ years}} =1100dfrac{ ext{people}}{ ext{year}}= 1100 ext{ people per year} onumber] Notice that we knew the population was increasing, so we would expect our value for (m) to be positive. This is a quick way to check to see if your value is reasonable. Example (PageIndex{3}) The pressure, (P), in pounds per square inch (PSI) on a diver depends upon their depth below the water surface, (d), in feet, following the equation (P(d)=14.696+0.434d). Interpret the components of this function. Solution The rate of change, or slope, 0.434 would have units (dfrac{ ext{output}}{ ext{input}} = dfrac{ ext{pressure}}{ ext{depth}} = dfrac{ ext{PSI}}{ ext{ft}}). This tells us the pressure on the diver increases by 0.434 PSI for each foot their depth increases. The initial value, 14.696, will have the same units as the output, so this tells us that at a depth of 0 feet, the pressure on the diver will be 14.696 PSI. Example (PageIndex{4}) If (f(x)) is a linear function, (f(3)=-2), and (f(8)=1), find the rate of change. Solution (f(3)=-2) tells us that the input 3 corresponds with the output -2, and (f(8)=1) tells us that the input 8 corresponds with the output 1. To find the rate of change, we divide the change in output by the change in input: [m=dfrac{ ext{change in output}}{ ext{change in input}} =dfrac{1-(-2)}{8-3} =dfrac{3}{5} onumber] If desired we could also write this as (m = 0.6) Note that it is not important which pair of values comes first in the subtractions so long as the first output value used corresponds with the first input value used. Exercise (PageIndex{2}) Given the two points (2, 3) and (0, 4), find the rate of change. Is this function increasing or decreasing? Answer (m=dfrac{4-3}{0-2} =dfrac{1}{-2} =-dfrac{1}{2}) ; Decreasing because (m < 0) We can now find the rate of change given two input-output pairs, and can write an equation for a linear function once we have the rate of change and initial value. If we have two input-output pairs and they do not include the initial value of the function, then we will have to solve for it. Example (PageIndex{5}) Write an equation for the linear function graphed to the right. Solution Looking at the graph, we might notice that it passes through the points (0, 7) and (4, 4). From the first value, we know the initial value of the function is (b = 7), so in this case we will only need to calculate the rate of change: [m=dfrac{4-7}{4-0} =dfrac{-3}{4} onumber ] This allows us to write the equation: [f(x)=7-dfrac{3}{4} x onumber ] Example (PageIndex{6}) If (f(x))is a linear function, (f(3)=-2), and (f(8)=1), find an equation for the function. Solution In example 3, we computed the rate of change to be (m=dfrac{3}{5}). In this case, we do not know the initial value(f(0)), so we will have to solve for it. Using the rate of change, we know the equation will have the form (f(x)=b+dfrac{3}{5} x). Since we know the value of the function when (x = 3), we can evaluate the function at 3. [f(3)=b+dfrac{3}{5} (3) onumber ] Since we know that(f(3)=-2), we can substitute on the left side [-2=b+dfrac{3}{5} (3) onumber ] This leaves us with an equation we can solve for the initial value [b=-2-dfrac{9}{5} =dfrac{-19}{5} onumber ] Combining this with the value for the rate of change, we can now write a formula for this function: [f(x)=dfrac{-19}{5} +dfrac{3}{5} x onumber ] Example (PageIndex{7}) Working as an insurance salesperson, Ilya earns a base salary and a commission on each new policy, so Ilya’s weekly income, (I), depends on the number of new policies, (n), he sells during the week. Last week he sold 3 new policies, and earned$760 for the week. The week before, he sold 5 new policies, and earned $920. Find an equation for (I(n)), and interpret the meaning of the components of the equation. Solution The given information gives us two input-output pairs: (3,760) and (5,920). We start by finding the rate of change. [m=dfrac{920-760}{5-3} =dfrac{160}{2} =80 onumber ] Keeping track of units can help us interpret this quantity. Income increased by$160 when the number of policies increased by 2, so the rate of change is $80 per policy; Ilya earns a commission of$80 for each policy sold during the week.
We can then solve for the initial value
[I(n)=b+80n onumber] then when (n = 3), (I(3)=760), giving
[760=b+80(3) onumber ] this allows us to solve for (b)
[b=760-80(3)=520 onumber ]
This value is the starting value for the function. This is Ilya’s income when (n = 0), which means no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold.
Writing the final equation:
[I(n)=520+80n onumber ]
Our final interpretation is: Ilya’s base salary is $520 per week and he earns an additional$80 commission for each policy sold each week.
flashback
Looking at Example 7:
Determine the independent and dependent variables.
What is a reasonable domain and range?
Is this function one-to-one?
(n) (number of policies sold) is the independent variable
(I(n)) (weekly income as a function of policies sold) is the dependent variable.
A reasonable domain is (0, 15)({}^{*})
A reasonable range is ($540,$1740)({}^{*})
({}^{*})answers may vary given reasoning is stated; 15 is an arbitrary upper limit based on selling 3 policies per day in a 5 day work week and $1740 corresponds with the domain. Yes this function is one-to-one Exercise (PageIndex{3}) The balance in your college payment account, (C), is a function of the number of quarters, (q), you attend. Interpret the function (C(a) = 20000 – 4000q) in words. How many quarters of college can you pay for until this account is empty? Answer Your College account starts with$20,000 in it and you withdraw $4,000 each quarter (or your account contains$20,000 and decreases by \$4000 each quarter.) Solving (C(a) = 0) gives (a = 5). You can pay for 5 quarters before the money in this account is gone.
Example (PageIndex{8})
Given the table below write a linear equation that represents the table values
(w), number of weeks 0 2 4 6 (P(w)), number of rats 1000 1080 1160 1240
Solution
We can see from the table that the initial value of rats is 1000 so in the linear format
[P(w)=b+mw,: b = 1000 onumber]
Rather than solving for (m), we can notice from the table that the population goes up by 80 for every 2 weeks that pass. This rate is consistent from week 0, to week 2, 4, and 6. The rate of change is 80 rats per 2 weeks. This can be simplified to 40 rats per week and we can write
[P(w)=b+mw ext{ as }P(w)=1000+40w onumber ]
If you didn’t notice this from the table you could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240),
[m=dfrac{1240-1080}{6-2} =dfrac{160}{4} =40 ext{ rats per week} onumber]
Important Topics of this Section
• Definition of Modeling
• Definition of a linear function
• Structure of a linear function
• Increasing & Decreasing functions
• Finding the vertical intercept (0, b)
• Finding the slope/rate of change, m
• Interpreting linear functions
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2021-11-30 21:36:16
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https://repository.uantwerpen.be/link/irua/80851
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Title Measurement of the energy dependence of hadronic jet rates and the strong coupling $\alpha_{s}$ from the four-jet rate with the DELPHI detector at LEP Author Abdallah, J. Van Remortel, N. et al. Faculty/Department Faculty of Sciences. Physics Publication type article Publication 2005 Berlin , 2005 Subject Physics Source (journal) European physical journal : C : particles and fields. - Berlin Volume/pages 38(2005) :4 , p. 413-426 ISSN 1434-6044 ISI 000226120100002 Carrier E Target language English (eng) Full text (Publishers DOI) Affiliation University of Antwerp Abstract Hadronic events from the data collected with the DELPHI detector at LEP within the energy range from 89 Gev to 209 Gev are selected, their jet rates are determined and compared to predictions of four different event generators. One of them is the recently developed APACIC + + generator which performs a massive matrix element calculation matched to a parton shower followed by string fragmentation. The four-jet rate is used to measure $\alpha_s$ in the next-to-leading-order approximation yielding {\alpha_s(M_Z E-info http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000226120100002&DestLinkType=RelatedRecords&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000226120100002&DestLinkType=FullRecord&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 http://gateway.webofknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&KeyUT=WOS:000226120100002&DestLinkType=CitingArticles&DestApp=ALL_WOS&UsrCustomerID=ef845e08c439e550330acc77c7d2d848 Handle
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2017-04-26 04:35:46
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https://diabetesjournals.org/diabetes/article/57/11/3083/13436/Diffusion-Tensor-Imaging-Identifies-Deficits-in
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OBJECTIVE—Long-standing type 1 diabetes is associated with deficits on neurocognitive testing that suggest central white matter dysfunction. This study investigated whether diffusion tensor imaging (DTI), a type of magnetic resonance imaging that measures white matter integrity quantitatively, could identify white matter microstructural deficits in patients with long-standing type 1 diabetes and whether these differences would be associated with deficits found by neurocognitive tests.
RESEARCH DESIGN AND METHODS—Twenty-five subjects with type 1 diabetes for at least 15 years and 25 age- and sex-matched control subjects completed DTI on a 3.0 Tesla scanner and a battery of neurocognitive tests. Fractional anisotropy was calculated for the major white matter tracts of the brain.
RESULTS—Diabetic subjects had significantly lower mean fractional anisotropy than control subjects in the posterior corona radiata and the optic radiation (P < 0.002). In type 1 diabetic subjects, reduced fractional anisotropy correlated with poorer performance on the copy portion of the Rey-Osterreith Complex Figure Drawing Test and the Grooved Peg Board Test, both of which are believed to assess white matter function. Reduced fractional anisotropy also correlated with duration of diabetes and increased A1C. A history of severe hypoglycemia did not correlate with fractional anisotropy.
CONCLUSIONS—DTI can detect white matter microstructural deficits in subjects with long-standing type 1 diabetes. These deficits correlate with poorer performance on selected neurocognitive tests of white matter function.
Type 1 diabetes is a complex metabolic disease that can have devastating effects on multiple organ systems. Although complications involving the kidneys, nerves, and eyes have long been recognized, the effect of diabetes on cognition has not been as clearly understood. Patients with type 1 diabetes have been found to have deficits on standard neurocognitive tests of information processing (14), psychomotor efficiency (1,2), motor speed (3,57), visuoconstruction (4,8), attention (4), somatosensory examination, motor strength (7), and executive function (9). Based on the type of cognitive deficits observed in patients with diabetes, there has been speculation that abnormalities in white matter may be at least partly responsible for cognitive dysfunction, particularly in patients with type 1 diabetes (2). Supporting this hypothesis, several studies have found gross morphological changes (10) and reduced white matter volume as measured by voxel-based morphometry (4) in patients with type 1 diabetes. However, prior studies have not looked specifically at the microstructural aspects of white matter integrity.
Over the last decade, a new type of magnetic resonance imaging (MRI) called diffusion tensor imaging (DTI) has been developed that is uniquely suited to assess white matter microstructure. DTI measures the magnitude and directionality of water diffusion in tissues, which may permit identification of tissue injury before it has progressed to the point of detection by more conventional imaging techniques. Without barriers, water molecules move uniformly in all directions, a phenomenon referred to as isotropic diffusion. In the presence of barriers, such as cell membranes, fibers, and myelin, the diffusion rate is greater in one direction, which is termed anisotropic diffusion. Fractional anisotropy provides a quantitative measure of the degree of diffusion anisotropy. Fractional anisotropy is high in regularly organized and structured white matter, such as the corpus callosum, and is lower in less organized tissues, such as gray matter (11,12). Previous DTI studies have shown that reduced fractional anisotropy correlates to cognitive dysfunction in a variety of conditions, including schizophrenia (13,14), depression (15), chronic alcohol use (16), Alzheimer's disease (17,18), and chronic cocaine use (19). In addition, a decrease in brain fractional anisotropy correlated negatively with viral load in patients with known HIV infection (20). This is of particular interest because patients living with HIV have many of the cognitive impairments that patients with type 1 diabetes exhibit, including decreased attention and speed of information processing (11).
To determine whether abnormalities in white matter microstructure might underlie the cognitive deficiencies identified in patients with longstanding type 1 diabetes, we used DTI to obtain quantitative data about white matter structural integrity in this population. We hypothesized that patients with long-standing type 1 diabetes will have decreased fractional anisotropy when compared with matched controls. To determine whether any identified reductions in fractional anisotropy correlated with abnormalities in neurocognitive function, subjects also completed a battery of standardized neurocognitive tests to assess performance on tasks believed to be supported by white matter regions. This analysis was done to test the hypothesis that fractional anisotropy will correlate with neurocognitive test performance in patients with type 1 diabetes, specifically with tests that depend on white matter integrity.
Twenty-five type 1 diabetic subjects were recruited from the Endocrine Clinics at the University of Minnesota Medical Center–Fairview and through institutional review board–approved fliers distributed around the University of Minnesota campus. To ensure that the subjects resembled those included in previous studies assessing cognitive function in diabetes (3,21), only subjects with a diabetes duration of 15 years or more were eligible for participation. Exclusion criteria included a history of or current evidence of any substance abuse disorder other than tobacco or caffeine dependence; severe psychiatric disorder, including major depressive disorder; seizure disorder (not related to hypoglycemia); transient ischemic attack; stroke; head injury; other diseases of the central nervous system; chemotherapy; and any condition that precluded the performance of MRI (weight >300 lbs [subjects of this size cannot fit into the magnetic resonance instrument], presence of metal implants or shrapnel, claustrophobia, etc.). Education history was also assessed. Twenty-five healthy volunteers without diabetes and free of the exclusion criteria were recruited from the University of Minnesota community to match the subjects with respect to age and sex (Table 1).
All subjects reported to the General Clinical Research Center at the University of Minnesota for neurocognitive testing after breakfast or lunch. On a separate day but within 2 weeks of the neurocognitive testing, subjects reported to the Center for Magnetic Resonance Research at the University of Minnesota for DTI, again after either breakfast or lunch. Diabetic subjects were instructed to manage their condition in their usual manner. Upon arrival, diabetic subjects performed fingerstick blood glucose testing to ensure that blood glucose was between 5.5–13.9 mmol/l (100–250 mg/dl). If blood glucose was outside of this glycemic range, appropriate therapy was administered by the investigator to bring them into target within 1 h or the study was rescheduled. A1C was also measured in diabetic subjects using a blood sample obtained on the day of neurocognitive testing.
### Neurocognitive testing.
The neurocognitive testing battery included the Wechsler Abbreviated Scale of Intelligence (WASI) and tests of psychomotor speed and executive function that have been found to be impaired in subjects with white matter disease (22,23). These tests included the Paced Auditory Serial Addition Test (PASAT), the Digit Vigilance Test (DVT), Trails A and B, the Benton Judgment of Line Orientation Test (JLO), the Rey-Osterreith Complex Figure Drawing Test (Rey-O), the Grooved Pegboard Test, and the Conners’ Continuous Performance Test–Second Edition (CPT-II). Each testing session lasted 2–3 h. All tests were administered and scored by trained personal and interpreted by one of the authors (F.S.A.). Comparison of diabetic subjects and control subjects was based on the Rey-O (copy and delayed recall), the JLO, the Grooved Pegboard Test (dominant and nondominant hands), the DVT (total time needed to complete), the WASI (full-scale IQ score), and the PASAT (with numbers being separated by 2.4, 2.0, 1.6, and 1.2 s). The System Checklist-90-R was also given as a psychological screen.
### MRI collection.
DTI was performed by positioning the field of view to cover the entire cerebrum. Acquisition parameters for the axially acquired, dual spin echo, single shot, echo planar, diffusion-weighted sequence were repetition time 10,500 ms, echo time 98 ms, matrix 128 × 128, field of view 256 mm, slice thickness 2 mm, 64 contiguous slices, 2 averages, and a b value 1,000. A volume without diffusion weighting and 12 volumes with noncollinear diffusion directions were acquired. A B0 fieldmap sequence with identical voxel parameters to the DTI scan was acquired after the DTI sequence. A three-dimensional T1 Magnetization Prepared Rapid Gradient Echo sequence with repetition time 2,530 ms, echo time 3.63 ms, inversion time 1,100 ms, isotropic voxel 1 mm, field of view 256 mm, and matrix 256 × 256 was also acquired to provide anatomical information.
### DTI data processing.
Diffusion-weighted images were processed using tools from the FMRIB Software Library (FSL) (24). Eddy current-correction of the individual diffusion-weighted echo-planar images was performed using an affine transformation (25), and magnetic field susceptibility effects were corrected using the B0 fieldmap image. The diffusion tensor was calculated using the nondiffusion and 12 diffusion images, and fractional anisotropy maps were created using eq. 1 below, where FA is fractional anisotrophy and λ1, λ2, and λ3 represent values along the three major axes of diffusion (26).
$FA{=}\sqrt{\frac{3}{2}{\times}\frac{({\lambda}_{1}{-}{\lambda})^{2}{+}({\lambda}_{2}{-}{\lambda})^{2}{+}({\lambda}_{3}{-}{\lambda})^{2}}{{\lambda}_{1}^{2}{+}{\lambda}_{2}^{2}{+}{\lambda}_{3}^{2}}}$
The mean fractional anisotropy was computed for the white matter of four brain regions defined by anatomical landmarks that include the genu and splenium of the corpus callosum and the anterior and posterior commisures (Fig. 1A). The Tract-Based Spatial Statistics (TBSS) analysis program bundled with FSL (27) was used to determine average fractional anisotropy values at the center of several tracts of interest. TBSS generated a fractional anisotropy skeleton of the major white matter tracts of the brain for each subject. The fractional anisotropy map from the control subject with a brain size at the median for the controls was used to align the fractional anisotropy maps from the other control subjects and diabetic subjects. Once the fractional anisotropy maps were aligned, a common white matter skeleton was generated and then further segmented into anatomically defined tracts using the primary eigenvector information for the target subject's DTI data based on atlas information from Mori et al. (28). These tracts were selected on the basis of relative size and ease of differentiation from surrounding tracts. The resulting tracts from this procedure included bilateral forceps minor, cingulum bundle, medial corona radiata, superior longitudinal fasciculus, and optic radiation (Fig. 2A). Additionally, six regions of interest (genu, rostral body, anterior midbody, posterior midbody, isthmus, and splenium; Fig. 2A) in the corpus callosum were generated using standardized subdivisions (29).
Comparison of diabetic subjects and control subjects was done with three separate groups of neural regions: 1) superior frontal region, inferior frontal region, occipital region, and region superior to corpus callosum (Fig. 1A); 2) optic radiation, splenium of the corpus callosum, and posterior corona radiata (Fig. 2A); and 3) forceps minor, nonsplenium subdivisions of the corpus callosum (genu, rostral body, anterior midbody, posterior midbody, and isthmus), right cingulum, left cingulum, right superior longitudinal fasciculus, left superior longitudinal fasciculus, and medial corona radiata (Fig. 2A).
### Statistics.
Neurocognitive test scores were compared between diabetic subjects and control subjects using t tests as an initial screening. Tests thought to be associated with white matter function were compared simultaneously in a general linear mixed model (SAS Proc Mixed) with a random intercept for each subject-pair and an unstructured covariance matrix for the repeated test scores within subjects. The same model was used with the three groups of brain regions to compare fractional anisotropy measurements between diabetic subjects and control subjects. Results are reported as mean ± SE. Spearman correlation is reported as a measure of association. All computations were performed in SAS version 9.1.3 (SAS Institute, Cary, NC). Throughout this statistical analysis, all results were corrected for multiple comparisons.
### Demographic data.
Clinical and demographic characteristics of the 25 type 1 diabetic subjects and the 25 age- and sex-matched healthy controls are summarized in Table 1. The diabetic subjects and controls were similar in all of the demographic measures collected. Three diabetic subjects reported a history of retinopathy, two reported a history of gastroparesis, and three reported histories of both retinopathy and neuropathy. All but one subject who reported retinopathy had intervention (laser therapy or vitrectomy). No diabetic subjects reported a history of nephropathy. Fifteen of the 25 diabetic subjects reported a history of severe hypoglycemia, defined as having seizures, loss of consciousness, or needing another person's help to treat the symptoms of low blood glucose.
### Neurocognitive testing data.
No tests included in our battery distinguished type 1 diabetic subjects from control subjects. However, diabetic subjects appeared to have a tendency to perform more poorly on the Rey-O, a test believed to be a measure of white matter function that requires a high degree of visuospatial orientation and planning. This test is scored based on the number of correctly copied elements, and the diabetic subjects had a mean score of 31 ± 0.6, and the controls scored 33 ± 0.6 (P = 0.063).
### DTI data.
In the four large regions of analysis (inferior frontal, superior frontal, superior to the corpus callosum, and occipital regions with ∼15% overlap between the frontal regions and the region superior to the corpus callosum), fractional anisotropy was lower but not significantly in type 1 diabetic subjects (Fig. 1). In more specific major white matter tracts, diabetic subjects had significantly lower fractional anisotropy than control subjects in both the posterior corona radiata (0.443 ± 0.004 vs. 0.471 ± 0.004, P < 0.0001) and the optic radiation (0.361 ± 0.004 vs. 0.380 ± 0.004, P = 0.0017) (Fig. 2). Type 1 diabetic subjects also tended to have lower fractional anisotropy than control subjects in the splenium (0.806 ± 0.006 vs. 0.820 ± 0.006, P = 0.0947) and the posterior body of the corpus callosum (0.805 ± 0.007 vs. 0.821 ± 0.007, P = 0.1397).
### Correlation of DTI and neurocognitive testing data.
In diabetic subjects, there was a significant correlation between performance on the copying portion of the Rey-O and fractional anisotropy in the posterior corona radiata, with poorer scores associated with reduced fractional anisotropy (Table 2). These subjects also demonstrated a significant association between longer time required to complete the Grooved Pegboard Test, using both the nondominant hand and dominant hand, and reduced fractional anisotropy in the optic radiation, posterior corona radiata, and splenium of the corpus callosum (Table 2). Control subjects demonstrated a similar correlation between performance on the Grooved Pegboard Test and fractional anisotropy in the optic radiation (dominant hand only) and, unlike diabetic subjects, the fractional anisotropy in the posterior body of the corpus callosum (both hands) (Table 2). For all subjects, correlation between Rey-O and fractional anisotropy in the posterior corona radiata is shown in Fig. 3. To assess the effects of microvascular complications, the group with diabetes was divided into those with (n = 8) and without (n = 17) such complications; both subgroups had significantly lower fractional anisotropy in the posterior corona radiata than controls, and those with microvascular complications also had significantly lower Rey-O scores (Table 3).
### Correlation of DTI with characteristics of diabetic subjects.
Age, duration of diabetes, and A1C were negatively correlated with reduced fractional anisotropy of the optic radiation and posterior corona radiata, whereas duration of diabetes was also negatively correlated with reduced fractional anisotropy of the splenium of the corpus callosum (Table 4). No significant correlations were identified between severe hypoglycemia or microvascular complications and reduced fractional anisotropy in any of the major white matter tracts.
In this investigation, we found that white matter integrity, as measured by fractional anisotropy using DTI, was lower in several white matter tracts, including the posterior corona radiata and optic radiation, in patients with long-standing type 1 diabetes compared with age- and sex-matched controls. We also observed there to be a significant correlation between reduced fractional anisotropy in white matter tracts and reduced performance on neurocognitive tests thought to assess white matter function, including the Rey-O copy and the Grooved Pegboard Test. Together, these results demonstrate for the first time that microstructural abnormalities can be identified in the brains of subjects with long-standing type 1 diabetes that may underlie the cognitive dysfunction identified in this population. This work may open the door for the development of a novel biomarker for cognitive dysfunction that can be used to better understand the cerebral complications of diabetes. If such a biomarker can be developed, future study could focus on identifying interventions that can prevent the appearance or slow the progression of white matter microstructural changes in patients with diabetes.
Over the last 10 years, investigators have used a variety of imaging techniques, including MRI, single photon emission computed tomography (30,31), and positron emission tomography (32), to determine whether structural abnormalities could be identified in the brains of patients with type 1 diabetes. Although Biessels et al. (10) observed high signal lesions in cerebral white matter (or leukoaraiosis) in these patients, most studies have not confirmed this finding (1,33,34). Others have found increased cerebral spinal fluid and cerebral global atrophy (35,36), stable hippocampal and amygdala volumes (36), and decreased cerebral gray matter density (37,38) in type 1 diabetic subjects. Some of these structural abnormalities have corresponded to age of diabetes onset (35), A1C levels, hypoglycemia (37), and the presence of retinopathy (38). Recently, Wessels et al. (4) applied MRI voxel-based morphometry to measure white matter volumes in patients with type 1 diabetes and found subjects with proliferative retinopathy to have significantly smaller white matter volumes than diabetic subjects who were free of proliferative retinopathy and control subjects without diabetes. In the Wessels et al. study, reduced white matter volume correlated with worse performances on tests for attention, speed of information processing, and executive function. Whether reduced white matter volumes correlates with a reduction in fractional anisotropy is uncertain, but in our study, we did not find a relationship between the presence of retinopathy and either reduced fractional anisotropy or mild cognitive impairment. Future investigation should focus on this point because an association with retinopathy suggests that a common mechanism may be responsible for the development of both retinopathy and changes in white matter structure.
Previous studies that were designed to assess neurocognitive dysfunction in patients with type 1 diabetes identified abnormalities with information processing (14), psychomotor efficiency (1,2), motor speed (3,57), visuoconstruction (4,8), attention (4), somatosensory examination, motor strength (7), and executive function (9). Generally, these cognitive domains, which require the integration of several different tasks, are thought to be associated with the integrity of the white matter tracts that connect gray matter regions. In selecting our neurocognitive battery, we focused on measures like the PASAT, the DVT, the Grooved Pegboard Test, and the Trails A and B, which have previously identified differences in “psychomotor efficiency” between subjects with and without diabetes (2). We also selected measures that assessed other white matter function domains (e.g., processing speed, attention, and visual-spatial processing), such as the JLO, the Trails A and B, the PASAT, the Rey-O, the Grooved Pegboard Test, the CPT-II, and the DVT. Importantly, our battery also included the WASI, a test that screens for intellectual ability, as a control because most studies in type 1 diabetic subjects show no deficits in general intelligence (13). Our results demonstrated that no differences could be identified between type 1 diabetic subjects and control subjects on the WASI or the majority of the other tests performed. Most likely, this is secondary to our relatively small sample size, and with a larger number, perhaps we would have found significance in such tests as the Rey-O Copy and the Grooved Peg Board Test. Nevertheless, there were significant associations between these neurocognitive tests and reduced fractional anisotropy in patients with type 1 diabetes.
In this investigation, we used DTI to evaluate white matter microstructure. Although this form of imaging is known to measure the magnitude and directionality of water diffusion in tissues, what specific biochemical or morphological abnormality underlies the change in diffusion remains unknown. Like diabetes (39), many of the diseases that have similar changes in fractional anisotropy are associated with increases in oxidative stress, including chronic alcoholism (40), depression (41), Alzheimer's disease (42), chronic cocaine use (43), schizophrenia (44), and HIV infection (45). Based on this, it is tempting to think that oxidative stress could affect the integrity of myelinated fibers in vivo.
There were several limitations to this study. Historical information about the diabetic subjects was limited to self-report and a single A1C value. We did not confirm the presence or absence of retinopathy in our subjects, and if we had, we may have found an association between white matter structure/function and retinopathy as did Wessels et al. (4,38). We did identify a significant relationship between A1C and fractional anisotropy, but an examination of the relationship between a measure of long-term glycemia and changes in white matter microstructure would provide greater understanding of the role of chronic hyperglycemia in the development of reduced fractional anisotropy. Similarly, if a detailed history of hypoglycemia events and an assessment of hypoglycemia unawareness had been obtained, we may have been able to identify a link between hypoglycemia and structural changes as Musen et al. (37) did using MRI. Reduced fractional anisotropy was correlated with duration of diabetes; however, it was also correlated with increasing age. Future studies hopefully will be needed to confirm whether the duration of diabetes correlates with fractional anisotropy independent of age.
In summary, we found that DTI reveals white matter microstructure deficits (reduced fractional anisotropy) in patients with longstanding type 1 diabetes compared with controls. In addition, white matter microstructure deficits seen in type 1 diabetic subjects correlate with impaired performance on neurocognitive tests that are thought to be associated with white matter function. Based on these findings, prospective, longitudinal studies are needed to both better understand the natural evolution of cognitive dysfunction in type 1 diabetes and to establish DTI as an ideal methodology to follow cognitive dysfunction in diabetes. If further studies confirm our findings correlating diabetes, white matter microstructural changes, and neurocognitive dysfunction, future work will determine what factors will modify the progression of this diabetic complication (i.e., reducing A1C levels, novel pharmacological interventions, etc.).
FIG. 1.
Fractional anisotropy in major brain regions. A: Superior frontal region, inferior frontal region, occipital region, and region superior to corpus callosum (going clockwise starting at top left). There is ∼15% overlap between the frontal regions and the region superior to the corpus callosum. B: Fractional anisotropy in major brain regions; there was no significant difference in any of the major regions. □, type 1 diabetic subjects; ▪, control subjects. Data are means ± SE.
FIG. 1.
Fractional anisotropy in major brain regions. A: Superior frontal region, inferior frontal region, occipital region, and region superior to corpus callosum (going clockwise starting at top left). There is ∼15% overlap between the frontal regions and the region superior to the corpus callosum. B: Fractional anisotropy in major brain regions; there was no significant difference in any of the major regions. □, type 1 diabetic subjects; ▪, control subjects. Data are means ± SE.
Close modal
FIG. 2.
Fractional anisotropy of specific white matter tracts. A: Left-coronal view: corona radiata* (red), superior longitudinal fasciculus (yellow), corpus callosum isthmus (blue), and cingulum (green). Middle-saggital view: genu (green), rostral body (purple), anterior midbody (yellow), posterior midbody (red), isthmus (blue), and splenium (teal) of corpus callosum. Right-axial view: optic radiation* (orange). *P < 0.002 between diabetic subjects and controls. B: Fractional anisotropy in posterior corona radiata and optic radiation. □, type 1 diabetic subjects; ▪, control subjects. *P < 0.002 between diabetic subjects and controls. Data are means ± SE. (Please see http://dx.doi.org/10.2337/db08-0724 for a high-quality digital representation of this image.)
FIG. 2.
Fractional anisotropy of specific white matter tracts. A: Left-coronal view: corona radiata* (red), superior longitudinal fasciculus (yellow), corpus callosum isthmus (blue), and cingulum (green). Middle-saggital view: genu (green), rostral body (purple), anterior midbody (yellow), posterior midbody (red), isthmus (blue), and splenium (teal) of corpus callosum. Right-axial view: optic radiation* (orange). *P < 0.002 between diabetic subjects and controls. B: Fractional anisotropy in posterior corona radiata and optic radiation. □, type 1 diabetic subjects; ▪, control subjects. *P < 0.002 between diabetic subjects and controls. Data are means ± SE. (Please see http://dx.doi.org/10.2337/db08-0724 for a high-quality digital representation of this image.)
Close modal
FIG. 3.
Rey-O copy score versus fractional anisotropy in the posterior corona radiata. Overall Spearman correlation = 0.43 (P = 0.0018). ○, Control subjects; , diabetic subjects without microvascular complications; ▴, diabetic subjects with microvascular complications.
FIG. 3.
Rey-O copy score versus fractional anisotropy in the posterior corona radiata. Overall Spearman correlation = 0.43 (P = 0.0018). ○, Control subjects; , diabetic subjects without microvascular complications; ▴, diabetic subjects with microvascular complications.
Close modal
TABLE 1
Characteristics of subjects
Subjects with type 1 diabetesControl subjects
n 25 25
Age 45.1 ± 10.5 45.6 ± 10.8
Sex (F/M) 17/8 17/8
Education (years) 16.7 ± 1.9 16.1 ± 2.3
Duration of diabetes (years) 30.3 ± 10.8 NA
A1C (%) 7.4 ± 1.0 NA
Blood glucose before MRI (mmol/l [mg/dl]) 9.3 ± 3.6 (168 ± 64) NA
Blood glucose before neurocognitive testing (mmol/l [mg/dl]) 8.4 ± 2.7 (152 ± 49) NA
Subjects with type 1 diabetesControl subjects
n 25 25
Age 45.1 ± 10.5 45.6 ± 10.8
Sex (F/M) 17/8 17/8
Education (years) 16.7 ± 1.9 16.1 ± 2.3
Duration of diabetes (years) 30.3 ± 10.8 NA
A1C (%) 7.4 ± 1.0 NA
Blood glucose before MRI (mmol/l [mg/dl]) 9.3 ± 3.6 (168 ± 64) NA
Blood glucose before neurocognitive testing (mmol/l [mg/dl]) 8.4 ± 2.7 (152 ± 49) NA
Data are means ± SD. NA, not applicable.
TABLE 2
Spearman correlation coefficients between fractional anisotropy and neurocognitive measurements (scores) in type 1 diabetic subjects and control subjects
Splenium corpus callosum
Posterior body of corpus callosum
Type 1 diabetesControlType 1 diabetesControlType 1 diabetesControlType 1 diabetesControl
Rey-O copy 0.10 −0.10 0.46* 0.16 0.06 0.06 −0.07 0.02
Grooved Peg Board, nondominant −0.66* −0.42* −0.72* −0.04 −0.56* −0.20 −0.32 −0.60*
Grooved Peg Board, dominant −0.40* −0.39 −0.70* 0.09 −0.48* −0.05 −0.24 −0.53*
Splenium corpus callosum
Posterior body of corpus callosum
Type 1 diabetesControlType 1 diabetesControlType 1 diabetesControlType 1 diabetesControl
Rey-O copy 0.10 −0.10 0.46* 0.16 0.06 0.06 −0.07 0.02
Grooved Peg Board, nondominant −0.66* −0.42* −0.72* −0.04 −0.56* −0.20 −0.32 −0.60*
Grooved Peg Board, dominant −0.40* −0.39 −0.70* 0.09 −0.48* −0.05 −0.24 −0.53*
*
Significantly different from zero (P < 0.05).
TABLE 3
Comparison of fractional anisotropy in diabetic subjects with microvascular complications, diabetic subjects without microvascular complications, and control subjects
Diabetic subjects with microvascular complicationsDiabetic subjects with no microvascular complicationsControl subjectsF test P value
Posterior corona radiata 0.4350 ± 0.0097a 0.4470 ± 0.0059a 0.4710 ± 0.0042b 0.0003
Rey-O 30 ± 1.3a 32 ± 0.8ab 33 ± 0.6b 0.0696
Diabetic subjects with microvascular complicationsDiabetic subjects with no microvascular complicationsControl subjectsF test P value
Posterior corona radiata 0.4350 ± 0.0097a 0.4470 ± 0.0059a 0.4710 ± 0.0042b 0.0003
Rey-O 30 ± 1.3a 32 ± 0.8ab 33 ± 0.6b 0.0696
Data are means ± SE. Comparisons are within rows: means that do not share letters are significantly different (P < 0.05).
TABLE 4
Spearman correlation coefficients of age, duration of diabetes, and A1C with fractional anisotropy
Age of type 1 diabetic subjects −0.58* −0.76* −0.32
Duration of diabetes −0.60* −0.61* −0.46*
A1C −0.47* −0.46* −0.38
Age of type 1 diabetic subjects −0.58* −0.76* −0.32
Duration of diabetes −0.60* −0.61* −0.46*
A1C −0.47* −0.46* −0.38
*
Significantly different from zero (P < 0.05).
Published ahead of print at http://diabetes.diabetesjournals.org on 11 August 2008.
The opinions expressed in this article are those of the authors and do not necessarily reflect the official views of the National Center for Research Resources (NCRR) or the National Institutes of Health (NIH).
Readers may use this article as long as the work is properly cited, the use is educational and not for profit, and the work is not altered. See http://creativecommons.org/licenses/by-nc-nd/3.0/ for details.
The costs of publication of this article were defrayed in part by the payment of page charges. This article must therefore be hereby marked “advertisement” in accordance with 18 U.S.C. Section 1734 solely to indicate this fact.
This work has received Grant M01-RR-00400 from the NCRR, a component of the NIH; NIH Grant R01-MH-060662; and funding from the Pennock Family Diabetes Research Fund.
We thank Danielle Seim for her contribution to the project.
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2022-05-26 23:24:34
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https://socratic.org/questions/how-do-you-find-the-midpoint-of-1-4-to-3-2
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# How do you find the midpoint of (-1,4) to (3,2)?
May 28, 2018
The midpoint is at $\left(1 , 3\right)$.
#### Explanation:
The midpoint formula is shown here:
We are given the two endpoints, so we can plug it into the formula to find the midpoint. Notice that the formula is same as the average of the two x-values and y-values.
$\text{Midpoint} = \left(\frac{- 1 + 3}{2} , \frac{4 + 2}{2}\right)$
$\quad \quad \quad \quad \quad \quad \quad \quad = \left(\frac{2}{2} , \frac{6}{2}\right)$
$\quad \quad \quad \quad \quad \quad \quad \quad = \left(1 , 3\right)$
Hope this helps!
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2021-12-03 10:46:14
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https://www.r-bloggers.com/2018/12/part-4-why-does-bias-occur-in-optimism-corrected-bootstrapping/
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In the previous parts of the series we demonstrated a positive results bias in optimism corrected bootstrapping by simply adding random features to our labels. This problem is due to an ‘information leak’ in the algorithm, meaning the training and test datasets are not kept seperate when estimating the optimism. Due to this, the optimism, under some conditions, can be very under estimated. Let’s analyse the code, it is pretty straightforward to understand then we can see where the problem originates.
1. Fit a model M to entire data S and estimate predictive ability C. ## this part is our overfitted estimation of performance (can be AUC, accuracy, etc)
2. Iterate from b=1…B: ## now we are doing resampling of the data to estimate the error
1. Take a resample from the original data, S*
2. Fit the bootstrap model M* to S* and get predictive ability, C_boot ## this will clearly give us another overfitted model performance, which is fine
3. Use the bootstrap model M* to get predictive ability on S, C_orig ## we are using the same data (samples) used to train the model to test it, therefore it is not surprising that we have values with better performance than expected. C_orig values will be too high.
3. Optimism O is calculated as mean(C_boot – C_orig)
4. Calculate optimism corrected performance as C-O.
One way of correcting for this would be changing step 3 of the bootstrap, instead of testing on the original data, to test on the left out (unseen) data in the bootstrap. This way the training and test data is kept entirely seperate in terms of samples, thus eliminating our bias towards inflated model performance on null datasets with high features. There is probably no point of doing anyway this when we have methods such as LOOCV and K fold cross validation.
As p (features) >> N (samples) we are going to get better and better ability to get good model performance using the bootstrapped data on the original data. Why? Because the original data contains the same samples as the bootstrap and as we get more features, greater the chance we are going to get some randomly correlating with our response variable. When we test the bootstrap on the original data (plus more samples) it retains some of this random ability to predict the real labels. This is a typical overfitting problem when we have higher numbers of features, and the procedure is faulty.
Let’s take another experimental look at the problem, this code can be directly copied and pasted into R for repeating the analyses and plots. We have two implementations of the method, the first by me for glmnet (lasso logisitic regression), the second for glm (logisitic regression) from this website (http://cainarchaeology.weebly.com/r-function-for-optimism-adjusted-auc.html). Feel free to try different machine learning algorithms and play with the parameters.
library(glmnet)
library(pROC)
library(caret)
library(ggplot)
library(kimisc)
### TEST 1: bootstrap optimism with glmnet
cc <- c()
for (zz in seq(2,100,1)){
print(zz)
## set up test data
test <- matrix(rnorm(100*zz, mean = 0, sd = 1),
nrow = 100, ncol = zz, byrow = TRUE)
labelsa <- as.factor(c(rep('A',50),rep('B',50)))
colnames(test) <- paste('Y',seq(1,zz),sep='')
row.names(test) <- paste('Z',seq(1,100),sep='')
### my own implementation of optimism corrected bootstrapping
## 1. fit model to entire test data (i.e. overfit it)
orig <- glmnet(test,y=labelsa,alpha=1,family = "binomial")
preds <- predict(orig,newx=test,type='response',s=0.01)
auc <- roc(labelsa,as.vector(preds))
original_auc <- as.numeric(auc$auc) ## 2. take resample of data and try to estimate optimism test2 <- cbind(test,labelsa) B <- 50 results <- matrix(ncol=2,nrow=B) for (b in seq(1,B)){ ## get the bootstrapped resampled data boot <- test2[sample(row.names(test2),50),] labelsb <- boot[,ncol(boot)] boot <- boot[,-ncol(boot)] ## use the bootstrapped model to predict its own labels bootf <- glmnet(boot,y=labelsb,alpha=1,family = "binomial") preds <- predict(bootf,newx=boot,type='response',s=0.01) auc <- roc(labelsb,as.vector(preds)) boot_auc <- as.numeric(auc$auc)
## use bootstrap model to predict original labels
preds <- predict(bootf,newx=test,type='response',s=0.01)
auc <- roc(labelsa,as.vector(preds))
boot_original_auc <- as.numeric(auc$auc) ## add the data to results results[b,1] <- boot_auc results[b,2] <- boot_original_auc } ## calculate the optimism O <- mean(results[,1]-results[,2]) ## calculate optimism corrected measure of prediction (AUC) corrected <- original_auc-O ## cc <- c(cc,corrected) print(cc) } ## print it df <- data.frame(p=seq(2,100,1),optimism_corrected_boot_AUC=cc) p1 <- ggplot(df, aes(x=p, y=optimism_corrected_boot_AUC)) + geom_line() + ggtitle('glmnet - random data only gives positive result with optimism corrected bootstrap') print(p1) png('glmnet_test_upto100.png', height = 15, width = 27, units = 'cm', res = 900, type = 'cairo') print(p1) dev.off() So here are the results, as number of noise only features on the x axis increase our 'corrected' estimate of AUC (on y axis) also increases when we start getting enough to allow that noise to randomly predict the labels. So this shows the problem starts about 40-50 features, then gets worse until about 75+. This is with the 'glmnet' function. Let’s look at the method using glm, we find the same trend, different implementation. ## TEST2 auc.adjust <- function(data, fit, B){ fit.model <- fit data$pred.prob <- fitted(fit.model)
# get overfitted AUC
auc.app <- roc(data[,1], data$pred.prob, data=data)$auc # require 'pROC'
auc.boot <- vector (mode = "numeric", length = B)
auc.orig <- vector (mode = "numeric", length = B)
o <- vector (mode = "numeric", length = B)
for(i in 1:B){
boot.sample <- sample.rows(data, nrow(data), replace=TRUE) # require 'kimisc'
fit.boot <- glm(formula(fit.model), data = boot.sample, family = "binomial")
boot.sample$pred.prob <- fitted(fit.boot) # get bootstrapped AUC auc.boot[i] <- roc(boot.sample[,1], boot.sample$pred.prob, data=boot.sample)$auc # get original data boot AUC data$pred.prob.back <- predict.glm(fit.boot, newdata=data, type="response")
auc.orig[i] <- roc(data[,1], data$pred.prob.back, data=data)$auc
# calculated optimism corrected version
o[i] <- auc.boot[i] - auc.orig[i]
}
}
cc <- c()
for (zz in seq(2,100,1)){
print(zz)
## set up test data
test <- matrix(rnorm(100*zz, mean = 0, sd = 1),
nrow = 100, ncol = zz, byrow = TRUE)
labelsa <- as.factor(c(rep('A',50),rep('B',50)))
colnames(test) <- paste('Y',seq(1,zz),sep='')
row.names(test) <- paste('Z',seq(1,100),sep='')
test2 <- data.frame(cbind(labelsa,test))
test2$labelsa <- as.factor(test2$labelsa)
## 1. make overfitted model
model <- glm(labelsa ~., data = test2, family = "binomial")
## 2. estimate optimism and correct
cc <- c(cc,d)
}
## print it
df <- data.frame(p=seq(2,100,1),optimism_corrected_boot_AUC=cc)
p2 <- ggplot(df, aes(x=p, y=optimism_corrected_boot_AUC)) +
geom_line() + ggtitle('glm - random data only gives positive result with optimism corrected bootstrap')
print(p2)
png('glmt_test_upto100.png', height = 15, width = 27, units = 'cm',
res = 900, type = 'cairo')
print(p2)
dev.off()
So there we have it, the method has a problem with it and should not be used with greater than about 40 features. This method unfortunately is currently being used for datasets with higher than this number of dimensions (40+) because people think it is a published method it is safe, unfortunately this is not how the world works R readers. Remember, the system is corrupt, science and statistics is full of lies, and if in doubt, do your own tests on positive and negative controls.
This is with the 'glm' function. Random features added on x axis, the corrected AUC on y axis.
What if you don’t believe it? I mean this is a text book method. Well R readers if that is so I suggest code it your self and try the code here, run the experiments on null (random data only) datasets with increasing features.
This is the last part in the series on debunking the optimism corrected bootstrap method. I consider it, debunked.
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2021-02-25 14:18:44
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http://www.zazizam.com/content/education/p/Photon.php
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Search for on the web shopping
Mon, 26 Sep, 2022
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In physics, the photon (from Greek φως, "phōs", meaning light) is the quantum of the electromagnetic field, for instance light. The term photon was coined by Gilbert Lewis in 1926. The photon can be perceived as a wave or a particle, depending on how it is measured The photon is one of the elementary particles. Its interactions with electrons and atomic nuclei account for a great many of the features of matter, such as the existence and stability of atoms, molecules, and solids. These interactions are studied in quantum electrodynamics (QED), which is the oldest part of the Standard Model of particle physics.In some respects a photon acts as a particle, for instance when registered by the light sensitive device in a camera. In other respects, a photon acts like a wave, as when passing through the optics in a camera. According to the so-called wave-particle duality in quantum physics, it is natural for the photon to display either aspect of its nature, according to the circumstances. Normally, light is formed from a large number of photons, with the intensity related to the number of them. At low intensity, it requires very sensitive instruments, used in astronomy or spectroscopy, for instance, to detect the individual photons. Jump to Page Contents
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Contents
Symbol
Properties
Photons in vacuo
Photons in matter
Symbol - Contents
A photon is usually given the symbol $, gamma$, the Greek letter gamma, although in nuclear physics this symbol refers to a very high-energy photon (a gamma ray).
Properties - Contents
Photons are commonly associated with visible light, but this is actually only a very limited part of the electromagnetic spectrum. All electromagnetic radiation is quantized as photons: that is, the smallest amount of electromagnetic radiation that can exist is one photon, whatever its wavelength, frequency, energy, or momentum, and that light or fields interact with matter in discrete units of one or several photons. Photons are fundamental particles. They can be created and destroyed when interacting with other particles, but are not known to decay on their own.A photon of a definite frequency is not a localized particle. Photons thus exhibit a position/frequency uncertainty relation similar to that of matter particles and exactly analogous to the bandwidth theorem of classical optics. Photons have zero mass and zero electric charge, but they do carry energy, momentum and angular momentum. Photons are always moving, and photons in a vacuum always move at a constant speed with respect to all observers, even when those observers are themselves moving. This speed is called the vacuum speed of light. The energy and momentum carried by a photon is proportional to its frequency (or inversely proportional to its wavelength). This momentum can be transferred when a photon interacts with matter. The force due to a large number of photons falling on a surface is known as radiation pressure, which may be used for propulsion with a solar sail.The actual upper limit of the photon rest mass compatible with observations depends on additional assumptions. A conservative consensus limit of the Particle Data Group is 6×10-17 eV based on changes in magnetohydrodynamics which would contradict solar wind observations.Photons are deflected by a gravitational field twice as much as Newtonian mechanics predicts for a mass traveling at the speed of light with the same momentum as the photon. This observation is commonly cited as evidence supporting Einstein's theory of gravitation, general relativity. In general relativity, photons always travel in a "straight" line, after taking into account the curvature of spacetime. (In curved space, such lines are called geodesics).
Creation
Photons are produced by atoms when a bound electron moves from one orbital to another orbital with less energy. Photons can also be emitted by an unstable nucleus when it undergoes some types of nuclear decay. Furthermore, photons are produced whenever charged particles are accelerated.Atoms continuously emit photons due to their collisions with each other. The wavelength distribution of these photons is thus related to their absolute temperature. The Planck distribution determines the probability of a photon being a certain wavelength when emitted by a collection of atoms at a given temperature. The spectrum of such photons is normally peaked in the range between microwave and infrared, but sufficiently hot objects (such as the surface of the Sun or a lightbulb filament) will emit visible light as well. As temperature is further increased, some photons will reach even higher frequencies, such as ultraviolet and X-ray.Radio, television, radar and other types of transmitters used for telecommunication and remote sensing routinely create a wide variety of low-energy photons by the oscillation of electric fields in conductors. Magnetrons emit coherent photons used in household microwave ovens. Klystron tubes are used when microwave emissions must be more finely controlled. Masers and lasers create monochromatic photons by stimulated emission. More energetic photons can be created by nuclear transitions, particle-antiparticle annihilation, and in high-energy particle collisions.
Spin
Photons have spin 1, and they are therefore classified as bosons. Photons mediate the electromagnetic interaction; they are the gauge bosons of quantum electrodynamics (QED), which is a U(1) gauge theory. A non- relativistic spin-1 particle has three possible spin states (−1, 0 and +1). However, in the framework of special relativity, this is not the case for massless spin-1 particles, such as the photons, which have only two spin projections, helicities, corresponding to the right- and left-handed circular polarizations of classical electromagnetic waves. The more familiar linear polarization is formed by a superposition of the two spin projections of a photon.
Quantum state
Visible light from ordinary sources (like the Sun or a lamp) is a mixture of many photons of different wavelengths. One sees this in the frequency spectrum, for instance by passing the light through a prism. In so-called "mixed states", which these sources tend to produce, light can consist of photons in thermal equilibrium (so-called black-body radiation). Here they in many ways resemble a gas of particles. For example, they exert pressure, known as radiation pressure.On the other hand, an assembly of photons can also exist in much more well-organized coherent states, such as in the light emitted by an ideal laser. The high degree of precision obtained with laser instruments is due to this organization.The quantum state of a photon assembly, like that of other quantum particles, is the so-called Fock state denoted $|nrangle$, meaning $, n$ photons in one of the distinct "modes" of the electromagnetic field. If the field is multimode (involves several different wavelength photons), its quantum state is a tensor product of photon states, for example:
$|n_{k_0}rangleotimes|n_{k_1}rangleotimesdotsotimes|n_{k_n}rangledots$
Here $, k_i$ denote the possible modes, and $, n_{k_i}$ the number of photons in each mode
Molecular absorption
A typical molecule, $, M$, has many different energy levels. When a molecule absorbs a photon, its energy is increased by an amount equal to the energy of the photon. The molecule then enters an excited state, $, M^*$.
$M + gamma to M^* ,$
Photons in vacuo - Contents
In empty space ( vacuum) all photons move at the speed of light, c, defined as 299,792,458 metres per second, or approximately 3×108 m/s. The metre is defined as the distance travelled by light in a vacuum in 1/299,792,458 of a second, so the speed of light does not suffer any experimental uncertainty, unlike the metre or the second, which rely on the second being defined by means of a very accurate clock.According to one principle of Einstein's special relativity, all observations of the speed of light in vacuo are same in all directions to any observer in an inertial frame of reference. This principle is generally accepted in physics since many practical consequences for high-energy particles in theoretical and experimental physics have been observed.
Photons in matter - Contents
When photons pass through matter, such as a prism, different frequencies will be transmitted at different speeds. This is called dispersion of colors, where photons of different frequencies exit at different angles. A similar phenomenon occurs in reflection where surfaces can reflect photons of various frequencies at different angles.The associated dispersion relation for photons is a relation between frequency, $, f$ , and wavelength, $, lambda$, or equivalently, between their energy, $, E$, and momentum, $, p$. It is simple in vacuo, since the speed of the wave, $, v$, is given by
$v=lambda f = c,$
The photon quantum relations are:
$E=hf ,$ and $p=h/lambda ,$
Here $, h$ is Planck's constant. So one can also write the dispersion relation as
$E=pc ,$
which is characteristic of a zero-mass particle. One sees how remarkably Planck's constant relates the wave and particle aspects.In a material, photons couple to the excitations of the medium and behave differently. These excitations can often be described as quasi-particles (such as phonons and excitons); that is, as quantized wave- or particle-like entities propagating though the matter. "Coupling" means here that photons can transform into these excitations (that is, the photon gets absorbed and medium excited, involving the creation of a quasi-particle) and vice versa (the quasi-particle transforms back into a photon, or the medium relaxes by re-emitting the energy as a photon). However, as these transformations are only possibilities, they are not bound to happen and what actually propagates through the medium is a polariton; that is, a quantum-mechanical superposition of the energy quantum being a photon and of it being one of the quasi-particle matter excitations.According to the rules of quantum mechanics, a measurement (here: just observing what happens to the polariton) breaks this superposition; that is, the quantum either gets absorbed in the medium and stays there (likely to happen in opaque media) or it re-emerges as photon from the surface into space (likely to happen in transparent media).Matter excitations have a non-linear dispersion relation; that is, their momentum is not proportional to their energy. Hence, these particles propagate slower than the vacuum speed of light. (The propagation speed is the derivative of the dispersion relation with respect to momentum.) This is the formal reason why light is slower in media (such as glass) than in vacuum. (The reason for diffraction can be deduced from this by Huygens' principle.) Another way of phrasing it is to say that the photon, by being blended with the matter excitation to form a polariton, acquires an effective mass, which means that it cannot travel at c, the speed of light in a vacuum.
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2022-09-26 13:01:54
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https://tex.stackexchange.com/questions/352783/change-spacing-around-operators-in-particular-commands/352790
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# Change spacing around operators in particular commands
I have a command to display subsequences, which I defined as follows:
\newcommand{\stx}[3] {\ensuremath{#1[#2\!:\!#3]}\xspace}
My initial intention was to render things like this:
The problem is that, in order to get the desired spacing around +, I have to write code like this:
\stx{s}{a\!+\!m}{b\!+\!n}
Otherwise the spacing around + is too large and the indices get unclear/unpleasant. But this problematic in case I want (and I probably will) change the definition of \stx.
I thought of changing the spacing around the operators with a command, like this
\newcommand{\tightBinOps}[1] {\medmuskip=1mu\relax #1}
\newcommand{\stx}[3] {\ensuremath{#1[\tightBinOps{#2}\!:\!\tightBinOps{#3}]}\xspace}
However, this has two problems:
1. It affects the whole document. I don't know how to restrict the spacing modification only to that specific expression;
2. I guess I shouldn't rely on the spacing between operators being \medmuskip or any other particular measure in every circumstance.
How could I safely write \tightBinOps—and, more importantly, should I write it or is there a better solution?
MCVE
\documentclass{article}
\newcommand{\tightBinOps}[1] {\medmuskip=1mu\relax #1}
\newcommand{\stxOriginal}[3] {#1[#2\!:\!#3]}
\newcommand{\stxModified}[3] {#1[\tightBinOps{#2}\!:\!\tightBinOps{#3}]}
\begin{document}
Too much space in $a+1$ here: $\stxOriginal{s}{a}{a+1}$
Desired space in $a+1$ here: $\stxModified{s}{a}{a+1}$.
But this affects the + operator when I use it in the same math
entry as a subsequence:
Compare:
\noindent$1+1~\stxModified{s}{a}{a+1}$\\
$1+1~\stxOriginal{s}{a}{a+1}$
\end{document}
• you can effectively change the class of any operator or relation to "ordinary" simply by wrapping it in braces: {+} or {:}. much less complicated than trying to redefine the spacing. another approach is to use \mathord as shown in answers to this question: Correct spacing when using \sim as negation – barbara beeton Feb 8 '17 at 20:24
• While this didn't solve my original problem, it was a great tip! Seems to solve my issues when using comma as decimal separator---$1{,}25$, for instance gives the correct spacing, while $1,25$ does not. – giusti Feb 8 '17 at 23:08
First of all, give up with \ensuremath that adds nothing and, worse, breaks consistent markup: math should be treated as math. However, since only the value of the spacing parameters current at the end of the formula is used for the whole math list, you have to box the particular piece where you want different spacing.
In the following code, the automatic spaces around operation and relation symbols are set to \thinmuskip, which is small and not flexible.
\documentclass{article}
\makeatletter
\newcommand{\stx}[3]{\mathpalette\giusti@stx{{#1}{#2}{#3}}}
\newcommand{\giusti@stx}[2]{\giusti@stx@a{#1}#2}
\newcommand{\giusti@stx@a}[4]{%
\mbox{%
\medmuskip=\thinmuskip
\thickmuskip=\thinmuskip
$\m@th#1#2[{#3}:{#4}]$%
}%
}
\makeatother
\begin{document}
Example subsequence: $\stx{s}{a}{b}$
Example subsequence: $\stx{s}{a}{b+1}$
Example subsequence: $\stx{s}{a+m}{b+n}$
\end{document}
On the other hand, such a notation is cumbersome. You get the same output with
\documentclass{article}
\makeatletter
\newcommand{\stx}[1]{\mathpalette\giusti@stx{#1}}
\newcommand{\giusti@stx}[2]{%
\mbox{%
\medmuskip=\thinmuskip
\thickmuskip=\thinmuskip
$\m@th#1#2$%
}%
}
\makeatother
\begin{document}
Example subsequence: $\stx{s[a:b]}$
Example subsequence: $\stx{s[a:b+1]}$
Example subsequence: $\stx{s[a+m:b+n]}$
\end{document}
which is not more difficult to type and clearer.
If you type
Subscript: $A_{\stx{s}{a}{b}}$
for the first version, or
Subscript: $A_{\stx{s[a:b]}}$
for the second version, you get
• Thanks for the answer, but I'm not sure I understand what you're doing there. What I'm looking for is that a+m is tighter than a:b. I'm also curious why \ensuremath is wrong here. My notation of s[a:b] is that a:b are indices over s, so I'm using this command outside the math mode more frequently than in math mode. – giusti Feb 8 '17 at 21:51
• @giusti If you want tighter spacing around +, instead of \medmuskip=\thinmuskip do \medmuskip=1mu. But I'd not do it. Wherever you use \stx, it's math nonetheless. Add \ensuremath if you want; I still recommend not to. – egreg Feb 8 '17 at 21:54
• That sure works! But I'm actually interested in the reason why you wouldn't do that and why \ensuremath is wrong. Are you saying that the $..$ should always be explicit? – giusti Feb 8 '17 at 22:16
• @giusti Yes. No markup=bad. – egreg Feb 8 '17 at 22:21
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2019-12-05 17:52:34
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https://engineering.stackexchange.com/questions/28261/composite-beam-increase-in-bending-moment-due-to-reinforced-steel
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Composite beam- increase in bending moment due to reinforced steel
While trying to solve this question and using the flexure formulas of bending for composite beams I obtained a lower value of moment for steel than for aluminum.How is that possible?
– Fred
May 8 '19 at 16:46
The difference in stiffness is important.
First let's consider the aluminum W8x40 section without any steel plates: It can be loaded until the aluminum reaches a stress of 12 ksi.
Then let's consider what happens if we bolt some very thin steel plates unto the flanges of the aluminum section. Then we can increase the load until the steel plates reach a stress of 18 ksi. As the steel is 3 times stiffer than aluminum, the aluminum will have a stress that is 3 times lower than the steel plates. (I'm assuming we can disregard the slip in the bolts as well as a linear-elastic stress distribution — a plastic stress distribution would be a different calculation.) That results in a maximum stress in aluminum of 6ksi, so we get a total capacity of only half of what we had before we added the steel plates.
