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https://guzintamath.com/textsavvy/the-distance-formula/
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# The Distance Formula
the distance formula
Using the Distance Formula
8-G.B.8
Click on the image at right to get the lesson app with instructor notes. Or you can install right from here by clicking the logo below:
Students are introduced to the Distance Formula, $$\mathtt{d = \sqrt{a^2 + b^2}}$$, as derived from the Pythagorean Theorem. In the lesson, a and b will be expanded to show that they are equal to |x1 – x2| and |y1 – y2|, respectively.
Students will determine distances on the coordinate plane, first using the complete right triangle, then using just the drawn distance. Students should use calculators throughout this lesson.
In the second module, students use what they have learned about the Distance Formula and apply it to determine the perimeters and areas of polygons such as triangles, parallelograms, and trapezoids on the coordinate plane. At the end, students see that the Distance Formula can be applied even when given only a set of ordered pairs for a polygon’s vertices.
Finally, students will solve some application and mathematical problems using the Distance Formula.
Module 1 Video
This video introduces students to the Distance Formula as derived from the Pythagorean Theorem. This is not the Distance Formula as you know it, but is kept in this form to show how it is a re-expression of the Pythagorean Theorem in another form.
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2020-01-26 08:22:14
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https://forum.qt.io/topic/3123/question-which-qt-tool-to-use-to-develop-this-project
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[Question] Which QT tool to use to develop this project
• Dear all,
I'm interested to create a project that can be written once and deployed for Nokia N8 (running Symbian^3) and Windows Mobile 6
I've downloaded QT SDK for Windows but it does not have QT version for Symbian when I create a new project in QT Creator (as in the tutorial "Creating a mobile application". Tried downloading Symbian SDK and does not seem to have an IDE like QT Creator.
• Hi HamikSu,
It is Qt and not QT (Quick Time)
You should get the Nokia Qt SDK from "forum nokia":http://www.forum.nokia.com/Develop/Qt/ which makes it quite easy for you to develop and deploy on to your mobile .....
And you can use creator as the IDE
• What that is confusing for you is just some notes before starting with Qt. There's no different between platforms to use with QtCreator. Then you need to have a little knowledge about configuring Qt for different platforms. Although you can download compiled binaries from http://qt.nokia.com
then you can set the QtCreator environment to your desired SDK at Tools->Options->Qt4
to compile for WinMobile and WinCE you'll need to have their SDK installed. After configuring your Qt for required version, you'll have to use QtAddIn under VS2008 to compile and deploy to your device.
Looks like your connection to Qt Forum was lost, please wait while we try to reconnect.
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2017-11-24 02:20:06
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https://encyclopediaofmath.org/index.php?title=Stokes_theorem&oldid=31286
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Stokes theorem
2010 Mathematics Subject Classification: Primary: 58A [MSN][ZBL]
The term refers, in the modern literature, to the following theorem.
Theorem 1 Let $M$ be a compact orientable differentiable manifold with boundary (denoted by $\partial M$) and let $k$ be the dimension of $M$. If $\omega$ is a differential $k-1$-form, then $$\label{e:Stokes_1} \int_M d \omega = \int_{\partial M} \omega$$ (see Integration on manifolds for the definition of integral of a form on a differentiable manifold).
The theorem can be considered as a generalization of the Fundamental theorem of calculus. The classical Gauss-Green theorem and the "classical" Stokes formula can be recovered as particular cases. The latter is also often called Stokes theorem and it is stated as follows.
Theorem 2 Let $\Sigma\subset \mathbb R^3$ be a compact regular $2$-dimensional surface $\Sigma$ that bounds the $C^1$ curve $\gamma$ and let $v$ be a $C^1$ vector field. Then $$\label{e:Stokes_2} \int_\Sigma (\nabla \times v) \cdot \nu = \int_\gamma \tau \cdot v\, ,$$ where
• $\nu$ is a continuous unit vector field normal to the surface $\Sigma$
• $\tau$ is a continuous unit vector field tangent to the curve $\gamma$, compatible with $\nu$
• $\nabla \times v$ is the curl of the vector field $v$.
The right hand side of \eqref{e:Stokes_2} is called the flow of $v$ through $\Sigma$, whereas the left hand side is called the circulation of $v$ along $\gamma$. The theorem can be easily generalized to surfaces whose boundary consists of finitely many curves: the right hand side of \eqref{e:Stokes_2} is then replaced by the sum of the integrals over the corresponding curves.
Both \eqref{e:Stokes_1} and \eqref{e:Stokes_2} are often called Stokes formula. If the vector field of Theorem 2 is given, in the coordinates $x_1, x_2, x_3$, by $(v_1, v_2, v_3)$ and we introduce the $1$-form $\omega = v_1 dx_1 + v_2 dx_2 + v_3 dx_3\, ,$ then the right hand side of \eqref{e:Stokes_1} is indeed $\int_\Sigma d \omega\, ,$ whereas the left hand side is $\int_{\partial \Sigma} \omega\, .$
Remark 3 The compatibility between the vector fields $\tau$ and $\nu$ in Theorem 2 can be expressed intuitively as follows. The normal $\nu$ identifies a "bottom" and a "top" on the surface $\Sigma$. To an observer which is standing on the top, $\tau$ gives a counterclockwise orientation to the curve $\gamma$. The precise mathematical definition is more cumbersome. Fix $p_0\in \gamma$, let $V\subset \mathbb R^3$ be an open neighborhood of $x_0$ and $U\subset \mathbb R^2$ the intersection of an open neighborhood of $0\in \mathbb R^2$ with the closed upper half plane $\{(x_1, x_2): x_2\geq 0\}$. Assume $\Phi: U\to V$ is a local parametrization of $\Sigma\cap V$ with $\Phi (0) = p_0$, namely that
• $\Phi$ is $C^1$ and $D\Phi$ has rank 2 at each point of $U$
• $\Phi$ is an homeomorphism between $U$ and $\Sigma \cap V$
• $\Phi$ maps $\{x_2=0\}\cap U$ onto $\gamma \cap V$.
Then the vector field $n := \frac{\partial \Phi}{\partial x_1} \times \frac{\partial \Phi}{\partial x_2}$ is a nonzero vector field normal to the surface $\Sigma$ and therefore the scalar product $n (x) \cdot \nu (\Phi (x))$ is either everywhere positive or everywhere negative. In the first case $\tau (x_0) = \left|\frac{\partial \Phi}{\partial x_1} (0)\right|^{-1} \frac{\partial \Phi}{\partial x_1} (0)\, ,$ otherwise $\tau (x_0) = - \left|\frac{\partial \Phi}{\partial x_1} (0)\right|^{-1} \frac{\partial \Phi}{\partial x_1} (0)\, .$
References
[Ap] T.M. Apostol, "Calculus" , I , Blaisdell (1967) MR0214705 Zbl 0148.28201 [Sp] M. Spivak, "Calculus on manifolds" , Benjamin (1965) MR0209411 Zbl 0141.05403
How to Cite This Entry:
Stokes theorem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Stokes_theorem&oldid=31286
This article was adapted from an original article by L.D. Kudryavtsev (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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2022-11-27 12:54:30
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https://proofwiki.org/wiki/Conic_Section_through_Five_Points
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# Conic Section through Five Points
## Theorem
Let $A, B, C, D, E$ be distinct points in the plane such that no $3$ of them are collinear.
Then it is possible to draw a conic section that passes through all $5$ points.
## Historical Note
The technique for constructing a conic section that passes through $5$ non-collinear points was demonstrated by Pappus of Alexandria.
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2019-03-24 01:26:31
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https://socratic.org/questions/596e39f07c014971add6a890
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# Can activation energy ever be negative in ordinary scenarios?
Jul 30, 2017
Absolutely not. Activation energy is defined as the energy INPUT needed to successfully carry out the reaction. The transition state is always higher in energy than the reactants, with DeltaG^(‡) prop E_a.
(Note: do not confuse DeltaG^(‡) as a thermodynamic property; although it is proportional to the activation energy, a kinetic property, $\Delta {G}_{r x n}$ is normally a thermodynamic property, relating the reactants to the products.)
Otherwise, there would exist a reaction that gets stuck in the transition state, which is quite unheard of. Transition states are notorious for being short-lived and thus difficult to detect by experiment.
On the other hand, it is possible that an intermediate is lower in energy than the reactants, but I've only seen that in a step in glycolysis.
At that point, one's body forms a LESS stable, thioester intermediate, to get the reaction to proceed.
However, that is NOT indicative of a negative activation energy. It just means the second activation energy is much larger than the first, both positive.
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2019-09-15 16:02:24
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https://itprospt.com/num/8487569/i-let-overrightarrow-v-1-be-the-velocity-of-the-buggy-after
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1
# (i) Let $overrightarrow{v_{1}}$ be the velocity of the buggy after both man jump off simultaneously. For the closed system (two men + buggy), from the conservation ...
## Question
###### (i) Let $overrightarrow{v_{1}}$ be the velocity of the buggy after both man jump off simultaneously. For the closed system (two men + buggy), from the conservation of linear momentum, or, $$egin{gathered} M overrightarrow{v_{1}}+2 mleft(vec{u}+overrightarrow{v_{1}} ight)=0 \ overrightarrow{v_{1}}=frac{-2 m vec{u}}{M+2 m} end{gathered}$$ (ii) Let $vec{v}^{prime}$ be the velocity of buggy with man, when one man jump off the buggy. For the closed system (buggy with one man $+$ other man) from th
(i) Let $overrightarrow{v_{1}}$ be the velocity of the buggy after both man jump off simultaneously. For the closed system (two men + buggy), from the conservation of linear momentum, or, $$egin{gathered} M overrightarrow{v_{1}}+2 mleft(vec{u}+overrightarrow{v_{1}} ight)=0 \ overrightarrow{v_{1}}=frac{-2 m vec{u}}{M+2 m} end{gathered}$$ (ii) Let $vec{v}^{prime}$ be the velocity of buggy with man, when one man jump off the buggy. For the closed system (buggy with one man $+$ other man) from the conservation of linear momentum : $$0=(M+m) vec{v}^{prime}+m(vec{u}+vec{v})$$ Let $overrightarrow{v_{2}}$ be the sought velocity of the buggy when the second man jump off the buggy; then from conservation of linear momentum of the system (buggy + one man): $$(M+m) vec{v}=M overrightarrow{v_{2}}+mleft(vec{u}+vec{v}_{2} ight)$$ Solving equations (2) and (3) we get $$overrightarrow{v_{2}}=frac{m(2 M+3 m) vec{u}}{(M+m)(M+2 m)}$$ From (1) and (4) $$frac{v_{2}}{v_{1}}=1+frac{m}{2(M+m)}>1$$ Hence $v_{2}>v_{1}$
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2022-08-15 16:22:59
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http://mathhelpforum.com/calculus/31082-some-integrals.html
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1. ## Some integrals
I have to take a test that's a review of single-variable calculus, which I haven't really done in years. Most of the sample questions I've been able to do, but these are sticking me:
$\int^{\pi/2}_{0}\frac{\sin(x)}{1+\cos^{2}(x)}dx$
Now I know that this should be $\tan^{-1}(\cos(x))$ evaluated at the endpoints, but that's because I just looked that up. Is there a way to figure this out without simply having to know the derivative of $\tan^{-1}$?
$\int^{2}_{1}|x^{2}\sin(\pi x^{3})|dx$
Absolute values? Yikes. Do I just integrate this as if the absolutes weren't there, and then apply them when evaluating at the endpoints?
Thanks
2. Originally Posted by Reisen
I have to take a test that's a review of single-variable calculus, which I haven't really done in years. Most of the sample questions I've been able to do, but these are sticking me:
$\int^{\pi/2}_{0}\frac{\sin(x)}{1+\cos^{2}(x)}dx$
Now I know that this should be $\tan^{-1}(\cos(x))$ evaluated at the endpoints, but that's because I just looked that up. Is there a way to figure this out without simply having to know the derivative of $\tan^{-1}$?
do a substitution. u = cos(x)
$\int^{2}_{1}|x^{2}\sin(\pi x^{3})|dx$
Absolute values? Yikes. Do I just integrate this as if the absolutes weren't there, and then apply them when evaluating at the endpoints?
no. the absolute values make a difference. you need to know how the graph without the absolute values behave. any interval that will give a negative integral will now give a positive one with the absolute values there. so you have to split the interval up. you have your work cut out for this one! see the graph below
3. Originally Posted by Jhevon
do a substitution. u = cos(x)
I meant in terms of figuring out that it's supposed to be $\tan^{-1}$ without having to memorize the trig derivatives, but now that I think about it, that's kind of a silly question since I'm not doing to be deriving that stuff in the middle of the test.
no. the absolute values make a difference. you need to know how the graph without the absolute values behave. any interval that will give a negative integral will now give a positive one with the absolute values there. so you have to split the interval up. you have your work cut out for this one! see the graph below
Eep. Let's see, so the function equals zero whenever $x^{3}$ is an integer. $1^{3}=1, 2^{3}=8$, so I can split this up into
$-\int_{1}^{\sqrt[3]{2}}f(x)+\int_{\sqrt[3]{2}}^{\sqrt[3]{3}}f(x) - \int_{\sqrt[3]{3}}^{\sqrt[3]{4}}f(x) + \int_{\sqrt[3]{4}}^{\sqrt[3]{5}}f(x) - \int_{\sqrt[3]{5}}^{\sqrt[3]{6}}f(x) + \int_{\sqrt[3]{6}}^{\sqrt[3]{7}}f(x) - \int_{\sqrt[3]{7}}^{2}f(x)$
and evaluate that. The antiderivative looks like it should be $-\frac{1}{3\pi}\cos(\pi x^{3})$, so this won't be super difficult, just tedious. Thanks!
4. Originally Posted by Reisen
I meant in terms of figuring out that it's supposed to be $\tan^{-1}$ without having to memorize the trig derivatives, but now that I think about it, that's kind of a silly question since I'm not doing to be deriving that stuff in the middle of the test.
nope, in this case, memorizing the trig derivative is the way to go. but i do not think your question is silly in general. it is good to know how to derive things. in general, it makes life easier for you, helps to cut down on how much memory you have to use. and if you're fast enough, it doesn't slow you down much in tests. at least, not as far as elementary calculus is concerned... usually
Eep. Let's see, so the function equals zero whenever $x^{3}$ is an integer. $1^{3}=1, 2^{3}=8$, so I can split this up into
$-\int_{1}^{\sqrt[3]{2}}f(x)+\int_{\sqrt[3]{2}}^{\sqrt[3]{3}}f(x) - \int_{\sqrt[3]{3}}^{\sqrt[3]{4}}f(x) + \int_{\sqrt[3]{4}}^{\sqrt[3]{5}}f(x) - \int_{\sqrt[3]{5}}^{\sqrt[3]{6}}f(x) + \int_{\sqrt[3]{6}}^{\sqrt[3]{7}}f(x) - \int_{\sqrt[3]{7}}^{2}f(x)$
and evaluate that. The antiderivative looks like it should be $-\frac{1}{3\pi}\cos(\pi x^{3})$, so this won't be super difficult, just tedious. Thanks!
yes
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2016-12-10 09:26:23
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http://mathhelpforum.com/trigonometry/96423-how-prove-identity-print.html
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# How to prove as an identity?
• July 29th 2009, 10:28 AM
crosser43
How to prove as an identity?
how do i prove that this is an identity?
$\frac{tan^2\theta}{1+tan^2\theta}+\frac{cot^3\thet a}{1+cot^2\theta}=\frac{1-2sin^2\theta cos^2\theta}{sin\theta cos\theta}$
• July 29th 2009, 11:23 AM
stapel
The left-hand side, having two fractions, is (or at least looks to be) more complicated than the right-hand side, so try working on the left-hand side.
A good start would probably be to convert everything to sines and cosines, convert to a common denominator, combine the two fractions, and see where that leads.... (Wink)
• July 29th 2009, 12:20 PM
crosser43
ok but does $1+cot\theta$ equal to $1-2sin^2\theta cos^2\theta$ ?
• July 29th 2009, 12:44 PM
e^(i*pi)
Quote:
Originally Posted by crosser43
ok but does $1+cot\theta$ equal to $1-2sin^2\theta cos^2\theta$ ?
$
1+cot(x) = 1+\frac{cos(x)}{sin(x)} = \frac{cos(x)+sin(x)}{sin(x)}$
You may recall that $1+tan^2(x) = sec^2(x)$ and $1+cot^2(x) = csc^2(x)$
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2015-02-01 04:33:53
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https://www.hackmath.net/en/math-problem/7897
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# Dozen
What is the product of 26 and 5? Write the answer in Arabic numeral. Add up the digits. How many of this is in a dozen?
Divide #114 by this
Correct result:
a = 130
b = 4
c = 3
d = 38
#### Solution:
$b=1+3+0=4$
$c=12\mathrm{/}b=12\mathrm{/}4=3$
$d=114\mathrm{/}c=114\mathrm{/}3=38$
We would be very happy if you find an error in the example, spelling mistakes, or inaccuracies, and please send it to us. We thank you!
Tips to related online calculators
Do you want to perform natural numbers division - find the quotient and remainder?
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2020-10-01 03:52:50
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https://www.physicsforums.com/threads/moving-water-electrostatic-generator.442723/
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# Moving water electrostatic generator
#### Stanley514
There is known effect when electric field bends water stream:
http://www.school-for-champions.com/science/static_induction.htm" [Broken]
Does that mean that moving water inversely affect electric field?If yes,could we make electric generator in which moving water would affect electric field and produce voltage or current?
Exhaust of engines is mostly water.
I mean something different from Kelvins generator because the latest one uses impure (ionized water).
Last edited by a moderator:
Related Classical Physics News on Phys.org
#### WhoWee
Do you have a specific application in mind?
#### Stanley514
Code:
Do you have a specific application in mind?`
Sure.Because engine exhaust is mostly water and carbon dioxide which are both poler molecules
we could try to build some kind of electrohydrodynamic generator.When water vapor will go past electric field it may influence this field somehow and generate electricity. Electrostatic engine.
Last edited:
"Moving water electrostatic generator"
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2019-04-20 20:47:30
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http://math.tutorcircle.com/calculus/how-to-do-trigonometric-functions.html
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?> How to Do Trigonometric Functions? | Math@TutorCircle.com
Sales Toll Free No: 1-855-666-7446
# How to Do Trigonometric Functions?
TopWhenever we solve Trigonometry problems we have to deal with trigonometric formula. let's learn How to Do Trigonometric Functions? For understating the concept of Trigonometric Functions, you must have the knowledge of all the basic functions of trigonometry so, for solving trigonometric function we need to go through some examples.
Example 1: Solve the equation given as cotx/cosecx + tan x?
For solving this type of equation, we need to convert the trigonometric variable into simplest form. = cotx/cosecx,
We can write cotx as 1/tanx,
= cosecx as 1/sinx, and
= tanx as sinx/cosx,
Put this value in the above equation and we will get:
= (1/tanx )/ (1/sinx) +sinx/cosx,
Now, tanx can be written as sinx/cosx ,on putting this value in the above equation as:
= 1/(sinx/cosx) / (1/sinx) +sinx/cosx
We can rewrite the equation as:
= cosx/sinx *sinx + sinx/cosx,
Now, sinx gets cancel with sinx and we will get the required answer as cosx.+sinx/cosx.
The required simplified equation is (cosx +sinx)/cosx.
In this way, we can simplify any expression.
Example 2: Solve the equation given as: (cotx +tanx ) *cosx ?
Solution: = (cotx +tanx ) *cosx,
= cotx can be written as 1/tanx, so we can rewrite the given equation as:
= (1/tanx +tanx)*cosx,
Now, tanx can be written as sinx /cosx,
= (1/(sinx/cosx) + (sinx /cosx))*cosx,
= (cosx/sinx +sinx/cosx )*cosx,
Now, on taking LCM we will get:
= (cos2x+sin2x/cosx*sinx)*cosx,
Now, we can write sin2x +cos2x=1
So, we can write above equation as:
= (1/cosx*sinx) *cosx,
= (1/cosx *1/sinx)*cosx,
= (secx *cosecx)*cosx,
Now, we can write cosx as 1/secx so, we will rewrite the equation as:
= secx *cosecx *1/secx,
So, secx gets cancel with secx and we will get the solution as:
= cosecx.
Example:
Solve the equation given as: cotx + 1 / cosecx ?
Solution: cotx can be written as 1/tanx,
tanx can be written as sinx/cosx,
cotx can be written as cosx/sinx, and
We can write cosecx as 1/sinx.
Now, putting all these values in the given equation we will get:
= cosx /sinx +1/1/sinx,
We can write this equation as:
= (cosx +sinx ) * sinnx *1/sinx,
So, sinx gets cancel with sinx and we will get:
= cosx + sinx,
So, the required equation can be simplified as cosx + sinx.
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2017-11-24 18:41:23
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http://superuser.com/questions/479983/how-can-i-get-libreoffice-to-number-footnotes-in-the-order-%c2%a7-etc?answertab=votes
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# How can I get LibreOffice to 'number' footnotes in the order *, †, ‡, § etc.?
With LaTeX, I can do:
\documentclass[10pt]{article}
\usepackage[symbol*]{footmisc}
\begin{document}
One\footnote{f1} Two \footnote{f2} Three \footnote{f3} Four \footnote{f4}
\end{document}
And get *, †, ‡, § ... as consecutive footnote markers. MS-Word has this feature too - an alternative footnote numbering scheme. How can I achieve the same with LibreOffice?
PS - Shouldn't the OpenOffice and LibreOffice tags be merged?
-
LibreOffice is a fork from OpenOffice, so they are 2 separate softwares and their tags shouldn't be merged imho. – m4573r Sep 26 '12 at 21:50
@m4573r: OpenOffice-tagged questions are probably about both, and LibreOffice seems to be the more popular. But, well, never mind. – einpoklum Sep 27 '12 at 7:05
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2015-01-31 08:50:57
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http://www.numdam.org/item/SPS_1991__25__354_0/
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An extension of Krein's inverse spectral theorem to strings with nonreflecting left boundaries
Séminaire de probabilités de Strasbourg, Tome 25 (1991) , pp. 354-373.
@article{SPS_1991__25__354_0,
author = {K\"uchler, Uwe and Neumann, Kirsten},
title = {An extension of Krein's inverse spectral theorem to strings with nonreflecting left boundaries},
journal = {S\'eminaire de probabilit\'es de Strasbourg},
pages = {354--373},
publisher = {Springer - Lecture Notes in Mathematics},
volume = {25},
year = {1991},
zbl = {0744.60086},
mrnumber = {1187794},
language = {en},
url = {http://www.numdam.org/item/SPS_1991__25__354_0/}
}
Küchler, Uwe; Neumann, Kirsten. An extension of Krein's inverse spectral theorem to strings with nonreflecting left boundaries. Séminaire de probabilités de Strasbourg, Tome 25 (1991) , pp. 354-373. http://www.numdam.org/item/SPS_1991__25__354_0/
[1] Dym, H.; Mckean, H.P., Gaussian processes, function-theory and the inverse spectral theorem, New York, Academic Press (1976). | Zbl 0327.60029
[2] Ito, K.; Mckean, H.P., Diffusion Processes and their Sample Paths, 2nd Printing, Springer, Berlin (1974). | MR 345224 | Zbl 0285.60063
[3] Kac, I.S.; Krein, M.G., On the spectral functions of the string, Amer. Math. Soc. Transl., (2) 103 (1974), 19-102. | Zbl 0291.34017
[4] Karlin, S.; Mcgregor, J., The differential equations of the birth- and death processes and the Stieltjes moment problem, Trans. Amer. Math. Soc. 85(1957), 489-546. | MR 91566 | Zbl 0091.13801
[5] Knight, F.B., Characterization of the Lévy measures of inverse local times of gap diffusion, Progress in Prob. Statist. 1, Birkhäuser, Boston, Mass. 1981. | MR 647781 | Zbl 0518.60083
[6] Kotani, S.; Watanabe, S., Krein's spectral theory of strings and generalized diffusion processes, Lecture Notes of Mathematics Vol. 923, (1981), 235-259. | MR 661628 | Zbl 0496.60080
[7] Küchler, U., Some Asymptotic Properties of the Transition Densities of One-Dimensional Quasidiffusion, Publ. RIMS, Kyoto-University, 16(1980), 245-268. | MR 574035 | Zbl 0443.60077
[8] Küchler, U., On sojourn times, excursions and spectral measures connected with quasidiffusions, J. Math. Kyoto University, 26(1986), 403-421. | MR 857226 | Zbl 0625.60093
[9] Küchler, U.; Salminen, P., On spectral measures of strings and excursions of quasidiffusions, Lecture Notes of Mathematics Vol. 1372, (1989), 490-502. | Numdam | MR 1022933 | Zbl 0731.60071
[10] Neumann, K., Asymptotische Eigenschaften von Quasidiffusionen und eine Verallgemeinerung des Kreinschen Spektralsatzes, Dissertation A, Humboldt-Universität Berlin, 1989.
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2021-09-27 16:03:12
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https://www.genetics.org/content/172/2/1337?ijkey=bc7fa7501631cdff19882dccae41c281e2e95336&keytype2=tf_ipsecsha
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The Effects of Dominance, Regular Inbreeding and Sampling Design on QST, an Estimator of Population Differentiation for Quantitative Traits | Genetics
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2021-11-27 21:31:24
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https://tvscheduleindia.com/channel/z-etc-bollywood
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# Z ETC Bollywood Today's TV Schedule
39 shows
TV listing of 39 shows for Z ETC Bollywood, today.
Programme Show start Show ends Description
Back To Beats12:00:00 AM12:30:00 AMA compilation of some of the most popular and ente...
Teleshopping12:30:00 AM01:00:00 AMA wide range of products manufactured by popular b...
Teleshopping01:00:00 AM02:00:00 AMA wide range of products manufactured by popular b...
Back To Beats02:00:00 AM06:00:00 AMA compilation of some of the most popular and ente...
Teleshopping06:00:00 AM07:00:00 AMA wide range of products manufactured by popular b...
Song Affair07:00:00 AM07:30:00 AMA comprehensive compilation of some of the most po...
Song Affair07:30:00 AM08:00:00 AMA comprehensive compilation of some of the most po...
Song Affair08:00:00 AM08:30:00 AMA comprehensive compilation of some of the most po...
Song Affair08:30:00 AM09:00:00 AMA comprehensive compilation of some of the most po...
Sunny Songs Up09:00:00 AM09:30:00 AMA comprehensive compilation of catchy songs from t...
Sunny Songs Up09:30:00 AM10:00:00 AMA comprehensive compilation of catchy songs from t...
Sunny Songs Up10:00:00 AM10:30:00 AMA comprehensive compilation of catchy songs from t...
Sunny Songs Up10:30:00 AM11:00:00 AMA comprehensive compilation of catchy songs from t...
Song Affair11:00:00 AM11:30:00 AMA comprehensive compilation of some of the most po...
Song Affair11:30:00 AM12:00:00 PMA comprehensive compilation of some of the most po...
Song Affair12:00:00 PM12:30:00 PMA comprehensive compilation of some of the most po...
Song Affair12:30:00 PM01:00:00 PMA comprehensive compilation of some of the most po...
Bollywood Central01:00:00 PM01:30:00 PMA comprehensive compilation of some of the famous ...
Back To Beats01:30:00 PM02:00:00 PMA compilation of some of the most popular and ente...
Back To Beats02:00:00 PM02:30:00 PMA compilation of some of the most popular and ente...
Back To Beats02:30:00 PM03:00:00 PMA compilation of some of the most popular and ente...
Back To Beats03:00:00 PM03:30:00 PMA compilation of some of the most popular and ente...
Bollywood Central03:30:00 PM04:00:00 PMA comprehensive compilation of some of the famous ...
Song Affair04:00:00 PM04:30:00 PMA comprehensive compilation of some of the most po...
Song Affair04:30:00 PM05:00:00 PMA comprehensive compilation of some of the most po...
Bring It On05:00:00 PM05:30:00 PMA comprehensive compilation of some of the most po...
Bring It On05:30:00 PM06:00:00 PMA comprehensive compilation of some of the most po...
Bring It On06:00:00 PM06:30:00 PMA comprehensive compilation of some of the most po...
Bring It On06:30:00 PM07:00:00 PMA comprehensive compilation of some of the most po...
Bring It On07:00:00 PM07:30:00 PMA comprehensive compilation of some of the most po...
Bring It On07:30:00 PM08:00:00 PMA comprehensive compilation of some of the most po...
Song Affair08:00:00 PM08:30:00 PMA comprehensive compilation of some of the most po...
Song Affair08:30:00 PM09:00:00 PMA comprehensive compilation of some of the most po...
Song Affair09:00:00 PM09:30:00 PMA comprehensive compilation of some of the most po...
Bollywood Central09:30:00 PM10:00:00 PMA comprehensive compilation of some of the famous ...
Song Affair10:00:00 PM10:30:00 PMA comprehensive compilation of some of the most po...
Song Affair10:30:00 PM11:00:00 PMA comprehensive compilation of some of the most po...
Song Affair11:00:00 PM11:30:00 PMA comprehensive compilation of some of the most po...
Song Affair11:30:00 PM12:00:00 AMA comprehensive compilation of some of the most po...
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2019-06-16 05:23:59
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https://zbmath.org/?q=an:0745.62008
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## A note on the large sample properties of linearization, jackknife and balanced repeated replication methods for stratified samples.(English)Zbl 0745.62008
Summary: D. Krewski and J. N. K. Rao [ibid. 9, 1010-1019 (1981; Zbl 0474.62013)] considered inference for a (nonlinear) function of a vector of finite population means $$\theta = g(\bar Y)$$. For a sequence of finite populations with increasing number of strata, they demonstrate that $$\hat \theta = g(\bar y)$$ is asymptotically normal, where $$\bar y$$ is the usual unbiased stratified estimator of $$\bar Y$$. Additionally, they demonstrate that $$(\hat \theta - \theta)/v^{1/2}(\hat \theta )$$ is asymptotically a standard normal distribution, where $$v(\hat \theta)$$ is a variance estimator obtained using linearization, jackknife or balanced repeated replication (BRR) methods.
We extend their results to when the partial first derivatives $$(g_ 1(\mu ),g_ 2(\mu ),\dots ,g_ p(\mu))\equiv 0$$, where $$\mu$$ is the limit of $$\bar Y$$ with increasing number of strata. We explore the asymptotic distribution of $$(\hat \theta - \theta)/v^{1/2}(\hat \theta )$$ and show (1) that it is no longer normal and (2) that it depends upon which variance estimator is used. We describe an application of these results to hypothesis testing using complex survey data.
### MSC:
62D05 Sampling theory, sample surveys 62F12 Asymptotic properties of parametric estimators
Zbl 0474.62013
Full Text:
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2022-05-17 02:30:51
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https://www.vedantu.com/question-answer/question-two-water-taps-together-can-fill-a-tank-class-8-maths-cbse-5ed6ab0d84c7204b9f978cb7
|
Question
# Question- Two water taps together can fill a tank in $9\dfrac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time (in hours) in which a tap of smaller diameter can separately fill the tank.(A) 25(B) $\dfrac{{ - 25}}{4}$ (C) 24(D) $\dfrac{{25}}{4}$
Verified
150.3k+ views
Hint- Here, we will be using the concept of how much part of the tank is filled in one hour.
Let us assume that the tap with smaller diameter can separately fill the tank in $x$ hours.
According to the problem statement it is given that the tap with larger diameter can separately fill the tank in $\left( {x - 10} \right)$ hours.
Also, given that both the taps can together fill the tank in $9\dfrac{3}{8} = \dfrac{{\left( {8 \times 9} \right) + 3}}{8} = \dfrac{{75}}{8}$ hours.