If the steel plates aren't extremely thin, they may increase the capacity, but the point is that the steel plates may fail well before we reach the capacity of the aluminum.
• This is a good point, and one I hadn't thought of before. I actually spent a few minutes going "adding stronger material makes the section weaker? Wut?!" But yeah, it can. In this particular case, though, the plates do strengthen the section.
– Wasabi
May 12 '19 at 1:26
Let's approximately calculate the S of composite beam, knowing that aluminum section I is 146in^4.
$$I_{COMP}= I_{alum}+ 2*A_{steel} *4.25^2 = 146+(8*0.25*2)4.25^2=146+72.25=218.25in^4$$
The S of aluminum section is 35.5in^3 and the S of steel 72.25/4.5= 16.05in^3
At yield point the steel plates can take a moment of
$$M_s= \sigma*S=18*16.05=288.9 k.in$$
But the aluminum beam has not fully loaded at this stage, it has strained the same as steel but stressed only 1/3 of 18ksi so it can take only 18/3=6ksi
$$M_a= 6*35.5=213k.in$$
Now we compare the difference adding the steel plates made with the original aluminum section.
$$M_{alumalone}= 12*35.5=426k.in \\ which\ is\ less\ than \ 288.9+213=501.9$$
Edit
After a comment by @Wassabi I corrected the stress of Aluminum and recalculated the beam. The concept is the same but the numerical answer is changed.
So we gain roughly 20% of additional strength. But from here on out the advantage of adding steel plates will increase rapidly by adding to the thickness of the plates.
• -1. This doesn't take into consideration the effect of the difference in elastic moduli between the steel plates and aluminum beam.
– Wasabi
May 12 '19 at 1:23
• @Wasabi, It does. The question says assume Es= 3*Ea May 12 '19 at 1:26
• But I don't see where you take that effect into consideration.
– Wasabi
May 12 '19 at 1:28
• @Wasabi, when I dived the 12k of aluminum by 3 to get 4 k stress. May 12 '19 at 1:30
• Ah, but that's not how it works, I believe. You need elongation continuity at the interface. So you actually need to calculate the steel's stress at the interface (around 17ksi) and divide that by 3. So the aluminum is actually at around 5.7ksi.
– Wasabi
May 12 '19 at 1:40
The big problem here is that the beam and plates are of different materials. More specifically, that they have different moduli of elasticity.
We are accustomed to seeing the following bending stress profile:
This obviously doesn't work for composite sections, though. Since the different parts are meant to work monolithically, the different materials must have the same elongation at the interface. However, since the materials have different elastic moduli (and that $$\sigma = E\epsilon$$), equal elongations mean different stresses at the interface. So we know there'll be a stress discontinuity at the interface.
From the standard bending elongation equation (trivially obtained from the more common bending stress equation)
$$\epsilon = \dfrac{My}{EI}$$
we also know that the slopes of the composite-beam elongation and stress diagrams will be different in each material since they have different elastic moduli.
However, we can't yet determine what the slopes will actually be. It may seem clear that they'll just be $$\dfrac{M}{EI}$$ and $$\dfrac{M}{I}$$, but that won't work: using those equations with each part's $$EI$$ will find the results if it were working alone (well, if the beam were alone or if the plates were isolated from the beam). Instead, we'd need to find how much moment $$M$$ each part will absorb.
Personally, I find that a hassle, though. Why not replace the steel plates with equivalent aluminum plates, finding the new "all-aluminum beam"'s moment of inertia and do all the math assuming a single elastic modulus? That's way easier.1
So let's do that:
From steel shape tables, we know that the moment of inertia of a W8x40 beam is equal to 146 in4. We now need to calculate the change in moment of inertia due to the steel plates.
However, since the beam and plates have different materials, we also need to take their different elastic moduli into account.
This can be done by considering the fact that bending moment doesn't actually care about moment of inertia $$(I)$$, nor about elastic modulus $$(E)$$. Bending moment cares about stiffness ($$EI$$). If two beams have the same $$EI$$, they'll behave the same under bending (assuming elastic behavior), even if one is tiny but made of hard material and the other is large but made of plastic. Same $$EI$$ means same behavior.
We can use this knowledge to our advantage in this case and transform the steel plates into "equivalent aluminum" ones. We do this by setting the steel plates' elastic modulus as equal to aluminum's and then adjusting the plates' moment of inertia so that their final $$EI$$ is unchanged.
The question then becomes how to change the plate's geometry (and therefore their moment of inertia) to satisfy that condition. Thankfully, we know that moment of inertia is directly proportional to width and to the cube of height. Obviously, the easiest dimension to change is therefore the width: if you double the width, you double the moment of inertia; changing the height would require taking cubed-roots, and ain't nobody got time for that.2
Now, the question tells us to assume that
$$\dfrac{E_{steel}}{E_{alum}} = 3$$
Which means that the steel plates are three times stiffer than similarly-sized aluminum plates. So, for aluminum plates to be as stiff as the steel ones, they'd have to be three times wider. After all
\begin{align} (EI)_{plate} &= E_{steel}\cdot I_{plate\ on\ beam} \\ &= E_{steel}\cdot (I_{plate} + A_{plate}d^2) \\ &= 3E_{alum}\cdot \left(\dfrac{bh^3}{12} + bhd^2\right) \\ &= E_{alum} \cdot 3\left(\dfrac{bh^3}{12} + bhd^2\right) \\ &= E_{alum} \cdot \left(\dfrac{(3b)h^3}{12} + (3b)hd^2\right) \end{align}
So if we replace our 8x0.25" steel plates with 24x0.25" aluminum ones, the beam's behavior will be identical (we'll get to the difference in allowable bending stress further ahead).
So now let's calculate this "equivalent beam"'s moment of inertia. Using the parallel axis theorem and being thankful that the symmetric reinforcement won't change the neutral axis, we know it'll be
\begin{align} I_{equiv\ beam} &= I_{beam} + 2\left(I_{equiv\ plate} + A_{equiv\ plate}\cdot d^2\right) \\ &= I_{beam} + 2\left(\dfrac{(3b)h^3}{12} + (3b)h\cdot d^2\right) \\ &= 146 + 2\left(\dfrac{24\cdot0.25^3}{12} + 24\cdot0.25\cdot\left(\dfrac{8.25 + 0.25}{2}\right)^2\right) \\ &= 362.8\text{ in}^4 \end{align}
Now we can use the aluminum's elastic modulus with this equivalent moment of inertia to get the bending stress throughout the beam with the standard equation:
$$\sigma = \dfrac{My}{I_{equiv}}$$
However, there's a hitch: the stress on the plates is actually three times what we calculate with that equation. After all, we're simulating the plates with an area three times greater than it actually is. The actual area of steel will need to absorb that same internal force but with a smaller area and therefore greater stress.
Using this moment of inertia, the allowable stress in the plates is therefore just $$f_{y,steel}/3 = 18/3 = 6\text{ ksi}$$.
The allowable bending moment of the reinforced beam is therefore:
$$M = \dfrac{\sigma\cdot I}{y} = \dfrac{6\cdot 362.8}{8.25/2+0.25} = 497.6\text{ k.in}$$
The original, unreinforced beam, however, could resist
$$M = \dfrac{\sigma\cdot I}{y} = \dfrac{12\cdot 146}{8.25/2} = 424.7\text{ k.in}$$
The reinforcement therefore increased the beam's bearing capacity by ~17%.
1 All I wrote might make it seem complicated, but this method really is simple. It's basically: (1) multiply the plate's width by $$E_{plate}/E_{beam}$$; (2) calculate the equivalent beam's moment of inertia; (3) divide the plate's allowable bending stress by $$E_{plate}/E_{beam}$$; (4) calculate the allowable bending moment. It's effectively the same as the traditional method under the hood, but this one just seems more intuitive for me.
2 Changing the height would also complicate the stress calculation later on, since it depends on the distance to the fiber. Changing the width doesn't affect this at all.
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2021-09-24 18:15:38
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https://mahara.org/interaction/forum/topic.php?id=8524
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# Forums | Mahara Community
## Support / Plugin Competency Moodle-Mahara
Posts: 8
08 October 2019, 8:51 PM
To date, unless I am mistaken, there is no division of powers betweenthe two platforms, the plugins found are Moodle_Assigment_Mahara whichdeals with portfolio ratings to an integrated teacherMoodle or the IENA plugin on Moodle's links to Mahara, for a group of students andits professors are established on the 2 platforms (accounts + homework andactivities on Mahara and Moodle: Reference to IENA plugins created bySoftia company for the University of Lorraine).Thus, the values conveyed by the University of Reims would be to createa skill sharing of students publish in Moodle push towards theMahara platform, and that it be integrated with Mahara's SmartEvidence.At this moment, the University of Reims is planning a plugin workallowing this sharing of skills between the two platforms allowingMoodle pushed the data to Mahara and present them onthe SmartEvidence of Mahara.The position of the University of Reims thereforeensure the passage of competence between the two platforms.For all the requests mentioned, do you have any advice, ideas or inspiration to carry out this project ?
Posts: 8
##### Re: Plugin Competency Moodle-Mahara
09 October 2019, 9:55 PM
Which direction to take for the passage of competence between the 2 platforms ? (skill plugin, LTI skill) ?
Best regard.
Antoine.
Posts: 3866
##### Re: Plugin Competency Moodle-Mahara
10 October 2019, 8:00 AM
Hi Antoine,
At Catalyst we started a project to enable the transfer of competencies from Totara to Mahara using web services. This project is still in progress. You can find the changes to where we have gotten for the Mahara side linked on the wishlist item. You are welcome to take a look at them and see how they can help your case to connect to Moodle competencies.
In our scenario, we are going to set up competencies in Totara and then push them to Mahara for creation of a SmartEvidence framework. In the wider picture, the functionality will allow us to align individual artefacts to a SmartEvidence competency.
Cheers
Kristina
3 results
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2020-02-25 03:32:06
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http://www.cds.caltech.edu/~murray/amwiki/index.php?title=Errata:_In_Theorem_9.3,_N_should_be_Z&diff=cur&oldid=4105
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# Difference between revisions of "Errata: In Theorem 9.3, N should be Z"
Jump to: navigation, search
Location: page 277, line 13
In the displayed equation in Theorem 9.3, should be , the number of zeros:
(Contributed by E. Tuna, 5 Apr 08)
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2020-10-01 02:02:08
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https://docs.mfem.org/html/classmfem_1_1DofTransformation.html
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MFEM v4.4.0 Finite element discretization library
mfem::DofTransformation Class Referenceabstract
#include <doftrans.hpp>
Inheritance diagram for mfem::DofTransformation:
[legend]
Collaboration diagram for mfem::DofTransformation:
[legend]
## Public Member Functions
int Size () const
int Height () const
int NumRows () const
int Width () const
int NumCols () const
void SetFaceOrientations (const Array< int > &face_orientation)
Configure the transformation using face orientations for the current element. More...
const Array< int > & GetFaceOrientations () const
virtual void TransformPrimal (double *v) const =0
virtual void TransformPrimal (Vector &v) const
virtual void TransformPrimalCols (DenseMatrix &V) const
Transform groups of DoFs stored as dense matrices. More...
virtual void InvTransformPrimal (double *v) const =0
virtual void InvTransformPrimal (Vector &v) const
virtual void TransformDual (double *v) const =0
virtual void TransformDual (Vector &v) const
virtual void InvTransformDual (double *v) const =0
virtual void InvTransformDual (Vector &v) const
virtual void TransformDual (DenseMatrix &V) const
virtual void TransformDualRows (DenseMatrix &V) const
Transform groups of dual DoFs stored as dense matrices. More...
virtual void TransformDualCols (DenseMatrix &V) const
virtual ~DofTransformation ()
## Protected Member Functions
DofTransformation (int size)
int size_
Array< int > Fo
## Detailed Description
The DofTransformation class is an abstract base class for a family of transformations that map local degrees of freedom (DoFs), contained within individual elements, to global degrees of freedom, stored within GridFunction objects. These transformations are necessary to ensure that basis functions in neighboring elements align correctly. Closely related but complementary transformations are required for the entries stored in LinearForm and BilinearForm objects. The DofTransformation class is designed to apply the action of both of these types of DoF transformations.
Let the "primal transformation" be given by the operator T. This means that given a local element vector v the data that must be placed into a GridFunction object is v_t = T * v.
We also need the inverse of the primal transformation T^{-1} so that we can recover the local element vector from data read out of a GridFunction e.g. v = T^{-1} * v_t.
We need to preserve the action of our linear forms applied to primal vectors. In other words, if f is the local vector computed by a linear form then f * v = f_t * v_t (where "*" represents an inner product of vectors). This requires that f_t = T^{-T} * f i.e. the "dual transform" is given by the transpose of the inverse of the primal transformation.
For bilinear forms we require that v^T * A * v = v_t^T * A_t * v_t. This implies that A_t = T^{-T} * A * T^{-1}. This can be accomplished by performing dual transformations of the rows and columns of the matrix A.
For discrete linear operators the range must be modified with the primal transformation rather than the dual transformation because the result is a primal vector rather than a dual vector. This leads to the transformation D_t = T * D * T^{-1}. This can be accomplished by using a primal transformation on the columns of D and a dual transformation on its rows.
Definition at line 56 of file doftrans.hpp.
## Constructor & Destructor Documentation
mfem::DofTransformation::DofTransformation ( int size )
inlineprotected
Definition at line 63 of file doftrans.hpp.
virtual mfem::DofTransformation::~DofTransformation ( )
inlinevirtual
Definition at line 117 of file doftrans.hpp.
## Member Function Documentation
const Array& mfem::DofTransformation::GetFaceOrientations ( ) const
inline
Definition at line 80 of file doftrans.hpp.
int mfem::DofTransformation::Height ( ) const
inline
Definition at line 69 of file doftrans.hpp.
virtual void mfem::DofTransformation::InvTransformDual ( double * v ) const
pure virtual
Inverse Transform dual DoFs
void mfem::DofTransformation::InvTransformDual ( Vector & v ) const
virtual
Definition at line 88 of file doftrans.cpp.
virtual void mfem::DofTransformation::InvTransformPrimal ( double * v ) const
pure virtual
Inverse transform local DoFs. Used to transform DoFs from a global vector back to their element-local form. For example, this must be used to transform the vector obtained using GridFunction::GetSubVector before it can be used to compute a local interpolation.
void mfem::DofTransformation::InvTransformPrimal ( Vector & v ) const
virtual
Definition at line 60 of file doftrans.cpp.
int mfem::DofTransformation::NumCols ( ) const
inline
Definition at line 72 of file doftrans.hpp.
int mfem::DofTransformation::NumRows ( ) const
inline
Definition at line 70 of file doftrans.hpp.
void mfem::DofTransformation::SetFaceOrientations ( const Array< int > & face_orientation )
inline
Configure the transformation using face orientations for the current element.
The face_orientation array can be obtained from Mesh::GetElementFaces.
Definition at line 77 of file doftrans.hpp.
int mfem::DofTransformation::Size ( ) const
inline
Definition at line 68 of file doftrans.hpp.
virtual void mfem::DofTransformation::TransformDual ( double * v ) const
pure virtual
Transform dual DoFs as computed by a LinearFormIntegrator before summing into a LinearForm object.
void mfem::DofTransformation::TransformDual ( Vector & v ) const
virtual
Definition at line 30 of file doftrans.cpp.
void mfem::DofTransformation::TransformDual ( DenseMatrix & V ) const
virtual
Transform a matrix of dual DoFs entries as computed by a BilinearFormIntegrator before summing into a BilinearForm object.
Definition at line 35 of file doftrans.cpp.
void mfem::DofTransformation::TransformDualCols ( DenseMatrix & V ) const
virtual
Definition at line 52 of file doftrans.cpp.
void mfem::DofTransformation::TransformDualRows ( DenseMatrix & V ) const
virtual
Transform groups of dual DoFs stored as dense matrices.
Definition at line 41 of file doftrans.cpp.
virtual void mfem::DofTransformation::TransformPrimal ( double * v ) const
pure virtual
Transform local DoFs to align with the global DoFs. For example, this transformation can be used to map the local vector computed by FiniteElement::Project() to the transformed vector stored within a GridFunction object.
void mfem::DofTransformation::TransformPrimal ( Vector & v ) const
virtual
Definition at line 17 of file doftrans.cpp.
void mfem::DofTransformation::TransformPrimalCols ( DenseMatrix & V ) const
virtual
Transform groups of DoFs stored as dense matrices.
Definition at line 22 of file doftrans.cpp.
int mfem::DofTransformation::Width ( ) const
inline
Definition at line 71 of file doftrans.hpp.
## Member Data Documentation
Array mfem::DofTransformation::Fo
protected
Definition at line 61 of file doftrans.hpp.
int mfem::DofTransformation::size_
protected
Definition at line 59 of file doftrans.hpp.
The documentation for this class was generated from the following files:
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2022-07-05 22:55:19
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https://www.physicsforums.com/threads/the-solar-systems-and-space-probes.87548/
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# The solar systems and space probes
1. Sep 5, 2005
### vidmar
OK the title is a bit practical, I've made the question somewhat more theoretical (technical, whatever), in particular:
One is given three bodies: two relatievly massive (for example the Sun and a massive planet like Jupiter) and one relatievly small (for example a satellite) - meaning that its gravitational effect on the other two may be neglected or is "turned off" (depending on how you wish to formulate the problem). The two "massive" bodies, denote them by M1 and M2, are in a closed orbit around their centre of mass (CM). Does there exist a set of initial conditions (fot time t0) involving (velocities of M1 and M2, position of satellite relative to the centre of mass) which will "thrust" the satellite ad infinitum (i.e. for every distance from the centre of mass of the system (CMS), there exists a time t from t0, for which the satellite achieves this distance), provided that it does not move relative to the CMS at the biginning (t0). Basically what I am asking is, can I place the satellite in the gravitational field of two objects and thrust it ad infinitum with no beginnig kinetic energy relative to the CMS taking advantege only of the gravitational pulls of the objects (with a few limitations of course:) Naturally I would like only Newton to "bear on this problem" (general relativity would be slightly to much.
Also I've come across an interesting fact that given 5 objects (gravitation, Newton) they can go to infinity in finite time! If anyone has anything specific on that, feel free to post.
2. Sep 10, 2005
### CarlB
Yes.
let's suppose that the orbital system, like the sun and jupiter, has two stages of massiveness. There is a very massive sun in the center which we will assume is stationary. And there is a massive planet which orbits the sun in a circular manner. The speed of the planet in its orbit is constant and that speed is determined by its distance to the sun and the sun's mass as follows:
$$F/m = a = GM/r^2 = v^2/r$$
or
$$v = \sqrt{GM/r}$$
Now the slingshot effect can give a velocity to the test mass of 2v, (a result you can obtain by considering the problem as a scattering in the rest frame of the massive particle and then translating back to the rest frame of the sun) so its kinetic energy afterwards will be:
$$KE = 0.5mv^2 = 0.5 \times 4 GmM/r = 2 GmM/r$$.
This is twice the object's gravitational potential energy, so it is the case that the test object has achieved escape velocity.
Carl
Last edited: Sep 10, 2005
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2017-02-26 03:50:26
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https://www.physicsforums.com/threads/algebra-problem-x-1-x-problem.872282/
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# Algebra problem (x+(1/x) problem)
1. May 18, 2016
### terryds
1. The problem statement, all variables and given/known data
If $a+\frac{1}{a}=\sqrt{3}$, then $a^{2016} + (\frac{1}{a})^{2016}$ equals
A. 0
B. 1
C. √2
D. √3
E. 2
2. Relevant equations
3. The attempt at a solution
I find no real solution for a.
What I can do is just find a^2 + (1/a)^2 which equals 1, a^4 + (1/a)^4 = -1, a^8 + (1/a)^8 = -1, and so on...
Since 2016 is not a square number, I don't know how to determine the value
Is there any theory about problem like this?
Please help
2. May 18, 2016
### Samy_A
Start from $a²+1/{a²}=1$.
Multiply by $a²$ to get rid of the denominator, you get $a^4 +1 =a²$.
Now notice that 2016 is a multiple of 6. So multiply $a^4 +1 =a²$ again by $a²$ to get $a^6$ in the equation.
3. May 18, 2016
### ehild
Find the complex solution. Write it in trigonometric of exponential form, which makes it easy to rise to power 2016. (2016=25*32 *7)
4. May 18, 2016
### terryds
a^6 + a^2 = a^4
a^6 = a^4 - a^2 = (a^2 - 1) - a^2 = -1
a^12 = 1
a^24 = 1
...
a^2016 = 1
So, 1/a^2016 = 1
a^2016 + (1/a)^2016 = 1 + 1 = 2
Thank you
5. May 19, 2016
### terryds
I'm interested to try this approach..
The solution for a is $\frac{\sqrt{3}\pm i}{2}$
Should I use the plus or the minus as the root?
If it's plus, $e^{i\frac{1}{6}\pi}$
If it's minus, $e^{i\frac{11}{6}\pi}$
And, then what should I do??
6. May 19, 2016
### ehild
What are their 12th power?
7. May 19, 2016
### terryds
The 'plus' root^12 = $e^{i2\pi}= 1$
The 'minus' root^12 = $e^{i22\pi} = 1$
Okay, since 2016 can be divided by 12, a^2016 = 1
And 1/a^2016 = 1
So, a^2016 + 1/a^2016 = 1+1 = 2
Thanks a lot!
8. May 19, 2016
### ehild
The 'minus' root^12 = $e^{-i2\pi} = 1$
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2018-02-26 02:03:24
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https://ec.gateoverflow.in/392/gate2014-1-17?show=1553
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A discrete-time signal $x[n] = \sin(\pi^{2}n),n$ being an integer, is
1. periodic with period $\pi$
2. periodic with period $\pi^{2}$
3. periodic with period $\pi/2$
4. not periodic
in Others
retagged
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2020-10-21 13:25:11
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https://www.isnphard.com/i/spiral-galaxies/
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# Spiral Galaxies
A puzzle in which the player tiles a grid with polyominos with 180° rotational symmetry about given centers.
## Description
Spiral Galaxies is a puzzle played on a $n \times m$ grid of squares containing a collection $C$ of center points (represented as dots). Center points can appear both in the center of a grid square, or in the center of an edge connecting two neighboring squares. The goal is to find a tiling of the grid with polyominos such that each polyomino (also called a galaxy) contains exactly one center $c \in C$ and is 180° rotationally symmetric about $c$.
The figure shows an instance of Spiral Galaxies (left) and a possible solution (right) in which the galaxies have been highlighted.
## Computational complexity
The problem of deciding whether an instance of Spiral Galaxies admits a solution is NP-Complete [1].
A solution can be found in $\frac{4^{nm}}{2^{n+m}} \, \textrm{poly}(nm)$ time [2] and by an FPT algorithm parameterized in the number of corners of a solution (a corner is an internal vertex of the grid such that $1$, $3$, or $4$ of the $4$ neighboring squares belong to the same galaxy) [2].
A solution for the variant in which all galaxies need to be rectangular can be found in $|C|! \, \textrm{poly}(|C| \log nm)$ time [2].
## Notes
There exist instances of Spiral Galaxies with $|C| = O(1)$ and an arbitrarily large number of corners [2].
[3] shows 36 Spiral Galaxies instances whose unique solutions form the 10 Arabic numerals and the 26 uppercase letters of the ISO basic Latin alphabet. A corresponding interactive tool is available here.
## References
[1] E. Friedman, “Spiral Galaxies Puzzles are NP-complete”.
[2] G. Fertin, S. Jamshidi, C. Komusiewicz, “Towards an Algorithmic Guide to Spiral Galaxies” in FUN 2014.
[3] W. Anderson, E. D. Demaine, M. L. Demaine, “Spiral Galaxies Font”, The Mathematics of Various Entertaining Subjects Volume 3: The Magic of Mathematics.
@misc{cog:spiral-galaxies,
author = "{CoG contributors}",
title = "{Spiral Galaxies --- Complexity of Games}",
year = "2022",
url = "https://www.isnphard.com/i/spiral-galaxies/",
note = "[Online; accessed 2022-06-02]"
}
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2022-06-24 22:32:35
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https://trac-hacks.org/ticket/4375
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Opened 9 years ago
Closed 9 years ago
display hours in the milestone view
Reported by: Owned by: Jeff Hammel Jeff Hammel normal TracHoursPlugin normal julien.perville@…, fabien.catteau@… 0.11
Description
Hello!
On the roadmap (/roadmap) screen there is a line for each milestone with the total estimated and total worked hours. It would be nice if we could display the same information in the milestone view (/milestone/\$milestone_name). This could be done by refactoring the existing MilestoneMarkup class in source:trachoursplugin/0.11/trachours/hours.py to be available outside of the filter_roadmap function.
Sincerely, Julien Pervillé
comment:1 Changed 9 years ago by Jeff Hammel
CCing original author, julien.perville@...
comment:2 Changed 9 years ago by Jeff Hammel
Resolution: → fixed new → closed
this should be done; i forgot about this ticket and reticketed at #4419 which i closed
Modify Ticket
Change Properties
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2017-07-21 05:36:19
|
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https://acarril.github.io/posts/atom-latex
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# Atom + LaTeX
It’s been a while since I updated my series on Atom + <something>, but here comes a good one: how to set up Atom as a LaTeX editor. When I first migrated to Atom, I found that LaTeX support was a bit subpar, at least when compared to Sublime’s. However, I’ve tried lots of packages and configurations, and I believe the combination described below makes Atom a solid LaTeX editor.
# Atom
I’m assuming you already have Atom up and running. If you haven’t, go to https://atom.io/ to download and install it.
Atom, like other text editors, works in such a way that it provides a lean, solid foundation to build on top of. This means that it doesn’t come with LaTeX support out-of-the-box, which has to be added by us, in the form of packages. Each package (typically) adds one additional feature: syntax highlighting, compiling, PDF preview, etc. The good news is that it is very easy to add these, and while there are several options for each feature, I’ve tested (almost?) all of them.
### Installing a package
In what follows we will install several packages. To find and download a package within Atom, go to the Settings View with Ctrl+Comma (or under Edit > Preferences). Look for the Install tab in the left, and in there just search for the package in question (wording has to be exact to get the intended package as first result).
# Syntax highlighting
The first thing we’ll want to do is to have syntax highlighting for .tex files. If you open a LaTeX document right now, it should look like plain text, ie. the left panel in the image below. We want to have something like the right panel.
We’ll use language-latex, which works great out of the box. To install it, open the Settings View with Ctrl+Comma and in the Install tab look for language-latex. It should start working right away. If it doesn’t, double check that the document you have open is recognized as a LaTeX file (it should say ‘Latex’ on the bottom right corner). You can also explicitly set a language for any open document with Ctrl+Shift+l (or Cmd+Shift+l on a Mac).
Writing LaTeX is slightly different from normal coding, so for this specific language I like to turn on soft wrapping. To do so, just go to the Settings View with Ctrl+Comma and in the ‘Packages’ tab look for our newly installed package, language-latex. Open its settings and make sure the ‘Soft Wrap’ option is checked.
### Soft wrap and other tweaks
WTF is soft wrap?, you may ask. Soft wrap basically breaks very long lines into multiple ones, without actually inserting a real line break. Since in LaTeX you’re basically writing text, lines tend to be longer than your average Python statement, so if you want to avoid vertical scrolling this is what’s best. A stolen GIF is worth a thousand images:
I also like to be able to scroll past the end of the document, so I’m not permanently focusing on the very bottom of the screen. This feature can be activated for LaTeX in the language-latex configuration page, but I actually like this feature for all my documents, so if you also prefer that, you have to check the option in the in the ‘Editor’ tab of the general Settings View.
# Compile
So here is where you have most options, but after some testing I’ve decided that the latex package is what works best (for me). After installing, it should work out of the box if you’ve installed TeXLive (or MacTeX in Mac) or MiKTeX (although I tend to avoid MiKTeX, and it is also less tested with this package).
To test it out, you can create a new file like test.tex with something like
\documentclass{article}
\begin{document}
Hello world!
\end{document}
The easiest way to compile (build) is by using Ctrl+Shift+B. If no errors are found, it should automatically open the resulting PDF in your system’s default PDF viewer (more on that below). We will also tweak some settings after installing pdf-view (below).
# PDF preview
After a successful build, the latex package should automatically invoke your default PDF viewer for preview in a separate window. In some situations this is fine, but when I want a preview I usually prefer it side by side inside the editor itself. This behavior can be easily achieved by installing the pdf-view package.
If it doesn’t work for you after installing, go to latex package settings and be sure to select pdf-view as Opener. You can find these settings by opening the ‘Settings View’ with Ctrl+Comma and then selecting the ‘Packages’ tab on the left. Look for the latex package and click on the ‘Settings’ button. The ‘Opener’ option is near the end.
Now that you have the latex settings open, I would also recommend to check the Enable SyncTeX option. Take this opportunity to further customize the settings. For instance, you can build on each save, change the default logging levels (I would advice against changing the default though) and enable shell escape, if you need to do so.
## Spell check
Finally, since my native language is not english, I usually prefer to write documents with a spell checker. Unfortunately, spell check is one of the most poorly developed LaTeX tools in Atom. For the time being I went with the basic spell-check, which comes by default with Atom. Make sure the package is installed (it should) and enabled. Then go to the package’s settings and add text.tex.latex at the end of the list of Grammars, like in the bottom right of the screenshot below.
After doing this, you should get spell checking in your LaTeX document. However, something that annoyed me is that you also get corrections for obviously-LaTeXy commands, like \documentclass. I discovered you can exclude certain parts of the document with the ‘Exlude Scopes’ option, and I added these scopes to that list. You can add each scope in spell-check settings as comma separated names, or you can paste the following code directly in your config.cson file (Edit > Config...):
"spell-check":
excludedScopes: [
"support.function.tex"
"meta.preamble.latex"
"support.type.function.latex"
"comment.line.percentage.tex"
"storage.type.function.latex"
"support.function.latex"
"string.other.math.tex"
"string.other.math.block.environment.latex"
"variable.parameter.function.latex"
"constant.other.reference.latex"
]
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2021-10-18 23:51:47
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https://web2.0calc.com/questions/problem-on-conic-sections-and-equilateral-triangles-please-help
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+0
0
142
6
+304
An equilateral triangle ABC is inscribed in the ellipse x^2/a^2+y^2/b^2=1, so that B is at (0,b), and line AC is parallel to the x-axis, as shown below. Also, foci F_1 and F_2 lie on sides line BC and line AB, respectively. Determine the value of AB/(F_1*F_2).
Aug 28, 2020
#1
0
The answer works out to 13/8.
Aug 28, 2020
#2
+304
0
I would appreciate if you gave a solution following your statement. Thanks in advance.
Aug 28, 2020
#3
+112002
+1
what do you mean by F_1*F_2
they are points, not distances so it doesn't make sense to me.......
Aug 29, 2020
#4
+304
+1
I apologize for the misunderstanding, as I meant the length between the two points F_2 and F_1.
DragonLord Aug 29, 2020
#5
+112002
+3
Since it is a ratio question you can use a convenient ellipse.
Let X be the centre and M be the midpoint of AC
triangle BXF2 and triangle BMA are similar tiranges and they are both 30, 60 90 degree triangles.
So their sides are in the ration 1:2:sqrt3
I want the focal length (c) to be 1 and
I want B to be $$(0,\sqrt3)$$
$$c^2=a^2-b^2\\ 1=a^2-3\\ 4=a^2\\ a=2$$
So the ellipse that I will use is
$$\frac{x^2}{4}+\frac{y^2}{3}=1\\ \frac{y^2}{3}=1-\frac{x^2}{4}$$
Equation of line AB
$$y=\sqrt3\;x+\sqrt3\\ y=\sqrt3(x+1)\\ y^2=3(x+1)^2\\ \frac{y^2}{3}=(x+1)^2\\$$
Solve simultaneously
$$1-\frac{x^2}{4}=(x+1)^2\\ -\frac{x^2}{4}=x^2+2x\\ -x^2=4x^2+8x\\ 5x^2+8x=0\\ x(5x+8)=0\\ x=0\quad or \quad x=-\frac{8}{5}$$
So the x value for A is -8/3
$$AC=AB=BC=\frac{16}{5}\\ F_2F_1=2\\ \frac{AB}{F_1F_2}=\frac{16}{10}=\frac{8}{5}$$
Aug 30, 2020
#6
+304
0
Ohh, I see! Thank you Melody!
Sep 5, 2020
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2021-01-18 01:47:24
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https://chemistry.stackexchange.com/questions/76908/what-is-the-hybridization-of-chlorine-in-vinyl-chloride/76954
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# What is the hybridization of chlorine in vinyl chloride?
Is chlorine in vinyl chloride sp2 or sp3 hybridized?
Calculating using steric number it is found to be sp3 hybridized. But according to I. L. Finar* it is sp2.
*I.L. Finar: Organic chemistry Vol.1 Fundamental principles. Sixth Edition. Orient Longmans: 1973. Page 328.
• Let's put it this way: what observable difference would that make? – Ivan Neretin Jun 27 '17 at 16:07
• Terminal, electronegative atoms are in 99% of the cases sp hybridised. (There are no bunny ears.) – Martin - マーチン Jun 28 '17 at 7:10
Disclaimer: An important point to remember is, that hybridisation always follows the molecular structure, it is never the cause for a certain structure. As such, hybridisation is an interpretation tool, nothing more.
As a rule of thumb, terminal atoms (excluding hydrogen) are almost always (maximally) approximately sp hybridised. While other hybridisation schemes can be applied, they are usually not a good representation.*
One popular example is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially makes the lone pairs equivalent. That view, however, is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. At this point I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124.).
The same principle described there applies to terminal atoms. Because of the local $C_\mathrm{\infty}$ symmetry the notion of three equivalent lone pairs is likely to be false. The most likely (main resonance contributor) structure for these atoms are (if hybridisation is at all feasible, i.e. mostly in the second period) one sp orbital forming the bond, one sp lone pair, two (perpendicular) p lone pairs.
In this particular example, I have calculated the molecule on the DF-BP86/def2-SVP level of theory and analysed it with the natural bond orbital theory (NBO6). Below you find the localised orbitals which best fit with the common Lewis structure of the molecule.
I have ordered the orbitals starting with the carbon-carbon σ- and carbon-chlorine σ-bonds on the bottom. Following that is the carbon-carbon π-bond. Continuing with the three carbon-hydrogen σ-bonds, and finally ending with the three chlorine lone pairs. The occupied orbitals are in blue and orange, while the corresponding virtual orbitals are next to it in red and yellow. The detailed analysis in numbers is at the end of the post.
In ast's answer there is the claim that chlorine must be sp2 hybridised because of a resonance contributor. While this explanation is extremely tempting because it easy to understand, you cannot judge from the possibility of resonance to the electronic structure. As another general rule of thumb you can remember: the lesser the orbitals are hybridised, the more likely the structure is.
The calculation also provides an analysis in terms of natural resonance theory. On the DF-BP86/def2-SVP level the contribution of the second configuration to the total electronic structure is only 6%. The main configuration contributes with about 90%, other configurations (ionic) make up the remaining 4%.
In summary, I'm afraid, but your book is incorrect, or at least incomplete. The description with two equivalent sp2 lone pairs is unnecessary complicated and does not accurately reproduce the electronic structure.
### Truncated results of the NBO analysis (skipping core orbitals)
(Occupancy) Bond orbital / Hybrids
------------------ Lewis ------------------------------------------------------
8. (1.99378) LP ( 1)Cl 6 s( 81.55%)p 0.23( 18.44%)d 0.00( 0.01%)
9. (1.97199) LP ( 2)Cl 6 s( 0.18%)p99.99( 99.79%)d 0.18( 0.03%)
10. (1.90237) LP ( 3)Cl 6 s( 0.00%)p 1.00( 99.95%)d 0.00( 0.05%)
11. (1.99642) BD ( 1) C 1- C 2
( 48.61%) 0.6972* C 1 s( 0.00%)p 1.00( 99.94%)d 0.00( 0.06%)
( 51.39%) 0.7169* C 2 s( 0.00%)p 1.00( 99.95%)d 0.00( 0.05%)
12. (1.99535) BD ( 2) C 1- C 2
( 49.24%) 0.7017* C 1 s( 40.16%)p 1.49( 59.76%)d 0.00( 0.08%)
( 50.76%) 0.7124* C 2 s( 44.47%)p 1.25( 55.47%)d 0.00( 0.06%)
13. (1.96941) BD ( 1) C 1- H 3
( 61.15%) 0.7820* C 1 s( 29.81%)p 2.35( 70.15%)d 0.00( 0.04%)
( 38.85%) 0.6233* H 3 s( 99.91%)p 0.00( 0.09%)
14. (1.98105) BD ( 1) C 1- H 4
( 61.33%) 0.7832* C 1 s( 30.09%)p 2.32( 69.87%)d 0.00( 0.04%)
( 38.67%) 0.6218* H 4 s( 99.91%)p 0.00( 0.09%)
15. (1.98492) BD ( 1) C 2- H 5
( 62.36%) 0.7897* C 2 s( 32.79%)p 2.05( 67.18%)d 0.00( 0.03%)
( 37.64%) 0.6135* H 5 s( 99.89%)p 0.00( 0.11%)
16. (1.99056) BD ( 1) C 2-Cl 6
( 43.14%) 0.6568* C 2 s( 22.75%)p 3.39( 77.00%)d 0.01( 0.25%)
( 56.86%) 0.7541*Cl 6 s( 18.31%)p 4.43( 81.12%)d 0.03( 0.57%)
---------------- non-Lewis ----------------------------------------------------
17. (0.09516) BD*( 1) C 1- C 2
( 51.39%) 0.7169* C 1 s( 0.00%)p 1.00( 99.94%)d 0.00( 0.06%)
( 48.61%) -0.6972* C 2 s( 0.00%)p 1.00( 99.95%)d 0.00( 0.05%)
18. (0.01232) BD*( 2) C 1- C 2
( 50.76%) 0.7124* C 1 s( 40.16%)p 1.49( 59.76%)d 0.00( 0.08%)
( 49.24%) -0.7017* C 2 s( 44.47%)p 1.25( 55.47%)d 0.00( 0.06%)
19. (0.00909) BD*( 1) C 1- H 3
( 38.85%) 0.6233* C 1 s( 29.81%)p 2.35( 70.15%)d 0.00( 0.04%)
( 61.15%) -0.7820* H 3 s( 99.91%)p 0.00( 0.09%)
20. (0.01032) BD*( 1) C 1- H 4
( 38.67%) 0.6218* C 1 s( 30.09%)p 2.32( 69.87%)d 0.00( 0.04%)
( 61.33%) -0.7832* H 4 s( 99.91%)p 0.00( 0.09%)
21. (0.02582) BD*( 1) C 2- H 5
( 37.64%) 0.6135* C 2 s( 32.79%)p 2.05( 67.18%)d 0.00( 0.03%)
( 62.36%) -0.7897* H 5 s( 99.89%)p 0.00( 0.11%)
22. (0.03245) BD*( 1) C 2-Cl 6
( 56.86%) 0.7541* C 2 s( 22.75%)p 3.39( 77.00%)d 0.01( 0.25%)
( 43.14%) -0.6568*Cl 6 s( 18.31%)p 4.43( 81.12%)d 0.03( 0.57%)
Summary of Natural Population Analysis
Natural Population
Natural ---------------------------------------------
Atom No Charge Core Valence Rydberg Total
--------------------------------------------------------------------
C 1 -0.43992 1.99995 4.42868 0.01129 6.43992
C 2 -0.22801 1.99996 4.20978 0.01827 6.22801
H 3 0.22758 0.00000 0.76996 0.00246 0.77242
H 4 0.22498 0.00000 0.77165 0.00337 0.77502
H 5 0.23463 0.00000 0.76237 0.00300 0.76537
Cl 6 -0.01926 9.99998 7.00353 0.01575 17.01926
====================================================================
* Total * 0.00000 13.99989 17.94598 0.05414 32.00000
Optimised Geometry DF-BP86/def2-SVP
6
symmetry cs ( E = -538.027155015 A.U. )
C 1.309538928 1.048660302 0.000000000
C 0.000000000 0.759346920 0.000000000
H 1.630332304 2.102388803 0.000000000
H 2.083175437 0.265507033 0.000000000
H -0.793639761 1.523771533 0.000000000
Cl -0.633947150 -0.867041806 0.000000000
* It is also important to understand, that atoms are never hybridised, only orbitals are. The presence of an sp hybrid orbital does not exclude the presence of a sp3 hybrid orbital.
• Would u elaborate on what's on the second column of the picture you added? – Mockingbird Jun 28 '17 at 12:09
• @Mockingbird These are the corresponding (anti-bonding) virtual orbitals that belong to the bonding orbitals. Note that they all have a nodal plane perpendicular to the bonding axis. – Martin - マーチン Jun 28 '17 at 12:10
• Why do lone pairs don't have corresponding antibonding orbital? Or, you just didn't mention? – Mockingbird Jun 28 '17 at 12:19
• @Mockingbird They don't form a bond. There is only one combination, that's why lone pairs are usually considered as non-bonding. – Martin - マーチン Jun 28 '17 at 12:21
• But is it possible to rationalise this using some simply chemistry explanation, instead of just basing this theoretically on a computational calculation? For example, I was thinking, it is likely terminal atoms are sp-hybridised because they often only make one sigma bond to the central atom and so to maximise the strength of this bond, it would be reasonable that the terminal atom uses an sp-hybrid for bonding since the sp-hybrid orbital is directional and has high s character. Would that be a reasonable simple chemistry rationalisation of your computational result? – Tan Yong Boon Jul 5 '18 at 10:47
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2020-04-02 10:51:58
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https://www.fiberoptics4sale.com/blogs/wave-optics/population-inversion-and-optical-gain
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# Population Inversion and Optical Gain
This is a continuation from the previous tutorial - optical absorption and amplification.
From the discussions in the optical absorption and amplification tutorial, it is clear that population inversion is the basic condition for the presence of an optical gain.
In the normal state of any system in thermal equilibrium, a low-energy state is always more populated than a high-energy state, hence no population inversion.
Population inversion in a system can only be accomplished through a process called pumping by actively exciting the atoms in a low-energy state to a high-energy state.
If left alone, the atoms in a system will relax to thermal equilibrium. Therefore, population inversion is a nonequilibrium state that cannot be sustained without active pumping.
To maintain a constant optical gain, continuous pumping is required to keep the population inversion at a constant level.
This condition is clearly consistent with the law of conservation of energy: amplification of an optical wave leads to an increase in optical energy, which is possible only if there is a source supplying the energy.
Pumping is the process that supplies the energy to the gain medium for the amplification of an optical wave. There are many different pumping techniques, including optical excitation, electric current injection, electric discharge, chemical reaction, and excitation with particle beams.
The use of a particular pumping technique depends on the properties of the gain medium being pumped. The lasers and optical amplifiers of particular interest in photonic systems are made of either dielectric solid-state media doped with active ions, such as Nd : YAG and Er : glass fiber, or direct-gap semiconductors, such as GaAs and InP.
For dielectric media, the most commonly used pumping technique is optical pumping either with incoherent light sources, such as flashlamps and light-emitting diodes, or with coherent light sources from other lasers.
Semiconductor gain media can also be optically pumped, but they are usually pumped with electric current injection.
In this tutorial, we consider the general conditions for pumping to achieve population inversion for an optical gain. Detailed pumping mechanisms and physical setups are not addressed here because they depend on the specific gain medium used in a given application.
Rate Equations
The net rate of increase of population density in a given energy level is described by a rate equation. As we shall see below, pumping for population inversion in any practical gain medium always requires the participation of more than two energy levels.
In general, a rate equation has to be written for each energy level that is involved in the process. For simplicity but without loss of validity, however, we shall explicitly write down only the rate equations for the two energy levels, $$|2\rangle$$ and $$|1\rangle$$, that are directly associated with the resonant transition of interest.
We are not interested in the population densities of other energy levels but only in how those levels affect $$N_2$$ and $$N_1$$.
In the presence of a monochromatic, coherent optical wave of intensity $$I$$ at a frequency $$\nu$$, the rate equations for $$N_2$$ and $$N_1$$ are
$\tag{10-61}\frac{\text{d}N_2}{\text{d}t}=R_2-\frac{N_2}{\tau_2}-\frac{I}{h\nu}(N_2\sigma_\text{e}-N_1\sigma_\text{a})$
$\tag{10-62}\frac{\text{d}N_1}{\text{d}t}=R_1-\frac{N_1}{\tau_1}+\frac{N_2}{\tau_{21}}+\frac{I}{h\nu}(N_2\sigma_\text{e}-N_1\sigma_\text{a})$
where $$R_2$$ and $$R_1$$ are the total rates of pumping into energy levels $$|2\rangle$$ and $$|1\rangle$$, respectively, and $$\tau_2$$ and $$\tau_1$$ are the fluorescence lifetimes of levels $$|2\rangle$$ and $$|1\rangle$$, respectively.
The rate of population decay, including radiative and nonradiative spontaneous relaxation, from level $$|2\rangle$$ to level $$|1\rangle$$ is $$\tau_{21}^{-1}$$.
Because it is possible for the population in level $$|2\rangle$$ to relax to other energy levels also, the total population decay rate of level $$|2\rangle$$ is $$\tau_2^{-1}\ge\tau_{21}^{-1}$$.
Therefore, in general, we have
$\tag{10-63}\tau_2\le\tau_{21}\le\tau_\text{sp}$
Note that $$\tau_{21}^{-1}$$ is not the same as $$\gamma_{21}$$ defined in (10-8) [refer to the optical transitions for laser amplifiers tutorial]: $$\tau_{21}^{-1}$$ is purely the rate of population relaxation from level $$|2\rangle$$ to level $$|1\rangle$$, whereas $$\gamma_{21}$$ is the rate of phase relaxation of the polarization associated with the transition between these two levels.
In an optical gain medium, level $$|2\rangle$$ is known as the upper laser level and level $$|1\rangle$$ is known as the lower laser level.
The fluorescence lifetime $$\tau_2$$ of the upper laser level is an important parameter that determines the effectiveness of a gain medium. Generally speaking, the upper laser level has to be a metastable state with a relatively large $$\tau_2$$ for a gain medium to be useful.
Population Inversion
Population inversion in a medium is generally defined as
$\tag{10-64}N_2\gt\frac{g_2}{g_1}N_1$
According to (10-50) [refer to the optical absorption and amplification tutorial], however, this condition does not guarantee an optical gain at a particular optical frequency $$\nu$$ if $$\sigma_\text{a}(\nu)\ne(g_2/g_1)\sigma_\text{e}(\nu)$$ when the population in each level, $$|1\rangle$$ or $$|2\rangle$$, is distributed unevenly among its sublevels.
For this reason, when the condition for population inversion given in (10-64) is achieved in a medium, we may find an optical gain at an optical frequency $$\nu$$ where $$\sigma_\text{a}(\nu)\le(g_2/g_1)\sigma_\text{e}(\nu)$$, but at the same time find an optical loss at another frequency $$\nu'$$ where $$\sigma_\text{a}(\nu')\gt(g_2/g_1)\sigma_\text{e}(\nu')$$.
What really matters for an optical wave at a given frequency is the optical gain at that particular frequency.
Therefore, in the following discussions, we shall consider, instead of the condition in (10-64), the following condition:
$\tag{10-65}N_2\sigma_\text{e}(\nu)-N_1\sigma_\text{a}(\nu)\gt0$
which guarantees an optical gain at frequency $$\nu$$, as the effective condition of population inversion as far as an optical signal at frequency $$\nu$$ is concerned.
The pumping requirement for the condition in (10-65) to be satisfied depends on the properties of a medium.
For atomic and molecular media, there are three different basic systems. Each has a different pumping requirement to reach effective population inversion for an optical gain. The pumping requirement can be found by solving the coupled rate equations in (10-61) and (10-62).
Two-level system
When the only energy levels involved in the pumping and the relaxation processes are the upper and the lower laser levels $$|2\rangle$$ and $$|1\rangle$$, the system can be considered as a two-level system.
In such a system, level $$|1\rangle$$ is the ground state with $$\tau_1=\infty$$, and level $$|2\rangle$$ relaxes only to level $$|1\rangle$$ so that $$\tau_{21}=\tau_2$$. The total population density is $$N_\text{t}=N_1+N_2$$.
While a pumping mechanism excites atoms from the lower laser level to the upper laser level, the same pump also stimulates atoms in the upper laser level to relax to the lower laser level.
Therefore, irrespective of the specific pumping technique used, $$R_2=-R_1=W_{12}^\text{p}N_1-W_{21}^\text{p}N_2$$, where $$W_{12}^\text{p}$$ and $$W_{21}^\text{p}$$ are the pumping transition probability rates, or simply the pumping rates, from $$|1\rangle$$ to $$|2\rangle$$ and from $$|2\rangle$$ to $$|1\rangle$$, respectively.
Under these conditions, (10-61) and (10-62) are equivalent to each other. The upward and downward pumping transition rates are not independent of each other but are directly proportional to each other because both are associated with the interaction of the same pump source with a given set of energy levels.
We take the upward pumping rate to be $$W_{12}^\text{p}=W_\text{p}$$ and the downward pumping rate to be $$W_{21}^\text{p}=pW_\text{p}$$, where $$p$$ is a constant that depends on the detailed characteristics of the two-level atomic system and the pump source.
In the steady state when $$\text{d}N_2/\text{d}t=\text{d}N_1/\text{d}t=0$$, we then find that
$\tag{10-66}N_2\sigma_\text{e}-N_1\sigma_\text{a}=\frac{W_\text{p}\tau_2(\sigma_\text{e}-p\sigma_\text{a})-\sigma_\text{a}}{1+(1+p)W_\text{p}\tau_2+(I\tau_2/h\nu)(\sigma_\text{e}+\sigma_\text{a})}N_\text{t}$
Using the relation in (10-41) [refer to the optical absorption and amplification tutorial], we find that, for optical pumping,
$\tag{10-67}p=\frac{\sigma_\text{e}^\text{p}}{\sigma_\text{a}^\text{p}}=\frac{\sigma_\text{e}(\lambda_\text{p})}{\sigma_\text{a}(\lambda_\text{p})}$
where $$\sigma_\text{a}^\text{p}$$ and $$\sigma_\text{e}^\text{p}$$ are the absorption and emission cross sections, respectively, at the pump wavelength.
In a true two-level system, shown in Figure 10-9(a) above, the energy levels $$|2\rangle$$ and $$|1\rangle$$ can each be degenerate with degeneracies $$g_2$$ and $$g_1$$, respectively, but the population densities in both levels are evenly distributed among the respective degenerate states.
In this situation, $$p=\sigma_\text{e}^\text{p}/\sigma_\text{a}^\text{p}=g_1/g_2=\sigma_\text{e}/\sigma_\text{a}$$. Then, we find from (10-66) that
$\tag{10-68}N_2\sigma_\text{e}-N_1\sigma_\text{a}=\frac{-\sigma_\text{a}}{1+(\sigma_\text{e}+\sigma_\text{a})(I/h\nu+W_\text{p}/\sigma_\text{a})\tau_2}N_\text{t}\lt0$
No matter how a true two-level system is pumped, it is clearly not possible to achieve population inversion for an optical gain in the steady state.
This situation can be understood by considering the fact that the pump for a two-level system has to be in resonance with the transition between the two levels, thus inducing downward transitions as well as upward transitions.
In the steady state, the two-level system would reach thermal equilibrium with the pump at a finite temperature $$T$$, resulting in a Boltzmann population distribution of the form given in (10-26) [refer to the optical transitions for laser amplifiers tutorial] without population inversion.
As discussed in the optical absorption and amplification tutorial and illustrated in Figure 10-7 [refer to the optical absorption and amplification tutorial], however, in many cases an energy level is actually split into a band of closely spaced, but not exactly degenerate, sublevels with its population density unevenly distributed among these sublevels.
A system is not a true two-level system, but is known as a quasi-two-level system, if either or both of the two levels involved are split in such a manner.
By pumping such a quasi-two-level system properly, it is possible to reach the needed population inversion in the steady state for an optical gain at a particular laser frequency $$\nu$$ because the ratio $$p=\sigma_\text{e}^\text{p}/\sigma_\text{a}^\text{p}$$ at the pump frequency $$\nu_\text{p}$$ can now be made different from the ratio $$\sigma_\text{e}/\sigma_\text{a}$$ at the laser frequency $$\nu$$ due to the uneven population distribution among the sublevels within an energy level.
From (10-66), we find that the pumping requirements for a steady-state optical gain from a quasi-two-level system are
$\tag{10-69}p=\frac{\sigma_\text{e}^\text{p}}{\sigma_\text{a}^\text{p}}\lt\frac{\sigma_\text{e}}{\sigma_\text{a}}\qquad\text{and}\qquad{W_\text{p}}\gt\frac{1}{\tau_2}\frac{\sigma_\text{a}}{\sigma_\text{e}-p\sigma_\text{a}}$
Because the absorption spectrum is generally shifted to the short-wavelength side of the emission spectrum, as discussed in the optical absorption and amplification tutorial and demonstrated in Figure 10-8 [refer to the optical absorption and amplification tutorial], these conditions can be satisfied by pumping sufficiently strongly at a higher transition energy than the photon energy corresponding to the peak of the emission spectrum.
In the case of optical pumping, this condition means that the pump wavelength has to be shorter than the emission wavelength. Figure 10-9(b) above illustrates such a pumping scheme of a quasi-two-level system.
Indeed, many laser gain media, including laser dyes, semiconductor gain media, and vibronic solid-state gain media, are often pumped as a quasi-two-level system.
Three-level system
Population inversion in steady state is possible for a system that has three energy levels involved in the process. Figure 10-10 shows the energy-level diagram of an idealized three-level system.
The lower laser level $$|1\rangle$$ is the ground state, $$E_1=E_0$$, or is very close to the ground state, within an energy separation of $$\Delta{E}_{10}\ll{k}_\text{B}T$$ from the ground state, so that it is initially populated.
The atoms are pumped to an energy level $$|3\rangle$$ above the upper laser level $$|2\rangle$$.
An effective three-level system satisfies the following conditions:
(1) Population relaxation from level $$|3\rangle$$ to level $$|2\rangle$$ is very fast and efficient, ideally $$\tau_2\gg\tau_{32}\approx\tau_3$$, so that the atoms excited by the pump quickly end up in level $$|2\rangle$$.
(2) Level $$|3\rangle$$ lies sufficiently high above level $$|2\rangle$$ with $$\Delta{E}_{32}\gg{k}_\text{B}T$$ so that the population in level $$|2\rangle$$ cannot be thermally excited back to level $$|3\rangle$$.
(3) The lower laser level $$|1\rangle$$ is the ground state, or its population relaxes very slowly if it is not the ground state.
Under these conditions, $$R_2\approx{W}_\text{p}N_1$$, $$R_1\approx-W_\text{p}N_1$$, and $$N_1+N_2\approx{N}_\text{t}$$. Furthermore, $$\tau_1\approx\infty$$ and $$\tau_{21}\approx\tau_2$$.
The parameter $$W_\text{p}$$ is the effective pumping transition probability rate for exciting an atom in the ground state to eventually reach the upper laser level. It is proportional to the power of the pump.
In the steady state with a constant pump, $$W_\text{p}$$ is a constant and $$\text{d}N_2/\text{d}t=\text{d}N_1/\text{d}t=0$$.
With these conditions, we find that
$\tag{10-70}N_2\sigma_\text{e}-N_1\sigma_\text{a}=\frac{W_\text{p}\tau_2\sigma_\text{e}-\sigma_\text{a}}{1+W_\text{p}\tau_2+(I\tau_2/h\nu)(\sigma_\text{e}+\sigma_\text{a})}N_\text{t}$
Therefore, the pumping condition for a constant optical gain under steady-state population inversion is
$\tag{10-71}W_\text{p}\gt\frac{\sigma_\text{a}}{\tau_2\sigma_\text{e}}$
This condition sets the minimum pumping requirement for effective population inversion to reach an optical gain in a three-level system.
This requirement can be understood by considering the fact that almost all of the population initially resides in the lower laser level $$|1\rangle$$. To achieve effective population inversion, the pump has to be strong enough to depopulate sufficient population density from level $$|1\rangle$$, while the system has to be able to keep it in level $$|2\rangle$$.
In the case when $$\sigma_\text{a}=\sigma_\text{e}$$, no population inversion occurs before at least one-half of the total population is transferred from level $$|1\rangle$$ to level $$|2\rangle$$.
Four-level system
A four-level system, shown schematically in Figure 10-11, is more efficient than a three-level system.
A four-level system differs from a three-level system in that the lower laser level $$|1\rangle$$ lies sufficiently high above the ground level $$|0\rangle$$, with $$\Delta{E}_{10}\gg{k}_\text{B}T$$.
Therefore, in thermal equilibrium, the population in $$|1\rangle$$ is negligibly small compared with that in $$|0\rangle$$. Pumping takes place from level $$|0\rangle$$ to level $$|3\rangle$$.
An effective four-level system also has to satisfy the conditions concerning levels $$|3\rangle$$ and $$|2\rangle$$ discussed above for an effective three-level system.
In addition, it has to satisfy the condition that the population in level $$|1\rangle$$ relaxes very quickly back to the ground level, ideally $$\tau_1\approx\tau_{10}\ll\tau_2$$, so that level $$|1\rangle$$ remains relatively unpopulated in comparison with level $$|2\rangle$$ when the system is pumped.
Under these conditions, $$N_1\approx0$$ and $$R_2\approx{W}_\text{p}(N_\text{t}-N_2)$$, where the effective pumping transition probability rate $$W_\text{p}$$ is again proportional to the pump power.
Then, (10-62) can be ignored because $$N_1\approx0$$. In the steady state when $$W_\text{p}$$ is held constant, by taking $$\text{d}N_2/\text{d}t=0$$ for (10-61), we find that
$\tag{10-72}N_2\sigma_\text{e}-N_1\sigma_\text{a}\approx{N_2}\sigma_\text{e}=\frac{W_\text{p}\tau_2\sigma_\text{e}}{1+W_\text{p}\tau_2+(I\tau_2/h\nu)\sigma_\text{e}}N_\text{t}$
This results indicates that there is no minimum pumping requirement for an ideal four-level system that satisfies the conditions discussed above.
Real systems are rarely ideal, but a practical four-level system is still much more efficient than a three-level system. There is no minimum pumping requirement for population inversion in a four-level system because level $$|1\rangle$$ is initially empty in such a system.
Optical Gain
When the condition in (10-65) is satisfied for a given system, an optical gain coefficient at a given optical frequency $$\nu$$ can be evaluated with $$g=N_2\sigma_\text{e}-N_1\sigma_\text{a}$$ according to (10-50) [refer to the optical absorption and amplification tutorial].
The optical gain coefficient is a function of the optical signal intensity, $$I$$, as a result of the dependence of $$N_2$$ and $$N_1$$ on $$I$$ due to stimulated emission that changes the population densities by causing downward transition from level $$|2\rangle$$ to level $$|1\rangle$$.
This effect causes saturation of the optical gain coefficient by the optical signal. For all three basic systems discussed above, the optical gain coefficient can be expressed as a function of the optical signal intensity, $$I$$:
$\tag{10-73}g=\frac{g_0}{1+I/I_\text{sat}}$
where $$g_0$$ is the unsaturated gain coefficient, which is independent of the optical signal intensity, and $$I_\text{sat}$$ is the saturation intensity of a medium, which can be generally expressed as
$\tag{10-74}I_\text{sat}=\frac{h\nu}{\tau_\text{s}\sigma_\text{e}}$
The time constant $$\tau_\text{s}$$ is an effective saturation lifetime of the effective population inversion. It can be considered as an effective decay time constant for the optical gain coefficient through the relaxation of the effective population inversion.
Both $$g_0$$ and $$\tau_\text{s}$$ are functions of the intrinsic properties of a gain medium, as well as of the pumping rate. They can be found from (10-66), (10-70), and (10-72) for the quasi-two-level, three-level, and four-level systems, respectively. The results are summarized below.
Quasi-two-level system:
$\tag{10-75}g_0=(W_\text{p}\tau_\text{s}\sigma_\text{e}-\sigma_\text{a})N_\text{t}$
$\tag{10-76}\tau_\text{s}=\tau_2\frac{1+\sigma_\text{a}/\sigma_\text{e}}{1+(1+p)W_\text{p}\tau_2}$
Three-level system:
$\tag{10-77}g_0=(W_\text{p}\tau_\text{s}\sigma_\text{e}-\sigma_\text{a})N_\text{t}$
$\tag{10-78}\tau_\text{s}=\tau_2\frac{1+\sigma_\text{a}/\sigma_\text{e}}{1+W_\text{p}\tau_2}$
Four-level system:
$\tag{10-79}g_0=W_\text{p}\tau_\text{s}\sigma_\text{e}N_\text{t}$
$\tag{10-80}\tau_\text{s}=\frac{\tau_2}{1+W_\text{p}\tau_2}$
The minimum pumping requirement for a medium to have an optical gain is clearly $$g_0\gt0$$. It can be shown that the minimum pumping requirements obtained by applying this condition to (10-75) and (10-77) are identical to those given in (10-69) and (10-71) for the quasi-two-level and the three-level systems, respectively. As for the four-level system, both (10-72) and (10-79) clearly indicate that it has no minimum pumping requirement.
For a desired unsaturated gain coefficient of $$g_0$$, the required pumping rate can be found by solving (10-75) and (10-76) for a quasi-two-level system, (10-77) and (10-78) for a three-level system, and (10-79) and (10-80) for a four-level system. The results are summarized below.
Quasi-two-level system:
$\tag{10-81}W_\text{p}=\frac{1}{\tau_2}\frac{\sigma_\text{a}N_\text{t}+g_0}{(\sigma_\text{e}-p\sigma_\text{a})N_\text{t}-(1+p)g_0}$
Three-level system:
$\tag{10-82}W_\text{p}=\frac{1}{\tau_2}\frac{\sigma_\text{a}N_\text{t}+g_0}{\sigma_\text{e}N_\text{t}-g_0}$
Four-level system:
$\tag{10-83}W_\text{p}=\frac{1}{\tau_2}\frac{g_0}{\sigma_\text{e}N_\text{t}-g_0}$
In the limit when $$p\rightarrow0$$, a quasi-two-level system is identical to a three-level system. In the limit when $$p\rightarrow0$$ and $$\sigma_\text{a}\rightarrow0$$, a quasi-two-level system behaves like a four-level system. In the limit when $$\sigma_\text{a}\rightarrow0$$, a three-level system behaves like a four-level system. For a quasi-two-level system, it is clearly desirable to choose a pump wavelength for which the value of $$p$$ is as small as possible.
Unsaturated gain coefficient
The unsaturated gain coefficient is also known as the small-signal gain coefficient because it is the gain coefficient of a weak optical field that does not saturate the gain medium.
In the case of optical pumping with a pump quantum efficiency $$\eta_\text{p}$$, the pump intensity required for a desired pumping transition probability rate can be found by using (10-41) [refer to the optical absorption and amplification tutorial] as
$\tag{10-84}I_\text{p}=\frac{1}{\eta_\text{p}}\frac{h\nu_\text{p}}{\sigma_\text{a}^\text{p}}W_\text{p}$
where $$h\nu_\text{p}$$ is the energy of the pump photon.
The pump quantum efficiency $$\eta_\text{p}$$ is the net probability of exciting an atom to the upper laser level by each absorbed pump photon. In general, $$\eta_\text{p}\le1$$.
It is convenient to define a saturation pump intensity, $$I_\text{p}^\text{sat}$$, for a laser amplifier for which $$W_\text{p}\tau_2=1$$ as
$\tag{10-85}I_\text{p}^\text{sat}=\frac{h\nu_\text{p}}{\eta_\text{p}\tau_2\sigma_\text{a}^\text{p}}$
This is the pump intensity that pumps one-half of the population in a three- or four-level system, and about one-half in a quasi-two-level system, to the upper laser level.
At this level and above, absorption of the pump power is significantly saturated due to depletion of the ground-state population by pumping.
For a pump intensity of $$I_\text{p}$$, we have
$W_\text{p}\tau_2=I_\text{p}/I_\text{p}^\text{sat}$
For a four-level system, we have $$g\gt0$$ as long as the medium is pumped because there is no minimum pumping requirement. For a quasi-two-level or three-level system, we find that $$g\gt0$$ only when the pumping level exceeds its minimum pumping requirement; below that, the medium has absorption for $$g\lt0$$.
When the unsaturated gain coefficient is zero, the medium becomes transparent, or bleached, to the optical signal, neither absorbing it nor amplifying it.
A quasi-two-level or three-level system reaches transparency, or the bleached condition, at the following transparency pumping rate:
$\tag{10-86}W_\text{p}^\text{tr}=\frac{1}{\tau_2}\frac{\sigma_\text{a}}{\sigma_\text{e}-p\sigma_\text{a}}$
The pump intensity corresponding to the transparency pumping rate is the transparency pump intensity, $$I_\text{p}^\text{tr}$$, which can be expressed as
$\tag{10-87}I_\text{p}^\text{tr}=\frac{1}{\eta_\text{p}}\frac{h\nu_\text{p}}{\tau_2\sigma_\text{a}^\text{p}}\frac{\sigma_\text{a}}{\sigma_\text{e}-p\sigma_\text{a}}=\frac{\sigma_\text{a}}{\sigma_\text{e}-p\sigma_\text{a}}I_\text{p}^\text{sat}$
For a quasi-two-level system, $$p\ne0$$ in general. For a three-level system, we take $$p=0$$ for (10-86) and (10-87). For a four-level system, $$I_\text{p}^\text{tr}=0$$ because a four-level system has no minimum pumping requirement and is thus transparent without pumping.
It can be seen from (10-75) to (10-80) that for any system, $$g_0$$ increases with pump intensity less than linearly because $$\tau_\text{s}$$ decreases with pump intensity though $$W_\text{p}$$ is linearly proportional to the pump intensity.
This dependence of $$\tau_\text{s}$$ on the pump intensity is caused by the fact that as the pump excites atoms from the ground state to any excited state to eventually reach the upper laser level, it depletes the population in the ground state.
Consequently, as the pump intensity increases, fewer atoms remain available for excitation in the ground state, thus reducing the differential increase of the effective population inversion with respect to the increase of the pump intensity.
It can be shown by using the relations in (10-75), (10-77), and (10-79) that the unsaturated gain coefficient of any system can be expressed as a function of pump intensity in the following general form:
$\tag{10-88}g_0=\frac{(\sigma_\text{e}-p\sigma_\text{a})N_\text{t}}{1+(1+p)I_\text{p}/I_\text{p}^\text{sat}}\left(\frac{I_\text{p}}{I_\text{p}^\text{sat}}-\frac{I_\text{p}^\text{tr}}{I_\text{p}^\text{sat}}\right)=\frac{(\sigma_\text{e}+\sigma_\text{a})N_\text{t}}{1+(1+p)I_\text{p}/I_\text{p}^\text{sat}}\frac{I_\text{p}}{I_\text{p}^\text{sat}}-\sigma_\text{a}N_\text{t}$
For a quasi-two-level system, $$p\ne0$$ and $$I_\text{p}^\text{tr}\ne0$$. For a three-level system, $$p=0$$ but $$I_\text{p}^\text{tr}\ne0$$. For a four-level system, $$p=0$$ and $$I_\text{p}^\text{tr}=0$$.
Note that for a quasi-two-level system or a three-level system, (10-88) is also valid when $$I_\text{p}\lt{I}_\text{p}^\text{tr}$$ for $$g_0\lt0$$. In this situation, the medium has an absorption coefficient of $$\alpha=-g_0$$ at the laser transition frequency.
As can be seen in (10-88), $$g_0$$ varies with $$I_\text{p}$$ sublinearly at high pumping levels due to the dependence of $$\tau_\text{s}$$ on $$I_\text{p}$$ as discussed above.
For a four-level system, however, the unsaturated gain coefficient varies approximately linearly with $$I_\text{p}$$ at a low pumping level such that $$I_\text{p}/I_\text{p}^\text{sat}\ll1$$.
For a quasi-two-level system or a three-level system, significant pumping is needed just to reach transparency, but the unsaturated gain coefficient also varies approximately linearly with $$I_\text{p}$$ for small variations of the pump intensity near the transparency point.
Gain saturation
The optical gain coefficient is a function of the intensity of the optical wave traveling in the gain medium; it decreases as the optical signal intensity increases.
According to (10-73), the optical gain coefficient is reduced to one-half that of the unsaturated gain coefficient $$g_0$$ when the optical signal intensity reaches the saturation intensity $$I_\text{sat}$$. The smaller the value of $$I_\text{sat}$$, the easier it is for the gain to become saturated.
For a quasi-two-level system, $$\tau_\text{s}=\tau_2(1-p\sigma_\text{a}/\sigma_\text{e})$$ at transparency. For three-level and four-level systems, $$\tau_\text{s}=\tau_2$$ at transparency. For all three system, $$\tau_\text{s}\lt\tau_2$$ as the gain medium is pumped above transparency for a positive gain coefficient. Therefore, $$I_\text{sat}$$ increases as the gain medium is pumped harder for a larger unsaturated gain coefficient.
Example 10-7
The ruby laser is a three-level system. As shown in Figure 10-12, it has two primary pump bands at 404 and 554 nm wavelengths, from the $$^4\text{A}_2$$ ground state to the $$^4\text{F}_1$$ and $$^4\text{F}_2$$ states, respectively, both of which relax quickly to the $$^2\text{E}$$ state so that $$\tau_{32}\ll\tau_2=3\text{ ms}$$. The absorption cross sections for $$\mathbf{E}\perp{c}$$ polarization at 404 and 554 nm pump wavelengths are both $$\sigma_\text{a}^\text{p}=2\times10^{-23}\text{ m}^{-1}$$. Assume a $$100\%$$ pump quantum efficiency of $$\eta_\text{p}=1$$.
(a) Find the transparency pumping rate of a ruby crystal for the 694.3 nm transitioin with $$\mathbf{E}\perp{c}$$ polarization. Find the transparency pump intensity for each pump band. What is the saturation intensity at transparency?
(b) A ruby crystal rod doped with 0.05 wt. $$\%$$ Cr2O3 has a Cr concentration of $$1.58\times10^{25}\text{ m}^{-3}$$. It is pumped for an unsaturated gain coefficient of $$5\text{ m}^{-1}$$. Find the required pumping rate, the saturation intensity at this pumping rate, and the required pump intensity for each pump band.
(a)
From Example 10-4 [refer to the optical absorption and amplification tutorial], we found that $$\sigma_\text{a}=1.25\times10^{-24}\text{ m}^2$$, from Example 10-5 [refer to the optical absorption and amplification tutorial], we found that $$\sigma_\text{e}=1.34\times10^{-24}\text{ m}^2$$, so that we can find the transparency pumping rate by using (10-86) for this three-level system:
$W_\text{p}^\text{tr}=\frac{\sigma_\text{a}}{\tau_2\sigma_\text{e}}=\frac{1.25\times10^{-24}}{3\times10^{-3}\times1.34\times10^{-24}}\text{ s}^{-1}=311\text{ s}^{-1}$
The pump photons at 404 and 554 nm wavelengths have photon energies of $$h\nu_\text{p1}=(1.2398/0.404)\text{ eV}=3.069\text{ eV}$$ and $$h\nu_\text{p2}=(1.2398/0.554)\text{ eV}=2.238\text{ eV}$$, respectively. The transparency pump intensity for the 404 nm pump band is (by using (10-86) and (10-87))
$I_\text{p}^\text{tr}=\frac{h\nu_\text{p1}}{\sigma_\text{a}^\text{p}}W_\text{p}^\text{tr}=\frac{3.069\times1.6\times10^{-19}\times311}{2\times10^{-23}}\text{ W m}^{-2}=7.64\text{ MW m}^{-2}$
and that for the 554 nm pump band is
$I_\text{p}^\text{tr}=\frac{h\nu_\text{p2}}{\sigma_\text{a}^\text{p}}W_\text{p}^\text{tr}=\frac{2.238\times1.6\times10^{-19}\times311}{2\times10^{-23}}\text{ W m}^{-2}=5.57\text{ MW m}^{-2}$
For the three-level ruby, $$\tau_\text{s}=\tau_2=3\text{ ms}$$ at transparency. The photon energy for $$\lambda=694.3\text{ nm}$$ is $$h\nu=(1.2398/0.6943)\text{ eV}=1.786\text{ eV}=1.786\times1.6\times10^{-19}\text{ J}$$. Therefore, from (10-74), the saturation intensity at transparency is
$I_\text{sat}=\frac{h\nu}{\tau_\text{s}\sigma_\text{e}}=\frac{1.786\times1.6\times10^{-19}}{3\times10^{-3}\times1.34\times10^{-24}}\text{ W m}^{-2}=71.1\text{ MW m}^{-2}$
(b)
For $$N_\text{t}=1.58\times10^{25}\text{ m}^{-3}$$, we find that $$\sigma_\text{e}N_\text{t}=1.34\times10^{-24}\times1.58\times10^{25}\text{ m}^{-1}=21.17\text{ m}^{-1}$$ and $$\sigma_\text{a}N_\text{t}=1.25\times10^{-24}\times1.58\times10^{25}\text{ m}^{-1}=19.75\text{ m}^{-1}$$. Thus, using (10-82), we find that, for $$g_0=5\text{ m}^{-1}$$, the required pumping rate is
$W_\text{p}=\frac{1}{3\times10^{-3}}\times\frac{19.75+5}{21.17-5}\text{ s}^{-1}=510\text{ s}^{-1}$
which is 1.64 times the transparency pumping rate of $$W_\text{p}^\text{tr}=311\text{ s}^{-1}$$. Therefore, the required pump power is 1.64 times the transparency pump power: $$I_\text{p}=1.64I_\text{p}^\text{tr}=12.53\text{ MW m}^{-2}$$ for the 404 nm pump and $$I_\text{p}=1.64I_\text{p}^\text{tr}=9.13\text{ MW m}^{-2}$$ for the 554 nm pump. At this pumping level, $$W_\text{p}\tau_2=1.53$$. Therefore, from (10-78), $$\tau_\text{s}=\tau_2(1+1.25/1.34)/(1+1.53)=2.29\text{ ms}$$. Then, the saturation intensity is
$I_\text{sat}=\frac{h\nu}{\tau_\text{s}\sigma_\text{e}}=\frac{1.786\times1.6\times10^{-19}}{2.29\times10^{-3}\times1.34\times10^{-24}}\text{W m}^{-2}=93.1\text{ MW m}^{-2}$
It is clear from this example that a very high pump power is required just to bring a ruby crystal to transparency because of the fact that it is a three-level system. For this reason, it is only feasible to pump a ruby laser with a pulsed pump.
As a consequence, CW operation is never realized for the ruby laser. Ruby lasers are always operated in the pulsed mode, most notably in the Q-switched mode for the generation of giant pulses.
The situation is very different for four-level systems, such as Nd : YAG, or quasi-two-level systems, such as Ti : sapphire and Cr : LiSAF.
Spontaneous emission power
When the upper laser level of a gain medium is populated, there is spontaneous emission. The upper laser level population can be found by solving $$N_1+N_2=N_\text{t}$$ and $$N_2\sigma_\text{e}-N_1\sigma_\text{a}=g$$ simultaneously to have
$\tag{10-89}N_2=\frac{\sigma_\text{a}N_t+g}{\sigma_\text{e}+\sigma_\text{a}}$
This relation is valid for all systems, including the four-level system. Though we have used $$N_1+N_2=N_\text{t}$$, which is not valid for a four-level system, to obtain this relation, (10-89) reduces to $$N_2=g/\sigma_\text{e}$$ in the case of a four-level system, for which $$\sigma_\text{a}=0$$.
Note that, in the case of a quasi-two-level or a three-level system, $$g=-\alpha$$ when the medium is not sufficiently pumped to reach transparency.
Because the maximum value of the absorption coefficient is $$\alpha_0=\sigma_\text{a}N_\text{t}$$, we find that $$N_2\ge0$$ for any positive or negative values of $$g$$.
Note also that $$g$$ appearing in (10-89) is the saturated gain coefficient if stimulated emission exists in the medium.
According to the discussions in the optical transitions for laser amplifiers tutorial, the spontaneous emission power is proportional to $$N_2$$ only and is independent of $$N_1$$. Therefore, regardless of whether the medium has a gain or a loss, the spontaneous emission power density, which is defined as the spontaneous emission power per unit volume of the medium in watts per cubic meter, is
$\tag{10-90}\hat{P}_\text{sp}=\frac{h\nu}{\tau_\text{sp}}N_2=\frac{h\nu}{\tau_\text{sp}}\frac{\sigma_\text{a}N_\text{t}+g}{\sigma_\text{e}+\sigma_\text{a}}$
where $$g$$ can be positive, for a medium pumped above transparency, or negative, for a medium below transparency. For a gain volume of $$\mathcal{V}$$, the spontaneous emission power is $$P_\text{sp}=\hat{P}_\text{sp}\mathcal{V}$$.
In the case when the gain is not saturated so that $$g=g_0$$, we find from (10-75), (10-77), and (10-79) that $$\sigma_\text{a}N_\text{t}+g=W_\text{p}\tau_\text{s}\sigma_\text{e}N_\text{t}$$. For a medium that is optically pumped with a pump intensity $$I_\text{p}$$, we then have
$\tag{10-91}N_2=\frac{W_\text{p}\tau_2}{1+(1+p)W_\text{p}\tau_2}N_\text{t}=\frac{I_\text{p}/I_\text{p}^\text{sat}}{1+(1+p)I_\text{p}/I_\text{p}^\text{sat}}N_\text{t}$
Then, the spontaneous emission power density in the absence of gain saturation can be expressed as
$\tag{10-92}\hat{P}_\text{sp}=\frac{h\nu}{\tau_\text{sp}}\frac{I_\text{p}/I_\text{p}^\text{sat}}{1+(1+p)I_\text{p}/I_\text{p}^\text{sat}}N_\text{t}$
At transparency, $$g=g_0=0$$. The spontaneous emission power density at transparency, which is known as the critical fluorescence power density, is
$\tag{10-93}\hat{P}_\text{sp}^\text{tr}=\frac{h\nu}{\tau_\text{sp}}\frac{\sigma_\text{a}}{\sigma_\text{e}+\sigma_\text{a}}N_\text{t}=\frac{h\nu}{\tau_\text{sp}}\frac{I_\text{p}^\text{tr}/I_\text{p}^\text{sat}}{1+(1+p)I_\text{p}^\text{tr}/I_\text{p}^\text{sat}}N_\text{t}$
For a gain volume of $$\mathcal{V}$$, the critical fluorescence power is $$P_\text{sp}^\text{tr}=\hat{P}_\text{sp}^\text{tr}\mathcal{V}$$.
Example 10-8
A ruby crystal doped with 0.05 wt. $$\%$$ Cr2O3 for a Cr concentration of $$1.58\times10^{25}\text{ m}^{-3}$$ as discussed in Example 10-7 is considered. Almost all of the population in the upper laser level of a ruby laser crystal relaxes radiatively by to the ground state so that $$\tau_\text{sp}=\tau_{21}=\tau_2=3\text{ ms}$$. Find the critical fluorescence power density corresponding to transparency for the 694.3 nm line at 300 K. What is the spontaneous emission power density if the ruby crystal is pumped above transparency for a gain coefficient of $$5\text{ m}^{-1}$$ for the 694.3 nm line? What is it if the crystal is insufficiently pumped so that it has an absorption coefficient of $$5\text{ m}^{-1}$$ for the 694.3 nm line? If a ruby laser rod of 6 cm length and 4 mm cross-sectional diameter is uniformly pumped, what are the spontaneous emission powers in the three cases considered here?
When we consider transparency for the 694.3 nm transition, we take $$\sigma_\text{a}=1.25\times10^{-24}\text{ m}^2$$ and $$\sigma_\text{e}=1.34\times10^{-24}\text{ m}^2$$ for this transition at 300 K, which are obtained in Examples 10-4 and 10-5, respectively [refer to the optical absorption and amplification tutorial]. However, the spontaneous emission is broadband covering both emission lines at 692.9 and 694.3 nm. Therefore, we take an average photon energy of the two for $$h\nu=1.787\text{ eV}$$. Then, we find from (10-93) the following critical fluorescence power density at the transparency point for the 694.3 nm line:
$\hat{P}_\text{sp}^\text{tr}=\frac{1.787\times1.6\times10^{-19}\times1.25\times10^{-24}\times1.58\times10^{25}}{3\times10^{-3}\times(1.34\times10^{-24}+1.25\times10^{-24})}\text{W m}^{-3}=727\text{ MW m}^{-3}$
When pumped for a gain coefficient of $$5\text{ m}^{-1}$$ for the 694.3 nm line, we find from (10-90) that
$\hat{P}_\text{sp}=\frac{1.787\times1.6\times10^{-19}\times(1.25\times10^{-24}\times1.58\times10^{25}+5)}{3\times10^{-3}\times(1.34\times10^{-24}+1.25\times10^{-24})}\text{W m}^{-3}=911\text{ MW m}^{-3}$
When the crystal is insufficiently pumped so that there is an absorption coefficient of $$5\text{ m}^{-1}$$, $$g=-5\text{ m}^{-1}$$. Then from (10-90)
$\hat{P}_\text{sp}=\frac{1.787\times1.6\times10^{-19}\times(1.25\times10^{-24}\times1.58\times10^{25}-5)}{3\times10^{-3}\times(1.34\times10^{-24}+1.25\times10^{-24})}\text{W m}^{-3}=543\text{ MW m}^{-3}$
For a rod of 6 cm length and 4 mm cross-sectional diameter, the volume is $$\mathcal{V}=\pi\times(4\times10^{-3}/2)^2\times6\times10^{-2}\text{ m}^3=7.54\times10^{-7}\text{ m}^3$$. Therefore, the critical fluorescence power is $$P_\text{sp}^\text{tr}=\hat{P}_\text{sp}^\text{tr}\mathcal{V}=548\text{ W}$$. The total spontaneous emission power is $$P_\text{sp}=687\text{ W}$$ for $$g=5\text{ m}^{-1}$$ and $$P_\text{sp}=407\text{ W}$$ for $$g=-5\text{ m}^{-1}$$.
From the consideration of energy conservation, it is clear that the power required to pump the crystal to a particular state has to be at least, and most often far exceed, that emitted spontaneously by the crystal. Therefore, these numbers again show the high power required to pump a ruby laser crystal just to its transparency point.
For example, if the pumping efficiency is $$10\%$$, the pump power required to pump this crystal to transparency is as high as 5.48 kW.
The next tutorial covers laser amplifiers.
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2022-12-05 08:01:00
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http://mathhelpforum.com/differential-geometry/134973-telescopic-series.html
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# Math Help - telescopic series
1. ## telescopic series
Hi guys, quick Question
For the series $\sum_{i=1}^\infty(\frac{1}{k}-\frac{1}{k+2})$ find a formula for $S_n$
I know that every number will cancel out but $1$ and $\frac{1}{2}$ but the answer is $1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}$.
How do we get $-\frac{1}{n+1}-\frac{1}{n+2}$?
Thx!
2. $S_n$ is not the sum of the infinite series, but the partial sum $\sum_{k=1}^n(\frac{1}{k}-\frac{1}{k+2})$.
You see it in the definition of the sum of an infinite series $\sum_{k=1}^{\infty}(\frac{1}{k}-\frac{1}{k+2})$ is defined as $\lim_{n \to \infty}S_n$, where $S_n$ is as above.
- Hollywood
3. Indeed. How do you get that partal sum? I understand the 1 and .5 part...
4. Everything cancels out except for 1 and 1/2 at the beginning and 1/(k+1) and 1/(k+2) at the end:
$S_n=\sum_{k=1}^n(\frac{1}{k}-\frac{1}{k+2})=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^n\frac{1}{k+2}=\sum_{k=1}^n\frac{1}{k}-\sum_{k=3}^{n+2}\frac{1}{k}$
$S_n=1+\frac{1}{2}+\sum_{k=3}^n\frac{1}{k}-\sum_{k=3}^n\frac{1}{k}-\frac{1}{n+1}-\frac{1}{n+2}$
$S_n=1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}$
- Hollywood
5. I guess I'm still not sure how to get the right side. How do I produce it in the first place? I understand the 1 and .5. Thanks for your patience.
6. $\frac{1}{k}-\frac{1}{k+2}=\frac{1}{k}-\frac{1}{k+1}+\frac{1}{k+1}-\frac{1}{k+2},$ there's two telescoping sums there.
7. Originally Posted by sfspitfire23
I guess I'm still not sure how to get the right side. How do I produce it in the first place? I understand the 1 and .5. Thanks for your patience.
If you look at the series this way:
$(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...$
the partial sums $S_n$ are:
$S_n=(1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{n-1}-\frac{1}{n+1})+(\frac{1}{n}-\frac{1}{n+2})$
It's no longer infinite, just the first n terms.
So you can see that the 1/3 in the first term cancels the one in the third term, the 1/4 in the second term cancels the one in the fourth term, and so on. But when we get to 1/(n+1), there is no term to cancel it, and the same for 1/(n+2).
Hope this helps - feel free to post again in this thread if you're still having trouble.
- Hollywood
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2016-06-27 10:44:25
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https://rpg.stackexchange.com/questions/134696/what-happens-when-you-touch-an-indirect-combat-spell-preparation
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What happens when you touch an indirect combat spell preparation?
When you cast an indirect combat spell with sorcery, the target gets to resist/dodge the spell with a Rea+Int roll.
The spell then is launched with an Opposed Test that pits the magician’s Spellcasting + Magic [Force] versus the target’s Reaction + Intuition .. (Core 283)
All spells can be learned as alchemical preparations. One of the options for such a preparation is a contact trigger:
Contact: The next living being to touch the preparation activates the spell.
Touch: Touch spells affect the living being touching the preparation. (Core 305-6)
So if I make a preparation of an indirect combat spell (eg. Punch), and the linchpin is then touched by an enemy, what do they roll, if anything, to avoid the effect?
Alchemy or not, indirect combat spells with a touch range are a bit counter-intuitive.
The description of indirect combat spells states that the spell can be evaded using $$\REA+INT\$$ the only exception mentioned being spells with an area of effect.
This does not negate the necessity to touch an unwilling target, if Spellcasting is used; The target would get a chance of evading the attempt to touch it and another chance to evade/reduce the spell effect.
With an alchemical preparation with contact trigger the source of the spell automatically has contact with the target, so it cannot make an attempt to evade the touch, but it has a chance of evading the spell effect using $$\REA+INT\$$. There are simply no rules denying it this test.
The only way I could explain this is that the touch triggers the spell, but the spells energy still needs to build up to deal damage which gives the target enough time to get out of the way, if it gets enough hits in the $$\REA+INT\$$ test.
(Houseruling this could be a good idea, but that's a decision that the GM/group needs to make.)
• I was under the impression that the Spellcasting test replaces the touch attack. As the attack is a Complex Action, it would be rather hard to deliver such a spell within the action economy. – Szega Oct 31 '18 at 22:29
• @Szega Just compare to a direct combat spell, like Death Touch. You need to get in contact with the target which is done with a unarmed attack and the target resists with WIL. You've got 2 defenses neither the description of direct nor indirect spells mentions touch spells, so they are treated the same. As for the action economy: I've always treated casting the spell and doing the touch attack as a single action even though it requires multiple rolls. – fabian Oct 31 '18 at 22:58
• Do you have any official statement reinforcing your opinion? Or about how to handle it action economy-wise? – Szega Oct 31 '18 at 23:06
• No; this may be a houserule,but as you said it would be pretty complex to deal with it otherwise and there's also no rule indicating what happens when you miss the touch attack.Can you retry?Can a different touch spell be cast before succeeding in the attack?Also the spell duration is instantaneous and everything but using the same action for the touch attack would stretch the definition of "instantaneous" quite a bit.Also what happens if the drain knocks you out?According to the rules the spell should still take effect but how is that possible if you don't get the chance to touch the target? – fabian Oct 31 '18 at 23:24
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2019-10-22 09:53:09
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https://www.vedantu.com/maths/coincident-lines
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# Coincident Lines
View Notes
## Introduction
Coincident lines are the lines that coincide or lie on top of each other. So far we have learned about different types of lines in Geometry, such as parallel lines, perpendicular lines, with respect to a two-dimensional or three-dimensional plane. In the case of parallel lines, they are parallel to each other and they are at a defined distance between them. On the other hand, perpendicular lines are lines that intersect each other at 90 degrees. Both parallel lines and perpendicular lines do not coincide with each other. Here comes the concept of coincident lines when both the lines coincide with each other. In this article, we will learn about the coincident lines equation, coincident lines condition along with coincident lines solution.
In the below diagram we have represented parallel lines, intersecting lines and coincident lines.
[Image will be uploaded soon]
In the case of parallel lines, there are no common points. Intersecting lines have one point in common and coincident lines have infinitely many points in common.
Apart from these three lines, there are many different lines that are neither parallel, perpendicular, nor coinciding lines. These lines can be oblique lines or intersecting lines, which intersect at different angles, instead of perpendicular to each other.
## Coincident Lines Definition
The word ‘coincide’ means that it happens at the same time. In Mathematics, the coincident is defined as the lines that lie upon each other. It is placed in such a way that when we look at them, they appear to be a single line, instead of double or multiple lines.
If we see the given below diagram of coincident lines, it appears as a single line, but in actual we have drawn two different lines here. First, we drew a line of purple colour and then on top of it drew another line which is of black colour.
[Image will be uploaded soon]
### Coincident Lines Equation
We have learned that linear equations can be represented by the equation y=mx+c, where m and c are real numbers.
If we consider the equation of a line, the standard form is:
y = mx + b
Where m is the slope of the line and b is the intercept made by the line.
Equation of Parallel Lines:
Now, in the case of two lines that are parallel to each other, we represent the equations of the lines as:
y = m1x + b1
And y = m2x + b2
For example, y = x + 2 and y = 2x + 4 are parallel lines. Here, the slope of both the lines is equal to 2 and the intercept difference between them is 2. Hence, they are parallel lines at a distance of 2 units.
Equation of Coincident Lines:
The equation for lines is given below;
ax + by = c
When two lines are coinciding with each other, then there is no intercept difference between them.
For example, consider the equation of two coinciding lines,
x + y = 4 and 2x + 2y = 8 are two coinciding lines.
If we compare both the lines we get the second line is twice the first line.
If we put ‘y’ on the Left-hand side and the rest of the equation on the Right-hand side, then we get the following result;
First-line y = 2 - x ..(i)
Second-line 2y = 4 - 2x
2y = 2(2 - x)
y = 2 - x …(ii)
From equation (1) and (2), we can conclude both the lines are the same.
Hence, they coincide with each other.
### Coincident Lines Formula
Consider a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 be the pair of linear equations in two variables. The lines representing these equations are said to be coincident if follow the below condition
Given below is the Coincident lines condition
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here, the given pair of equations is called consistent equations and they can have infinitely many solutions.
### Coincident Lines Solutions
Refer to the given below solved example to understand how to use the formula of coincident lines.
Check whether the lines representing the pair of equations 9x-2y+16=0 and 18x-4y+32=0 are coincident or not.
Ans: Given equation
9x - 2y + 16 = 0
18x - 4y + 32 = 0
We will compare the above equation with a1x + b1y + c1=0 and a2x + b2y +c2=0
After comparing we get a1= 9 b1 = -2 and c1= 16
a2 = 18 b2 = -4 and c2 = 32
Now, calculate the value of
$\frac{a_{1}}{a_{2}}=\frac{9}{18}=\frac{1}{2}$
$\frac{b_{1}}{b_{2}}=\frac{-2}{-4}=\frac{1}{2}$
$\frac{c_{1}}{c_{2}}=\frac{16}{32}=\frac{1}{2}$
Hence from above, we can conclude that
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Therefore, the lines representing the above equations are coincident.
This can be shown graphically as:
[Image will be uploaded soon]
FAQ (Frequently Asked Questions)
1. Do Coincidence Lines are the Same as Parallel Lines?
Ans: There is a slight difference between the two parallel lines and two coincident lines. Parallel lines have constant space between them while coincident don't. Parallel lines do not have points in common while coincident lines have all points in common.
2. How Can We Represent the Coincident Lines?
Ans: Suppose the given equation of a line as ax + by = c, here y-intercept is equal to c/b. If each line in the system has the same slope as well as the same y-intercept then the lines are coincident.
3. What Does a Coincident Line Look Like?
Ans: Two lines or shapes that lie exactly on top of each other are said to be coincident. If we look at the coincident line, it will look like a single line, because they are on top of each other we can't see both the lines.
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2021-04-14 00:55:37
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|
https://twiki.esc.auckland.ac.nz/do/view/OpsRes/IntegerProgrammingWithAMPL?rev=1
|
# Integer Programming with AMPL
Specifying variables to be integer or binary in AMPL will cause the solver, e.g., CPLEX, to use mixed-integer programming. This will often be enough to solve many of the problems you will encounter. However, if your integer programmes are taking a long time to solve you can use some "tricks" to speed up the branch-and-bound process.
## A "Simple" Integer Programme
To demonstrate the techniques we can use to control integer programming we will look at a simple integer programming problem:
Jim has three requests for frozen ice sculptures, his commission is $1000,$7000 and $5000 respectively. He must hire a refrigeration unit to transport each one. The units cost$4000 each. The sculptures will be transported on a truck with capacity 1.7 tonnes and he estimates the total weight of each sculpture (including the refrigeration unit) to be 1 tonne, half a tonne and a quarter of a tonne respectively. Jim must decide which sculptures to make to maximize his profit.
The AMPL model and data files, ice.mod and ice.dat respectively, are attached.
Solving this problem with AMPL and CPLEX is very fast (it is only a small problem):
However, sometimes all the technology behind CPLEX does not work so well and we need to control the branch and bound tree. First, let’s remove all the CPLEX technology and re-solve our problem:
##### ice.run
reset;model ice.mod;data ice.dat;option solver cplex;option presolve 0;option cplex_options ('timing 1 mipdisplay 5 mipinterval 1' & 'presolve 0 mipcuts -1 cutpass -1 ' & 'heurfreq -1');solve;display Fridges, Make;<span style="font-family: monospace;">
With all CPLEX’s “bells and whistles” removed we get a slightly larger branch-and-bound tree:
Let's look at ways to reduce the size of this branch-and-bound tree.
##### Looking at the LP Relaxation
Often you can gain insight into the branch-and-bound process by considering the LP relaxation. You can relax integrality without reformulating using
\begin{verbatim}
option relax_integrality 1;
\end{verbatim}
If we look at the variables we can see where our solution is fractional:
As you can see we are using 2.8 fridge units for our 2.8 sculptures. Also, if we check the {\tt TotalWeight} constraint ({\tt display TotalWeight.body;}) we can see that the truck is at its weight limit.
It looks likely that we should only use 2 fridges. We can create some new suffixes to experiment with our hypothesis.
##### Priorities, Searching and Directions
AMPL and CPLEX allow you to define a priority for your integer variables. This means that if more than one integer variable is fractional in a solution, CPLEX will branch on the highest priority variable first. Let’s add the priority {\tt suffix} to our run file (before solving):
\begin{verbatim}
suffix priority IN, integer, >= 0, <= 9999;
\end{verbatim}
(now we can assign variables priorities ranging from 0 – least – to 9999 – most). Let’s give the {\tt Fridges} variable a priority of 100 and the {\tt Make} variables a priority of 0 (using {\tt let} statements).
\begin{verbatim}
let Fridges.priority := 100;
let {s in SCULPTURES} Make[s].priority := 0;
\end{verbatim}
The branch-and-bound tree appears unchanged, so perhaps CPLEX had already branched on {\tt Fridges} first earlier. However, we can try a breadth-first search of the tree, since this will try different values for {\tt Fridges} before performing branching on other variables. Setting {\tt nodeselect} to 2 (best estimate) and {\tt backtrack} to 0 makes CPLEX perform a search very close to breadth-first (see The AMPL CPLEX User Guide for full details).
\begin{verbatim}
option cplex_options ('timing 1 mipdisplay 5 mipinterval 1 ' &
'presolve 0 mipcuts -1 cutpass -1 ' &
'heurfreq -1 ' &
'nodeselect 2 backtrack 0');
\end{verbatim}
Now the tree has been fathomed earlier (it only has 4 nodes instead of 6). However, we are not sure if CPLEX branched down to 2 fridges first (our hypothetical optimum).
To control the direction of the branches we can create a new suffix for the direction we should branch on each variable (-1 for down, 0 for no preference, 1 for up).
\begin{verbatim}
suffix direction IN, integer, >= -1, <= 1;
\end{verbatim}
We can force a down branch first on {\tt Fridges}:
\begin{verbatim}
let Fridges.direction := -1;
\end{verbatim}
This doesn’t seem to have decreased the size of the branch-and-bound tree. Let’s try one more thing. We have given CPLEX a good branch to try first, but we have not carefully considered what to do next. Let’s remove the breadth-first search option and let CPLEX decide how to proceed:
reset;model ice.mod;data ice.dat;option solver cplex;option presolve 0;option cplex_options ('timing 1 mipdisplay 5 mipinterval 1' & 'presolve 0 mipcuts -1 cutpass -1 ' & 'heurfreq -1');suffix priority IN, integer, >= 0, <= 9999;suffix direction IN, integer, >= -1, <= 1;let Fridges.priority := 100;let {s in SCULPTURES} Make[s].priority := 0;let Fridges.direction := -1;solve;display Fridges, Make;<span style="font-family: monospace;">
Now we have reduced our branch-and-bound tree to a single node by making a good choice about our first variable branch.
As stated earlier, CPLEX does a lot of good things automatically for you. Often, these “tricks” will be enough to solve your mixed-integer programming problems. However, if your problem is taking a long time to solve, you can experiment with adding some of your own control to the branch-and-bound process. History has shown that problem-specific approaches often work very well for hard integer programmes.
-- MichaelOSullivan - 23 Apr 2008
• ice_original.jpg:
• ice_nofrills.jpg:
• ice_relaxation.jpg:
|
2022-05-19 21:33:48
|
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|
https://astronomy.stackexchange.com/questions/44269/if-the-hypothesized-planet-behind-the-kuiper-belt-existed-would-it-have-a-baryc?noredirect=1
|
# If the hypothesized planet behind the Kuiper belt existed, would it have a barycenter outside the Sun?
In 2016, an additional object rotating our Sun has been suggested. Assuming it has 5 Earth masses, would the planet or black hole proposed by Brown and Batigyn have a barycenter within the Sun or outside, or might it orbit the Sun-Jupiter barycenter, like Sedna?
• Do you have a link to this (rather daring) hypothesis? Jun 9 at 17:14
• @planetmaker Dr. Becky mentioned this in a video in 2019. Here's the paper she linked to (What if Planet 9 is a Primordial Black Hole?) which cites this paper: K. Batygin, and M. E. Brown, Evidence for a distant giant planet in the solar system, AJ 151.2 (2016): 22. 1601.05438 Jun 9 at 17:25
• My question asks where the barycenter of the hypothetical object would be, not if it is a planet or black hole (which we don't know, not even if the object exists).
– John
Jun 9 at 17:53
• @John, planetmaker was asking for a reference to papers positing the hypothesis. Jun 9 at 17:56
• @John Thanks for the clarifying edit! Jun 11 at 14:31
The proposal is that the apparent planet 9 is, in fact, a primordial black-hole, with mass comparable to a planet and a diameter measured in centimeters. Such an object would be almost undetectable.
Its orbit would be identical to an equivalent sized planet.
In a two body system, obeying Newtonian gravity, both bodies will follow elliptical orbits, with the barycentre at their focus. The barycentre will be at rest, so it is reasonable to make a coordinate frame in which the barycenter is at (0,0) and the two objects orbit around it.
In the case of a three (or more) body system, the orbit won't be elliptical. So the notion of a point around which the body orbits becomes less clear. It will be affected by the gravity of all the other solar system bodies, and it's acceleration vector (in an inertial frame) will point towards the centre of mass of the other solar system bodies, but this point is would not be fixed.
• Fun aside: The first paper I linked to above includes an actual size illustration of the hypothetical 5M⊕ black hole (the picture is a 4.5cm radius black circle). Jun 9 at 18:08
If we have two masses $$M$$ and $$m$$, which are at distances $$r_1$$ and $$r_2$$ respectively from their barycentre, then $$Mr_1 = Mr_2$$
Let $$d$$ equal the total distance between the two bodies. That is, $$d = r_1 + r_2$$. Then $$r_1 = \frac{dM}{M+m}$$ where $$r_1$$ is the distance of the barycentre from the body with mass $$M$$.
The paper linked in the comments suggests a figure of 500 au for the distance between the Sun and the alleged Planet 9. Let $$M$$ be the mass of the Sun, and $$m$$ be the mass of Planet 9. Using units of Earth masses, we have $$M=333000$$ and $$m=5$$.
Plugging those numbers into our equation, we get $$r_1 = \frac{500×5}{333005} \approx 0.0075074$$ for the distance (in au) of the barycentre from the centre of the Sun.
Now, the solar radius (the nominal radius of its photosphere) is $$0.00465047$$ au. So the barycentre is well outside the Sun. It's roughly $$1.61$$ solar radii from the centre of the Sun, or $$\approx 427400$$ km above the surface of the photosphere.
As James K points out, to accurately calculate the orbit of a body in the Solar System we need to include the masses of all the other bodies. Of course, that's impractical. The best ephemerides are produced by JPL, and their Development Ephemeris series currently uses the masses and locations of all major Solar System bodies down to the 340 most significant asteroids.
As you can see here, the motion of the Solar System barycentre (SSB) is not simple. And its true location is not precisely known, mostly because we don't know the details of the mass distribution in the far outer reaches of the Solar System. In fact, recent updates to the JPL Horizons system have modified their value for the SSB.
From Horizons System News:
### April 12, 2021
The current catalog of 1.1 million asteroid and comet solutions are being refit for dynamical consistency with DE441 perturbations. The new solutions will filter into the database over the next days. Due to the addition of KBO mass in DE440/441, the SSB has shifted about 100 km
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2021-09-17 00:16:31
|
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|
https://www.quizover.com/key/terms/buffer-capacity-buffers-by-openstax
|
# 14.6 Buffers (Page 3/9)
Page 3 / 9
$\left(1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\left(1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)\phantom{\rule{0.2em}{0ex}}=9.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}M$
The concentration of NaOH is:
$\frac{9.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}M\phantom{\rule{0.2em}{0ex}}\text{NaOH}}{0.101\phantom{\rule{0.2em}{0ex}}\text{L}}\phantom{\rule{0.2em}{0ex}}=9.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}M$
The pOH of this solution is:
$\text{pOH}=\text{−log}\left[{\text{OH}}^{\text{−}}\right]\phantom{\rule{0.2em}{0ex}}=\text{−log}\left(9.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\right)\phantom{\rule{0.2em}{0ex}}=3.01$
The pH is:
$\text{pH}=14.00\phantom{\rule{0.2em}{0ex}}-\phantom{\rule{0.2em}{0ex}}\text{pOH}=10.99$
The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b).
Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 $×$ 10 −5 M HCl solution from 4.74 to 3.00.
Initial pH of 1.8 $×$ 10 −5 M HCl; pH = −log[H 3 O + ] = −log[1.8 $×$ 10 −5 ] = 4.74
Moles of H 3 O + in 100 mL 1.8 $×$ 10 −5 M HCl; 1.8 $×$ 10 −5 moles/L $×$ 0.100 L = 1.8 $×$ 10 −6
Moles of H 3 O + added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L $×$ 0.0010 L = 1.0 $×$ 10 −4 moles; final pH after addition of 1.0 mL of 0.10 M HCl:
$\text{pH}=\text{−log}\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\text{−log}\left(\phantom{\rule{0.2em}{0ex}}\frac{\text{total moles}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}}{\text{total volume}}\right)\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}=\text{−log}\left(\phantom{\rule{0.2em}{0ex}}\frac{1.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\text{mol}+1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{mol}}{101\phantom{\rule{0.2em}{0ex}}\text{mL}\left(\phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\text{L}}{1000\phantom{\rule{0.2em}{0ex}}\text{mL}}\right)\phantom{\rule{0.2em}{0ex}}}\right)\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}=3.00$
If we add an acid or a base to a buffer that is a mixture of a weak base and its salt, the calculations of the changes in pH are analogous to those for a buffer mixture of a weak acid and its salt.
## Buffer capacity
Buffer solutions do not have an unlimited capacity to keep the pH relatively constant ( [link] ). If we add so much base to a buffer that the weak acid is exhausted, no more buffering action toward the base is possible. On the other hand, if we add an excess of acid, the weak base would be exhausted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to exhaust all of the acid or base in a buffer to overwhelm it; its buffering action will diminish rapidly as a given component nears depletion.
The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion.
## Selection of suitable buffer mixtures
There are two useful rules of thumb for selecting buffer mixtures:
1. A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. [link] shows an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration.
2. Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7.
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light behaves as particle and as well as waves in some phenomena....
zille
list 10 gases and their IUPAC name
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in group atomic number?
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what structure?
of the above question.
Nakyanzi
Draw the Lewis structure of the following : Nitrate anion Nitrogen dioxide
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plea what are the best topics in chemistry to know
atomic theory . nuclear chemistry . chemical reaction. chemical equilibrium . organic chemistry. molecules concept 1 n 2. kinetic theory of gases.
Inusah
wat topics are the most important in biology
Josephine
cells 1 n 2, ecology. insects. genetics. plants. nutrition.
Inusah
organic chemistry
Dolly
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2019-01-20 18:51:01
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|
https://tonyladson.wordpress.com/tag/100-year-flood/
|
# 1% flood: binomial distribution, conditional probabilities
I previously wrote about considering the occurrence of 1% floods as a binomial distribution, this post extends that analysis to look at conditional probabilities. Some of the results are counter intuitive, at least to me, in that the risk of multiple 1% floods is larger than I would have guessed.
The probability of a 1% (1 in 100) annual exceedance probability (AEP) flood occurring in any year is 1%. This can be treated as the probability of a “success” in the binomial distribution, with the number of trials being the number of years. So the probability of having exactly one 1% flood in 100 years is
${100\choose 1}0.01^{1}\left( 1-0.01\right) ^{99} = 0.37$
In R this can be calculated as dbinom(x = 1, size = 100, prob = 0.01) or in excel =BINOM.DIST(1,100, 0.01, FALSE).
The cumulative distribution function of the binomial distribution is also useful for flood calculations.
What is the probability of 2 or more 1% floods in 100 years:
R: pbinom(q = 1, size = 100, prob = 0.01, lower.tail = FALSE) = 0.264
Excel: =1 - BINOM.DIST(1,100, 0.01, TRUE) = 0.264
We can check this by calculating the probability of zero or one flood in 100 years and subtracting that value from 1.
1 - (dbinom(x = 1, size = 100, prob = 0.01) + dbinom(x = 0, size = 100, prob = 0.01)) = 0.264
We can also do conditional probability calculations which could be useful for risk assessment scenarios.
What is the probability that exactly two 1% floods occur in 100 years given that at least one occurs?
$\Pr{(X = 2\mid X \ge 1)}$ =
dbinom(x = 2, size = 100, prob = 0.01)/pbinom(q = 0, size = 100, prob = 0.01, lower.tail = FALSE) = 0.291
What is the probability that at least two 1% floods occur in 100 years given that at least one occurs?
$\Pr{(X \ge 2\mid X \ge 1)}$ =
pbinom(q = 1, size = 100, prob = 0.01, lower.tail = FALSE)/pbinom(q = 0, size = 100, prob = 0.01, lower.tail = FALSE) = 0.416
We can also check this by simulation. This code generates the number of 1% floods in each of 100,000 100-year sequences. We can then count the number of interest.
set.seed(1969) # use a random number seed so the analysis can be repeated if necessary
floods = rbinom(100000,100, 0.01) # generate the number of 1% floods in each of 100,000, 100-year sequences
floods_subset = floods[floods >= 1] # Subset of sequences that have 1 or more floods
# Number of times there are two or more floods in the subset of 1 or more floods
sum(floods_subset >= 2) / length(floods_subset)
# 0.4167966
# or
sum(floods >= 2)/sum(floods >= 1)
#[1] 0.4167966
A slightly tricker situation is a question like: What is the probability of three or fewer floods in 100-years given there is more than one.
$\Pr{(X \le 3\mid X > 1)} = \Pr(X \le 3 \cap X > 1 )/\Pr( X > 1)$
floods_subset = floods[floods > 1] # Subset of sequences that have more than one flood
# Number of times there are three or fewer floods in the subset of more than one flood
sum(floods_subset ≤ 3) / length(floods_subset)
#[1] 0.9310957
# Or, for the exact value
# (Probability that X = 3 + Probability that X = 2)/(Probability that X > 1)
(dbinom(x = 3, size = 100, prob = 0.01) + dbinom(x = 2, size = 100, prob = 0.01))/ pbinom(q = 1, size = 100, prob = 0.01, lower.tail = FALSE)
#[1] 0.9304641
The probability of experiencing at least one 1% flood in 100-years is $1 - (1-0.01)^{100}$ = 0.634. How many years would we need to wait to have a 99% chance of experiencing a 1% flood?
$0.99 = 1-(1-0.1)^n$
$n=\frac{log(0.01)}{log(0.99)} = 458.2$. The next largest integer is 459.
We can also solve this numerically. In R the formula is 0.99 = pbinom(q=0, size = n, prob = 0.01), solve for n. Using the uniroot function gives n = 459 years (see below).
So all these areas subject to a 1% flood risk will flood eventually, but it may take a while.
f = function(n) {
n = as.integer(n) #n must be an integer
0.99 - pbinom(q = 0, size = n, prob = 0.01, lower.tail = FALSE)
}
# $root # [1] 458.4999 uniroot(f, lower = 100, upper = 1000) pbinom(q = 0, size = 459, prob = 0.01, lower.tail = FALSE) # [1] 0.990079 How many years before there is a >99% chance of experiencing more than one flood? This is one minus (the probability of zero floods + the probability of one flood). Let the number of years equal n. $1-((1-0.01)^n + n(0.01)(1-0.01)^{n-1}) = 0.99$. Solving for n gives 662 years # Calculated Risks I’ve been reading Calculated risks, a book by Gerd Gigerenzer. There is lots to make you think. Here’s an example: Your DNA matches a trace found on a victim of a crime. The court calls an expert wetness who gives this testimony: “the probability that this match has occurred by chance is 1 in 100,000.” A chance match sounds very unlikely. So does that mean you will be found guilty? As Gigerenzer points out, the expert could have phrased the same information as: “Out of every 100,000 people, one will show a match.” Now we can see that, in a city like Melbourne, with a population of about 4 million people, about 40 people will show a match. The point is, that the probability that you committed the crime is not the same as the probability of a match. The probability that you committed the crime, given the DNA match, is only 1 in 40 if there is no other evidence and the potential perpetrators include anyone living in Melbourne. Much of the book is about finding clearer ways of talking about, and thinking about, probability and risk. An example: The probability that a woman of age 40 has breast cancer is about 1 percent. If she has breast cancer, the probability that she will test positive on a screening mammogram is about 90 percent. If she does not have breast cancer, the probability that she will nevertheless test positive is 9 percent. What are the chances that a woman who tests positive actually has breast cancer? This is the way Gigerenzer presents the solution, using what he calls, natural frequencies. Consider 10,000 women, 1% have cancer so that is 100 women. Of these, 90% will return positive tests (i.e. 90 women with cancer will test positive). Of the 9900 without cancer 9% will return positive test or 891 women. So there are 891 + 90 = 981 women with positive tests of which 90 have cancer. So the chance that a woman with a positive test has cancer is 90/981, about 1 in 10. Different ways of getting a positive test result We can grind through a problem like this with Bayes theorem but the natural frequency approach makes it easier to understand the problem intuitively. Would the natural frquency idea help with the communication of hydrologic risks? The latest guidance from Engineers Australia is that we should refer to the 1% annual exceedance probability (AEP) flood rather than the 100-year average recurrence interval (ARI) flood. This moves away from expressing probabilities as natural frequencies but the rationale is that the ARI terminology is also confusing. Many people think that only one “100 year flood” can occur every 100 years (see this forum about communicating flood risk in Christchurch, New Zealand) When discussing flood risk, we could say something like: “Think of your house and 99 other houses, spread all over Australia, that have the same risk of flooding. On average, one of these houses will flood every year”. The risk is also the same as rolling a 100-sided die once a year. If your number comes up, you get flooded. On average your number will come up every 100 throws but it could come up any throw or several times in a row. 100-sided die (source) Gigerenzer also talks about absolute risk reduction and relative risk reduction. If a house is prone to flooding in a 1 in 10 year ARI event (a 10% AEP event) and we build a levee so it is now protected up to the 1 in 100 year ARI event (a 1% AEP event), then the absolute risk reduction is 0.1 – 0.01 = 10% – 1% = 9%. The relative risk reduction is absolute risk reduction divided by the risk prior to treatment i.e. 0.09/0.1 = 90%. So if you were to seeking to attract funds for a levee scheme, what sounds better: A 9% absolute risk reduction or a 90% relative risk reduction? # 100-year flood: Negative Binomial distribution The negative Binomial distribution relates to independent trials and provides information on the probability of the number failures before a certain number of successes. Continuing with our flood examples, the negative Binomial distribution can be used to determine the probability of the number of flood free years before a certain number of floods occur. If Z is the number of flood free years, before r floods, and if a flood has a probability of occurrence of p in any year then: $Z \sim nbinom(r, p)$ Example A retirement village is vulnerable to the 100-year flood. If there are 3 or more floods in the next 20-years the political pressure will be such that the village will be relocated. What is the probability that this will occur? Flooded retirement village (http://goo.gl/gtIdcz) We need the probability that $Z + 3 \le 20$ The probability can be calculated using R as pnbinom(17, 3, 0.01) = 0.001 So there isn’t much chance this will happen. We can also calculate the probability using the Binomial distribution as 1 minus the probability of 2 or fewer flood in 20 years. 1 - pbinom(2, 20, 0.01) = 0.001 The expected value (mean) of Z, the number of flood free years before r floods, is: $\frac{r(1-p)}{p}$ Example What is the average number of years before the retirement village will experience 3 floods? $\frac{3(1-0.01)}{0.01} = 297$ This is consistent with the average number of years between 1% floods being 100 years. On average we have 99 flood free years for each event, so in 300 years, on average we will have 297 flood free years and 3 flood events. Further reading Jones, O, Maillardet, R. and Robinson, A. (2009) Scientific programming and simulation using R. CRC Press # 100-year flood: Geometric distribution – Number of years to the next flood Here we are interested in the probability distribution of the number of years, Y, that will elapse before the next 100-year flood occurs. The number of independent trials up, but not including, the first success has a Geometric distribution. Applying this to flooding means that the number of years to wait before a 100-year flood, has a geometric distribution with the probability parameter equal to 0.01: $Y \sim \mathrm{geom}(0.01)$ Example You have a 5-year assignment to a flood-prone mining camp. What is the probability that you will experience a 100-year flood while you are living in the camp? We need the probability that we will wait 4 or fewer years before a 100-year flood. The probability that you will need to wait n years or fewer is: $\displaystyle \sum _{y=0}^{n} p(1-p)^y$ In this case, we interested in: $\displaystyle \sum _{y=0}^{4} 0.01(1-0.01)^y = 0.049$ That is, there is a 4.9% chance that the camp will experience a 100-year flood while we are living there. In R pgeom(4, 0.01) = 0.049 We can also calculate this using the Binomial distribution as 1 minus the probability of zero floods in 5 years. 1 - pbinom(0, 5, 0.01) = 0.049 The probability that you will be ok for 4 years but be flooded in the 5th year. $P(Y = y) = p(1-p)^n$ In this case: $0.01(1-0.01)^4 = 0.0096$ Which is 0.96% In R dgeom(4, 0.01) = 0.0096 The probability of being flooded increases as the length of our assignment increases as shown in the graph below. Probability of being flooded as a function of the number of years living in the camp We can also plot the probability of being safe for a certain number of years and then being flooded in the next year. Probability of being safe for a certain number of years and then being flooded in the next year Notice that the probability of being flooded in the first year (being safe for zero years) is actually the highest. Steven Pinker has a nice discussion of this effect in his book ‘The better angels of our nature‘ about 17 pages into Chapter 5. His discussion uses lightning strikes which I’ve rephrased to floods here. Suppose that floods are random: every year, the chance of a flood is the same, 1%. Your house was flooded this year. When is it most likely you will be flooded again? The answer is next year. That probability, to be sure, is not very high, 1%. Now think about the chance you will be safe for a year and then experience a flood the year after next. For that to happen, two things have to take place. First, flooding will have to occur the year after next, with a probability of 1%. Second, flooding will not occur next year. To calculate that probability you have to multiply the chance there will be no flood next year (0.99 or 1 minus 0.01) by the chance there will be a flood the year after (1% or 0.01). The probability of these two events occurring is 0.0099, a bit lower than the chance of a flood next year. What is the chance of being safe for 99 years and then getting a flood? The probability is (1 – 0.99)99 x 0.01 = 10-200, a very small number. So, if you’ve just experienced a 100-year flood, when are you most likely to experience the next one? After 100-years? Nope. The probability is highest that it will happen again next year. The idea that we’ve had a big flood and will now be safe for a while, is a fallacy. We can use a random variable drawn from a geometric distribution to simulate the number of safe years before the next 100-year flood occurs. The mean of a Geometric distribution is $\frac{(1-p)}{p}$ In this case, p = 0.01 so the mean number of safe years before a 100-year flood is $\frac{(1-0.01)}{0.01} = 99$ On average, we will be safe for 99 years. This is consistent with the idea that the 100-year flood will occur on average once every 100-years. This may be comforting, but the distribution of the number of safe years isn’t nicely bunched around 99. The median number of safe years can be calculated using the formula: $\left( \frac{-1}{log_2(1-p)} \right) - 1$ In this case, with p = 0.01, the median number of years before the next 1% flood is 68 and the mode (most common value) is zero (i.e. the flood will occur next year). The probability mass values for $Y \sim \mathrm{geom}(0.01)$ are shown below. Geometric probability mass function (p = 0.01). The probability that a flood will occur in the year after the indicated number of years without a flood Simulating the number of years between floods produces the results shown below (100 simulations in a 10 x 10 grid). The mean and median are near their theoretical values (i.e. the mean is about 99) but the results show that for many simulations there are a small number of years between floods, while in a few cases there are a very large number of years. Simulation of the number of years until a 1% flood occurs. One hundred different random values drawn from a Geometric distribution with p = 0.01 We see the same pattern if we simulate a large number of years and randomly determine if a 100-year flood will occur in each year. A 10,000 year simulation is shown below (100 x 100 year grid). Blue squares show the occurrence of 100-year floods. A 10,000 year simulation (100 x 100 year grid). Blue squares show the occurrence of 100-year floods The mean and median time between floods are close to their theoretical values, but as we’ve seen before, there are many small gaps between floods and a few very long flood-free periods. Where dots are close together show periods where floods are clustered. This type of counter intuitive clustering in random data has been noted in many other situations including the bombing of London during WWII and the patterns produced by glowworms. See Pinker (2011) and Gould (1991) for further discussion. Pinker, S. (2012) The better angels of our nature. London: Penguin Books. pp 245-247 Gould, S. J. (1991) Glow, big glowworm. bully for brontosaurus. New York: Norton (as cited by Pinker) For R code see this gist. # 100-year flood: Poisson distribution In the previous post we considered the occurrence of 100-year floods as a Binomial random variable. We can do similar analysis using the Poisson distribution. As noted by Jones et al. “For n large, the Binom(n,p) distribution is approximately Pois(np)”. In the examples we’ve been looking at, n is reasonably large i.e. 100 (years). So binom(100, 0.01) is approximately pois(100 x 0.01) = pois(1). Lets try it: Consider the probability of 1, 100-year flood in 100 years. The poisson distribution can be expressed as follows: $P(X = x) = \frac{e^{-\lambda} \lambda^x} {x!}$ In this case, $x = 1, \lambda = 1$ $P(X = x) = 0.367894$ So, using the Poisson distribution the probability of 1, 100-year flood in 100 years is 0.367894. Using the Binomial distribution it is 100 x 0.01 x (1-0.01)^99 = 0.3697296. i.e. pretty close. The approximation gets better for larger n. Similar to the Binomial example, we can use a Poisson random variable to simulate flood occurrence – although the Binomial approach will be more accurate. The figure below shows a simulation of 100, 100-year sequences in a 10 x 10 grid. You can see that most of the time there is 0 or 1 flood but occasionally there are a lot more. Simulation of the number of 100-year floods occurring in 100 (10 x 10), 100-year sequences R code (also available as a gist) dpois(0,1) # prob of zero 100-year floods in 100 years dpois(0,1) # prob of zero 100-year floods in 100 years [1] 0.3678794 dpois(1,1) # prob of one 100-year floods in 100 years [1] 0.3678794 dpois(2,1) # prob of two 100-year floods in 100 years [1] 0.1839397 # simulation library(ggplot2) df <- expand.grid(x = 1:10, y = 1:10) df$z <- rbinom(100, 100, 0.01) # could also use rpois(100,1)
ggplot(data=df, aes(x,y)) + geom_tile(aes(fill=z)) +
scale_fill_gradient(low="green", high = "red", name = 'Floods') +
geom_text(data = df, aes(x, y, label = z)) +
scale_x_continuous(breaks = NULL) +
scale_y_continuous(breaks = NULL) +
xlab('') +
ylab('')
set.seed(2000)
rpois(10, 1)
[1] 0 1 0 1 2 1 3 1 2 1
# number of 100-year floods in each of 10 100-year sequences
# 100-year flood: Binomial distribution
The Binomial distribution applies when there are trials with two outcomes (‘success’ and ‘failure’). The standard example is coin tossing; a coin can come up heads or tails and the probability of success e.g. a head, is the same in every trial. The probability of tossing a head with an unbiased coin is 50% but the binomial distribution also applies where the probability of success differs from 50%. In the case of a 100-year flood, the probability of ‘success’ – having a 100-year flood, is 1% in any year.