So, the tap with a smaller diameter can fill $\dfrac{1}{x}$ part of the tank in 1 hour. Similarly, the tap with a larger diameter can fill $\dfrac{1}{{\left( {x - 10} \right)}}$ part of the tank in 1 hour. Also, both the taps fill $\dfrac{8}{{75}}$ part of the tank in 1 hour.
Then, $\dfrac{1}{x} + \dfrac{1}{{\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow \dfrac{{x - 10 + x}}{{x\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow \dfrac{{2x - 10}}{{x\left( {x - 10} \right)}} = \dfrac{8}{{75}} \Rightarrow 75\left( {2x - 10} \right) = 8x\left( {x - 10} \right) \\ \Rightarrow 150x - 750 = 8{x^2} - 80x \Rightarrow 8{x^2} - 230x + 750 = 0 \Rightarrow 4{x^2} - 115x + 375 = 0 \\ \Rightarrow 4{x^2} - 100x - 15x + 375 = 0 \Rightarrow 4x\left( {x - 25} \right) - 15\left( {x - 25} \right) = 0 \Rightarrow \left( {x - 25} \right)\left( {4x - 15} \right) = 0 \\$
Either $x = 25$ or $4x = 15 \Rightarrow x = \dfrac{{15}}{4}$
When $x = 25$, $\left( {x - 10} \right) = 25 - 10 = 15$ and when $x = \dfrac{{15}}{4}$, $\left( {x - 10} \right) = \dfrac{{15}}{4} - 10 = \dfrac{{15 - 40}}{4} = - \dfrac{{25}}{4}$
Since, the time can never be negative so $x = \dfrac{{15}}{4}$ is rejected.
Hence, $x = 25$ and $\left( {x - 10} \right) = 15$
The tap with smaller diameter can separately fill the tank in 25 hours and the tap with larger diameter can separately fill the tank in 15 hours.
Therefore, option A is correct.
Note- In these types of problems, the concept of part of the work done (here it is filling the tank) in one hour is utilized to obtain an equation in one variable so that we can solve for it.
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2021-10-25 17:50:47
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http://www.mscroggs.co.uk/blog/tags/Chalkdust
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mscroggs.co.uk
mscroggs.co.uk
subscribe
# Blog
## Archive
Show me a Random Blog Post
▼ show ▼
2017-01-13 04:32:41
## Is MEDUSA the New BODMAS?
I wrote this post with, and after much discussion with Adam Townsend. It also appeared on the Chalkdust Magazine blog.
Recently, Colin "IceCol" Beveridge blogged about something that's been irking him for a while: those annoying social media posts that tell you to work out a sum, such as $$3-3\times6+2$$, and state that only $n$% of people will get it right (where $$n$$ is quite small). Or as he calls it "fake maths".
A classic example of "fake maths".
This got me thinking about everyone's least favourite primary school acronym: BODMAS (sometimes known as BIDMAS, or PEMDAS if you're American). As I'm sure you've been trying to forget, BODMAS stands for "Brackets, (to the power) Of, Division, Multiplication, Addition, Subtraction" and tells you in which order the operations should be performed.
Now, I agree that we all need to do operations in the same order (just imagine trying to explain your working out to someone who uses BADSOM!) but BODMAS isn't the order mathematicians use. It's simply wrong. Take the sum $$4-3+1$$ as an example. Anyone can tell you that the answer is 2. But BODMAS begs to differ: addition comes first, giving 0!
The problem here is that in reality, we treat addition and subtraction as equally important, so sums involving just these two operations are calculated from left-to-right. This caveat is quite a lot more to remember on top of BODMAS, but there's actually no need: Doing all the subtractions before additions will always give you the same answer as going from left-to-right. The same applies to division and multiplication, but luckily these two are in the correct order already in BODMAS (but no luck if you're using PEMDAS).
So instead of BODMAS, we should be using BODMSA. But that's unpronounceable, so instead we suggest that from now on you use MEDUSA. That's right, MEDUSA:
• Mabano (brackets in Swahili)
• Exponentiation
• Division
• Subtraction
This is big news. MEDUSA vs BODMAS could be this year's pi vs tau... Although it's not actually the biggest issue when considering sums like $$3-3\times6+2$$.
The real problem with $$3-3\times6+2$$ is that it is written in a purposefully confusing and ambiguous order. Compare the following sums:
$$3-3\times6+2$$ $$3+2-3\times6$$ $$3+2-(3\times6)$$
In the latter two, it is much harder to make a mistake in the order of operations, because the correct order is much closer to normal left-to-right reading order, helping the reader to avoid common mistakes. Good mathematics is about good communication, not tricking people. This is why questions like this are "fake maths": real mathematicians would never ask them. If we take the time to write clearly, then I bet more than $$n$$% of people will be able get the correct answer.
### Similar Posts
Comments in green were written by me. Comments in blue were not written by me.
To prove you are not a spam bot, please type "apple" in the box below (case sensitive):
2016-12-20 09:21:56
## Christmas Card 2016
Last week, I posted about the Christmas card I designed on the Chalkdust blog.
The card looks boring at first glance, but contains 12 puzzles. Converting the answers to base 3, writing them in the boxes on the front, then colouring the 1s green and 2s red will reveal a Christmassy picture.
If you want to try the card yourself, you can download this pdf. Alternatively, you can find the puzzles below and type the answers in the boxes. The answers will be automatically converted to base 3 and coloured...
# Answer (base 10) Answer (base 3) 1 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 4 0 0 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 0 0 7 0 0 0 0 0 0 0 0 0 8 0 0 0 0 0 0 0 0 0 9 0 0 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0 0 0 11 0 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 0
1. The square number larger than 1 whose square root is equal to the sum of its digits.
2. The smallest square number whose factors add up to a different square number.
3. The largest number that cannot be written in the form $$23n+17m$$, where $$n$$ and $$m$$ are positive integers (or 0).
4. Write down a three-digit number whose digits are decreasing. Write down the reverse of this number and find the difference. Add this difference to its reverse. What is the result?
5. The number of numbers between 0 and 10,000,000 that do not contain the digits 0, 1, 2, 3, 4, 5 or 6.
6. The lowest common multiple of 57 and 249.
7. The sum of all the odd numbers between 0 and 66.
8. One less than four times the 40th triangle number.
9. The number of factors of the number $$2^{756}$$×$$3^{12}$$.
10. In a book with 13,204 pages, what do the page numbers of the middle two pages add up to?
11. The number of off-diagonal elements in a 27×27 matrix.
12. The largest number, $$k$$, such that $$27k/(27+k)$$ is an integer.
### Similar Posts
Comments in green were written by me. Comments in blue were not written by me.
2016-12-20
Thank you for the prompt response! It makes sense now and perhaps I should have read a little closer!
Dan Whitman
2016-12-20
Find the difference between the original number and the reverse of the original. Call this difference $$a$$. Next add $$a$$ to the reverse of $$a$$...
Matthew
2016-12-20
In number 4 what are we to take the difference between? Do you mean the difference between the original number and its reverse? If so when you add the difference back to the reverse you simply get the original number, which is ambiguous. I am not sure what you are asking us to do here.
Dan Whitman
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2017-02-26 21:13:40
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https://studyadda.com/solved-papers/jamia-millia-islamia/jamia-millia-islamia-solved-paper-2005/287
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# Solved papers for JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2005
### done Jamia Millia Islamia Solved Paper-2005
• question_answer1) The inductance between A and D is:
A) 3.66 H
B) 9 H
C) 0.66 H
D) 1 H
• question_answer2) A ball whose kinetic energy is E, is projected at an angle of$45{}^\circ$to the horizontal. The kinetic energy of the ball at the highest point of its flight will be:
A) E
B) $E/\sqrt{2}$
C) E/2
D) zero
• question_answer3) From a building two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically). If${{v}_{A}}$and${{v}_{B}}$are their respective velocities on reaching the ground, then
A) ${{v}_{B}}>{{v}_{A}}$
B) ${{v}_{A}}={{v}_{B}}$
C) ${{v}_{A}}>{{v}_{B}}$
D) their velocities depends on their masses
• question_answer4) If a body loses half of its velocity on penetrating 3cm in a wooden block, then how much will it penetrate more before coming to rest?
A) 1 cm
B) 2 cm
C) 3 cm
D) 4 cm
• question_answer5) If suddenly the gravitational force of attraction between earth and a satellite revolving around it becomes zero, then the satellite will:
A) continue to move in its orbit with same velocity
B) move tangentially to the original orbit with the same velocity
C) become stationary in its orbit
D) move towards the earth
• question_answer6) If an ammeter is to be used in place of a voltmeter, then we must connect with the ammeter a:
A) low resistance in parallel
B) high resistance in parallel
C) high resistance in series
D) low resistance in series
• question_answer7) If in a circular coil A of radius R, current$i$is flowing and in another coil B of radius 2R a current$2i$is flowing, then the ratio of the magnetic fields,${{B}_{A}}$and${{B}_{B}}$produced by them will be:
A) 1
B) 2
C) 1/2
D) 4
• question_answer8) If two mirrors are kept at$60{}^\circ$to each other, then the number of images formed by them is:
A) 5
B) 6
C) 7
D) 8
• question_answer9) A wire when connected to 220V mains supply has power dissipation${{P}_{1}}$. Now the wire is cut into two equal pieces which are connected in parallel to the same supply. Power dissipation in this case is${{P}_{2}}$. Then${{P}_{2}}:{{P}_{1}}$is
A) 1
B) 4
C) 2
D) 3
• question_answer10) If 13.6eV energy is required to ionize the hydrogen atom, then the energy required to remove an electron from$n=2$is:
A) 10.2 eV
B) 0 eV
C) 3.4 eV
D) 6.8 eV
• question_answer11) Tube A has both ends open while tube B has one end closed, otherwise they are identical. The ratio of fundamental frequency of tubes A and B is:
A) $1:2$
B) $1:4$
C) $2:1$
D) $4:1$
• question_answer12) A tuning fork arrangement (pair) produces 4 beats/sec with one fork of frequency 288 .cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The frequency of the unknown folk is:
A) 286 cps
B) 292 cps
C) 294 cps
D) 288 cps
• question_answer13) A wave$y=a\sin (\omega t-kx)$on a string meets with another wave producing a node at$x=0$. Then the equation of the unknown wave is:
A) $y=a\sin (\omega t+kx)$
B) $y=-a\sin (\omega t+kx)$
C) $y=a\sin (\omega t-kx)$
D) $y=-a\sin (\omega t-kx)$
• question_answer14) On moving a charge of 20 coulombs by 2 cm, 2 J of work is done, then the potential difference between the points is:
A) 0.1 V
B) 8 V
C) 2 V
D) 0.5 V
• question_answer15) If an electron and a proton having same momenta enter perpendicularly to a magnetic field, then:
A) curved path of electron and proton will be same (ignoring the sense of revolution)
B) they will move under flecked
C) curved path of electron is more curved than that of proton
D) path of proton is more curved
• question_answer16) Energy required to move a body of mass m from an orbit of radius 2R to 3R is:
A) $GMm/12{{R}^{2}}$
B) $GMm/3{{R}^{2}}$
C) $GMm/8R$
D) $GMm/6R$
• question_answer17) If a spring has time period T, and is cut into n equal parts, then the time period of each part will be:
A) $T\sqrt{n}$
B) $T/\sqrt{n}$
C) $nT$
D) $T$
• question_answer18) A charged particle q is placed at the centre$O$ of cube of length L (ABCDEFGH). Another same charge q is placed at a distance L from Then the electrons flux through ABCD is:
A) $q/4\pi {{\varepsilon }_{0}}L$
B) zero
C) $q/2\pi {{\varepsilon }_{0}}L$
D) $q/3\pi {{\varepsilon }_{0}}L$
• question_answer19) If in the circuit, power dissipation is 150 W, then R is:
A) $2\,\Omega$
B) $6\,\Omega$
C) $5\,\Omega$
D) $4\,\Omega$
• question_answer20) Wavelength of light used in an optical instrument are${{\lambda }_{1}}=4000\,\overset{o}{\mathop{\text{A}}}\,$and${{\lambda }_{2}}=5000\,\overset{o}{\mathop{\text{A}}}\,,$then ratio of their respective resolving powers (corresponding to${{\lambda }_{1}}$and${{\lambda }_{2}}$) is:
A) $16:25$
B) $9:1$
C) $4:5$
D) $5:4$
• question_answer21) Two identical particles move towards each other with velocity 2v and v respectively. The velocity of centre of mass is
A) v
B) v/3
C) v/2
D) zero
• question_answer22) If a current is passed through a spring then the spring will:
A) expand
B) compress
C) remain same
D) none of these
• question_answer23) Heat given to a body which raises its temperature by$1{}^\circ C$is:
A) water equivalent
B) thermal capacity
C) specific heat
• question_answer24) At absolute zero, Si acts as:
A) non-metal
B) metal
C) insulator
D) none of these
• question_answer25) Electromagnetic waves are transverse in nature is evident by:
A) polarization
B) interference
C) reflection
D) diffraction
• question_answer26) Which of the following is used in optical fibres?
A) Total internal reflection
B) Scattering
C) Diffraction
D) Refraction
• question_answer27) The escape velocity of a body depends upon mass as:
A) ${{m}^{0}}$
B) ${{m}^{1}}$
C) ${{m}^{2}}$
D) ${{m}^{3}}$
• question_answer28) Which of the following are not electro- magnetic waves?
A) Cosmic-rays
B) y rays
C) $\beta -rays$
D) X-rays
• question_answer29) Identify the pair whose dimensions are equal:
A) torque and work
B) stress and energy
C) force and stress
D) force and work
• question_answer30) If${{\theta }_{i}}$ is the inversion temperature,${{\theta }_{n}}$is the neutral temperature,${{\theta }_{c}}$. is the temperature of the cold junction then:
A) ${{\theta }_{i}}+{{\theta }_{c}}={{\theta }_{n}}$
B) ${{\theta }_{i}}-{{\theta }_{c}}=2{{\theta }_{n}}$
C) $\frac{{{\theta }_{i}}-{{\theta }_{c}}}{2}={{\theta }_{n}}$
D) ${{\theta }_{c}}-{{\theta }_{i}}=2{{\theta }_{n}}$
A) spectrometer
B) pyrometer
C) nanometer
D) photometer
• question_answer32) If${{N}_{0}}$is the original mass of the substance of half-life period${{t}_{1/2}}=5$years, then the amount of substance left after 15 years is:
A) ${{N}_{0}}/8$
B) ${{N}_{0}}/16$
C) ${{N}_{0}}/2$
D) ${{N}_{0}}/4$
• question_answer33) By increasing the temperature, the specific resistance of a conductor and a semiconductor:
A) increases for both
B) decreases for both
C) increases, decreases respectively
D) decreases, increases respectively
• question_answer34) If there are n capacitors in parallel connected to V volt source, then the energy stored is equal to:
A) $CV$
B) $\frac{1}{2}nC{{V}^{2}}$
C) $C{{V}^{2}}$
D) $\frac{1}{2n}C{{V}^{2}}$
• question_answer35) Which of the following is more close to a black body?
A) Black board paint
B) Green leaves
C) Black holes
D) Red roses
• question_answer36) Which statement is incorrect?
A) All reversible cycles have same efficiency
B) Reversible cycle has more efficiency than an irreversible one
C) Carnot cycle is a reversible one
D) Carnot cycle has the maximum efficiency in all cycles
• question_answer37) Length of a string tied to two rigid supports is 40cm. Maximum length (wavelength in cm) of a stationary wave produced on it, is:
A) 20
B) 80
C) 40
D) 120
• question_answer38) The power factor of an A.C. circuit having resistance R and inductance L (connected in series) and an angular velocity$\omega$is:
A) $R/\omega L$
B) $R{{({{R}^{2}}+{{\omega }^{2}}{{L}^{2}})}^{1/2}}$
C) $\omega L/R$
D) $R/{{({{R}^{2}}-{{\omega }^{2}}{{L}^{2}})}^{1/2}}$
• question_answer39) An astronomical telescope has a large aperture to:
A) reduce spherical aberration
B) have high resolution
C) increase span of observation
D) have low dispersion
• question_answer40) The kinetic energy needed to project a body of mass m from the earths surface (radius R) to infinity is:
A) $mgR/2$
B) $2mgR$
C) $mgR$
D) $mgR/4$
• question_answer41) Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will:
A) increase
B) decrease
C) remain same
D) decrease for some, while increase for others
• question_answer42) When temperature increases, the frequency of a tuning fork:
A) increases
B) decreases
C) remains same
D) increases or decreases depending on the material
• question_answer43) If mass-energy equivalence is taken into account, when water is cooled to form ice, the mass of water should:
A) increase
B) remain unchanged
C) decrease
D) first increase then decrease
• question_answer44) The energy band gap is maximum in:
A) metals
B) superconductors
C) insulators
D) semiconductors
• question_answer45) The part of a transistor which is most heavily doped to produce large number of majority carriers is:
A) emitter
B) base
C) collector
D) can be any of the above three
• question_answer46) In a simple harmonic oscillator, at the mean position:
A) kinetic energy is minimum, potential energy is maximum
B) both kinetic and potential energies are maximum
C) kinetic energy is maximum, potential energy is minimum
D) both kinetic and potential energies are minimum
• question_answer47) Initial angular velocity of a circular disc of mass M is${{\omega }_{1}}$. Then two small spheres of mass m are attached gently to two diametrically opposite points on the edge of the disc. What is the final angular velocity of the disc?
A) $\left( \frac{M+m}{M} \right){{\omega }_{1}}$
B) $\left( \frac{M+m}{m} \right){{\omega }_{1}}$
C) $\left( \frac{M}{M+4m} \right){{\omega }_{1}}$
D) $\left( \frac{M}{M+2m} \right){{\omega }_{1}}$
• question_answer48) The minimum velocity (in$m{{s}^{-1}}$) with which a car driver must traverse a flat curve of radius 150m and coefficient of friction 0.6 to avoid skidding is:
A) 60
B) 30
C) 15
D) 25
• question_answer49) A cylinder of height 20 m is completely filled with water. The velocity of efflux of water (in $m{{s}^{-1}}$) through a small hole on the side wall of the cylinder near its bottom, is:
A) 10
B) 20
C) 25.5
D) 5
• question_answer50) A spring of force constant 800 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is:
A) 16 J
B) 8 J
C) 32 J
D) 24 J
• question_answer51) A child swinging on a swing in sitting position, stands up, then the time period of the swing will:
A) increase
B) decrease
C) remain same
D) increase if the child is long and decrease if the child is short
• question_answer52) A lift is moving down with acceleration a. Aman in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively:
A) $g,g$
B) $g-a.\text{ }g-a$
C) $g-a,g$
D) $a,g$
• question_answer53) The mass of a product liberated on anode in an electrochemical cell depends on:
A) ${{(It)}^{1/2}}$
B) $It$
C) $I/t$
D) ${{I}^{2}}t$
• question_answer54) At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at$47{}^\circ C$?
A) 80 K
B) $-73\text{ }K$
C) 3 K
D) 20 K
• question_answer55) The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its:
A) speed
B) mass
C) charge
D) magnetic induction
• question_answer56) Which of the following is a redox reaction?
A) $NaCl+KN{{O}_{3}}\xrightarrow[{}]{{}}NaN{{O}_{3}}+KCl$
B) $Ca{{C}_{2}}{{O}_{4}}+2HCl\xrightarrow[{}]{{}}CaC{{l}_{2}}+{{H}_{2}}{{C}_{2}}{{O}_{4}}$
C) $Ca{{(OH)}_{2}}+2N{{H}_{4}}Cl\xrightarrow[{}]{{}}$$CaC{{l}_{2}}+2N{{H}_{3}}+2{{H}_{2}}O$
D) $2K[Ag{{(CN)}_{2}}]+Zn\xrightarrow[{}]{{}}$ $2Ag+{{K}_{2}}[Zn{{(CN)}_{4}}]$
• question_answer57) For an ideal gas, number of mol per litre in terms of its pressure P, temperature T and gas constant R is:
A) PT/R
B) PRT
C) P/RT
D) RT/P
• question_answer58) Number of$P-O$bonds in${{P}_{4}}{{O}_{10}}$is:
A) 17
B) 16
C) 15
D) 6
• question_answer59) $K{{O}_{2}}$is used in space and submarines because it:
A) absorbs$C{{O}_{2}}$and increases${{O}_{2}}$concentration
B) absorbs moisture
C) absorbs$C{{O}_{2}}$
D) produces ozone
• question_answer60) Which of the following ions has the maximum magnetic moment?
A) $M{{n}^{2+}}$
B) $F{{e}^{2+}}$
C) $T{{i}^{2+}}$
D) $C{{r}^{2+}}$
• question_answer61) Acetylene does not react with:
A) Na
B) ammonical$AgN{{O}_{3}}$
C) $HCl$
D) $NaOH$
• question_answer62) Compound A given below is:
A) antiseptic
B) antibiotic
C) analgesic
D) pesticide
• question_answer63) For the following cell with hydrogen electrodes at two different pressures${{p}_{1}}$and${{p}_{2}}$$\underset{{{p}_{1}}}{\mathop{pt({{H}_{2}})}}\,|\underset{1\,M}{\mathop{{{H}^{+}}(aq).}}\,|\underset{{{p}_{2}}}{\mathop{Pt({{H}_{2}})}}\,$emf is given by:
A) $\frac{RT}{F}{{\log }_{e}}\frac{{{p}_{1}}}{{{p}_{2}}}$
B) $\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{1}}}{{{p}_{2}}}$
C) $\frac{RT}{F}{{\log }_{e}}\frac{{{p}_{2}}}{{{p}_{1}}}$
D) $\frac{RT}{2F}{{\log }_{e}}\frac{{{p}_{2}}}{{{p}_{1}}}$
• question_answer64) Acetylene reacts with hypochlorous acid to form:
A) $C{{l}_{2}}CHCHO$
B) $ClC{{H}_{2}}COOH$
C) $C{{H}_{3}}COCl$
D) $ClC{{H}_{2}}CHO$
• question_answer65) On heating benzyl amine with chloroform and ethanolic KOH, product obtained is:
A) benzyl alcohol
B) benzaldehyde
C) benzonitrile
D) benzyl isocyanide
• question_answer66) Which of the following reaction is possible at anode?
A) ${{F}_{2}}+2{{e}^{-}}\xrightarrow[{}]{{}}2{{F}^{-}}$
B) $2{{H}^{+}}+\frac{1}{2}{{O}_{2}}+2{{e}^{-}}\xrightarrow[{}]{{}}{{H}_{2}}O$
C) $2Cr_{2}^{3+}+7{{H}_{2}}O\xrightarrow[{}]{{}}C{{r}_{2}}O_{7}^{2-}+14{{H}^{+}}+6{{e}^{-}}$
D) $F{{e}^{2+}}\xrightarrow[{}]{{}}F{{e}^{3+}}+{{e}^{-}}$
• question_answer67) Which of the following concentration factor is affected by change in temperature?
A) Molarity
B) Molality
C) Mol fraction
D) Weight fraction
• question_answer68) Cyanide process is used for the extraction of:
A) barium
B) silver
C) boron
D) zinc
• question_answer69) Following reaction${{(C{{H}_{3}})}_{3}}CBr+{{H}_{2}}O\xrightarrow[{}]{{}}{{(C{{H}_{3}})}_{3}}COH+HBr$is an example of:
A) elimination reaction
C) nucleophilic substitution
D) electrophilic substitution
• question_answer70) A metal M forms water soluble$MS{{O}_{4}}$ and inert $MO.\text{ }MO$in aqueous solution forms insoluble $M{{(OH)}_{2}}$soluble in$NaOH$. Metal M is:
A) Be
B) Mg
C) Ca
D) Si
• question_answer71) Half life of a substance A following first order kinetics is 5 days. Starting with 100 g of A, amount left after 15 days is:
A) 25 g
B) 50 g
C) 12.5 g
D) 6.25 g
• question_answer72) The most stable ion is:
A) ${{[Fe{{(OH)}_{5}}]}^{3-}}$
B) ${{[FeC{{l}_{6}}]}^{3-}}$
C) ${{[Fe{{(CN)}_{6}}]}^{3-}}$
D) ${{[Fe{{({{H}_{2}}O)}_{6}}]}^{3+}}$
• question_answer73) A substance forms zwitter ion. It can have functional groups:
A) $-N{{H}_{2}},-COOH$
B) $-N{{H}_{2}},-S{{O}_{3}}H$
C) both (a) and (b)
D) none of these
• question_answer74) If$F{{e}^{3+}}$and$C{{r}^{3+}}$both are present in group III of qualitative analysis, then distinction can be made by:
A) addition of$N{{H}_{4}}OH$in presence of$N{{H}_{4}}Cl$when only$Fe{{(OH)}_{3}}$is precipitated
B) addition of$N{{H}_{4}}OH$in presence of$N{{H}_{4}}Cl$when$Cr{{(OH)}_{3}}$and$Fe{{(OH)}_{3}}$both are precipitated and on adding$B{{r}_{2}}$water and $NaOH,Cr{{(OH)}_{3}}$dissolves
C) precipitate of$Cr{{(OH)}_{3}}$and$Fe{{(OH)}_{3}}$as obtained in are treated! with cone.$HCl$ when only$Fe{{(OH)}_{3}}$dissolves
D) and (b) the (c) correct.
• question_answer75) In an organic compound of molar mass 108 g $mo{{l}^{-1}}C,H$ and N atoms are present in $9:1:3.5$by weight. Molecular formula can be:
A) ${{C}_{6}}{{H}_{8}}{{N}_{2}}$
B) ${{C}_{7}}{{H}_{10}}N$
C) ${{C}_{5}}{{H}_{6}}{{N}_{3}}$
D) ${{C}_{4}}{{H}_{18}}{{N}_{3}}$
• question_answer76) Solubility of$Ca{{(OH)}_{2}}$is$S\text{ }mol\text{ }litr{{e}^{-1}}$. The solubility product$({{K}_{sp}})$under the same condition is:
A) $4{{S}^{3}}$
B) $3{{S}^{4}}$
C) $4{{S}^{2}}$
D) ${{S}^{3}}$
• question_answer77) Heat required to raise the temperature of 1 mol of a substance by$1{}^\circ$is called:
A) specific heat
B) molar heat capacity
C) water equivalent
D) specific gravity
• question_answer78) $\beta -$particle is emitted in a radioactive reaction when:
A) a proton changes to neutron
B) a neutron changes to proton
C) a neutron changes to electron
D) an electron changes to neutron
• question_answer79) In a mixture of A and B, components show negative deviation when:
A) $A-B$ interaction is stronger than$A-A$ and$B-B$interaction
B) $A-B$interaction is weaker than$A-A$and $B-B$interaction
C) $\Delta {{V}_{mix}}>0,\Delta {{S}_{mix}}>0$
D) $\Delta {{V}_{mix}}=0,\Delta {{S}_{mix}}>0$
• question_answer80) Refining of impure copper with zinc impurity is to be done by electrolysis using electrons as: Cathode Anode
A) pure copper pure zinc
B) pure zinc pure copper
C) pure copper impure copper
D) pure zinc impure zinc
• question_answer81) Aluminium is extracted by the electrolysis of:
A) alumina
B) bauxite
C) molten cryolite
D) alumina mixed with molten cryolite
• question_answer82) For an aqueous solution, freezing point is$-0.186{}^\circ C$. Elevation of the boiling point of the same solution is (${{K}_{f}}=1.86{}^\circ mo{{l}^{-1}}kg$and ${{K}_{f}}=0.512{}^\circ mo{{l}^{-1}}kg$):
A) $0.186{}^\circ$
B) $0.0512{}^\circ$
C) $1.86{}^\circ$
D) $5.12{}^\circ$
• question_answer83) Underlined carbon is$s{{p}^{3}}$hybridised in:
A) $C{{H}_{3}}\underline{C}H=C{{H}_{2}}$
B) $C{{H}_{3}}\underline{C}{{H}_{2}}N{{H}_{2}}$
C) $C{{H}_{3}}\underline{C}ON{{H}_{2}}$
D) $C{{H}_{3}}C{{H}_{2}}\underline{C}N$
• question_answer84) Bond angle of$109{}^\circ 28$is found in:
A) $N{{H}_{3}}$
B) ${{H}_{2}}O$
C) $C{{H}_{5}}$
D) $N{{H}_{4}}$
• question_answer85) For a reaction$A+2B\xrightarrow[{}]{{}}C,$rate is given by $+\frac{d[C]}{dt}=k[A][B],$hence the order of the reaction is:
A) 3
B) 2
C) 1
D) 0
• question_answer86) $C{{H}_{3}}MgI$is an organometallic compound due to:
A) $Mg-I$bond
B) $C-I$bond
C) $C-Mg$bond
D) $C-H$bond
• question_answer87) One of the following species acts as both Bronsted acid and base:
A) ${{H}_{2}}PO_{2}^{-}$
B) $HPO_{3}^{2-}$
C) $HPO_{4}^{2-}$
D) all of the above
• question_answer88) Hybridisation of the underline atom changes in:
A) $\underline{A}l{{H}_{3}}$changes to$AlH_{4}^{-}$
B) ${{H}_{2}}\underline{O}$changes to${{H}_{3}}{{O}^{+}}$
C) $\underline{N}{{H}_{3}}$changes to$NH_{4}^{+}$
D) in all cases
• question_answer89) Racemic mixture is formed by mixing two:
A) isomeric compounds
B) chiral compounds
C) meso compounds
D) enantiomers with chiral carbon
• question_answer90) The number of lone pairs on$Xe$in$Xe{{F}_{2}},$$Xe{{F}_{4}}$ and$Xe{{F}_{6}}$respectively are:
A) 3, 2, 1
B) 2, 4, 6
C) 1, 2, 3
D) 6, 4, 2
• question_answer91) An aqueous solution of$1\,M\,NaCl$and$1M\text{ }HCl$is:
A) not a buffer but$pH<7$
B) not a buffer but$pH>7$
C) a buffer with$pH<7$
D) a buffer with$pH>7$
• question_answer92) Consider following two reactions$A\xrightarrow[{}]{{}}Product-\frac{d[A]}{dt}={{k}_{1}}{{[A]}^{0}}$$B\xrightarrow[{}]{{}}Product-\frac{d[B]}{dt}={{k}_{2}}[B]$]${{k}_{1}}$and${{k}_{2}}$ are expressed in terms of molarity$(mol\text{ }{{L}^{-1}})$and time$(se{{c}^{-1}})$as:
A) $se{{c}^{-1}},M\,{{\sec }^{-1}}{{L}^{-1}}$
B) $M\,se{{c}^{-1}},M\,{{\sec }^{-1}}$
C) $se{{c}^{-1}},{{M}^{-1}}\,{{\sec }^{-1}}$
D) $M\,se{{c}^{-1}},{{L}^{-1}}\,{{\sec }^{-1}}$
A) ribose sugar and thymine
B) ribose sugar and uracil
C) deoxyribose sugar and uracil
D) deoxyribose sugar and thymine
• question_answer94) For a cell given below: $\underset{-}{\mathop{Ag|A{{g}^{+}}}}\,||\underset{+}{\mathop{C{{u}^{2+}}|Cu}}\,$$A{{g}^{+}}+{{e}^{-}}\xrightarrow[{}]{{}}Ag$ $E{}^\circ =x$$C{{u}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Cu,$ $E{}^\circ =y$E? cell is:
A) $x+2y$
B) $2x+y$
C) $y-x$
D) $y-2x$
• question_answer95) Based on kinetic theory of gases following laws can be proved:
A) Boyles law
B) Charles law
D) all of these
• question_answer96) $Mn{{O}_{4}}$is a good oxidising agent in different medium changing to $MnO_{4}^{-}\xrightarrow[{}]{{}}M{{n}^{2+}}$ $\xrightarrow[{}]{{}}MnO_{4}^{2-}$ $\xrightarrow[{}]{{}}Mn{{O}_{2}}$ $\xrightarrow[{}]{{}}M{{n}_{2}}{{O}_{3}}$ Changes in oxidation number respectively are:
A) 1,3,4,5
B) 5,4,3,2
C) 5,1,3,4
D) 2,6,4,3
• question_answer97) For the reaction:${{H}_{2}}+{{I}_{2}}\xrightarrow[{}]{{}}2HI,$the differential rate law is:
A) $-\frac{d[{{H}_{2}}]}{dt}=-\frac{d[{{I}_{2}}]}{dt}=2\frac{d[HI[}{dt}$
B) $-2\frac{d[{{H}_{2}}]}{dt}=-2\frac{d[{{I}_{2}}]}{dt}=\frac{d[HI]}{dt}$
C) $-\frac{d[{{H}_{2}}]}{dt}=-\frac{d[{{I}_{2}}]}{dt}=\frac{d[HI]}{dt}$
D) $-\frac{d[{{H}_{2}}]}{2dt}=-\frac{d[{{I}_{2}}]}{2dt}=\frac{d[HI]}{dt}$
• question_answer98) Number of atoms in 560 g of Fe (atomic mass$56\text{ }g\text{ }mo{{l}^{-1}}$) is:
A) is twice that of 70 g N
B) is half that of 20 g H
C) both are correct
D) none is correct
• question_answer99) Geometrical isomerism is not shown by:
A) 1, 1-dichloro-l-pentene
B) 1, 2-dichloro-l-pentene
C) 1, 3-dichloro-2-pentene
D) 1, 4-dichloro-2-pentene
• question_answer100) Number of atoms in the unit cell of Na (BCC type crystal) and Mg (FCC type crystal) are respectively:
A) 4, 4
B) 4, 2
C) 2, 4
D) 1, 1
• question_answer101) Which of the following compounds has incorrect IUPAC nomenclature?