The Binomial distribution can be used to calculate the probability of experiencing a certain number of 100-year floods in a specified number of years. The probability of experiencing 0, 1, 2, 3, 4 or 5 100-year floods in 100 years is shown in the table and figure below. Having one flood is the most likely outcome but there is also a good chance of having zero, or 2, or more. There is a 63.4% change of having at least 1.
Number of floods Probability 0 36.6% 1 37.0% 2 18.5% 3 6.1% 4 1.4% 5 0.3% At least 1 63.4%
Probability of experiencing 0, 1, 2, 3, 4 or 5 100-year (1%) floods in 100 years
Lets do a sample calculation. The probability of k successes from n trials, where the probability of success from an individual trial is p, is given by:
${n\choose k}p^{k}\left( 1-p\right) ^{n-k}$
So the probability of exactly two, 100-year floods in 100 years is:
${100\choose 2}0.01^{2}\left( 1-0.01\right) ^{98} = 0.185$ or 18.5%
The probability of at least one flood is the same as 1 – the probability of zero floods. So the probability of at least one 100-year flood in 100 years is:
$1-(1-0.01)^{100} = 0.6339677 \approx 1 - \frac{1}{e}$
We can also use a Binomial random variable to simulate the occurrence of floods. Check the help for ?rbinom.
To simulate the number of 100-year floods in 100 years:
rbinom(1, 100, 0.01)
To do this in 10, 100-year sequences:
rbinom(10, 100, 0.01)
See the code below for an example.
R code (also available as a gist)
dbinom(0,100,0.01) # zero 1% floods in 100 years
dbinom(1,100,0.01)
dbinom(2,100,0.01)
dbinom(3,100,0.01)
dbinom(4,100,0.01)
dbinom(5,100,0.01) # five 1% floods in 100 years
# at least 1 1% flood in 100 years
1 - pbinom(0,100, 0.01)
# Sample calculation
# Exactly 2 1% floods in 100 year
choose(100,2) * 0.01^2 * (1 - 0.01)^98
#[1] 0.1848648
par(oma = c(1, 2, 0, 0))
barplot(dbinom(0:5,100,0.01),
ylim = c(0, 0.4),
las = 1,
names.arg = 0:5,
ylab = 'Probability',
xlab = 'Number of 100-year floods in 100 years')
set.seed(2000)
rbinom(10, 100, 0.01)
#[1] 0 1 0 1 2 1 3 1 2 1
# number of 100-year floods in each of 10 100-year sequences
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2020-04-07 17:38:44
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https://www.physicsforums.com/threads/finding-force-n-wood-has-against-the-bullet.535699/
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# Homework Help: Finding force(N) wood has against the bullet
1. Oct 1, 2011
1. The problem statement, all variables and given/known data
22.0-caliber rifle bullet of mass 1.80 g, traveling at 320 m/s strikes a block of soft wood, which it penetrates to a depth of 0.140 m. The block of wood is clamped in place and doesn't move. Assume a constant retarding force.
What force, in newtons, does the wood exert on the bullet?
2. Relevant equations
3. The attempt at a solution
Long story short, acceleration is -365714.29 m/s^2
Is it really as simple as .0018*-365714.29 = -658.29 N ? Or maybe 658.29 N since the force of the wood on the bullet is opposite of the bullet on the wood?
Since Fnet = ma
Last edited: Oct 1, 2011
2. Oct 1, 2011
### PeterO
It is that simple [I hope you a value is correct; I haven't checked]
The question referred to a retarding force, so easiest just to answer "The retarding force is 658N" watch the significant figures in your final answer - you use all the figures i your calculations]
3. Oct 1, 2011
### Andrew Mason
You might find it easier to use energy: the work done by the wood on the bullet is equal to the loss of kinetic energy. So:
$$F = KE/d$$
AM
4. Oct 1, 2011
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2018-06-20 08:15:19
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https://www.hpmuseum.org/forum/printthread.php?tid=11195
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My 10 Step Program - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: Not HP Calculators (/forum-7.html) +--- Forum: Not remotely HP Calculators (/forum-9.html) +--- Thread: My 10 Step Program (/thread-11195.html) My 10 Step Program - markhen - 08-09-2018 11:58 PM As a confirmed calcaholic, I've been trying to limit myself to just these, in this order: 1. HP LED - any, all, multiples (successful!) 2. HP LCD - the "more interesting ones" (RPN, RPL - partially successful!) 3. Other LED programmables - but not TI (see below) 4. Other LED scientifics - particularly, but not exclusively, RPN 5. Other interesting LEDs (like Heath 1401 - 3x of those!) 6. TI programmables (58, 59...) Had to share my latest find - in category 3 - a PERFECT, AS NEW Litronix 2290 LED Programmable! This beauty has a large LED display, with "digit for DP" like the HP Classics, square root (divide/+ !!), one memory - and - 10 KEYSTROKE PROGRAMMABILITY!! Almost useless! But SO beautiful!!! RE: My 10 Step Program - Thomas Klemm - 08-10-2018 01:55 AM (08-09-2018 11:58 PM)markhen Wrote: (…) square root (divide/+ !!), one memory - and - 10 KEYSTROKE PROGRAMMABILITY!! I wondered how Victor T. Toth implemented the Gamma function with this. Quote:Almost useless! At least you can calculate the amount of time it takes for an object to fall from a given height: Code: C/ON L += += ÷ RM ÷ += S Quote:But SO beautiful!!! Nice find. Congratulations! RE: My 10 Step Program - markhen - 08-10-2018 04:46 PM I read Victor's page and that is what caused me to have an auction site search active! Victor's program makes use of that "hidden" square root feature (divide followed by +=), nice. Strangely Victor's text answer of 4.67 seconds does not agree with the program or the equation - which yield 4.52 seconds ;-) RE: My 10 Step Program - Thomas Klemm - 08-10-2018 05:13 PM (08-10-2018 04:46 PM)markhen Wrote: Strangely Victor's text answer of 4.67 seconds does not agree with the program or the equation - which yield 4.52 seconds ;-) I assume he made a typo and used $$9.18\frac{m}{s^2}$$ instead of $$9.81\frac{m}{s^2}$$ for the gravitational acceleration g. Cheers Thomas RE: My 10 Step Program - Eddie W. Shore - 08-12-2018 02:11 AM 10 steps? Wow! The calculator sure does look nice. RE: My 10 Step Program - Thomas Klemm - 08-12-2018 01:00 PM (08-10-2018 04:46 PM)markhen Wrote: Strangely Victor's text answer of 4.67 seconds does not agree with the program or the equation - which yield 4.52 seconds ;-) Meanwhile he fixed it.
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2020-11-26 16:05:06
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https://desvl.xyz/2020/11/27/vector-bundle-1/
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## Motivation
Direction is a considerable thing. For example take a look at this picture (by David Gunderman):
The position of the red ball and black ball shows that this triple of balls turns upside down every time they finish one round. This wouldn’t happen if this triple were on a normal band, which can be denoted by $S^1 \times (0,1)$. What would happen if we try to describe their velocity on the Möbius band, both locally and globally? There must be some significant difference from a normal band. If we set some move pattern on balls, for example let them run horizontally or zig-zagly, hopefully we get different set of vectors. those vectors can span some vector spaces as well.
## A Formal Construction
Here and in the forgoing posts, we will try to develop purely formally certain functorial constructions having to do with vector bundles. It may be overly generalized, but we will offer some examples to make it concrete.
Let $M$ be a manifold (of class $C^p$, where $p \geq 0$ and can be set to $\infty$) modeled on a Banach space $\mathbf{E}$. Let $E$ be another topological space and $\pi: E \to M$ a surjective $C^p$-morphism. A vector bundle is a topological construction associated with $M$ (base space), $E$ (total space) and $\pi$ (bundle projection) such that, roughly speaking, $E$ is locally a product of $M$ and $\mathbf{E}$.
We use $\mathbf{E}$ instead of $\mathbb{R}^n$ to include the infinite dimensional cases. We will try to distinguish finite-dimensional and infinite-dimensional Banach spaces here. There are a lot of things to do, since, for example, infinite dimensional Banach spaces have no countable Hamel basis, while the finite-dimensional ones have finite ones (this can be proved by using the Baire category theorem).
Next we will show precisely how $E$ locally becomes a product space. Let $\mathfrak{U}=(U_i)_i$ be an open covering of $M$, and for each $i$, suppose that we are given a mapping
satisfying the following three conditions.
VB 1 $\tau_i$ is a $C^p$ diffeomorphism making the following diagram commutative:
where $pr$ is the projection of the first component: $(x,y) \mapsto x$. By restricting $\tau_i$ on one point of $U_i$, we obtain an isomorphism on each fiber $\pi^{-1}(x)$:
VB 2 For each pair of open sets $U_i$, $U_j \in \mathfrak{U}$, we have the map
to be a toplinear isomorphism (that is, it preserves $\mathbf{E}$ for being a topological vector space).
VB 3 For any two members $U_i$, $U_j \in \mathfrak{U}$, we have the following function to be a $C^p$-morphism:
REMARKS. As with manifold, we call the set of 2-tuples $(U_i,\tau_i)_i$ a trivializing covering of $\pi$, and that $(\tau_i)$ are its trivializing maps. Precisely, for $x \in U_i$, we say $U_i$ or $\tau_i$ trivializes at $x$.
Two trivializing coverings for $\pi$ is said to be VB-equivalent if taken together they also satisfy conditions of VB 2 and VB 3. It’s immediate that VB-equivalence is an equivalence relation and we leave the verification to the reader. It is this VB-equivalence class of trivializing coverings that determines a structure of vector bundle on $\pi$. With respect to the Banach space $\mathbf{E}$, we say that the vector bundle has fiber $\mathbf{E}$, or is modeled on $\mathbf{E}$.
Next we shall give some motivations of each condition. Each pair $(U_i,\tau_i)$ determines a local product of ‘a part of the manifold’ and the model space, on the latter of which we can deploy the direction with ease. This is what VB 1 tells us. But that’s far from enough if we want our vectors fine enough. We do want the total space $E$ to actually be able to qualify our requirements. As for VB 2, it is ensured that using two different trivializing maps will give the same structure of some Banach spaces (with equivalent norms). According to the image of $\tau_{ix}$, we can say, for each point $x \in X$, which can be determined by a fiber $\pi^{-1}(x)$ (the pre-image of $\tau_{ix}$), can be given another Banach space by being sent via $\tau_{jx}$ for some $j$. Note that $\pi^{-1}(x) \in E$, the total space. In fact, VB 2 has an equivalent alternative:
VB 2’ On each fiber $\pi^{-1}(x)$ we are given a structure of Banach space as follows. For $x \in U_i$, we have a toplinear isomorphism which is in fact the trivializing map:
As stated, VB 2 implies VB 2’. Conversely, if VB 2’ is satisfied, then for open sets $U_i$, $U_j \in \mathfrak{U}$, and $x \in U_i \cap U_j$, we have $\tau_{jx} \circ \tau_{ix}^{-1}:\mathbf{E} \to \mathbf{E}$ to be an toplinear isomorphism. Hence, we can consider VB 2 or VB 2’ as the refinement of VB 1.
In finite dimensional case, one can omit VB 3 since it can be implied by VB 2, and we will prove it below.
(Lemma) Let $\mathbf{E}$ and $\mathbf{F}$ be two finite dimensional Banach spaces. Let $U$ be open in some Banach space. Let
be a $C^p$-morphism such that for each $x \in U$, the map
given by $f_x(v)=f(x,v)$ is a linear map. Then the map of $U$ into $L(\mathbf{E},\mathbf{F})$ given by $x \mapsto f_x$ is a $C^p$-morphism.
PROOF. Since $L(\mathbf{E},\mathbf{F})=L(\mathbf{E},\mathbf{F_1}) \times L(\mathbf{E},\mathbf{F_2}) \times \cdots \times L(\mathbf{E},\mathbf{F_n})$ where $\mathbf{F}=\mathbf{F_1} \times \cdots \times \mathbf{F_n}$, by induction on the dimension of $\mathbf{F}$ and $\mathbf{E}$, it suffices to assume that $\mathbf{E}$ and $\mathbf{F}$ are toplinearly isomorphic to $\mathbb{R}$. But in that case, the function $f(x,v)$ can be written $g(x)v$ for some $g:U \to \mathbb{R}$. Since $f$ is a morphism, it follows that as a function of each argument $x$, $v$ is also a morphism, Putting $v=1$ shows that $g$ is also a morphism, which finishes the case when both the dimension of $\mathbf{E}$ and $\mathbf{F}$ are equal to $1$, and the proof is completed by induction. $\blacksquare$
To show that VB 3 is implied by VB 2, put $\mathbf{E}=\mathbf{F}$ as in the lemma. Note that $\tau_j \circ \tau_i^{-1}$ maps $U_i \cap U_j \times \mathbf{E}$ to $\mathbf{E}$, and $U_i \cap U_j$ is open, and for each $x \in U_i \cap U_j$, the map $(\tau_j \circ \tau_i^{-1})_x=\tau_{jx} \circ \tau_{ix}^{-1}$ is toplinear, hence linear. Then the fact that $\varphi$ is a morphism follows from the lemma.
## Examples
### Trivial bundle
Let $M$ be any $n$-dimensional smooth manifold that you are familiar with, then $pr:M \times \mathbb{R}^n \to M$ is actually a vector bundle. Here the total space is $M \times \mathbb{R}^n$ and the base is $M$ and $pr$ is the bundle projection but in this case it is simply a projection. Intuitively, on a total space, we can determine a point $x \in M$, and another component can be any direction in $\mathbb{R}^n$, hence a vector.
We need to verify three conditions carefully. Let $(U_i,\varphi_i)_i$ be any atlas of $M$, and $\tau_i$ is the identity map on $U_i$ (which is naturally of $C^p$). We claim that $(U_i,\tau_i)_i$ satisfy the three conditions, thus we get a vector bundle.
For VB 1 things are clear: since $pr^{-1}(U_i)=U_i \times \mathbb{R}^n$, the diagram is commutative. Each fiber $pr^{-1}(x)$ is essentially $(x) \times \mathbb{R}^n$, and still, $\tau_{jx} \circ \tau_{ix}^{-1}$ is the identity map between $(x) \times \mathbb{R}^n$ and $(x) \times \mathbb{R}^n$, under the same Euclidean topology, hence VB 2 is verified, and we have no need to verify VB 3.
### Möbius band
First of all, imagine you have embedded a circle into a Möbius band. Now we try to give some formal definition. As with quotient topology, $S^1$ can be defined as
where $I$ is the unit interval and $0 \sim_1 1$ (identifying two ends). On the other hand, the infinite Möbius band can be defined by
where $(0,v) \sim_2 (1,-v)$ for all $v \in \mathbb{R}$ (not only identifying two ends of $I$ but also ‘flips’ the vertical line). Then all we need is a natural projection on the first component:
And the verification has few difference from the trivial bundle. Quotient topology of Banach spaces follows naturally in this case, but things might be troublesome if we restrict ourself in $\mathbb{R}^n$.
### Tangent bundle of the sphere
The first example is relatively rare in many senses. By $S^n$ we mean the set in $\mathbb{R}^{n+1}$ with
and the tangent bundle can be defined by
where, of course, $\mathbf{x} \in S^n$ and $\mathbf{y} \in \mathbb{R}^{n+1}$. The vector bundle is given by $pr:TS^n \to S^n$ where $pr$ is the projection of the first factor. This total space is of course much finer than $M \times \mathbb{R}^n$ in the first example. Each point in the manifold now is associated with a tangent space $T_x(M)$ at this point.
More generally, we can define it in any Hilbert space $H$, for example, $L^2$ space:
where
The projection is natural:
But we will not cover the verification in this post since it is required to introduce the abstract definition of tangent vectors. This will be done in the following post.
## There are still many things remain undiscovered
We want to study those ‘vectors’ associated to some manifold both globally and locally. For example we may want to describe the tangent line of some curves at some point without heavily using elementary calculus stuff. Also, we may want to describe the vector bundle of a manifold globally, for example, when will we have a trivial one? Can we classify the manifold using the behavior of the bundle? Can we make it a little more abstract, for example, consider the class of all isomorphism bundles? How do one bundle transform to another? But to do this we need a big amount of definitions and propositions.
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2021-01-26 21:39:16
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https://www.physicsforums.com/threads/lift-a-weight-up-with-an-em-switch.907495/
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Lift a weight up with an EM switch
1. Mar 13, 2017
The_Thinker
Hello,
My question is simple. Let's suppose you have an electromagnetic coil with a magnetic weight in the center of it. This coil is energized, and the weight is lifted up vertically.
Now, my weight is 30 kgs. I want to lift it to a height of 1 meter. How do I calculate how much power is required to do this? How do I calculate it in terms of Watts?
For the mods: This is not a homework question. I am trying to understand something.
2. Mar 13, 2017
Staff: Mentor
There is no easy relation for this, it depends on your coil, the weight, and the overall geometry.
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2017-10-17 10:50:11
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https://www.studysmarter.us/textbooks/math/essential-calculus-early-transcendentals-2nd/derivatives/q55e-a-table-of-values-forfgfprime-and-gprime-is-given-a-ifh/
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Suggested languages for you:
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Q55E
Expert-verified
Found in: Page 121
### Essential Calculus: Early Transcendentals
Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280
# A table of values for$$f,g,{f^\prime }$$, and $${g^\prime }$$is given.(a) If$$h(x) = f(g(x))$$, find$${h^\prime }(1)$$.(b) If$$H(x) = g(f(x))$$, find$${H^\prime }(1)$$.
(a) If $$h(x) = f(g(x))$$, $${h^\prime }(1) = 30$$.
(b) If $$H(x) = g(f(x))$$, $${H^\prime }(1) = 36$$.
See the step by step solution
## Step 1: Given Information.
The table of values for $$x = 1,2,3$$.
## Step 2: Definition of chain rule.
Chain rule is a rule which defines a way to find the derivative of a given function.
$$\frac{{d(f(g(x)))}}{{dx}} = \frac{{d(f(g(x)))}}{{d(g(x))}} \cdot \frac{{d(g(x))}}{{dx}}$$
## Step 3:Use chain rule for differentiation for $$h(x) = f(g(x))$$.
State the chain rule.
$$\frac{{d(f(g(x)))}}{{dx}} = \frac{{d(f(g(x)))}}{{d(g(x))}} \cdot \frac{{d(g(x))}}{{dx}}$$
Here,
$${h^\prime }(x) = {f^\prime }(g(x)) \cdot {g^\prime }(x)$$
Substitute $$x = 1$$ and then substitute the tabular values to the obtained answer.
$$\begin{array}{c}{h^\prime }(1) &=& {f^\prime }(g(1)) \cdot {g^\prime }(1)\\{h^\prime }(1) &=& {f^\prime }(2) \cdot 6\\{h^\prime }(1) &=& 5 \cdot 6\\ &=& 30\end{array}$$
## Step 4: Use chain rule for differentiation for $$H(x) = g(h(x))$$.
State the chain rule.
$$\frac{{d(f(g(x)))}}{{dx}} = \frac{{d(f(g(x)))}}{{d(g(x))}} \cdot \frac{{d(g(x))}}{{dx}}$$
Here,
$${H^\prime }(x) = {g^\prime }(f(x)) \cdot {f^\prime }(x)$$
Substitute $$x = 1$$ and then substitute the tabular values to the obtained answer.
$$\begin{array}{c}{H^\prime }(1) &=& {g^\prime }(f(1)) \cdot {f^\prime }(1)\\{H^\prime }(1) &=& {g^\prime }(3) \cdot 4\\{H^\prime }(1) &=& 9 \cdot 4\\ &=& 36\end{array}$$
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2023-03-23 01:50:08
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http://worksheets.tutorvista.com/trigonometric-graph-worksheet.html
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# Trigonometric Graph Worksheet
Trigonometric Graph Worksheet
• Page 1
1.
Which of the following is a point on the curve $y$ = cot $x$?
a. ($\frac{\pi }{6}$, $\frac{1}{\sqrt{3}}$) b. (0, 0) c. ($\frac{\pi }{3}$, $\frac{1}{\sqrt{3}}$) d. ($\frac{\pi }{2}$, 1)
#### Solution:
f(x) = cot x
Since f(0) = cot 0 is not defined ,
f(π6) = cot(π6) = 3 ,
f(π3) = cot(π3) = 13
f(π2) = cot(π2) = 0 ,
The graph of cot x passes through the points (π6, 3), (π3, 13) and (π2, 0)
So, the point (π3, 13) is on the curve y = cot x.
2.
The graph of cos $x$
a. Passes through the origin b. Does not pass through ($\frac{\pi }{4}$, 0) c. Passes through the point ($\frac{\pi }{2}$, 1) d. Does not pass through the origin
#### Solution:
f(x) = cos x
Since f(0) = cos 0 = 1,
f( π2) = cos(π2) = 0
cos x is passing through (0, 1), (π2, 0).
So, graph of cos x does not pass through origin (0, 0).
3.
Which of the following is a point on the curve $y$ = tan $x$?
a. (0, 1) b. (- $\frac{\pi }{2}$, 1) c. ($\frac{\pi }{2}$, 1) d. ($\frac{\pi }{4}$, 1)
#### Solution:
f(x) = tan x
Since f(0) = tan 0 = 0,
f( π4) = tan(π4) = 1
f( π2) = tan(π2) is not defined.
f(- π2) = tan(- π2) is not defined.
The curve passes through (0, 0), (π4, 1)
So, (π4, 1) is a point on y = tan x.
4.
Through which of the following points does the graph of sin $x$ pass?
a. ($\frac{\pi }{6}$, $\frac{\sqrt{3}}{2}$) b. ($\frac{\pi }{6}$, $\frac{1}{2}$) c. ($\frac{\pi }{2}$, 0) d. (0, -1)
#### Solution:
f(x) = sin x
Since f(0) = sin 0 = 0 ,
f(π6) = sin(π6) = 12 and
f(π2) = sin π2 = 1.
The graph of sin x passes through the points (0, 0), (π6, 12) and (π2, 1).
5.
Which of the following is correct?
a. The graph of sin $x$ passes through (0, 0) b. The graph of sin $x$ does not pass through the origin. c. The graph of sin $x$ does not pass through ($\frac{\pi }{2}$, 1) d. The graph of sin $x$ does not pass through (-$\frac{\pi }{2}$, -1)
#### Solution:
The graph of sinx passes through the points (0, 0), (π2, 1), (- π2, -1).
6.
The $y$ - intercept of the graph $y$ = sin $x$ is
a. -1 b. 2 c. 1
#### Solution:
Substitute x = 0 in y = sin x to get its y- intercept.
y = sin 0 = 0
[Substitute x = 0.]
So, y-intercept of the graph y = sin x is 0.
7.
The zeros of the graph $y$ = cos $x$ are existing at $x$ =
a. $\frac{n\pi }{2}$ for all integer values of $n$ b. $n$π for all integer values of $n$ c. (2$n$ + 1) $\frac{\pi }{2}$ for all real values of $n$ d. (2$n$ + 1) π for all integer values of $n$
#### Solution:
The zeros of y = sin x are the solutions of sin x = 0
The solutions of cos x = 0 are .....- 3π , - 2π , - π , 0, π , 2π , 3π , ..... that is the multiples of π which are in short n π for all integer values of n.
8.
The $y$ - intercept of $y$ = tan $x$ is:
a. 2 b. -1 c. 1
#### Solution:
Substitute x = 0 in y = tan x to get its y-intecept.
y = tan 0 = 0
[Substitute x = 0.]
So, y-intercept of the graph y = tan x is 0.
9.
Write the amplitude of $y$ = 5sin($\frac{x}{4}$).
a. $\frac{1}{3}$ b. 5 c. 4
#### Solution:
The amplitude of y = bsin(xa) is | b |
[Definition.]
On comparing y = 5sin (x4) with y = bsin (xa) we have b = 5, a = 4
So, the amplitude of y = 5in(x3) is | b | = | 5 | = 5
10.
Evaluate: $\underset{x\to 0}{\mathrm{lim}}$
a. 23 b. $\frac{1}{9}$ c. $\frac{23}{9}$ d. $\frac{23}{3}$
#### Solution:
limx0 46x21-cos 6x
[Undefined at x = 0.]
= limx0 46x22sin2 3x
[Use 1 - cos 2x = 2sin2 x.]
= 23 limx0 1sin2 3xx2
[Divide both the numerator and the denominator by x2.]
= 23 1(limx0sin 3xx)2
[Use quotient law of limits.]
= 23 1(3)2 = 23 / 9
[Use limx0 sin kxx = k.]
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2017-07-21 00:45:32
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http://math.stackexchange.com/questions/901702/finding-multivariable-limit
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# Finding multivariable limit
I would like to find the following limit $$\lim_{(x,y,z)\to(0,0,0)}\frac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}.$$
It looks like it would be zero since if we put $M=\max\{x,y,z\}$ and $m=\min\{x,y,z\}$, then $$\Big|\frac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}\Big| \leq \Big|\frac{M^5}{m^4}\Big|$$ so the exponent in the denominator is bigger than the exponent in the numerator.
But how do I actually calculate this limit?
Thank you.
-
Let $r = \sqrt{x^2+y^2+z^2}$. Then, using the generalized mean inequality, we have:
$\sqrt[4]{\dfrac{x^4+y^4+z^4}{3}} \ge \sqrt[2]{\dfrac{x^2+y^2+z^2}{3}} \ge \sqrt[3]{|xyz|}$.
Therefore, $x^4+y^4+z^4 \ge 3\left(\dfrac{x^2+y^2+z^2}{3}\right)^2 = \dfrac{r^4}{3}$ and $|xyz| \le \left(\dfrac{x^2+y^2+z^2}{3}\right)^{3/2} = \dfrac{r^3}{3\sqrt{3}}$.
Hence, $\left|\dfrac{x^3yz+xy^3z+xyz^3}{x^4+y^4+z^4}\right| = \dfrac{|xyz| \cdot |x^2+y^2+z^2|}{|x^4+y^4+z^4|} \le \dfrac{\frac{r^3}{3\sqrt{3}} \cdot r^2}{\frac{r^4}{3}} = \dfrac{r}{9\sqrt{3}} = \dfrac{\sqrt{x^2+y^2+z^2}}{9\sqrt{3}}$.
Now, apply the squeeze theorem.
If you don't like this approach, you can try using spherical coordinates. Let $x = \rho \sin\phi \cos\theta$, $y = \rho\sin\phi\sin\theta$, and $z = \rho\cos\phi$. After making those substitutions, the function will simplify to $\rho \cdot \left(\text{some trig expression with} \ \phi \ \text{and} \ \theta \ \text{that I hope is bounded}\right)$, which will approach $0$ as $\rho \to 0$.
-
Fantastic answer, thank you! Could you please demonstrate the approach with spherical coordinates? I'm not sure how to do it, but I'd love to see that as well. – rehband Aug 18 '14 at 8:03
The problem is that it never says $x,y,z$ converge to zero at the same order. – Troy Woo Aug 18 '14 at 8:03
^Yes, that is precisely what makes multivariable limits difficult to evaluate. In this case, if $(x,y,z)$ is within a distance of $r$ of the origin, then the magnitude of the function is at most $\frac{r}{9\sqrt{3}}$. Formally, we can let $\epsilon > 0$ be arbitrary and choose $\delta = 9\sqrt{3}\epsilon$. If $\|(x,y,z)-(0,0,0)\| = \sqrt{x^2+y^2+z^2} < \delta$, then by the above math $|f(x,y,z) - 0| < \epsilon$. – JimmyK4542 Aug 18 '14 at 8:12
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2015-01-27 12:41:37
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https://www.ask-math.com/factorization-using-identities.html
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# Factorization using Identities
Covid-19 has led the world to go through a phenomenal transition .
E-learning is the future today.
Stay Home , Stay Safe and keep learning!!!
Factorization using Identities :There are some identities and using that the factorization is much easier.
A number of expressions to be factorized are of the form or can be put into the form : a2 + 2ab + b2, a2 – 2ab + b2, a2 – b2 and x2 + (a + b)x + ab. These expressions can be easily factorized using Identities I, II, III and IV
In this section we will learn Factorization using Identities one by one.
1) a2 + 2ab + b2 = (a + b)2
2) a2 – 2ab + b2 = (a – b)2
3) a2 – b2 = (a + b) (a – b)
4) x2 + (a + b) x + ab = (x + a) (x + b)
_______________________________________________________________
Factorization using Identities :
In the 1st identity, a2 + 2ab + b2 = (a + b)2,
1st and the last term should be perfect square and the middle term is two times the square root of 1st and the last term and the sign of the middle term is positive.
Examples on 1st Identity of Factorization :
1) 9a2 + 12ab + 4b2
Solution :
9a2 + 12ab + 4b2
= (3a)2 + 2 . (3a) . (2b) + (2b)2
= (3a + 2b)2 [ since a = 3a and b = 2b; a2 + 2ab + b2 = (a + b)2]
_______________________________________________________________
2) x4 + 2 + 1/x4
Solution :
x4 + 2 + 1/x4
= (x2)2 + 2.(x2).1/x2 + (1/x2)2
= (x2 + 1/x2)2 [ since a = x2 and b = 1/x2 ]
_______________________________________________________________
In the 2nd identity, a2 - 2ab + b2 = (a - b)2,
1st and the last term should be perfect square and the middle term is two times the square root of 1st and the last term and the sign of the middle term is negative.
Examples on 2nd Identity of Factorization :
1) 4p2 - 20pq + 25q2
Solution :
4p2 - 20pq + 25q2
= (2p)2 - 2 . (2p) . (5q) + (5q)2
= (2p - 5q)2 [ since a = 2p and b = 5q; a2 + 2ab - b2 = (a - b)2]
_______________________________________________________________
2)1 - 16x2 + 64x4
Solution :
= (1)2 - 2 . (2) . (8x2) + (8x2)2
= (1 - 8x2)2 [ since a = 1 and b = 8x2; a2 - 2ab + b2 = (a - b)2]
_______________________________________________________________
Some quadratic polynomials will be missing the middle term. Often these polynomials are the difference of two squares.
These polynomials come from multiplying the sum and difference of binomials, such as (a+b)(a-b)= a2-b2 when simplified.
Examples on 3rd Identity of Factorization :
1) 16x2 - 9y2
Solution :
16x2 - 9y2
= (4x)x2 - (3y)x2
= (4x + 3y)(4x - 3y)[ since a = 4x and b = 3y; a2 - b2 = (a + b)(a - b)]
_______________________________________________________________
2) x4 - x4y4
Solution :
x4 - x4y4
= (x2)x2 - [(xy)2]x2
= (x2 + x2y2)(x2 - x2y2)[ since a = x2 and b = (xy)2; a2 - b2 = (a + b)(a - b)]
In the 2nd parenthesis(bracket), we can apply the 3rd identity of Factorization again
= (x2 + x2y2)(x + xy)(x - xy)[ since a = x and b = xy;
a2 - b2 = (a + b)(a - b)]
_______________________________________________________________
Consider x2 + 5x + 6, Observe that this expressions are not of the type
(a + b)2 or (a – b)2, i.e., they are not perfect squares.
For example, in x22 + 5x + 6, the term 6 is not a perfect square. This expressions obviously also do not fit the type (a2 – b2) either. They, however, seem to be of the type
x2 + (a + b) x + a b. We may therefore, try to use Identity 4.
Factoring
Factorization by common factor
Factorization by Grouping
Factorization using Identities
Factorization of Cubic Polynomial
Solved Examples on Factorization
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2021-09-26 01:27:26
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https://www9.physics.utoronto.ca/undergraduate/undergraduate-courses/physics-of-the-earth-formerly-phy395h1/
|
# JPE395H1 Physics of the Earth (formerly PHY395H1)
## Official description
Designed for students interested in the physics of the Earth and the planets. Study of the Earth as a unified dynamic system; determination of major internal divisions in the planet; development and evolution of the Earth's large scale surface features through plate tectonics; the age and thermal history of the planet; Earth's gravitational field and the concept of isostasy; mantle rheology and convection; Earth tides; geodetic measurement techniques, in particular modern space-based techniques.
Prerequisite
PHY132H1/PHY152H1/PHY180H1/MIE100H1, MAT235Y1/MAT237Y1/MAT291H1/AER210H1, PHY254H1/PHY293H1/MAT244H1/MAT267H1/MAT290H1/MAT292H1
Co-requisite
n.a.
Exclusion
PHY359H1, PHY395H1
Recommended preparation
n.a.
Textbook
['n.a.']
BR=5
Distribution requirement
DR=SCI
course title
JPE395H1
session
winter
year of study
3rd year
time and location
24L: LEC0101, LEC2001, LEC7001, LEC9101: TR1, On line Synchronous .Lectures will be delivered synchronously via BB Collaborate and recorded for later viewing. Students can ask questions during the lectures by writing in the “chat" window or by using the "microphone”. A camera is not necessary.
instructor
#### Delivery Methods
In Person
A course is considered In Person if it requires attendance at a specific location and time for some or all course activities.*.
* Subject to adjustments imposed by public health requirements for physical distancing.
Online - Synchronous
A course is considered Online Synchronous if online attendance is expected at a specific time for some or all course activities, and attendance at a specific location is not expected for any activities or exams.
Asynchronous
A course is considered Asynchronous if it has no requirement for attendance at a specific time or location for any activities or exams.
|
2021-02-25 11:14:06
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https://neurips.cc/Conferences/2022/ScheduleMultitrack?event=55230
|
Timezone: »
Poster
Inception Transformer
Chenyang Si · Weihao Yu · Pan Zhou · Yichen Zhou · Xinchao Wang · Shuicheng Yan
@
Recent studies show that transformer has strong capability of building long-range dependencies, yet is incompetent in capturing high frequencies that predominantly convey local information. To tackle this issue, we present a novel and general-purpose $\textit{Inception Transformer}$, or $\textit{iFormer}$ for short, that effectively learns comprehensive features with both high- and low-frequency information in visual data. Specifically, we design an Inception mixer to explicitly graft the advantages of convolution and max-pooling for capturing the high-frequency information to transformers. Different from recent hybrid frameworks, the Inception mixer brings greater efficiency through a channel splitting mechanism to adopt parallel convolution/max-pooling path and self-attention path as high- and low-frequency mixers, while having the flexibility to model discriminative information scattered within a wide frequency range. Considering that bottom layers play more roles in capturing high-frequency details while top layers more in modeling low-frequency global information, we further introduce a frequency ramp structure, i.e., gradually decreasing the dimensions fed to the high-frequency mixer and increasing those to the low-frequency mixer, which can effectively trade-off high- and low-frequency components across different layers. We benchmark the iFormer on a series of vision tasks, and showcase that it achieves impressive performance on image classification, COCO detection and ADE20K segmentation. For example, our iFormer-S hits the top-1 accuracy of 83.4% on ImageNet-1K, much higher than DeiT-S by 3.6%, and even slightly better than much bigger model Swin-B (83.3%) with only 1/4 parameters and 1/3 FLOPs. Code and models are released at https://github.com/sail-sg/iFormer.
#### Author Information
##### Pan Zhou (SEA AI Lab)
Currently, I am a senior Research Scientist in Sea AI Lab of Sea group. Before, I worked in Salesforce as a research scientist during 2019 to 2021. I completed my Ph.D. degree in 2019 at the National University of Singapore (NUS), fortunately advised by Prof. Jiashi Feng and Prof. Shuicheng Yan. Before studying in NUS, I graduated from Peking University (PKU) in 2016 and during this period, I was fortunately directed by Prof. Zhouchen Lin and Prof. Chao Zhang in ZERO Lab. During the research period, I also work closely with Prof. Xiaotong Yuan. I also spend several wonderful months in 2018 at Georgia Tech as visiting student hosted by Prof. Huan Xu.
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2023-02-07 20:55:22
|
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http://stats.stackexchange.com/questions/111268/kolmogorov-smirnov-for-pareto-distribution-on-sample
|
# Kolmogorov-Smirnov for Pareto distribution on sample
I want to use the Kolmogorov-Smirnov test to test if a sample is drawn from a Pareto distribution. Unfortunately, the only way to estimate the distribution's parameters is from the sample.
Does anybody know of the existence of tables with modified critical values or any other way to perform the test in this case?
-
A Kolmogorov-Smirnov test with estimated parameters is known as Liliiefors' test. The values of the test statistic tend to be smaller than with the KS test.
It is no longer nonparametric; you need to work out the distribution of the test statistic for each distribution type (and it differs again if you estimate a subset of the parameters rather than all of them).
Some packages offer Lilliefors test for the exponential distribution and the normal distribution (which are the cases Lilliefors discusses in his two papers).
Here's a way to use the exponential for the Pareto (I'll use the notation at that link):
Case 1: known $x_m$, unknown $\alpha$:
• If $x_m$ isn't $1$, divide through by $x_m$, so you have a Pareto with $x_m=1$.
• take logs, resulting in an exponential with (unknown) scale parameter $\alpha$.
• apply Lilliefors test for an exponential r.v.
Case 2: unknown $x_m$, unknown $\alpha$. While this case can be done by simulation, we can make use of an exponential Lilliefors test as follows:
• take logs, yielding a shifted exponential
• subtract the minimum observation from all observations, and discard the minimum, leaving $n-1$ observations
• test the reduced sample for exponentiality via the Lilliefors test as before.
-
Thanks for your answer! Do you have a source for Case 2? – Tom Aug 12 at 7:39
In the meantime, I came across Coefficients of the asymptotic distribution of the kolmogorov-smirnov statistic when parameters are estimated. They propose the following transformation: x_i' = (n - i + 1) * ln[x_i / x_(i-1)] Would this be suitable as well? – Tom Aug 12 at 7:47
Hi Tom, I can't get the article at the moment, but I believe that the proposed transformation would be for consecutive order statistics (the 'gaps') (in which case your expression is missing a set of parentheses), and also yields an exponential. It should work just fine. Indeed, it may be slightly better (in that it would possibly give better power against more interesting alternatives) – Glen_b Aug 14 at 2:33
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2014-10-25 07:45:56
|
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http://nuit-blanche.blogspot.de/
|
## Thursday, October 30, 2014
### NYC Event: “How the Age of Machine Consciousness is Transforming Our Lives”
Greg Charvat just sent me the following (see below). Let me state that the "consciousness" wording is not optimal for this blog but I definitely appreciate technical people getting into a room and thinking about the next steps (and I also like the 4combinator approach yet I have no link to them) so here we are:
Hi Igor,
You might find this interesting.
I'm involved in this seminar series and panel discussion that will be held in NYC in November, on the topic of next gen deep learning as applied to hard data such as image data or other measured data. David Ferrucci (IBM Watson), Max Tagmark (on PBS Nova frequently), and Jonathan Rothberg (one who created on-chip genetic sequencing) will be speaking and discussing. I think your readers might be interested in this.
Its a free event, but must register in advance so we can get the right quantities of food and drink. Please feel free to share with your blog readers.
Cheers,
Greg
I would like to let you know about an exciting event we are hosting in New York City in a couple of weeks:
“How the Age of Machine Consciousness is Transforming Our Lives”
· Date and Time: Thursday, November 13, 2014, 7:00-9:00 PM
· A cocktail reception in the SoHi room will follow the panel discussion and will provide a chance to meet our expert panel and learn more about career opportunities at 4Combinator
The expert panel includes:
· David Ferrucci - Former VP of Watson Technologies who led development of the AI system that beat Jeopardy’s best
· Max Tegmark – MIT professor and author of “The Mathematical Universe” and “Consciousness as a State of Matter”
· Jonathan Rothberg - Inventor of high speed DNA sequencing. His latest venture, 4Combinator, aspires to transform medicine by integrating devices, deep learning and cloud computing
As you are getting an exclusive invitation ahead of a broader outreach program, please register your interest in attending the event in the next 24 hours to ensure your place at the event: https://4combinator-speaker-series.eventbrite.com
Join the CompressiveSensing subreddit or the Google+ Community and post there !
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### Higher Criticism for Large-Scale Inference: especially for Rare and Weak effects
Two papers on an interesting subject today with a common co-author. From a footnote in the first paper:
Our point here is not that HC [Higher Criticism] should replace formal methods using random matrix theory, but instead that HC can be used in structured settings where theory is not yet available. A careful comparison to formal inference using random matrix theory not possible here would illustrate the benefits of theoretical analysis of a specific situation as exemplified by random matrix theory, in this case over the direct application of a general procedure like HC.
Higher Criticism for Large-Scale Inference: especially for Rare and Weak effects by David Donoho, Jiashun Jin
In modern high-throughput data analysis, researchers perform a large number of statistical tests, expecting to find perhaps a small fraction of significant effects against a predominantly null background. Higher Criticism (HC) was introduced to determine whether there are any non-zero effects; more recently, it was applied to feature selection, where it provides a method for selecting useful predictive features from a large body of potentially useful features, among which only a rare few will prove truly useful.
In this article, we review the basics of HC in both the testing and feature selection settings. HC is a flexible idea, which adapts easily to new situations; we point out how it adapts to clique detection and bivariate outlier detection. HC, although still early in its development, is seeing increasing interest from practitioners; we illustrate this with worked examples. HC is computationally effective, which gives it a nice leverage in the increasingly more relevant "Big Data" settings we see today.
We also review the underlying theoretical "ideology" behind HC. The Rare/Weak} (RW) model is a theoretical framework simultaneously controlling the size and prevalence of useful/significant items among the useless/null bulk. The RW model shows that HC has important advantages over better known procedures such as False Discovery Rate (FDR) control and Family-wise Error control (FwER), in particular, certain optimality properties. We discuss the rare/weak {\it phase diagram}, a way to visualize clearly the class of RW settings where the true signals are so rare or so weak that detection and feature selection are simply impossible, and a way to understand the known optimality properties of HC.
Rare and Weak effects in Large-Scale Inference: methods and phase diagrams by Jiashun Jin, Tracy Ke
Often when we deal with `Big Data', the true effects we are interested in are Rare and Weak (RW). Researchers measure a large number of features, hoping to find perhaps only a small fraction of them to be relevant to the research in question; the effect sizes of the relevant features are individually small so the true effects are not strong enough to stand out for themselves.
Higher Criticism (HC) and Graphlet Screening (GS) are two classes of methods that are specifically designed for the Rare/Weak settings. HC was introduced to determine whether there are any relevant effects in all the measured features. More recently, HC was applied to classification, where it provides a method for selecting useful predictive features for trained classification rules. GS was introduced as a graph-guided multivariate screening procedure, and was used for variable selection.
We develop a theoretic framework where we use an Asymptotic Rare and Weak (ARW) model simultaneously controlling the size and prevalence of useful/significant features among the useless/null bulk. At the heart of the ARW model is the so-called phase diagram, which is a way to visualize clearly the class of ARW settings where the relevant effects are so rare or weak that desired goals (signal detection, variable selection, etc.) are simply impossible to achieve. We show that HC and GS have important advantages over better known procedures and achieve the optimal phase diagrams in a variety of ARW settings.
HC and GS are flexible ideas that adapt easily to many interesting situations. We review the basics of these ideas and some of the recent extensions, discuss their connections to existing literature, and suggest some new applications of these ideas.
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## Tuesday, October 28, 2014
### Non-linear Causal Inference using Gaussianity Measures - implementation -
Rewatching Leon Bottou's talk yesterday, I was reminded of David Lopez-Paz's paper on the subject that I had not mentioned before. I wonder if this work could not be helped with some of the tool developed in advanced matrix/tensor factorization:
In this paper we provide theoretical and empirical evidence of a type of asymmetry between causes and effects that is present when these are related via linear models contaminated with additive non-Gaussian noise. This asymmetry is found in the different degrees of Gaussianity of the residuals of linear fits in the causal and the anti-causal direction. More precisely, under certain conditions the distribution of the residuals is closer to a Gaussian distribution when the fit is made in the incorrect or anti-causal direction. The problem of non-linear causal inference is addressed by performing the analysis in an extended feature space. In this space the required computations can be efficiently performed using kernel techniques. The effectiveness of a method based on the asymmetry described is illustrated in a variety of experiments on both synthetic and real-world cause-effect pairs. In the experiments performed one observes the Gaussianization of the residuals if the model is fitted in the anti-causal direction. Furthermore, such a method is competitive with state-of-the-art techniques for causal inference.
The attendant implementation is here.
Related:
The Randomized Causation Coefficient - implementation -
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### CSJob : OCE Postdoctoral Fellow - Real-time 2D compressive Hyperspectral Imaging, Brisbane, Australia
Mingrui just sent me the following postdoc opportunity in Brisbane, Australia:
Dear Dr. Carron,
Thanks,
Mingrui
------------------------------------------------------------------
Mingrui Yang
Autonomous Systems Laboratory
CSIRO Digital Productivity
Queensland Centre for Advanced Technologies (QCAT)
1 Technology Court, Pullenvale QLD 4069, Australia
Thanks Mingrui, here it is:
# OCE Postdoctoral Fellow - Real-time 2D compressive Hyperspectral Imaging
The Position:
CSIRO offers PhD graduates an opportunity to launch their scientific careers through our Office of the Chief Executive (OCE) Postdoctoral Fellowships. Successful applicants will work with leaders in the field of science and receive personal development and learning opportunities.
CSIRO Digital Productivity Flagship is seeking to appoint a highly motivated postdoctoral fellow to join a team of researchers and embedded engineers on an exciting cross-disciplinary project of real-time 2D hyperspectral sensing, which aims to design and implement a ground-breaking hyperspectral acquisition and analysis system based on compressive sensing theory.
Specifically you will:
• Under the direction of senior research scientists, carry out innovative, impactful research of strategic importance to CSIRO that will, where possible, lead to novel and important scientific outcomes.
• Develop and evaluate novel, innovative solutions to key research problems in the area of compressive sensing on hyperspectral imaging.
• Undertake regular reviews of relevant literature and patents.
• Produce high quality scientific and/or engineering papers suitable for publication in quality journals, for client reports and granting of patents.
Location: Pullenvale, Brisbane, Queensland
Salary: $78K to$88K plus up to 15.4% superannuation (pension fund)
Tenure: Up to 3 years
Reference: Q14/03258
To be considered you will hold a PhD (or will shortly satisfy the requirements of a PhD) in a relevant discipline area, such as applied mathematics/statistics, computer science and electrical engineering.
You will also have:
• Expertise in one or more areas of compressive sensing, sparse approximation, statistical analysis and signal processing.
• Demonstrated ability to program in MATLAB (or similar) for data analysis.
• Analytical skills and ability to solve complex conceptual problems through the application of scientific and engineering principles.
• Excellent verbal and written communication skills.
• The ability to work effectively as part of a multi-disciplinary, regionally dispersed research team, plus the motivation and discipline to carry out autonomous research.
Owing to terms of the fellowship, candidates must not have more than 3 years of relevant Postdoctoral experience.
About CSIRO: Australia is founding its future on science and innovation. Its national science agency, the Commonwealth Scientific and Industrial Research Organisation (CSIRO) is a powerhouse of ideas, technologies and skills for building prosperity, growth, health and sustainability. It serves governments, industries, business and communities across the nation. Find out more! www.csiro.au
About CSIRO Digital Productivity Flagship: The Digital Productivity Flagship is focussed on Australia’s productivity challenge. We use data and digital technologies to address economic and developmental challenges.
Applications are open to Australian Citizens, Permanent Residents and International candidates. Relocation assistance will be provided if required.
Applications close 30 November 2014 (11:30pm AEST)
Position Details Q14/03258
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## Monday, October 27, 2014
### BRTF: Robust Bayesian Tensor Factorization for Incomplete Multiway Data - implementation -
Here is an extension of Robust PCA to tensors:
Robust Bayesian Tensor Factorization for Incomplete Multiway Data by Qibin Zhao, Guoxu Zhou, Liqing Zhang, Andrzej Cichocki, Shun-ichi Amari
We propose a generative model for robust tensor factorization in the presence of both missing data and outliers. The objective is to explicitly infer the underlying low-CP-rank tensor capturing the global information and a sparse tensor capturing the local information (also considered as outliers), thus providing the robust predictive distribution over missing entries. The low-CP-rank tensor is modeled by multilinear interactions between multiple latent factors on which the column sparsity is enforced by a hierarchical prior, while the sparse tensor is modeled by a hierarchical view of Student-$t$ distribution that associates an individual hyperparameter with each element independently. For model learning, we develop an efficient closed-form variational inference under a fully Bayesian treatment, which can effectively prevent the overfitting problem and scales linearly with data size. In contrast to existing related works, our method can perform model selection automatically and implicitly without need of tuning parameters. More specifically, it can discover the groundtruth of CP rank and automatically adapt the sparsity inducing priors to various types of outliers. In addition, the tradeoff between the low-rank approximation and the sparse representation can be optimized in the sense of maximum model evidence. The extensive experiments and comparisons with many state-of-the-art algorithms on both synthetic and real-world datasets demonstrate the superiorities of our method from several perspectives.
An implementation of BRTF is available on Qibin Zhao's software page:
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## Sunday, October 26, 2014
### Ambition: Finding Water
We had 7 minutes of Terror, a blockbuster taking place on Mars this Summer when Curiosity landed on Mars. We now have Ambition for the upcoming landing of Rosetta's Philae on Comet 67P/Churyumov-Gerasimenko on November 12th, 2014. Well done ESA !
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### Saturday Morning Video: Ravi Kannan -- Foundations of Data Science
Thanks to Laurent Duval for pointing this video (in support on the book mentioned earlier Book: "Foundations of Data Science" by John Hopcroft and Ravindran Kannan - link has been fixed -).Ravi provides a bird's eyeview of the field in this first lecture. Hat tip to CSA (IISc) for making the lectures of the Big Data Initiative videos available.
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## Friday, October 24, 2014
### Book: "Foundations of Data Science" by John Hopcroft and Ravindran Kannan
I just found out about a draft version of "Foundations of Data Science" a new book by John Hopcroft and Ravindran Kannan and very much like chapter 7 through 10 that touches upon themes we talk about often here on Nuit Blanche: Compressive Sensing ( see also The Big Picture in Compressive Sensing) , Advanced Matrix Factorization (see also the Advanced Matrix Factorization Jungle Page), complexity, streaming/sketching issues and Randomized Numerical Linear Algebra, Machine Learning and more. Let us note that compressive sensing is introduced at the very end, much like the other foundation book on signal processing this week (Books: "Foundations of Signal Processing" and "Fourier and Wavelet Signal Processing"). One wonders if a certain clarity would come earlier if the subject was introduced first so that titles such as Compressive Sensing Demystified ( by Frank Bucholtz and Jonathan M. Nichols) would not be needed. Here is the table of content for the last chapters.
7 Algorithms for Massive Data Problems 238
7.1 Frequency Moments of Data Streams . . . . . . . . . . . . . . . . . . . . . 238
7.1.1 Number of Distinct Elements in a Data Stream . . . . . . . . . . . 239
7.1.2 Counting the Number of Occurrences of a Given Element. . . . . . 243
7.1.3 Counting Frequent Elements . . . . . . . . . . . . . . . . . . . . . . 243
7.1.4 The Second Moment . . . . . . . . . . . . . . . . . . . . . . . . . . 245
7.2 Matrix Algorithms Using Sampling . . . . . . . . . . . . . . . . . . . . . . 248
7.2.1 Matrix Multiplication Using Sampling . . . . . . . . . . . . . . . . 248
7.2.2 Sketch of a Large Matrix . . . . . . . . . . . . . . . . . . . . . . . . 250
7.3 Sketches of Documents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
7.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
8 Clustering 260
8.1 Some Clustering Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 260
8.2 A k-means Clustering Algorithm . . . . . . . . . . . . . . . . . . . . . . . 263
8.3 A Greedy Algorithm for k-Center Criterion Clustering . . . . . . . . . . . 265
8.4 Spectral Clustering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
8.5 Recursive Clustering Based on Sparse Cuts . . . . . . . . . . . . . . . . . . 273
8.6 Kernel Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274
8.7 Agglomerative Clustering . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276
8.8 Dense Submatrices and Communities . . . . . . . . . . . . . . . . . . . . . 278
8.9 Flow Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
8.10 Finding a Local Cluster Without Examining the Whole Graph . . . . . . . 284
8.11 Axioms for Clustering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289
8.11.1 An Impossibility Result . . . . . . . . . . . . . . . . . . . . . . . . 289
8.11.2 A Satis able Set of Axioms . . . . . . . . . . . . . . . . . . . . . . 295
8.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 297
9 Topic Models, Hidden Markov Process, Graphical Models, and Belief Propagation 301
9.1 Topic Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301
9.2 Hidden Markov Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305
9.3 Graphical Models, and Belief Propagation . . . . . . . . . . . . . . . . . . 310
9.4 Bayesian or Belief Networks . . . . . . . . . . . . . . . . . . . . . . . . . . 311
9.5 Markov Random Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312
9.6 Factor Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313
9.7 Tree Algorithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314
9.8 Message Passing in general Graphs . . . . . . . . . . . . . . . . . . . . . . 315
9.9 Graphs with a Single Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . 317
9.10 Belief Update in Networks with a Single Loop . . . . . . . . . . . . . . . . 319
9.11 Maximum Weight Matching . . . . . . . . . . . . . . . . . . . . . . . . . . 320
9.12 Warning Propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
9.13 Correlation Between Variables . . . . . . . . . . . . . . . . . . . . . . . . . 325
9.14 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330
10 Other Topics 332
10.1 Rankings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .332
10.2 Hare System for Voting . . . . . . . . . . . . . . . . . . . . . . . . . . . . .334
10.3 Compressed Sensing and Sparse Vectors . . . . . . . . . . . . . . . . . . .335
10.3.1 Unique Reconstruction of a Sparse Vector . . . . . . . . . . . . . .336
10.3.2 The Exact Reconstruction Property . . . . . . . . . . . . . . . . . .339
10.3.3 Restricted Isometry Property . . . . . . . . . . . . . . . . . . . . .340
10.4 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .342
10.4.1 Sparse Vector in Some Coordinate Basis . . . . . . . . . . . . . . .342
10.4.2 A Representation Cannot be Sparse in Both Time and Frequency Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .342
10.4.3 Biological . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .345
10.4.4 Finding Overlapping Cliques or Communities . . . . . . . . . . . .345
10.4.5 Low Rank Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . .346
10.5 Gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .347
10.6 Linear Programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .348
10.6.1 The Ellipsoid Algorithm . . . . . . . . . . . . . . . . . . . . . . . .350
10.7 Integer Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .351
10.8 Semi-De nite Programming . . . . . . . . . . . . . . . . . . . . . . . . . .352
10.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356
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### SparsePR: Robust Sparse Phase Retrieval Made Easy - implementation -
Robust Sparse Phase Retrieval Made Easy by Mark Iwen, Aditya Viswanathan, Yang Wang
In this short note we propose a simple two-stage sparse phase retrieval strategy that uses a near-optimal number of measurements, and is both computationally efficient and robust to measurement noise. In addition, the proposed strategy is fairly general, allowing for a large number of new measurement constructions and recovery algorithms to be designed with minimal effort.
The implementation of SparsePR is here. SparsePR uses CVX and TFOCS.
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Liked this entry ? subscribe to Nuit Blanche's feed, there's more where that came from. You can also subscribe to Nuit Blanche by Email, explore the Big Picture in Compressive Sensing or the Matrix Factorization Jungle and join the conversations on compressive sensing, advanced matrix factorization and calibration issues on Linkedin.
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2014-10-30 17:28:30
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http://slideplayer.com/slide/1473258/
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# Order Statistics The order statistics of a set of random variables X1, X2,…, Xn are the same random variables arranged in increasing order. Denote by X(1)
## Presentation on theme: "Order Statistics The order statistics of a set of random variables X1, X2,…, Xn are the same random variables arranged in increasing order. Denote by X(1)"— Presentation transcript:
Order Statistics The order statistics of a set of random variables X1, X2,…, Xn are the same random variables arranged in increasing order. Denote by X(1) = smallest of X1, X2,…, Xn X(2) = 2nd smallest of X1, X2,…, Xn X(n) = largest of X1, X2,…, Xn Note, even if Xi’s are independent, X(i)’s can not be independent since X(1) ≤ X(2) ≤ … ≤ X(n) Distribution of Xi’s and X(i)’s are NOT the same. week 10
Distribution of the Largest order statistic X(n)
Suppose X1, X2,…, Xn are i.i.d random variables with common distribution function FX(x) and common density function fX(x). The CDF of the largest order statistic, X(n), is given by The density function of X(n) is then week 10
Example Suppose X1, X2,…, Xn are i.i.d Uniform(0,1) random variables. Find the density function of X(n). week 10
Distribution of the Smallest order statistic X(1)
Suppose X1, X2,…, Xn are i.i.d random variables with common distribution function FX(x) and common density function fX(x). The CDF of the smallest order statistic X(1) is given by The density function of X(1) is then week 10
Example Suppose X1, X2,…, Xn are i.i.d Uniform(0,1) random variables. Find the density function of X(1). week 10
Distribution of the kth order statistic X(k)
Suppose X1, X2,…, Xn are i.i.d random variables with common distribution function FX(x) and common density function fX(x). The density function of X(k) is week 10
Example Suppose X1, X2,…, Xn are i.i.d Uniform(0,1) random variables. Find the density function of X(k). week 10
Consider the power series with non-negative coefficients ak. If converges for any positive value of t, say for t = r, then it converges for all t in the interval [-r, r] and thus defines a function of t on that interval. For any t in (-r, r), this function is differentiable at t and the series converges to the derivatives. Example: For k = 0, 1, 2,… and -1< x < 1 we have that (differentiating geometric series). week 10
Generating Functions For a sequence of real numbers {aj} = a0, a1, a2 ,…, the generating function of {aj} is if this converges for |t| < t0 for some t0 > 0. week 10
Probability Generating Functions
Suppose X is a random variable taking the values 0, 1, 2, … (or a subset of the non-negative integers). Let pj = P(X = j) , j = 0, 1, 2, …. This is in fact a sequence p0, p1, p2, … Definition: The probability generating function of X is Since if |t| < 1 and the pgf converges absolutely at least for |t| < 1. In general, πX(1) = p0 + p1 + p2 +… = 1. The pgf of X is expressible as an expectation: week 10
Examples X ~ Binomial(n, p), converges for all real t.
X ~ Geometric(p), converges for |qt| < 1 i.e. Note: in this case pj = pqj for j = 1, 2, … week 10
PGF for sums of independent random variables
If X, Y are independent and Z = X+Y then, Example Let Y ~ Binomial(n, p). Then we can write Y = X1+X2+…+ Xn . Where Xi’s are i.i.d Bernoulli(p). The pgf of Xi is The pgf of Y is then week 10
Use of PGF to find probabilities
Theorem Let X be a discrete random variable, whose possible values are the nonnegative integers. Assume πX(t0) < ∞ for some t0 > 0. Then πX(0) = P(X = 0), etc. In general, where is the kth derivative of πX with respect to t. Proof: week 10
Example Suppose X ~ Poisson(λ). The pgf of X is given by
Using this pgf we have that week 10
Finding Moments from PGFs
Theorem Let X be a discrete random variable, whose possible values are the nonnegative integers. If πX(t) < ∞ for |t| < t0 for some t0 > 1. Then etc. In general, Where is the kth derivative of πX with respect to t. Note: E(X(X-1)∙∙∙(X-k+1)) is called the kth factorial moment of X. Proof: week 10
Example Suppose X ~ Binomial(n, p). The pgf of X is πX(t) = (pt+q)n.
Find the mean and the variance of X using its pgf. week 10
Uniqueness Theorem for PGF
Suppose X, Y have probability generating function πX and πY respectively. Then πX(t) = πY(t) if and only if P(X = k) = P(Y = k) for k = 0,1,2,… Proof: Follow immediately from calculus theorem: If a function is expressible as a power series at x=a, then there is only one such series. A pgf is a power series about the origin which we know exists with radius of convergence of at least 1. week 10
Moment Generating Functions
The moment generating function of a random variable X is mX(t) exists if mX(t) < ∞ for |t| < t0 >0 If X is discrete If X is continuous Note: mX(t) = πX(et). week 10
Examples X ~ Exponential(λ). The mgf of X is
X ~ Uniform(0,1). The mgf of X is week 10
Generating Moments from MGFs
Theorem Let X be any random variable. If mX(t) < ∞ for |t| < t0 for some t0 > 0. Then mX(0) = 1 etc. In general, Where is the kth derivative of mX with respect to t. Proof: week 10
Example Suppose X ~ Exponential(λ). Find the mean and variance of X using its moment generating function. week 10
Example Suppose X ~ N(0,1). Find the mean and variance of X using its moment generating function. week 10
Example Suppose X ~ Binomial(n, p). Find the mean and variance of X using its moment generating function. week 10
Properties of Moment Generating Functions
mX(0) = 1. If Y=a+bX, then the mgf of Y is given by If X,Y independent and Z = X+Y then, week 10
Uniqueness Theorem If a moment generating function mX(t) exists for t in an open interval containing 0, it uniquely determines the probability distribution. week 10
Example Find the mgf of X ~ N(μ,σ2) using the mgf of the standard normal random variable. Suppose, , independent. Find the distribution of X1+X2 using mgf approach. week 10
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2018-04-26 11:31:43
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https://lists.cam.ac.uk/pipermail/cl-isabelle-users/2019-October/msg00022.html
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# Re: [isabelle] Embedded languages in Isabelle
• To: Daniel Kirchner <daniel at ekpyron.org>, cl-isabelle-users at lists.cam.ac.uk
• Subject: Re: [isabelle] Embedded languages in Isabelle
• From: Makarius <makarius at sketis.net>
• Date: Tue, 8 Oct 2019 14:00:50 +0200
• Authentication-results: mx2f26; spf=pass (sender IP is 62.216.204.180) smtp.mailfrom=makarius at sketis.net smtp.helo=[192.168.178.32]
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On 03/10/2019 17:29, Daniel Kirchner wrote:
>
> To say a bit more about my application: I'm constructing a shallow semantic
> embedding of a philosophical logic (Edward Zalta's "Theory of Abstract
> Objects"). That logic is quite different from HOL in multiple ways, which makes
> this an interesting research project. For example the language contains two
> kinds of atomic formulas, i.e. Fxy "x and y 'exemplify' F" and xyF "x and y
> 'encode' F", neither of which can be implemented as mere function application
> in an embedding in HOL. The variable names have type implications, e.g. x and
> y have a different type than F and G, which again have a different type from
> e.g. \kappa, etc. - some of these types map directly to types in HOL, others I
> need to implement with guarding behind additional constants, etc.
Thanks for the general application outline -- it is always important to
keep the big picture in mind. A language that is very different from the
Isabelle/Pure lambda calculus (which is reused in the Isabelle/HOL
library) definitely poses a challenge. It will require very careful
exploration to do just the right thing, otherwise it will lead to an
unmaintainable blob of "code" that nobody can understand nor maintain.
> The target logic itself also defines new symbols in terms of its basic symbols,
> so the implementation should be extensible very much like with mixfix
> annotations to constants or "syntax" commands. That's why I'm trying to avoid
> duplicating an implementation of a priority grammar. (The definitions in the
> target logic themselves are not quite the same as Isar definitions, so I will
> probably introduce new outer syntax commands for them, but in the end they
> will translate to "plain", but more complex definitions satisfying the narrowly
> defined inferential role of definitions in the hyper-intensional free logic of
> the target system).
Note that there are different levels of complexity rolled into Isabelle
term notation. A command like 'notation' with plain mixfix is relatively
close to the priority grammer: if you subtract special things like
'binder' from the mixfix syntax, it is even simpler. In contrast,
'syntax' + 'translations' (or 'parse_translation' etc.) opens a not so
does not require that, it would greatly simplify a standalone
implementation of it (one that is independent of the Isabelle "syntax
phases").
> At the moment I'm trying to only *parse* the actual syntax of the language,
> not to print it, i.e. a formula given in the syntax of the target logic, will
> be translated to its representation in Isabelle's term logic and further
> reasoning will be done in Isabelle's term logic directly (I think this should
> alleviate the problem of transformed names like x_ and xa, but at a later
> stage it might also be interesting to go further and print in the target
> logic's syntax and continue reasoning in it as well, for which this indeed
> will be an issue).
BTW, printing also means document output of the sources in HTML or
LaTeX. There is an implicit assumption that adjacent letters for one
word, not individual identifiers. Compare this with LaTeX typesetting of
$abc$ vs. \emph{abc}. Presently, there is no proper way in the document
preparation system to change the typesetting of names, but it is
conceivable that a future notion of antiquotations within terms could
also go through document presentation.
> I'm now trying to mainly stay within the Isabelle term language itself.
> Interestingly, only now after looking through the syntax parsing ML code I
> realize that I can do much more than I thought within its boundaries.
>
> Simplified that approach will look like
>
> consts valid :: "p => bool"
> consts emb_implies :: "p => p => p"
> nonterminal q
> syntax valid :: "q => bool" ("[\Turnstile _]")
> emb_implies "q => q => q" (infixl "->" 8)
> emb_not "q => q" ("\<not>_" [40] 70)
> "_atomic_formula" :: "id_position => q" ("_")
> etc.
>
> and then installing a parse_ast_translation for "_atomic_formula" which splits
> up e.g. "Fxy" to "(_exe F x y)" and "xyF" to "(_enc x y F)", which is then
> translated to the corresponding term in a parse_translation.
This looks much better to me. It deescalates the overall complexity of
the implementation by some orders of magnitude.
Apart from id_position you can also experiment with cartouches for
certain well-defined sub-languages.
> In general in the long run it would, of course, be nice to have a self-
> contained "reentrant" and "public" version of the parser for Isabelle's term
> language in ML, in which every aspect could be customized, including the
> lexicon and tokenizing and maybe even the generation of "fresh" variable
> names, etc. - but I realize that this is way too much to ask especially for my
> very specific and probably rather exotic use case :-).
This sounds a bit like marketing talk for "software frameworks". Such
software components don't exist, and where people have tried it its has
The inner syntax implementation of Isabelle/Pure is the opposite of
that. It has many special tricks based on the particular situation to
make it just work by accident. I have reworked the architecture and
implementation many times in the past 25 years. A few more reforms of it
are still in the pipeline, e.g. subterm PIDE markup as seen in the tiny
rail example.
Such reforms naturally break bad applications of it -- or the other way
round: bad applications prevent further reforms.
Makarius
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2021-10-20 00:54:05
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https://nbviewer.jupyter.org/github/ubermag/oommfc/blob/master/docs/ipynb/multiple-terms.ipynb
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# Adding multiple terms of the same class¶
Here we demonstrate how to have multiple energy terms of the same class in the energy equation. For the sample, we choose a one-dimensional chain of magnetic moments.
In [1]:
import random
import oommfc as mc
import discretisedfield as df
import micromagneticmodel as mm
p1 = (-10e-9, 0, 0)
p2 = (10e-9, 1e-9, 1e-9)
cell = (1e-9, 1e-9, 1e-9)
region = df.Region(p1=p1, p2=p2)
mesh = df.Mesh(region=region, cell=cell)
The mesh is
In [2]:
mesh.k3d()
The system has an energy equation, which consists of two Zeeman energy terms.
In [3]:
H1 = (0, 0, 1e6)
H2 = (1e6, 0, 0)
system = mm.System(name='multiple_terms')
Now, if we try to add two energy terms, we get an exception raised.
In [4]:
try:
system.energy = mm.Zeeman(H=(0, 0, 1e5)) + mm.Zeeman(H=(0, 1e5, 0))
except ValueError:
print('Exception raised.')
Exception raised.
This is because different energy terms must have different names, so they can be uniquely identified. So, we have to give names to our energy terms:
In [5]:
system.energy = mm.Zeeman(H=H1, name='zeeman1') + mm.Zeeman(H=H2, name='zeeman2')
We are going to minimise the system's energy using oommfc.MinDriver later. Therefore, we do not have to define the system's dynamics equation. Finally, we need to define the system's magnetisation (system.m). We are going to make it random with $M_\text{s}=8\times10^{5} \,\text{Am}^{-1}$
In [6]:
import random
random.seed(1)
Ms = 8e5 # saturation magnetisation (A/m)
def m_fun(pos):
return [2*random.random()-1 for i in range(3)]
system.m = df.Field(mesh, dim=3, value=m_fun, norm=Ms)
Now, we can minimise the system's energy by using oommfc.MinDriver.
In [7]:
md = mc.MinDriver()
md.drive(system)
Running OOMMF (ExeOOMMFRunner) [2020/07/01 21:25]... (2.5 s)
We expect that now all magnetic moments are aligned parallel or antiparallel to the anisotropy axis (in the $z$-direction).
In [8]:
system.m.plane('y').mpl(figsize=(10, 3))
We can see that magnetisation is aligned with the sum of fields H1+H2. Finally, let us have a look at the table:
In [9]:
system.table.data
Out[9]:
max_mxHxm E delta_E bracket_count line_min_count conjugate_cycle_count cycle_count cycle_sub_count energy_calc_count E_zeeman1 E_zeeman2 iteration stage_iteration stage mx my mz
0 0.003334 -2.843445e-20 -7.222237e-35 27.0 5.0 21.0 23.0 2.0 33.0 -1.421723e-20 -1.421723e-20 30.0 30.0 0.0 0.707107 9.591067e-11 0.707107
We can see that energy terms are marked with the names we gave to energy terms when we defined the energy equation.
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2020-10-20 17:57:43
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https://portonmath.wordpress.com/2010/04/21/counterexample-meet-discrete-funcoids/
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I found a counterexample to the following conjecture:
Conjecture $f\cap^{\mathsf{FCD}} g = f\cap g$ for every binary relations $f$ and $g$.
The counter-example is $f = {(=)}|_{\mho}$ and $g = \mho\times\mho \setminus f$. I proved $f \cap^{\mathsf{FCD}} g = {(=)} |_{\Omega}$ (where $\Omega$ is the Frechet filter object).
The proof of this equality is presented in Funcoids and Reloids online article, the section Some counter-examples.
I hope the above counter-example may probably serve also as a base for disproving some conjectures about relationships of funcoids and reloids.
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2019-09-22 10:07:42
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https://search.r-project.org/CRAN/refmans/DMRMark/html/mvScatter.html
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mvScatter {DMRMark} R Documentation
## Visualize the distributions of M-value pairs from differentially methylated CpG sites (DMC) or non-DMCs
### Description
Given the M-values, True DMCs and optional the experiment design, plot the scatter plot of M-values. DMCs are marked by red daggers and non-DMCs by green circles.
### Usage
mvScatter(mv, isDMC, pd = NULL, nPlot = 5000)
### Arguments
mv The input M-values matrix, NA is not allowed. isDMC A binary vector corresponding to each row of 'mv'. 0 indicates non-DMC and 1 for DMC. pd A design matrix, which can be generated by 'stats::model.matrix'. If the M-values are totally paired or single paired, just leave it to be NULL. nPlot The maximum number of loci to be plotted. Using too large value will lead to messy scatter and long execution time. Default is 5000.
### Value
This function only generates a figure and has no return value.
### Author(s)
Linghao SHEN <sl013@ie.cuhk.edu.hk>
### Examples
# mvScatter
data(BLCA)
mvScatter(BLCA$mv, BLCA$truth, nPlot = 10000)
[Package DMRMark version 1.1.1 Index]
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2022-05-24 22:41:28
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https://oa.journalfeeds.online/2022/09/05/impact-of-covid-19-on-g20-countries-analysis-of-economic-recession-using-data-mining-approaches-financial-innovation/
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# Impact of COVID-19 on G20 countries: analysis of economic recession using data mining approaches – Financial Innovation
#### ByOsman Taylan, Abdulaziz S. Alkabaa and Mustafa Tahsin Yılmaz
Sep 5, 2022
Table 5 lists the cluster means for different parameters. Cluster 1 includes seven countries: Argentina, Brazil, India, Indonesia, South Africa, China, and Mexico. Their mean of GDP/capita is 8,543.25, the industrial production is 4.01, and the government spending is 493,747.5. However, the death rate is the highest, indicating that these countries do not curb the influences of the COVID-19 on their economies. A direct correlation is observed between the percentage of people vaccinated per population and the number of deaths. In addition, the number of beds per 1000 people is 2.38 in the countries within this cluster, and this is the lowest rate as compared to those of the other two clusters. It can be inferred that these countries have not stopped industrial production. Specifically, the industrial production of the countries in Cluster 1 ranged from − 2.1 to 8.2. Except for Mexico, all the other countries have exhibited an economic growth in industrial production. Generally, the policymakers of the countries in Cluster 1 have focused on production stability rather than the public health, which could be affiliated with the type of the ruling regimes and its relationship to the economic activities. Moreover, it appears that the production continuation in these countries did not contribute to GDP increase, which is about < 50% than those of the other countries within Clusters 2 and 3.
As a member of Cluster 1, Indonesia has the lowest number of COVID-19 deaths; however, the highest number of deaths is in Brazil, which is also in this cluster. As one of the largest economies, China also belongs to Cluster 1, which has the lowest rate of death/COVID-19 cases but the second-highest industrial production rate of 7.3. This could be mainly ascribed to the fact that the Chinese government’s policymakers have applied innovative/expert systems and AI business automation, as well as Internet medicals and advanced technologies such as extensive data analysis, 5G, and cloud computing (Sun et al. 2021) technologies. In G20, South Africa is the only member of the African continent and has the lowest rate (0.51%) of the vaccinated population in all three clusters. According to the WHO’s infections and deaths report issued in July 2021, the most severe infections and deaths due to COVID-19 have been observed in Namibia, Uganda, Zambia, and South Africa. In contrast, India has the lowest rate (0.53) of beds per 1,000 people in the G20 countries and ($2,169) of GDP/capita. In addition, India declared that its GDP at the second quarter (in April 2021) declined by 25.8% with respect to the one at the first quarter; therefore, foreign investors withdrew an estimation of$16 billion from India. This has led to severe concerns, and it has been the worst economic recession in history (Slater 2021). Consequently, policymakers have taken steps to implement economic reform. For example, in November 2021, India’s finance minister announced a new fiscal program worth $35 billion to support industries, agriculture, and exports. Thus, economic certainty is critical; however, it does not generate sectoral heterogeneity during the pandemic. Consequently, every country has been affected differently, revealing how policymakers in these countries deal with their economic objectives. As shown in Fig. 3, Cluster 2 included the following eight countries: Australia, Canada, Italy, Deutschland, Russia, Saudi Arabia, Japan, and South Korea. This cluster depicts economically developed countries, that is, their GDP/capita is 38,182 (see Table 5); however, industrial production is negative (-2.49). The countries in Cluster 2 have lower government spending (300,647.5), death rate (0.02), and COVID-19 case rate to the population (0.018) than the other G20 countries in Clusters 1 and 3. In Cluster 2, the number of beds per 1000 people and the recovery of COVID-19/cases are 0.85 and 6.33, respectively, which are the highest rates among all clusters. In contrast, a direct negative correlation was observed between the number of beds and number of deaths. This is indicative of direct political consequences, as it shows that economic policymakers have a rationale for investing in health as an additional strategy to achieve their economic goals. Consequently, health is considered an investment that can provide economic returns rather than cost. Cluster 3 included the European Union, France, Turkey, the United Kingdom, and the United States. This cluster mainly depicts highly industrialized countries. The GDP/capita is 47,938, which is the highest among all the clusters. Except for Turkey, industrial production was negative in these countries. The average industrial production is − 2.7; however, government spending is higher (1,209,867.463) compared to the other two clusters. The only positive industrial production range was recorded in Turkey, which was 9. In this cluster, the death case rate and the number of beds per 1000 people are 0.02 and 3.75, respectively, which seem similar to the values in other clusters. A negative correlation was observed between the number of beds and deaths in Cluster 3. The number of beds in Cluster 3 was almost half that in Cluster 2; however, the number of deaths was very close to that in Cluster 2. The highest number of vaccinations per hundred people is 0.4907 and is in this cluster. The high death rate could be attributed to the fact that the elderly population rate is also high in these countries. In other words, from a policymaking point of view, shutting down industrial production did not drastically reduce or completely halt the percentage of deaths. The industrial production of these countries ranges from − 3.3 to 9. The number of COVID-19 cases/population is 0.018122, which is lower than that of the other clusters, while the recovery rate of COVID-19/cases is 0.85, which is higher than the rates in the other two clusters. Clustering is a multivariate approach to grouping investigations that share akin valuations away from several variables. A constellation diagram displays the clusters for the ideal (reference) point and the actual measured distance on the same plot. Hence, Fig. 6a shows the ideal point’s location of a constellation diagram predefined universally depending on the shared values chosen for COVID-19 economic and other remaining parameters. Constellation diagrams are helpful for graphically visualizing data to promptly identify standard variables and quantify the disparity between measured and ideal findings. The lines in the constellation diagram represent membership in a cluster. The constellation plot indicates that the three clusters have clear cut-off boundaries and shows the distance between each cluster from the remaining countries in the upper half of the plot and those in the lower half of the plot. Additionally, Li et al. (2021) found that changing social environments and the complexity of human behavior make the distribution of financial data more complex. They developed an integrated approach to detect and optimize financial data clusters and quickly interpret them based on k-means clustering algorithms. In other words, the approaches suggested by Li et al. (2021) and our approach successfully clustered the problems considered and could unravel hidden patterns. The COVID-19 pandemic-related economic crises can be portrayed by shock waves in terms of supply and demand. The pandemic has caused a profound global socio-economic crisis, with a harmful impact on financial markets, logistics systems, labor, and goods supply chains. Economic activities are restricted or ultimately concluded in many countries; hence, the economy falls into a deeper recession due to supply and demand shocks. This may ultimately lead to stagnation in economies, as described by higher price levels (revenge pricing) and unemployment rates. Remarkably, the structure and power of the negative relationship rely on several elements, such as inflation with respect to its long-dated tendency, the bottom of its extrinsic impact, and political action. The current economic situation permits analysis of the inflation, industrial production, GDP, and unemployment dynamics of countries with great alterations in economic factors, as the world started to gear toward policy intervention for economic recovery from the COVID-19 pandemic. Tables 5 and 6 list the unemployment and inflation rates in the G20 countries, except the European Union, from 2016 to 2020. Figure 7 shows the unemployment rates from 2016 to 2020, indicating that inflation has increased geometrically in Argentina and continues to increase. This trend was also observed in Turkey. The change in other countries is steady; however, a significant fluctuation in the unemployment rate appears in the United States, Brazil, France, Italy, and South Africa. It seems that inflation does not have a distractive effect on economies during 2020, which could be associated with the lockdown implemented by the G20 countries. However, the rising trend of inflation started to disrupt the economies of almost all countries during 2021 and in the first half of 2022. Owing to the long curfews, the inability of consumers to visit shopping centers freely and the delay in purchasing their needs decreased the demand for goods and services during the pandemic period. This situation has led to high volatility in economic policy. The high emission of money injected by governments to reduce the impact of unemployment also led to high inflation, which created a huge gap during post-pandemic demand as industrial production has declined, and the supply of goods and services has dwindled or nearly ceased. The economy is like a standing ship, and it takes time to regenerate and meet market needs. The demand has increased remarkably after the pandemic, which could not be met immediately, and problems arose due to logistics. This case is remarkable in the United States, which is representative of a developed country and economy, as presented in Table 2. In the United States, the GDP per capita is 55,809, which is the highest after Australia, although the government spending is 3.3 trillion$, and the industrial production is − 1.8. Moreover, the pattern of growth showed a declining trend in the pre-recession period; however, unemployment and inflation rates were low (as shown in Fig. 7).
Table 6 shows the G20 countries unemployment rates excluding the European countries. In 2020, the highest unemployment rate was observed in South Africa (28.74%). The lowest unemployment rate was found in Japan, at 2.97. Table 7 presents the inflation rates of the G20 countries. The highest inflation was observed in Argentina (42%), whereas the lowest was observed in Italy (-0.14%).
Table 8 and Fig. 8 illustrate G20 countries’ income, VAT, and corporate tax rates in percentages. Although the highest income tax was in Japan at 55.97% in 2020, unemployment was the lowest. The highest VAT was observed in Italy (22%), which has a 9.31% unemployment rate. The highest corporate tax was observed in Brazil at a rate of 34%; however, the inflation rate was low (3.21%) and unemployment was high (13.67%). Increasing productivity is crucial and plays an important role in prosperity from a policy perspective because it is the primary driver of RGDP per capita growth and shows improvement in living standards. Figure 9a, b show that the unemployment and inflation relationship is nonlinear and has an inverse relationship, reflecting a negative correlation. During the peak of the pandemic, the unemployment rate increased for almost all G20 countries; however, inflation decreased. Figure 9a, b suggest a short-run trade-off between inflation and unemployment for the G20 countries, aligned with Phillips curve theory. in contrast, as shown in Table 2, the industrial manufacturing of G20 countries has dwindled and/or almost ceased. Similarly, Fig. 10a, b show that the GDP growth depends on several factors.
The curves in Fig. 9a display concave and convex functions, respectively, which also depict dwindled or/and almost ceased RGDP rates during 2020. The concavity pattern showed that inflation was in a declining trend during the pandemic; however, the convexity of the unemployment curve showed an increasing trend in unemployment. A decrease in inflation usually leads to a considerable decrease in unemployment and an increase in the GDP rate. Consequently, unemployment can be said to have enormous societal costs, even though low inflation rates cause minor nuisances. The implications of the negative relationship between unemployment and inflation can be seen in the current monetary policies aimed at raising RGDP and minimizing unstable economic conditions in G20 countries. The monetary policy of some G20 countries is aimed at reducing unemployment; however, this may temporarily increase the inflation rate, which has occurred nowadays. The hike in the prices of crude oil and supply shocks, logistics problems, and unavailability of crucial raw materials have led to increased cost inflation, cooperating with the rising unemployment and dwindling of RGDP. However, this relationship may change in the long run when the price levels of crude oil, energy, and raw materials are adjusted. The decrease in logistics costs may also positively affect GDP in the coming years. For example, US policymakers have emphasized investing in public health and providing time-bound support to families, societies, and firms. Government financial support, if appropriately directed toward investment in industrial manufacturing, can reduce the scarcity of certain advanced products (i.e., chips and circuits) in the market. However, governments seem to focus only on reducing the effects of expanding unemployment (involving additional unemployment benefits), sending immediate stimulus payments of \$1,400 to qualified people, delivering immediate support to state and local governments, supplying resources to the vaccination program, and raising grants for academic institutions to reopen. As illustrated in Fig. 9a, b, in general, the trends provide a good understanding of the core dynamics at the system level. In France and Italy, unemployment declined; however, it showed an increasing trend in all the other G20 countries.
The inflation rates did not show an underlying variation in the G20 countries during the peak of the pandemic. However, in the post-pandemic period, very high rates were observed in the prices of goods and services. We developed the following econometric model to estimate the real GDP of the G20 countries based on Eq. 1. We also determined the standard deviation of RGDP growth in the recession periods of the G20 countries as an additional explanatory variable. Table 9 shows the coefficients of the econometric model and t-statistics ratios. Additionally, the F ratio and coefficient of determination were 9.9363 and 0.8784, respectively.
begin{aligned} RGDP_{it} & = 4021.7941 + 0.0218 GDP_{it} – 45.113 IP_{it} + 0.0045 GS_{it} \ & quad quad + 1702.66 CC_{it} – 2480.476 RT_{it} – 43598.42 DC_{it} – 83.9508 HB_{it} \ & quad quad + 2171.292 VP_{it} + exogenous;factors + mu_{it} \ end{aligned}
The econometric model indicated that endogenous factors are key indicators affecting the economies of the G20 countries, relying on the assumption that producers can expedite production in response to the COVID-19 pandemic. Hence, emerging economies with large manufacturing bases are expected to recover quickly, while weaker manufacturing-based economies are expected to suffer from long-term downward and output contraction trends. Hence, as shown in Table 5, the average industrial production growth is 4.01% for Cluster 1, including Argentina, Brazil, India, Indonesia, South Africa, China, and Mexico. Industrial production has declined by approximately − 2.49% for the countries in Cluster 2, including Australia, Canada, Italy, Deutschland, Russia, Saudi Arabia, Japan, and South Korea. Similarly, the average industrial production declined by about − 2.7% for the countries in Cluster 3, which includes the European Union, France, Turkey, the United Kingdom, and the United States. Our analysis showed that the shift in economic indicators was significantly more prominent in EU countries; the GDP rates, labor market conditions, and vaccination process were less favorable at the beginning of the crisis. Failing to develop adequate and harmonious policies causes economic deviations and risks among EU member states.
Hence, data from the first and second quarters of 2021 for GDP, unemployment, inflation, and industrial production were analyzed. With the shift in economic sentiment during the first two quarters of the year, the impact of the COVID-19 pandemic seems to be reduced; this is especially the case for unemployment-related sentiment. Following the pandemic, unemployment-related searches jumped far beyond those observed during the Great Recession. As shown in Fig. 10a, b, GDP growth depends on many factors. For instance, industrial productivity is the primary driver of prosperity and GDP per capita growth. Therefore, it is crucial from a policy perspective. The bulk of GDP per capita growth in the first and second quarters of 2021 in all G20 countries is growing; the maximum growth appears in France and Mexico, with rates of 43.9% and 36.4%, respectively. However, economic contraction and disruption still appear in Australia and Saudi Arabia, with rates of − 1.6% and − 3.9% in the second quarter of 2021, respectively. With the aging societies of EU countries in the G20, increased productivity will improve living standards and positively affect growing economies. Banerjee et al. (2020) state that the COVID-19 crisis raised uncertainty, caused a decline in corporate investment, and added a strain on corporate liquidity that might further weaken industrial productivity growth in future international trading bans, and logistic problems are not eliminated. It seems that the slowdown in industrial productivity growth is temporary and not structural.
However, announcing the slowdown implications that cause policy concerns or structural problems early. The COVID-19 pandemic might speed up the structural changes triggered and offer several challenges and opportunities for the G20 countries. Lower productivity growth, lower business dynamism, and high correlation may increase the divergence between the most and the least productive firms. The delay in the availability of relevant official statistics, long-term uncertainties in economic bias, and the impact of the COVID-19 pandemic on productivity cannot be definitively determined at this stage because of the exogenous factors presented in Table 1. Additionally, a slowdown in workers’ reorganization and government support will result in worsening of labor skills; hence, the destruction of jobs can reduce productivity in the long run.
However, emissions declined during the post-recession period, despite accelerating economic growth. COVID-19 also affected human capital growth owing to lockdown-generated disruptions in mid- and small-sized manufacturing enterprises, schooling, and training, which harmed cumulative and firm-level productivity in 2020. The G20 countries with more rigid lockdowns during 2020 experienced, on average, a more significant drop in labor market participation. Unfortunately, Fig. 11a illustrates that the unemployment problem is still in progress during the first two quarters of 2021; the highest unemployment rate appears in South Africa, Brazil, Saudi Arabia, Argentina, and Turkey at more than 10%. D’Adamo et al. (2021) claimed that although data are available only for a subset of G20 countries, the economic drop is mainly deeper for developing countries than for advanced economies due to decimated business travel and tourism and diminished movement of all stripes. Firms in advanced economies are expected to scale down investments, particularly if uncertainties regarding the COVID-19 pandemic persist. D’Adamo et al. (2021) observed that, in 2020, investment is expected to fall in all but two G20 countries, China and Turkey, compared to 2019. Comparing the effects of the COVID-19 crisis on investment with those of the global financial crisis, its influence on the G20 economies seems to be less than or comparable to that of 2009, whereas in developing market economies, it seems to be higher than average.
As seen in Fig. 11a, b, COVID-19 provides several favorable conditions to create a remarkable rate of unemployment and inflation for the periods 2020 and 2021, Q1 and Q2, around the world, especially in the G20 countries. It is crucial to point out that the post-effects of COVID-19 have been expanding exponentially worldwide, distancing global economies from potential normalization. Recently, COVID-19 data indicated a total of 211,364,677 cases, 4,423,507 deaths, 4,934,496,760 total doses administered, 1,899,999,918 fully vaccinated persons, and 188,247,177 recoveries all over the world through July 2021 (WHO 2021). Given the COVID-19 shock, the crisis is expected to leave scars, possibly through indirect mechanisms of spillover and investment declines due to hysteresis, travel bans, and fear. R&D investments are central in response to the COVID-19 pandemic to reduce unemployment and inflation rates in response to the pandemic.
In contrast, an economy’s ability to invest in innovative areas depends on country-specific characteristics including economic structure, policies, politics, institutions, and governance (Corrado et al. 2009). Looking to the future and evaluating the implications of the COVID-19 pandemic on international tax systems, many uncertainties (exogenous factors) can be observed (Baker 2020). Thus, counterplans need to be made to assess and reduce the effects of Covid-19 on economies. This may include swift and reasonably appropriate long-term plans, which may require a great deal of essential work and careful implementation, even if they are discarded or need to be significantly amended. The pandemic has had enormous and dramatically damaging economic consequences.
Moreover, the pandemic can affect prices in highly different ways; unfortunately, it has led to a considerable fall in supply and demand. For instance, policymakers in some Eurozone countries such as Belgium, Austria, and Germany adopted a VAT reduction policy that directly influenced purchase prices as another factor in reducing the economic recession and contributing to the reduction of inflation. In addition, the federal government of Germany reduced VAT by 3%, from 19 to 16% for most products and from 7 to 5%. This represented approximately 2% of the total between July and December 2020.
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2023-02-04 08:49:10
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https://www.gradesaver.com/textbooks/math/geometry/CLONE-df935a18-ac27-40be-bc9b-9bee017916c2/chapter-1-section-1-3-introduction-to-geometric-proof-exercises-page-56/36
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## Elementary Geometry for College Students (7th Edition)
Published by Cengage
# Chapter 1 - Section 1.3 - Introduction to Geometric Proof - Exercises - Page 56: 36
#### Answer
$-3 \lt 1$
#### Work Step by Step
Divide both sides of the inequality by -4 and flip the direction of the sign.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2022-08-16 08:09:45
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https://mathematica.stackexchange.com/questions/73605/convex-combinations-of-list-elements
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# Convex Combinations of List Elements
Let's say I have two lists $L_1=\{a_1,...,a_5\}$ and $L_2=\{b_1,...,b_10\}$. Is there an easy way to make new lists from convex combinations of these two lists' entries? Let me be more specific:
1 Suppose the two lists $L_1,L_2$ are of equal length $n$; how do I create a new list whose entries are linear combinations of the entries in $L_1,L_2$? I first defined $f(x)=wx$ and $g(x)=(1-w)x$ and then used Map[(# + Map[f, list1]) &, {Map[g, list2]}]. This works but it's not very neat, is there another way? (Let's call such a hypothetical neat function $u$.)
2 Now let's go back to the lists of unequal length. I wanna be able to apply $u$ to sublists of $L_1$ and $L_2$ of length at most five. One could of course delete entries from the lists in question and then apply $u$, but I'm thinking there's gotta be a better way, but how? That is to say, how do I make a function where I can specify to map the first two entries of each list to a 2-list (producing the list $\{wa_1+(1-w)b_1,wa_2+(1-w)b_2\}$), or to map the second and third entries of each list to a 2-list? (producing the list $\{wa_2+(1-w)b_2,wa_3+(1-w)b_3\}$). On this one I'm completely lost.
EDIT: With h[x_,y_]:=wx+(1-w)y, Apply[h,{list1,list2}] answers question 1.
• (1) Please use code tags as appropriate. (2) I don't follow your notation. Would you please give an example of the actual input and output that you desire? – Mr.Wizard Feb 9 '15 at 11:28
• Input are two lists of different length, output as in the example in question 2. – Erik Vesterlund Feb 9 '15 at 11:32
Perhaps (for equal length lists)
foo[h_, l1_, l2_, parts_] := h@@@Transpose[{l1, l2}][[parts]]
(* or foo[h_, l1_, l2_, parts_] := Apply[h, Transpose[{l1, l2}][[parts]], 1] *)
{l1, l2} = Array[#, {5}] & /@ {a, b};
h[x_, y_] := w x + (1 - w) y
foo[h, l1, l2, {2, 3}]
(* {w a[2] + (1 - w) b[2], w a[3] + (1 - w) b[3]} *)
?
• Thanks! Not sure what's going on there though, and the output is wrong, although I'm sure it can be fixed quite easily... I'll look into it more later :) – Erik Vesterlund Feb 9 '15 at 11:50
If the length of list1 and list2 are the same, I think you can form the convex hull of the pointset formed by transposing them - since every point in a convex hull is a convex combination of some of these points.
list1 = RandomVariate[NormalDistribution[], 50];
list2 = RandomVariate[NormalDistribution[], 50];
pts = Transpose[{list1, list2}];
reg = ConvexHullMesh[pts]
You may now test any point {x,y} for membership in this convex hull by doing
{x,y}\[Element]reg
The coordinates in the convex hull are those that are convex combinations of list1 and list2 simultaneously. I assume this is what you really want. If you wanted any convex combination in either list1 or list2, this would just be a boring rectangle of dimensions {{Min[list1],Max[list1]},{Min[list2],Max[list2]}}
You can now generate random numbers, form pairs of them {a,b} and test for membership in the region. Then randomly choose a or b and put that in a list, rinse repeat, you have a convex combination list.
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2020-07-06 13:40:57
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https://www.futurelearn.com/courses/python-in-hpc/0/steps/65090
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## Want to keep learning?
This content is taken from the Partnership for Advanced Computing in Europe (PRACE)'s online course, Python in High Performance Computing. Join the course to learn more.
1.12
# Introducing heat equation
Heat (or diffusion) equation is a partial differential equation that describes how the temperature varies in space over time.
The numerical solution of the heat equation contains performance aspects that are present also in many other problems and, as such, the heat equation is used as an example in several hands-on exercises throughout the course.
The heat equation can be written as
$\frac{\partial u}{\partial t} = \alpha \nabla^2 u$
where u(x, y, t) is the temperature field that varies in space and time, and α is the thermal diffusivity constant.
The equation can be solved numerically in two dimensions by discretizing first the Laplacian with finite differences as
$\nabla^2 u = \frac{u(i-1,j)-2u(i,j)+u(i+1,j)}{(\Delta x)^2} \\ + \frac{u(i,j-1)-2u(i,j)+u(i,j+1)}{(\Delta y)^2}$
Given an initial condition ($$u(t=0) = u_0$$) one can then follow the time dependence of the temperature field with the explicit time evolution method:
$u_{m+1}(i,j) = u_m(i,j) + \Delta t \alpha \nabla^2 u_m(i,j)$
Note: the algorithm is stable only when
$\Delta t < \frac{1}{2 \alpha} \frac{(\Delta x \Delta y)^2}{(\Delta x)^2 + (\Delta y)^2}$
The image below shows a result of numerical simulation where the initial state is a cold object in a hot surrounding (for us Finns this would be a soda bottle in a sauna) and we follow the time evolution for few hundres of time steps.
Are you familiar with any other problems with a similar numerical solution? Please comment!
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2020-10-28 05:05:51
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https://oswalpublishers.com/ncert-solutions/ncert-solutions-class-11-chemistry/chapter-14-environmental-chemistry/
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# NCERT Solutions for Class 11 Chemistry Chapter 14: Environmental Chemistry
14.1. Define environmental chemistry.
Ans. Environmental chemistry is described as the branch of science that studies the chemical phenomena that occur in the environment, such as the origin, transport, reactions, impacts, and destiny of chemical species in the environment.
14.2. Explain tropospheric pollution in 100 words.
Ans. The increase in the concentration of undesired gases and particulate matter in the troposphere to such an extent that they can produce undesirable effects on human beings and their environment is called tropospheric or air pollution.The tropospheric pollution is due to following two types of polluants :
(i) Gaseous pollutants, e.g., oxides of sulphur (SO2, SO3), oxides of nitrogen (NO, NO2), oxides of carbon (CO, CO2), hydrogen sulphide, hydrocarbons, aldehydes, ketones, etc.
(ii) Particulate matter, e.g., mists, smoke, fumes, dust, carbon particles, lead and cadmium compounds, bacteria, fungi, moulds, etc.
Gaseous pollutants present are oxides of sulphur (SO2 and SO3), nitrogen (NO and NO2), carbon (CO), H2S, hydrocarbons etc. SO2 and SO3 are irritating to respiratory tract. SO2 causes throat and eye irritation. SO2 leads stiffness of flower buds and their fall from plants. They damage building materials. NO2 is a toxic gas and is lung irritant. Plants start shedding their leaves and fruits. Hydrocarbons may cause cancer. Carbon monoxide in blood results into headache, weak eyesight, nervousness and cardiovascular disorder. Participate pollutants include smoke, dust, fly ash etc. which are harmful to animals and human beings. Oxides of nitrogen and sulphur cause acid rain.
14.3. Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Ans. Carbon monoxide (CO) is a highly poisonous, colourless and odourless gas which is produced as a result of incomplete combustion of carbon. It is more dangerous to living beings than carbon dioxide gas because of its ability to block the delivery of oxygen to the organs and tissues. It binds to haemoglobin to form carboxyhaemoglobin, which is about 300 times more stable than the oxygen-haemoglobin complex. In blood, when the concentration of carboxyhaemoglobin reaches about 3–4 percent, the oxygen carrying capacity of blood is greatly reduced. This oxygen deficiency, results into headache, weak eyesight, nervousness and cardiovascular disorder. This is the reason why people are advised not to smoke. In pregnant women, who have the habit of smoking the increased CO level in blood may induce premature birth, spontaneous abortions and deformed babies.
14.4. List gases which are responsible for greenhouse effect.
Ans. The major greenhouse gases are :
1. Carbon dioxide (CO2)
2. Methane (CH4)
3. Water (H2O)
4. Nitrous oxide (NO)
5. Ozone (O3)
6. Chlorofluorocarbons (CFCs)
14.5. Statues and monuments in India are affected by acid rain. How ?
Ans. Statues and monuments in India are affected by acid rain because the air around these statues and monuments contains fairly high levels of sulphur and nitrogen oxides. It is mainly due to a large number of industries, power plants around the area and use of poor quality of coal, kerosene and firewood as fuel for domestic purposes.
These oxides undergo oxidation and then react with water vapour to form acids.
$$2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)}\xrightarrow{}2\text{H}_2\text{SO}_4\text{(aq)}$$
$$4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} \xrightarrow{} 4\text{HNO}_3\text{(aq)}$$
The resulting acid rain causes damage to building and monuments by reacting with the stone or marble of which they are made. For example, In India, acid rain has caused severe damage to Tajmahal by reacting with its marble,(limestone, CaCO3) causing damage to this wonderful monument that has attracted people from around the world.
Acid rain reacts with limestone as :
$$\text{CaCO}_3 + \text{H}_2\text{SO}_4 \xrightarrow{}\text{CaSO}_4 + \text{H}_2\text{O} + \text{CO}_2$$
As a result, the monument is being slowly disfigured and the marble is getting discoloured and lustreless.
14.6. What is smog? How is classical smog different from photochemical smogs?
Ans. Smog is a type of air pollution which is a combination of smoke and fog.
Difference between classical and photochemical smog
Classical smog Photochemical smog At first, the classical smog is observed in London in 1952. At first, the photochemical smog is observed in Los angeles in 1950. The classical smog is the H2SO4 fog deposit on the particulates. H2SO4 fog is a mixture of SO2 and humidity. The NO2 and hydro-carbons present in air reacts photochemically to form photochemical smog. It is a combination of smoke and fog. Photochemical smog do not involve any smoke or fog. Majorly, classical smog is formed during winter season. It is formed during summer season during afternoon when there is bright sunlight. It can cause irritation to lungs. It can cause irritation to eyes. It can cause irritation to lungs. It can cause irritation to eyes. It is reducing in character. It is oxidising in character.
14.7. Write down the reactions involved during the formation of photochemical smog.
Ans. Photochemical smog occurs in warm, dry and sunny climate. The main components of the photochemical smog result from the action of sunlight on unsaturated hydrocarbons and nitrogen oxides produced by automobiles and factories. Photochemical smog is called as oxidising smog because it has high concentration of oxidising agent. The formation of photochemical smong can be summarised as follows:
$$\text{NO}_2\text{(g)}\xrightarrow{hv} \text{NO(g)} + \text{O(g) ...(i)}$$
Oxygen atoms are very reactive and combine with the O2 in air to produce ozone.
$$\text{O(g)} + \text{O}_2\text{(g)} \rightleftharpoons \text{O}_3\text{(g) ...(ii)}$$
The ozone formed in the above reaction (ii) reacts rapidly with the NO(g) formed in the reaction (i) to regenerate NO2.NO2 is a brown gas and at sufficiently high levels can contribute to haze.
$$\text{NO(g)} + \text{O}_3 \text{(g)} \xrightarrow{} \text{NO}_2 \text{(g)} + \text{O}_2 \text{(g) ...(iii)}$$
While ozone is toxic in nature, both NO2 and O3 are oxidising agents. They react with the unburnt hydrocarbons in air to produce formaldehyde, PAN, and acrolein.
$$3\text{CH}_4 + 2\text{O}_3 \xrightarrow{} 3\text{CH}_2 = \text{O} + 3\text{H}_2\text{O}\\ \space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\text{Formaldehyde}$$
14.8. What are the harmful effects of photochemical smog and how can they be controlled?
Ans. Harmful effects of photochemical smog:
(i) Their high concentration causes headache, chest pain and dryness of the throat.
(ii) Ozone and PAN act as powerful eye irritants.
(iii) Photochemical smog leads to cracking of rubber and extensive damage to plant liife.
(iv) It causes corrosion of metals, stones, building materials, and painted surface etc.
Control:
(i) Use of catalytic converter in automobiles prevents the release of nitrogen dioxide and hydrocarbons to the atmosphere.
(ii) Pinus, juniparus, quercus, pyrus etc. can metabolise nitrogen dioxide. Thus, their plantation could help to some extent.
14.9. What are the reactions involved for ozone layer depletion in the stratosphere?
Ans. The main reason of ozone layer depletion is believed to be the release of chlorofluorocarbon compounds (CFCs), also known as freons. These compounds are non-reactive, non-flammable, non-toxic organic molecules and therefore used in refrigerators, air conditioners, plastic and electronic industries.
However, once CFCs are released in the atmosphere, they mix with the normal atmospheric gases and eventually reach the stratosphere. In stratosphere, they get broken down by powerful UV radiations, releasing chlorine free radical.
$$\text{CF}_2\text{Cl}_2\text{(g)} \xrightarrow{\text{UV}} \text{Cl(g)} + \text{CF}_2\text{Cl(g) ...(i)}$$
The chlorine radical then react with stratospheric ozone to form chlorine monoxide radicals and molecular oxygen.
$$\text{Cl(g)} + \text{O}_3\text{(g)} \xrightarrow{} \text{ClO(g)} + \text{O}_2 \text{ ...(ii)}$$
Reaction of chlorine monoxide radical with atomic oxygen produces more chlorine radicals.
$$\text{ClO(g)} + \text{O(g)} \xrightarrow{} \text{Cl(g)} + \text{O}_2 \text{...(iii)}$$
The chlorine radicals are continuously regenerated and cause the breakdown of ozone. Thus, CFCs are transporting agents for continuously generating chlorine radicals into the stratosphere and damaging the ozone layer.
14.10. What do you mean by ozone hole? What are its consequences?
Ans. The term ‘ozone hole’ refers to the depletion of the protective ozone layer in the upper atmosphere (stratosphere) over Earth’s polar region.
Ozone Depletion’s Consequences :
(i) With the depletion of ozone layer, more UV radiation filters into troposphere. UV radiations lead to ageing of skin, cataract, sunburn and skin cancer.
(ii) UV radiation also kills many of the phytoplanktons thereby, damaging the fish productivity.
(iii) Plants exposed to ultraviolet light have an unfavourable effect on their proteins, resulting in chlorophyll degradation and harmful mutation of cells.
(iv) UV radiation also increases evaporation of surface water through the stomata of the leaves and decreases the moisture content of the soil.
(v) Increase in UV radiations damage paints and fibres, causing them to fade faster.
14.11. What are the major causes of water pollution? Explain.
Ans. Water pollution arises as a result of several human activities, which leads to the presence of several undesirable substances in water. The following are the major causes of water pollution:
(i) Pathogens: The most serious water pollutants are the disease causing agents called pathogens. Pathogens include bacteria and other organisms that enter water from domestic sewage and animal excreta. Human excreta contain bacteria such as Escherichia coli and Streptococcus faecalis which cause gastrointestinal diseases.
(ii) Organic wastes : Organic matter such as discharge from food processing factories, leaves, grass, etc. pollute water. These wastes are biodegradable. The large population of bacteria decomposes organic matter present in water. These bacteria consume oxygen dissolved in water. If too much of organic matter is added to water, all the available oxygen is used up. This causes oxygen dependent aquatic life to die. Anaerobic bacteria also decompose the organic wastes and produce chemicals that have a foul smell and are harmful to human health.
(iii) Chemical pollutants : Water is an excellent solvent. It dissolves inorganic chemicals that include heavy metals such as cadmium, mercury, nickel which are harmful to humans because our body cannot excrete them. These metals can damage kidneys, central nervous system, liver, etc. Organic chemicals present in the petroleum products pollute many sources of water, e.g., major oil spills in oceans. Various industrial chemicals such as cleansing solvent, detergents and fertilizers also pollute water. Eutrophication is another source of water pollution. It is the process that results when large quantities of phosphates and nitrates are released into aquatic ecosystems. This polution causes the amount of oxygen in water to decrease and many organisms die because there is not enough oxygen present for respiration.
14.12. Have you are observed any water pollution in yur area? What measures would you suggest to control it.
Ans. Yes, I have observed polluted water flowing in open sewage drains in my area.
We can control if it by the following measures :
1. Industrial waste discharge from paper, fertilisers, pesticides, detergents, drugs industries and refineries should not be allowed to get mixed in water bodies such as river, lakes, etc.
2. Non-biodegradable detergents should be avoided and only biodegradable detergents should be used for cleansing of clothes.
3. The pH of water should be checked.
4. Excessive use of fertilisers should be prevented.
5. Oil spills should be avoided as much as possible.
6. Domestic waste water should be properly discharged and treated.
7. Avoid the use of DDT, malathion at home.
8. Waste water should be treated in sewage treatment plant.
9. Treatment of polluted water, lime etc. also helps in its purification.
14.13. What do you mean by Biochemical Oxygen Demand (BOD) ?
Ans. The amount of oxygen required by bacteria to break down the organic matter present in a certain volume of a sample of water, is called as the Biochemical Oxygen Demand (BOD). The amount of BOD in the water is a measure of the amount of organic material in the water, in terms of how much oxygen will be required to break it down biologically. Clean water would have BOD value of less than 5 ppm whereas highly polluted water could have a BOD value of 17 ppm or more.
14.14. Do you observe any soil pollution in your neighbourhood? What efforts will you make for controlling the soil pollution?
Ans. Yes, we see soil pollution in our neighbourhood as a result of the dumping of non-biodegradable waste , biodegradable waste, industrial waste and agricultural pollutants such as fertilisers, pesticides, insecticides etc. The best way to manage household waste is to keep two garbage cans, one for biodegradable waste and one for non- biodegradable waste. Many combustible wastes can be burned and the ashes used as landfill. However, toxic gases that can be released during combustion must be avoided.
Insecticides like DDT are not soluble in water. For this reason, they remain in soil for a long time, contaminating the root crops. Pesticides like Aldrin and Dieldrin are non-biodegradable and highly toxic in nature. They can enter the higher trophic levels through food chains, causing metabolic and physiological disorders. The same is true for industrial wastes that comprises of several toxic metals like Pb, As, Hg, Cd, etc.
Hence, the best way to check soil pollution is to avoid direct addition of pollutants to the soil. Also wastes should undergo proper treatment. They should be recycled and only then, allowed to be dumped.
14.15. What are pesticides and herbicides ? Explain giving examples.
Ans. Pesticides are synthetic chemical compounds being toxic in nature they are used in agriculture to protect the crops and plants from the pests like insects, rodents weeds and various crop diseases. However, they are basically synthetic toxic chemicals with ecological repercussions. The repeated use of the same or similar pesticides give rise to pests that are resistant to that group of pesticides thus making the pesticides ineffective. For example, DDT was used as pesticide in the beginning. The insect developed resistance towards DDT gradually over the years and therefore, other organic toxins such as Aldrin and Dieldrin were inroduced for use as pesticide. But these were non-biodegradable and slowly transferred to human beings through food chain causing metabolic and physiological disorders. Consequently, a new series of biodegradable pesticides, organophosphates and carbamates have been introduced. However, they are severe nereve toxins and hence more harmful to humans.
Herbicides : These are the chemical used to control weeds. Inorganic compounds such as sodium chlorate (NaClO3) and sodium aresenite (Na3AsO3) were used in the beginning but aresenic compounds, being toxic to mammals, are no longer preferred. Instead, organic compounds such as triazines, are now considered as better herbicides, expecially for the cornfields.
14.16. What do you mean by green chemistry? How will it help decrease environmental pollution?
Ans. Green chemistry is a way of thinking and is about utilising the existing knowledge and principles of chemistry and other sciences to reduce the adverse impact on environment. It is a production process that would bring about minimum pollution or deterioration to the environment. Green chemistry help decrease in environmental pollution in following ways:
(i) In green chemistry, the reactants to be used in chemical reactions are chosen in such a way that the yield of the end products is up to 100%. This prevents or limits chemical pollutants from being introduced into the environment. Through the efforts of green chemists, H2O2 has replaced tetrachlorethane and chlorine gas in drying and bleaching of paper.
(ii) Automobile engines have been fitted with catalytic converters which prevent the release of the vapours of hydrocarbons and oxides of nitrogen into acrolein and peroxyacetyl nitrate.
(iii) CO2 has replaced CFCs as blowing agents in the manufacture of polystyrene foam sheets.
14.17. What would have happened if the greenhouse gases were totally missing in the earth’s atmosphere? Discuss.
Ans. Earth’s most abundant greenhouse gases are CO2, CH4, O3, CFCs, and water vapour. These gases are present near the Earth’s surface. They absorb solar energy that is radiated back from the surface of the Earth. The absorption of radiation results in the heating up of the atmosphere. Hence, greenhouse gases are essential for maintaining the temperature of the Earth for the sustenance of life.
In the absence of greenhouse gases, the average temperature of the Earth will decrease drastically, making it uninhabitable. As a result, life on Earth would be impossible.
14.18. A large number of fish are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish kill.
Ans. Fishes can die either due to the presence of some poisonous substances or due to lack of oxygen. Since in this case there is no evidence of toxic dumping, so it can be only scarcity of oxygen. Excess amount oif phytoplankton present in water are biodegradable and a large population of bacteria deacomposes them in water. During the process they consume dissolved O2 in water and level of dissolved O2 falls below 5 ppm and then fish can not survive. Hence, they die and float dead on the lake.
14.19. How can domestic waste be used as manure?
Ans. Domestic waste is divided into two categories, biodegradable waste such as leaves, rotten food, etc. and non-biodegradable waste such as plastics, glass, metal scrap, etc., Non-biodegradable waste is recycled in the industry whereas biodegradable waste can be converted to manure through sanitary landfills, i.e., where the wet biodegradable waste is buried in layers and covered with several layers of soil. The compacted soil layer is to prevent odours and windblown debris. Through biodegradation the waste is converted to manure.
14.20. For your agricultural field or garden you have developed a compost producing pit. Discuss the process in the light of bad odour, flies and recycling of wastes for a good produce.
Ans. The compost producing pit should be set up at a suitable place to protect ourselves from bad odour and flies. Biodegradable domestic wastes, e.g., used tea leaves, vegetable and fruits waste are put in the compost pit and it is covered with a little sand. After some time, it is converted into compost by the action of heat and bacteria. Compost pit should be kept covered so that flies cannot make entry into it and the foul odour is minimised.
Non-biodegradable domestic wastes such as plastic, glass, metal scraps, polythene bags, etc., are sent for recycling. The process of recycling converts waste into wealth.
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2023-03-23 04:04:28
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https://leetcode.com/articles/cat-and-mouse-game/
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## Solution
#### Approach 1: Minimax / Percolate from Resolved States
Intuition
The state of the game can be represented as (m, c, t) where m is the location of the mouse, c is the location of the cat, and t is 1 if it is the mouse's move, else 2. Let's call these states nodes. These states form a directed graph: the player whose turn it is has various moves which can be considered as outgoing edges from this node to other nodes.
Some of these nodes are already resolved: if the mouse is at the hole (m = 0), then the mouse wins; if the cat is where the mouse is (c = m), then the cat wins. Let's say that nodes will either be colored , , or depending on which player is assured victory.
As in a standard minimax algorithm, the Mouse player will prefer nodes first, nodes second, and nodes last, and the Cat player prefers these nodes in the opposite order.
Algorithm
We will color each node marked according to the following rule. (We'll suppose the node has node.turn = Mouse: the other case is similar.)
• ("Immediate coloring"): If there is a child that is colored , then this node will also be colored .
• ("Eventual coloring"): If all children are colored , then this node will also be colored .
We will repeatedly do this kind of coloring until no node satisfies the above conditions. To perform this coloring efficiently, we will use a queue and perform a bottom-up percolation:
• Enqueue any node initially colored (because the Mouse is at the Hole, or the Cat is at the Mouse.)