A) $\underset{ethylbutanoate}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}\overset{\begin{smallmatrix} O \\ |\,| \end{smallmatrix}}{\mathop{C}}\,O{{C}_{2}}{{H}_{5}}}}\,$
B) $\underset{3-methyl\text{ }butanal}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,C{{H}_{2}}CHO}}\,$
C) $\underset{2-methyl-3-pentanone}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH\overset{\begin{smallmatrix} O \\ |\,| \end{smallmatrix}}{\mathop{C}}\,}}\,C{{H}_{2}}C{{H}_{3}}}}\,$
D) $\underset{2-methyl-3-butanol}{\mathop{C{{H}_{3}}\underset{\begin{smallmatrix} | \\ {{H}_{3}}C \end{smallmatrix}}{\mathop{C}}\,H\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,C{{H}_{3}}}}\,$
• question_answer102) End product of the following reaction is:$C{{H}_{3}}C{{H}_{2}}COOH\xrightarrow[red\,p]{C{{l}_{2}}}$ $\xrightarrow[{}]{alcoholic\text{ }KOH}$
A) $C{{H}_{3}}\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,HCOOH$
B) $\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,C{{H}_{2}}COOH$
C) $C{{H}_{2}}=CHCOOH$
D) $\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,COOH$
• question_answer103) For the following reaction in gaseous phase$CO+\frac{1}{2}{{O}_{2}}\xrightarrow[{}]{{}}C{{O}_{2}}$${{K}_{c}}/{{K}_{p}}$is:
A) ${{(RT)}^{1/2}}$
B) ${{(RT)}^{-1/2}}$
C) $(RT)$
D) ${{(RT)}^{-1}}$
• question_answer104) Energy of H-atom in the ground state is$-13.6$ eV, hence energy in the second excited state is:
A) $-6.8eV$
B) $-3.4eV$
C) $-1.51eV$
D) $-4.53eV$
• question_answer105) A square planar complex is formed by hybridisation of the following atomic orbitals:
A) $s,{{p}_{x}},{{p}_{y}},{{p}_{z}}$
B) $s,\text{ }{{p}_{x}}\text{ }{{p}_{y}},\text{ }{{p}_{z}}\text{, }d$
C) $d,\text{ s, }\,\text{ }{{p}_{x}},\text{ }{{p}_{y}}$
D) $\text{s, }{{p}_{x}},\text{ }{{p}_{y}},{{p}_{z}}\,d,\,d$
• question_answer106) Type of isomerism shown by$[Cr{{(N{{H}_{3}})}_{5}}N{{O}_{2}}]C{{l}_{2}}$is:
A) optical
B) ionization
C) geometrical
• question_answer107) One of the following equilibria is not affected by change in volume of the flask:
A) $PC{{l}_{5}}(g)PC{{l}_{3}}(g)+C{{l}_{2}}(g)$
B) ${{N}_{2}}(g)+3{{H}_{2}}(g)2N{{H}_{3}}(g)$
C) ${{N}_{2}}(g)+{{O}_{2}}(g)2NO(g)$
D) $S{{O}_{2}}C{{l}_{2}}(g)S{{O}_{2}}(g)+C{{l}_{2}}(g)$
• question_answer108) Uncertainty in position of a particle of 25 g in space is${{10}^{-5}}m$. Hence uncertainty in velocity $(m{{s}^{-1}})$is (Plancks constant$h=6.6\times {{10}^{-34}}Js$):
A) $2.1\times {{10}^{-28}}$
B) $2.1\times {{10}^{-34}}$
C) $0.5\times {{10}^{-34}}$
D) $5.0\times {{10}^{-24}}$
• question_answer109) Consider the following reactions at$1100{}^\circ C$ (I) $2C+{{O}_{2}}\xrightarrow[{}]{{}}2CO,$$\Delta G{}^\circ =-460kJmo{{l}^{-1}}$ (II) $2Zn+{{O}_{2}}\xrightarrow[{}]{{}}2ZnO,$$\Delta G{}^\circ =-360kJ\,mo{{l}^{-1}}$ Based on these, select correct alternate:
A) zinc can be oxidised by$CO$
B) zinc oxide can be reduced by carbon
C) both are correct
D) none is correct
• question_answer110) A reaction is non-spontaneous at the freezing point of water but is spontaneous at the boiling point of water then: $\Delta H$ $\Delta S$
A) $+ve~~~~~~~~~~~+ve$
B) $-ve~~~~~~~~~~~-ve$
C) $-ve~~~~~~~~~~~+ve$
D) $+ve~~~~~~~~~~~-ve$
• question_answer111) If$\alpha \ne \beta$and${{\alpha }^{2}}=5\alpha -3,\text{ }{{\beta }^{2}}=5\beta -3,$then the equation having$\alpha /\beta$and$\beta /\alpha$as its roots is:
A) $3{{x}^{2}}+19x+3=0$
B) $3{{x}^{2}}-19x+3=0$
C) $3{{x}^{2}}-19x-3=0$
D) ${{x}^{2}}-16x+1=0$
• question_answer112) If$y={{(x+\sqrt{1+{{x}^{2}}})}^{n}},$then $(1+{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}$is:
A) ${{n}^{2}}y$
B) $-{{n}^{2}}y$
C) $-y$
D) $2{{x}^{2}}y$
• question_answer113) If$1,{{\log }_{3}}\sqrt{({{3}^{1-x}}+2)},{{\log }_{3}}({{4.3}^{x}}-1)$are in A.P, then$x$equals:
A) $lo{{g}_{3}}4$
B) $1-lo{{g}_{3}}4$
C) $1-lo{{g}_{4}}3$
D) $lo{{g}_{4}}3$
• question_answer114) A problem in mathematics is given to three students A, B, C and their respective probability of solving the problem is$\frac{1}{2},\frac{1}{3}$and$\frac{1}{4}$. Probability that the problem is solved is:
A) $\frac{3}{4}$
B) $\frac{1}{2}$
C) $\frac{2}{3}$
D) $\frac{1}{3}$
• question_answer115) The period of${{\sin }^{2}}\theta$is:
A) ${{\pi }^{2}}$
B) $\pi$
C) $2\pi$
D) $\frac{\pi }{2}$
• question_answer116) $l,m,n$are the${{p}^{th}},{{q}^{th}}$and${{r}^{th}}$term of an G.P. and all positive, then $\left| \begin{matrix} \log l & p & 1 \\ \log m & q & 1 \\ \log n & r & 1 \\ \end{matrix} \right|$equals:
A) 3
B) 2
C) 1
D) zero
• question_answer117) $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1-\cos 2x}}{\sqrt{2}x}$is:
A) $\lambda$
B) $-1$
C) zero
D) does not exist
• question_answer118) A triangle with vertices$(4,0),(-1,-1),(3,5)$is:
A) isosceles and right angled
B) isosceles but not right angled
C) right angled but not isosceles
D) neither right angled nor isosceles
• question_answer119) In a class of 100 students there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls?:
A) 73
B) 65
C) 68
D) 74
• question_answer120) ${{\cot }^{-1}}(\sqrt{\cos \alpha })={{\tan }^{-1}}(\sqrt{\cos \alpha })=x,$then$\sin x$is equal to:
A) ${{\tan }^{2}}\left( \frac{\alpha }{2} \right)$
B) $co{{t}^{2}}\left( \frac{\alpha }{2} \right)$
C) $\tan \alpha$
D) $\cot \left( \frac{\alpha }{2} \right)$
• question_answer121) The order and degree of the differential equation${{\left( 1+3\frac{dy}{dx} \right)}^{2/3}}=4\frac{{{d}^{2}}y}{d{{x}^{3}}}$are:
A) $\left( 1,\frac{2}{3} \right)$
B) $(3,1)$
C) $(3,3)$
D) $(1,2)$
• question_answer122) A plane which passes through the point (3,2,0) and the line$\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}$is:
A) $x-y+z=1$
B) $x+y+z=5$
C) $x+2y-z=1$
D) $2x-y+z=5$
• question_answer123) The solution of the equation$\frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{-2x}}$is:
A) $\frac{{{e}^{-2x}}}{4}$
B) $\frac{{{e}^{-2x}}}{4}+cx+d$
C) $\frac{1}{4}{{e}^{-2x}}+c{{x}^{2}}+d$
D) $\frac{1}{4}{{e}^{-2x}}+c+d$
• question_answer124) $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{{{x}^{2}}+5x+3}{{{x}^{2}}+x+2} \right)}^{x}}$is equal to:
A) ${{e}^{4}}$
B) ${{e}^{2}}$
C) ${{e}^{3}}$
D) $e$
• question_answer125) The domain of${{\sin }^{-1}}[{{\log }_{3}}(x/3)]$is:
A) $[1,9]$
B) $[-1,9]$
C) $[-9,1]$
D) $[-9,-1]$
• question_answer126) The value of${{2}^{1/4}}{{.4}^{1/8}}{{.8}^{1/6}}....\infty$is:
A) 1
B) 2
C) 3/2
D) 4
• question_answer127) Fifth term of an G.P. is 2, then the product of its 9 terms is
A) 256
B) 512
C) 1024
D) none of these
• question_answer128) $\int_{0}^{10x}{|\sin x|dx}$is:
A) 20
B) 8
C) 10
D) 18
• question_answer129) ${{I}_{n}}=\int_{0}^{\pi /4}{{{\tan }^{n}}x}dx,$then$\underset{n\to \infty }{\mathop{\lim }}\,n[{{I}_{n}}+{{I}_{n+2}}]$equals:
A) $\frac{1}{2}$
B) 1
C) $\infty$
D) zero
• question_answer130) $\int_{0}^{2}{[{{x}^{2}}]}\,dx$is:
A) $2-\sqrt{2}$
B) $2+\sqrt{2}$
C) $\sqrt{2}-1$
D) $-\sqrt{2}-\sqrt{3}+5$
• question_answer131) .$\int_{-\pi }^{\pi }{\frac{2x(1+\sin x)}{1+{{\cos }^{2}}x}}dx$is:
A) $\frac{{{\pi }^{2}}}{4}$
B) ${{\pi }^{2}}$
C) zero
D) $\frac{\pi }{2}$
• question_answer132) The period of the function$f(x)={{\sin }^{4}}x+{{\cos }^{4}}x$is:
A) $\pi$
B) $\frac{\pi }{2}$
C) $2\pi$
D) none of these
• question_answer133) The domain of definition of the function$f(x)=\sqrt{{{\log }_{10}}\left( \frac{5x-{{x}^{2}}}{4} \right)}$is:
A) [1, 4]
B) [1, 0]
C) [0, 5]
D) [5, 0]
• question_answer134) If$\sin y=a\sin (x+y),$then$\frac{dy}{dx}$is:
A) $\frac{\sin a}{{{\sin }^{2}}(a+y)}$
B) $\frac{{{\sin }^{2}}(a+y)}{\sin a}$
C) $\sin a{{\sin }^{2}}(a+y)$
D) $\frac{{{\sin }^{2}}(a-y)}{\sin a}$
• question_answer135) If${{x}^{y}}={{e}^{x-y}}$then$\frac{dy}{dx}$is:
A) $\frac{1+x}{1+\log x}$
B) $\frac{1-\log x}{1+\log x}$
C) not defined
D) $\frac{logx}{(1+\log x)}$
• question_answer136) The two curves${{x}^{3}}-3x{{y}^{2}}+2=0$and$3{{x}^{2}}y-{{y}^{3}}-2=0$:
A) cut at right angles
B) touch each other
C) cut at an angle$\frac{\pi }{3}$
D) cut at an angle$\frac{\pi }{4}$
• question_answer137) The function$f(x)={{\cot }^{-1}}x+x$increases in the interval:
A) $(1,\infty )$
B) $(-1,\infty )$
C) $(-\infty ,\infty )$
D) $(0,\infty )$
• question_answer138) The greatest value of$f(x)={{(x+1)}^{1/3}}-{{(x-1)}^{1/3}}$on [0, 1] is:
A) 1
B) 2
C) 3
D) 1/3
• question_answer139) Evaluate$\int_{0}^{\pi /2}{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}}dx$:
A) $\frac{\pi }{4}$
B) $\frac{\pi }{2}$
C) zero
D) 1
• question_answer140) $\int{\frac{dx}{x({{x}^{n}}+1)}}$is equal to:
A) $\frac{1}{n}\log \left( \frac{{{x}^{n}}}{{{x}^{n}}+1} \right)+c$
B) $\frac{1}{n}\log \left( \frac{{{x}^{n}}+1}{{{x}^{n}}} \right)+c$
C) $\log \left( \frac{{{x}^{n}}}{{{x}^{n}}+1} \right)+c$
D) none of these
• question_answer141) The area bounded by the curve$y=2x-x$ and the straight line$y=-x$is given by:
A) $\frac{9}{2}$
B) $\frac{43}{6}$
C) $\frac{35}{6}$
D) none of these
• question_answer142) The differential equation of all non-vertical lines in a plane is:
A) $\frac{{{d}^{2}}y}{d{{x}^{2}}}=0$
B) $\frac{{{d}^{2}}x}{d{{y}^{2}}}=0$
C) $\frac{dy}{dx}=0$
D) $\frac{dx}{dy}=0$
• question_answer143) Given two vectors$\hat{i}-\hat{j}$and$\hat{i}+2\hat{j}$the unit vector coplanar with the two vectors and perpendicular to first is:
A) $\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
B) $\frac{1}{\sqrt{5}}(2\hat{i}+\hat{j})$
C) $\pm \frac{1}{\sqrt{2}}(\hat{i}+\hat{k})$
D) none of these
• question_answer144) The vector$\hat{i}+x\hat{j}+3\hat{k}$is rotated through an angle$\theta$and doubled in magnitude, then it becomes$4\hat{i}+(4x-2)\hat{i}+2\hat{k}$. The value of$x$ is:
A) $\left( -\frac{2}{3},2 \right)$
B) $\left( \frac{1}{3},2 \right)$
C) $\left( \frac{2}{3},0 \right)$
D) $(2,7)$
• question_answer145) A paralleiopiped is formed by planes drawn through the points (2,3,5) and (5,9,7), parallel to the coordinate planes. The length of a diagonal of the parallelepiped to piped is:
A) $7$
B) $\sqrt{38}$
C) $\sqrt{155}$
D) None of these
• question_answer146) The equation of the plane containing the line $\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}$is$a(x-{{x}_{1}})+b(y-{{y}_{1}})+c(z-{{z}_{1}})=0,$where:
A) $a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}=0$
B) $al+bm+cn=0$
C) $\frac{a}{l}=\frac{b}{m}=\frac{c}{n}$
D) $l{{x}_{1}}+m{{y}_{1}}+n{{z}_{1}}=0$
• question_answer147) A and B play a game where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is:
A) $\frac{1}{25}$
B) $\frac{24}{25}$
C) $\frac{2}{25}$
D) None of these
• question_answer148) If A and B are two mutually exclusive events, then
A) $P(A)<P(\overline{B})$
B) $P(A)>P(\overline{B})$
C) $P(A)<P(B)$
D) None of these
• question_answer149) The equation of the directrix of the parabola ${{y}^{2}}+4v+4x+2=0$is:
A) $x=-1$
B) $x=1$
C) $x=-\frac{3}{2}$
D) $x=\frac{3}{2}$
• question_answer150) Let${{T}_{n}}$denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If${{T}_{n+1}}-{{T}_{n}}=21$then$n$ equals:
A) 5
B) 7
C) 6
D) 4
• question_answer151) In a triangle ABC, $2ca\sin \frac{A-B+C}{2}$ is equal to:
A) ${{a}^{2}}+{{b}^{2}}-{{c}^{2}}$
B) ${{c}^{2}}+{{a}^{2}}-{{b}^{2}}$
C) ${{b}^{2}}-{{c}^{2}}-{{a}^{2}}$
D) ${{c}^{2}}-{{a}^{2}}-{{b}^{2}}.$
• question_answer152) For$x\in R\,\,\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{x-3}{x+2} \right)}^{x}}$is equal to:
A) $e$
B) ${{e}^{-1}}$
C) ${{e}^{-5}}$
D) ${{e}^{5}}$
• question_answer153) The in centre of the triangle with vertices $(1,\sqrt{3}),(0,0)$and$(2,0)$is:
A) $\left( 1,\frac{\sqrt{3}}{2} \right)$
B) $\left( \frac{2}{3},\frac{1}{\sqrt{3}} \right)$
C) $\left( \frac{2}{3},\frac{\sqrt{3}}{2} \right)$
D) $\left( 1,\frac{1}{\sqrt{3}} \right)$
• question_answer154) If the vectors$\overrightarrow{a},\overrightarrow{b}$and$\overrightarrow{c}$from the sides BC, CA and AB respectively, of a triangle ABC then:
A) $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{b}=0$
B) $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\times \overrightarrow{a}$
C) $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$
D) $\overrightarrow{a}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=0$
• question_answer155) If$\omega$is an imaginary cube root of unity then ${{(1+\omega +{{\omega }^{2}})}^{7}}$equals:
A) $128\omega$
B) $-128\omega$
C) $128\,{{\omega }^{2}}$
D) $-128\,{{\omega }^{2}}$
• question_answer156) If$\left| \begin{matrix} 6i & -3i & 1 \\ 4 & 3i & -1 \\ 20 & 3 & i \\ \end{matrix} \right|=x+iy$then:
A) $x=3,y=1$
B) $x=1,y=3$
C) $x=0,\text{ y}=3$
D) $x=0,\text{ }y=0$
• question_answer157) ${{\sin }^{2}}\theta =\frac{4xy}{{{(x+y)}^{2}}}$is true if and only if:
A) $x+y\ne 0$
B) $x=y,x\ne 0,y\ne 0$
C) $x=y$
D) $x\ne 0,y\ne 0$
• question_answer158) The radius of the circle passing through the foci of the ellipse $\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{9}=1$and having its centre at (0, 3) is:
A) 4
B) 3
C) $\sqrt{12}$
D) $\frac{7}{2}$
• question_answer159) The probability of India winning a test match against West-Indies is$\frac{1}{2}$assuming independence from match to match the probability that in a match series Indias second win occurs at the third test is:
A) $\frac{1}{8}$
B) $\frac{1}{4}$
C) $\frac{1}{2}$
D) $\frac{2}{3}$
• question_answer160) If$(\omega \ne 1)$is a cubic root of unity, then$\left| \begin{matrix} 1 & 1+i+{{\omega }^{2}} & {{\omega }^{2}} \\ 1-i & -1 & {{\omega }^{2}}-1 \\ -i & -1+\omega -i & -1 \\ \end{matrix} \right|$equals:
A) zero
B) 1
C) $i$
D) $\omega$
• question_answer161) A biased coin with probability$p,0<p<1$of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is$\frac{2}{5},$then p equals:
A) $\frac{1}{3}$
B) $\frac{2}{3}$
C) $\frac{2}{5}$
D) $\frac{3}{5}$
• question_answer162) A fair die is tossed eight times. The probability that a third six is observed on the eight throw is:
A) $\frac{^{7}{{C}_{2}}\times {{5}^{5}}}{{{6}^{7}}}$
B) $\frac{^{7}{{C}_{2}}\times {{5}^{5}}}{{{6}^{8}}}$
C) $\frac{^{7}{{C}_{2}}\times {{5}^{5}}}{{{6}^{6}}}$
D) none of these
• question_answer163) Let$f(2)=4$and$f(2)=4$Then$\underset{x\to 2}{\mathop{\lim }}\,\frac{xf(2)-2f(x)}{x-2}$is given by:
A) 2
B) $-2$
C) $-4$
D) 3
• question_answer164) Three straight lines$2x+11y-5=0,$ $24x+7y-20=0$and$4x-3y-2=0$:
A) form a triangle
B) arc only concurrent
C) are concurrent with on line bisecting the angle between the other two
D) none of these
• question_answer165) A straight line through the point (2, 2) intersects the lines$\sqrt{3}x+y=0$and $\sqrt{3}x-y=0$at the points A and B. The equation to the line AB so that the triangle OAB is equilateral is:
A) $x-2=0$
B) $y-2=0$
C) $x+y-4=0$
D) none of these
• question_answer166) The greatest distance of the point$P(10,7)$from the circle${{x}^{2}}+{{y}^{2}}-4x-2y-20=0$is:
A) $10$
B) 15
C) 5
D) none of these
• question_answer167) The equation of the tangent to the circle${{x}^{2}}+{{y}^{2}}+4x-4y+4=0$which make equal intercepts on the positive coordinate axes is:
A) $x+y=2$
B) $x+y=2\sqrt{2}$
C) $x+y=4$
D) $x+y=8$
• question_answer168) The equation of the ellipse whose foci are $(\pm ,2,0)$and eccentricity$\frac{1}{2}$is:
A) $\frac{{{x}^{2}}}{12}+\frac{{{y}^{2}}}{16}=1$
B) $\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{12}=1$
C) $\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{8}=1$
D) none of these
• question_answer169) The equation of the chord joining two points $({{x}_{1}},{{y}_{2}})$and$({{x}_{2}},{{y}_{2}})$on the rectangular hyperbola$xy={{c}^{2}}$is:
A) $\frac{x}{{{x}_{1}}+{{x}_{2}}}+\frac{y}{{{y}_{1}}+{{y}_{2}}}=1$
B) $\frac{x}{{{x}_{1}}-{{x}_{2}}}+\frac{y}{{{y}_{1}}-{{y}_{2}}}=1$
C) $\frac{x}{{{y}_{1}}+{{y}_{2}}}+\frac{y}{{{x}_{1}}+{{x}_{2}}}=1$
D) $\frac{x}{{{y}_{1}}-{{y}_{2}}}+\frac{y}{{{x}_{1}}-{{x}_{2}}}=1$
• question_answer170) If the vectors$\overrightarrow{c},\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{j}$and$\hat{b}=\hat{j}$are such that $\overrightarrow{a},\overrightarrow{c}$ and$\overrightarrow{b}$form a right handed system then$\overrightarrow{c}$is:
A) $z\hat{j}-x\hat{k}$
B) $\overrightarrow{0}$
C) $y\hat{j}$
D) $-z\hat{i}+x\hat{k}$
|
2019-01-19 18:42:22
|
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https://brilliant.org/problems/super-efficient/
|
# Super-Efficient!
Computer Science Level 2
A program can have an efficiency of $$O(1)$$.
True or False?
×
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2016-10-22 17:55:12
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https://calendar.math.illinois.edu/?year=2020&month=02&day=26&interval=day
|
Department of
Mathematics
Seminar Calendar
for events the day of Wednesday, February 26, 2020.
.
events for the
events containing
Questions regarding events or the calendar should be directed to Tori Corkery.
January 2020 February 2020 March 2020
Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa
1 2 3 4 1 1 2 3 4 5 6 7
5 6 7 8 9 10 11 2 3 4 5 6 7 8 8 9 10 11 12 13 14
12 13 14 15 16 17 18 9 10 11 12 13 14 15 15 16 17 18 19 20 21
19 20 21 22 23 24 25 16 17 18 19 20 21 22 22 23 24 25 26 27 28
26 27 28 29 30 31 23 24 25 26 27 28 29 29 30 31
Wednesday, February 26, 2020
2:00 pm in 447 Altgeld Hall,Wednesday, February 26, 2020
Introduction to moduli spaces of sheaves
Sungwoo Nam (Illinois Math)
Abstract: This talk will be an introduction to moduli spaces of sheaves. We will see some motivating questions that lead to the study of moduli spaces of sheaves, and discuss examples telling us why the notion of stability is needed, even in the simplest case of vector bundles on curves. Then I will survey some results on moduli spaces of sheaves on surfaces, especially those of K3 and abelian surfaces and applications to holomorphic symplectic geometry.
4:00 pm in 245 Altgeld Hall,Wednesday, February 26, 2020
Coble
Frank Calegari (University of Chicago)
Abstract: Coble is known (in part) for his work on invariant theory and the geometry of certain of exceptional moduli spaces in low dimension. We discuss the quest to find explicit equations for one particular family of moduli spaces. An important role is played by a number of exceptional geometrical coincidences and also the theory of complex reflection groups. This will be a colloquium style talk and will be independent of the first two talks.
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2021-01-19 22:24:20
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http://hal.in2p3.fr/in2p3-00185471
|
# On the measurement of B(E2, 0$^{+}$_{1} -> 2^{+}$_{1}$) using intermediate-energy Coulomb excitation
Abstract : Coulomb excitation is a standard method used to extract quadrupole excitation strengths of even–even nuclei. In typical analyzes the reaction is assumed to be one step, Coulomb only, and is treated within a semi-classical model. In this work, fully quantal coupled-channel calculations are performed for three test cases in order to determine the importance of multi-step effects, nuclear contributions, feeding from other states and corrections to the semi-classical approximation. We study the excitation of $^{30}$S, $^{58}$Ni and $^{78}$Kr on $^{197}$Au at ≈50 A MeV. We find that nuclear effects may contribute more than 10% and that feeding contributions can be larger than 15%. These corrections do not alter significantly the published B(E2) values; however, an additional theoretical error of up to 13% should be added to the experimental uncertainty if the semi-classical model is used. This theoretical error is reduced to less than 7% when performing a quantal coupled-channel analysis.
Document type :
Journal articles
http://hal.in2p3.fr/in2p3-00185471
Contributor : Sandrine Guesnon <>
Submitted on : Tuesday, November 6, 2007 - 10:56:48 AM
Last modification on : Tuesday, November 24, 2020 - 11:30:12 AM
### Citation
F. Delaunay, F.M. Nunes. On the measurement of B(E2, 0$^{+}$_{1} -> 2^{+}$_{1}$) using intermediate-energy Coulomb excitation. Journal of Physics G Nuclear Physics, Institute of Physics (IOP), 2007, 34, pp.2207-2213. ⟨10.1088/0954-3899/34/10/010⟩. ⟨in2p3-00185471⟩
Record views
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2021-06-13 05:07:08
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https://www.llvm.org/doxygen/MSFBuilder_8h_source.html
|
LLVM 17.0.0git
MSFBuilder.h
Go to the documentation of this file.
1//===- MSFBuilder.h - MSF Directory & Metadata Builder ----------*- C++ -*-===//
2//
3// Part of the LLVM Project, under the Apache License v2.0 with LLVM Exceptions.
6//
7//===----------------------------------------------------------------------===//
8
9#ifndef LLVM_DEBUGINFO_MSF_MSFBUILDER_H
10#define LLVM_DEBUGINFO_MSF_MSFBUILDER_H
11
16#include "llvm/Support/Error.h"
17#include <cstdint>
18#include <utility>
19#include <vector>
20
21namespace llvm {
22class FileBufferByteStream;
23namespace msf {
24
25struct MSFLayout;
26
28public:
29 /// Create a new MSFBuilder.
30 ///
31 /// \param BlockSize The internal block size used by the PDB file. See
32 /// isValidBlockSize() for a list of valid block sizes.
33 ///
34 /// \param MinBlockCount Causes the builder to reserve up front space for
35 /// at least MinBlockCount blocks. This is useful when using MSFBuilder
36 /// to read an existing MSF that you want to write back out later. The
37 /// original MSF file's SuperBlock contains the exact number of blocks used
38 /// by the file, so is a good hint as to how many blocks the new MSF file
39 /// will contain. Furthermore, it is actually necessary in this case. To
40 /// preserve stability of the file's layout, it is helpful to try to keep
41 /// all streams mapped to their original block numbers. To ensure that this
42 /// is possible, space for all blocks must be allocated beforehand so that
43 /// streams can be assigned to them.
44 ///
45 /// \param CanGrow If true, any operation which results in an attempt to
46 /// locate a free block when all available blocks have been exhausted will
47 /// allocate a new block, thereby growing the size of the final MSF file.
48 /// When false, any such attempt will result in an error. This is especially
49 /// useful in testing scenarios when you know your test isn't going to do
50 /// anything to increase the size of the file, so having an Error returned if
51 /// it were to happen would catch a programming error
52 ///
53 /// \returns an llvm::Error representing whether the operation succeeded or
54 /// failed. Currently the only way this can fail is if an invalid block size
55 /// is specified, or MinBlockCount does not leave enough room for the
56 /// mandatory reserved blocks required by an MSF file.
58 uint32_t BlockSize,
59 uint32_t MinBlockCount = 0,
60 bool CanGrow = true);
61
62 /// Request the block map to be at a specific block address. This is useful
63 /// when editing a MSF and you want the layout to be as stable as possible.
66 void setFreePageMap(uint32_t Fpm);
67 void setUnknown1(uint32_t Unk1);
68
69 /// Add a stream to the MSF file with the given size, occupying the given
70 /// list of blocks. This is useful when reading a MSF file and you want a
71 /// particular stream to occupy the original set of blocks. If the given
72 /// blocks are already allocated, or if the number of blocks specified is
73 /// incorrect for the given stream size, this function will return an Error.
75
76 /// Add a stream to the MSF file with the given size, occupying any available
77 /// blocks that the builder decides to use. This is useful when building a
78 /// new PDB file from scratch and you don't care what blocks a stream occupies
79 /// but you just want it to work.
81
82 /// Update the size of an existing stream. This will allocate or deallocate
83 /// blocks as needed to match the requested size. This can fail if CanGrow
84 /// was set to false when initializing the MSFBuilder.
86
87 /// Get the total number of streams in the MSF layout. This should return 1
88 /// for every call to addStream.
89 uint32_t getNumStreams() const;
90
91 /// Get the size of a stream by index.
92 uint32_t getStreamSize(uint32_t StreamIdx) const;
93
94 /// Get the list of blocks allocated to a particular stream.
96
97 /// Get the total number of blocks that will be allocated to actual data in
98 /// this MSF file.
100
101 /// Get the total number of blocks that exist in the MSF file but are not
102 /// allocated to any valid data.
104
105 /// Get the total number of blocks in the MSF file. In practice this is equal
106 /// to getNumUsedBlocks() + getNumFreeBlocks().
108
109 /// Check whether a particular block is allocated or free.
110 bool isBlockFree(uint32_t Idx) const;
111
112 /// Finalize the layout and build the headers and structures that describe the
113 /// MSF layout and can be written directly to the MSF file.
115
116 /// Write the MSF layout to the underlying file.
118
120
121private:
122 MSFBuilder(uint32_t BlockSize, uint32_t MinBlockCount, bool CanGrow,
123 BumpPtrAllocator &Allocator);
124
125 Error allocateBlocks(uint32_t NumBlocks, MutableArrayRef<uint32_t> Blocks);
126 uint32_t computeDirectoryByteSize() const;
127
128 using BlockList = std::vector<uint32_t>;
129
131
132 bool IsGrowable;
133 uint32_t FreePageMap;
134 uint32_t Unknown1 = 0;
135 uint32_t BlockSize;
137 BitVector FreeBlocks;
138 std::vector<uint32_t> DirectoryBlocks;
139 std::vector<std::pair<uint32_t, BlockList>> StreamData;
140};
141
142} // end namespace msf
143} // end namespace llvm
144
145#endif // LLVM_DEBUGINFO_MSF_MSFBUILDER_H
This file defines the BumpPtrAllocator interface.
This file implements the BitVector class.
Returns the sub type a function will return at a given Idx Should correspond to the result type of an ExtractValue instruction executed with just that one unsigned Idx
uint64_t Size
Basic Register Allocator
ArrayRef - Represent a constant reference to an array (0 or more elements consecutively in memory),...
Definition: ArrayRef.h:41
Allocate memory in an ever growing pool, as if by bump-pointer.
Definition: Allocator.h:66
Lightweight error class with error context and mandatory checking.
Definition: Error.h:156
Tagged union holding either a T or a Error.
Definition: Error.h:470
MutableArrayRef - Represent a mutable reference to an array (0 or more elements consecutively in memo...
Definition: ArrayRef.h:305
StringRef - Represent a constant reference to a string, i.e.
Definition: StringRef.h:50
uint32_t getNumStreams() const
Get the total number of streams in the MSF layout.
Definition: MSFBuilder.cpp:224
Request the block map to be at a specific block address.
Definition: MSFBuilder.cpp:62
ArrayRef< uint32_t > getStreamBlocks(uint32_t StreamIdx) const
Get the list of blocks allocated to a particular stream.
Definition: MSFBuilder.cpp:230
Error setDirectoryBlocksHint(ArrayRef< uint32_t > DirBlocks)
Definition: MSFBuilder.cpp:87
uint32_t getTotalBlockCount() const
Get the total number of blocks in the MSF file.
Definition: MSFBuilder.cpp:151
uint32_t getNumFreeBlocks() const
Get the total number of blocks that exist in the MSF file but are not allocated to any valid data.
Definition: MSFBuilder.cpp:149
BumpPtrAllocator & getAllocator()
Definition: MSFBuilder.h:119
Error setStreamSize(uint32_t Idx, uint32_t Size)
Update the size of an existing stream.
Definition: MSFBuilder.cpp:192
Expected< FileBufferByteStream > commit(StringRef Path, MSFLayout &Layout)
Write the MSF layout to the underlying file.
Definition: MSFBuilder.cpp:337
Expected< MSFLayout > generateLayout()
Finalize the layout and build the headers and structures that describe the MSF layout and can be writ...
Definition: MSFBuilder.cpp:250
bool isBlockFree(uint32_t Idx) const
Check whether a particular block is allocated or free.
Definition: MSFBuilder.cpp:153
void setFreePageMap(uint32_t Fpm)
Definition: MSFBuilder.cpp:83
uint32_t getStreamSize(uint32_t StreamIdx) const
Get the size of a stream by index.
Definition: MSFBuilder.cpp:226
static Expected< MSFBuilder > create(BumpPtrAllocator &Allocator, uint32_t BlockSize, uint32_t MinBlockCount=0, bool CanGrow=true)
Create a new MSFBuilder.
Definition: MSFBuilder.cpp:50
uint32_t getNumUsedBlocks() const
Get the total number of blocks that will be allocated to actual data in this MSF file.
Definition: MSFBuilder.cpp:145
void setUnknown1(uint32_t Unk1)
Definition: MSFBuilder.cpp:85
Expected< uint32_t > addStream(uint32_t Size, ArrayRef< uint32_t > Blocks)
Add a stream to the MSF file with the given size, occupying the given list of blocks.
Definition: MSFBuilder.cpp:155
This is an optimization pass for GlobalISel generic memory operations.
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2023-02-06 11:59:31
|
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|
http://math.stackexchange.com/questions/442849/series-expansion-of-jacobizeta-in-both-arguments
|
# Series Expansion of JacobiZeta in Both Arguments
I am looking for a series expansion of the JacobiZeta function at the following argument values:
$$JacobiZeta[~ArcSin[1+a \epsilon]~,~1-b\epsilon~]_{\epsilon\approx 0}=?$$
where $a$ and $b$ are complex parameters and $\epsilon$ is going to zero. The main problem is due to the same power of $\epsilon$ appearing in both arguments at the respective places, since it causes conventional series for one argument not to terminate due to balancing of $\epsilon$ terms from the other argument after insertion.
Any help or suggestion how to proceed is appreciated!
-
What do you mean that one argument does not terminate? – Igor Rivin Jul 13 '13 at 14:05
By "conventional series for one argument" I mean series expansions around a specific value of one single argument, not both. But I actually obtained a solution for above problem an hour ago, will post it as an answer soon. – Kagaratsch Jul 13 '13 at 15:47
## migrated from mathoverflow.netJul 13 '13 at 17:00
This question came from our site for professional mathematicians.
Basically, the question asks for the behaviour of the Jacobi Zeta function for $c\to0$
$$\mathcal{Z}(\arcsin(1+ac)|1-bc)_{c\approx0}=?$$
where $a$ and $b$ are parameters. Consider the definition:
$$\mathcal{Z}(z,m)=\mathbb{E}(z|m)-\frac{\mathbb{E}(m)}{\mathbb{K}(m)}F(z|m)$$
where $\mathbb{E}(z,m)$ is the incomplete elliptic integral of the first kind, $\mathbb{E}(m)=\mathbb{E}(\frac{\pi}{2}|m)$ is the complete version, $F(z|m)$ is the incomplete elliptic integral of the second kind and $\mathbb{K}(m)=F(\frac{\pi}{2}|m)$ is again the complete version. Series expansions for $\mathbb{E}(m)$ and $\mathbb{K}(m)$ around $m\approx 1$ are readily available in the literature. Similarly, to get the leading contribution one can just straightforwardly evaluate
$$\mathbb{E}(\arcsin(1)|1)=1.$$
So the main difficulty lies with the evaluation of $F(z|m)$. Consider the following expansion of $F(z|m)$ which can be found on the Wolfram functions website:
$$F(z|m)=\sum_{k=0}^\infty\frac{(-1)^k{\left(\frac{1}{2}\right)_k}^2}{(k!)^2(m-1)^{-k}}\left(\ln(\sec(z)+\tan(z))+\frac{1}{2}\csc(z)\sum_{j=1}^k\frac{(-1)^j(j-1)!\tan^{2j}(z)}{{\left(\frac{1}{2}\right)_j}}\right)$$
where $\left(a\right)_n=\Gamma(a+n)/\Gamma(a)$ is the Pochhammer function. This expansion, untruncated describes the exact function and is valid for $|Re(z)|\leq\pi/2$. Let us first consider the $\ln$ term in the big parenthesis. For $z=\arcsin(1+ac)$ it behaves as
$$\log(\sec(z)+\tan(z))\approx\ln\left(-i\sqrt{2/ac}\right)+O(c)$$
while the factor $(m-1)^k$ in front goes as $(m-1)^k=(-bc)^k$. Therefore, even though the $\ln$ term diverges logarithmically, it gets suppressed by $(-bc)^k$ for any value of $k$ other than zero when we take $c\to 0$. Therefore, the $\ln$ term contributes only for $k=0$ and for the rest of the $k$ values we can throw it away.
At $k>0$ the second term in the big parenthesis kicks in. Here we always have the suppressing factor $(m-1)^k=(-bc)^k$ in front. For the functions of $z$ involved we observe:
$$\csc(z)\approx 1+O(c)$$ $$\tan^2(z)\approx -\frac{1}{2ac}-\frac{3}{4}+O(c)$$
Therefore, we realize that only terms where $j=k$ survive $c\to0$, since then we get a finite contribution from $(m-1)^k\tan^{2k}(z)$. Keeping only $j=k$ summands in the second sum we can perform the first infinite sum over $k$ to find:
$$\sum_{k=1}^\infty\frac{\left(\frac{1}{2}\right)_k}{k!~2 k}\left(\frac{b}{2a}\right)^k=\ln\left(\frac{4\sqrt{a}}{2\sqrt{a}+\sqrt{4a-2b}}\right)$$
Combining the leading divergence and this correction we arrive at:
$$F(\arcsin(1+ac)|1-bc)_{c\approx0}=\ln\left(-i\frac{4\sqrt{2}}{\sqrt{c}(2\sqrt{a}+\sqrt{4a-2b})}\right)+O(c)$$
With this the concrete behaviour of $\mathcal{Z}(\arcsin(1+ac)|1-bc)_{c\approx0}$ follows straightforwardly.
-
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2014-04-18 15:42:27
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https://www.gamedev.net/forums/topic/660579-intrinsics-to-improve-performance-of-interpolation-mix-functions/
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# Intrinsics to improve performance of interpolation / mix functions
This topic is 1400 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hello all
In pixel shaders (SM3 to 5) I often do some mixing / interpolation between two or more vectors (e.g. colors) or scalars (e.g. luminances).
A simple example of a common code as used for example in gaussion-blur-like implementation, looks like this
float4 mixColor = (tex2D(colorSampler, uv-blur)+tex2D(colorSampler, uv+blur)) / 2.f;
I would like to improve that, performance- and instruction-limits-wise, making use of hardware-supported HLSL intrinsic functions, but I'm not sure what would best here. lerp(), for example? I think I'd basically need the opposite of the mad() command. Are there better ways?
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It is pretty difficult to make a simple average of two values faster. Your compiler probably changes the division by 2 to multiply by 0.5 and that's like the end of your optimization possibilities - at least with so short code.
Gaussian blur can be implemented better with a compute shader - with a different algorithm.
Cheers!
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I don't think there's anything available that the compiler won't already being using, and even if there were there's no guarantee that it would actually map to a single instruction once it's JIT compiled for your GPU.
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Thanks to you both.
I'm doings loads of such averages, not just one single. That was just an example. Typically I'm processing e.g 128 to 1024 iterations in a for-loop (depending on the effect I implement). That's why performance considerations come into play. Compute shaders are out of scope in my case, I have to do everything in plain old HLSL pixel shaders.
As for what the compiler already optimizes, this is primarily what interests me. My question might have been not clear enough. I'm less interested in what theoretically could improve performance but what command / functions or best practices lead to the best optimized compiled code.
I am well aware that different GPU hardware (or firmware) might JIT compile intermediate code differently, but I'm interesting to know what *most* common GPU / systems will do.
E.g. it wouldn't help to use a mad() in my shaders when most systems would still compile that command into two instructions (mul,add) instead of a single madd. Similar goes for lerp() and the like.The noise() function for example is not supported by most GPUs although present in the HLSL standard.
Of course I can read what Microsoft writes on their HLSL intrinsic functions pages. But since reality differs from standards definitions such as HLSL, I'm interesting to learn how to write my HLSL shaders to benefit from hardware supported features as much as possible.
Edited by Meltac
Cool, thanks!
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Thanks I've already sort of known that but it was a good reminder
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On the topic: Is there a simple and fast way to get and compare the average of the components of a vector, e.g. luminance of a color? I'm often doing things like this:
float lum1 = (col1.r+col1.g+col1.b)/3.f;
float lum2 = (col2.r+col2.g+col2.b)/3.f;
if (lum2-lum1 > threshold) { ... }
Which of cause generates loads of instrucstions for a simple average computation and comparison. Is there a simpler / leaner / more elegant way?
Edited by Meltac
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Well, you can easily sum the components (and do the division) in a single instruction:
float lum1 = dot(col1.rgb, float3(1.0f / 3.0f, 1.0f / 3.0f, 1.0f / 3.0f));
...
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2018-07-21 12:24:44
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https://www.mathway.com/popular-problems/Algebra/283537
|
# Algebra Examples
Simplify - cube root of x^9
Rewrite as .
Pull terms out from under the radical, assuming real numbers.
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2022-05-21 05:04:02
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https://crypto.stackexchange.com/questions/linked/12956
|
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### Multiplicative Inverse in AES [duplicate]
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I actually do want to know how to generate the multiplication inverse (M.I) table, I mean the calculation how the M.I been generated.. such as the M.I of 22 is 5A regarding to the table.. Do you know ...
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### How to find multiplicative inverse of a hexadecimal number in the finite field GF(2^8)? [duplicate]
I am new to cryptography. As I was reading a note regarding S-box construction, I found a step mentioning multiplicative inverse of a number in the finite field GF(2^8). I couldn't understand it!! Can ...
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### Calculating the inverse for a Rijndael s-box [duplicate]
I want to calculate a Rijndael S-box but something went wrong. How can I calculate the inverse? In the following is my approach: given: $f(x)=x^8+x^4+x^3+x+1 \rightarrow$ binary: 100011011 0x12 = ...
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### Multiplication/Division in Galois Field (2^8)
I'm attempting to implement multiplication and division in $GF(2^8)$ using log and exponential tables. I'm using the exponent of 3 as my generator, using instructions from here. However I'm having ...
4k views
### Calculating the multiplicative inverse of a number in $GF(2^n)$ where $n > 8$
Suppose that: We have a polynomial $g(x)$ of degree $n$. $n > 8$. $q$ is the multiplicative inverse of $p$ in $G(2^n)$ modulo $g(x)$. If $p = 0$, then $q = 0$. This could be used: As a non-...
380 views
### Efficient pen-and-paper calculation of the Galois Field multiplication?
I have been reading on multiplication over GF for the Inverse Matrix Step in AES, and I feel like converting the hex numbers to binary, multiplying them and then doing the modifications required for ...
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### How are calculated numbers in AES S-box?
I have a problem calculating the 0x02 value of the AES S-Box. I followed the steps descibed in “How are the AES S-Boxes calculated?”, but my answer is not correct. ...
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### How to do Hexadecimal multiplication in GF(2^8)
I'm trying to do a hexadecimal multiplication in GF(2^8). Now, I know the technique of shifting left the binary numbers in some cases, for example: 02 * 9E 0000 0010 * 1001 1110 you shift one time ...
I'd like to use the Itoh-Tsujii algorithm for a dynamic substitution table, but I do not get the following line: $$r\ \gets\ (p^m - 1)\,/\,(p - 1)$$ And why can $r$ be used to calculate the ...
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2020-07-09 04:23:44
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https://tex.stackexchange.com/questions/541986/table-columns-not-correctly-separated/541988#541988
|
# Table columns not correctly separated
The following code produces a table. On the right half (frogs) the columns are not correctly distributed as on the left half (birds). How to make it work correctly?
\documentclass{article}
\usepackage{multirow}
\begin{document}
\begin{table}
\centering
\begin{tabular}{c|c| c|c| c|c| c|c |c}
\multirow{2}{*}{AAA} & \multicolumn{4}{c|}{birds} & \multicolumn{4}{c}{frogs} \\ \cline{2-9}
& \multicolumn{2}{c|}{w/o huge hat} & \multicolumn{2}{c|}{w/ huge hat} & \multicolumn{2}{c|}{w/o huge hat} & \multicolumn{2}{c}{w/ huge hat} \\
\hline
$\frac{1}{3}$ & 2.75 & 3.6 & 2.75 & 3.68 & 5 & 5 & 5 & 5 \\ \hline
1 & 8.3 & 10 & 8.2 & 10 & 15 & 15 & 15 & 15 \\ \hline
3 & 24.25 & 28 & 24.25 & 28.5 & 44 & 48 & 45 & 45 \\ \hline\hline
$\frac{p}{q}$ & $\frac{31}{97}$ & $\frac{53}{157}$ & $\frac{31}{97}$ & $\frac{53}{157}$ & $\frac{31}{97}$ & $\frac{53}{157}$ & $\frac{31}{97}$ & $\frac{53}{157}$ \\
\end{tabular}
\end{table}
\end{document}
Similar to @Zarko's answer, but omitting all vertical lines and using fewer but well-spaced horizontal lines in order to create a more open "look".
\documentclass{article}
\usepackage{tabularx,booktabs,ragged2e}
\newcolumntype{C}{>{\Centering}X}
\begin{document}
\begin{table}
\begin{tabularx}{\textwidth}{@{} c *{8}{C} @{}}
AAA & \multicolumn{4}{c}{birds} & \multicolumn{4}{c@{}}{frogs} \\
\cmidrule(lr){2-5} \cmidrule(l){6-9}
& \multicolumn{2}{c}{w/o huge hat} & \multicolumn{2}{c}{w/ huge hat}
& \multicolumn{2}{c}{w/o huge hat} & \multicolumn{2}{c@{}}{w/ huge hat} \\
\cmidrule(lr){2-3} \cmidrule(lr){4-5} \cmidrule(lr){6-7} \cmidrule(l){8-9}
$1/3$ & 2.75 & 3.6 & 2.75 & 3.68 & 5 & 5 & 5 & 5 \\
1 & 8.3 & 10 & 8.2 & 10 & 15 & 15 & 15 & 15 \\
3 & 24.25 & 28 & 24.25 & 28.5 & 44 & 48 & 45 & 45 \\
\midrule
$p/q$ & $\frac{31}{97}$ & $\frac{53}{157}$ & $\frac{31}{97}$ & $\frac{53}{157}$
& $\frac{31}{97}$ & $\frac{53}{157}$ & $\frac{31}{97}$ & $\frac{53}{157}$ \\
\end{tabularx}
\end{table}
\end{document}
• Thank you! Is it possible to add a vertical line after the 'AAA' column while not extending to the 'p/q' row? May 3 '20 at 18:13
• @xiaohuamao - Short answer: No. Longer answer: The whole point of not using any vertical lines and using fewer, but well spaced horizontal lines is to give the table an open and inviting look. If, however, you prefer a closed "look", then this solution is definitely not the right one for you.
– Mico
May 3 '20 at 18:26
• @xiaohuamao - Even longer answer: If the purpose of the vertical bar between the first and second column is not to reintroduce a closed look but just to provide a bit more visual separation between the header column and the 8 data columns, then no vertical bar is needed. Instead, just replace \begin{tabularx}{\textwidth}{@{} c *{8}{C} @{}} in the answer shown above with, say, \begin{tabularx}{\textwidth}{@{} c @{\qquad} *{8}{C} @{}}. Give it a try.
– Mico
May 3 '20 at 18:36
The width of the multi column cell withw/o huge hat is bigger than sum of widths of the spanned columns. To have equal width that columns you need to define their widths (for example as p{<width>}) or simpler use tabularx with all columns with equal widths:
\documentclass{article}
\usepackage{multirow, tabularx}
\newcolumntype{C}{>{\centering\arraybackslash}X}
\begin{document}
\begin{table}
\centering
\begin{tabularx}{\linewidth}{C|C| C|C| C|C| C|C |C}
\multirow{2}{*}{AAA} & \multicolumn{4}{c|}{birds} & \multicolumn{4}{c}{frogs} \\ \cline{2-9}
& \multicolumn{2}{c|}{w/o huge hat} & \multicolumn{2}{c|}{w/ huge hat} & \multicolumn{2}{c|}{w/o huge hat} & \multicolumn{2}{c}{w/ huge hat} \\
\hline
$\frac{1}{3}$ & 2.75 & 3.6 & 2.75 & 3.68 & 5 & 5 & 5 & 5 \\ \hline
1 & 8.3 & 10 & 8.2 & 10 & 15 & 15 & 15 & 15 \\ \hline
3 & 24.25 & 28 & 24.25 & 28.5 & 44 & 48 & 45 & 45 \\ \hline\hline
$\frac{p}{q}$ & $\frac{31}{97}$ & $\frac{53}{157}$ & $\frac{31}{97}$ & $\frac{53}{157}$ & $\frac{31}{97}$ & $\frac{53}{157}$ & $\frac{31}{97}$ & $\frac{53}{157}$ \\
\end{tabularx}
\end{table}
\end{document}
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2022-01-23 19:20:16
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|
https://books.compclassnotes.com/rothphys110-2e/2021/06/30/section-11-1-v2/
|
# Chapter 11: Torque
If you are attempting to push open a door, you do not push on it right next to the hinges. Even if you push with equal force near to the hinges, and then far from the hinges, you’ll find the door accelerates more quickly in one case than the other. (I encourage you to try pushing on the door close to the hinges to see how it feels.) When you are causing rotational motion, both the force and the where you apply the force are important.
## 11.1 Torque
Torque is what causes angular acceleration, just as force causes linear acceleration. The magnitude of any particular torque is given by the expression
$\tau = Fr\sin\theta \tag{11.1}$
where $$F$$ is the magnitude of the force, and $$r$$ is the distance from where the force is applied to the axis of rotation. If you imagine $$r$$ running in a straight line from the axis of rotation to the point where the force is applied, $$\theta$$ is the angle between that line and the force, as shown below.
The unit of torque is the netwon-meter (N · m). Although this is the same unit as for energy, we do not refer to torque in units of joules. This is because of the difference in what that meter measures: for work and energy, the distance (we used $$d$$) is a distance traveled, or a displacement; for torque, the distance $$r$$ is simply a measurement from one point to another on the object that is rotating—$$r$$ does not measure any sort of displacement.
Torque is a vector quantity, so it needs a direction. We use the same convention as for rotation (chapter 5): torque is positive if it causes a counterclockwise angular acceleration, and negative if it causes a clockwise angular acceleration.
#### Example 11.1
Consider a ship whose wheel has a radius of 0.5 m. There are two people fighting for control of the ship. Nyneave is applying a force of 35 N tangentially to the wheel, and Lan is applying a force of 42 N at an angle of 30° relative to the radial direction. Nyneave is applying her force to make the wheel rotate in a clockwise direction; Lan is applying his force to oppose Nyneave. What is the net torque on the wheel? Which person gains control of the ship?
The net torque exerted on the wheel is the sum of the torque exerted by each person:
$\tau_\textit{net} = \tau_1 + \tau_2$
Call the wheel’s radius $$R$$, and note how the direction of the applied force affects the torque.
\begin{align*} \tau_\textit{net} &= \tau_1 + \tau_2 \\ &= -RF_1\sin(90^\circ) + RF_2\sin(30^\circ) \\ &= -7\ \textrm{N}\cdot\textrm{m} \end{align*}
Since the net torque is negative, the wheel rotates clockwise. Nyneave wins, even though she is applying a smaller force—the direction relative to the axis of rotation is very important!
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2023-03-29 10:49:08
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|
http://physics.stackexchange.com/questions/56789/pressure-loss-in-a-syringe
|
Pressure loss in a syringe
I'm currently working on a problem which is really giving me some issues.
The problem concerns the force required to expel water from a syringe. We have a 20 ml syringe (which is $2\times10^{-5}$ meters cubed) with a diameter of 1 cm, full of water. The needle of the syringe is 40 mm in length and has a diameter of 0.2 mm. All of the water must be expelled from the syringe in 20 s. How much force must be applied to the syringe head to achieve this?
Ordinarily this is fine, but we have to include the pressure loss as a result of the friction in the needle. I'm using the Darcy–Weisbach equation to determine this. I calculated the speed the fluid needs to flow at by dividing the flow rate by the cross-sectional area of the needle. I've used a Moody chart to get $f_D$ as 0.046, and I'm using $\rho = 998.21$. I'm guessing the pressure loss in the needle is therefore $$0.046\times\frac{0.04}{0.002}\times\frac{998.21\times31.8^2}{2} = 4.64\,\mathrm{MPa}$$ Is that correct? In which case, how do I now get to the force from here?
-
What is the relation between force and pressure? – Bernhard Mar 14 '13 at 6:35
I got a slightly different friction coefficient. But I used the fact that in case of a laminar flow the friction coefficient is equal to $64/Re$. And I also got quite a different speed (so no 31.8 m/s) how did you got this? And the rest of the syringe will also ad some drop in pressure, however a lot less than the needle (maybe check if this can be neglected?) and you could also add some transition factors due to the sudden transition from the wide diameter to the smaller one of the needle. – fibonatic Jun 4 '13 at 15:14
You really should add units to make it readable - it may even give you hints to the error. – Volker Siegel Apr 23 at 19:51
Did not check the pressure is correct, but assuming is is:
You want to create the pressure in the syringe by applying the pressure to it's piston, right? Pressure is force per area, so:
$f=p\times a$
You need to add the friction of the piston (a force) to the force you need.
$f=p\times a + f_p$
-
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2014-10-20 09:44:47
|
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https://www.pims.math.ca/scientific-event/130312-pdsml
|
## PIMS Distinguished Speaker: Mark Lewis
• Date: 03/12/2013
• Time: 15:30
Lecturer(s):
Mark Lewis, University of Alberta
Location:
University of Victoria
Topic:
Mathematics behind stream population dynamics
Description:
Human activities change the natural flow regimes in streams and rivers and this impacts ecosystems. In this talk I will mathematically investigate the impact of changes in water flow on biological populations. The approach I will take is to develop process-oriented advection-diffusion-reaction equations that couple hydraulic flow to population growth, and then to analyze the equations so as to assess the effect of impacts of water flow on population dynamics. The mathematical framework is based on new theory for the net reproductive rate $R_0$ as applied to advection-diffusion-reaction equations. I will then connect the theory to populations in rivers under various flow regimes.
This work lays the groundwork for connecting $R_0$ to more complex models of spatially structured and interacting populations, as well as more detailed habitat and hydrological data. This is achieved through explicit numerical simulation of two dimensional depth-averaged models for river population dynamics.
|
2021-12-06 04:43:09
|
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https://github.com/adrelino/idp-graph-algorithms
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Files
Failed to load latest commit information.
Type
Name
Commit time
push-relabel algorithm
to solve the maximum flow problem
label-setting algorithm
to solve the shortest path problem with resource constraints
An interdisciplinary project by Adrian Haarbach
Live Demo
Development
Open index.html in Firefox or Safari, or run Google Chrome from the command line with the flag --allow-file-access-from-files
Alternatively, install Node.js and run npm install && grunt serve which will install all needed packages using npm, start a local webserver using Grunt, and open index.html in the default browser.
All needed library files except for MathJax are already included. Since the MathJax library consists of a lot of individual files, we did not want to include it into the repository. Instead we implemented a fallback mechanism, meaning that if you do not copy a MathJax installation into the library folder, we fallback to a Content Delivery Network (CDN) to load the required files needed to render LaTeX math equations.
If you want to be able to see the rendered LaTeX equations even when not having an internet connection, download and install mathjax into the directory: implementation/library/js/mathjax/
Installation
Copy index.html and the implementation folder into the document root of your webserver.
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2020-08-06 23:36:00
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http://math.stackexchange.com/questions/408132/find-functions-f-such-that-ffx-xfx1
|
find functions f such that $f(f(x))=xf(x)+1$,
let $f:R\longrightarrow R$, and $f$ is continous,and such that $f(f(x))=xf(x)+1$,
find all this $f$?
follow is my some idea:(but I don't have solution)
We have $f(f(0)) = 1$, so there is your $c = f(0)$, such that $f(c) = 1$. Assume there exists $v$ such that $f(v) = 0$. Then $f(0) = f(f(v)) = vf(v) + 1 = 1$, meaning $c=1$. Now, $f(0) = f(1) = 1$, so $1 = f(f(0)) = f(f(1)) = f(1)+1 = 2$, absurd.
So $f(x) \neq 0$ for all $x$, thus $f$ takes constant sign, being continuous. Assume now $f(x) = f(y) = t \neq 0$, so $xt+1 = f(f(x)) = f(f(y))= yt + 1$, whence $(x-y)t=0$, thus $x=y$. This means $f$ is injective, therefore monotonous, being continuous. Moreover, assume $x=f(x)$, so $x = f(x) = f(f(x)) = xf(x) + 1 = x^2+1$, thus $x^2-x+1 = 0$, but this has no real roots, so $f(x) \neq x$ for all $x$. Then either $f(x) > x$ for all $x$, or $f(x) < x$ for all $x$, since $f(x)-x$ is continuous
mark: this problem is my found,come from this probelm,if when $f:N\longrightarrow N$,and add $f(1)=1$then this problem is equivalent follow problem $$a_{n+1}=na_{n}+1,a_{1}=1$$
we can find $a_{n}=[e(n-1)!]$,
Thank you everyone can help
-
today I see this problem:math.stackexchange.com/questions/408004/… – math110 Jun 1 '13 at 1:06
Your problem is really different from that one, I think. – Eric Stucky Jun 1 '13 at 1:08
Hahaha,Thank you @EricStucky – math110 Jun 1 '13 at 1:16
The recurrence of the title should be $f(x+1)=xf(x)+1$ if it is to match that of your recurrence for $a_n$ with $f(n)=a_n$. Also you might mention that $a_n=[e(n-1)!]$ gives $2$ at $n=1$, so you are really only saying this $a_n$ satisfies your recurrence provided you define $a_1=1$ separately, not using the formula. – coffeemath Jun 1 '13 at 1:35
I don't have an answer, but here is a fact that may be useful: The function is decreasing and positive, and hence must decrease to some $L \geq 0$ as $x \rightarrow -\infty$. Since $$f(x) = \frac{f(f(x)) - 1}{x},$$ we see that the LHS converges to the number $L$ while the RHS converges to 0, and so $L = 0$. – Zach L. Jun 1 '13 at 2:49
From the OP and Zach L's comment to the OP, we can continuously extend $f$ to a function on the extended real numbers by setting $f(-\infty) = 0$ and $f(+\infty) = +\infty$.
Define a sequence $a_n$ of extended real numbers for all natural numbers by $a_0 = -\infty$, and $a_{n+1} = f(a_n)$. Observe that $f$ is a bijective function $[a_n, a_{n+1}] \to [a_{n+1}, a_{n+2}]$.
Let $g$ be the inverse of $f$ (with domain the non-negative extended real numbers)
The sequence $a_n$ is monotonic and increasing, and therefore has a limit $L$ in the extended real numbers. This satisfies
\begin{align} L &= \lim_{n \to +\infty} a_n \\&= \lim_{n \to +\infty} a_{n+1} \\&= \lim_{n \to +\infty} f(a_n) \\&= f(\lim_{n \to +\infty} a_n) \\&= f(L) \end{align}
The OP has already shown that $f$ has no finite fixed points, so therefore $L = +\infty$.
This means the intervals $[a_n, a_{n+1}]$ cover the entire range $[-\infty, +\infty)$.
The fact that $f(f(x)) = x f(x) + 1$ means that the value of $f$ on $[a_{n+1}, a_{n+2}]$ is determined by its values on $[a_n, a_{n+1}]$ (by considering $x \in [a_n, a_{n+1}]$).
Therefore, $f$ is completely determined by its values on $[a_0, a_1] = [-\infty, 0]$.
Conversely, I assert that if you choose any continuous, monotonically increasing function $f_0$ on $[-\infty, 0]$ such that $f_0(-\infty) = 0$ and $0 < f_0(0) < 1$, then we have a n increasing sequence (converging to $+\infty$) recursively defined by
• $a_0 = -\infty$
• $a_1 = 0$
• $a_2 = f_0(0)$
• $a_{n+2} = a_{n+1} a_n + 1$
and a sequence of invertible functions $f_n : [a_n, a_{n+1}] \to [a_{n+1}, a_{n+2}]$ recursively defined by
• $f_{n+1}(x) = f_n^{-1}(x) x + 1$
and then the function
$$f(x) = \begin{cases} f_n(x) & x \in [a_n, a_{n+1}] \\ +\infty & x = +\infty \end{cases}$$
is (the continuous extension to the extended real numbers of) a solution to the problem.