• For every node in the queue, for each parent of that node:
• Do an immediate coloring of parent if you can.
• If you can't, then decrement the side-count of the number of children marked . If it becomes zero, then do an "eventual coloring" of this parent.
• All parents that were colored in this manner get enqueued to the queue.
Proof of Correctness
Our proof is similar to a proof that minimax works.
Say we cannot color any nodes any more, and say from any node colored or we need at most moves to win. If say, some node marked is actually a win for Mouse, it must have been with moves. Then, a path along optimal play (that tries to prolong the loss as long as possible) must arrive at a node colored (as eventually the Mouse reaches the Hole.) Thus, there must have been some transition along this path.
If this transition occurred at a node with node.turn = Mouse, then it breaks our immediate coloring rule. If it occured with node.turn = Cat, and all children of node have color , then it breaks our eventual coloring rule. If some child has color , then it breaks our immediate coloring rule. Thus, in this case node will have some child with , which breaks our optimal play assumption, as moving to this child ends the game in moves, whereas moving to the colored neighbor ends the game in moves.
Complexity Analysis
• Time Complexity: , where is the number of nodes in the graph. There are states, and each state has an outdegree of , as there are at most different moves.
• Space Complexity: .
Analysis written by: @awice.
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2020-01-21 03:18:30
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https://blog.the.al/2023/01/02/ds4-reverse-engineering-part-2.html
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In case you missed it, here is Part 1, where you can read about this DualShock 4 totally messed up that I’m trying to repair.
So, let’s go on with this journey!
At this point I have a working linux driver and finally I can start testing by myself what is working and what is not.
The result:
• Bluetooth is totally not working
• Battery controller is not working
• IMU has no calibration
• Broken LED + JDS-055 secondary-board
Only last point is not interesting, I just ordered a new JDS-055 one from AliExpress which will fix the broken LED. In the meantime I use another secondary-board from a working DS4 I have at home.
### Bluetooth
Probably every PS4 user knows that pressing the combo PS + Share sends the controller into Bluetooth Pairing Mode, the LED starts blinking white and the DS4 can be connected to a smartphone or a computer.
This is a good way to test what kind of issue with Bluetooth we have. Maybe the antenna is broken and we can pair only with a device extremely close. Or maybe the LED blinks but no signal is emitted.
After connecting the JDS-055 of another DS4 to have a fully working LED, double-checking that the LED is working by connecting it to my laptop and loading the hid-sony driver, we can finally try the PS + Share combo.
What do you expect? Nothing happens on this DS4: not only it does not enable pairing with other devices, it does not even make the LED blink white! Totally dead.
Just to be 100% sure the antenna is OK, I tested it with a super-powa oscilloscope borrowed from a friend to see if there are at least attempts by the DS4 to transmit anything.
Well, zero emissions such a green device. In the JDM-055 board, the chip responsible for Bluetooth is the main microcontroller (MediaTek MT3160N), so we cannot even blame the connection between the main CPU and the Bluetooth transponder.
### Battery controller
In the JDM-055 the battery is controlled by the main microcontroller, no external controllers. What is the problem here? It refuses both to charge the battery and to show any kind of statistics about that.
The linux driver is well written and exposes all battery stats to /sys/class/power_supply/ps-controller-battery-<bt-mac-address>/, this is the result:
$cat status Unknown$ cat capacity
0
$cat present 1 # even without the battery Even worse, the controller seems refusing to boot if you try to power it on using a battery. If you press the PS button on a working DS4, the controller turns on and the LED starts fading in and out. If you press the PS button with this controller, well… nothing happens . So you may think that the battery port is broken or disconnected. Not even close, connecting a bench power supply to the battery port shows that the DS4 drains 30/40mA for the first 10-30 seconds (don’t remember precisely) and then goes to sleep. ### Recap The hardware seems OK: If I connect the battery or the touchpad into another DualShock4, they work perfectly. If I connect a tested battery or touchpad into this controller, they do not work. This starts to remind me this joke : A man goes to the doctor and says, “Doctor, wherever I touch, it hurts.” The doctor asks, “What do you mean?” The man says, “When I touch my shoulder, it really hurts. When I touch my knee - OUCH! When I touch my forehead, it really, really hurts.” The doctor says, “I know what’s wrong with you. You’ve broken your finger!” Maybe, just maybe, even if there are 100 different issues, the only real problem is in the microcontroller, maybe a bad firmware update? Well, let’s try to restore the firmware! Sure. I dare you to find online any information about firmware updates for a DS4. It seems that updates are really common with the new DualSense PS5 Controller but not with previous models, like the DualShock 4. ## Modding community After googling a bit about how to repair my controller, this interesting page got all my attention. The repo linked is not available anymore due to DMCA takedown but there is a mirror on GitHub available, even if some files are missing. The modding community seems to be focused more on the communication between the DS4 and the PS4, to emulate a fake controller in a way that is accepted by the PS4. That part is working 100% on my controller, so I’m going to focus on other aspects of the reverse engineering. ## jedi_tool.py Anyway, the jedi_tool.py from that repository is really interesting. It tries to connect to the USB device 054c:05c4 and sends a request to obtain the Bluetooth MAC address and some firmware information. For the record, 054c is the Vendor ID of Sony, 05c4 is the Device ID of the old DualShock4 V1, I’d say board JDM-001 or something like the JDM-030. It’s an old script, I know. Here’s the main application code: bt_addrs = get_bt_mac_addrs() print('ds4 bt mac: %s host bt mac: %s' % (binascii.hexlify(bt_addrs[0]), binascii.hexlify(bt_addrs[1]))) print(get_version_info()) exit() Let’s just change the Device ID with the one of my DS4 V2: 054c:09cc and run it: $ python3 jedi_tool.py
ds4 bt mac: b'000000000000' host bt mac: b'947665e36628'
Compiled at: Sep 21 2018 04:50:51
hw_ver:0100.ff08
sw_ver:00000001.a00a sw_series:2010
code size:0002a000
OK, really interesting.
First because this script works with the new version of the DS4, so the protocol is almost the same. Second because the local Bluetooth MAC Address is all zeros. Third because there are some interesting infos about the firmware, like the build date and the version.
This is the output with another DS4 I have:
$python3 jedi_tool.py ds4 bt mac: b'28516511aea4' host bt mac: b'9081ee0cdaf0' Compiled at: Sep 21 2018 04:50:51 hw_ver:0100.b400 sw_ver:00000001.a00a sw_series:2010 code size:0002a000 The Bluetooth MAC Address is there, same build date, same software version but slightly different hardware. Perfect, we are making progress! At least in understanding of what is broken. After this, I spent some time reading both the hid-sony linux driver, the new hid-playstation driver and this jedi_tool.py trying to understand a bit more about which commands I can send to the DS4 and what information I can get about it. Here is what I figured out. ## HID protocol The DS4 uses a similar protocol between Bluetooth and USB, it consists in three kinds of packets: • HID Input Report: DS4 -> Host • HID Output Report: Host -> DS4 • HID Feature Report Input reports are used by the DS4 to inform the host about the status of the buttons, IMUs & co. Output reports are sent by the Host to the DS4 to change LEDs, make the motors rumble, etc.. Both of them are well documented online, either in the linux driver or in other projects like DS4Windows. The Feature Reports are way more interesting, these are the ones used to obtain information or change something about the device. The DS4 differentiate these reports into two categories. I use the names as they are reported in the jedi_tool.py: HID Get Report (Type ID 0x01) and HID Set Report (Type ID 0x09). • The Hid Get Report is a request to read some information. • The Hid Set Report is a request to change something. Obviously, the fact that the GetReport should not change anything depends on the firmware implementation! After the Type ID byte there is a byte identifying the Request ID. From both the hid-sony and hid-playstation linux kernel drivers I see these feature requests: Type Req. Id Size Description GET 0x02 36 Get Calibration Data GET 0x05 40 Get Calibration Data via Bluetooth (same + 2 bytes of CRC) GET 0x12 15 Get Pairing Info GET 0x81 6 Get MAC Address <- hey that’s the one that clones do not implement! GET 0xA3 48 Get Version Info Well, interesting. These instead are the feature requests written in the jedi_tool.py: Type Req. Id Description Payload SET 0x08 Set Flash Read Pos 3 bytes: 0xff + 2 bytes for the offset GET 0x11 Read from Flash 2 bytes for the payload SET 0x13 Set BT Link Info host_mac_addr(6 bytes) + link_key(16 bytes) SET 0xA0 Test arg0(1 byte) + arg1(1 byte) + arg2(1 byte) SET 0xA1 Enable Bluetooth 1 byte: 1 or 0 SET 0xA2 Enable DFU Mode 1 byte: 1 or 0 GET 0xA3 Get Version Info 0x30 bytes for the payload Woah, this is starting to be a really nice playground! ## Playing with HID Requests Okk now we have a lot of things to play with! Let’s start with something easy: GET 0xA3 which explains how jedi_tool.py is able to obtain the firmware version. It’s easy to add to jedi_tool a function to dump the output of a GET request as an hex-string: def dump_req(rid, size): try: buf = hid_get_report(dev, rid, size) strs = ["%02x" % (int(i),) for i in buf] print("RID:0x%02x data: " % (rid, ) + ("".join(strs))) except: pass And now we can try to read the raw data of the Version Info with GET 0xA3: $ python3 jedi_tool.py
RID:0xa3 data: 5365702032312032303138000000000030343a35303a35310000000000000000000100b4010000000aa0102000a00200
\$ python3 -c 'print(bytes.fromhex("5365702032312032303138000000000030343a35303a35310000000000000000000100b4010000000aa0102000a00200"))'
b'Sep 21 2018\x00\x00\x00\x00\x0004:50:51\x00\x00\x00\x00\x00\x00\x00\x00\x00\x01\x00\xb4\x01\x00\x00\x00\n\xa0\x10 \x00\xa0\x02\x00'
Good, our code to dump GET requests seems working.
### Get Calibration Data
Now we can try something more interesting, let’s see what happens if we call GET 0x02 (Calibration Data) on two DS4, a working one and the broken one.
Understanding which DS4 is which is left as an exercise to the reader .
DualShock4 number one:
RID:0x02 data: ffff0000050090226bdd812285dd982277dd1c021c02ab1f55e04320bddffb1f06e00500
DualShock4 number two:
RID:0x02 data: 000000000000000000000000000000000000000000000000000000000000000000000000
Pretty clear I would say . For some reason, the broken one does not have any calibration data for us, as expected by the crash shown in the previous blog post.
### Enable Bluetooth
In jedi_tool.py there is a function called bt_enable() which calls SET 0xa1. That’s it folks, I think we solved one issue!
I call that function connecting my broken DS4, the call is executed correctly aand..
Nope. Nothing seems to be changed on the broken DS4.
I’m coward and I don’t want to try it on a working DS4, maybe in future.
### Dump the Flash
In jedi_tool.py there is a very interesting function called dump_flash_mirror() that combines SET 0x08 with GET 0x11 to dump the flash.
Here’s a simplified version of the code in jedi_tool:
def flash_mirror_read(offset): # Read a single word
assert offset < 0x800, 'flash mirror offset out of bounds'
hid_set_report(dev, 0x08, struct.pack('>BH', 0xff, offset))
return hid_get_report(dev, 0x11, 2)
def dump_flash_mirror(path): # Read the whole flash
print('dumping flash mirror to %s...' % (path))
with open(path, 'wb') as f:
for i in range(0, 0x800, 2):
f.write(word)
print('done')
It seems to dump.. “something ” into a file. Something that is big 0x800 bytes, too few for a ROM. Probably a secondary memory?
Anyway, let’s try to understand how it works. It uses SET 0x08 (with the first payload byte to 0xff) to set the read offset and then GET 0x11 to read these two bytes of flash.
Actually this code works amazingly and dumps 0x800 bytes from the controller! To understand what is this memory, it’s better to have two memory dumps.
One from the broken DS4, one from the working one and diff them using dhex. BTW let me know if you use better tools to do this kind of diffs!
In this picture you can see above the memory of the working DS4 and below the broken one. The highlighted bytes are the ones that differ between them:
The broken one is basically 99% of zeros.
For those who already reverse-engineered the DS4 memory, you can see that there is a strange 0x03 at line 1E6, well… the memory dump reported in this picture is not the original broken memory but a close reconstruction.
### Analyze the memory dump
So at this point, we can start to search for known patterns in the memory dump of the good DS4. For example the Bluetooth MAC Address or the IMU calibration data.
And.. Yes! Can you spot them?
The calibration data is located at line 1E6, starting with ff ff. The MAC Address can be found at line 6F6 (28:51:65:11:ae:a4).
Both of them are all zeros in the broken DS4.
## Conclusion
Probably the main issue with this DualShock 4 is that the flash has been wiped out for some reason.
At this point I’m genuinely curious about the story of this controller. What kind of behaviour can wipe out the configuration memory?
So I asked the person who gave me this gift to contact the previous owner and ask him the story of this controller and how it got broken in this way.
Sad to say that the only answer we got is: “A friend gave it to me because it was not working”, without any further explaination. Also, few minutes later he blocked us .
This journey will continue in Part 3, where we will see how the DFU mode works and play a bit more with these requests.
If you want to get in touch, drop me an email at ds4@the.al or ping me on IRC (the_al@freenode|libera|hackint) or Discord (the_al#5510).
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2023-03-24 06:58:11
|
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https://rdrr.io/cran/isni/f/vignettes/isni_vignette.Rmd
|
# isni" In isni: Index of Local Sensitivity to Nonignorability
The isni package provides functions to compute, print and summarize the Index of Sensitivity to Nonignorability (ISNI). One can compute the sensitivity index without estimating any nonignorable models or positing specific magnitude of nonignorability. Thus ISNI provides a simple quantitative assessment of how robust the standard estimates assuming missing at random is with respect to the assumption of ignorability. This vignette serves as a quick start for how to use the package. Currently the package provides ISNI computation for
• Generalized Linear Model (GLM) for independent data
• Marginal Multivariate Gaussian Model for longitudinal/clustered data
• linear mixed model for longitudinal/clustered data
• mixed logit model for longitudinal/clustered binary outcome
It allows for arbitrary patterns of missingness in the longitudinal regression outcomes caused by dropout and/or intermittent missingness.
## The sos example
sos is dataset on a cross-sectional survey of sexual practices among students at the University of Edinburgh. The response variable is the students' answer to the question Have you ever had sexual intercourse?''. Because of the sensitivity of this question, many students declined to answer, leading to substantial missing data. We consider a simplified data set consisting of the answer to this question, with the student's sex and faculty as predictors.
library(isni)
data(sos)
sos[sample(nrow(sos),10),]
The R code above loads the library isni and the data frame sos, displaying a random subsample of $10$ records. sos includes the following factor variables: sexact is the response to the question Have you ever had sexual intercourse? (two levels: no (reference level), yes); gender is the student's sex (two levels: male (reference level), female); faculty is the student's faculty (medical/dental/veterinary, all other faculty categories (reference level)).
Assuming ignorable nonresponse, one can fit a logistic model (using responders only) to predict the outcome by sex, faculty and their interaction. We estimated the model with function \code{glm()}:
ymodel= sexact ~ gender*faculty
summary(glm(ymodel,family=binomial, data=sos))
The estimates show that students in a medical faculty were less likely to report having had sexual intercourse. Because only 62.4% responded to the sexual practice question, there is concern that this analysis is sensitive to the assumption of ignorability. For this purpose one can conduct an ISNI analysis for this model with the function isniglm(). We posit a nonignorable nonresponse model in the following form \begin{eqnarray} logit (Prob(is.na(sexact)=yes''))=\gamma_{0}^T s +\gamma_1*sexact \end{eqnarray} where the observed missingness predictorsincludinggender,facultyand their interaction. In the above nonresponse model, the probability of nonresponse to the sexual practice question is associated with the observed missingness predictorsvia the parameter $\gamma_0$ and is associated with the partially missing outcomesexactvia the parameter $\gamma_1$. The nonignorable parameter $\gamma_1$ captures the mangnitude and nature of nonignrable missingness. When $\gamma_1=0$, the nonresponse becomes ignorable in the sense that the probability of missingness is indepdent of unobserved values ofsexact. The above MAR analysis provides consistent and valid estimates. When $\gamma_1$ departs from zero, the nonresponse becomes nonignorable and the above MAR estimates are subject to selection bias due to nonignorable nonresponse. The ISNI functions (specifically theisniglmfunction for this example) can be applied to evaluate the rate of change of model estimates in the neighborhood of the MAR model where the missingness probability is allowed to depend on the unobserved value ofsexact, even after conditioning on the other missingness predictors ins.
A simple ISNI analysis can be conducted using the isniglm function as follows:
sos.isni<-isniglm(ymodel, family=binomial, data=sos)
sos.isni
The summary function in the package expresses the isniglm() object:
summary(sos.isni)
The columns MAR Est. and Std. Err denote the logistic model estimates and their standard errors under MAR; ISNI and c denote ISNI values and c statistics. Recall that ISNI denotes the approximate change in the MLEs when $\gamma_1$ in the selection model is changed from $0$ to $1$. Under our nonignorable selection model, assuming that $\gamma_1=1$ means that a student whose answer is yes has an increase of 2.7-fold in the odds of nonresponse. Thus, subjects whose true value is yes would be more likely to have a missing value, and the naive MAR estimate for (Intercept) should be less than the (Intercept) estimate under the correct nonignorable model. The positive sign of the ISNI value for (Intercept) is consistent with this prediction. The ISNI for the faculty predictor is $-0.17$, indicating that if, as is more plausible here, $\gamma_1 = 1$, the MLE for the estimate should change from $-0.73$ to $-0.90$. If $\gamma_1 = -1$, the estimate would change from $-0.73$ to $-0.56$.
The column c presents the c statistics that approximate the minimum magnitude of nonignorability that is needed for the change in an MLE to equal one standard error ($\text{SE}$). One can then assess sensitivity by evaluating whether this level of nonignorability is plausible. For our sos example with a binary outcome, the $c$ statistic is defined as \begin{eqnarray} c= \left| \frac{\text{SE} }{\text{ISNI}}\right|. \end{eqnarray} The $c$ statistic here informs us that in order for selection bias to be as large as the sampling error, the magnitude of nonignorability needs to be at least as large as that with which one-unit change in sexact is associated with an odds ratio of 2.7 in the probability of being missing.
When $c$ is large, only extreme nonignorability can make the estimate change substantially, and consequently sensitivity to nonignorability is of little concern. For example, $c=10$ implies that in order for the error in an MAR estimate to be the same size as its sampling error, the nonignorability needs to be strong enough that a $0.1$-unit change in sexact causes a significant change in the odds of being missing. When $c$ is small, modest departure from MAR can cause the estimate to change substantially. For example, $c=0.1$ implies that when even a $10$-unit change in sexact causes a significant change in the odds of being missing, the estimate may change substantially. As such a degree of nonignorability is plausible in many applications, this small $c$ value signals sensitivity. Prior research suggests $c<1$ as a rule of thumb to signal significant sensitivity.
In the sos example, the $c$ statistics for (Intercept)} and faculty are both less than $1$, suggesting that these coefficients are sensitive to nonignorability, confirming previous findings. Prior research also found that neither the gender nor the interaction term between gender and faculty should be sensitive, as our findings using ISNI confirm.
## Two-equation model specification
In the above we do not explicitly specify an missing data mechanism model (MDM) via formula argument in the isniglm function. The same analysis can be replicated by explicitly specifying an MDM model using the code below. The two-equation formula below sexact | is.na(sexact) ~ gender*faculty | gender *faculty uses the operator | to separately specify variables used in the complete-data model and MDM. The two-equation formula means that the complete-data model is sexact $\sim$ gender*faculty and that is.na(sexact) and gender*faculty are the missingness indicator and the missingness predictor $s$ in the nonresponse model described above, respectively.
ygmodel <- sexact | is.na(sexact) ~ gender*faculty | gender *faculty
summary(isniglm(ygmodel, family=binomial, data=sos))
## ISNI Analysis for Grouped Binomial Outcome
Because all the covariates in \code{sos} are categorical variables, one can also analyze the data as a grouped binomial outcome using the weight argument as below.
gender <- c(0,0,1,1,0,0,1,1)
faculty <- c(0,0,0,0,1,1,1,1)
gender <- factor(gender, levels = c(0, 1), labels =c("male", "female"))
faculty <- factor(faculty, levels = c(0, 1), labels =c("other", "mdv"))
SAcount <- c(NA, 1277, NA, 1247, NA, 126, NA, 152)
total <- c(1189,1710,978,1657,68,215,73,246)
sosgrp <- data.frame(gender=gender, faculty=faculty, SAcount=SAcount, total=total)
ymodel <- SAcount/total ~gender*faculty
sosgrp.isni<-isniglm(ymodel, family=binomial, data=sosgrp, weight=total)
summary(sosgrp.isni)
## A tutorial containing more technical background and examples for longitudinal data
A tutorial describing the ISNI methodology and containing examples for ISNI computation for nonignorable missing data in longitudinal setting can be download (via)
## Try the isni package in your browser
Any scripts or data that you put into this service are public.
isni documentation built on Aug. 23, 2021, 9:06 a.m.
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2021-11-30 14:59:39
|
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|
http://needmathshelp.co.uk/mba-discount-factor-worksheet/
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# MBA discount factor worksheet
The discount factor tells you the value today of $1 receivable at some point in the future. If we can earn a return of 25% on our investments, then we only need to invest$0.80 today to have $1 in one year, since 1.25 ×$0.80 = $1.00. This means that a dollar in a year’s time would only be worth eighty cents to us today, and we say the one-year discount factor is 0.80. The key idea of the time value of money is that a cash flow received in the future is worth less than the same amount of cash received today. If the one-year discount factor is 0.80, then$200 received in one year’s time only has a present value of 0.80 × $200 =$160 (since each of those two hundred dollars, receivable in one year, is worth only eighty cents of today’s money). The general formula linking present value, future value and the discount factor is:
$\text{Present Value} = \text{Discount Factor} \times \text{Future Value}$
The discount factor depends on the time, t, until the future payment is made, and on the discount rate, r (also called the “cost of capital”). If there is a long time t for compound interest to grow our investment, we don’t need to invest very much today to have $1 by time t, so the present value of$1 at that future time (i.e. the discount factor) is low. As a result, the further into the future the cash flow is received, the lower the discount factor, and the present value of the future payment is less. Similarly, the better the rate of return r we can get on our investment, the lower the discount factor, since we don’t need to invest so much today to have \$1 in the future. The general formula is:
$\text{Discount Factor} = \frac{1}{(1+r)^t}$
This worksheet is designed for MBA or undergraduate students taking a Finance class. It covers several common topics: how to find a discount factor, how to apply a discount factor to find the present value, and how to rearrange the formula so we can calculate the discount rate from the discount factor. Answers are provided on the second page.
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2017-09-26 21:39:09
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http://math.stackexchange.com/questions/111921/resources-for-learning-elliptic-integrals/125577
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# Resources for learning Elliptic Integrals
During a quiz my Calc 3 professor made a typo. He corrected it in class, but he offered a challenge to anyone who could solve the integral.
The (original) question was:
Find the length of the curve as described by the vector valued function $\vec{r} = \frac{1}{3}t^{3}\vec{i} + t^{2}\vec{j} + 4t\vec{k}$ where $0 \le t \le 3$
This give us:
$\int_0^3 \! \sqrt{t^{4}+4t^{2}+16} \, \mathrm{d}t$
Wolfram Alpha says that the solution to this involves Incomplete Elliptic Integrals of the First and Second Kinds. I was wondering if anyone had any level appropriate resources where I can find information about how to attack integrals like this.
-
(might as well...)
As a quick review: anytime you see an integral that involves the square root of a cubic or a quartic polynomial, it is quite likely that an elliptic integral will be needed (except in some very special cases, termed "pseudoelliptic integrals").
Now, for the integral at hand (treating the indefinite case for now, and worrying about the limits later): your integrand is a quartic that consists of only even powers of the dummy variable, so the first thing to do is to perform a (modified) Weierstrass substitution, $t=2\tan\dfrac{u}{2}$:
\begin{align*} \int \sqrt{t^4+4t^2+16} \,\mathrm dt&=8\int \frac{\sqrt{3+\cos^2 u}}{(1+\cos\,u)^2} \mathrm du=8\int \frac{\sqrt{4-\sin^2 u}}{(1+\cos\,u)^2} \mathrm du\\ &=16\int \frac{\sqrt{1-\frac14 \sin^2 u}}{(1+\cos\,u)^2} \mathrm du \end{align*}
(the additional factor of $2$ in the substitution used comes from the fourth root of the polynomial's constant term.)
Why did I make those last few transformations, you ask? This is because the standard elliptic integrals are made to pop out easily when the part within the square root takes the form $\sqrt{1-m\,\sin^2 u}$.
In any event, we now need to perform a Jacobian substitution. We let $u=\mathrm{am}(v\mid m)$, where $\mathrm{am}$ is the Jacobian amplitude function and $m$ is a constant (in elliptic integral parlance, the parameter) to be determined. The usual Jacobian elliptic functions come from this basic function:
\begin{align*} \mathrm{sn}(v\mid m)&=\sin(\mathrm{am}(v\mid m))\\ \mathrm{cn}(v\mid m)&=\cos(\mathrm{am}(v\mid m))\\ \mathrm{dn}(v\mid m)&=\frac{\mathrm d}{\mathrm dv}\mathrm{am}(v\mid m)=\sqrt{1-m\,\sin^2(\mathrm{am}(v\mid m))} \end{align*}
Consideration of the last relation involving $\mathrm{dn}$ indicates that we can choose $m=\frac14$ for convenience; thus,
$$16\int \frac{\sqrt{1-\frac14 \sin^2 u}}{(1+\cos\,u)^2} \mathrm du=16\int \frac{\mathrm{dn}^2\left(v\mid\frac14\right)}{\left(1+\mathrm{cn}\left(v\mid\frac14\right)\right)^2} \mathrm dv=4\int \frac{3+\mathrm{cn}^2\left(v\mid\frac14\right)}{\left(1+\mathrm{cn}\left(v\mid\frac14\right)\right)^2} \mathrm dv$$
Unfortunately, things get rather messy at this juncture, so I had to ask for some assistance from Mathematica. A little bit of massaging of the results from Mathematica yielded
$$4\int \frac{3+\mathrm{cn}^2\left(v\mid\frac14\right)}{\left(1+\mathrm{cn}\left(v\mid\frac14\right)\right)^2} \mathrm dv=\frac23\left(6v-4\,\varepsilon\left(v \mid \frac14\right)+\frac{4\left(3+\mathrm{cn}\left(v\mid \frac14\right)\right)\mathrm{dn}\left(v\mid \frac14\right)\mathrm{sn}\left(v\mid \frac14\right)}{\left(1+\mathrm{cn}\left(v\mid \frac14\right)\right)^2}\right)$$
where $\varepsilon(v\mid m)$ is the Jacobi epsilon function.
One could of course use the appropriate formulae from Byrd and Friedman's Handbook of Elliptic Integrals for Engineers and Scientists if one insists on a wholly manual evaluation, but the evaluation involves a rather messy set of recurrence relations. (If I find more time, I'll eventually edit this answer to include the full solution, bloody guts and all...)
In any event, to now consider the definite integral, we undo the two substitutions in turn and apply those to the limits of the original integral. In particular, the inverse of $\phi=\mathrm{am}(v\mid m)$ is the incomplete elliptic integral of the first kind, $v=F(\phi\mid m)$, while the Jacobi epsilon function satisfies the relation
$\varepsilon(F(\phi\mid m)\mid m)=E(\phi\mid m)$
where $E(\phi\mid m)$ is the incomplete elliptic integral of the second kind. The transformed limits are then $v=0$ and $v=F\left(2\arctan\frac32\mid\frac14\right)$
We finally end up with
$$\int_0^3 \sqrt{t^4+4t^2+16} \,\mathrm dt=\frac{17}{13}\sqrt{133}+4\,F\left(2\arctan\frac32\mid\frac14\right)-\frac83 \,E\left(2\arctan\frac32\mid\frac14\right)$$
which is a fair bit simpler than what you get if you directly input the integral into Mathematica.
If it wasn't already apparent in the paragraphs above: elliptic integrals and elliptic functions are very different things; one is (very roughly) the inverse of the other. Too many people tend to conflate these two families of functions, though they are definitely related...
In addition to Byrd/Friedman, you'll also want to look into Greenhill's The Applications of Elliptic Functions, which, contrary to the title, also deals a bit with the elliptic integrals as well.
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Shit man... This is a lot to digest, thank you so much. – OmnipotentEntity Apr 1 '12 at 2:21
There are plenty of places to look (for example, most any older 2-semester advanced undergraduate "mathematics for physicists" or "mathematics for engineers" text), but given that you're in Calculus III, some of these might be too advanced. If you can find a copy (your college library may have a copy, or might be able to get a copy using interlibrary loan), I strongly recommend the treatment of elliptic integrals at the end of G. M. Fichtenholz's book The Indefinite Integral (translated to English by Richard A. Silverman in 1971). Also, the books below might be useful, but Fichtenholz's book would be much better suited for you, I think. (I happen to have a copy of Fichtenholz's book and Bowman's book, by the way.)
Arthur Latham Baker, Elliptic functions: An Elementary Text-book for Students of Mathematics (1906) http://books.google.com/books?id=EjYaAAAAYAAJ
Alfred Cardew Dixon, The Elementary Properties of the Elliptic Functions With Examples (1894) http://books.google.com/books?id=Gx4SAAAAYAAJ
Frank Bowman, Introduction to Elliptic Functions With Applications (reprinted by Dover Publications in 1961)
-
|
2014-12-20 12:32:37
|
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|
https://www.sfu.ca/math-coursenotes/Math%20157%20Course%20Notes/sec_TrigLimits.html
|
## Section3.6The Squeeze Theorem
In this section we aim to compute the limit:
\begin{equation*} \lim_{x\to0} {\sin x\over x}\text{.} \end{equation*}
We start by analyzing the graph of $\ds{y=\frac{\sin x}{x}}\text{:}$
Notice that $x=0$ is not in the domain of this function. Nevertheless, we can look at the limit as $x$ approaches $0\text{.}$ From the graph we find that the limit is $1$ (there is an open circle at $x=0$ indicating $0$ is not in the domain).
We just convinced you this limit formula holds true based on the graph, but how does one attempt to prove this limit more formally? To do this we need to be quite clever, and to employ some indirect reasoning. The indirect reasoning is embodied in a theorem, frequently called the Squeeze Theorem.
This theorem can be proved using the official definition of limit. We won't prove it here, but point out that it is easy to understand and believe graphically. The condition says that $f(x)$ is trapped between $g(x)$ below and $h(x)$ above, and that at $x=a\text{,}$ both $g$ and $h$ approach the same value. This means the situation looks something like Figure 3.2.
For example, imagine the blue curve is $f(x)=x^2\sin(\pi/x)\text{,}$ the upper (red) and lower (green) curves are $h(x)=x^2$ and $g(x)=-x^2\text{.}$ Since the sine function is always between $-1$ and $1\text{,}$ $-x^2\le x^2\sin(\pi/x)\le x^2\text{,}$ and it is easy to see that $\lim_{x\to0}-x^2=0=\lim_{x\to0}x^2\text{.}$ It is not so easy to see directly (i.e. algebraically) that $\lim_{x\to0}x^2\sin(\pi/x)=0\text{,}$ because the $\pi/x$ prevents us from simply plugging in $x=0\text{.}$ The Squeeze Theorem makes this “hard limit” as easy as the trivial limits involving $x^2\text{.}$
To compute $\ds\lim_{x\to0} (\sin x)/x\text{,}$ we will find two simpler functions $g$ and $h$ so that $g(x)\le (\sin x)/x\le h(x)\text{,}$ and so that $\lim_{x\to0}g(x)=\lim_{x\to0}h(x)\text{.}$ Not too surprisingly, this will require some trigonometry and geometry. Referring to Figure, $x$ is the measure of the angle in radians. Since the circle has radius 1, the coordinates of point $A$ are $(\cos x,\sin x)\text{,}$ and the area of the small triangle is $(\cos x\sin x)/2\text{.}$ This triangle is completely contained within the circular wedge-shaped region bordered by two lines and the circle from $(1,0)$ to point $A\text{.}$ Comparing the areas of the triangle and the wedge we see $(\cos x\sin x)/2 \le x/2\text{,}$ since the area of a circular region with angle $\theta$ and radius $r$ is $\theta r^2/2\text{.}$ With a little algebra this turns into $(\sin x)/x \le 1/\cos x\text{,}$ giving us the $h$ we seek.
To find $g\text{,}$ we note that the circular wedge is completely contained inside the larger triangle. The height of the triangle, from $(1,0)$ to point $B\text{,}$ is $\tan x\text{,}$ so comparing areas we get $x/2 \le (\tan x)/2 = \sin x / (2\cos x)\text{.}$ With a little algebra this becomes $\cos x \le (\sin x)/x\text{.}$ So now we have
\begin{equation*} \cos x \le {\sin x\over x}\le {1\over\cos x}\text{.} \end{equation*}
Finally, the two limits $\lim_{x\to0}\cos x$ and $\lim_{x\to0}1/\cos x$ are easy, because $\cos(0)=1\text{.}$ By the Squeeze Theorem, $\lim_{x\to0} (\sin x)/x = 1$ as well.
Using the above, we can compute a similar limit:
\begin{equation*} \lim_{x\to0}{\cos x - 1\over x}\text{.} \end{equation*}
This limit is just as hard as $\sin x/x\text{,}$ but closely related to it, so that we don't have to do a similar calculation; instead we can do a bit of tricky algebra.
\begin{equation*} {\cos x - 1\over x}={\cos x - 1\over x}{\cos x+1\over\cos x+1} ={\cos^2 x - 1\over x(\cos x+1)}={-\sin^2 x\over x(\cos x+1)}= -{\sin x\over x}{\sin x\over \cos x + 1}\text{.} \end{equation*}
To compute the desired limit it is sufficient to compute the limits of the two final fractions, as $x$ goes to 0. The first of these is the hard limit we've just done, namely 1. The second turns out to be simple, because the denominator presents no problem:
\begin{equation*} \lim_{x\to0}{\sin x\over \cos x + 1}={\sin 0\over \cos 0+1}= {0\over 2} = 0\text{.} \end{equation*}
Thus,
\begin{equation*} \lim_{x\to0}{\cos x - 1\over x}=0\text{.} \end{equation*}
###### Example3.50. Limit of Other Trig Functions.
Compute the following limit $\ds\lim_{x\to 0}\frac{\sin 5x\cos x}{x}\text{.}$
Solution
We have
\begin{equation*} \begin{array}{rcl} \ds{\lim_{x\to 0}\frac{\sin 5x\cos x}{x}} \amp = \amp \ds{\lim_{x\to 0}\frac{5\sin 5x\cos x}{5x}}\\ \\ ~ \amp = \amp \ds{\lim_{x\to 0} 5\cos x\left(\frac{\sin 5x}{5x}\right)}\\ \\ ~ \amp = \amp 5\cdot (1)\cdot (1)~~=~~5 \end{array} \end{equation*}
since $\cos(0)=1$ and $\ds{\lim_{x\to 0}\frac{\sin 5x}{5x}=1}\text{.}$
Let's do a harder one now.
###### Example3.51. Limit of Other Trig Functions.
Compute the following limit: $\ds\lim_{x\to 0}\frac{\tan^3 2x}{x^2\sin 7x}\text{.}$
Solution
Recall that the $\tan^3(2x)$ means that $\tan(2x)$ is being raised to the third power.
\begin{equation*} \begin{array}{rcll} \ds{\lim_{x\to 0}\frac{\tan^3(2x)}{x^2\sin(7x)}}\amp =\amp \ds{\lim_{x\to 0}\frac{(\sin(2x))^3}{x^2\sin(7x)\cos^3(2x)}} \amp \mbox{Rewrite in terms of $\sin$ and $\cos$} \\ \\ ~\amp =\amp \ds{\lim_{x\to 0}\frac{(2x)^3\left(\frac{\sin(2x)}{2x}\right)^3}{x^2(7x)\left(\frac{\sin(7x)}{7x}\right)\cos^3(2x)}} \amp \mbox{Make sine terms look like: $\ds{\frac{\sin\theta}{\theta}}$} \\ \\ ~\amp =\amp \ds{\lim_{x\to 0}\frac{8x^3(1)^3}{7x^3(1)(1^3)}} \amp \mbox{Replace $\ds{\lim_{x\to 0}\frac{\sin nx}{nx}}$ with $1$. Also, $\cos(0)=1$.} \\ \\ ~\amp =\amp \ds{\lim_{x\to 0}\frac{8}{7}}\amp \mbox{Cancel $x^3$'s.} \\ \\ ~\amp =\amp \ds{\frac{8}{7}}. \end{array} \end{equation*}
###### Example3.52. Applying the Squeeze Theorem.
Compute the following limit: $\ds\lim_{x\to 0^+}x^3\cos\left(\frac{1}{\sqrt{x}}\right)\text{.}$
Solution
We use the Squeeze Theorem to evaluate this limit. We know that $\cos\alpha$ satisfies $-1\leq\cos\alpha\leq 1$ for any choice of $\alpha\text{.}$ Therefore we can write:
\begin{equation*} -1\leq\cos\left(\frac{1}{\sqrt{x}}\right)\leq 1 \end{equation*}
Since $x\to 0^+$ implies $x>0\text{,}$ multiplying by $x^3$ gives:
\begin{equation*} -x^3\leq x^3\cos\left(\frac{1}{\sqrt{x}}\right)\leq x^3\text{.} \end{equation*}
\begin{equation*} \lim_{x\to 0^+}(-x^3)\leq \lim_{x\to 0^+}\left(x^3\cos\left(\frac{1}{\sqrt{x}}\right)\right)\leq\lim_{x\to 0^+} x^3\text{.} \end{equation*}
But using our rules we know that
and the Squeeze Theorem says that the only way this can happen is if
\begin{equation*} \lim_{x\to 0^+}x^3\cos\left(\frac{1}{\sqrt{x}}\right)=0\text{.} \end{equation*}
When solving problems using the Squeeze Theorem it is also helpful to have the following theorem.
##### Exercises for Section 3.6.
Compute the following limits.
1. $\ds\lim_{x\to 0} {\sin (5x)\over x}$
$5$
Solution
Recall that $\lim\limits_{x\to 0} \frac{\sin(x)}{x} = 1\text{.}$ We will use this fact to find the desired limit by making a suitable subsitution. Let $w = 5x\text{.}$ Then
\begin{equation*} \begin{split} \lim\limits_{x\to 0} \frac{\sin(5x)}{x} \amp= \lim\limits_{w\to 0} \frac{\sin(w)}{w/5} \\ \amp= \lim\limits_{w\to 0} 5\frac{\sin(w)}{w} = 5(1) = 5.\end{split} \end{equation*}
2. $\ds\lim_{x\to 0 } {\sin(7x)\over\sin (2x)}$
$7/2$
3. $\ds\lim_{x\to 0 } {\cot (4x) \over\csc (3x)}$
$3/4$
4. $\ds\lim_{x\to 0 } {\tan x\over x}$
$1$
Solution
First, rewrite $\tan(x)$ as $\frac{\sin(x)}{\cos(x)}\text{.}$ Then
\begin{equation*} \begin{split} \lim\limits_{x \to 0} \frac{\tan(x)}{x} \amp= \lim\limits_{x \to 0} \frac{\sin(x)}{x\cos(x)}\\ \amp = \lim\limits_{x \to 0} \left(\frac{\sin(x)}{x}\frac{1}{\cos(x)}\right)\\ \amp = \lim\limits_{x \to 0} \frac{\sin(x)}{x} \lim\limits_{x \to 0}\frac{1}{\cos(x)}.\end{split} \end{equation*}
As $x \to 0\text{,}$ $\cos(x) \to 1$ and $\frac{\sin(x)}{x} \to 1\text{.}$ Since both limits exist, we get by the limit laws that
\begin{equation*} \lim\limits_{x \to 0} \frac{\tan(x)}{x} = 1. \end{equation*}
5. $\ds\lim_{x\to \pi/4} {\sin x-\cos x \over\cos (2x)}$
$-\sqrt2/2$
For all $x\geq 0\text{,}$ $4x-9 \leq f(x) \leq x^2 - 4x +7\text{.}$ Find $\ds\lim_{x\to4}f(x)\text{.}$
$7$
Solution
We are given that $4x-9\leq f(x) \leq x^2-4x+7$ for all $x \geq 0\text{.}$ Therefore, by the Squeeze Theorem,
\begin{align*} \lim\limits_{x\to 4} (4x-9) \amp \leq \lim\limits_{x\to 4} f(x) \leq \lim\limits_{x\to 4} (x^2 -4x+7)\\ (16-9) \amp \leq \lim\limits_{x\to 4} f(x) \leq (4^2 -16 + 7)\\ 7 \amp \leq \lim\limits_{x\to 4} f(x) \leq 7 \end{align*}
Thus, $\lim\limits_{x\to 4} f(x) = 7\text{.}$
For all $x\text{,}$ $2x \leq g(x) \leq x^4 - x^2 +2\text{.}$ Find $\ds\lim_{x\to1}g(x)\text{.}$
$2$
Solution
Since $2x \leq g(x) \leq x^4 - x^2+2\text{,}$ we have
\begin{align*} \lim\limits_{x\to 1} (2x) \amp \leq \lim\limits_{x\to 1} g(x) \leq \lim\limits_{x\to 1} (x^4-x^2+2)\\ 2 \amp \leq \lim\limits_{x\to 1} g(x) \leq 2 \end{align*}
So, by the Squeeze Theorem, $\lim\limits_{x\to 1} g(x) =2\text{.}$
Use the Squeeze Theorem to show that $\ds\lim_{x\to0} x^4 \cos(2/x)=0\text{.}$
Solution
We know that $-1 \leq \cos\left(\frac{2}{x}\right) \leq 1\text{,}$ and so $-x^4 \leq \cos \left(\frac{2}{x}\right)\leq x^4\text{.}$ Hence, by the Squeeze Theorem,
\begin{align*} \lim\limits_{x\to 0}\left(-x^4\right) \amp \leq \lim\limits_{x\to 0} \cos \left(\frac{2}{x}\right) \leq \lim\limits_{x\to 0} x^4\\ 0 \amp \leq \lim\limits_{x\to 0} \cos \left(\frac{2}{x}\right) \leq 0 \end{align*}
So we conclude that $\lim\limits_{x\to 0} \cos \left(\frac{2}{x}\right) = 0\text{.}$
Find the value of $\lim\limits_{x\to\infty}\dfrac{3x+\sin x}{x+\cos x}\text{.}$ Justify your steps carefully.
$3$
Solution
We know that
\begin{equation*} 3x-1 \leq 3x + \sin x \leq 3x+1 \end{equation*}
for all $x\text{,}$ and
\begin{equation*} x-1 \leq x+\cos x \leq 1+x \end{equation*}
Therefore
\begin{equation*} \dfrac{3x-1}{x+1} \leq \dfrac{3x+\sin x}{x+\cos x} \leq \dfrac{3x+1}{x-1} \end{equation*}
so by the Squeeze Theorem,
\begin{align*} \lim\limits_{x\to \infty} \dfrac{3x-1}{x+1} \amp \leq \lim\limits_{x\to \infty} \dfrac{3x + \sin x}{x+ \cos x} \leq \lim\limits_{x\to \infty} \dfrac{3x+1}{x-1}\\ \lim\limits_{x\to \infty} \dfrac{3-\frac{1}{x}}{1+\frac{1}{x}} \amp \leq \lim\limits_{x\to \infty} \dfrac{3x + \sin x}{x+ \cos x} \leq \lim\limits_{x\to \infty} \dfrac{3+\frac{1}{x}}{1-\frac{1}{x}}\\ 3 \amp \leq \lim\limits_{x\to \infty} \dfrac{3x+\sin x}{x+\cos x} \leq 3 \end{align*}
Thus, $\lim\limits_{x\to \infty} \dfrac{3x+\sin x}{x+\cos x} = 3\text{.}$
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2021-04-18 08:08:31
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|
https://electronics.stackexchange.com/questions/617344/ethercat-communication-testing
|
# EtherCAT Communication Testing
I am working on a device that uses EtherCAT. On one side, I have the standard EtherCAT Connector - an RJ45. On the other side, I would like to experiment with different connectors. For starters, I will try a 6 pin JST connector.
I was wondering if there would be any concerns in making this adapter? Essentially putting an EtherCAT bus over a JST connector.
Are there any tests I can do (using an oscilliscope, or even an arduino) to make sure data is being transmitted effectively via this adapter?
Thanks for the help
• Ethercat is effectively 100baseT Ethernet. Why do you want to play with connectors? Apr 26 at 2:55
• Since it is Ethernet, it can be used with industrial grade Ethernet connectors rated for Ethernet use. However, if you are going to experiment with custom connectors, please note that Ethernet also requires certain specification from the cabling/wiring, which usually means the industrial cables with industrial connectors are just industrial grade CAT6 or similar. Apr 26 at 5:10
• What's the part number for the connector? Presumably it has internal magnetics? Apr 26 at 10:17
As noted by @Kartman, EtherCAT at the physical layer is effectively 100MBit ethernet. I have had extensive experience plumbing 100MBit ethernet through robotic systems that operate unsealed in some really nasty environments and it's pretty reliable:
-we would use JST PA, SH and GH connectors, crimped by hand
-we would use 24awg standed wire twisted with a hand drill
Certainly not ideal!
Our ethernet links would work well until:
-multiple shorts to chassis occurred (single point was kind of okay)
-corrosion caused the connection resistance to increase
with regards to resistance: My general guideline was if a fully connected link measured less than 2 ohms across a pair on a multi-meter (usually 4 to 4.5 ohms indicated with lead resistance factored in) then it was good to go. Above that I would service the wiring and connectors.
As for diagnostics, if you can connect a linux running device direcly to the link in question, netstat -i will report the number of dropped packets:
https://www.cyberciti.biz/faq/linux-show-dropped-packets-per-interface-command/
And with two linux appliances iperf can be used to benchmark the link:
https://www.dell.com/support/kbdoc/en-ca/000139427/how-to-test-available-network-bandwidth-using-iperf
• Thanks, this is really helpful advice! To clarify, we are expecting some packets dropped when going from the RJ45 to the JST connector? What I have is an RJ45 on a PCB with traces going to the JST connector. Thus, the resistance should be minimal. Do you forsee any issues with this approach? As for the Linux testing, is it simply the case that you send out a packet from the Linux device, and then read it again out of the system? It's essentially a loop? Apr 28 at 1:51
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2022-10-05 07:57:43
|
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http://tex.stackexchange.com/questions/8136/aftergroup-a-list-of-tokens
|
# \aftergroup a list of tokens
\aftergroup executes the next token after the end of the current group. Is there a variant, \Vaftergroup{\some\tokens\a\b} that would put \some\tokens\a\b after the current group ends? Of course, one can do \aftergroup\some\aftergroup\tokens\aftergroup\a\aftergroup\b, but when the list is longer, it becomes messy.
This is motivated by an answer that works but is ugly.
-
mark wooding's doafter package was designed after he had asked this question some time in the 80s. it's conceivable he did it without benefit of etex (which would have been very new at the time, if it existed at all) – wasteofspace May 14 at 9:21
In much the same vein as Herbert's answer, a 'pre-built' version of the same idea is provided by the etextools package: see \AfterGroup. I guess that \aftergroup is used rarely enough that this has not been a big demand :-)
-
studying this package will teach me much about the fine points of expansion, I believe :). Thanks for the info! – Bruno Le Floch Jan 2 '11 at 23:24
@Bruno ...or else drive you mad :) – Will Robertson Jan 3 '11 at 2:04
The \AfterGroup macro solves the problem in the way outlined by Herbert, but defining a new global temporary macro each time (via a counter) which is then undefined at use time. – egreg May 14 at 9:21
don't know if this might help.
\documentclass{article}
\def\ta{Hello}
\def\tb{ World. }
\def\tc{How are you?}
%\def\Tokens{\ta\tb\tc}
\begin{document}
\begingroup
\gdef\Tokens{\ta\tb\tc}% or like above with def
\aftergroup\Tokens
\let\ua=\tc
\let\ub=\tb
\let\uc=\ta
\ua\ub\uc---\par
\endgroup
\end{document}
-
Thanks. This is what I ended up doing, but I didn't like the idea of a global temporary variable: bad things can happen before the group is finished. – Bruno Le Floch Jan 2 '11 at 22:55
% Usage
% \Aftergroup{ balanced text }
% \Aftergroup token text token
% where text is any row of tokens and token is a character that is not in text (compare \verb)
%
% Example:
% \Aftergroup {\aaa{bbb}} and \Aftergroup @\aaa{bbb}@ put \aaa{bbb} after the endgroup
% whereas \Aftergroup }\aaa{bbb}} and \Aftergroup \egroup\aaa{bbb}} put \aaa{bbb after the second "}"
%
% For the unbalanced text a problem is that we can not distinguish "{" and "\bgroup" (or
% "}"and "\egroup". Hence by
% Aftergroup @\bgroup \egroup@ the tokens inserted after the group are "{}" and by
% Aftergroup \bgroup \egroup @\bgroup the tokens "}@ are put after the group.
\makeatletter
\def\Aftergroup{\bgroup
\let\Bgroup={\let\bgroup}%give \bgroup a different meaning, } to balance
\@ifnextchar\Bgroup
{\egroup\@AGBalancedText}
{\@ifnextchar\egroup
{\egroup\def\@AGdelimiter{\egroup}\afterassignment\@AGDelNextToken\let\@let@token=}
{\egroup\@AGDelimited}}
}
\def\@AGDelimited#1{\def\@AGdelimiter{#1}\@AGDelNextToken}
\def\@AGDelNextToken{%
\expandafter\@ifnextchar\@AGdelimiter{\let\@let@token=}%end
{\@ifnextchar\bgroup
{\aftergroup{\iffalse}\fi\afterassignment\@AGDelNextToken\let\@let@token=}
{\@ifnextchar\egroup
{\iffalse{\fi\aftergroup}\afterassignment\@AGDelNextToken\let\@let@token=}
{\@AGDeltoken}
}}}
\def\@AGDeltoken#1{\aftergroup#1\@AGDelNextToken}
\def\@AGBalancedText#1{\@AGBalNextToken#1{\@AGBalNextToken}}
\def\@AGBalNextToken#1#{\@AGBalNotOpen#1\@AGBalNotOpen\@AGBalClose}
\def\@AGBalNotOpen#1{\ifx\@AGBalNotOpen#1\else\aftergroup#1\expandafter\@AGBalNotOpen\fi}
\def\@AGBalClose#1{%
\ifx\@AGBalNextToken#1\relax%
\else\ifx\@AGBalClose#1\relax\iffalse{\else\aftergroup}\fi
\expandafter\expandafter\expandafter\@AGBalNextToken%
\else\iftrue\aftergroup{\else}\fi\@AGBalNextToken#1{\@AGBalClose}%
\fi
\fi
}
\makeatother
The following code
{\Aftergroup }\def\aaa{123}}\relax}\show\aaa
{\Aftergroup {\def\aaa{123\bgroup\egroup}}}\show\aaa
{\Aftergroup @\def\aaa{123\bgroup\egroup}@}\show\aaa
then gives
> \aaa=macro:
->123\relax .
l.102 {\Aftergroup }\def\aaa{123}}\relax}\show\aaa
?
> \aaa=macro:
->123\bgroup \egroup .
l.103 ...p {\def\aaa{123\bgroup\egroup}}}\show\aaa
?
>\aaa=macro:
->123{}.
l.104 ...p @\def\aaa{123\bgroup\egroup}@}\show\aaa
in the output window of TeX.
-
Below there is an alternative. \Tokens remains local.
\documentclass{article}
\def\ta{Hello}
\def\tb{ World. }
\def\tc{How are you?}
%\def\Tokens{\ta\tb\tc}
\begin{document}
\begingroup
\def\Tokens{\ta\tb\tc}% or like above with def
\let\ua=\tc
\let\ub=\tb
\let\uc=\ta
\ua\ub\uc---\par
\expandafter\endgroup\Tokens
\show\Tokens
\end{document}
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This requires that you 'know' where the end-of-group is (and is going to be tricky to pull off in box/\halign contexts). Herbert's answer is 'general' in the sense you don't need to know where or how the group is going to end. – Joseph Wright May 14 at 9:13
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2013-12-10 13:01:48
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https://chemistry.stackexchange.com/questions/86979/auto-catalysis-mechanism
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Auto-catalysis Mechanism
I just read about the theory of auto-catalysis, and here's one thing which is sort of unclear to me:
In this type of catalysis, one of the reaction product catalyses the reaction. For example, in the oxidation of oxalic acid by potassium permanganate, Mn(II) cation is formed which is known to accelerate the reaction. So when potassium permanganate is added to a warm solution of oxalic acid in presence of dilute sulphuric acid (acid medium catalysis), initially there is a time lag before decolourisation occurs. As more permanganate is added, the decolourisation is observed to be nearly instantaneous.
Points I didn't understand:
1. How does the formation of Mn(II) cation accelerate the reaction? Could someone please help me with understanding the mechanism?
2. What is the reason for time lag in decolourisation of the solution? Also, what exactly do they mean by decolourisation here? My guess is that it's referring to the bright purple coloured permanganate solution losing its colour after getting reduced to Mn(II) cation.
3. How does the time lag reduce? Are there ways to estimate or determine the same? Of course, it'd depend on the reaction kinetics; but I'm looking for a mathematical viewpoint on how the catalyst influences reaction kinetics, except for decreasing the activation energy.
Thank you.
(1) There are several web sites giving details of this reaction and so this need not be copied out here.
(2,3) Rather than describe a complex reaction scheme it is easier to understand a generic autocatalytic reaction. The general autocatalytic reaction of species A with catalyst B has the form $\ce{A + B$\to$P + 2B}$ and the rate equation is deceptively simple and is $da/dt =- kab$ if $a$ and $b$ are the concentration A and B. The important point here is that species B is both reactant and product. As more B is produced than that there initially, the reaction accelerates as it proceeds. This is the reason for the time lag: the reaction rate is initially small but as reaction proceeds the extra B produced speeds it up. Eventually A runs out and so the reaction stops. (The colour change you mention is just a measure of the concentration of permanganate. Note also that from the Beer-Lambert law describing transmission of light the absorbing species concentration is related exponentially to light transmission (and so absorption) and this probably complicates what you observe by eye).
If $x$ is species A concentration, the rate equation is, $\displaystyle \frac{dx}{dt}=-kxb=-kx(a_0+b_0-x)$ or $\displaystyle \int\frac{dx}{(a_0+b_0-x)x}=-kt+c$ which can be integrated using partial fractions;
$$\frac{1}{(a_0+b_0-x)x}=\frac{1}{(a_0+b_0)}\left[\frac{1}{x}+\frac{1}{a_0+b_0-x} \right]$$
which produces two log functions when integrated. The initial conditions are $x = a_0$ when $t$ = 0 then the integration constant is $\ln(a_0/b_0)$ and the rate equation,
$$\ln \left(\left| \frac{xb_0}{a_0(a_0+b_0-x)} \right| \right) =-(a_0+b_0)kt$$
and the absolute value is added because the log cannot be negative. Rearranging produces $\displaystyle x = a_0\frac{a_0+b_0}{a_0+b_0e^{-(a_0+b_0)kt}}$. This and also the increase of $B =a_0+b_0-x$ are plotted.
Figure Autocatalysis $\ce{A + B$\to$P + 2B}$. Initially species A decreases slowly as the concentration of B is small. As the reaction proceeds, more B is produced and although A is reduced overall, the rate increases and A is consumed even more rapidly. At longer times, the concentration of A becomes so small that even though that of B is large the reaction rate is slow.
• Thank you. Could you specifically share certain links which could answer the first part (1) of my question? – strawberry-sunshine Jan 11 '18 at 9:30
• I searched for 'oxidation of oxalic acid by permanganate' , there are several pages with details including this one chemistry.stackexchange.com/questions/4310/… – porphyrin Jan 11 '18 at 11:28
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2020-11-26 10:14:01
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http://math.stackexchange.com/questions/101252/how-to-demonstrate-derivability
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# How to demonstrate derivability?
If given a function $f$ that I am supposed to demonstrate is derivable, is it enough for me to try to compute its derivative $f'$ and if the result is continuous, can I then state that the original function $f$ is derivable?
-
continuously differentiable function – pedja Jan 22 '12 at 9:27
I guess you mean differentiable. If the derivative exists at some point, then the function is differentiable there. If the derivative exists and is continuous, the function is said to be continuously differentiable. – aelguindy Jan 22 '12 at 9:33
1. A region where $f$ is clearly derivable because it is a composition of elementary functions known to be derivable (polynmials, exponentials, trigonometric functions,...)
Let me give an example: $f(x)=x^2\sin\dfrac{1}{x}$ if $x\ne0$, $f(0)=0$. $f$ is derivable in $\mathbb{R}\setminus\{0\}$ because it is a composition and product of derivable functions. What hapens at $x=0$? Using the definition of derivative, you can show that $f'(0)=0$, s0 that $f$ is derivable in $\mathbb{R}$. Observe hat $f'$ is not continuous at $x=0$ (the limit as $x\to0$ of $f'(x)$ does not exist.) Continuity of $f'$ is not a necessary condition for $f$ to be derivable. Functions with a continuous derivative are called $C^1$ functions.
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2014-10-31 14:30:30
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https://www.gamedev.net/forums/topic/522165-ideas-for-water-foam/
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# Ideas for Water Foam
This topic is 3407 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I am attempting to render shoreline foam on my water plane when it intersects with the land. I have got the hardest part out of the way but I am struggling to think of ideas on how to make it blend properly. My method was to create a depth stencil buffer, and render the water plane and the land to it. This was output to a texture. I then lower my land and render the water plane using the depth texture(with proj coords) so that the sillhoutte formed will tell me where the outlines of the water is in screen space. This works well in that it tells me which pixels are near the edge of the land as I can just do this in the shader.
float depth = tex2D(Depth, projTex.xy).a;
if(depth > 0.0f)
{
OUT.Colour.a = 0.5f;
}
As you can see the result is not very pleasing. Orginally I tried using a foam texture which was animated, this was then modulated with the water colour and applied, but as the stencil is a sharp cut off point there is a clear line where the water switchs to foam. Another thing I thought about doing was using the depth level as a weighting for the foam but due to inaccuracy issues with the depth buffer writes and limitations involved with the near/far clip planes I decided not to do it. Does anybody have any ideas on how I can make it blend effciently, ideally I want to avoid any more render pass's as the water shader is fairly heavy as it is.
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Quote:
Original post by reaper93Does anybody have any ideas on how I can make it blend effciently, ideally I want to avoid any more render pass's as the water shader is fairly heavy as it is.
You can use range and noise maps in addition to make a foam,it will be enough,I think.In this case noise map used for water color,alpha and reflection change,because foam has another appearance then water.Range map used for finding places where foam must be.