-
While there are many continuous solutions, there is probably a unique analytic solution. – Hurkyl Jun 1 '13 at 9:25
The definition implies that we can take $x$ from $f(f(x))$. That means that there exists a function that is the inverse of $f$. Let's call that function $g$ with $g(f(x)) = x$.
We would have:
$f(f(x)) = g(f(x))f(x) +1$
Right now we only have defined the case $f(f(x))$. Since we have to find functions that satisfy that case, but there are no restriction upon other cases, we can safely define the general case $f(x)$ as:
$f(x) = g(x)x +1$
By simple algebra, we can define $g$ as:
$g(x) = (f(x) - 1)/x$
Now, suppose we have $f(z_1) = z_2$, then we can use $g(f(x)) = x$ to find $g(z_2)$, and use that to find $f(z_2)$. We can repeat this method indefinitely, since OP proved that $f(x) \not= x$, for all $x$. We still have to find a suitable values for $z_1$ and $z_2$
-
Your functional equation $f(x) = g(x)x + 1$ is only valid for $x$ in the range of $f$. You cannot use it to deduce $f(0)=1$ since $0$ is not in the range of $f$ (proved by the OP already). – Erick Wong Jun 1 '13 at 4:12
You are absolutely right Erick. I edited it so that it might be at least partially useful. – jmj Jun 1 '13 at 6:24
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2016-07-26 14:09:45
|
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https://www.omnicalculator.com/physics/black-hole-temperature
|
# Black Hole Temperature Calculator
Created by Álvaro Díez
Reviewed by Bogna Szyk and Jack Bowater
Last updated: Nov 10, 2022
The black hole temperature calculator allows you to calculate the black body temperature of a black hole in a dark, cold environment.
A black hole is a massive object with very special properties, one of which is the relationship between its temperature and its mass when considered as a black body.
We explore this relationship in the calculator, and we hope this simple tool will help you better understand the properties of black holes.
## Black hole as an almost perfect black body
To understand this black hole temperature calculator, we first need to explain what is a black body.
A black body is a theoretical object in physics that absorbs all radiation that gets to it. The temperature of this black body depends on the amount of absorbed radiation, which in turn determines the frequency and intensity of the electromagnetic radiation it emits.
A black hole is an almost perfect black body and, as such, follows similar rules of emission and absorption.
As a black body, a black hole has an emission pattern in the shape of a Stefan-Boltzmann distribution, as shown in the image below, which also depends on temperature.
Since the temperature of a black hole (or its first approximation) depends on its mass, you can predict how a black hole will emit radiation, the so-called Hawking radiation, just by knowing its mass and using the Stefan-Boltzmann law calculator.
That calculator was built precisely for this purpose, allowing you to obtain a black hole's temperature (and thus emission spectra) by just knowing its mass.
💡 Converting temperatures between different units is a breeze with our temperature conversion.
## Black body temperature, Hawking radiation, and consequences
The emission mechanism of a black hole is a very complex process that was first theorized by Stephen Hawking. It involves complex quantum mechanics, the uncertainty principle, and the spontaneous creation of particle pairs. Don't worry, we don't fully understand it either. We can, however, provide a very simple and quick explanation.
Imagine a pair of particles created on the edge of the event horizon so that one of the particles can escape the black hole whilst the other gets trapped inside the black hole's gravitational field. The one that escapes is Hawking radiation.
To learn more about this process and the event horizon, please have a look at the Wikipedia articles on and the , as well as at the Schwarzschild radius calculator, for its explanation on how the event horizon is defined and its importance.
One of the surprising facts about Hawking radiation is that it increases as the size of the black hole decreases.
This result constitutes one of the most mind-boggling pieces of Hawking's work – the conservation of energy, together with the emission/radiation process, leads to only one possible conclusion (again, in a very simple and very incomplete statement): "All black holes evaporate, and they evaporate faster as they get smaller".
## How the black hole temperature calculator works
This black hole temperature calculator provides a very quick and simple way of obtaining the temperature of a black hole from its mass or vice versa:
$T = \frac{\hbar c^3}{8 \pi G M k_{\rm B}}$
The relationship between the mass and temperature of a black hole is very simple and is set out in the above equation, where all values, except for the mass ($M$) and the temperature ($T$), are universal constants.
Let's see what each of the parameters refers to, as well as the values of the constants and typical values for the free parameters:
1. $\hbar$ is the reduced Planck constant $h/2 \pi$, and has a value of 1.0545718001 × 10−34 J·s in SI units.
2. $c$ is the speed of light, equal to 299,792,458 m/s.
3. $\pi$ is the mathematical constant with a value of 3.141592...
4. $k_{\rm B}$ is the Boltzmann constant; in SI units, its value is 1.380649 × 10−23 J/K.
5. $G$ is the universal gravitational constant with a value of 6.67408 × 10−11 Nm2/kg2 in SI units.
6. $T$ is the black hole temperature/ black body temperature. For example, a black hole with a mass of $\small M = 1\ \rm Solar\ Mass$ has a temperature of 0.06172 × 10-6 K.
7. $M$ is the black hole mass, commonly expressed in solar masses due to their typical size in the universe.
Using the black hole temperature calculator is as simple as inputting either the mass or the temperature of the black hole, which will give you the other parameter. It is important to note that typical black holes have a mass of at least several solar masses. Still, it is also interesting to see what would happen in the case of much smaller micro black holes, which have been theorized but never experimentally observed because of how fast they evaporate (amongst many other reasons).
As we can see from the example above, the black body temperature of a typical black hole is very low; that is, they emit a very small amount of Hawking radiation, which explains why black holes in the universe seem to exist forever.
It is also important to realize that this temperature-mass relationship only holds when there is no radiation absorbed by the black hole, i.e., in a dark and cold environment very similar to the darkest parts of the universe. In reality, many of the black holes found in the universe do absorb and interact with other massive objects. Hence, their real temperature is different from the expected black body temperature.
Álvaro Díez
Mass
Suns
Temperature
°F
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2022-12-07 22:18:35
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https://www.phphotoimages.co.uk/photo_6133870.html
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# Pity Those That Find Her
Model Nikkia. Set made out of the usual foam and foamboard. Image really was an exercise in mimicing the patterns of light on the pillar on the person, which I think I've acheived. Image copyright Paul Holroyd.
Ref:
Date:
Location:
Photographer:
# Pity Those That Find Her
Model Nikkia. Set made out of the usual foam and foamboard. Image really was an exercise in mimicing the patterns of light on the pillar on the person, which I think I've acheived. Image copyright Paul Holroyd.
Ref:
Date:
Location:
Photographer:
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2018-11-17 06:38:41
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http://openstudy.com/updates/4ddd3835ee2c8b0baea03fe8
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• julie
How exactly do these procedures work? (define (square x) (exp (double (log x)))) (define (double x) (+ x x))
MIT 6.001 Structure and Interpretation of Computer Programs, Spring 2005
Looking for something else?
Not the answer you are looking for? Search for more explanations.
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2017-04-24 07:39:21
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https://cn.maplesoft.com/support/help/maple/view.aspx?path=RegularChains%2FEquations
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Equations - Maple Help
RegularChains
Equations
list of equations of the regular chain
Calling Sequence Equations(rc, R)
Parameters
rc - regular chain of R R - polynomial ring
Description
• The command Equations(rc,R) returns the list of the polynomials of rc by decreasing order of main variable.
• This command is part of the RegularChains package, so it can be used in the form Equations(..) only after executing the command with(RegularChains). However, it can always be accessed through the long form of the command by using RegularChains[Equations](..).
Examples
> $\mathrm{with}\left(\mathrm{RegularChains}\right):$$\mathrm{with}\left(\mathrm{ChainTools}\right):$
> $R≔\mathrm{PolynomialRing}\left(\left[x,y,z\right]\right)$
${R}{≔}{\mathrm{polynomial_ring}}$ (1)
In the polynomial ring the variables are ordered as follows: $x>y>z$.
> $\mathrm{rc}≔\mathrm{Empty}\left(R\right)$
${\mathrm{rc}}{≔}{\mathrm{regular_chain}}$ (2)
> $\mathrm{rc}≔\mathrm{Chain}\left(\left[{z}^{2}+1,{y}^{2}-z,x+y+z\right],\mathrm{rc},R\right)$
${\mathrm{rc}}{≔}{\mathrm{regular_chain}}$ (3)
> $\mathrm{Equations}\left(\mathrm{rc},R\right)$
$\left[{x}{+}{y}{+}{z}{,}{{y}}^{{2}}{-}{z}{,}{{z}}^{{2}}{+}{1}\right]$ (4)
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2022-12-08 23:00:36
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http://www.global-sci.org/intro/article_detail/aamm/118.html
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Volume 4, Issue 2
Exact Vibration Solutions of Nonhomogeneous Circular, Annular and Sector Membranes
Adv. Appl. Math. Mech., 4 (2012), pp. 250-258.
Published online: 2012-04
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• Abstract
In this paper, exact vibration frequencies of circular, annular and sector membranes with a radial power law density are presented for the first time. It is found that in general, the sequence of modes may not correspond to increasing az- imuthal mode number n. The normalized frequency increases with the absolute value of the power index |\nu|. For a circular membrane, the fundamental frequency occurs at n = 0 where n is the number of nodal diameters. For an annular mem- brane, the frequency increases with respect to the inner radius b. When b is close to one, the width 1 - b is the dominant factor and the differences in frequencies are small. For a sector membrane, n - 1 is the number of internal radial nodes and the fundamental frequency occurs at n = 1. Increased opening angle \beta increases the frequency.
• Keywords
Membrane vibration non-homogeneous exact circular annular sector
74.K15 74.H45
• BibTex
• RIS
• TXT
@Article{AAMM-4-250, author = {Chang Yi Wang and Wang Chien Ming}, title = {Exact Vibration Solutions of Nonhomogeneous Circular, Annular and Sector Membranes}, journal = {Advances in Applied Mathematics and Mechanics}, year = {2012}, volume = {4}, number = {2}, pages = {250--258}, abstract = {
In this paper, exact vibration frequencies of circular, annular and sector membranes with a radial power law density are presented for the first time. It is found that in general, the sequence of modes may not correspond to increasing az- imuthal mode number n. The normalized frequency increases with the absolute value of the power index |\nu|. For a circular membrane, the fundamental frequency occurs at n = 0 where n is the number of nodal diameters. For an annular mem- brane, the frequency increases with respect to the inner radius b. When b is close to one, the width 1 - b is the dominant factor and the differences in frequencies are small. For a sector membrane, n - 1 is the number of internal radial nodes and the fundamental frequency occurs at n = 1. Increased opening angle \beta increases the frequency.
}, issn = {2075-1354}, doi = {https://doi.org/10.4208/aamm.10-m1135}, url = {http://global-sci.org/intro/article_detail/aamm/118.html} }
TY - JOUR T1 - Exact Vibration Solutions of Nonhomogeneous Circular, Annular and Sector Membranes AU - Chang Yi Wang & Wang Chien Ming JO - Advances in Applied Mathematics and Mechanics VL - 2 SP - 250 EP - 258 PY - 2012 DA - 2012/04 SN - 4 DO - http://dor.org/10.4208/aamm.10-m1135 UR - https://global-sci.org/intro/aamm/118.html KW - Membrane KW - vibration KW - non-homogeneous KW - exact KW - circular KW - annular KW - sector AB -
In this paper, exact vibration frequencies of circular, annular and sector membranes with a radial power law density are presented for the first time. It is found that in general, the sequence of modes may not correspond to increasing az- imuthal mode number n. The normalized frequency increases with the absolute value of the power index |\nu|. For a circular membrane, the fundamental frequency occurs at n = 0 where n is the number of nodal diameters. For an annular mem- brane, the frequency increases with respect to the inner radius b. When b is close to one, the width 1 - b is the dominant factor and the differences in frequencies are small. For a sector membrane, n - 1 is the number of internal radial nodes and the fundamental frequency occurs at n = 1. Increased opening angle \beta increases the frequency.
Chang Yi Wang & Wang Chien Ming. (1970). Exact Vibration Solutions of Nonhomogeneous Circular, Annular and Sector Membranes. Advances in Applied Mathematics and Mechanics. 4 (2). 250-258. doi:10.4208/aamm.10-m1135
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2020-07-05 09:55:36
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https://www.aimsciences.org/article/doi/10.3934/jimo.2014.10.1225
|
American Institute of Mathematical Sciences
• Previous Article
Optimal dividend and capital injection strategy with fixed costs and restricted dividend rate for a dual model
• JIMO Home
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Hedging strategies for discretely monitored Asian options under Lévy processes
October 2014, 10(4): 1225-1234. doi: 10.3934/jimo.2014.10.1225
Lower semicontinuity of the solution mapping to a parametric generalized vector equilibrium problem
1 College of Sciences, Chongqing Jiaotong University, Chongqing, 400074 2 College of Mathematics and Statistics, Chongqing University, Chongqing, 401331
Received March 2013 Revised September 2013 Published February 2014
This paper deals with the lower semicontinuity of the solution mapping to a parametric generalized vector equilibrium problem. Under new assumptions, which do not contain any information about solution mappings, we establish the lower semicontinuity of the solution mapping to a parametric generalized vector equilibrium problem by using a scalarization method. These results improve the corresponding ones in recent literature. Some examples are given to illustrate our results.
Citation: Qilin Wang, Shengji Li. Lower semicontinuity of the solution mapping to a parametric generalized vector equilibrium problem. Journal of Industrial & Management Optimization, 2014, 10 (4) : 1225-1234. doi: 10.3934/jimo.2014.10.1225
References:
show all references
References:
[1] Liping Tang, Ying Gao. Some properties of nonconvex oriented distance function and applications to vector optimization problems. Journal of Industrial & Management Optimization, 2021, 17 (1) : 485-500. doi: 10.3934/jimo.2020117 [2] Shasha Hu, Yihong Xu, Yuhan Zhang. Second-Order characterizations for set-valued equilibrium problems with variable ordering structures. Journal of Industrial & Management Optimization, 2020 doi: 10.3934/jimo.2020164 [3] Christopher S. Goodrich, Benjamin Lyons, Mihaela T. Velcsov. Analytical and numerical monotonicity results for discrete fractional sequential differences with negative lower bound. Communications on Pure & Applied Analysis, 2021, 20 (1) : 339-358. doi: 10.3934/cpaa.2020269 [4] Ying Lin, Qi Ye. Support vector machine classifiers by non-Euclidean margins. Mathematical Foundations of Computing, 2020, 3 (4) : 279-300. doi: 10.3934/mfc.2020018 [5] Wen Li, Wei-Hui Liu, Seak Weng Vong. Perron vector analysis for irreducible nonnegative tensors and its applications. Journal of Industrial & Management Optimization, 2021, 17 (1) : 29-50. doi: 10.3934/jimo.2019097 [6] Anna Abbatiello, Eduard Feireisl, Antoní Novotný. Generalized solutions to models of compressible viscous fluids. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 1-28. doi: 10.3934/dcds.2020345 [7] Qianqian Han, Xiao-Song Yang. Qualitative analysis of a generalized Nosé-Hoover oscillator. Discrete & Continuous Dynamical Systems - B, 2020 doi: 10.3934/dcdsb.2020346 [8] Yi An, Bo Li, Lei Wang, Chao Zhang, Xiaoli Zhou. Calibration of a 3D laser rangefinder and a camera based on optimization solution. Journal of Industrial & Management Optimization, 2021, 17 (1) : 427-445. doi: 10.3934/jimo.2019119 [9] Gunther Uhlmann, Jian Zhai. Inverse problems for nonlinear hyperbolic equations. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 455-469. doi: 10.3934/dcds.2020380 [10] Monia Capanna, Jean C. Nakasato, Marcone C. Pereira, Julio D. Rossi. Homogenization for nonlocal problems with smooth kernels. Discrete & Continuous Dynamical Systems - A, 2020 doi: 10.3934/dcds.2020385 [11] Shun Zhang, Jianlin Jiang, Su Zhang, Yibing Lv, Yuzhen Guo. ADMM-type methods for generalized multi-facility Weber problem. Journal of Industrial & Management Optimization, 2020 doi: 10.3934/jimo.2020171 [12] Zhilei Liang, Jiangyu Shuai. Existence of strong solution for the Cauchy problem of fully compressible Navier-Stokes equations in two dimensions. Discrete & Continuous Dynamical Systems - B, 2020 doi: 10.3934/dcdsb.2020348 [13] Thabet Abdeljawad, Mohammad Esmael Samei. Applying quantum calculus for the existence of solution of $q$-integro-differential equations with three criteria. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020440 [14] Hai-Feng Huo, Shi-Ke Hu, Hong Xiang. Traveling wave solution for a diffusion SEIR epidemic model with self-protection and treatment. Electronic Research Archive, , () : -. doi: 10.3934/era.2020118 [15] Sihem Guerarra. Maximum and minimum ranks and inertias of the Hermitian parts of the least rank solution of the matrix equation AXB = C. Numerical Algebra, Control & Optimization, 2021, 11 (1) : 75-86. doi: 10.3934/naco.2020016 [16] Vieri Benci, Sunra Mosconi, Marco Squassina. Preface: Applications of mathematical analysis to problems in theoretical physics. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020446 [17] Leilei Wei, Yinnian He. A fully discrete local discontinuous Galerkin method with the generalized numerical flux to solve the tempered fractional reaction-diffusion equation. Discrete & Continuous Dynamical Systems - B, 2020 doi: 10.3934/dcdsb.2020319 [18] Aihua Fan, Jörg Schmeling, Weixiao Shen. $L^\infty$-estimation of generalized Thue-Morse trigonometric polynomials and ergodic maximization. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 297-327. doi: 10.3934/dcds.2020363 [19] Lihong Zhang, Wenwen Hou, Bashir Ahmad, Guotao Wang. Radial symmetry for logarithmic Choquard equation involving a generalized tempered fractional $p$-Laplacian. Discrete & Continuous Dynamical Systems - S, 2020 doi: 10.3934/dcdss.2020445 [20] Thomas Frenzel, Matthias Liero. Effective diffusion in thin structures via generalized gradient systems and EDP-convergence. Discrete & Continuous Dynamical Systems - S, 2021, 14 (1) : 395-425. doi: 10.3934/dcdss.2020345
2019 Impact Factor: 1.366
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2020-12-05 12:29:09
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https://gitane-rhapsody.com/1b3h3g/0351d0-mutation-selection-balance-calculator
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B {\displaystyle spq} Supposing there is selection against a deleterious allele. μ respectively, where µ = frequency of new mutant alleles per locus per generation µ = 10 -6 : 1 in 1,000,000 gametes has new mutant {\displaystyle \mu } {\displaystyle q={\sqrt {\mu /s}}} [3] Nevertheless, the concept is still widely used in evolutionary genetics, e.g. {\displaystyle 1-hs} q . Mutation–selection balance was originally proposed to explain how genetic variation is maintained in populations, although several other ways for deleterious mutations to persist are now recognized, notably balancing selection. {\displaystyle 1-\mu } Mutation–selection balance is an equilibrium in the number of deleterious alleles in a population that occurs when the rate at which deleterious alleles are created by mutation equals the rate at which deleterious alleles are eliminated by selection. At that equilibrium, µ € µ=sq2. s [3] Thus, provided that the mutant allele is not weakly deleterious (very small balance between selective loss of variation and creation of variation by beneficial mutations).[6]. p / p {\displaystyle q} = s {\displaystyle q} Putting these two pieces together, we can write the expression for the change in allele frequency that is due to BOTH gene flow and selection: Dq = -m(qx t- qy t) - sqx2(1-qx). p and B s B − p B p {\displaystyle h=0} 2 indicates that A is completely dominant while h is small). h of normal alleles A increases at rate / As a simple example of mutation-selection balance, consider a single locus in a haploid population with two possible alleles: a normal allele A with frequency MUTATION-SELECTION BALANCE. 0 μ ; thus In a diploid population, a deleterious allele B may have different effects on individual fitness in heterozygotes AB and homozygotes BB depending on the degree of dominance of the normal allele A. / Mutation-selection balance can maintain a genetic polymorphism. , and so the frequency of deleterious alleles is Herron, JC and S Freeman. A Learn how and when to remove this template message, "De Novo Rearrangements Found in 2% of Index Patients with Spinal Muscular Atrophy: Mutational Mechanisms, Parental Origin, Mutation Rate, and Implications for Genetic Counseling", "Beneficial Mutation–Selection Balance and the Effect of Linkage on Positive Selection", https://en.wikipedia.org/w/index.php?title=Mutation–selection_balance&oldid=976844169, Short description is different from Wikidata, Wikipedia articles that are too technical from September 2010, Creative Commons Attribution-ShareAlike License, This page was last edited on 5 September 2020, at 11:13. {\displaystyle \mu } , and the reverse beneficial mutation from B to A occurs rarely enough to be negligible (e.g. q The degree of dominance affects the relative importance of selection on heterozygotes versus homozygotes. h p B {\displaystyle \mu } − Mutation–selection balance occurs when these forces cancel and 1 . Evolutionary Analysis, 5th Edition. ≈ is not close to zero), then deleterious mutations are primarily removed by selection on heterozygotes because heterozygotes contain the vast majority of deleterious B alleles (assuming that the deleterious mutation rate , and a mutated deleterious allele B with frequency Pearson. q {\displaystyle \mu p} {\displaystyle 1-s} is constant from generation to generation, implying A / 1 q by an amount {\displaystyle p_{AA}} μ A How can we calculate the equilibrium gene frequencies that result from mutation-selection balance. s q [1] This equilibrium frequency is potentially substantially larger than for the case of partial dominance, because a large number of mutant alleles are carried in heterozygotes and are shielded from selection. = 1 p A In this case, we can work out the equilibrium frequency of the mutation: the equilibrium is between the rate at which the mutant gene arises by recurrent mutation, and its elimination by natural selection. Eventually, it will be lost from the population. μ {\displaystyle q} when rate of replacement (by mutation) balances rate of removal (by selection). + be the frequencies of the corresponding genotypes. s {\displaystyle p} B because the mutation rate is so low that {\displaystyle q\approx \mu /hs} 2014. q by an amount 1 = μ , which has a small relative fitness disadvantage of Selection against deleterious dominant Increase in frequency due to mutation Because selection and mutation are opposing forces, they balance each other to create an equilibrium Or, , If we assume that the mutant is rare, then q 2 is very small and all term with q 2 go to zero Also, if mutant is rare, then q m is vanishingly small indicates no dominance). . p For simplicity, suppose that mating is random. and {\displaystyle 1} h measuring the degree of dominance ( , while mutation creates more deleterious alleles increasing {\displaystyle 0} 1 {\displaystyle s} New alleles will be formed by mutation at some rate µ per generation. h and selection acts on heterozygotes with selection coefficient = s The frequency {\displaystyle q} ) and the mutation rate is not very high, the equilibrium frequency of the deleterious allele will be small.
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2021-01-16 06:40:49
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https://learn-staging.adacore.com/courses/intro-to-ada/chapters/standard_library.html
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# Standard library¶
Section author: Gustavo A. Hoffmann
## Containers¶
Ada's standard library includes support for containers (such as vectors and sets). We present an introduction to them here. For a list of all containers available in Ada, see Appendix B.
### Vectors¶
In the following sections, we present a general overview of vectors, including instantiation, initialization, and operations on vector elements and vectors.
#### Instantiation¶
Here's an example showing the instantiation and declaration of a vector V:
with Ada.Containers.Vectors; procedure Show_Vector_Inst is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); V : Integer_Vectors.Vector; begin null; end Show_Vector_Inst;
Containers are based on generic packages, so we can't simply declare a vector as we would declare an array of a specific type:
A : array (1 .. 10) of Integer;
Instead, we first need to instantiate one of those packages. We with the container package (Ada.Containers.Vectors in this case) and instantiate it to create an instance of the generic package for the desired type. Only then can we declare the vector using the type from the instantiated package. This instantiation needs to be done for any container type from the standard library.
In the instantiation of Integer_Vectors, we indicate that the vector contains elements of Integer type by specifying it as the Element_Type. By setting Index_Type to Natural, we specify that the allowed range includes all natural numbers. We could have used a more restrictive range if desired.
#### Initialization¶
One way to initialize a vector is from a concatenation of elements. We use the & operator, as shown in the following example:
with Ada.Containers; use Ada.Containers; with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Vector_Init is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; V : Vector := 20 & 10 & 0 & 13; begin Put_Line ("Vector has " & Count_Type'Image (V.Length) & " elements"); end Show_Vector_Init;
We specify use Integer_Vectors, so we have direct access to the types and operations from the instantiated package. Also, the example introduces another operation on the vector: Length, which retrieves the number of elements in the vector. We can use the dot notation because Vector is a tagged type, allowing us to write either V.Length or Length (V).
#### Appending and prepending elements¶
You add elements to a vector using the Prepend and Append operations. As the names suggest, these operations add elements to the beginning or end of a vector, respectively. For example:
with Ada.Containers; use Ada.Containers; with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Vector_Append is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; V : Vector; begin Put_Line ("Appending some elements to the vector..."); V.Append (20); V.Append (10); V.Append (0); V.Append (13); Put_Line ("Finished appending."); Put_Line ("Prepending some elements to the vector..."); V.Prepend (30); V.Prepend (40); V.Prepend (100); Put_Line ("Finished prepending."); Put_Line ("Vector has " & Count_Type'Image (V.Length) & " elements"); end Show_Vector_Append;
This example puts elements into the vector in the following sequence: (100, 40, 30, 20, 10, 0, 13).
The Reference Manual specifies that the worst-case complexity must be:
• O($$log N$$) for the Append operation, and
• O($$N log N$$) for the Prepend operation.
#### Accessing first and last elements¶
We access the first and last elements of a vector using the First_Element and Last_Element functions. For example:
with Ada.Containers; use Ada.Containers; with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Vector_First_Last_Element is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; function Img (I : Integer) return String renames Integer'Image; function Img (I : Count_Type) return String renames Count_Type'Image; V : Vector := 20 & 10 & 0 & 13; begin Put_Line ("Vector has " & Img (V.Length) & " elements"); -- Using V.First_Element to retrieve first element Put_Line ("First element is " & Img (V.First_Element)); -- Using V.Last_Element to retrieve last element Put_Line ("Last element is " & Img (V.Last_Element)); end Show_Vector_First_Last_Element;
You can swap elements by calling the procedure Swap and retrieving a reference (a cursor) to the first and last elements of the vector by calling First and Last. A cursor allows us to iterate over a container and process individual elements from it.
With these operations, we're able to write code to swap the first and last elements of a vector:
with Ada.Containers; use Ada.Containers; with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Vector_First_Last_Element is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; function Img (I : Integer) return String renames Integer'Image; V : Vector := 20 & 10 & 0 & 13; begin -- We use V.First and V.Last to retrieve cursor for first and -- last elements. -- We use V.Swap to swap elements. V.Swap (V.First, V.Last); Put_Line ("First element is now " & Img (V.First_Element)); Put_Line ("Last element is now " & Img (V.Last_Element)); end Show_Vector_First_Last_Element;
#### Iterating¶
The easiest way to iterate over a container is to use a for E of Our_Container loop. This gives us a reference (E) to the element at the current position. We can then use E directly. For example:
with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Vector_Iteration is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; function Img (I : Integer) return String renames Integer'Image; V : Vector := 20 & 10 & 0 & 13; begin Put_Line ("Vector elements are: "); -- -- Using for ... of loop to iterate: -- for E of V loop Put_Line ("- " & Img (E)); end loop; end Show_Vector_Iteration;
This code displays each element from the vector V.
Because we're given a reference, we can display not only the value of an element but also modify it. For example, we could easily write a loop to add one to each element of vector V:
for E of V loop
E := E + 1;
end loop;
We can also use indices to access vector elements. The format is similar to a loop over array elements: we use a for I in <range> loop. The range is provided by V.First_Index and V.Last_Index. We can access the current element by using it as an array index: V (I). For example:
with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Vector_Index_Iteration is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; V : Vector := 20 & 10 & 0 & 13; begin Put_Line ("Vector elements are: "); -- -- Using indices in a "for I in ..." loop to iterate: -- for I in V.First_Index .. V.Last_Index loop -- Displaying current index I Put ("- [" & Extended_Index'Image (I) & "] "); Put (Integer'Image (V (I))); -- We could also use the V.Element (I) function to retrieve the -- element at the current index I New_Line; end loop; end Show_Vector_Index_Iteration;
Here, in addition to displaying the vector elements, we're also displaying each index, I, just like what we can do for array indices. Also, we can access the element by using either the short form V (I) or the longer form V.Element (I) but not V.I.
As mentioned in the previous section, you can use cursors to iterate over containers. For this, use the function Iterate, which retrieves a cursor for each position in the vector. The corresponding loop has the format for C in V.Iterate loop. Like the previous example using indices, you can again access the current element by using the cursor as an array index: V (C). For example:
with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Vector_Cursor_Iteration is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; V : Vector := 20 & 10 & 0 & 13; begin Put_Line ("Vector elements are: "); -- -- Use a cursor to iterate in a loop: -- for C in V.Iterate loop -- Using To_Index function to retrieve index -- for the cursor position Put ("- [" & Extended_Index'Image (To_Index (C)) & "] "); Put (Integer'Image (V (C))); -- We could use Element (C) to retrieve the vector -- element for the cursor position New_Line; end loop; -- Alternatively, we could iterate with a while-loop: -- -- declare -- C : Cursor := V.First; -- begin -- while C /= No_Element loop -- some processing here... -- -- C := Next (C); -- end loop; -- end; end Show_Vector_Cursor_Iteration;
Instead of accessing an element in the loop using V (C), we could also have used the longer form Element (C). In this example, we're using the function To_Index to retrieve the index corresponding to the current cursor.
As shown in the comments after the loop, we could also use a while ... loop to iterate over the vector. In this case, we would start with a cursor for the first element (retrieved by calling V.First) and then call Next (C) to retrieve a cursor for subsequent elements. Next (C) returns No_Element when the cursor reaches the end of the vector.
You can directly modify the elements using a reference. This is what it looks like when using both indices and cursors:
-- Modify vector elements using index
for I in V.First_Index .. V.Last_Index loop
V (I) := V (I) + 1;
end loop;
-- Modify vector elements using cursor
for C in V.Iterate loop
V (C) := V (C) + 1;
end loop;
The Reference Manual requires that the worst-case complexity for accessing an element be O($$log N$$).
Another way of modifing elements of a vector is using a process procedure, which takes an individual element and does some processing on it. You can call Update_Element and pass both a cursor and an access to the process procedure. For example:
with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Vector_Update is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; procedure Add_One (I : in out Integer) is begin I := I + 1; end Add_One; V : Vector := 20 & 10 & 12; begin -- -- Use V.Update_Element to process elements -- for C in V.Iterate loop V.Update_Element (C, Add_One'Access); end loop; end Show_Vector_Update;
#### Finding and changing elements¶
You can locate a specific element in a vector by retrieving its index. Find_Index retrieves the index of the first element matching the value you're looking for. Alternatively, you can use Find to retrieve a cursor referencing that element. For example:
with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Find_Vector_Element is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; V : Vector := 20 & 10 & 0 & 13; Idx : Extended_Index; C : Cursor; begin -- Using Find_Index to retrieve index of element with value 10 Idx := V.Find_Index (10); Put_Line ("Index of element with value 10 is " & Extended_Index'Image (Idx)); -- Using Find to retrieve cursor for element with value 13 C := V.Find (13); Idx := To_Index (C); Put_Line ("Index of element with value 13 is " & Extended_Index'Image (Idx)); end Show_Find_Vector_Element;
As we saw in the previous section, we can directly access vector elements by using either an index or cursor. However, an exception is raised if we try to access an element with an invalid index or cursor, so we must check whether the index or cursor is valid before using it to access an element. In our example, Find_Index or Find might not have found the element in the vector. We check for this possibility by comparing the index to No_Index or the cursor to No_Element. For example:
-- Modify vector element using index
if Idx /= No_Index then
V (Idx) := 11;
end if;
-- Modify vector element using cursor
if C /= No_Element then
V (C) := 14;
end if;
Instead of writing V (C) := 14, we could use the longer form V.Replace_Element (C, 14).