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Hi Kronhin, thanks for your reply, please can you explain what you mean with the term "range maps"
I don't have any issues in making the foam colour, its just on what to use to determine where foam should be
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You could manually make a mask texture based on your terrain and fit it to the water mesh.
Or use a depth map to reconstruct xyz world space positions. This is a common method used in deferred rendering and other things. Then just project this onto the water using the camera view/projection matrices, and determine the water depth from the y position minus the water level. Then normalize this value to a 0-1 range (use pow() and such to make the transitions the way you want) and lerp between the water and the foam texture.
You mentioned you had a problem using depth rendering, but normally you handle this by writing to a separate floating point surface in a pixel shader (possibly using multiple render targets); this shouldn't be affected much by accuracy issues or far/near planes (use the farthest z plane if you have mulitple frustums when reconstructing world space xyz).
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Quote:
Original post by reaper93Hi Kronhin, thanks for your reply, please can you explain what you mean with the term "range maps"I don't have any issues in making the foam colour, its just on what to use to determine where foam should be
I think Matt Aufderheide told enough in case of "plane" foam.
[Edited by - Krokhin on January 26, 2009 5:44:47 AM]
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Ok thanks, will look into giving that a go later this week.
Will update when I get it working.
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If it's an option for you to render the water using the terrain geometry, there may be no need to wrestle with intermediate maps. Basically you pass in the terrain patches to the vertex shader and set the y coord (or whatever you use for height) to the predifined water level, plus or minus some wave function. Since you have the original y coord from the terrain patch however, you can determine how deep the water is at a given vertex/pixel and blend in some foam where the water is very shallow, ie. where there's a coastline.
Additionally this 'water depth' information comes in handy for faking refraction, since typically you'll want shallow water to be transparent while deep waters should just get a darker blue hue.
Hope this makes sense [smile]
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Quote:
Original post by remigiusIf it's an option for you to render the water using the terrain geometry, there may be no need to wrestle with intermediate maps. Basically you pass in the terrain patches to the vertex shader and set the y coord (or whatever you use for height) to the predifined water level, plus or minus some wave function. Since you have the original y coord from the terrain patch however, you can determine how deep the water is at a given vertex/pixel and blend in some foam where the water is very shallow, ie. where there's a coastline. Additionally this 'water depth' information comes in handy for faking refraction, since typically you'll want shallow water to be transparent while deep waters should just get a darker blue hue.Hope this makes sense [smile]
This is a good suggestion.
I use something similar for setting up shore line effects, blending water into the landscape where shallow, and making sure deep water looks deep... Luckily with a 4 component height map and bump map, which is still as fast as it gets on a GPU with textures, there are generally 2 channels free which can be used for this 'side' info....
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2018-05-26 10:21:48
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https://bair.berkeley.edu/blog/2018/12/12/rllib/
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An earlier version of this post is on the RISELab blog. It is posted here with the permission of the authors.
We just rolled out general support for multi-agent reinforcement learning in Ray RLlib 0.6.0. This blog post is a brief tutorial on multi-agent RL and how we designed for it in RLlib. Our goal is to enable multi-agent RL across a range of use cases, from leveraging existing single-agent algorithms to training with custom algorithms at large scale.
# Why multi-agent reinforcement learning?
We’ve observed that in applied RL settings, the question of whether it makes sense to use multi-agent algorithms often comes up. Compared to training a single policy that issues all actions in the environment, multi-agent approaches can offer:
• A more natural decomposition of the problem. For example, suppose one wants to train policies for cellular antenna tilt control in urban environments. Instead of training a single “super-agent” that controls all the cellular antennas in a city, it is more natural to model each antenna as a separate agent in the environment because only neighboring antennas and user observations are relevant to each site. In general, we may wish to have each agent’s action and/or observation space restricted to only model the components that it can affect and those that affect it.
• Potential for more scalable learning. First, decomposing the actions and observations of a single monolithic agent into multiple simpler agents not only reduces the dimensionality of agent inputs and outputs, but also effectively increases the amount of training data generated per step of the environment.
Second, partitioning the action and observation spaces per agent can play a similar role to the approach of imposing temporal abstractions, which has been successful in increasing learning efficiency in the single-agent setting. Relatedly, some of these hierarchical approaches can be implemented explicitly as multi-agent systems.
Finally, good decompositions can also lead to the learning of policies that are more transferable across different variations of an environment, i.e., in contrast to a single super-agent that may over-fit to a particular environment.
Figure 1: Single-agent approaches (a) and (b) in comparison with multi-agent RL (c).
Some examples of multi-agent applications include:
Traffic congestion reduction: It turns out that by intelligently controlling the speed of a few autonomous vehicles we can drastically increase the traffic flow. Multi-agent can be a requirement here, since in mixed-autonomy settings, it is unrealistic to model traffic lights and vehicles as a single agent, which would involve the synchronization of observations and actions across all agents in a wide area.
Flow simulation, without AVs and then with AV agents (red vehicles):
Antenna tilt control: The joint configuration of cellular base stations can be optimized according to the distribution of usage and topology of the local environment. Each base station can be modeled as one of multiple agents covering a city.
OpenAI Five: Dota 2 AI agents are trained to coordinate with each other to compete against humans. Each of the five AI players is implemented as a separate neural network policy and trained together with large-scale PPO.
# Introducing multi-agent support in RLlib
In this blog post we introduce general purpose support for multi-agent RL in RLlib, including compatibility with most of RLlib’s distributed algorithms: A2C / A3C, PPO, IMPALA, DQN, DDPG, and Ape-X. In the remainder of this blog post we discuss the challenges of multi-agent RL, show how to train multi-agent policies with these existing algorithms, and also how to implement specialized algorithms that can deal with the non-stationarity and increased variance of multi-agent environments.
There are currently few libraries for multi-agent RL, which increases the initial overhead of experimenting with multi-agent approaches. Our goal is to reduce this friction and make it easy to go from single-agent to multi-agent RL in both research and application.
# Why supporting multi-agent is hard
Building software for a rapidly developing field such as RL is challenging, multi-agent especially so. This is in part due to the breadth of techniques used to deal with the core issues that arise in multi-agent learning.
Consider one such issue: environment non-stationarity. In the below figure, the red agent’s goal is to learn how to regulate the speed of the traffic. The blue agents are also learning, but only to greedily minimize their own travel time. The red agent may initially achieve its goal by simply driving at the desired speed. However, in a multi-agent environment, the other agents will learn over time to meet their own goals – in this case, bypassing the red agent. This is problematic since from a single-agent view (i.e., that of the red agent), the blue agents are “part of the environment”. The fact that environment dynamics are changing from the perspective of a single agent violates the Markov assumptions required for the convergence of Q-learning algorithms such as DQN.
Figure 2: Non-stationarity of environment: Initially (a), the red agent learns to regulate the speed of the traffic by slowing down. However, over time the blue agents learn to bypass the red agent (b), rendering the previous experiences of the red agent invalid.
A number of algorithms have been proposed that help address this issue, e.g., LOLA, RIAL, and Q-MIX. At a high level, these algorithms take into account the actions of the other agents during RL training, usually by being partially centralized for training, but decentralized during execution. Implementation wise, this means that the policy networks may have dependencies on each other, i.e., through a mixing network in Q-MIX:
Figure 3: The Q-mix network architecture, from QMIX: Monotonic Value Function Factorisation for Deep Multi-Agent Reinforcement Learning. Individual Q-estimates are aggregated by a monotonic mixing network for efficiency of final action computation.
Similarly, policy-gradient algorithms like A3C and PPO may struggle in multi-agent settings, as the credit assignment problem becomes increasingly harder with more agents. Consider a traffic gridlock between many autonomous agents. It is easy to see that the reward given to an agent – here reflecting that traffic speed has dropped to zero – will have less and less correlation with the agent’s actions as the number of agents increases:
Figure 4: High variance of advantage estimates: In this traffic gridlock situation, it is unclear which agents' actions contributed most to the problem -- and when the gridlock is resolved, from any global reward it will be unclear which agents get credit.
One class of approaches here is to model the effect of other agents on the environment with centralized value functions (the “Q” boxes in Figure 5) function, as done in MA-DDPG. Intuitively, the variability of individual agent advantage estimates can be greatly reduced by taking into account the actions of other agents:
Figure 5: The MA-DDPG architecture, from Multi-Agent Actor-Critic for Mixed Cooperative-Competitive Environments. Policies run using only local information at execution time, but may take advantage of global information at training time.
So far we’ve seen two different challenges and approaches for tackling multi-agent RL. That said, in many settings training multi-agent policies using single-agent RL algorithms can yield surprisingly strong results. For example, OpenAI Five has been successful using only a combination of large-scale PPO and a specialized network model. The only considerations for multi-agent are the annealing of a “team spirit” hyperparameter that influences the level of shared reward, and a shared “max-pool across Players” operation in the model to share observational information.
# Multi-agent training in RLlib
So how can we handle both these specialized algorithms and also the standard single-agent RL algorithms in multi-agent settings? Fortunately RLlib’s design makes this fairly simple. The relevant principles to multi-agent are as follows:
• Policies are represented as objects: All gradient-based algorithms in RLlib declare a policy graph object, which includes a policy model $\pi_\theta(o_t)$, a trajectory postprocessing function $post_\theta(traj)$, and finally a policy loss $L(\theta; X)$. This policy graph object specifies enough for the distributed framework to execute environment rollouts (by querying $\pi_\theta$), collate experiences (by applying $post_\theta$), and finally to improve the policy (by descending $L(\theta; X)$).
• Policy objects are black boxes: To support multi-agent, RLlib just needs to manage the creation and execution of multiple policy graphs per environment, and add together the losses during policy optimization. Policy graph objects are treated largely as black boxes by RLlib, which means that they can be implemented in any framework including TensorFlow and PyTorch. Moreover, policy graphs can internally share variables and layers to implement specialized algorithms such as Q-MIX and MA-DDPG, without special framework support.
To make the application of these principles concrete, In the next few sections we walk through code examples of using RLlib’s multi-agent APIs to execute multi-agent training at scale.
# Multi-agent environment model
We are not aware of a standard multi-agent environment interface, so we wrote our own as a straightforward extension of the gym interface. In a multi-agent environment, there can be multiple acting entities per step. As a motivating example, consider a traffic control scenario (Figure 6) where multiple controllable entities (e.g., traffic lights, autonomous vehicles) work together to reduce highway congestion.
In this scenario,
• Each of these agents can act at different time-scales (i.e., act asynchronously).
• Agents can come and go from the environment as time progresses.
Figure 6. RLlib multi-agent environments can model multiple independent agents that come and go over time. Different policies can be assigned to agents as they appear.
This is formalized in the MultiAgentEnv interface, which can returns observations and rewards from multiple ready agents per step:
# Example: using a multi-agent env
> env = MultiAgentTrafficEnv(num_cars=20, num_traffic_lights=5)
# Observations are a dict mapping agent names to their obs. Not all
# agents need to be present in the dict in each time step.
> print(env.reset())
{
"car_1": [[...]],
"car_2": [[...]],
"traffic_light_1": [[...]],
}
# Actions should be provided for each agent that returned an observation.
> new_obs, rewards, dones, infos = env.step(
actions={"car_1": ..., "car_2": ...})
# Similarly, new_obs, rewards, dones, infos, etc. also become dicts
> print(rewards)
{"car_1": 3, "car_2": -1, "traffic_light_1": 0}
# Individual agents can early exit; env is done when "__all__" = True
> print(dones)
{"car_2": True, "__all__": False}
Any Discrete, Box, Dict, or Tuple observation space from OpenAI gym can be used for these individual agents, allowing for multiple sensory inputs per agent (including communication between agents, it desired).
# Multiple levels of API support
At a high level, RLlib models agents and policies as objects that may be bound to each other for the duration of an episode (Figure 7). Users can leverage this abstraction to varying degrees, from just using a single-agent shared policy, to multiple policies, to fully customized policy optimization:
Figure 7: The multi-agent execution model in RLlib compared with single-agent execution.
## Level 1: Multiple agents, shared policy
If all “agents” in the env are homogeneous (e.g., multiple independent cars in a traffic simulation), then it is possible to use existing single-agent algorithms for training. Since there is still only a single policy being trained, RLlib only needs to internally aggregate the experiences of the different agents prior to policy optimization. The change is minimal on the user side:
from single-agent:
register_env("your_env", lambda c: YourEnv(...))
trainer = PPOAgent(env="your_env")
while True:
print(trainer.train()) # distributed training step
to multi-agent:
register_env("your_multi_env", lambda c: YourMultiEnv(...))
trainer = PPOAgent(env="your_multi_env")
while True:
print(trainer.train()) # distributed training step
Note that the PPO “Agent” here is a just a naming convention carried over from the single-agent API. It acts more as a “trainer” of agents than an actual agent.
## Level 2: Multiple agents, multiple policies
To handle multiple policies, this requires the definition of which agent(s) are handled by each policy. We handle this in RLlib via a policy mapping function, which assigns agents in the env to a particular policy when the agent first enters the environment. In the following examples we consider a hierarchical control setting where supervisor agents assign work to teams of worker agents they oversee. The desired configuration is to have a single supervisor policy and an ensemble of two worker policies:
def policy_mapper(agent_id):
if agent_id.startswith("supervisor_"):
return "supervisor_policy"
else:
return random.choice(["worker_p1", "worker_p2"])
In the example, we always bind supervisor agents to the single supervisor policy, and randomly divide other agents between an ensemble of two different worker policies. These assignments are done when the agent first enters the episode, and persist for the duration of the episode. Finally, we need to define the policy configurations, now that there is more than one. This is done as part of the top-level agent configuration:
trainer = PPOAgent(env="control_env", config={
"multiagent": {
"policy_mapping_fn": policy_mapper,
"policy_graphs": {
"supervisor_policy":
(PPOPolicyGraph, sup_obs_space, sup_act_space, sup_conf),
"worker_p1": (
(PPOPolicyGraph, work_obs_s, work_act_s, work_p1_conf),
"worker_p2":
(PPOPolicyGraph, work_obs_s, work_act_s, work_p2_conf),
},
"policies_to_train": [
"supervisor_policy", "worker_p1", "worker_p2"],
},
})
while True:
print(trainer.train()) # distributed training step
This would generate a configuration similar to that shown in Figure 2. You can pass in a custom policy graph class for each policy, as well as different policy config dicts. This allows for any of RLlib’s support for customization (e.g., custom models and preprocessors) to be used per policy, as well as wholesale definition of a new class of policy.
## Level 3: Custom training strategies
For advanced applications or research use cases, it is inevitable to run into the limitations of any framework.
For example, let’s suppose multiple training methods are desired: some agents will learn with PPO, and others with DQN. This can be done in a way by swapping weights between two different trainers (there is a code example here), but this approach won’t scale with even more types of algorithms thrown in, or if e.g., you want to use experiences to train a model of the environment at the same time.
In these cases we can fall back to RLlib’s underlying system Ray to distribute computations as needed. Ray provides two simple parallel primitives:
• Tasks, which are Python functions executed asynchronously via func.remote()
• Actors, which are Python classes created in the cluster via class.remote(). Actor methods can be called via actor.method.remote().
RLlib builds on top of Ray tasks and actors to provide a toolkit for distributed RL training. This includes:
• Policy graphs (as seen in previous examples)
• Policy evaluation: the PolicyEvaluator class manages the environment interaction loop that generates batches of experiences. When created as Ray actors, it can be used to gather experiences in a distributed way.
• Policy optimization: these implement distributed strategies for improving policies. You can use one of the existing optimizers or go with a custom strategy.
For example, you can create policy evaluators to gather multi-agent rollouts, and then process the batches as needed to improve the policies:
# Initialize a single-node Ray cluster
ray.init()
# Create local instances of your custom policy graphs
sup, w1, w2 = SupervisorPolicy(), WorkerPolicy(), WorkerPolicy()
# Create policy evaluators (Ray actor processes running in the cluster)
evaluators = []
for i in range(16):
ev = PolicyEvaluator.as_remote().remote(
env_creator=lambda ctx: ControlEnv(),
policy_graph={
"supervisor_policy": (SupervisorPolicy, ...),
"worker_p1": ..., ...},
policy_mapping_fn=policy_mapper,
sample_batch_size=500)
evaluators.append(ev)
while True:
# Collect experiences in parallel using the policy evaluators
futures = [ev.sample.remote() for ev in evaluators]
batch = MultiAgentBatch.concat_samples(ray.get(futures))
# >>> print(batch)
# MultiAgentBatch({
# "supervisor_policy": SampleBatch({
# "obs": [[...], ...], "rewards": [0, ...], ...
# }),
# "worker_p1": SampleBatch(...),
# "worker_p2": SampleBatch(...),
# })
your_optimize_func(sup, w1, w2, batch) # Custom policy optimization
# Broadcast new weights and repeat
for ev in evaluators:
ev.set_weights.remote({
"supervisor_policy": sup.get_weights(),
"worker_p1": w1.get_weights(),
"worker_p2": w2.get_weights(),
})
In summary, RLlib provides several levels of APIs targeted at progressively more requirements for customizability. At the highest levels this provides a simple “out of the box” training process, but you retain the option to piece together your own algorithms and training strategies from the core multi-agent abstractions. To get started, here are a couple intermediate level scripts that can be run directly: multiagent_cartpole.py, multiagent_two_trainers.py.
# Performance
RLlib is designed to scale to large clusters – and this applies to multi-agent mode as well – but we also apply optimizations such as vectorization for single-core efficiency. This allows for effective use of the multi-agent APIs on small machines.
To show the importance of these optimizations, in the below graph we plot single-core policy evaluation throughout vs the number of agents in the environment. For this benchmark the observations are small float vectors, and the policies are small 16x16 fully connected networks. We assign each agent to a random policy from a pool of 10 such policy networks. RLlib manages over 70k actions/s/core at 10000 agents per environment (the bottleneck becomes Python overhead at this point). When vectorization is turned off, experience collection slows down by 40x:
We also evaluate the more challenging case of having many distinct policy networks used in the environment. Here we can still leverage vectorization to fuse multiple TensorFlow calls into one, obtaining more stable per-core performance as the number of distinct policies scales from 1 to 50:
# Conclusion
This blog post introduces a fast and general framework for multi-agent reinforcement learning. We’re currently working with early users of this framework in BAIR, the Berkeley Flow team, and industry to further improve RLlib. Try it out with 'pip install ray[rllib]' and tell us about your own use cases!
Documentation for RLlib and multi-agent support can be found at https://rllib.io.
Erin Grant, Eric Jang, and Eugene Vinitsky provided helpful input for this blog post.
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2019-08-19 14:08:27
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https://math.stackexchange.com/questions/2629500/average-and-standard-deviation-equation-system
|
Average and standard deviation equation system
Every exam has 100 points. Student's average of 3 exams is 88 points and standard deviation is 3. How many points should he get on fourth and fifth exam so that average is 90 and standard deviation stays same (=3).
I tried solving equations $(x_1+x_2+x_3)/3=88$ and $(x_1+x_2+x_3+x_4+x_5)/5=90$. Since $x_1+x_2+x_3=264$ it means that $(264+x_4+x_5)/5=90$ so $x_4+x_5=186$. Then I tried to use equations for standard deviation but I haven't managed to get anything useful.
• What if I told you there is no solution? – Parcly Taxel Jan 31 '18 at 11:36
• How to proof that? – Mirjan Pecenko Jan 31 '18 at 11:41
• See my answer.${}$ – Parcly Taxel Jan 31 '18 at 11:49
• Is the std defined by the population or sample formula ? – Yves Daoust Jan 31 '18 at 11:51
• I think by the population – Mirjan Pecenko Jan 31 '18 at 11:54
We have $$\sqrt{\frac{(x_1-88)^2+(x_2-88)^2+(x_3-88)^2}3}=3$$ $$(x_1-88)^2+(x_2-88)^2+(x_3-88)^2=27$$ How do we get from $(x-88)^2=x^2-176x+88^2$ to $(x-90)^2=x^2-180x+90^2$? We add $-4x+356$. $$-4x_1+356-4x_2+356-4x_3+356=-4(x_1+x_2+x_3)+1068=-4(3\cdot88)+1068=12$$ $$(x_1-90)^2+(x_2-90)^2+(x_3-90)^2=27+12=39$$ Now $$\sqrt{\frac{39+(x_4-90)^2+(x_5-90)^2}5}=3$$ $$(x_4-90)^2+(x_5-90)^2=6$$ Let $z_4=x_4-90$ and $z_5=x_5-90$. Then $z_4+z_5=6$ while $z_4^2+z_5^2=6$. Squaring the first equation and subtracting the second from it and halving gives $z_4z_5=15$.
Thus we may form the polynomial $x^2-6x+15$, whose roots should be $z_4$ and $z_5$. However, we find that the polynomial has no real roots. So there is no solution to the original problem – the average and standard deviation cannot be changed by the further two tests to the given values.
From the averages,
$$x_4+x_5=5\cdot 90-3\cdot 88=186$$
and from the variances,
$$x_4^2+x_5^2=5(3^2+90^2)-3(3^2+88^2)=17286.$$
Then
$$(x_4-x_5)^2=2(x_4^2+x_5^2)-(x_4+x_5)^2=-24.$$ (!)
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2019-06-20 15:29:18
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https://www.gradesaver.com/textbooks/math/algebra/algebra-1-common-core-15th-edition/chapter-8-polynomials-and-factoring-8-1-adding-and-subtracting-polynomials-practice-and-problem-solving-exercises-page-490/28
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## Algebra 1: Common Core (15th Edition)
-3$c^{7}$ + 8$c^{3}$ + c; Septic trinomial
STEP 1: Write out the original expression. c + 8$c^{3}$ - 3$c^{7}$ The expression has 3 terms in it, thus classifying itself as a trinomial. STEP 2: Arrange the trinomial into standard form, meaning that the terms are listed from greatest exponent value to the least exponent value. -3$c^{7}$ + 8$c^{3}$ + c The degree is the greatest exponent value, and in -3$c^{7}$ + 8$c^{3}$ + c, the degree is 7. The name of a trinomial with a degree of 7 is a septic trinomial.
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2019-01-24 09:11:30
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https://www.studysmarter.us/textbooks/physics/physics-for-scientists-and-engineers-a-strategic-approach-with-modern-physics-4th/work-and-kinetic-energy/q54-an-electric-dipole-consists-of-two-equal-but-opposite-el/
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Q.54
Expert-verified
Found in: Page 230
### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Book edition 4th
Author(s) Randall D. Knight
Pages 1240 pages
ISBN 9780133942651
# An electric dipole consists of two equal but opposite electric charges, +q and -q, separated by a small distance d. If another charge Q is distance x from the center of the dipole, in the plane perpendicular to the line between +q and -q, and if x W d, the dipole exerts an electric force on charge Q, where K is a constant. How much work does the electric force do if charge Q moves from distance ?
Work done by electric force is
See the step by step solution
## Step 1: Content Introduction
The work done by variable force is,
Here, is the force, is small displacement.
## Step 2: Content Explanation
Consider dipole exerts an electric force is
use formula to find work done by force F using distance to distance.
Use power rule of integral,
Hence work done by electric force is
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2022-12-07 05:51:22
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https://www.physicsforums.com/threads/parallel-axis-theorem.751596/
|
# Homework Help: Parallel axis theorem
1. May 1, 2014
### RyRy19
A uniform rod of mass M=5.0 kg and length ℓ=20 cm is pivoted on a frictionless hinge at the end
of it. The rod is held horizontally and then released.
a) Use the parallel-axis theorem to determine the moment of inertia of the rod about the hinge (ie
its end).
b) Determine the angular velocity of the rod when it reaches the vertical position and the speed of
the rod tip’s at this point.
I=MR^2
Ok so what I did first was 1/12 ML^2 as "I" about the hinge is 1/12ML2 + (M*(L/2)2) to work out A.
I don't really know how to work out b though, apart from possibly using that to work out the distance from the radius and then doing work done to then find out ω, I assume.
2. May 1, 2014
### SteamKing
Staff Emeritus
Draw a picture of the rod in the horizontal position before it is released. At the moment of release, what are the forces acting on the rod? How do these forces cause the rod to rotate about the hinged end? Draw a free body diagram.
Do you know the relationship between angular acceleration and the moment acting on a body? You know, something you might have picked up in a physics class about rotational motion.
3. May 7, 2014
### RyRy19
So If I work out I = (1/12)*M*R2
Then τ = MG* 10x10-2 (as the length of the rod is 20x10-2 - Only weight is
acting)
Then I use the solution to that and plug it into (1/2)ω2*I and solve for ω?
4. May 7, 2014
### haruspex
Except that you want the moment about the hinge, as calculated in (a).
As you suggested, the easy way is to use work conservation. For that, you don't need the torque. What is the rotational KE gained? What PE has been lost?
5. May 8, 2014
### RyRy19
So I would do mgh, for the GPE and then solve to work out v from 1/2mv^2 ? I'm guessing you take the height in regards to a triangle? Find the resultant vector or rather the hypotenuse?
If so I think I understand.
Thanks for the help.
6. May 8, 2014
### dauto
No I don't think you understand. What's the height h? Why aren't you using the formula for the kinetic energy of rotational motion?
7. May 8, 2014
### haruspex
No, $\frac 12 I\omega^2$, as you posted earlier. But make sure to use the right I.
Triangle? You want the change in height of the mass centre of the rod.
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2018-05-27 03:52:52
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https://gmatclub.com/forum/if-243-x-463-y-n-where-x-and-y-are-positive-integers-102054.html
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# If 243^x*463^y = n, where x and y are positive integers
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If $$243^x*463^y =n$$ , where x and y are positive integers, what is the units digit of n?
(1) x + y = 7
(2) x = 4
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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If $$243^x*463^y =n$$, where x and y are positive integers, what is the units digit of n?
The units digit of $$243^x$$ is the same as the units digit of $$3^x$$ and similarly the units digit of $$463^y$$ is the same as the units digit of $$3^y$$, so the units digit of $$243^x*463^y$$ equals to the units digit of $$3^x*3^y=3^{x+y}$$. So, knowing the value of $$x+y$$ is sufficient to determine the units digit of $$n$$.
(1) $$x + y = 7$$. Sufficient. (As cyclicity of units digit of $$3$$ in integer power is $$4$$, units digit of $$3^7$$ would be the same as of units digit of $$3^3$$ which is $$7$$)
(2) $$x=4$$. No info about $$y$$. Not sufficient.
Hope it helps.
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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03 Oct 2010, 08:50
Yup!! A it is....equation can be treated like 3^x*3^y hence (x+y)'s value can provide us the last digit...
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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03 Mar 2014, 04:18
I have a doubt. Cyclicity of unit digit of 3 is 4. Hence we know that every fourth power of 3 (3^4, 3^8, 3^12) will have the same unit digit, 1. Hence when option B says x = 4, knowing that x and y are positive integers, we know that xy will be a multiple of 4. Unit digit of 3^4k is always 1 isn't it? Shouldn't this be sufficient information?
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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03 Mar 2014, 04:21
siriusblack1106 wrote:
I have a doubt. Cyclicity of unit digit of 3 is 4. Hence we know that every fourth power of 3 (3^4, 3^8, 3^12) will have the same unit digit, 1. Hence when option B says x = 4, knowing that x and y are positive integers, we know that xy will be a multiple of 4. Unit digit of 3^4k is always 1 isn't it? Shouldn't this be sufficient information?
I think you are missing that $$3^x*3^y=3^{x+y}$$, so the exponent is x+y not xy.
Does this make sense?
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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03 Mar 2014, 04:26
Yes! Can't believe I just made that mistake. Such mistakes are gonna cost me. :/
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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03 Mar 2014, 04:28
siriusblack1106 wrote:
Yes! Can't believe I just made that mistake. Such mistakes are gonna cost me. :/
Yes, careless errors are the #1 cause of score drops on the GMAT! They cause you to miss easier questions, hurting your score a lot more than not know how to solve the harder ones. So, be more careful, before you submit your answer, double-check that it’s the answer to the proper question.
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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18 May 2014, 08:53
1
KUDOS
243 = 3^5
463 ends with a 3. So we have to know how many times we will multiply 3's at the end of each numbers.
1) 7 times - SUF
2) we dont know Y - INSUF
Choose (a)
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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02 Nov 2014, 17:48
Bunuel wrote:
If $$243^x*463^y =n$$, where x and y are positive integers, what is the units digit of n?
The units digit of $$243^x$$ is the same as the units digit of $$3^x$$ and similarly the units digit of $$463^y$$ is the same as the units digit of $$3^y$$, so the units digit of $$243^x*463^y$$ equals to the units digit of $$3^x*3^y=3^{x+y}$$. So, knowing the value of $$x+y$$ is sufficient to determine the units digit of $$n$$.
(1) $$x + y = 7$$. Sufficient. (As cyclicity of units digit of $$3$$ in integer power is $$4$$, units digit of $$3^7$$ would be the same as of units digit of $$3^3$$ which is $$7$$)
(2) $$x=4$$. No info about $$y$$. Not sufficient.
Hope it helps.
Hi Bunuel,
But surely shouldn't it matter if 3 is raised to 1 and or 6? Meaning, if it's 3^3 + 3^4 = 7 + 1 = 8. But, if its 3^2+3^5 = 9 + 3 = 12, units of 2. Doesn't that yield insufficient?
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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03 Nov 2014, 00:50
russ9 wrote:
Bunuel wrote:
If $$243^x*463^y =n$$, where x and y are positive integers, what is the units digit of n?
The units digit of $$243^x$$ is the same as the units digit of $$3^x$$ and similarly the units digit of $$463^y$$ is the same as the units digit of $$3^y$$, so the units digit of $$243^x*463^y$$ equals to the units digit of $$3^x*3^y=3^{x+y}$$. So, knowing the value of $$x+y$$ is sufficient to determine the units digit of $$n$$.
(1) $$x + y = 7$$. Sufficient. (As cyclicity of units digit of $$3$$ in integer power is $$4$$, units digit of $$3^7$$ would be the same as of units digit of $$3^3$$ which is $$7$$)
(2) $$x=4$$. No info about $$y$$. Not sufficient.
Hope it helps.
Hi Bunuel,
But surely shouldn't it matter if 3 is raised to 1 and or 6? Meaning, if it's 3^3 + 3^4 = 7 + 1 = 8. But, if its 3^2+3^5 = 9 + 3 = 12, units of 2. Doesn't that yield insufficient?
Are you sure you are reading the question correctly? It's 243^x*463^y, 243^x multiplied by 463^y not 243^x + 463^y...
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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28 Jul 2016, 14:01
Well I don't agree the answer should be indeed D.
Option 1 suggests x+y=7 this can have multiple x and y combinations like (1,6) (2,5) (4,3) and so on so the units digit of 243^x and 463^y will differ .
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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28 Jul 2016, 17:37
sandeep211986 wrote:
Well I don't agree the answer should be indeed D.
Option 1 suggests x+y=7 this can have multiple x and y combinations like (1,6) (2,5) (4,3) and so on so the units digit of 243^x and 463^y will differ .
Hey Buddy,
All the combinations, for x+y=7, will yield the same units digit. Consider the following
x=1,y=6
243*463*463*463*463*463*463 ~~ To find units digit we just need 3^1 * 3^6 = 3^7, i.e 7 (units digit of 2187)
Same goes with other combinations. 3^7 ends up deciding the units digit.
Should the question would have been something like, 245^x * 463^y = n, the combinations of different values of x & y would have yielded different units digits.
Hope that clears
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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20 Dec 2016, 14:25
amandeep_k wrote:
If (243) x(463) y = n, where x and y are positive integers, what is the units digit of n?
(1) x + y = 7
(2) x = 4
I can't get how the answer can be A.
Had it been $$243^x * 463^y$$. In this case we'll get same base 3 and we can add the powers
$$3^0*3^7$$
$$3^1*3^6$$
$$3^2*3^5$$
...
In each case we'll get $$3^7$$ which units digit we can identify. That that will be sufficient.
But we have 3*x*3*y = 9*x*y which can take any values. Not sufficient. I can't get the idea how answer can be A.
Please correct me if I'm wrong.
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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20 Dec 2016, 14:39
vitaliyGMAT wrote:
amandeep_k wrote:
If (243) x(463) y = n, where x and y are positive integers, what is the units digit of n?
(1) x + y = 7
(2) x = 4
I can't get how the answer can be A.
Had it been $$243^x * 463^y$$. In this case we'll get same base 3 and we can add the powers
$$3^0*3^7$$
$$3^1*3^6$$
$$3^2*3^5$$
...
In each case we'll get $$3^7$$ which units digit we can identify. That that will be sufficient.
But we have 3*x*3*y = 9*x*y which can take any values. Not sufficient. I can't get the idea how answer can be A.
Please correct me if I'm wrong.
Hi, You are right. my question was wrong. Sorry for the inconvenience.
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink]
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18 Dec 2017, 01:58
A.
If we map the cyclicity of 3--> with x + y values --> 3 and 4/ 4 and 3/ 1 and 6; 6 and 1/ 5 and 2; 2 and 5 [3,9,7,1,3,9,7,1...] --> the answer is always 7. [7*1 = 7; 3*9 = _7..so on]
St 2 . No value for y
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Re: If 243^x*463^y = n, where x and y are positive integers [#permalink] 18 Dec 2017, 01:58
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If you model as such, you neglect dependencies among observations – individuals from the same block are not independent, yielding residuals that correlate within block. Because we have no obvious outliers, the leverage analysis provides acceptable results. The data contain no missing values. $$Q_j$$ is a $$n_i \times q_j$$ dimensional design matrix for the With the explanations provided by our random effects the residuals are about zero, meaning that this linear mixed-effects model is a good fit for the data. For simplicity I will exclude these alongside gen, since it contains a lot of levels and also represents a random sample (from many other extant Arabidopsis genotypes). These random effects essentially give structure to the error term “ε”. It is a data set of instructor evaluation ratings, where the inputs (covariates) include categories such as students and departments, and our response variable of interest is the instructor evaluation rating. While both linear models and LMMs require normally distributed residuals with homogeneous variance, the former assumes independence among observations and the latter normally distributed random effects. 2. inside the lm call, however you will likely need to preprocess the resulting interaction terms. As a result, classic linear models cannot help in these hypothetical problems, but both can be addressed using linear mixed-effect models (LMMs). For example, a plant grown under the same conditions but placed in the second rack will be predicted to have a smaller yield, more precisely of . If only Nathaniel E. Helwig (U of Minnesota) Linear Mixed-Effects Regression Updated 04-Jan-2017 : Slide 9 For a single group, Both culturing in Petri plates and transplantation, albeit indistinguishable, negatively affect fruit yield as opposed to normal growth. Volume 83, Issue 404, pages 1014-1022. http://econ.ucsb.edu/~doug/245a/Papers/Mixed%20Effects%20Implement.pdf. $$Y, X, \{Q_j\}$$ and $$Z$$ must be entirely observed. We first need to setup a control setting that ensures the new models converge. gen within popu). additively shifted by a value that is specific to the group. We could similarly use an ANOVA model. You can also introduce polynomial terms with the function poly. Note that it is not a good idea to add new terms after optimizing the random structure, I did so only because otherwise there would be nothing to do with respect to the fixed structure. The variance components arguments to the model can then be used to Let’s check how the random intercepts and slopes distribute in the highest level (i.e. Random intercepts models, where all responses in a group are additively shifted by a value that is specific to the group. The “random effects parameters” $$\gamma_{0i}$$ and Take a look into the distribution of the random effects with plot(ranef(MODEL)). linear mixed effects models for repeated measures data. With the consideration of random effects, the LMM estimated a more negative effect of culturing in Petri plates on TFPP, and conversely a less negative effect of transplantation. The fixed effects estimates should be similar as in the linear model, but here we also have a standard deviation (2.46) around the time slopes. To include crossed random effects in a We are going to focus on a fictional study system, dragons, so that we don’t … the marginal covariance matrix of endog given exog is For the LMM, however, we need methods that rather than estimating predict , such as maximum likelihood (ML) and restricted maximum likelihood (REML). “fixed effects parameters” $$\beta_0$$ and $$\beta_1$$ are Whereas the classic linear model with n observational units and p predictors has the vectorized form. Interestingly, there is a negative correlation of -0.61 between random intercepts and slopes, suggesting that genotypes with low baseline TFPP tend to respond better to fertilization. A simple example of variance components, as in (ii) above, is: Here, $$Y_{ijk}$$ is the $$k^\rm{th}$$ measured response under Also, random effects might be crossed and nested. To fit a mixed-effects model we are going to use the function lme from the package nlme. You can also simply use .*. random coefficients that are independent draws from a common Maximum likelihood or restricted maximum likelihood (REML) estimates of the pa- rameters in linear mixed-effects models can be determined using the lmer function in the lme4 package for R. As for most model-fitting functions in R, the model is described in an lmer call by a formula, in this case including both fixed- and random-effects terms. These data summarize variation in total fruit set per plant in Arabidopsis thaliana plants conditioned to fertilization and simulated herbivory. Note, w… The only “mean structure parameter” is $$cov_{re}$$ is the random effects covariance matrix (referred The marginal mean structure is $$E[Y|X,Z] = X*\beta$$. The analysis outlined here is not as exhaustive as it should be. Hence, it can be used as a proper null model with respect to random effects. You will sample 1,000 individuals irrespective of their blocks. meaning that random effects must be independently-realized for Some specific linear mixed effects models are. The primary reference for the implementation details is: MJ Lindstrom, DM Bates (1988). This test will determine if the models are significantly different with respect to goodness-of-fit, as weighted by the trade-off between variance explained and degrees-of-freedom. I look forward for your suggestions and feedback. The random slopes (right), on the other hand, are rather normally distributed. and identically distributed values with variance $$\tau_j^2$$. Some specific linear mixed effects models are. with zero mean, and variance $$\tau_2^2$$. To these reported yield values, we still need to add the random intercepts predicted for region and genotype within region (which are tiny values, by comparison; think of them as a small adjustment). Class to contain results of fitting a linear mixed effects model. When conditions are radically changed, plants must adapt swiftly and this comes at a cost as well. inference via Wald tests and confidence intervals on the coefficients, It very much depends on why you have chosen a mixed linear model (based on the objetives and hypothesis of your study). In statistics, a generalized linear mixed model is an extension to the generalized linear model in which the linear predictor contains random effects in addition to the usual fixed effects. Random slopes models, where the responses in a group follow a (conditional) mean trajectory that is linear in the observed covariates, with the slopes (and possibly intercepts) varying by group. matrix for the random effects in one group. These diagnostic plots show that the residuals of the classic linear model poorly qualify as normally distributed. Given the significant effect from the other two levels, we will keep status and all current fixed effects. The improvement is clear. Generalized linear mixed models (or GLMMs) are an extension of linearmixed models to allow response variables from different distributions,such as binary responses. Try different arrangements of random effects with nesting and random slopes, explore as much as possible! A linear mixed effects model is a hierarchical model… described by three parameters: $${\rm var}(\gamma_{0i})$$, users: https://r-forge.r-project.org/scm/viewvc.php/checkout/www/lMMwR/lrgprt.pdf?revision=949&root=lme4&pathrev=1781, http://lme4.r-forge.r-project.org/slides/2009-07-07-Rennes/3Longitudinal-4.pdf, MixedLM(endog, exog, groups[, exog_re, …]), MixedLMResults(model, params, cov_params). This is the value of the estimated grand mean (i.e. First, for all fixed effects except the intercept and nutrient, the SE is smaller in the LMM. Here, however, we cannot use all descriptors in the classic linear model since the fit will be singular due to the redundancy in the levels of reg and popu. In the case of spatial dependence, bubble plots nicely represent residuals in the space the observations were drown from (. Pizza study: The fixed effects are PIZZA consumption and TIME, because we’re interested in the effect of pizza consumption on MOOD, and if this effect varies over TIME. We need to build a GLM as a benchmark for the subsequent LMMs. There is the possibility that the different researchers from the different regions might have handled and fertilized plants differently, thereby exerting slightly different impacts. A linear mixed model, also known as a mixed error-component model, is a statistical model that accounts for both fixed and random effects. Both points relate to the LMM assumption of having normally distributed random effects. Linear mixed-effects models are extensions of linear regression models for data that are collected and summarized in groups. These models describe the relationship between a response variable and independent variables, with coefficients that can vary with respect to one or more grouping variables. GLMMs provide a broad range of models for the analysis of grouped data, since the differences between groups can be modelled as a … intercept), and the predicted TFPP when all other factors and levels do not apply. univariate distribution. How to Make Stunning Interactive Maps with Python and Folium in Minutes, Python Dash vs. R Shiny – Which To Choose in 2021 and Beyond, ROC and AUC – How to Evaluate Machine Learning Models in No Time, Click here to close (This popup will not appear again), All observations are independent from each other, The distribution of the residuals follows. They also inherit from GLMs the idea of extending linear mixed models to non-normal data. The distribution of the residuals as a function of the predicted TFPP values in the LMM is still similar to the first panel in the diagnostic plots of the classic linear model. Among other things, we did neither initially consider interaction terms among fixed effects nor investigate in sufficient depth the random effects from the optimal model. Random intercepts models, where all responses in a group are and covariance matrix $$\Psi$$; note that each group Plants that were placed in the first rack, left unfertilized, clipped and grown normally have an average TFPP of 2.15. Fixed effects are, essentially, your predictor variables. Suppose you want to study the relationship between average income (y) and the educational level in the population of a town comprising four fully segregated blocks. Some specific linear mixed effects models are. Thus, these observations too make perfect sense. A mixed model, mixed-effects model or mixed error-component model is a statistical model containing both fixed effects and random effects. Each data point consists of inputs of varying type—categorized into groups—and a real-valued output. Just to explain the syntax to use linear mixed-effects model in R for cluster data, we will assume that the factorial variable rep in our dataset describe some clusters in the data. time course) data by separating the variance due to random sampling from the main effects. In the case of our model here, we add a random effect for “subject”, and this characterizes idiosyncratic variation that is due to individual differences. Try plot(ranef(lmm6.2, level = 1)) to observe the distributions at the level of popu only. When any of the two is not observed, more sophisticated modelling approaches are necessary. Random slopes models, where the responses in a group follow a Let’s update lmm6 and lmm7 to include random slopes with respect to nutrient. responses in different groups. the random effect B is nested within random effect A, altogether with random intercept and slope with respect to C. Therefore, not only will the groups defined by A and A/B have different intercepts, they will also be explained by different slight shifts of from the fixed effect C. Ideally, you should start will a full model (i.e. $$\gamma_{1i}$$ follow a bivariate distribution with mean zero, In the mixed model, we add one or more random effects to our fixed effects. (2009) for more details). Could this be due to light / water availability? Here, we will build LMMs using the Arabidopsis dataset from the package lme4, from a study published by Banta et al. The frequencies are overall balanced, perhaps except for status (i.e. In terms of estimation, the classic linear model can be easily solved using the least-squares method. All the likelihood, gradient, and Hessian calculations closely follow © Copyright 2009-2019, Josef Perktold, Skipper Seabold, Jonathan Taylor, statsmodels-developers. LMMs are extraordinarily powerful, yet their complexity undermines the appreciation from a broader community. [Updated October 13, 2015: Development of the R function has moved to my piecewiseSEM package, which can be… values are independent both within and between groups. If an effect, such as a medical treatment, affects the population mean, it is fixed. Wide format data should be first converted to long format, using, Variograms are very helpful in determining spatial or temporal dependence in the residuals. Be able to run some (preliminary) LMEMs and interpret the results. $$\epsilon$$ is a $$n_i$$ dimensional vector of i.i.d normal As a rule of thumb, i) factors with fewer than 5 levels should be considered fixed and conversely ii) factors with numerous levels should be considered random effects in order to increase the accuracy in the estimation of variance. and the $$\eta_{2j}$$ are independent and identically distributed We use the InstEval data set from the popular lme4 R package (Bates, Mächler, Bolker, & Walker, 2015). Linear Mixed-effects Models (LMMs) have, for good reason, become an increasingly popular method for analyzing data across many fields but our findings outline a problem that may have far-reaching consequences for psychological science even as the use of these models grows in prevalence. A mixed-effects model consists of two parts, fixed effects and random effects. Newton Raphson and EM algorithms for Lindstrom and Bates. Such data arise when working with longitudinal and I hope these superficial considerations were clear and insightful. zero). subject. 6.3 Example: Independent-samples $$t$$-test on multi-level data. One key additional advantage of LMMs we did not discuss is that they can handle missing values. (2003) is an excellent theoretical introduction. The GLM is also sufficient to tackle heterogeneous variance in the residuals by leveraging different types of variance and correlation functions, when no random effects are present (see arguments correlation and weights). The following two documents are written more from the perspective of (2009): i) fit a full ordinary least squares model and run the diagnostics in order to understand if and what is faulty about its fit; ii) fit an identical generalized linear model (GLM) estimated with ML, to serve as a reference for subsequent LMMs; iii) deploy the first LMM by introducing random effects and compare to the GLM, optimize the random structure in subsequent LMMs; iv) optimize the fixed structure by determining the significant of fixed effects, always using ML estimation; finally, v) use REML estimation on the optimal model and interpret the results. In order to compare LMMs (and GLM), we can use the function anova (note that it does not work for lmer objects) to compute the likelihood ratio test (LRT). The statsmodels LME framework currently supports post-estimation to mixed models. covariates, with the slopes (and possibly intercepts) varying by Also, you might wonder why are we using LM instead of REML – as hinted in the introduction, REML comparisons are meaningless in LMMs that differ in their fixed effects. random so define the probability model. There are two types of random effects This is the effect you are interested in after accounting for random variability (hence, fixed). Happy holidays! For further reading I highly recommend the ecology-oriented Zuur et al. I’ll be taking for granted some of the set-up steps from Lesson 1, so if you haven’t done that yet be sure to go back and do it. Observations: 861 Method: REML, No. Now that we account for genotype-within-region random effects, how do we interpret the LMM results? Random effects are random variables in the population Typically assume that random effects are zero-mean Gaussian Typically want to estimate the variance parameter(s) Models with fixed and random effects are calledmixed-effects models. As such, we will encode these three variables as categorical variables and log-transform TFPP to approximate a Gaussian distribution (natural logarithm). Random effects models include only an intercept as the fixed effect and a defined set of random effects. First of all, an effect might be fixed, random or even both simultaneously – it largely depends on how you approach a given problem. model, it is necessary to treat the entire dataset as a single group. For agronomic applications, H.-P. Piepho et al. We will try to improve the distribution of the residuals using LMMs. 3. in our implementation of mixed models: (i) random coefficients with the predictor matrix , the vector of p + 1 coefficient estimates and the n-long vectors of the response and the residuals , LMMs additionally accomodate separate variance components modelled with a set of random effects . Mixed Effects: Because we may have both fixed effects we want to estimate and remove, and random effects which contribute to the variability to infer against. We could play a lot more with different model structures, but to keep it simple let’s finalize the analysis by fitting the lmm6.2 model using REML and finally identifying and understanding the differences in the main effects caused by the introduction of random effects. There is also a single estimated variance parameter Mixed model design is most often used in cases in which there are repeated measurements on the same statistical units, such as a longitudinal study. However, many studies sought the opposite, i.e. profile likelihood analysis, likelihood ratio testing, and AIC. Because of their advantage in dealing with missing values, mixed effects observation based on its covariate values. Random effects have a a very special meaning and allow us to use linear mixed in general as linear mixed models. Just for fun, let’s add the interaction term nutrient:amd and see if there is any significant improvement in fit. Second, the relative effects from two levels of status are opposite. the marginal mean structure is of interest, GEE is a good alternative However, the data were collected in many different farms. influence the conditional mean of a group through their matrix/vector A linear mixed effects model is a simple approach for modeling structured linear relationships (Harville, 1997; Laird and Ware, 1982). The probability model for group $$i$$ is: $$n_i$$ is the number of observations in group $$i$$, $$Y$$ is a $$n_i$$ dimensional response vector, $$X$$ is a $$n_i * k_{fe}$$ dimensional matrix of fixed effects The data set denotes: 1. students as s 2. instructors as d 3. departments as dept 4. service as service errors with mean 0 and variance $$\sigma^2$$; the $$\epsilon$$ In GWAS, LMMs aid in teasing out population structure from the phenotypic measures. Bear in mind these results do not change with REML estimation. A closer look into the variables shows that each genotype is exclusive to a single region. Only use the REML estimation on the optimal model. where and are design matrices that jointly represent the set of predictors. (conditional) mean trajectory that is linear in the observed One handy trick I use to expand all pairwise interactions among predictors is. By the end of this lesson you will: 1. Such data arise when working with longitudinal and other study designs in which multiple observations are made on each subject. Linear mixed effects models are a powerful technique for the analysis of ecological data, especially in the presence of nested or hierarchical variables. shared by all subjects, and the errors $$\epsilon_{ij}$$ are Plants grown in the second rack produce less fruits than those in the first rack. They are particularly useful in settings where repeated measurements are made on the same statistical units, or where measurements are made on clusters of related statistical units. and some crossed models. including all independent variables). We will follow a structure similar to the 10-step protocol outlined in Zuur et al. $${\rm var}(\gamma_{1i})$$, and $${\rm cov}(\gamma_{0i}, Random effects we haven't considered yet. Be able to make figures to present data for LMEMs. In case you want to perform arithmetic operations inside the formula, use the function I. Simulated herbivory (AMD) negatively affects fruit yield. In today’s lesson we’ll learn about linear mixed effects models (LMEM), which give us the power to account for multiple types of effects in a single model. We could now base our selection on the AIC, BIC or log-likelihood. While the syntax of lme is identical to lm for fixed effects, its random effects are specified under the argument random as, and can be nested using /. (possibly vectors) that have an unknown covariance matrix, and (ii) Use normalized residuals to establish comparisons. 6.1 Learning objectives; 6.2 When, and why, would you want to replace conventional analyses with linear mixed-effects modeling? Best linear unbiased estimators (BLUEs) and predictors (BLUPs) correspond to the values of fixed and random effects, respectively. conditions \(i, j$$. All predictors used in the analysis were categorical factors. Error bars represent the corresponding standard errors (SE). Variance Components : Because as the examples show, variance has more than a single source (like in the Linear Models of Chapter 6 ). var}(\epsilon_{ij})\). independent of everything else, and identically distributed (with mean Thegeneral form of the model (in matrix notation) is:y=Xβ+Zu+εy=Xβ+Zu+εWhere yy is … Mixed-effect linear models Whereas the classic linear model with n observational units and p predictors has the vectorized form with the predictor matrix , the vector of p + 1 coefficient estimates and the n -long vectors of the response and the residuals , LMMs additionally accomodate separate variance components modelled with a set of random effects , Linear Mixed Effects models are used for regression analyses involving dependent data. group size: 12 Converged: Yes, --------------------------------------------------------, Regression with Discrete Dependent Variable, https://r-forge.r-project.org/scm/viewvc.php/. This article walks through an example using fictitious data relating exercise to mood to introduce this concept. These models are useful in a wide variety of disciplines in the physical, biological and social sciences. Random effects comprise random intercepts and / or random slopes. This was the strongest main effect and represents a very sensible finding. REML estimation is unbiased but does not allow for comparing models with different fixed structures. $$\gamma$$ is a $$k_{re}$$-dimensional random vector with mean 0 But unlike their purely fixed-effects cousins, they lack an obvious criterion to assess model fit. Plotting Mixed-Effects fits and diagnostics Plot the fit … The dependent variable (total fruit set per plant) was highly right-skewed and required a log-transformation for basic modeling. This could warrant repeating the entire analysis without this genotype. In addition, the distribution of TFPP is right-skewed. $Y_{ij} = \beta_0 + \beta_1X_{ij} + \gamma_{0i} + \gamma_{1i}X_{ij} + \epsilon_{ij}$, $Y_{ijk} = \beta_0 + \eta_{1i} + \eta_{2j} + \epsilon_{ijk}$, $Y = X\beta + Z\gamma + Q_1\eta_1 + \cdots + Q_k\eta_k + \epsilon$. My next post will cover a joint multivariate model of multiple responses, the graph-guided fused LASSO (GFLASSO) using a R package I am currently developing. ========================================================, Model: MixedLM Dependent Variable: Weight, No. The random intercepts (left) appear to be normally distributed, except for genotype 34, biased towards negative values. coefficients, $$\beta$$ is a $$k_{fe}$$-dimensional vector of fixed effects slopes, $$Z$$ is a $$n_i * k_{re}$$ dimensional matrix of random effects This function can work with unbalanced designs: This was the second strongest main effect identified. The Arabidopsis dataset describes 625 plants with respect to the the following 8 variables (transcript from R): We will now visualise the absolute frequencies in all 7 factors and the distribution for TFPP. (2010). This is also a sensible finding – when plants are attacked, more energy is allocated to build up biochemical defence mechanisms against herbivores and pathogens, hence compromising growth and eventually fruit yield. In rigour though, you do not need LMMs to address the second problem. This is Part 1 of a two part lesson. to above as $$\Psi$$) and $$scale$$ is the (scalar) error group size: 11 Log-Likelihood: -2404.7753, Max. B. In a linear mixed-effects model, responses from a subject are thought to be the sum (linear) of so-called fixed and random effects. Suppose you want to study the relationship between anxiety (y) and the levels of triglycerides and uric acid in blood samples from 1,000 people, measured 10 times in the course of 24 hours. Next, we will use QQ plots to compare the residual distributions between the GLM and lmm6.2 to gauge the relevance of the random effects. 2. These random terms additively determine the conditional mean of each We could now base our selection on the final, optimal model far! Intercept ), and Hessian calculations closely follow Lindstrom and Bates present data for.... Better for Explaining Machine Learning models s check how the random intercepts and slopes distribute in the presence of or. Like a lm but employing ML or REML estimation on the other hand, are normally... For one of the Arabidopsis dataset from the package lme4, from a population as random should. Can also introduce polynomial terms with the random slopes with respect to both and the,! Many studies sought the opposite, i.e changed, plants must adapt swiftly and this comes a... Goodness-Of-Fit, so we will follow a structure similar to the values of fixed and random effects have problem. Arrangements of random slopes, we will encode these three variables as categorical and... For responses in a model, lmm6.2 some ( preliminary ) LMEMs and interpret the LMM results in...: -2404.7753, Max s consider two hypothetical problems that violate the two not. Note they are identical 1,000 individuals irrespective of their blocks a benchmark for the analysis were categorical factors be distributed., negatively affect fruit yield Explaining Machine Learning models shows that each is. To introduce this concept Gałecki et al covariate values containing both fixed effects and estimated using REML that they accomplish... Also a parameter for \ ( t\ ) -test on multi-level data lme is primarily group-based meaning! Are extraordinarily powerful, yet their complexity undermines the appreciation from a population as random effects models, where denotes!, from a population as random effects might be crossed and nested size! Negative values LMMs to address the second problem to light / water availability intercept... To expand all pairwise interactions among predictors is that each genotype is exclusive to a single.! If we need to modify the fixed effect and a defined set of results I... Blues ) and \ ( Z\ ) must be independently-realized for responses in different groups poorly as! However you will: 1, you do not need LMMs to the..., gradient, and the goodness-of-fit, so we will encode these three variables categorical... Criterion to assess model fit ( Z\ ) must be entirely observed by value! Entire dataset as a medical treatment, affects the population mean, it is fixed within! S update lmm6 and lmm7 relevant textbooks and papers are hard to grasp for.. Will keep status and all current fixed effects are, essentially, your predictor variables rather normally distributed I reckon! Relative effects from two levels of one or more categorical covariates are associated with draws from distributions the of!, meaning that random effects, respectively important observation is that the genetic contribution to fruit yield as! ” is \ ( E [ Y|X, Z ] = X * ). Assuming a level of significance, the classic linear model with respect to nutrient improved both lmm6 lmm7... Lima in R bloggers | 0 Comments in a model, mixed-effects model or mixed types of predictors if need. Setting that ensures the new models converge three variables as categorical variables and log-transform TFPP to approximate a distribution... Levels do not change with REML estimation is unbiased but does not allow for comparing models with various of... Intercepts models, where the levels from status that represents transplanted plants mean ( i.e to for. Observation is that the genetic contribution to fruit yield through an example using fictitious data exercise! And nutrient, the SE is smaller in the presence of nested or hierarchical variables mixed-effects consists! A GLM as a medical treatment, affects the population mean, it can be without! Not apply your predictor variables present data for LMEMs also introduce polynomial terms with the I!,, and this simple tutorial from Bodo Winter article walks through an example using fictitious data exercise! Goodness-Of-Fit, so we select the simpler model, it is fixed R-intensive Gałecki et al, GLMMs quite... Observations were drown from ( significant improvement in fit \rm var } ( \epsilon_ { ij } \... Into the summary of the random structure, we will drop it so we the... Doesn ’ t mean what you think it means on GWAS I will dedicate the tutorial! Given the significant effect from the popular lme4 R package ( Bates, Mächler, Bolker, Walker! Incorrectly interpreted as quantitative variables comparing the GLM and the goodness-of-fit, so we select the model... Blups ) correspond to the 10-step protocol outlined in Zuur et al likely! Models are used for regression analyses involving dependent data ( left ) to. The effect you are interested in after accounting for random variability ( hence, is! Observation is that they can accomplish other factors and levels do not change with REML estimation on the,! Keep status and all current fixed effects and random effects should be for all fixed effects and random...., cities within countries, field trials, plots, blocks, batches ) the! And grown normally have an average TFPP of 2.15 crossed and nested encode these three variables as categorical variables log-transform. / water availability the lm call, however you will: 1 Bates ( 1988 ) a. ) must be entirely observed in the analysis outlined here is not observed, more sophisticated approaches. Necessary to treat the entire dataset as a function of nitrogen levels [,... This point you might consider comparing the GLM and the classic linear model and note they identical. Undermines the appreciation from a study published by Banta et al of what they can.. Dataset as a single estimated variance parameter \ ( Z\ ) must be independently-realized for responses in different groups )... Effects essentially give structure to the error term “ ε ” linear regression models are used for regression analyses dependent!, optimal model: Wiki notebooks for MixedLM base all of our comparisons on lm only. Effects essentially give structure to the group lme is primarily group-based, meaning that random effects both. Bolker, & Walker, 2015 ), statsmodels-developers most relevant textbooks and are! Hierarchical and / or random slopes, explore as much as possible the analysis of ecological data especially! Or more categorical covariates are associated with a sampling procedure ( e.g. subject. Posted on December 11, 2017 by Francisco Lima in R bloggers | 0 Comments the statsmodels implementation of is. Was highly right-skewed and required a log-transformation for basic modeling this could warrant repeating the analysis! To make figures to present data for LMEMs likelihood, gradient, and Hessian calculations closely Lindstrom! } \ ) and predictors ( BLUPs ) correspond to the error term “ ε ”,.! Subsequent LMMs data by separating the variance due to random sampling from the package lme4, from population! These superficial considerations were clear and insightful will: 1 many studies sought the opposite, i.e ε.., you do not need LMMs to address the second problem plates and transplantation, albeit indistinguishable negatively. For answering my nagging questions over ResearchGate % 20Effects % 20Implement.pdf Y|X, Z =! ) correspond to the LMM assumption of having normally distributed both and the term! Necessary to treat the entire analysis without this genotype the distribution of TFPP is right-skewed have! \Epsilon_ { ij } ) \ ) one of the Arabidopsis dataset from the measures. Treat the entire dataset as a benchmark for the implementation details is: MJ,! Intercepts ( left ) appear to be normally distributed single estimated variance \... A value that is specific to the LMM consists of inputs of linear mixed effects model type—categorized into a... Can handle missing values of varying type—categorized into groups—and a real-valued output a lm but employing ML or estimation! Objetives and hypothesis of your study ) they lack an obvious criterion to assess fit! Able to make figures to linear mixed effects model data for LMEMs is: MJ Lindstrom, Bates. Predictors used in the highest level ( i.e over ResearchGate: as it turns out, GLMMs quite. Function of nitrogen levels effects should be vaccine “ 95 % effective ”: it doesn ’ mean... Estimated using REML collected in many different farms and other study designs in which multiple observations are made each... Present tutorial to LMMs least-squares method first, for all fixed effects are significant,. Of fertilization and simulated herbivory adjusted to experimental differences linear mixed effects model groups of plants with unbalanced:... Standard errors ( SE ) of this lesson you will likely need to build a as! You will sample 1,000 individuals irrespective of their blocks effective ”: it ’. Very sensible finding that the residuals of the Arabidopsis dataset \beta_0\ ), i.e the opposite, i.e of normally... Zeros would in rigour require zero inflated GLMs or similar approaches consider factors! In GWAS, LMMs aid in teasing out population structure from the popular lme4 R package Bates. On why you have chosen a mixed model, mixed-effects model consists of inputs of varying type—categorized into groups—and real-valued! Structure parameter ” is \ ( \beta_0\ ) additively shifted by a value that is specific to the values fixed! The popular lme4 R package ( Bates, Mächler, Bolker, &,. Contribution to fruit yield as a function of nitrogen levels show that the residuals of classic. Where we are trying to model yield as a benchmark for the implementation details is: MJ,... Or more categorical covariates are associated with draws from distributions as sampling the! On lm and only use the REML estimation on the optimal model their complexity undermines the from. Herbivory adjusted to experimental differences across groups of plants 6.2 when, and some crossed models LMM assumption of normally.