#### Inserting elements¶
In the previous sections, we've seen examples of how to add elements to a vector:
• using the concatenation operator (&) at the vector declaration, or
• calling the Prepend and Append procedures.
You may want to insert an element at a specific position, e.g. before a certain element in the vector. You do this by calling Insert. For example:
with Ada.Containers; use Ada.Containers; with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Vector_Insert is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; procedure Show_Elements (V : Vector) is begin New_Line; Put_Line ("Vector has " & Count_Type'Image (V.Length) & " elements"); if not V.Is_Empty then Put_Line ("Vector elements are: "); for E of V loop Put_Line ("- " & Integer'Image (E)); end loop; end if; end Show_Elements; V : Vector := 20 & 10 & 12; C : Cursor; begin Show_Elements (V); New_Line; Put_Line ("Adding element with value 9 (before 10)..."); -- -- Using V.Insert to insert element into vector -- C := V.Find (10); if C /= No_Element then V.Insert (C, 9); end if; Show_Elements (V); end Show_Vector_Insert;
In this example, we're looking for an element with the value of 10. If we find it, we insert an element with the value of 9 before it.
#### Removing elements¶
You can remove elements from a vector by passing either a valid index or cursor to the Delete procedure. If we combine this with the functions Find_Index and Find from the previous section, we can write a program that searches for a specific element and deletes it, if found:
with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Remove_Vector_Element is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; V : Vector := 20 & 10 & 0 & 13 & 10 & 13; Idx : Extended_Index; C : Cursor; begin -- Use Find_Index to retrieve index of element with value 10 Idx := V.Find_Index (10); -- Checking whether index is valid if Idx /= No_Index then -- Removing element using V.Delete V.Delete (Idx); end if; -- Use Find to retrieve cursor for element with value 13 C := V.Find (13); -- Check whether index is valid if C /= No_Element then -- Remove element using V.Delete V.Delete (C); end if; end Show_Remove_Vector_Element;
We can extend this approach to delete all elements matching a certain value. We just need to keep searching for the element in a loop until we get an invalid index or cursor. For example:
with Ada.Containers; use Ada.Containers; with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Remove_Vector_Elements is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); use Integer_Vectors; procedure Show_Elements (V : Vector) is begin New_Line; Put_Line ("Vector has " & Count_Type'Image (V.Length) & " elements"); if not V.Is_Empty then Put_Line ("Vector elements are: "); for E of V loop Put_Line ("- " & Integer'Image (E)); end loop; end if; end Show_Elements; V : Vector := 20 & 10 & 0 & 13 & 10 & 14 & 13; begin Show_Elements (V); -- -- Remove elements using an index -- declare E : constant Integer := 10; I : Extended_Index; begin New_Line; Put_Line ("Removing all elements with value of " & Integer'Image (E) & "..."); loop I := V.Find_Index (E); exit when I = No_Index; V.Delete (I); end loop; end; -- -- Remove elements using a cursor -- declare E : constant Integer := 13; C : Cursor; begin New_Line; Put_Line ("Removing all elements with value of " & Integer'Image (E) & "..."); loop C := V.Find (E); exit when C = No_Element; V.Delete (C); end loop; end; Show_Elements (V); end Show_Remove_Vector_Elements;
In this example, we remove all elements with the value 10 from the vector by retrieving their index. Likewise, we remove all elements with the value 13 by retrieving their cursor.
#### Other Operations¶
We've seen some operations on vector elements. Here, we'll see operations on the vector as a whole. The most prominent is the concatenation of multiple vectors, but we'll also see operations on vectors, such as sorting and sorted merging operations, that view the vector as a sequence of elements and operate on the vector considering the element's relations to each other.
We do vector concatenation using the & operator on vectors. Let's consider two vectors V1 and V2. We can concatenate them by doing V := V1 & V2. V contains the resulting vector.
The generic package Generic_Sorting is a child package of Ada.Containers.Vectors. It contains sorting and merging operations. Because it's a generic package, you can't use it directly, but have to instantiate it. In order to use these operations on a vector of integer values (Integer_Vectors, in our example), you need to instantiate it directly as a child of Integer_Vectors. The next example makes it clear how to do this.
After instantiating Generic_Sorting, we make all the operations available to us with the use statement. We can then call Sort to sort the vector and Merge to merge one vector into another.
The following example presents code that manipulates three vectors (V1, V2, V3) using the concatenation, sorting and merging operations:
with Ada.Containers; use Ada.Containers; with Ada.Containers.Vectors; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Vector_Ops is package Integer_Vectors is new Ada.Containers.Vectors (Index_Type => Natural, Element_Type => Integer); package Integer_Vectors_Sorting is new Integer_Vectors.Generic_Sorting; use Integer_Vectors; use Integer_Vectors_Sorting; procedure Show_Elements (V : Vector) is begin New_Line; Put_Line ("Vector has " & Count_Type'Image (V.Length) & " elements"); if not V.Is_Empty then Put_Line ("Vector elements are: "); for E of V loop Put_Line ("- " & Integer'Image (E)); end loop; end if; end Show_Elements; V, V1, V2, V3 : Vector; begin V1 := 10 & 12 & 18; V2 := 11 & 13 & 19; V3 := 15 & 19; New_Line; Put_Line ("---- V1 ----"); Show_Elements (V1); New_Line; Put_Line ("---- V2 ----"); Show_Elements (V2); New_Line; Put_Line ("---- V3 ----"); Show_Elements (V3); New_Line; Put_Line ("Concatenating V1, V2 and V3 into V:"); V := V1 & V2 & V3; Show_Elements (V); New_Line; Put_Line ("Sorting V:"); Sort (V); Show_Elements (V); New_Line; Put_Line ("Merging V2 into V1:"); Merge (V1, V2); Show_Elements (V1); end Show_Vector_Ops;
The Reference Manual requires that the worst-case complexity of a call to Sort be O($$N^2$$) and the average complexity be better than O($$N^2$$).
### Sets¶
Sets are another class of containers. While vectors allow duplicated elements to be inserted, sets ensure that no duplicated elements exist.
In the following sections, we'll see operations you can perform on sets. However, since many of the operations on vectors are similar to the ones used for sets, we'll cover them more quickly here. Please refer back to the section on vectors for a more detailed discussion.
#### Initialization and iteration¶
To initialize a set, you can call the Insert procedure. However, if you do, you need to ensure no duplicate elements are being inserted: if you try to insert a duplicate, you'll get an exception. If you have less control over the elements to be inserted so that there may be duplicates, you can use another option instead:
• a version of Insert that returns a Boolean value indicating whether the insertion was successful;
• the Include procedure, which silently ignores any attempt to insert a duplicated element.
To iterate over a set, you can use a for E of S loop, as you saw for vectors. This gives you a reference to each element in the set.
Let's see an example:
#### Operations on elements¶
In this section, we briefly explore the following operations on sets:
• Delete and Exclude to remove elements;
• Contains and Find to verify the existence of elements.
To delete elements, you call the procedure Delete. However, analogously to the Insert procedure above, Delete raises an exception if the element to be deleted isn't present in the set. If you want to permit the case where an element might not exist, you can call Exclude, which silently ignores any attempt to delete a non-existent element.
Contains returns a Boolean value indicating whether a value is contained in the set. Find also looks for an element in a set, but returns a cursor to the element or No_Element if the element doesn't exist. You can use either function to search for elements in a set.
Let's look at an example that makes use of these operations:
with Ada.Containers; use Ada.Containers; with Ada.Containers.Ordered_Sets; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Set_Element_Ops is package Integer_Sets is new Ada.Containers.Ordered_Sets (Element_Type => Integer); use Integer_Sets; procedure Show_Elements (S : Set) is begin New_Line; Put_Line ("Set has " & Count_Type'Image (S.Length) & " elements"); Put_Line ("Elements:"); for E of S loop Put_Line ("- " & Integer'Image (E)); end loop; end Show_Elements; S : Set; begin S.Insert (20); S.Insert (10); S.Insert (0); S.Insert (13); S.Delete (13); -- Calling S.Delete (13) again raises Constraint_Error -- because the element is no longer present -- in the set, so it can't be deleted. -- We can call V.Exclude instead: S.Exclude (13); if S.Contains (20) then Put_Line ("Found element 20 in set"); end if; -- Alternatively, we could use S.Find instead of S.Contains if S.Find (0) /= No_Element then Put_Line ("Found element 0 in set"); end if; Show_Elements (S); end Show_Set_Element_Ops;
In addition to ordered sets used in the examples above, the standard library also offers hashed sets. The Reference Manual requires the following average complexity of each operation:
Operations Ordered_Sets Hashed_Sets
• Insert
• Include
• Replace
• Delete
• Exclude
• Find
O($$(log N)^2)$$ or better $$O(log N)$$
Subprogram using cursor O($$1$$) O($$1$$)
#### Other Operations¶
The previous sections mostly dealt with operations on individual elements of a set. But Ada also provides typical set operations: union, intersection, difference and symmetric difference. In contrast to some vector operations we've seen before (e.g. Merge), here you can use built-in operators, such as -. The following table lists the operations and its associated operator:
Set Operation Operator
Union and
Intersection or
Difference -
Symmetric difference xor
The following example makes use of these operators:
with Ada.Containers; use Ada.Containers; with Ada.Containers.Ordered_Sets; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Set_Ops is package Integer_Sets is new Ada.Containers.Ordered_Sets (Element_Type => Integer); use Integer_Sets; procedure Show_Elements (S : Set) is begin Put_Line ("Elements:"); for E of S loop Put_Line ("- " & Integer'Image (E)); end loop; end Show_Elements; procedure Show_Op (S : Set; Op_Name : String) is begin New_Line; Put_Line (Op_Name & "(set #1, set #2) has " & Count_Type'Image (S.Length) & " elements"); end Show_Op; S1, S2, S3 : Set; begin S1.Insert (0); S1.Insert (10); S1.Insert (13); S2.Insert (0); S2.Insert (10); S2.Insert (14); S3.Insert (0); S3.Insert (10); New_Line; Put_Line ("---- Set #1 ----"); Show_Elements (S1); New_Line; Put_Line ("---- Set #2 ----"); Show_Elements (S2); New_Line; Put_Line ("---- Set #3 ----"); Show_Elements (S3); New_Line; if S3.Is_Subset (S1) then Put_Line ("S3 is a subset of S1"); else Put_Line ("S3 is not a subset of S1"); end if; S3 := S1 and S2; Show_Op (S3, "Union"); Show_Elements (S3); S3 := S1 or S2; Show_Op (S3, "Intersection"); Show_Elements (S3); S3 := S1 - S2; Show_Op (S3, "Difference"); Show_Elements (S3); S3 := S1 xor S2; Show_Op (S3, "Symmetric difference"); Show_Elements (S3); end Show_Set_Ops;
### Indefinite maps¶
The previous sections presented containers for elements of definite types. Although most examples in those sections presented Integer types as element type of the containers, containers can also be used with indefinite types, an example of which is the String type. However, indefinite types require a different kind of containers designed specially for them.
We'll also be exploring a different class of containers: maps. They associate a key with a specific value. An example of a map is the one-to-one association between a person and their age. If we consider a person's name to be the key, the value is the person's age.
#### Hashed maps¶
Hashed maps are maps that make use of a hash as a key. The hash itself is calculated by a function you provide.
In other languages
Hashed maps are similar to dictionaries in Python and hashes in Perl. One of the main differences is that these scripting languages allow using different types for the values contained in a single map, while in Ada, both the type of key and value are specified in the package instantiation and remains constant for that specific map. You can't have a map where two elements are of different types or two keys are of different types. If you want to use multiple types, you must create a different map for each and use only one type in each map.
When instantiating a hashed map from Ada.Containers.Indefinite_Hashed_Maps, we specify following elements:
• Key_Type: type of the key
• Element_Type: type of the element
• Hash: hash function for the Key_Type
• Equivalent_Keys: an equality operator (e.g. =) that indicates whether two keys are to be considered equal.
• If the type specified in Key_Type has a standard operator, you can use it, which you do by specifing using that operator as the value of Equivalent_Keys.
In the next example, we'll use a string as a key type. We'll use the Hash function provided by the standard library for strings (in the Ada.Strings package) and the standard equality operator.
You add elements to a hashed map by calling Insert. If an element is already contained in a map M, you can access it directly by using its key. For example, you can change the value of an element by calling M ("My_Key") := 10. If the key is not found, an exception is raised. To verify if a key is available, use the function Contains (as we've seen above in the section on sets).
Let's see an example:
with Ada.Containers.Indefinite_Hashed_Maps; with Ada.Strings.Hash; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Hashed_Map is package Integer_Hashed_Maps is new Ada.Containers.Indefinite_Hashed_Maps (Key_Type => String, Element_Type => Integer, Hash => Ada.Strings.Hash, Equivalent_Keys => "="); use Integer_Hashed_Maps; M : Map; -- Same as: M : Integer_Hashed_Maps.Map; begin M.Include ("Alice", 24); M.Include ("John", 40); M.Include ("Bob", 28); if M.Contains ("Alice") then Put_Line ("Alice's age is " & Integer'Image (M ("Alice"))); end if; -- Update Alice's age -- Key must already exist in M. -- Otherwise an exception is raised. M ("Alice") := 25; New_Line; Put_Line ("Name & Age:"); for C in M.Iterate loop Put_Line (Key (C) & ": " & Integer'Image (M (C))); end loop; end Show_Hashed_Map;
#### Ordered maps¶
Ordered maps share many features with hashed maps. The main differences are:
• A hash function isn't needed. Instead, you must provide an ordering function (< operator), which the ordered map will use to order elements and allow fast access, $$O(log n)$$, using a binary search.
• If the type specified in Key_Type has a standard < operator, you can use it in a similar way as we did for Equivalent_Keys above for hashed maps.
Let's see an example:
with Ada.Containers.Indefinite_Ordered_Maps; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Ordered_Map is package Integer_Ordered_Maps is new Ada.Containers.Indefinite_Ordered_Maps (Key_Type => String, Element_Type => Integer); use Integer_Ordered_Maps; M : Map; begin M.Include ("Alice", 24); M.Include ("John", 40); M.Include ("Bob", 28); if M.Contains ("Alice") then Put_Line ("Alice's age is " & Integer'Image (M ("Alice"))); end if; -- Update Alice's age -- Key must already exist in M M ("Alice") := 25; New_Line; Put_Line ("Name & Age:"); for C in M.Iterate loop Put_Line (Key (C) & ": " & Integer'Image (M (C))); end loop; end Show_Ordered_Map;
You can see a great similarity between the examples above and from the previous section. In fact, since both kinds of maps share many operations, we didn't need to make extensive modifications when we changed our example to use ordered maps instead of hashed maps. The main difference is seen when we run the examples: the output of a hashed map is usually unordered, but the output of a ordered map is always ordered, as implied by its name.
#### Complexity¶
Hashed maps are generally the fastest data structure available to you in Ada if you need to associate heterogeneous keys to values and search for them quickly. In most cases, they are slightly faster than ordered maps. So if you don't need ordering, use hashed maps.
The Reference Manual requires the following average complexity of operations:
Operations Ordered_Maps Hashed_Maps
• Insert
• Include
• Replace
• Delete
• Exclude
• Find
O($$(log N)^2)$$ or better $$O(log N)$$
Subprogram using cursor O($$1$$) O($$1$$)
## Dates & Times¶
### Date and time handling¶
The standard library supports representing and handling dates and times. This is part of the Ada.Calendar package. Let's look at a simple example:
This example displays the current date and time, which is retrieved by a call to the Clock function. We call the function Image from the Ada.Calendar.Formatting package to get a String for the current date and time. We could instead retrieve each component using the Split function. For example:
with Ada.Calendar; use Ada.Calendar; with Ada.Text_IO; use Ada.Text_IO; procedure Display_Current_Year is Now : Time := Clock; Now_Year : Year_Number; Now_Month : Month_Number; Now_Day : Day_Number; Now_Seconds : Day_Duration; begin Split (Now, Now_Year, Now_Month, Now_Day, Now_Seconds); Put_Line ("Current year is: " & Year_Number'Image (Now_Year)); Put_Line ("Current month is: " & Month_Number'Image (Now_Month)); Put_Line ("Current day is: " & Day_Number'Image (Now_Day)); end Display_Current_Year;
Here, we're retrieving each element and displaying it separately.
#### Delaying using date¶
You can delay an application so that it restarts at a specific date and time. We saw something similar in the chapter on tasking. You do this using a delay until statement. For example:
with Ada.Calendar; use Ada.Calendar; with Ada.Calendar.Formatting; use Ada.Calendar.Formatting; with Ada.Calendar.Time_Zones; use Ada.Calendar.Time_Zones; with Ada.Text_IO; use Ada.Text_IO; procedure Display_Delay_Next_Specific_Time is TZ : Time_Offset := UTC_Time_Offset; Next : Time := Ada.Calendar.Formatting.Time_Of (Year => 2018, Month => 5, Day => 1, Hour => 15, Minute => 0, Second => 0, Sub_Second => 0.0, Leap_Second => False, Time_Zone => TZ); -- Next = 2018-05-01 15:00:00.00 (local time-zone) begin Put_Line ("Let's wait until..."); Put_Line (Image (Next, True, TZ)); delay until Next; Put_Line ("Enough waiting!"); end Display_Delay_Next_Specific_Time;
In this example, we specify the date and time by initializing Next using a call to Time_Of, a function taking the various components of a date (year, month, etc) and returning an element of the Time type. Because the date specified is in the past, the delay until statement won't produce any noticeable effect. However, if we passed a date in the future, the program would wait until that specific date and time arrived.
Here we're converting the time to the local timezone. If we don't specify a timezone, Coordinated Universal Time (abbreviated to UTC) is used by default. By retrieving the time offset to UTC with a call to UTC_Time_Offset from the Ada.Calendar.Time_Zones package, we can initialize TZ and use it in the call to Time_Of. This is all we need do to make the information provided to Time_Of relative to the local time zone.
We could achieve a similar result by initializing Next with a String. We can do this with a call to Value from the Ada.Calendar.Formatting package. This is the modified code:
with Ada.Calendar; use Ada.Calendar; with Ada.Calendar.Formatting; use Ada.Calendar.Formatting; with Ada.Calendar.Time_Zones; use Ada.Calendar.Time_Zones; with Ada.Text_IO; use Ada.Text_IO; procedure Display_Delay_Next_Specific_Time is TZ : Time_Offset := UTC_Time_Offset; Next : Time := Ada.Calendar.Formatting.Value ("2018-05-01 15:00:00.00", TZ); -- Next = 2018-05-01 15:00:00.00 (local time-zone) begin Put_Line ("Let's wait until..."); Put_Line (Image (Next, True, TZ)); delay until Next; Put_Line ("Enough waiting!"); end Display_Delay_Next_Specific_Time;
In this example, we're again using TZ in the call to Value to adjust the input time to the current time zone.
In the examples above, we were delaying to a specific date and time. Just like we saw in the tasking chapter, we could instead specify the delay relative to the current time. For example, we could delay by 5 seconds, using the current time:
with Ada.Calendar; use Ada.Calendar; with Ada.Text_IO; use Ada.Text_IO; procedure Display_Delay_Next is D : Duration := 5.0; -- seconds Now : Time := Clock; Next : Time := Now + D; -- use duration to -- specify next point in time begin Put_Line ("Let's wait " & Duration'Image (D) & " seconds..."); delay until Next; Put_Line ("Enough waiting!"); end Display_Delay_Next;
Here, we're specifying a duration of 5 seconds in D, adding it to the current time from Now, and storing the sum in Next. We then use it in the delay until statement.
### Real-time¶
In addition to Ada.Calendar, the standard library also supports time operations for real-time applications. These are included in the Ada.Real_Time package. This package also include a Time type. However, in the Ada.Real_Time package, the Time type is used to represent an absolute clock and handle a time span. This contrasts with the Ada.Calendar, which uses the Time type to represent dates and times.
In the previous section, we used the Time type from the Ada.Calendar and the delay until statement to delay an application by 5 seconds. We could have used the Ada.Real_Time package instead. Let's modify that example:
with Ada.Text_IO; use Ada.Text_IO; with Ada.Real_Time; use Ada.Real_Time; procedure Display_Delay_Next_Real_Time is D : Time_Span := Seconds (5); Next : Time := Clock + D; begin Put_Line ("Let's wait " & Duration'Image (To_Duration (D)) & " seconds..."); delay until Next; Put_Line ("Enough waiting!"); end Display_Delay_Next_Real_Time;
The main difference is that D is now a variable of type Time_Span, defined in the Ada.Real_Time package. We call the function Seconds to initialize D, but could have gotten a finer granularity by calling Nanoseconds instead. Also, we need to first convert D to the Duration type using To_Duration before we can display it.
#### Benchmarking¶
One interesting application using the Ada.Real_Time package is benchmarking. We've used that package before in a previous section when discussing tasking. Let's look at an example of benchmarking:
with Ada.Text_IO; use Ada.Text_IO; with Ada.Real_Time; use Ada.Real_Time; procedure Display_Benchmarking is procedure Computational_Intensive_App is begin delay 0.5; end Computational_Intensive_App; Start_Time, Stop_Time : Time; Elapsed_Time : Time_Span; begin Start_Time := Clock; Computational_Intensive_App; Stop_Time := Clock; Elapsed_Time := Stop_Time - Start_Time; Put_Line ("Elapsed time: " & Duration'Image (To_Duration (Elapsed_Time)) & " seconds"); end Display_Benchmarking;
This example defines a dummy Computational_Intensive_App implemented using a simple delay statement. We initialize Start_Time and Stop_Time from the then-current clock and calculate the elapsed time. By running this program, we see that the time is roughly 5 seconds, which is expected due to the delay statement.
A similar application is benchmarking of CPU time. We can implement this using the Execution_Time package. Let's modify the previous example to measure CPU time:
with Ada.Text_IO; use Ada.Text_IO; with Ada.Real_Time; use Ada.Real_Time; with Ada.Execution_Time; use Ada.Execution_Time; procedure Display_Benchmarking_CPU_Time is procedure Computational_Intensive_App is begin delay 0.5; end Computational_Intensive_App; Start_Time, Stop_Time : CPU_Time; Elapsed_Time : Time_Span; begin Start_Time := Clock; Computational_Intensive_App; Stop_Time := Clock; Elapsed_Time := Stop_Time - Start_Time; Put_Line ("CPU time: " & Duration'Image (To_Duration (Elapsed_Time)) & " seconds"); end Display_Benchmarking_CPU_Time;
In this example, Start_Time and Stop_Time are of type CPU_Time instead of Time. However, we still call the Clock function to initialize both variables and calculate the elapsed time in the same way as before. By running this program, we see that the CPU time is significantly lower than the 5 seconds we've seen before. This is because the delay statement doesn't require much CPU time. The results will be different if we change the implementation of Computational_Intensive_App to use a mathematical functions in a long loop. For example:
with Ada.Text_IO; use Ada.Text_IO; with Ada.Real_Time; use Ada.Real_Time; with Ada.Execution_Time; use Ada.Execution_Time; with Ada.Numerics.Generic_Elementary_Functions; procedure Display_Benchmarking_Math is procedure Computational_Intensive_App is package Funcs is new Ada.Numerics.Generic_Elementary_Functions (Float_Type => Long_Long_Float); use Funcs; X : Long_Long_Float; begin for I in 0 .. 1_000_000 loop X := Tan (Arctan (Tan (Arctan (Tan (Arctan (Tan (Arctan (Tan (Arctan (Tan (Arctan (0.577)))))))))))); end loop; end Computational_Intensive_App; procedure Benchm_Elapsed_Time is Start_Time, Stop_Time : Time; Elapsed_Time : Time_Span; begin Start_Time := Clock; Computational_Intensive_App; Stop_Time := Clock; Elapsed_Time := Stop_Time - Start_Time; Put_Line ("Elapsed time: " & Duration'Image (To_Duration (Elapsed_Time)) & " seconds"); end Benchm_Elapsed_Time; procedure Benchm_CPU_Time is Start_Time, Stop_Time : CPU_Time; Elapsed_Time : Time_Span; begin Start_Time := Clock; Computational_Intensive_App; Stop_Time := Clock; Elapsed_Time := Stop_Time - Start_Time; Put_Line ("CPU time: " & Duration'Image (To_Duration (Elapsed_Time)) & " seconds"); end Benchm_CPU_Time; begin Benchm_Elapsed_Time; Benchm_CPU_Time; end Display_Benchmarking_Math;
Now that our dummy Computational_Intensive_App involves mathematical operations requiring significant CPU time, the measured elapsed and CPU time are much closer to each other than before.
## Strings¶
We've used strings in many previous examples. In this section, we'll cover them in more detail.
### String operations¶
Operations on standard strings are available in the Ada.Strings.Fixed package. As mentioned previously, standard strings are arrays of elements of Character type with a fixed-length. That's why this child package is called Fixed.
One of the simplest operations provided is counting the number of substrings available in a string (Count) and finding their corresponding indices (Index). Let's look at an example:
with Ada.Strings.Fixed; use Ada.Strings.Fixed; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Find_Substring is S : String := "Hello" & 3 * " World"; P : constant String := "World"; Idx : Natural; Cnt : Natural; begin Cnt := Ada.Strings.Fixed.Count (Source => S, Pattern => P); Put_Line ("String: " & S); Put_Line ("Count for '" & P & "': " & Natural'Image (Cnt)); Idx := 0; for I in 1 .. Cnt loop Idx := Index (Source => S, Pattern => P, From => Idx + 1); Put_Line ("Found instance of '" & P & "' at position: " & Natural'Image (Idx)); end loop; end Show_Find_Substring;
We initialize the string S using a multiplication. Writing "Hello" & 3 * " World" creates the string Hello World World World. We then call the function Count to get the number of instances of the word World in S. Next we call the function Index in a loop to find the index of each instance of World in S.
That example looked for instances of a specific substring. In the next example, we retrieve all the words in the string. We do this using Find_Token and specifying whitespaces as separators. For example:
with Ada.Strings; use Ada.Strings; with Ada.Strings.Fixed; use Ada.Strings.Fixed; with Ada.Strings.Maps; use Ada.Strings.Maps; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Find_Words is S : String := "Hello" & 3 * " World"; F : Positive; L : Natural; I : Natural := 1; Whitespace : constant Character_Set := To_Set (' '); begin Put_Line ("String: " & S); Put_Line ("String length: " & Integer'Image (S'Length)); while I in S'Range loop Find_Token (Source => S, Set => Whitespace, From => I, Test => Outside, First => F, Last => L); exit when L = 0; Put_Line ("Found word instance at position " & Natural'Image (F) & ": '" & S (F .. L) & "'"); -- & "-" & F'Img & "-" & L'Img I := L + 1; end loop; end Show_Find_Words;
We pass a set of characters to be used as delimitators to the procedure Find_Token. This set is a member of the Character_Set type from the Ada.Strings.Maps package. We call the To_Set function (from the same package) to initialize the set to Whitespace and then call Find_Token to loop over each valid index and find the starting index of each word. We pass Outside to the Test parameter of the Find_Token procedure to indicate that we're looking for indices that are outside the Whitespace set, i.e. actual words. The First and Last parameters of Find_Token are output parameters that indicate the valid range of the substring. We use this information to display the string (S (F .. L)).
The operations we've looked at so far read strings, but don't modify them. We next discuss operations that change the content of strings:
Operation Description
Insert Insert substring in a string
Overwrite Overwrite a string with a substring
Delete Delete a substring
Trim Remove whitespaces from a string
All these operations are available both as functions or procedures. Functions create a new string but procedures perform the operations in place. The procedure will raise an exception if the constraints of the string are not satisfied. For example, if we have a string S containing 10 characters, inserting a string with two characters (e.g. "!!") into it produces a string containing 12 characters. Since it has a fixed length, we can't increase its size. One possible solution in this case is to specify that truncation should be applied while inserting the substring. This keeps the length of S fixed. Let's see an example that makes use of both function and procedure versions of Insert, Overwrite, and Delete:
In this example, we look for the index of the substring World and perform operations on this substring within the outer string. The procedure Display_Adapted_String uses both versions of the operations. For the procedural version of Insert and Overwrite, we apply truncation to the right side of the string (Right). For the Delete procedure, we specify the range of the substring, which is replaced by whitespaces. For the function version of Delete, we also call Trim which trims the trailing whitespace.
### Bounded and unbounded strings¶
Using fixed-length strings is usually good enough for strings that are initialized when they are declared. However, as seen in the previous section, procedural operations on strings cause difficulties when done on fixed-length strings because fixed-length strings are arrays of characters. The following example shows how cumbersome the initialization of fixed-length strings can be when it's not performed in the declaration:
with Ada.Text_IO; use Ada.Text_IO; procedure Show_Char_Array is S : String (1 .. 15); -- Strings are arrays of Character begin S := "Hello "; -- Alternatively: -- -- #1: -- S (1 .. 5) := "Hello"; -- S (6 .. S'Last) := (others => ' '); -- -- #2: -- S := ('H', 'e', 'l', 'l', 'o', others => ' '); Put_Line ("String: " & S); Put_Line ("String Length: " & Integer'Image (S'Length)); end Show_Char_Array;
In this case, we can't simply write S := "Hello" because the resulting array of characters for the Hello constant has a different length than the S string. Therefore, we need to include trailing whitespaces to match the length of S. As shown in the example, we could use an exact range for the initialization ( S (1 .. 5)) or use an explicit array of individual characters.
When strings are initialized or manipulated at run-time, it's usually better to use bounded or unbounded strings. An important feature of these types is that they aren't arrays, so the difficulties presented above don't apply. Let's start with bounded strings.
#### Bounded strings¶
Bounded strings are defined in the Ada.Strings.Bounded.Generic_Bounded_Length package. Because this is a generic package, you need to instantiate it and set the maximum length of the bounded string. You can then declare bounded strings of the Bounded_String type.
Both bounded and fixed-length strings have a maximum length that they can hold. However, bounded strings are not arrays, so initializing them at run-time is much easier. For example:
with Ada.Strings; use Ada.Strings; with Ada.Strings.Bounded; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Bounded_String is package B_Str is new Ada.Strings.Bounded.Generic_Bounded_Length (Max => 15); use B_Str; S1, S2 : Bounded_String; procedure Display_String_Info (S : Bounded_String) is begin Put_Line ("String: " & To_String (S)); Put_Line ("String Length: " & Integer'Image (Length (S))); -- String: S'Length => ok -- Bounded_String: S'Length => compilation error -- bounded strings are not arrays! Put_Line ("Max. Length: " & Integer'Image (Max_Length)); end Display_String_Info; begin S1 := To_Bounded_String ("Hello"); Display_String_Info (S1); S2 := To_Bounded_String ("Hello World"); Display_String_Info (S2); S1 := To_Bounded_String ("Something longer to say here...", Right); Display_String_Info (S1); end Show_Bounded_String;
By using bounded strings, we can easily assign to S1 and S2 multiple times during execution. We use the To_Bounded_String and To_String functions to convert, in the respective direction, between fixed-length and bounded strings. A call to To_Bounded_String raises an exception if the length of the input string is greater than the maximum capacity of the bounded string. To avoid this, we can use the truncation parameter (Right in our example).
Bounded strings are not arrays, so we can't use the 'Length attribute as we did for fixed-length strings. Instead, we call the Length function, which returns the length of the bounded string. The Max_Length constant represents the maximum length of the bounded string that we set when we instantiated the package.