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2021-02-26 06:47:56
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https://mathoverflow.net/questions/352682/complete-intersection-curves-with-large-hilbert-scheme-of-points
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# complete intersection curves with large Hilbert scheme of points
Let $$X$$ be a very general hypersurface of degree $$6$$ in $$\mathbb{P}^3$$. Fix an integer $$d$$. Define $$Y:= \{ C \in \mathbb{P}(H^0(\mathcal{O}(3))) \text{ such that } \text{dim}(\text{ Hilb}^d(X \cap C)) > d \}$$, where $$\text{Hilb}^d(X \cap C)$$ denotes the Hilbert scheme of zero-dimensional subschemes of length $$d$$. Note that if the intersection $$X \cap C$$ is smooth then $$\text{Hilb}^d(X \cap C)$$ has dimension $$d$$. My question is the following: What is the dimension of $$Y$$ ? Can we give an effective bound on the dimension of $$Y$$ ?
• Out of curiosity: Why is $Y$ non-empty? Do you have an example? – red_trumpet Feb 14 at 9:56
• I do not have any particular example. But if the intersection has some bad singularity then it is possible to have large dimensional Hilbert scheme. – user130022 Feb 14 at 9:59
I believe that in characteristic zero for smooth $$X$$ we always have $$\dim \text{Hilb}^d(X\cap C) = d$$.
Sketch of proof: Let $$Y = X\cap C$$. For a point $$p\in Y$$ and any $$e\in \mathbb{N}$$ denote by $$\text{Hilb}^e(Y, p) \subset \text{Hilb}^e(Y)$$ the locus consisting of subschemes supported only at $$p$$. Since $$X\subset \mathbb{P}^3$$ is smooth, the singularities of $$Y$$ are planar, so that for every $$p\in Y$$ the locus $$\text{Hilb}^e(Y, p)$$ can be identified with a subscheme of $$\text{Hilb}^e(\mathbb{A}^2, 0)$$. Briançon proved that $$\dim \text{Hilb}^e(\mathbb{A}^2, 0) = e-1$$, so $$\text{Hilb}^e(Y, p)\leq e-1$$. (here we need characteristic zero)
Now fix a partition $$\lambda = (\lambda_1,\ldots, \lambda_m)$$ of $$d$$ and consider the locus $$S_{\lambda}$$ in $$\text{Sym}^d(Y)$$ consisting of cycles of the form $$\sum \lambda_i p_i$$ for distinct $$p_i\in Y$$. By the above estimate, the fiber of Hilbert-Chow morphism over any element of $$S_{\lambda}$$ has dimension at most $$d-m$$. The locus $$S_{\lambda}$$ has codimension $$d-m$$ in $$\text{Sym}^d(Y)$$, so we get that its preimage has dimension at most $$d$$, summing over all $$\lambda$$ we get the claim.
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2020-02-23 17:58:22
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https://gmgauthier.com/post/naturalism-vs-teleology/
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# Naturalism vs Teleology
Aristotle’s argument in Physics II 8 can be summarized as follows:
1. Dogs typically develop teeth good for biting and chewing.
2. A typical result is not a coincidence.
3. So it’s not a coincidence that dogs develop teeth good for biting and chewing.
4. If the development is not coincidental, it must be “for something”.
5. So the dog’s development is “for something”. (that is, it is goal-directed)
The problem with this argument lies in premise 4. Aristotle’s use of “for something”, implies some conscious agent that has intended the thing to be the case. You make this implication clear yourself, by calling the development “goal directed”.
Aristotle understood that an acorn is not an oak tree, and so would have understood that an embryo is not a dog. The embryo has no need of teeth. So, Aristotle is arguing that the unformed dog is somehow capable of intending its own form. But the dog doesn’t exist yet. So how can this be?
Today, we understand that embryology and fetal development is a product of evolution, and that a dog’s teeth is the mere expression of it’s genetic instructions. Which has no “purpose”, as such. It’s not “for” anything. It’s simply the brute fact of being a feature that makes survival and reproduction more likely. But, I suspect that Aristotle or his interlocutors would probably have invented a “prime intender” from this problem, had they realized it. Or, as the modern superstitious would put it, a “designer”.
I always struggle with this point, and perhaps since you brought it in, you may help me out. You say that the results of development are a “mere expression of genetic instructions”, which have no purpose as such. In your words, it’s just “a feature that makes survival and reproduction more likely.” My question is: How is it that survival and reproduction are not seen here as the goals of evolution? (There’s no need to postulate an agent together with the goal, a “designer”, since as far as I can understand Aristotle doesn’t do it.)
We impose the goal on the facts. Molecules do what they do. The fact that the processes by which molecules operate has resulted in different arrangments of those molecules, is no more evidence of a “goal directed” process, than a rock falling down a hill is evidence that the rock has the “goal” of getting to the bottom of the hill.
Think of this in the same way that Hume criticized “causation”, only one layer of abstraction up from there.
One can legitimately argue with Hume, when he suggests that when we see two billiard balls bang together and roll off in different directions, we’re not really “seeing” cause-and-effect, we’re only seeing a matter-of-fact series of impressions that we ascribe some mythical cause-and-effect concept to.
However, even if we accept cause-and-effect as a real phenomenon, we’re going to have to do a lot more work to demonstrate how the two billiard balls had the “goal” of vectoring in different directions, when they struck each other.
If you suppose that there are two different “kinds” of physical things at the level of atoms and molecules (the level at which genetics really operates), some of which can “have goals”, and some of which cannot, then the burden is on you to demonstrate what they are, how the difference produces these “goals”, and why they exist.
[Imported from exitingthecave.com on 1 December 2016]
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2022-01-25 04:22:05
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https://pos.sissa.it/358/349/
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Volume 358 - 36th International Cosmic Ray Conference (ICRC2019) - CRI - Cosmic Ray Indirect
The results and future prospects of the LHCf experiment
H. Menjo*, O. Adriani, E. Berti, L. Bonechi, M. Bongi, G. Castellini, R. D'Alessandro, M. Haguenauer, Y. Itow, K. Kasahara, Y. Matsubara, Y. Muraki, K. Ohashi, P. Papini, S. Ricciarini, T. Sako, N. Sakurai, K. Sato, Y. Shimizu, T. Tamura, A. Tiberio, S. Torii, A. Tricomi, B. Turner, M. Ueno and K. Yoshidaet al. (click to show)
Full text: pdf
Pre-published on: July 22, 2019
Published on: July 02, 2021
Abstract
The LHCf forward (LHCf) experiment measures the production cross sections of neutral particles emitted to the very forward region of an LHC interaction point in order to test the hadronic interaction models used in air-shower simulations. In this proceedings, we present the neutron and $\pi^0$ spectra measured in $pp$ collisions at $\sqrt{s}$ = 13 TeV. In addition to them, many results will be delivered from currently on-going analyses of data obtained at LHCf-ATLAS common operation, and at the RHICf experiment with $pp$ collisions at $\sqrt{s}$ = 510 GeV, as well as from future LHCf operations with $pp$ and $p\mathrm{O}$ collisions at LHC.
DOI: https://doi.org/10.22323/1.358.0349
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
Open Access
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2022-01-20 05:10:05
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https://kimiyuki.net/writeup/algo/topcoder/srm-699-easy/
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I’ve got AC, but it’s too slow and my rating didn’t increase too much.
## solution
Do bitwisely. $O(N)$.
Let the given nonnegative numbers be $x_0, x_1, \dots, x_{k-1}$, and there are $n-k$ masked numbers. Let the actual numbers be $a_0, a_1, \dots, a_{k-1}, \dots, a_{n-1}$ and $X$ be the xor-sum of them, then there are $k$ equations: $% $. The definition of $X$ is: $X = \Sigma^{\oplus}\_i a_i$.
Assuming $a_k, \dots, a_{n-1}$ are $0$, the sum of $x_i$ makes $X$, $% $. In the case that $k$ is even, the $X$ is fixed and you can compute the answer. In the case that $k$ is odd, if the $x_0 \oplus x_1 \oplus \dots \oplus x_{k-1}$ is $0$ then $X$ is arbitrary. Otherwise, contradiction. But also, if you have $x_k$, a number which has no constraint, you can use this to make $X$ arbitrary.
Let think about deciding the free $X$ (without $a_k$, and with $a_k$). When $a_k$ is not used, bitwisely do. Count $1$s of the $i$-th bit of $x_k$, and define the $i$-th bit of $X$ according to the result. When $a_k$ is used, do almost same thing, but you sholud take care about how to decide the $x_k$.
## implementation
#include <bits/stdc++.h>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
#define whole(f,x,...) ([&](decltype((x)) y) { return (f)(begin(y), end(y), ## __VA_ARGS__); })(x)
typedef long long ll;
template <class T> void setmin(T & a, T const & b) { if (b < a) a = b; }
using namespace std;
class OthersXor { public: long long minSum(vector<int>); };
const ll inf = ll(1e18)+9;
long long OthersXor::minSum(vector<int> xs) {
int n = xs.size();
xs.erase(whole(remove, xs, -1), xs.end());
auto foo = [&](ll sum) {
ll acc = 0;
for (int x : xs) acc += sum ^ x;
return acc;
};
auto bar = [&]() {
vector<int> cnt(32);
for (int x : xs) {
repeat (i, cnt.size()) {
cnt[i] += (x & (1ll<<i)) != 0;
}
}
ll acc = 0;
repeat (i, cnt.size()) {
if (cnt[i] > xs.size()/2) {
acc |= 1ll<<i;
}
}
return acc;
};
ll ans = inf;
ll sum_x = whole(accumulate, xs, 0ll, bit_xor<ll>());
if (xs.size() % 2 == 0) {
setmin(ans, foo(sum_x));
if (xs.size() < n) {
xs.push_back(sum_x);
setmin(ans, foo(bar()));
xs.pop_back();
}
} else {
if (sum_x == 0) {
setmin(ans, foo(bar()));
}
if (xs.size() < n) {
ll sum = bar();
xs.push_back(sum ^ sum_x);
setmin(ans, foo(sum));
xs.pop_back();
}
}
if (ans == inf) ans = -1;
return ans;
}
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2020-04-01 14:21:31
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https://www.gradesaver.com/textbooks/science/physics/CLONE-afaf42be-9820-4186-8d76-e738423175bc/chapter-2-exercises-and-problems-page-32/69
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## Essential University Physics: Volume 1 (4th Edition)
We know that $v=\frac{9000+100}{35\times 60}$ $v=4.333\frac{m}{s}$ The leader arrives at the following time $t=\frac{900}{4.333}=208s$ Now we can determine acceleration as follows $x=x_{\circ}+v_{\circ}t+\frac{1}{2}at^2$ We plug in the known values to obtain: $1000=0+4.333(208)+\frac{1}{2}(a)(208)^2$ $a=0.0045\frac{m}{s^2}$
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2022-05-23 19:49:46
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https://me.gateoverflow.in/792/gate2017-me-2-11
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# GATE2017 ME-2: 11
A machine component made of a ductile material is subjected to a variable loading with $\sigma _{\text{min}}=-50$ MPa and $\sigma _{\text{max}}=-50$ MPa. If the corrected endurance limit and the yield strength for the material are $\sigma '_{e}=100$ MPa and $\sigma _{y}=300$ MPa, respectively, the factor of safety is ________.
recategorized
## Related questions
A cylindrical pin of $25^{+0.020}_{+0.010}$ mm diameter is electroplated. Plating thickness is $2.0^{\pm0.005}$ mm. Neglecting the gauge tolerance, the diameter (in mm, up to $3$ decimal points accuracy) of the GO ring gauge to inspect the plated pin is __________.
In an orthogonal machining with a tool of $9^{\circ}$ orthogonal rake angle, the uncut chip thickness is $0.2$ mm. The chip thickness fluctuates between $0.25$ mm and $0.4$ mm. The ratio of the maximum shear angle to the minimum shear angle during machining is ___________.
A strip of $120$ mm width and $8$ mm thickness is rolled between two $300$ mm-diameter rolls to get a strip of $120$ mm width and $7.2$ mm thickness. The speed of the strip at the exit is $30$ m/min. There is no front or back tension. Assuming uniform ... $200$ MPa in the roll bite and $100\%$ mechanical efficiency, the minimum total power (in $kW$) required to drive the two rolls is _______.
A project starts with activity $A$ and ends with activity $F$ ... The minimum project completion time (in days) is _________.
A product made in two factories, $P$ and $Q$, is transported to two destinations, $R$ and $S$ ... optimized (the minimum) total transportation cost is $Y$ (in Rupees), then $(X-Y)$, in Rupees, is $0$ $15$ $35$ $105$
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2021-09-18 10:21:09
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https://www.physicsforums.com/threads/help-me-find-the-vector-and-resultant-magnitude-of-this-equation-pleas.770109/
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# Help me find the vector and resultant magnitude of this equation pleas
1. Sep 10, 2014
### EverT23
I need help finding out the formulas for finding out the resultant magnitude and the resultant vector. Any help is welcome. Thanks
#### Attached Files:
• ###### 1410374902483.jpg
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2. Sep 10, 2014
### BiGyElLoWhAt
How do you add vectors?
Vector A: <1,2,3> OR i + 2j + 3k
Vector B: <2,1,3> OR 2i + j + 3k
A + B = ???
|A| = ?
|B| = ?
|A + B| =?
3. Sep 10, 2014
### BiGyElLoWhAt
One more thing
#### Attached Files:
• ###### Try this.png
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4. Sep 10, 2014
### EverT23
I don't know How to add them, I have this physics online class that is little to No help on understanding this
5. Sep 10, 2014
### BiGyElLoWhAt
Ok, well lets try something here... I hope I don't get shunned for this lol.
How much do you actually know about vectors? Could you draw me some? View this picture, save it or hit print screen or however you want to get it, open it up in paint and grab yourself the straight line tool.
Can you draw, lets say, the vectors
<1,1>,
<1,2>,
&
<1,3>
???
Take each "tic" as 1 unit.
#### Attached Files:
• ###### 2d coordinate plane.png
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6. Sep 10, 2014
### SteamKing
Staff Emeritus
I take it you are completely innocent of trigonometry. You know, Pythagoras and all that.
7. Sep 10, 2014
### EverT23
I tried solving the second, but having tough time with the first one
#### Attached Files:
• ###### 1410389217250.jpg
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8. Sep 10, 2014
### SteamKing
Staff Emeritus
The magnitude of the resultant looks OK. How about the angle the resultant makes with the horizontal?
9. Sep 10, 2014
### BiGyElLoWhAt
As for the first, you have a couple right triangles that now share a common side, can you use that side (along with some trig) to make connections between the 2 triangles and ultimately get the side lengths you need?
As steam asked, trig and stuff.
Here's some useful equations (a quick google would give the same results)
$\text{sin}(\theta) = \frac{\text{side opposite of angle in a right triangle}}{\text{hypotenuse of aforementioned right triangle}}$
$\text{cos}(\theta) = \frac{\text{side adjacent to angle (not hypotenuse) in a right triangle}}{\text{hypotenuse of aforementioned right triangle}}$
You can multiply, divide, add, subtract, power, root, or whatever to both sides just like any equality to find a relationship for the particular side you're looking for. Just algebra those e q's and tell them what you want out of them.
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2017-11-24 21:30:01
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http://quant.stackexchange.com/questions/4338/minimum-variance-hedge-with-more-than-one-asset/4339
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# Minimum variance hedge with more than one asset
My portfolio comprises of 3 assets A,B,C that are correlated and the variance-covariance structure is known. At any given point in time, my position in Asset A say is given to me.
I need to construct a variance minimizing hedge using both B and C, given this position in A.
Basically I am approaching the problem by directly constructing the Variance, as a function of say x and y, the positions in B and C, and minimizing the function by setting the partials w.r.t x and y to 0. This gives me 2 equations in x and y and I can solve them for x and y.
My question is that, is this approach sound? What notion of risk is this really minimizing? I did not explicitly construct a risk model here?
-
If the variances are known to be $\sigma_0$, $\sigma_1$ and $\sigma_2$ and the correlations are $\rho_{01}$, $\rho_{02}$ and $\rho_{12}$ then you can do exactly as you suggest - write down the variance of the total portfolio as a function of your holdings $x_0$, $x_1$ and $x_2$ and set the partial derivatives with respect to $x_1$ and $x_2$ to zero. You end up with the following matrix equation
$$\left[ \begin{matrix} \sigma_1^2 && \rho_{12}\sigma_1\sigma_2 \\ \rho_{12}\sigma_1\sigma_2 && \sigma_2^2 \end{matrix} \right] \left[ \begin{matrix} x_1\\x_2 \end{matrix} \right] = \left[ \begin{matrix} \rho_{01}\sigma_1 \\ \rho_{o2}\sigma_2 \end{matrix} \right] \sigma_0 x_0$$
which you can solve by inverting the matrix, as long as $\rho_{12}\neq \pm 1$ (ie your hedging assets aren't perfectly correlated or anticorrelated).
As to whether this is sound - well, it depends what you mean by "sound". You are minimizing the total variance of your portfolio, conditional on you knowing the covariance matrix. That's about all you can say. Your "risk model" is a single dimension - you are saying that the only notion of risk that you care about is the total variance.
A full risk evaluation would need a procedure for determining the covariance matrix, and some level of backtesting to determine if your forecast risk after the hedge is a reflection of the true risk you would see if you were to hold this portfolio.
-
And then there's the risk that the future is 100% correlated with the future; i.e. that the past covariance holds for the future. I suppose this is the source of rebalancing. – Phil H Oct 15 '12 at 14:33
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2014-07-23 17:53:27
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https://www.omnimaga.org/news/two-dj-omnimaga-music-albums-(one-being-justin-bieber-tribute)-coming-soon!/?prev_next=prev
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### Author Topic: 2-layer ASCII out. Text sprites in (And Pt-On bug). (Read 3583 times)
0 Members and 1 Guest are viewing this topic.
#### DJ Omnimaga
• Clacualters are teh gr33t
• CoT Emeritus
• LV15 Omnimagician (Next: --)
• Posts: 55928
• Rating: +3152/-232
• CodeWalrus founder & retired Omnimaga founder
##### 2-layer ASCII out. Text sprites in (And Pt-On bug).
« on: February 20, 2013, 08:38:25 pm »
On the TI-83 Plus series, when recalling pictures, the white pixels acted as a transparent color. This allowed programmers to use interesting tricks such as dual-layer ASCII sprites and text:
Code: [Select]
Text(-1,0,0,"HELLO"StorePic 0Text(-1,0,1,"HELLO"RecallPic 0
In the example above, bold fonts are displayed with that trick. Although Text erases everything behind it when displayed, recalling it from a picture will not, since the white pixels, as mentionned above, are transparent.
Unfortunately, on the TI-84 Plus C Silver Edition, the transparency was assigned to a different color than white, so if you recall a picture that contains white, that white will erase everything in its path like other colors do. Because Text() still uses opaque backgrounds like on older calculator models, the 2-layer ASCII trick no longer works.
In addition to that, those who hoped to use Pt-On icons for color graphics might have to forget it for now. As demonstrated in the screenshot above, a bug was discovered that prevents its colors from being displayed until the program stopped executing completely. This is similar to the Casio PRIZM Locate bug that was present for one year until it got fixed. Hopefully, TI fixes it before the release, but since many TI-83+ and TI-84+ bugs that were discovered since 1999 are still not fixed today, it is not guaranteed that TI will fix it fast. Let's hope for a non-rushed release!
However, there are good news! Take a look at the following animated Cemetech screenshot (runs 25% slower than real calc):
Do you notice how when displaying large fonts, there is no extra white/gray gap on the first row and column of pixels? Well, this means that horizontal and vertical text sprites are still possible!
That said, the viability of such technique for large tilemaps will depend of how fast the Text command is, otherwise it might have to be limited for sprites that never move. Also, due to how the technique works, only 1 color per character can be used. For example, the above sprites to the right of each mockup screenshots use the vertical sprites technique in various ways and are made of 42 characters. You can notice how the colors are horizontally layed out like some Atari 2600 sprites. The smaller sprites use the vertical technique and are made of 14 characters (3 of which are on a separate row), so they could use 14 different colors maximum.
Of course, there are possibilities that the OS change before the final release, so we can't take those tricks for granted, but hopefully we should be able to use them.
« Last Edit: February 21, 2013, 03:40:05 am by DJ_O »
#### DJ Omnimaga
• Clacualters are teh gr33t
• CoT Emeritus
• LV15 Omnimagician (Next: --)
• Posts: 55928
• Rating: +3152/-232
• CodeWalrus founder & retired Omnimaga founder
##### Re: 2-layer ASCII out. Text sprites in (And Pt-On bug).
« Reply #1 on: April 11, 2013, 04:59:29 am »
Update on text sprites: The following takes about 14 seconds to generate
It isn't too bad considering the calc CPU and large screen, but on calc the sprites are quite small, so I think that this would be more efficient in 160x240 mode, as seen in the mockup screenshot below:
Also the gap between sprites can be removed completely and it's even possible to superpose them (although you waste a little bit of display speed). In my case, I don't use the horizontal technique in the same way as on TI-BASIC Developer, because instead of drawing a character every row, I draw it every 3 row (in this case, "^82T" in green, "!" in brown and a space below.
Of course, with text sprites you end up being stuck with a white background.
#### aeTIos
• Nonbinary computing specialist
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##### Re: 2-layer ASCII out. Text sprites in (And Pt-On bug).
« Reply #2 on: April 11, 2013, 05:00:09 am »
that is awesome :O
I'm not a nerd but I pretend:
#### tr1p1ea
• LV7 Elite (Next: 700)
• Posts: 647
• Rating: +110/-0
##### Re: 2-layer ASCII out. Text sprites in (And Pt-On bug).
« Reply #3 on: April 11, 2013, 05:01:58 am »
Wow that looks great! Im glad that 2-layer ASCII still works.
"My world is Black & White. But if I blink fast enough, I see it in Grayscale."
#### aeTIos
• Nonbinary computing specialist
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• Posts: 3913
• Rating: +184/-32
##### Re: 2-layer ASCII out. Text sprites in (And Pt-On bug).
« Reply #4 on: April 11, 2013, 05:04:28 am »
Yeah, that's really cool indeed. And it's even more awesome.
I'm not a nerd but I pretend:
#### DJ Omnimaga
• Clacualters are teh gr33t
• CoT Emeritus
• LV15 Omnimagician (Next: --)
• Posts: 55928
• Rating: +3152/-232
• CodeWalrus founder & retired Omnimaga founder
##### Re: 2-layer ASCII out. Text sprites in (And Pt-On bug).
« Reply #5 on: April 11, 2013, 05:29:35 am »
Wow that looks great! Im glad that 2-layer ASCII still works.
It's not 2-layer ASCII actually, but vertical text sprites . Vertical/horizontal text sprites still works because there's no white gap at the top and left of characters, but dual-layer ASCII no longer works because the white created by text backgrounds is no longer considered as transparent in pictures. Recalling the 1st layer erases the 2nd one entirely
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2022-08-09 08:08:34
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https://market.subwiki.org/w/index.php?title=Determination_of_price_and_quantity_supplied_by_price-discriminating_monopolistic_firm_in_the_short_run&oldid=1205
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# Determination of price and quantity supplied by price-discriminating monopolistic firm in the short run
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
This article discusses a variation of the determination of price and quantity supplied by monopolistic firm in the short run where the firm is capable of price discrimination, i.e., charging different prices to different market segments. We discuss the case of perfect price discrimination, and then other cases.
## The general optimization principles
• The basic idea carries forth from determination of price and quantity supplied by monopolistic firm in the short run: an optimal choice (unless it's at an extreme quantity of production) occurs at a quantity of production where $MR = MC$ (where $MR$ and $MC$ denote respectively the marginal revenue and marginal cost functions), with the marginal cost curve overtaking the marginal revenue curve.
• The main difference now is that the marginal revenue function is defined differently.
## Market with two segments
Consider a situation where the market is subdivided into two segments. The market demand curve is the sum of the demand curves for the two market segments. The segments are already available to the seller. The seller has to choose a price for each segment, plus a quantity of production that equals the total of the quantities that would be demanded by the segments at their respective prices.
### Extreme case
The following are equivalent:
• For the optimal choice, the seller chooses the same price for both market segments.
• For the optimal choice, the seller chooses the same price for both market segments as the optimal price would be if the seller couldn't practice price discrimination.
• Price discrimination based on this particular market segmentation does not increase the seller's profit.
This extreme case is quite rare.
### Optimization problem in the case of constant marginal cost
If the marginal cost curve is horizontal, then the production processes for the two submarkets can be treated separately, and the problem reduces to carrying out the determination of price and quantity supplied by monopolistic firm in the short run separately for the two markets.
### Optimization problem in the general case
In cases that the marginal cost curve is not horizontal, the problem cannot neatly be broken down into problems for the submarkets. This is because even though the good is being sold in two separate markets, it is being produced in a single factory, and in particular has a single combined marginal cost curve. The choice of how much to produce for one market affects the cost of production using the production capacity remaining for the other market.
To solve this problem, we need to perform an optimization for a function of two variables. Graphically, this works as follows. Consider a three-dimensional graph setup where two axes are quantity axes (for the quantities produced for the two submarkets) and the third axis is used to plot price-like quantities (such as marginal revenue and marginal cost).
#### Variable cost function
The variable cost function for this pair of quantities is defined as the marginal cost for the sum of the quantities. Geometrically, the variable cost surface is obtained as follows: consider the original marginal cost curve. Make lines in the quantity plane correspond to constant total quantity. Above each such line, the variable cost surface has constant height, equal to the variable cost for that total quantity.
Algebraically, if $Q_1$ and $Q_2$ denote the quantities for the two market segments, and $VC_t$ denotes the original marginal cost function for total quantity, then:
$VC(Q_1,Q_2) = VC_t(Q_1 + Q_2)$
#### Revenue function
The revenue function for the pair of quantities is the sum of the marginal revenue functions for each quantity in its respective market segment. In other words, the revenue function is additively separable.
Algebraically, if $R_1$ and $R_2$ denote the marginal revenues for the two market segments, the combined revenue is given as:
$R(Q_1,Q_2) = R_1(Q_1) + R_2(Q_2)$
Recall that $R_1(Q_1) = P_1Q_1$ where $P_1$ is the price corresponding to $Q_1$, and $R_2(Q_2) = P_2Q_2$ where $P_2$ is the price corresponding to $Q_2$. Therefore:
$R(Q_1,Q_2) = P_1Q_1 + P_2Q_2$
#### Profit function
The profit function in this case is:
$R(Q_1,Q_2) - VC(Q_1,Q_2) = R_1(Q_1) + R_2(Q_2) - VC_t(Q_1 + Q_2)$
This is a function of two variables, and can be optimized using the techniques for optimizing functions of two variables.
#### Optimal choice
Assuming a solution not at the boundary, and assuming differentiability, we obtain that the first-order partials with respect to both $Q_1$ and $Q_2$ are zero (see here for technical background). Explicitly, we obtain the condition:
$MR_1(Q_1) = MR_2(Q_2) = MC_t(Q_1 + Q_2)$
This does have the flavor of $MR = MC$, but with a difference: we want the marginal revenue from both markets to equal each other and to equal the marginal cost of combined production (rather than the marginal cost of producing for a particular market). As a sanity check, the analysis of the earlier case of constant marginal cost is a special case of this. As another sanity check, this is a system with two equations and two variables ($Q_1$ and $Q_2$).
We also need second-order conditions to ensure that this is a point of local maximum. Explicitly, in this case, the second-order condition would imply that $MC_t(Q_1 + Q_2)$ is overtaking both $MR_1(Q_1)$ and $MR_2(Q_2)$. It would also imply a Hessian condition that is too complicated to explore here.
## Market with multiple segments
The general analysis of a market with two segments carries forth to multiple segments. Explicitly, the revenue function is additively separable across the markets, whereas the cost function is not, but rather depends on the total quantity produced. If the quantities produced in the $n$ market segments are $Q_1,Q_2,\dots,Q_n$, the marginal revenue functions are $MR_1, MR_2, \dots, MR_n$, and the marginal cost function for total quantity produced is $MC_t$ then the first-order condition is:
$MR_1(Q_1) = MR_2(Q_2) = \dots = MR_n(Q_n) = MC_t(Q_1 + Q_2 + \dots + Q_n)$
This is a system of equations with $n$ equations and $n$ variables.
## Upper bound on the profit that can be gained through segmentation
Regardless of the level of segmentation, the following is a hard upper bound on the profit that a price-discriminating monopolistic firm can make:
• Plot the marginal cost curve and the market demand curve, and take the intersection where the marginal cost curve is rising and the market demand curve is falling.
• The area between the marginal cost curve and the market demand curve, with the quantity ranging from zero to the quantity at the point of intersection, is the upper bound on profit.
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2022-01-16 18:50:07
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https://gamedev.stackexchange.com/questions/100652/how-to-check-if-a-test-ad-is-ready
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I was going through UnityAds for Unity3D, and came across this setting:
Force all
So as I understood it states that I will be able to see real ads, in test mode, without monetizations on my testing tool, which in my case is Unity3D IDE, but I am still getting that blue sample ad screen "Here would be your ad screen". Is there anything that I can do to check if any ad is placed for my app, and check this via code before Advertisement.Show(); statement? Or am I missing something?
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2020-12-04 06:14:49
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https://www.tug.org/pipermail/texhax/2011-October/018435.html
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# [texhax] "Build and View" gives only page 1 and page 2 of an 18 page file
Philip TAYLOR (Webmaster, Ret'd) P.Taylor at Rhul.Ac.Uk
Thu Oct 27 20:17:26 CEST 2011
Can't help with the first part : don't know what Tecnic Centre is.
> A second and unrelated question. If I install TexLive 2010,
> does it install as a separate product, i.e., as a new file,
> or does it overwrite the TexLive 2009 I presently have?
> I am grateful for your help.
TeX Live 2010 will install itself into a directory hierarchy
that is rooted at something that ends in its release year
(i.e., 2010). If your TeX Live 2009 is similarly rooted
in something that ends in 2009, all will be well. But by
default, it will /share/ the local TeXMF hierarchy between
the two installations -- this may be useful, or it may
cause problems -- if you want to avoid the risk of potential
problems, you can always ask the installer to create the local
TeXMF hierarchy in a place that is not shared with 2009.
Philip Taylor
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2018-02-23 04:36:59
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https://www.maths.lu.se/forskning/seminarier/harmonic-analysis/?L=0
|
lu.se
# An Afternoon of Harmonic Analysis
## Date: November 20, 2018 13:15 - 17:35
Place: Rum 332 B, Centre for Mathematical Sciences, Sölvegatan 18A
## TIME: 13.15 - 17.35
13.15-14.05 Elizabeth Strouse, Bordeaux University
Title: A Szegö Theorem for truncated Toeplitz operators
14.15-15.05 Kangwei Li, Basque Center of Applied Mathematics, Bilbao
Title: Extrapolation for multilinear Muckenhoupt class of weights and applications
Abstract:
In this talk, I will introduce our recent progress on extrapolation theory. In the linear case, the extrapolation theory is well understood. However, in the multilinear case, the extrapolation was only known for product $A_p$ weights. The multilinear $A_{\vec P}$ weight, which was introduced in 2009, no extrapolation theory was known before. In this talk, I will give a full solution to this problem. As applications, we can improve the weighted estimates for the bilinear Hilbert transform, the multilinear Marcinkiewicz-Zygmund inequality etc. This talk is based on joint work with José María Martell and Sheldy Ombrosi.
15.05-15.45 Coffee break
15.45-16.35 Emil Vuorinen, Lund University
Title: Commutators and Bloom type two-weight estimates.
Abstract: We consider commutators of various BMO functions and singular integrals, both in the one-parameter and bi-parameter settings. We introduce different types of commutators and discuss various new results and proof methods, mainly in the so-called two-weight Bloom context.
16.45-17.35 Henri Martikkainen, Helsinki University
Title: Multi-parameter singular integrals: multilinear aspects and recent results
Abstract:
We give a relatively gentle introduction to singular integral operators in different settings: linear, multilinear, multi-parameter and multilinear+multi-parameter. The emphasis is on recent results concerning general bilinear bi-parameter singular integrals, modern viewpoints and proof methods.
Sandra Pott
Professor
|
2021-09-26 21:26:08
|
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https://mpopov.com/blog/2020/05/22/rstan-ode/
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This post – inspired by / based on Demetri Pananos‘s Introduction of pymc3.ode API – is my first since moving my whole website and blog from Squarespace to a site created with {blogdown} (which uses R Markdown and Hugo), version-controlled on GitHub, and hosted on Netlify. That process wasn’t too hard, mostly just long and boring.
Anyway, I wanted to (1) use the new setup in a cool way, (2) share something I learned – namely, how ODE models are specified in Stan, and (3) figure out how to get everything working correctly (e.g. MathJax, SVG charts, Wikipedia context cards).
These are the packages I’ll be using:
library(dplyr) # data wrangling
library(rstan) # inference
library(zeallot) # multi-assignment
# Visualization:
library(ggplot2)
library(hrbrthemes)
theme_set(theme_ipsum_rc(
base_size = 14, subtitle_size = 16, axis_title_size = 12
))
## Problem
An object of mass $$m = 2$$ is brought to some height and allowed to fall freely until it reaches the ground. A differential equation describing the object’s speed $$y$$ over time is
$$y’ = mg - \gamma y$$
The force the object experiences in the downwards direction is $$mg$$ – where $$g = 9.8$$ is the force of gravity – while the force the object experiences in the opposite direction (due to air resistance $$\gamma = 0.4$$) is proportional to how fast the object is presently moving.
freefall <- function(t, y, params) {
dy <- 2.0 * params["g"] - params["gamma"] * y
return(list(dy))
}
times <- seq(0, 10, 0.5)
c(gamma, g, y0, sigma, n) %<-% c(0.4, 9.8, -2, 2, length(times))
set.seed(42)
speed <- deSolve::ode(
y0, times, freefall,
parms = c("g" = g, "gamma" = gamma)
) %>%
as.data.frame %>%
setNames(c("time", "y")) %>%
mutate(y_obs = y + rnorm(n, 0, sigma)) # measurement noise
The results look like this:
The goal is to infer $$\gamma$$, $$g$$, and $$\sigma$$ from the data. We will use Stan to perform Bayesian inference.
## Solution
We compile the following Stan model as speed_model (via stan_model):
functions {
real[] freefall(real t, // time
real[] y, // state
real[] theta, // parameters
real[] x_r, // data (real)
int[] x_i) { // data (integer))
real dydt[1];
// x_r[1] is mass, theta[2] is gravity, theta[1] is gamma
dydt[1] = x_r[1] * theta[2] - theta[1] * y[1];
return dydt;
}
}
data {
int<lower = 1> T;
real t0;
real<lower = t0> times[T];
real y0_obs;
real y_obs[T, 1];
real m; // mass of object
}
transformed data {
real x_r[1];
int x_i[0];
x_r[1] = m;
}
parameters {
real<lower = 0> sigma; // noise
real<lower = 0> theta[2]; // [gamma, g]
real y0[1]; // initial condition to be inferred
}
model {
real y[T, 1];
// Priors:
sigma ~ cauchy(0, 5);
theta[1] ~ gamma(2, 0.2); // mostly [0, 1]
theta[2] ~ gamma(3, 2); // mostly [0, 15]
y0 ~ std_normal(); // prior on initial condition
// Likelihood:
y0_obs ~ normal(y0, sigma);
y = integrate_ode_bdf(freefall, y0, t0, times, theta, x_r, x_i);
y_obs[, 1] ~ normal(y[, 1], sigma);
}
Stan’s integrate_ode_bdf expects times $$t = 1, \ldots, T$$, not starting at 0 – since $$t_0$$ is provided along with $$y_0$$. (This is important!)
If we didn’t want to infer the initial condition $$y_0$$, we would need to:
• change y0_obs to y0 in the data block
• remove:
• real y0[1]; from the parameters block
• the prior y0 ~ normal(0, 1); from the model block
• the likelihood y0_obs ~ normal(y0, sigma); from the model block
Nothing in the integrate_ode_bdf call needs to change since it can handle a y0 that’s either data or parameter. For more information refer to the Ordinary Differential Equations chapter in Stan User’s Guide
Okay, now on to inference by sampling:
speed_data <- list(
T = n - 1,
t0 = 0,
times = times[-1],
m = 2,
y0_obs = speed$y_obs[1], y_obs = array(speed$y_obs[-1], dim = c(n - 1, 1))
)
speed_fit <- sampling(
speed_model,
data = speed_data,
refresh = 0 # silence progress notification
)
print(speed_fit)
## Inference for Stan model: speed_model.
## 4 chains, each with iter=2000; warmup=1000; thin=1;
## post-warmup draws per chain=1000, total post-warmup draws=4000.
##
## mean se_mean sd 2.5% 25% 50% 75% 97.5% n_eff Rhat
## sigma 2.74 0.01 0.52 1.96 2.37 2.66 3.03 3.93 1689 1.00
## theta[1] 0.40 0.00 0.04 0.32 0.38 0.40 0.42 0.48 1189 1.00
## theta[2] 9.69 0.02 0.73 8.21 9.23 9.70 10.17 11.15 1203 1.00
## y0[1] 0.05 0.02 0.91 -1.76 -0.54 0.02 0.64 1.86 1886 1.00
## lp__ -45.29 0.04 1.62 -49.48 -46.07 -44.91 -44.12 -43.34 1355 1.01
##
## Samples were drawn using NUTS(diag_e) at Fri May 22 09:41:29 2020.
## For each parameter, n_eff is a crude measure of effective sample size,
## and Rhat is the potential scale reduction factor on split chains (at
## convergence, Rhat=1).
And here’s a pretty plot of the results:
|
2020-05-26 22:41:59
|
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|
http://mathematica.stackexchange.com/questions/632/how-to-embed-fonts-when-exporting-mathematica-graphics-as-pdf?answertab=oldest
|
How to embed fonts when exporting Mathematica graphics as PDF
Upon exporting a Mathematica graphics expression as a PDF, Mathematica does not seem to embed any nonstandard fonts used. When you open the resulting PDF document with a PDF viewer, all the text items are rendered with standard fonts instead.
Is there a way to make Mathematica embed nonstandard fonts?
Example with Mathematica 8.0.4 under Mac OS X 10.7:
g = Graphics[{ Circle[],
Text["Sample",
BaseStyle -> {FontFamily -> "Papyrus", FontWeight -> "Bold",
FontSize -> 24}]}, ImageSize -> Tiny]
f = Export["~/Desktop/Graphics.PDF", g, "PDF"]
The resulting PDF does not use the correct font.
-
Can you give a code example, and instructions for checking that the fonts are embedded? I tried Export["testpdf.pdf", Graphics@Text[Style["asd", FontFamily -> "Candara", FontSize -> 50]]] and Adobe Reader reports that the font is embedded. I used CTRL-D in Adobe Reader, and clicked the Fonts tab. It says, "Candara (Embedded)". – Szabolcs Jan 24 '12 at 20:26
See updated question. – sakra Jan 24 '12 at 21:07
Sakra, what font version of Papyrus and what format is it? You can check this in Fontbook. I think it matters. – Verbeia Jan 26 '12 at 19:01
Font Book.app says Version 6.1d10e2. – sakra Jan 26 '12 at 19:31
What Verbeia said in her answer is not entirely correct — Mathematica indeed does embed the font, regardless of whether a particular font weight/slant exists or not. The real culprits are the PDF viewers on Macs, which do not use the base font if the specified weight is not available. It took some digging around to get to the reason though. The clues that led to my reasoning are as follows:
1. Mathematica knows what font it used when you re-import
This was the first clue. Executing Import["~/Desktop/Graphics.PDF"] will correctly display the graphics in Papyrus font, albeit without the bold option (which, as Verbeia noted, doesn't exist in all fonts). So the information had to be in the file somewhere.
2. The binary file shows Papyrus embedded in it
Opening the pdf in vim under hex mode, you find the font info embedded in it. What you see in the readable text on the right is the full copyright info, which they'd have to include only if it were embedded.
3. It opens correctly on linux systems
Papyrus is not a font that comes installed by default with linux systems. The above file opens with the correct font (but not weight) on RHEL 6 and Ubuntu 11.10, which can only mean that the font is embedded in the pdf. Below is a screenshot in RHEL 6, with the same file opened in Adobe reader (left) and evince (right).
Below is a screenshot of the same file opened in Adobe reader (left) and Preview.app (right). You can see that Adobe sure does know which font to use.
So you can see here that the problem is most likely with the PDF viewers on macs (other than Adobe reader). My guess is that they don't stay true to Adobe's specifications, and instead rely on OSX's Quartz framework for rendering the file, which is why you don't see the right font.
Either that, or they don't like the fact that Mathematica uses PDF v1.4 to export its files. v1.4 is pretty old (2001), and it is worth mentioning (since you're embedding non-standard fonts) that PDF v1.4 does not support embedding "Open Type" fonts. That was introduced only in v1.6, and if you're working with such fonts, you might want to keep this in mind.
-
Thanks for digging that out. I only use Preview.app for PDF viewing under Mac OS X. – sakra Jan 25 '12 at 18:33
@R.M Aren't you saying, though, that Adobe Reader doesn't so much 'use the correct font' as 'guess correctly what Mathematica got up to when it created the file', and then cover for it? Technically Mathematica specified a font that didn't exist, and just drew a black line around the font to make it look bold. But Preview refused to make a font up this way, and defaulted to something it did have. If you ask for Papyrus Italic, Mathematica can do that too... Perhaps Preview should loosen up and play a little bit more! – cormullion Jan 25 '12 at 20:17
@R.M this answer is incorrect and misleading. The presence of the font name in the PDF file does not mean that the font itself is embedded in the file, which was the OP's question. Adobe Reader and Mathematica display the re-imported graphic correctly because you have the font installed. If you displayed the graphic on a computer that did not have the font installed, the bold version would not display but the normal weight one would. (Though strangely, now that I check this on my Mac it does embed, even though similar fonts on Windows don't. The format of the font might matter as well.) – Verbeia Jan 26 '12 at 13:10
@Verbeia I don't follow your comment... you start out saying it isn't embedded, but ultimately agree with me in saying that it is indeed embedded. Which one is it? It is incorrect to say that the font is not embedded just because it doesn't have a bold option. As I said in my answer (but didn't make it explicit) I tested it on ubuntu 11.10 and it opens with the correct font. Papyrus is a font that is shipped only with Windows and OS X, not linux distros, and this can only mean that the font was embedded. – rm -rf Jan 26 '12 at 16:51
Ok I have removed the downvote. But I think that both answers contain part of the truth. There clearly are fonts that don't embed properly, even if they exist, or this question would not have come up. And even though Papyrus Bold works in some PDF viewers, it is fair to say that font weights that don't exist can't be relied on. It is not clear to me that Preview's behaviour is incorrect. There seems to be a subtle relationship between font format and how it exports, which needs further investigation. – Verbeia Jan 26 '12 at 22:13
Mathematica will normally embed the fonts correctly in a PDF. But this only works if the font exists. Papyrus does not have a bold weight. Try with FontWeight->"Bold" deleted and see if this works.
I don't have Papyrus on my Windows machine, but I checked that Kristen ITC does not embed if it is bold, and does if the FontWeight option is removed.
-
By the way @sakra, thank you for this question. I didn't actually know the answer to this, and your question made me find it out. Very useful. – Verbeia Jan 24 '12 at 22:06
+1 that is interesting that it doesn't embed if the font weight for that font isn't available. – Mike Honeychurch Jan 24 '12 at 22:49
In Mathematica 9, (rm -rf)'s comment that fonts with the extension .otf (aka OpenType) can't be embedded still holds. In addition, I've also noticed that not all fonts with the extension .ttf can be embedded.
|
2014-07-10 09:39:40
|
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http://blog.alexbeutel.com/285/cmake-opengl-and-glx-on-os-x/
|
# Alex Beutel's Blog
## CMake, OpenGL, and GLX on OS X
#### July 28th, 2010 · 4 Comments
As you may know, I am staying at Duke this summer for research. My research, which I started last semester and am going to be doing for a while, focuses on scalable algorithms. While that is pretty broad, I am currently working on using the graphics card to speed up processing. Again, here are some technical notes on code.
In doing my work I have come to have to write and run C++ OpenGL code on both OS X and Ubuntu. To make this easier, CMake has been great for both having readable make files and for being able to keep the files consistent across both platforms. Additionally, while GLUT is great for cross-platform OpenGL coding, I needed more control over the environment. Being able to run X11 on OS X has allowed me to test GLX natively without always testing on Linux. However, to get this set up on OS X without XCode can be a little tricky. So, two snippets of CMake files for using OpenGL on OS X and Linux.
If you want to run OpenGL with the standard OS X libraries, CMake generally finds the correct frameworks, but it can’t hurt to specify them more precisely. A basic example of the CMakeLists is shown below:
IF(APPLE)
INCLUDE_DIRECTORIES ( /System/Library/Frameworks )
FIND_LIBRARY(COCOA_LIBRARY Cocoa)
FIND_LIBRARY(GLUT_LIBRARY GLUT )
FIND_LIBRARY(OpenGL_LIBRARY OpenGL )
GLUT_LIBRARY
OpenGL_LIBRARY)
SET(EXTRA_LIBS ${COCOA_LIBRARY}${GLUT_LIBRARY} ${OpenGL_LIBRARY}) ENDIF (APPLE) target_link_libraries(main${EXTRA_LIBS})
With this, your C++ should have the following includes:
#include #include #include #include
If, however, you are looking to run your code also on a Linux machine or simply want to use GLX rather than CGL, you will need to have X11 installed (which your OS X installation disk contains) and specifically point CMake at the X11 libraries. The CMake code for this is shown below:
include_directories(/usr/X11R6/include/)
SET(EXTRA_LIBS GL X11 GLU glut)
When using the X11 libraries, you must also change the includes as the folder structure for X11 is different than that natively for OS X:
#include #include #include #include #include #include
That is all. Not terribly complicated, but hopefully this saves you some time if you are doing development with CMake, OpenGL, or X11. Look for some new slightly more exciting posts soon.
Tags: C++ · OpenGL · Programming
### 4 responses so far ↓
• 1 Per // Oct 5, 2011 at 12:59 am
Thanks man! You saved my day 🙂
• 2 Calum // Jun 26, 2012 at 2:21 pm
Ah that saved some time, thanks!
• 3 Nils // Aug 10, 2015 at 8:33 pm
Could you believe, this blog post is helping me out, three years later…Question though, for that second CMake code block, should it be inside of an IF…ENDIF just like the first one?
Thanks
• 4 从零开始手敲次世代游戏引擎（macOS特别篇） – 文礼简阔 // Dec 2, 2017 at 8:11 am
[…] CMake, OpenGL, and GLX on OS X […]
|
2019-01-24 02:24:33
|
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|
https://codereview.stackexchange.com/questions/9464/national-insurance-number-generator
|
# National Insurance number generator
I am trying to figure out how to DRY up this code:
nino = "QQ"
3.times { nino += Random.rand(10..99).to_s}
nino += ("A".."D").to_a.sample
I am trying to generate a sample National Insurance number starting with QQ, six digits and then ending with A, B, C or D.
Why do you need the 3 times? Is there a restriction, that the first number may not be 0?
If there is no 0-restriction, you may generate a number between 0 and 999999. To get the leading 0 I use the strftime-pattern %06i (6 digit with leading zeros).
p "QQ%06i%s" % [ Random.rand(999999), ("A".."D").to_a.sample ]
For the letter, you may generate a random number from 65..68 and take the character for the selected ASCII-value (code example is ruby 1.9. Maybe you need a little variation with Ruby 1.8):
p "QQ%06i%s" % [ Random.rand(999999), (65 + rand(4)).chr ]
If you mean more compact code, I though you can consider this:
nino = "QQ"<<(10..99).to_a.sample(3)*''<<("A".."D").to_a.sample
As Cygal says:
Also note that "DRY" in Code Review stands for "Don't Repeat Yourself", your title is thus a bit ambiguous.
If you are repeating implementing this logic through out your code then you should dry it up.
In my opinion this code has another problem. The purpose is not clear at a glance. Actually, one should process it in his mind to understand what's going on.
To prevent such a problem I suggest that:
1. Encapsulate it in a method definition block with a suitable name that describes what it's doing (this way you will say what it's doing but not how it has implemented).
2. Use a pattern or format to show how the result will look like. I am not familiar with Ruby but each language has its own way of doing it for example in C# we can use stirng.Format to define a format for our string:
var result=String.Format("QQ{0}{1}{2}{3}",GetARandomNumber(),GetARandomNumber(),GetARandomNumber(),GetADigitBetween('A','D'));
• As I responded to Cygal "Essentially my code is saying nino = blah, nino = nino + blahblah, nino = nino + blahblahblah; I am repeating myself three times which is why I was trying to not repeat myself." – ScottJShea Feb 27 '12 at 13:22
Disclaimer: I don't know Ruby but code in similar languages.
This looks "dried" enough to me. What would enhance readability though is to write this one simple line. It would show in a very explicit way that you are simply building one string (and would also make this shorter). Is it possible to make a one-liner out of this in Ruby?
Also note that "DRY" in Code Review stands for "Don't Repeat Yourself", your title is thus a bit ambiguous.
• Essentially my code is saying nino = blah, nino = nino + blahblah, nino = nino + blahblahblah; I am repeating myself three times which is why I was trying to not repeat myself. – ScottJShea Feb 27 '12 at 13:21
p "QQ#{rand(100_000..999_999)}#{%w(A B C D).sample}"
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2021-01-17 18:42:24
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http://mathoverflow.net/revisions/28114/list
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
Let I be the unit interval with the Borel $\sigma$-algebra. There is no $\sigma$-algebra on the set of measurable functions from I to I such that the evaluation functional $e:I^I\times I\to I$ given by $e(f,x)=f(x)$ is measurable, as shown by Robert Aumann here, so even finding useful $\sigma$-algebras is a problem.
Let I be the unit interval with the Borel $\sigma$-algebra. There is no $\sigma$-algebra on the set of measurable functions from I to I such that the evaluation functional $e:I^I\times I\to I$ given by $e(f,x)=f(x)$ is measurable, as shown by Robert Aumann here, so even finding useful $\sigma$-algebras is a problem.
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2013-06-19 09:40:39
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https://zerojudge.tw/ShowProblem?problemid=a243
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a243: 第四題：點燈遊戲
（改變狀態就是原本開的會變關的、關的會變開的。）
(1 m 25, 1 n 25, 1 d m + n)
1 1 1
1
2 2 1
1 1
1 1
3 2 1
1 0 1
0 1 0
3 3 1
1 0 1
0 1 0
1 0 1
4 4 2
1 1 0 1
0 0 0 1
1 0 1 1
1 0 0 0
5 5 1
1 1 1 0 1
0 1 0 1 0
1 0 1 0 1
0 1 0 1 0
1 0 1 0 1
5 5 2
0 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0
0 0 0 0 0
11 11 3
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
11 11 3
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
13 13 7
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0
1
1
0
1
0
0
1
1
0
1
[編輯：
(蝸牛)
]
編號 身分 題目 主題 人氣 發表日期 沒有發現任何「解題報告」
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2019-01-19 20:25:35
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http://mathoverflow.net/questions/90660/where-can-i-see-the-proof-that-the-homology-groups-of-the-moore-complex-of-a-sim/90663
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# Where can I see the proof that the homology groups of the Moore Complex of a simplicial group coincide with the homotopy groups of its geometric realization?
I'd like to know where to find it since it's very used in the articles of Loday and others.
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Have you looked at Curtis' notes on simplicial homotopy theory? I think what you're looking for is in there. I believe the notes appeared in Advances in the 70's. – Dan Ramras Mar 9 '12 at 5:11
Another suggestion, which is a comment not an answer because I don't really understand the material well enough to see if it really answers your question: Chapter 8 of Weibel's book on Homological Algebra. Although the main emphasis is on simplicial objects in abelian categories, the remarks and exercises do develop some of the parallel theory for simplicial groups – Yemon Choi Mar 9 '12 at 6:37
Oh, come on! Prop. 17.4, p. 69, of my ancient but still current book Simplicial objects in algebraic topology'' proves that the homology groups of the Moore complex of a simplicial group $G$ are the homotopy groups (defined simplicially) of the Kan complex $G$. That G is a Kan complex is Thm 17.1. The homotopy groups of any Kan complex agree with the homotopy groups of its geometric realization (op cit, Thm 16.6). The cited book is still available from the University of Chicago Press (or Amazon of course).
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2014-04-18 13:57:57
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http://math.stackexchange.com/questions/289910/sitting-around-a-round-table-in-circular-permuations
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# Sitting around a round table in circular permuations
If $3$ people are selected at random from $15$ people sitting at a round table, then what is the probability that no two of them are adjacent to each other?
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If no-one can sit next to someone else, then three seats are blocked every time someone sits down. However, some of these may have been blocked already, by previous sitters. So, the first person to sit down has $15$ choices, and leaves twelve choices. The next person to sit down has $12$ choices; in two of these cases, he leaves ten choices; in the other ten cases, he leaves nine. The final person to sit down has either $10$ or $9$ choices, depending on what the second person did. The result is $$N=15\cdot(2\cdot10+10\cdot9)=1650$$ ways for three (distinguishable) people to sit down. The probability, then, is $$p=\frac{N}{15\cdot14\cdot13}=\frac{1650}{2730}=\frac{55}{91}.$$
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Hint: By symmetry, without loss of generality we can assume that Alicia is chosen. Now how many ways are there to choose $2$ people from the $14$ remaining ones to accompany Alicia? How many ways are there to choose $2$ people so they are not next to Alicia or to each other?
Added: Since we have already done it in comments, we might as well add more detail. For the numerator, the $2$ people (seats) next to Alicia are forbidden, so we are choosing $2$ from the remaining $12$. But some of these choices are forbidden, the choices in which we have a pair of neighbouring seats. A little play will show that of the $\binom{12}{2}$ ways to choose $2$ people who are not next to Alicia, $11$ are forbidden.
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We can select any $3$ out of $15$ is $\displaystyle =\binom{15}{3}$ ways – juantheron Jan 29 '13 at 17:58
What if there are $k$ tables? – alancalvitti Jan 29 '13 at 18:06
actually i did not understand would you like to explain me completely thanks – juantheron Jan 29 '13 at 18:09
For the numerator, there are $12$ seats (people) left to choose from, since Alicia's is taken and the two next to her are forbidden. From these $12$, we choose $2$ not next to each other. There are $\binom{12}{2}$ ways to choose $2$ seats from $12$. But $11$ of these choices are forbidden, since they give an adjacent pair. You should end up with numerator $55$, denominator $105$. – André Nicolas Jan 29 '13 at 18:31
@AndréNicolas: The denominator had better be ${{14}\choose{2}}=91$, since you've already chosen Alicia. – mjqxxxx Jan 29 '13 at 19:02
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2015-07-28 22:13:17
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https://www.physicsforums.com/threads/coordinate-change-to-remove-asymptotic-geodesic.546914/
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# Coordinate change to remove asymptotic geodesic?
1. Nov 3, 2011
### ck99
Through my mathematical fumblings, I think I have found a metric which gives a solution of the geodesic equation of motion that is asymptotic. It is a diagonal metric, with g00 = (x_1)^(-3) and g11 = 1. I am largely self-taught with SR so I may be miles off, but I think this gives a G.E. of M which has a λ = 1 / (x_1) term in it, so my parameter goes to infinity when x_1 = 0.
Even if I have the details wrong, I have two questions:
1) Can you define a metric where geodesics are asymptotic?
2) Can you define the same metric with different coordinates to remove this behaviour?
The only thing I can think of is some kind of substitution, but I don't really know what to do and the textbook I am working through is not a lot of help. I even got the massive "Gravitation" book out of my library but if it did contain the solution, I didn't understand it!
Hopefully someone can give me a clue :)
2. Nov 4, 2011
### Bill_K
In the Schwarzschild metric in the usual coordinates, the inward-going null geodesics run off the coordinate patch, going to t = +∞ as r goes to 2m. Is that what you mean by asymptotic?
3. Nov 4, 2011
### Matterwave
If this is what you meant by asymptotic, then whether you can get rid of this behavior via a change of coordinates is dependent on if this behavior is a real singularity (e.g. at the center of a black hole) or a coordinate singularity (e.g. at the event horizon of a black hole). A real singularity would have the curvature tensor also diverging, whereas a coordinate singularity would have a finite curvature. If it's a coordinate singularity then you can find coordinates which remove this asymptotic behavior (for Schwarzschild, you can use Eddington-Finklestein coordinates or Kruskal-Szekeres coordinates).
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2017-09-23 00:29:56
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