After initializing a bounded string, we can manipulate it. For example, we can append a string to a bounded string using Append or concatenate bounded strings using the & operator. Like so:
with Ada.Strings; use Ada.Strings; with Ada.Strings.Bounded; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Bounded_String_Op is package B_Str is new Ada.Strings.Bounded.Generic_Bounded_Length (Max => 30); use B_Str; S1, S2 : Bounded_String; begin S1 := To_Bounded_String ("Hello"); -- Alternatively: A := Null_Bounded_String & "Hello"; Append (S1, " World"); -- Alternatively: Append (A, " World", Right); Put_Line ("String: " & To_String (S1)); S2 := To_Bounded_String ("Hello!"); S1 := S1 & " " & S2; Put_Line ("String: " & To_String (S1)); end Show_Bounded_String_Op;
We can initialize a bounded string with an empty string using the Null_Bounded_String constant. Also, we can use the Append procedure and specify the truncation mode like we do with the To_Bounded_String function.
#### Unbounded strings¶
Unbounded strings are defined in the Ada.Strings.Unbounded package. This is not a generic package, so we don't need to instantiate it before using the Unbounded_String type. As you may recall from the previous section, bounded strings require a package instantiation.
Unbounded strings are similar to bounded strings. The main difference is that they can hold strings of any size and adjust according to the input string: if we assign, e.g., a 10-character string to an unbounded string and later assign a 50-character string, internal operations in the container ensure that memory is allocated to store the new string. In most cases, developers don't need to worry about these operations. Also, no truncation is necessary.
Initialization of unbounded strings is very similar to bounded strings. Let's look at an example:
with Ada.Strings; use Ada.Strings; with Ada.Strings.Unbounded; use Ada.Strings.Unbounded; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Unbounded_String is S1, S2 : Unbounded_String; procedure Display_String_Info (S : Unbounded_String) is begin Put_Line ("String: " & To_String (S)); Put_Line ("String Length: " & Integer'Image (Length (S))); end Display_String_Info; begin S1 := To_Unbounded_String ("Hello"); -- Alternatively: A := Null_Unbounded_String & "Hello"; Display_String_Info (S1); S2 := To_Unbounded_String ("Hello World"); Display_String_Info (S2); S1 := To_Unbounded_String ("Something longer to say here..."); Display_String_Info (S1); end Show_Unbounded_String;
Like bounded strings, we can assign to S1 and S2 multiple times during execution and use the To_Unbounded_String and To_String functions to convert back-and-forth between fixed-length strings and unbounded strings. However, in this case, truncation is not needed.
And, just like for bounded strings, you can use the Append function and the & operator for unbounded strings. For example:
with Ada.Strings.Unbounded; use Ada.Strings.Unbounded; with Ada.Text_IO; use Ada.Text_IO; procedure Show_Unbounded_String_Op is S1, S2 : Unbounded_String := Null_Unbounded_String; begin S1 := S1 & "Hello"; S2 := S2 & "Hello!"; Append (S1, " World"); Put_Line ("String: " & To_String (S1)); S1 := S1 & " " & S2; Put_Line ("String: " & To_String (S1)); end Show_Unbounded_String_Op;
## Files and streams¶
This section presents the different options available in Ada for file input/output (I/O).
### Text I/O¶
In most parts of this course, we used the Put_Line procedure to display information on the console. However, this procedure also accepts a File_Type parameter. For example, you can select between standard output and standard error by setting this parameter explicitly:
with Ada.Text_IO; use Ada.Text_IO; procedure Show_Std_Text_Out is begin Put_Line (Standard_Output, "Hello World #1"); Put_Line (Standard_Error, "Hello World #2"); end Show_Std_Text_Out;
You can also use this parameter to write information to any text file. To create a new file for writing, use the Create procedure, which initializes a File_Type element that you can later pass to Put_Line (instead of, e.g., Standard_Output). After you finish writing information, you can close the file by calling the Close procedure.
You use a similar method to read information from a text file. However, when opening the file, you must specify that it's an input file (In_File) instead of an output file. Also, instead of calling the Put_Line procedure, you call the Get_Line function to read information from the file.
Let's see an example that writes information into a new text file and then reads it back from the same file:
with Ada.Text_IO; use Ada.Text_IO;
procedure Show_Simple_Text_File_IO is
F : File_Type;
File_Name : constant String := "simple.txt";
begin
Create (F, Out_File, File_Name);
Put_Line (F, "Hello World #1");
Put_Line (F, "Hello World #2");
Put_Line (F, "Hello World #3");
Close (F);
Open (F, In_File, File_Name);
while not End_Of_File (F) loop
Put_Line (Get_Line (F));
end loop;
Close (F);
end Show_Simple_Text_File_IO;
In addition to the Create and Close procedures, the standard library also includes a Reset procedure, which, as the name implies, resets (erases) all the information from the file. For example:
with Ada.Text_IO; use Ada.Text_IO;
procedure Show_Text_File_Reset is
F : File_Type;
File_Name : constant String := "simple.txt";
begin
Create (F, Out_File, File_Name);
Put_Line (F, "Hello World #1");
Reset (F);
Put_Line (F, "Hello World #2");
Close (F);
Open (F, In_File, File_Name);
while not End_Of_File (F) loop
Put_Line (Get_Line (F));
end loop;
Close (F);
end Show_Text_File_Reset;
By running this program, we notice that, although we've written the first string (Hello World #1) to the file, it has been erased because of the call to Reset.
In addition to opening a file for reading or writing, you can also open an existing file and append to it. Do this by calling the Open procedure with the Append_File option.
When calling the Open procedure, an exception is raised if the specified file isn't found. Therefore, you should handle exceptions in that context. The following example deletes a file and then tries to open the same file for reading:
with Ada.Text_IO; use Ada.Text_IO;
procedure Show_Text_File_Input_Except is
F : File_Type;
File_Name : constant String := "simple.txt";
begin
-- Open output file and delete it
Create (F, Out_File, File_Name);
Delete (F);
-- Try to open deleted file
Open (F, In_File, File_Name);
Close (F);
exception
when Name_Error =>
Put_Line ("File does not exist");
when others =>
Put_Line ("Error while processing input file");
end Show_Text_File_Input_Except;
In this example, we create the file by calling Create and then delete it by calling Delete. After the call to Delete, we can no longer use the File_Type element. After deleting the file, we try to open the non-existent file, which raises a Name_Error exception.
### Sequential I/O¶
The previous section presented details about text file I/O. Here, we discuss doing file I/O in binary format. The first package we'll explore is the Ada.Sequential_IO package. Because this package is a generic package, you need to instantiate it for the data type you want to use for file I/O. Once you've done that, you can use the same procedures we've seen in the previous section: Create, Open, Close, Reset and Delete. However, instead of calling the Get_Line and Put_Line procedures, you'd call the Read and Write procedures.
In the following example, we instantiate the Ada.Sequential_IO package for floating-point types:
with Ada.Text_IO;
procedure Show_Seq_Float_IO is
package Float_IO is new Ada.Sequential_IO (Float);
use Float_IO;
F : Float_IO.File_Type;
File_Name : constant String := "float_file.bin";
begin
Create (F, Out_File, File_Name);
Write (F, 1.5);
Write (F, 2.4);
Write (F, 6.7);
Close (F);
declare
Value : Float;
begin
Open (F, In_File, File_Name);
while not End_Of_File (F) loop
end loop;
Close (F);
end;
end Show_Seq_Float_IO;
We use the same approach to read and write complex information. The following example uses a record that includes a Boolean and a floating-point value:
with Ada.Text_IO;
procedure Show_Seq_Rec_IO is
type Num_Info is record
Valid : Boolean := False;
Value : Float;
end record;
procedure Put_Line (N : Num_Info) is
begin
if N.Valid then
Ada.Text_IO.Put_Line ("(ok, " & Float'Image (N.Value) & ")");
else
end if;
end Put_Line;
package Num_Info_IO is new Ada.Sequential_IO (Num_Info);
use Num_Info_IO;
F : Num_Info_IO.File_Type;
File_Name : constant String := "float_file.bin";
begin
Create (F, Out_File, File_Name);
Write (F, (True, 1.5));
Write (F, (False, 2.4));
Write (F, (True, 6.7));
Close (F);
declare
Value : Num_Info;
begin
Open (F, In_File, File_Name);
while not End_Of_File (F) loop
Put_Line (Value);
end loop;
Close (F);
end;
end Show_Seq_Rec_IO;
As the example shows, we can use the same approach we used for floating-point types to perform file I/O for this record. Once we instantiate the Ada.Sequential_IO package for the record type, file I/O operations are performed the same way.
### Direct I/O¶
Direct I/O is available in the Ada.Direct_IO package. This mechanism is similar to the sequential I/O approach just presented, but allows us to access any position in the file. The package instantiation and most operations are very similar to sequential I/O. To rewrite the Show_Seq_Float_IO application presented in the previous section to use the Ada.Direct_IO package, we just need to replace the instances of the Ada.Sequential_IO package by the Ada.Direct_IO package. This is the new source code:
with Ada.Text_IO;
procedure Show_Dir_Float_IO is
package Float_IO is new Ada.Direct_IO (Float);
use Float_IO;
F : Float_IO.File_Type;
File_Name : constant String := "float_file.bin";
begin
Create (F, Out_File, File_Name);
Write (F, 1.5);
Write (F, 2.4);
Write (F, 6.7);
Close (F);
declare
Value : Float;
begin
Open (F, In_File, File_Name);
while not End_Of_File (F) loop
end loop;
Close (F);
end;
end Show_Dir_Float_IO;
Unlike sequential I/O, direct I/O allows you to access any position in the file. However, it doesn't offer an option to append information to a file. Instead, it provides an Inout_File mode allowing reading and writing to a file via the same File_Type element.
To access any position in the file, call the Set_Index procedure to set the new position / index. You can use the Index function to retrieve the current index. Let's see an example:
with Ada.Text_IO;
procedure Show_Dir_Float_In_Out_File is
package Float_IO is new Ada.Direct_IO (Float);
use Float_IO;
F : Float_IO.File_Type;
File_Name : constant String := "float_file.bin";
begin
-- Open file for input / output
Create (F, Inout_File, File_Name);
Write (F, 1.5);
Write (F, 2.4);
Write (F, 6.7);
-- Set index to previous position and overwrite value
Set_Index (F, Index (F) - 1);
Write (F, 7.7);
declare
Value : Float;
begin
-- Set index to start of file
Set_Index (F, 1);
while not End_Of_File (F) loop
end loop;
Close (F);
end;
end Show_Dir_Float_In_Out_File;
By running this example, we see that the file contains 7.7, rather than the previous 6.7 that we wrote. We overwrote the value by changing the index to the previous position before doing another write.
In this example we used the Inout_File mode. Using that mode, we just changed the index back to the initial position before reading from the file (Set_Index (F, 1)) instead of closing the file and reopening it for reading.
### Stream I/O¶
All the previous approaches for file I/O in binary format (sequential and direct I/O) are specific for a single data type (the one we instantiate them with). You can use these approaches to write objects of a single data type that may be an array or record (potentially with many fields), but if you need to create and process files that include different data types, or any objects of an unbounded type, these approaches are not sufficient. Instead, you should use stream I/O.
Stream I/O shares some similarities with the previous approaches. We still use the Create, Open and Close procedures. However, instead of accessing the file directly via a File_Type element, you use a Stream_Access element. To read and write information, you use the 'Read or 'Write attributes of the data types you're reading or writing.
Let's look at a version of the Show_Dir_Float_IO procedure from the previous section that makes use of stream I/O instead of direct I/O:
with Ada.Text_IO;
procedure Show_Float_Stream is
F : File_Type;
S : Stream_Access;
File_Name : constant String := "float_file.bin";
begin
Create (F, Out_File, File_Name);
S := Stream (F);
Float'Write (S, 1.5);
Float'Write (S, 2.4);
Float'Write (S, 6.7);
Close (F);
declare
Value : Float;
begin
Open (F, In_File, File_Name);
S := Stream (F);
while not End_Of_File (F) loop
end loop;
Close (F);
end;
end Show_Float_Stream;
After the call to Create, we retrieve the corresponding Stream_Access element by calling the Stream function. We then use this stream to write information to the file via the 'Write attribute of the Float type. After closing the file and reopening it for reading, we again retrieve the corresponding Stream_Access element and processed to read information from the file via the 'Read attribute of the Float type.
You can use streams to create and process files containing different data types within the same file. You can also read and write unbounded data types such as strings. However, when using unbounded data types you must call the 'Input and 'Output attributes of the unbounded data type: these attributes write information about bounds or discriminants in addition to the object's actual data.
The following example shows file I/O that mixes both strings of different lengths and floating-point values:
with Ada.Text_IO;
procedure Show_String_Stream is
F : File_Type;
S : Stream_Access;
File_Name : constant String := "float_file.bin";
procedure Output (S : Stream_Access;
FV : Float;
SV : String) is
begin
String'Output (S, SV);
Float'Output (S, FV);
end Output;
procedure Input_Display (S : Stream_Access) is
SV : String := String'Input (S);
FV : Float := Float'Input (S);
begin
Ada.Text_IO.Put_Line (Float'Image (FV) & " --- " & SV);
end Input_Display;
begin
Create (F, Out_File, File_Name);
S := Stream (F);
Output (S, 1.5, "Hi!!");
Output (S, 2.4, "Hello world!");
Output (S, 6.7, "Something longer here...");
Close (F);
Open (F, In_File, File_Name);
S := Stream (F);
while not End_Of_File (F) loop
Input_Display (S);
end loop;
Close (F);
end Show_String_Stream;
When you use Stream I/O, no information is written into the file indicating the type of the data that you wrote. If a file contains data from different types, you must reference types in the same order when reading a file as when you wrote it. If not, the information you get will be corrupted. Unfortunately, strong data typing doesn't help you in this case. Writing simple procedures for file I/O (as in the example above) may help ensuring that the file format is consistent.
Like direct I/O, stream I/O supports also allows you to access any location in the file. However, when doing so, you need to be extremely careful that the position of the new index is consistent with the data types you're expecting.
## Numerics¶
The standard library provides support for common numeric operations on floating-point types as well as on complex types and matrices. This section presents a brief introduction to them.
### Elementary Functions¶
The Ada.Numerics.Elementary_Functions package provides common operations for floating-point types, such as square root, logarithm, and the trigonometric functions (e.g., sin, cos). For example:
with Ada.Text_IO; use Ada.Text_IO; with Ada.Numerics; use Ada.Numerics; with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions; procedure Show_Elem_Math is X : Float; begin X := 2.0; Put_Line ("Square root of " & Float'Image (X) & " is " & Float'Image (Sqrt (X))); X := e; Put_Line ("Natural log of " & Float'Image (X) & " is " & Float'Image (Log (X))); X := 10.0 ** 6.0; Put_Line ("Log_10 of " & Float'Image (X) & " is " & Float'Image (Log (X, 10.0))); X := 2.0 ** 8.0; Put_Line ("Log_2 of " & Float'Image (X) & " is " & Float'Image (Log (X, 2.0))); X := Pi; Put_Line ("Cos of " & Float'Image (X) & " is " & Float'Image (Cos (X))); X := -1.0; Put_Line ("Arccos of " & Float'Image (X) & " is " & Float'Image (Arccos (X))); end Show_Elem_Math;
Here we use the standard e and Pi constants from the Ada.Numerics package.
The Ada.Numerics.Elementary_Functions package provides operations for the Float type. Similar packages are available for Long_Float and Long_Long_Float types. For example, the Ada.Numerics.Long_Elementary_Functions package offers the same set of operations for the Long_Float type. In addition, the Ada.Numerics.Generic_Elementary_Functions package is a generic version of the package that you can instantiate for custom floating-point types. In fact, the Elementary_Functions package can be defined as follows:
package Elementary_Functions is new
### Random Number Generation¶
The Ada.Numerics.Float_Random package provides a simple random number generator for the range between 0.0 and 1.0. To use it, declare a generator G, which you pass to Random. For example:
with Ada.Text_IO; use Ada.Text_IO; with Ada.Numerics.Float_Random; use Ada.Numerics.Float_Random; procedure Show_Float_Random_Num is G : Generator; X : Uniformly_Distributed; begin Reset (G); Put_Line ("Some random numbers between " & Float'Image (Uniformly_Distributed'First) & " and " & Float'Image (Uniformly_Distributed'Last) & ":"); for I in 1 .. 15 loop X := Random (G); Put_Line (Float'Image (X)); end loop; end Show_Float_Random_Num;
The standard library also includes a random number generator for discrete numbers, which is part of the Ada.Numerics.Discrete_Random package. Since it's a generic package, you have to instantiate it for the desired discrete type. This allows you to specify a range for the generator. In the following example, we create an application that displays random integers between 1 and 10:
with Ada.Text_IO; use Ada.Text_IO; with Ada.Numerics.Discrete_Random; procedure Show_Discrete_Random_Num is subtype Random_Range is Integer range 1 .. 10; package R is new Ada.Numerics.Discrete_Random (Random_Range); use R; G : Generator; X : Random_Range; begin Reset (G); Put_Line ("Some random numbers between " & Integer'Image (Random_Range'First) & " and " & Integer'Image (Random_Range'Last) & ":"); for I in 1 .. 15 loop X := Random (G); Put_Line (Integer'Image (X)); end loop; end Show_Discrete_Random_Num;
Here, package R is instantiated with the Random_Range type, which has a constrained range between 1 and 10. This allows us to control the range used for the random numbers. We could easily modify the application to display random integers between 0 and 20 by changing the specification of the Random_Range type. We can also use floating-point or fixed-point types.
### Complex Types¶
The Ada.Numerics.Complex_Types package provides support for complex number types and the Ada.Numerics.Complex_Elementary_Functions package provides support for common operations on complex number types, similar to the Ada.Numerics.Elementary_Functions package. Finally, you can use the Ada.Text_IO.Complex_IO package to perform I/O operations on complex numbers. In the following example, we declare variables of the Complex type and initialize them using an aggregate:
with Ada.Text_IO; use Ada.Text_IO; with Ada.Numerics; use Ada.Numerics; with Ada.Numerics.Complex_Types; use Ada.Numerics.Complex_Types; with Ada.Numerics.Complex_Elementary_Functions; use Ada.Numerics.Complex_Elementary_Functions; with Ada.Text_IO.Complex_IO; procedure Show_Elem_Math is package C_IO is new Ada.Text_IO.Complex_IO (Complex_Types); use C_IO; X, Y : Complex; R, Th : Float; begin X := (2.0, -1.0); Y := (3.0, 4.0); Put (X); Put (" * "); Put (Y); Put (" is "); Put (X * Y); New_Line; New_Line; R := 3.0; Th := Pi / 2.0; X := Compose_From_Polar (R, Th); -- Alternatively: -- X := R * Exp ((0.0, Th)); -- X := R * e ** Complex'(0.0, Th); Put ("Polar form: " & Float'Image (R) & " * e**(i * " & Float'Image (Th) & ")"); New_Line; Put ("Modulus of "); Put (X); Put (" is "); Put (Float'Image (abs (X))); New_Line; Put ("Argument of "); Put (X); Put (" is "); Put (Float'Image (Argument (X))); New_Line; New_Line; Put ("Sqrt of "); Put (X); Put (" is "); Put (Sqrt (X)); New_Line; end Show_Elem_Math;
As we can see from this example, all the common operators, such as * and +, are available for complex types. You also have typical operations on complex numbers, such as Argument and Exp. In addition to initializing complex numbers in the cartesian form using aggregates, you can do so from the polar form by calling the Compose_From_Polar function.
The Ada.Numerics.Complex_Types and Ada.Numerics.Complex_Elementary_Functions packages provide operations for the Float type. Similar packages are available for Long_Float and Long_Long_Float types. In addition, the Ada.Numerics.Generic_Complex_Types and Ada.Numerics.Generic_Complex_Elementary_Functions packages are generic versions that you can instantiate for custom or pre-defined floating-point types. For example:
with Ada.Numerics.Generic_Complex_Types;
procedure Show_Elem_Math is
package Complex_Types is new
use Complex_Types;
package Elementary_Functions is new
use Elementary_Functions;
package C_IO is new Ada.Text_IO.Complex_IO (Complex_Types);
use C_IO;
X, Y : Complex;
R, Th : Float;
### Vector and Matrix Manipulation¶
The Ada.Numerics.Real_Arrays package provides support for vectors and matrices. It includes common matrix operations such as inverse, determinant, eigenvalues in addition to simpler operators such as matrix addition and multiplication. You can declare vectors and matrices using the Real_Vector and Real_Matrix types, respectively.
The following example uses some of the operations from the Ada.Numerics.Real_Arrays package:
with Ada.Text_IO; use Ada.Text_IO; with Ada.Numerics.Real_Arrays; use Ada.Numerics.Real_Arrays; procedure Show_Matrix is procedure Put_Vector (V : Real_Vector) is begin Put (" ("); for I in V'Range loop Put (Float'Image (V (I)) & " "); end loop; Put_Line (")"); end Put_Vector; procedure Put_Matrix (M : Real_Matrix) is begin for I in M'Range (1) loop Put (" ("); for J in M'Range (2) loop Put (Float'Image (M (I, J)) & " "); end loop; Put_Line (")"); end loop; end Put_Matrix; V1 : Real_Vector := (1.0, 3.0); V2 : Real_Vector := (75.0, 11.0); M1 : Real_Matrix := ((1.0, 5.0, 1.0), (2.0, 2.0, 1.0)); M2 : Real_Matrix := ((31.0, 11.0, 10.0), (34.0, 16.0, 11.0), (32.0, 12.0, 10.0), (31.0, 13.0, 10.0)); M3 : Real_Matrix := ((1.0, 2.0), (2.0, 3.0)); begin Put_Line ("V1"); Put_Vector (V1); Put_Line ("V2"); Put_Vector (V2); Put_Line ("V1 * V2 ="); Put_Line (" " & Float'Image (V1 * V2)); Put_Line ("V1 * V2 ="); Put_Matrix (V1 * V2); New_Line; Put_Line ("M1"); Put_Matrix (M1); Put_Line ("M2"); Put_Matrix (M2); Put_Line ("M2 * Transpose(M1) ="); Put_Matrix (M2 * Transpose (M1)); New_Line; Put_Line ("M3"); Put_Matrix (M3); Put_Line ("Inverse (M3) ="); Put_Matrix (Inverse (M3)); Put_Line ("abs Inverse (M3) ="); Put_Matrix (abs Inverse (M3)); Put_Line ("Determinant (M3) ="); Put_Line (" " & Float'Image (Determinant (M3))); Put_Line ("Solve (M3, V1) ="); Put_Vector (Solve (M3, V1)); Put_Line ("Eigenvalues (M3) ="); Put_Vector (Eigenvalues (M3)); New_Line; end Show_Matrix;
Matrix dimensions are automatically determined from the aggregate used for initialization when you don't specify them. You can, however, also use explicit ranges. For example:
M1 : Real_Matrix (1 .. 2, 1 .. 3) :=
((1.0, 5.0, 1.0),
(2.0, 2.0, 1.0));
The Ada.Numerics.Real_Arrays package implements operations for the Float type. Similar packages are available for Long_Float and Long_Long_Float types. In addition, the Ada.Numerics.Generic_Real_Arrays package is a generic version that you can instantiate with custom floating-point types. For example, the Real_Arrays package can be defined as follows:
package Real_Arrays is new
`
|
2019-09-15 09:54:42
|
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|
http://opentradingsystem.com/quantNotes/Spectrum_of_compact_and_symmetric_operator_.html
|
I. Basic math.
II. Pricing and Hedging.
III. Explicit techniques.
IV. Data Analysis.
V. Implementation tools.
VI. Basic Math II.
1 Real Variable.
2 Laws of large numbers.
3 Characteristic function.
4 Central limit theorem (CLT) II.
5 Random walk.
6 Conditional probability II.
7 Martingales and stopping times.
8 Markov process.
9 Levy process.
10 Weak derivative. Fundamental solution. Calculus of distributions.
11 Functional Analysis.
A. Weak convergence in Banach space.
B. Representation theorems in Hilbert space.
C. Fredholm alternative.
D. Spectrum of compact and symmetric operator.
E. Fixed point theorem.
F. Interpolation of Hilbert spaces.
G. Tensor product of Hilbert spaces.
12 Fourier analysis.
13 Sobolev spaces.
14 Elliptic PDE.
15 Parabolic PDE.
VII. Implementation tools II.
VIII. Bibliography
Notation. Index. Contents.
## Spectrum of compact and symmetric operator.
roposition
(Spectrum of compact operator) Let be a Hilbert space, and is a linear compact operator. Then
1. .
2.
3. Either is finite or it is a sequence converging to 0.
Proposition
(Eigenvalues of compact symmetric operator) Let be a separable Hilbert space and is a linear compact operator. Then there exists a countable orthonormal basis of consisting of eigenvectors of .
Notation. Index. Contents.
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2018-08-20 08:45:48
|
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http://sinapeyman.com/rates-of-interest-for-the-a-scene-with-defaults/
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+982188870249_50
# EuroDate visitors
Rates of interest for the a scene with Defaults and you can recognized Rising cost of living
Rising prices is the standard desire from cost to improve throughout the years fundamentally. Inflation is measured by keeping tabs on the values regarding a basket of goods and you can qualities. The brand new rising cost of living is generally stated because the a compounded annual price and you can possibly a continuous speed of come back.
The rise into the cost of products and qualities reduces the to invest in fuel of money. Which is, extra cash must get a lot fewer goods (this was illustrated in the last chapter)
## Hence, the new borrower’s fees is actually calculated since the mortgage payment, incase zero inflation increased of the proportion regarding directory observed during the the beginning of the loan identity to that particular observed at the avoid of one’s mortgage term
Moreover, rising cost of living impacts the new incentive and you will capability to borrow cash at the a good offered price. Regarding the visibility away from rising cost of living rates, the money out-of a debtor (eg, wages) increase, which means that the latest borrower pays an even more huge amount so you can eurodate provider that loan than if there’s no inflation.
Including, envision a trader whom takes a-two-seasons financing out-of 10,100000, which is said to be paid off as the a lump sum during the the termination of couple of years from the an annual energetic appeal off 3%. Now, if there is absolutely no rising cost of living, the newest investor will pay 1, $$(=step one000\left(step 1.03\right)^2)$$, thus the newest buyer will pay focus from (=step 1,-1,000). Now think that discover dos% rising cost of living per year. Then the buyer would have to pay step one, $$(=step one000\left(1.03\right)^2\left(step 1.02\right)^2)$$. Note it is equivalent to per year effective interest off 5.06%. Plainly, in this case, the brand new borrower pays an appeal regarding (=step one,-1,000).
The new analogy over certainly implies that, on the visibility out-of inflation, loan providers consult individuals to blow way more rates of interest to get to possess losing the newest to buy stamina when you look at the loan name. At the same time, this new consumers can spend the money for desire as his or her income will additionally boost in the presence of rising cost of living.
Fundamentally, think a trader willing to lend $$P_0$$ to have t many years and that the attention on financing try reduced at the conclusion of the loan title. If you have no inflation (and you can suspicion from mortgage cost is known as), upcoming payment of your own mortgage is provided of the:
Today, think that you will find a rising prices from i expressed just like the an enthusiastic yearly continued speed. Now once the inflation has an effect on both the wages and you may rates, the fresh borrower, in this case, are willing to spend:
They, thus, comes after whenever interest levels try cited once the continued per annum, then your interest rate when you look at the a scene or particular rising prices and you will default risk is offered from the:
The above mentioned research takes on the rising cost of living speed is well known inside the improve, and though basic, it is a significant concept inside examining brand new determinants of great interest costs.
Actually, not, brand new inflation price are not known ahead, and dealers deal with so it uncertainty in a few indicates
1. Finance which have Rising cost of living Security.
## Usually, speed list and individual price index are widely used to level rising cost of living
In this instance, that loan is decided in a manner that the degree of payment from the the fresh new debtor grabs the true rising cost of living rates over the mortgage label because the measured because of the a certain site directory for instance the individual rate list. This basically means, the mortgage bargain, in such a case, determine the interest rate that is paid, provided that there is no inflation and you can a supply away from an list to regulate the borrowed funds payments to provide rising prices.
If a trader (lender) refuses to use the chance embedded throughout the suspicion away from inflation from indices, the fresh new compensation try gotten in two suggests:
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2022-07-03 23:48:22
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https://www.ques10.com/p/40113/separation-of-boundary-layer-1/
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Page: Separation of boundary layer
0
1) When a solid body is immersed in a flowing fluid, a thin layer of fluid is formed, and this thin layer is called as boundary layer, which is adjacent to the solid body.
2) In this layer formed, the velocity varies from 0 to freestream velocity in the direction normal to solid body point. Thickness of the boundary layer increases along the length of the solid body.
3) This fluid layer for formed has to work against surface friction at the expense of kinetic energy.
4) This loss of kinetic energy is recovered from the immediate fluid layer through the momentum exchange process. Thus decreasing the velocity of layer.
5) If the boundary layer cannot provide kinetic energy to overcome the resistance offered by the solid body, then at certain point a stage may occur where the boundary layer may not be able to keep sticking to the solid body.
6) This phenomena is called as boundary layer separation.
7) The point on the body where the boundary layer is on the verge of separation from the surface is called as point of separation.
Methods of preventing the separation of boundary layer
Methods:
1) Suction of slow moving fluid by a suction slot.
2) Supplying additional energy from a blower.
3) Providing a bypass in the slotted wing.
4) Providing guide blades in a bend.
5) Providing small divergence in a diffuser.
Introduction of flow around submerged bodies:
1) When a body is stationary and a fluid is flowing over it, some force is exerted by the flowing fluid on the body.
2) Similarly when a body is moving through liquid which is stationary, some force is exerted by the fluid on the body.
3) Similarly when both the fluid and the body are moving (at different velocities) even then some force is exerted by the fluid on the body.
More examples:
a) Flow of air over buildings.
b) Flow of water over bridges
c) Submarines, ships, airplanes and automobiles moving through water or air.
Drag and lift:
Drag:
a) The component of total force, (which is denoted as 'FR'), in the direction of motion is called as drag.
b) Drag is denoted by '$F_D$'
c) Thus it can be defined as, the force exerted by the fluid in the direction of motion.
Lift:
a) The component of total force '$F_T$', in the direction of perpendicular to the motions direction is known as lift.
b) It is denoted by '$F_L$'.
c) When the axis of the body is inclined to the direction of fluid flow, only then lift occurs. If the axis is parallel (lift=0).
d) Thus, lift can be defined as the force exerted by the fluid in the direction perpendicular to the direction of motion.
*Note:- * When the fluid is assumed ideal and the body is symmetrical i.e., sphere or cylinder, both the drag and lift are zero.
Drag - $F_D=C_DA\dfrac {\rho V^2}2$
where $C_D$ is the coefficient of drag
Lift - $F_L=C_LA\dfrac {\rho V^2}2$
where $C_L$ is the coefficient of lift
and A = largest projected area of immersed body.
### Numericals
Q1. A flat plate is 1.5m$\times$1.5 m moves at 50 km per hour in stationary air of density $1.15kg/m^3$. If the coefficient of drag and lift are 0.15 and 0.75 respectively, determine
(i) The lift force
(ii) The drag force
(iii) The resultant force
(iv) The power required to keep the plate in motion
Solution:- Given: -
Area of plate, $A=1.5\times 1.5=2.25m^2$
Velocity of plate, $V=50km/hr=\dfrac {50\times 1000}{60\times 60}=13.89m/s$
Density of air, $\rho =1.15kg/m^3$
Coefficient of drag, $C_D=0.15$
Coefficient of lift, $C_L=0.75$
(i) Lift Force, ($F_L$):-
$F_L=C_LA\times \dfrac {\rho V^2}2$
$F_L=0.75\times2.25\times \dfrac {1.5\times 13.89^2}2$
$F_L=187.20N$
(ii) Drag Force ($F_D$): -
$F_D=C_DA\times \dfrac {\rho V^2}2$
$F_D=0.15\times 2.25\times \dfrac {1.15\times 13.89^2}2$
$F_D=37.44N$
(iii) Resultant Force, ($F_R$): -
$F_R=\sqrt{F_D^2+F_L^2}$
$F_R=\sqrt{37.44^2+187.20^2}$
$F_R=\sqrt{1400+35025}$
$F_R=190.85N$
(iv) Power required to keep the plate in motion: -
$P=\dfrac {F_D\times V}{1000}$
$P=\dfrac {37.425\times 13.89}{1000}$
$P=0.519KW$
Q2) A truck having a projected area of 6.5 square metres travelling at 70 km per hour has a total resistance of 2000N. Of this 20% is due to rolling friction and 10% is due to surface friction. The rest is due to form drag. Calculate the coefficient of drag. Take $\rho=1.25kg/m^3$ of air.
Solution:- Given: -
Area of truck, $A=6.5m^2$
Speed of truck, $V=70km/hr=\dfrac {70\times1000}{60\times60}=19.44m/s$
Total resistance, $F_T=200N$
Rolling friction resistance 20% of total force, i.e.,
$F_C=\dfrac {20}{100}\times2000=400N$
Surface friction resistance is 10% of total force, i.e.,
$F_S=\dfrac {10}{100}\times2000=200N$
Therefore, form drag, $F_D=200-F_C-F_S=2000-400-200=1400N$
Where if $F_D$ is the form drag,
Then $C_D$ is the coefficient of form drag
$1400=C_D\times6.5\times1.25×\dfrac{19.44^2}2$
$∴ C_D=\dfrac {1400\times2}{6.5\times1.25\times19.44\times19.44}$
$C_D=0.912$
page fluid mechanics 2 fm2 • 69 views
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2019-10-18 07:23:25
|
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https://quant.stackexchange.com/questions/51494/how-is-radon-nikodym-derivative-different-from-the-likelihood-ratio/51511
|
# How is Radon-Nikodym derivative different from the likelihood ratio?
I see that the Radon-Nikodym derivative is the ratio of probability measures, $$dP/dQ$$. How is this different, in general, from a likelihood ratio of two continuous distributions? I understand the RN-definition broadly applies for discrete/continuous/mixture densities, but beyond that is there any difference?
## 2 Answers
If $$dx$$ is Lebesgue measure, then it dominates both measures because they correspond with continuous random variables, and one of the properties of RN derivatives is $$\frac{dP}{dQ} = \frac{\frac{dP}{dx}}{\frac{dQ}{dx}}.$$ The numerator is the density of $$P$$, and the denominator is the density of $$Q$$. This is the second property on wikipedia.
So yes, the likelihood ratio is just a particular case. If these two measures were for discrete random variables, then you would replace $$dx$$ with the counting measure, and you would get a ratio of probability mass functions.
In Probability Theory, density functions are generally defined as Radon-Nikodym derivatives themselves, $$\frac{dP}{dQ}$$.
The likelihood function interprets these densities (R-N derivatives) as a function of the parameters, given some observed outcome. More explicitly, let $$X$$ be an absolutely continuous random variable. Then, $$\mathcal{L}(\theta|x\in X) = f(x|\theta) = \mathbb{P}(x\in X|\theta)$$ In other words, the likelihood function measures the probability of observing $$x$$ given the parameters $$\theta$$.
The likelihood ratio is meant to assess the goodness-of-fit of two statistical models (with different parameters) given the same set of observations $$x$$, not two entirely different distributions. More explicitly, let $$\Theta$$ be the set of all possible parameters, and consider some subsets $$\Theta_0, \Theta_1 \in \Theta$$. The likelihood ratio is then, $$\mathcal{L(\Theta_0,\Theta_1)} = -2\log\frac{\sup_{\Theta_0\in\Theta} \mathcal{L}(\theta)}{\sup_{\Theta_1\in\Theta}\mathcal{L}(\theta)}$$ which tests for the null hypothesis $$\theta\in\Theta_0$$.
• "Radon-Nikodym derivatives themselves" yes but then you should switch the notation instead of implicitly calling $Q$ Lebesgue measure – Taylor Apr 13 at 16:14
• Also, densities are not probability mass functions. They cannot be interepreted that way – Taylor Apr 13 at 16:14
• finally, the last expression is not the likelihood ratio, but it is a function of the likelihood ratio. The reason it is transformed is that when it is written in thsi way, it's asymptotically $\chi^2$ distributed--but this is irrelevant at the moment. – Taylor Apr 13 at 16:15
|
2020-07-12 11:53:41
|
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|
https://academy.vertabelo.com/course/ms-sql-basic-reporting/multi-level-aggregation/basic/explanation
|
Deals Of The Week - hours only!Up to 80% off on all courses and bundles.-Close
Introduction
Basic multi-level aggregation
3. Multi-level aggregation – explanation
Multi-level aggregation in groups
Multi-level aggregation with custom classification
Three or more aggregation levels
Summary
## Instruction
The code in the previous exercise used an SQL concept known as a CTE, or a common table expression. Think of it as a temporary set of rows that you define and use later in the same query. CTEs are similar to subqueries.
The most basic syntax of any common table expression looks like this:
WITH some_name AS (
)
SELECT
...
FROM some_name
You need to give your CTE a name (we used some_name in the example) and define it within a pair of parentheses. Then, once you close the bracket, you can select columns from the CTE as if it were a table. We will refer to the CTE as the "inner query" and the part after it as the "outer query." Note that you need to define your CTE first – i.e., before the outer query's SELECT.
Back to our example:
WITH OrderTotalPrices AS (
SELECT
O.OrderID,
SUM(UnitPrice * Quantity) AS TotalPrice
FROM Orders O
JOIN OrderItems OI
ON O.OrderID = OI.OrderID
GROUP BY O.OrderID
)
SELECT
AVG(TotalPrice) AS AvgTotalPrice
FROM OrderTotalPrices;
Here, the CTE is called OrderTotalPrices and allows us to access two columns: OrderID and TotalPrice. In the outer query, we aggregate the TotalPrice column using AVG(TotalPrice). As a result, we'll get a single number in the report.
## Exercise
The template code contains the query from the previous exercise. Your task is to modify it to show the average total price after discount. Rename the AvgTotalPrice column to AvgTotalDiscountedPrice.
### Stuck? Here's a hint!
Change the inner query. Calculate the total price for an order as:
SUM(UnitPrice * Quantity * (1 - Discount))
|
2022-09-26 21:46:56
|
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|
https://codereview.stackexchange.com/questions/220989/simple-function-that-simulates-survey-results-based-on-sample-size-and-probabili
|
# Simple function that simulates survey results based on sample size and probability
What is this:
This is a simple function, part of a basic Monte Carlo simulation. It takes sample size and probability as parameters. It returns the simulation result (positive answers) plus the input parameters in a tuple.
I'm trying to avoid using temporary variables, I have two questions.
1. Do I really save memory by avoiding storing interim results?
2. How could I improve readability without adding variables?
def simulate_survey(sample_size, percent_subscribes):
return (
sample_size,
percent_subscribes,
round(
(
sum([
r.random() < percent_subscribes
for _ in range(sample_size)
]) / sample_size
),
2
)
)
• As I discovered recently, summing a lot of booleans, where the chance that the value is False is not negligible, can be surprisingly slow.
So I would change your survey result calculation to:
sum([1 for _ in range(sample_size) if r.random() < percent_subscribes])
This allows sum to use its faster integer implementation and you do not sum a bunch of zeros.
• Alternatively, you could look at this problem as an application of the binomial distribution. You have some chance that a certain result is obtained and you want to know how often that chance was true for some population. For this you can use numpy.random.binomial:
import numpy as np
def simulate_survey(sample_size, percent_subscribes):
subscribers = np.random.binomial(sample_size, percent_subscribes)
return sample_size, percent_subscribes, round(subscribers / sample_size, 2)
Using numpy here may also speed up your process in other places. If you need to run this function many times, you probably want to use the third argument to generate multiple values at once.
IMO, the readability is also greatly increased by using one temporary variable here, instead of your many levels of parenthesis.
• I am not a fan of your function returning its inputs. The values of those should already be available in the scope calling this function, so this seems unnecessary. One exception would be that you have other, similar, functions which actually return different/modified values there.
• You should add a docstring describing what your function does.
I think avoiding temporary variables, when we have no strict memory limit, is a bad idea. There is no way to have a readable code without using variables. So let's create a version of your code with temp variables:
def simulate_survey(sample_size, percent_subscribes):
sum_result = sum([x for x in [True] * sample_size if r.random() < percent_subscribes])
third_value = round(sum_result / sample_size, 2)
return (
sample_size,
percent_subscribes,
third_value
)
It's not the most readable version of your code, But it's clearly more readable (I changed the way you created the sum value. I'm programming with Python for years, but that syntax is so strange to me. I hope my code do what your code did).
So Is there a huge memory usage gap between those programs? We now that Python does not remove temporary variables as a part of its optimization process (you can read more about it here). So obviously, my program should use more memory than yours. But how much?
I used resource module for comparing them. You can use this too if you are working on a UNIX based os.
Here is the code that I tried in both programs for measuring memory usage:
print(simulate_survey(64, 0.5))
Your variable-less program shows values around 11860 KB, But my program with temporary variables used almost 12008 KB. There is 200 KB difference, but don't forget that my code is not completely the same as your code and I changed how it creates third value.
So let's change the third value to the way you creates that:
def simulate_survey(sample_size, percent_subscribes):
sum_result = sum([
r.random() < percent_subscribes
for _ in range(sample_size)
])
third_value = round(sum_result / sample_size, 2)
return (
sample_size,
percent_subscribes,
third_value
)
So what happens if we test memory usage of this code that has the exact same logic as the first version? The result is around 11896 KB. Only between 10 to 30 KB more than the first version (Because each time we create a process, does not exactly same things happen, memory usage values are different each time).
So, as a conclusion, if you are not working on a machine with very tiny memory (something like embedded programming that is not common using python), I really recommend you that always use things like temporary variables to make your code readable.
|
2021-04-23 09:18:01
|
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|
https://www.encyclopediaofmath.org/index.php?title=Decision_problem&oldid=16725
|
# Decision problem
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
algorithmic problem
The question of the existence of an effective computational procedure or algorithm for deciding the truth of falsity of any instance of a parametric statement. Simple examples of decision problems are: addition of two given decimal numbers, multiplication of two given numbers, the verification of whether or not a given integer is prime, to find the derivative of a given function, to expand a given function into a power series, etc.
If no algorithm exists for a problem at hand, this problem is said to be undecidable or unsolvable.
The problem of finding an algorithm which solves a given decision problem is sometimes also called the recognition problem or solvability problem. This last somewhat unfortunate term historically first appeared in connection with the problem of deciding for a well-formed formula from the first-order predicate calculus whether the formula is valid. Generally speaking, when considering the solvability of a given decision problem it is natural to consider this problem in terms of the existence or non-existence of an algorithm for it.
Cf. also Algorithmic problem.
Decision problems are meaningful only when the notion of an effective computational procedure is suitably formalized, as in the theory of algorithms.
A positive solution to a decision problem consists of giving an algorithm for solving it, the problem is then said to be decidable or solvable.
Examples of solvable decision problems are: i) the problem of deciding whether or not a given integer is prime; and ii) the problem of deciding for any polynomial with integer coefficients whether or not it has a real root. Examples of unsolvable decision problems are: 1) the problem of deciding for any effectively given real number whether or not it is zero; 2) the problem of deciding for any multi-variable polynomial with integer coefficients whether or not that polynomial has all integer roots ( "Hilbert 10th problemHilbert's 10-th problem" ); 3) the problem of deciding for any well-formed formula from the first-order predicate calculus whether or not that formula is valid (the "EntscheidungsproblemEntscheidungsproblem" ); and 4) the problem of deciding for any group given in terms of generators and relations and any string of generators whether or not that string can be reduced to the identity in the group (the "word problemword problem" for groups).
The term "mass problemmass problem" , the literal translation of the original title of this article, rarely occurs in the literature.
#### References
[a1] M. Davis, "Computability and unsolvability" , McGraw-Hill (1958) [a2] M. Davis (ed.) , The undecidable , Raven Press (1965) [a3] H. Hermes, "Enumerability, decidability, computability" , Springer (1965) [a4] M. Minsky, "Computation: finite and infinite machines" , Prentice-Hall (1972) [a5] H. Rogers jr., "Theory of recursive functions and effective computability" , McGraw-Hill (1967) pp. 164–165
How to Cite This Entry:
Decision problem. Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Decision_problem&oldid=16725
This article was adapted from an original article by S.I. Adyan (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
|
2018-12-15 21:30:16
|
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|
http://www.kalidioscope.com/eric-bristow-agimwx/c50db6-how-to-tell-if-a-graph-is-connected-or-disconnected
|
Graph Connectivity: If each vertex of a graph is connected to one or multiple vertices then the graph is called a Connected graph whereas if there exists even one vertex which is not connected to any vertex of the graph then it is called Disconnect or not connected graph. ... Graphs can be connected or disconnected based on the arrangement of its nodes. Therefore the above graph is a 2-edge-connected graph. The following graph (Assume that there is a edge from to.) That is called the connectivity of a graph. A closed interval [a,b] is connected. How to tell if a group is cyclic? Prove or disprove: The complement of a simple disconnected graph must be connected. A k-vertex-connected graph or k-edge-connected graph is a graph in which no set of k − 1 vertices (respectively, edges) exists that, when removed, disconnects the graph. Let Gbe a simple disconnected graph and u;v2V(G). (The nodes are sometimes called vertices and the edges are sometimes called arcs. Given a directed graph. If an edge e is connected to v, then v is said to be incident on e. Also, the edge e is said to be incident on v. A graph G is connected if there exists path between every pair of distinct nodes… The task is to check if the given graph is connected or not. Graphs are a generalization of trees. To show this, suppose that it was disconnected. See the answer. If v and u are in different components of G, then certainly they're connected by an edge in G'. This implies, in G, there are 2 kinds of vertices. A directed graph that allows self loops? Connected and Disconnected Graph. A null graph of more than one vertex is disconnected (Fig 3.12). Here are the following four ways to disconnect the graph by removing two edges: 5. Check if Graph is Bipartite – Adjacency List using Depth-First Search(DFS). (Roseman, 1999) Definition A topological space X is connected if it is not disconnected. It has, in this case, three. From the edge list it is easy to conclude that the graph has three unique nodes, A, B, and C, which are connected by the three listed edges. Given a graph, determine whether the graph is connected. In an undirected graph G, two vertices u and v are called connected if G contains a path from u to v. Otherwise, they are called disconnected. A graph is connected if some vertex is connected to all other vertices. As of R2015b, the new graph and digraph classes have a method for computing connected components. PATH. Please use ide.geeksforgeeks.org, In Exercise, determine whether the graph is connected or disconnected. Ralph Tindell, in North-Holland Mathematics Studies, 1982. A graph G is disconnected, if it does not contain at least two connected vertices. NOTE: In an undirected graph G, the vertices u and v are said to be connected when there is a path between vertex u and vertex v. otherwise, they are called disconnected graphs. It is clear: counting the edges does not tell us much about the graph being connected. That's because an Euler circuit is only required to traverse every edge of the graph, it's not required to visit every vertex; so isolated vertices are not a problem. In an undirected graph, a connected component is a set of vertices in a graph that are linked to each other by paths. Definition A graph isconnectedif any two vertices are connected by a series of edges. Example 5.3.7. Dirac's and Ore's Theorem provide a … Graph Databases is a NoSQL database based on Graph Theory and it consists of objects called nodes, properties, and edges (relationships) to represent, store, … vertices the original graph G has. Figure 8 We can always find if an undirected is connected or not by finding all reachable vertices from any vertex. In the second, there is a way to get from each of the houses to each of the other houses, but it's not necessarily … (adsbygoogle = window.adsbygoogle || []).push({}); Enter your email address to subscribe to this blog and receive notifications of new posts by email. If this count is equal to no of vertices means all vertices are traveled during DFS implies graph is connected if the count is not equal to no of vertices implies all the vertices are not traveled means graph is not connected or disconnected. Lemma: A simple connected graph is a tree if and only if there is a unique path between any two vertices. Solution The statement is true. Prove or disprove: The complement of a simple disconnected graph must be connected. The following graph is an example of a Disconnected Graph, where there are two components, one with ‘a’, ‘b’, ‘c’, ‘d’ vertices and another with ‘e’, ’f’, ‘g’, ‘h’ vertices. You can use network X to find the connected components of an undirected graph by using the function number_connected_components and give it, the graph, its input and it would tell you how many. isDisconnected:: UGraph v e -> Bool Source # Tell if a 'UGraph is disconnected | An Undirected Graph is disconnected when its not connected. (a) (b) (c) View Answer Calculate the forward discount or premium for the following spot and three-month forward rates: (a) SR = $2.00/£1 and FR =$2.01/£1 (b) SR = \$2.00/£1 and FR = … From every vertex to any other vertex, there should be some path to traverse. Bridge A bridge is an edge whose deletion from a graph increases the number of components in the graph. Definition: A tree is a connected undirected graph with no cycles. Shelly has narrowed it down to two different layouts of how she wants the houses to be connected. If $T$ is a tree, then it has no cycles. An Eulerian path for the connected graph is also an Eulerian path for the graph with the added edge-free vertices (which clearly add no edges that need to be traversed). Determining if a Graph is Hamiltonian. We could have a square. How do you tell if a graph is connected? A Connected Graph A graph is said to be connected if any two of its vertices are joined by a path. code. If count of reachable vertices is equal to number of vertices in graph, then the graph is connected else not. A graph is not connected if there exists two vertices where I can’t find a path between these two vertices. It possible to determine with a simple algorithm whether a graph is connected: Choose an arbitrary node x of the graph G as the starting point. 2. If v is a cut of a graph G, then we know we can find two more vertices w and x of G where v is on every path between w and v. We know this because a graph is disconnected if there are two vertices in the graph … I realize this is an old question, but since it's still getting visits, I have a small addition. You can verify this yourself by trying to find an Eulerian trail in both graphs. Then Determine How Many Components The Graph Has. Don’t stop learning now. We assume that all graphs are simple. Vertex 2. If a graph is not connected, which means there exists a pair of vertices in the graph that is not connected by a path, then we call the graph disconnected. Make all visited vertices v as vis1[v] = true. A graph with multiple disconnected vertices and edges is said to be disconnected. EDIT: Perhaps you'd like a proof of this. )However, graphs are more general than trees: in a graph, a node can have any number of incoming edges (in a tree, the root node cannot have any incoming edges and the other nodes can only have one incoming edge). علمی O Disconnected о Connected. Now what to look for in a graph to check if it's Biconnected. Details. Simple, directed graph? Tell if a 'UGraph is connected | An Undirected Graph is connected when there is a path between every pair | of vertices. Dr. James Burk Introduction to Graph Theory Graph Theory - Some Properties Any graph is either connectedor disconnected. Minimum Increments to make all array elements unique, Add digits until number becomes a single digit, Add digits until the number becomes a single digit. Check if a directed graph is connected or not, Convert undirected connected graph to strongly connected directed graph, Check if a given directed graph is strongly connected | Set 2 (Kosaraju using BFS), Minimum edges required to make a Directed Graph Strongly Connected, Check if incoming edges in a vertex of directed graph is equal to vertex itself or not, Check if a given Graph is 2-edge connected or not, Convert the undirected graph into directed graph such that there is no path of length greater than 1, Print Nodes which are not part of any cycle in a Directed Graph, Check if a graph is strongly connected | Set 1 (Kosaraju using DFS), Check if there exists a connected graph that satisfies the given conditions, Check if a graph is Strongly, Unilaterally or Weakly connected, Queries to check if vertices X and Y are in the same Connected Component of an Undirected Graph, Check if every vertex triplet in graph contains two vertices connected to third vertex, Check if longest connected component forms a palindrome in undirected graph, Find if there is a path between two vertices in a directed graph, Shortest path with exactly k edges in a directed and weighted graph, Assign directions to edges so that the directed graph remains acyclic, Detect Cycle in a directed graph using colors, All Topological Sorts of a Directed Acyclic Graph, Hierholzer's Algorithm for directed graph, Determine whether a universal sink exists in a directed graph, Number of shortest paths in an unweighted and directed graph, Data Structures and Algorithms – Self Paced Course, We use cookies to ensure you have the best browsing experience on our website. Each member of a tuple being a vertex/node in the graph. In this case the graph is connected but no vertex is connected to every other vertex. If the two vertices are additionally connected by a path of length 1, i.e. A graph $$G = (V,E)$$ is said to be connected if for all $$u, v \in V(G)\text{,}$$ there is a $$u$$-$$v$$ path joining them. Writing code in comment? While "not connected'' is pretty much a dead end, there is much to be said about "how connected'' a connected graph is. Yes, a disconnected graph can have an Euler circuit. When I right click on this graph and edit the data, it still shows me the excel where the data is coming from. Method based eigenvalues return 15 as number of connected components while method based on graph search (depth-first / breadth-first) returns 1. By now it is said that a graph is Biconnected if it has no vertex such that its removal increases the number of connected components in the graph. Like trees, graphs have nodes and edges. The simplest approach is to look at how hard it is to disconnect a graph by removing vertices or edges. A Disconnected Graph. Determine the set A of all the nodes which can be reached from x. See | isConnected TODO: An edgeles graph with two or more vertices is disconnected. Start DFS at the vertex which was chosen at step 2. If is disconnected, then its complement is connected (Skiena 1990, p. 171; Bollobás 1998). A graph is called connected if given any two vertices, there is a path from to. Tarjan's strongly connected components algorithm (or Gabow's variation) will of course suffice; if there's only one strongly connected component, then the graph is strongly connected.. Graph is not connected due to point mentioned above. Connectedness wins, since the complement of any disconnected graph is connected. Show transcribed image text. For an extension exercise if you want to show off when you tell the teacher they're wrong, how many edges do you need to guarantee connectivity (and what's the maximum number of edges) in a. Unless I am not seeing something. Start at a random vertex v of the graph G, and run a DFS(G, v). Objective: Given an undirected graph, Write an algorithm to determine whether its tree or not. If this count is equal to no of vertices means all vertices are traveled during DFS implies graph is connected if the count is not equal to no of vertices implies all the vertices are not traveled means graph is not connected or disconnected. A directed graph is strongly connected if there is a directed path from any two vertices in the graph. Either those that belong to the same connected component of G, or those that are in different components. (Type A Whole Number.) A cut is a vertex in a graph that, when removed, separates the graph into two non-connected subgraphs. This problem has been solved! If a graph G is disconnected, then every maximal connected subgraph of G is called a connected component of the graph G. Vertex 1. is a connected graph. Therefore, by definition,. Yet the graph is not connected. As we can see graph G is a disconnected graph and has 3 connected components. A disconnected graph is made up of connected subgraphs that are called components. If any vertex v has vis1[v] = false and vis2[v] = false then the graph is not connected. 17622 Advanced Graph Theory IIT Kharagpur, Spring Semester, 2002Œ2003 Exercise set 1 (Fundamental concepts) 1. The edges of the graph represent a specific direction from one vertex to another. The Graph Is The Graph Ha (Type A Whole Disconnected Connected Determine Whether The Graph Is Connected Or Disconnected. 6.2.1 A Perron-Frobenius style result for the Laplacian What does the Laplacian tell us about the graph? Now reverse the direction of all the edges. In any graph, the sum of the degrees of the vertices equals twice the number of edges. The graph which has self-loops or an edge (i, j) occurs more than once (also called multiedge and graph is called multigraph) is a non-simple graph. You said that if it gets disconnected from the core it is automatically unparented from it? As with a normal depth first search, you track the status of each node: new, seen but still open (it's in the call stack), and seen and finished. If G is connected then we look at the number of the G i which are disconnected. Or a graph is said to be connected if there exist atleast one path between each and every pair of vertices in graph G, otherwise it is disconnected. How can I protect this file as I am about the share the power point to public, yet would like to keep the raw data confidential. Continuous and discrete graphs visually represent functions and series, respectively. Determine whether the graph is that of a function. If the graph had disconnected nodes, they would not be found in the edge list, and would have to be specified separately. They are useful in mathematics and science for showing changes in data over time. A bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint sets U and V such that every edge connects a vertex in U to one in V. It is possible to test whether a graph is bipartite or not using DFS algorithm. Check If Given Undirected Graph is a tree, Given Graph - Remove a vertex and all edges connect to the vertex, Graph – Depth First Search in Disconnected Graph, Graph Implementation – Adjacency Matrix | Set 3, Graph Implementation – Adjacency List - Better| Set 2, Count number of subgraphs in a given graph, Breadth-First Search in Disconnected Graph, Graph – Find Number of non reachable vertices from a given vertex, Articulation Points OR Cut Vertices in a Graph, Maximum number edges to make Acyclic Undirected/Directed Graph, Check if given an edge is a bridge in the graph, Graph – Count all paths between source and destination, Graph – Detect Cycle in an Undirected Graph using DFS. Tell if a Graph is connected | An Undirected Graph is connected when there is a path between every pair | of vertices. Consider an example given in the diagram. What is Directed Graph. For an extension exercise if you want to show off when you tell the teacher they're wrong, how many edges do you need to guarantee connectivity (and what's the maximum number of edges) in a. If not, the graph isdisconnected. If uand vbelong to different components of G, then the edge uv2E(G ). An orientation of an undirected graph G is totally cyclic if and only if it is a strong orientation of every connected component of G. Robbins' theorem states that a graph has a strong orientation if and only if it is 2-edge-connected; disconnected graphs may have totally cyclic orientations, but only if … You can find the Laplacian matrix of the graph and check the multiplicity of eigenvalue zero of the Laplacian matrix, if the multiplicity of zero is one then graph is connected, if multiplicity of eigenvalue zero of Laplacian matrix of the graph is two or more then it is disconnected. Objective: Given an undirected graph, write an algorithm to find out whether the graph is connected or not. Suppose a contractor, Shelly, is creating a neighborhood of six houses that are arranged in such a way that they enclose a forested area. Proof: To prove the statement, we need to realize 2 things, if G is a disconnected graph, then, i.e., it has more than 1 connected component. Solution The statement is true. Disconnected Graph. Q16. Both are linear time. Answer to Connected or Disconnected? Otherwise, it is called a weakly connected graph if every ordered pair of vertices in the graph is weakly connected. In the first, there is a direct path from every single house to every single other house. B is degree 2, D is degree 3, and E is degree 1. A connected graph is such that a path exists between any two given nodes. by a single edge, the vertices are called adjacent. A graph is disconnected if at least two vertices of the graph are not connected by a path. Since the complement G ¯ of a disconnected graph G is spanned by a complete bipartite graph it must be connected. Therefore this part is false. A disconnected graph consists of two or more connected graphs. A graph G is said to be disconnected if there is no edge between the two vertices or we can say that a graph which is not connected is said to be disconnected. U V = 0; U V = S. A set S (not necessarily open) is called disconnected if there are two open sets U and V such that (U S) # 0 and (V S) # 0(U S) (V S) = 0(U S) (V S) = SIf S is not disconnected it is called connected. When there is an edge representation as (V1, V2), the direction is from V1 to V2. A graph is said to be disconnected if it is not connected, i.e., if there exist two nodes in such that no path in has those nodes as endpoints. Example 1. A graph is said to be connected if there is a path between every pair of vertex. Connectivity on directed graph. A directed graph that allows self loops? It is denoted by K(G). It is possible that if we remove the vertex, we are left with one subgraph consisting of a single vertex and a large graph, in which case we call the cut point trivial. The nodes of a graph can also be said as it's vertices. An open circle indicates that the point does not belong to the graph. An undirected graph is a tree if it has properties 1. Create a boolean visited [] array. When a graph has an ordered pair of vertexes, it is called a directed graph. Is there a way I can just quickly look at an adjacency matrix and determine if the graph is a "connected graph" or not? By using our site, you A directed graph is connected, or weakly connected, if the correpsonding undirected graph (obtained by ignoring the directions of edges) is connected. Semi-Eulerian … A directed graphs is said to be strongly connected if every vertex is reachable from every other vertex. Because any two points that you select there is path from one to another. brightness_4 Disconnected Graph. Proof. A graph is said to be connectedif there exist at least one path between every pair of vertices otherwise graph is said to be disconnected. Connected or Disconnected Graph: A graph G is said to be connected if for any pair of vertices (Vi, Vj) of a graph G are reachable from one another. A topological space X is disconnected if X=A B, where A and B are disjoint, nonempty, open subsets of X. If uand vbelong to different components of G, then the edge uv2E(G ). So the graph is not Biconnected. Now reverse the direction of all the edges. Fig 3.9(a) is a connected graph where as Fig 3.13 are disconnected graphs. Start DFS at the vertex which was chosen at step 2. Vertex Connectivity. Below is the implementation of the above approach: edit Directed Graph, Graph, Nonlinear Data Structure, Undirected Graph. To check whether a graph is connected based on its adjacency matrix A, use Given a graph, determine if given graph is bipartite graph using DFS. Attention reader! We already know that we can tell if G is connected or not. Experience. A simpler solution is to remove the edge, check if graph remains connect after removal or not, finally add the edge back. Make all visited vertices v as vis2[v] = true. And these are the three connected components in this particular graph. Question: Determine Whether The Graph Is Connected Or Disconnected. Start DFS from any vertex and mark the visited vertices in the visited[] array. The numbers of disconnected simple unlabeled graphs on , 2, ... nodes are 0, 1, 2, 5, 13, 44, 191, ...(OEIS A000719).. This graph contains two vertices with odd degree (D and E) and three vertices with even degree (A, B, and C), so Euler’s theorems tell us this graph has an Euler path, but not an Euler circuit. See the answer. Unlike determining whether or not a graph is Eulerian, determining if a graph is Hamiltonian is much more difficult. If A is equal to the set of nodes of G, the graph is connected; otherwise it is disconnected. 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( Depth-First / breadth-first ) returns 1 t [ /math ] is a path between every pair of,. Was chosen at step 2 Theory - some properties any graph, write an algorithm to determine whether the graph. Disconnected from the core it is strongly connected or not is Full or not a graph in power that... Graph into two non-connected subgraphs G ' its complement is connected to every other vertex 1999 ) definition a space! 'D like a proof of this is made up of connected components while method based eigenvalues 15! Came from an excel the original graph G is connected to every other vertex ] t [ /math ] connected...
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https://www.physicsforums.com/threads/tensors-summation-convention.428076/
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# Tensors (summation convention)
1. Sep 11, 2010
### bremvil
Hi everyone,
I recently started a course on continuum mechanics. It started with the mathematical background of transforming tensors with contravariant and/or covariant indices. There is one thing I don't understand and it should be really straight forward. I hope you can give me a hint.
http://s35.photobucket.com/albums/d174/Brasempje/?action=view¤t=tensor.jpg
In equation 5 on the page that I linked above I do not see how I can get from term 3 to term 4. I end up with 3 kronecker delta's instead of just 1. Since there is a double index 'm' summation can be performed and it should lead to the same result as when you use a 'shortcut'. I can show what I do using 'latex' notation, with ^ = superscript, _ = subscript, d = kronecker delta, t = theta
'term 3' in Equation (5) reads:
(dt^i/dx^m) * (dx^m/dt^j)
in case I decide to do summation this equation will turn into:
(dt^i/dx^1) * (dx^1/dt^j) + (dt^i/dx^2) * (dx^2/dt^j) +
(dt^i/dx^3) * (dx^3/dt^j)
Each component of vector x is a function of all three components of vector theta. And each component of vector theta is a function of all three components of vector x. By the chain rule the last expression would become
dt^i/dt^j + dt^i/dt^j + dt^i/dt^j
this is:
d^i_j + d^i_j + d^i_j = 3 * d^i_j
so in case I decide to do the summation I end up with something different than I
would expect. 3 kronecker delta's instead of 1! Is there any objection to using a sum in this
case?
with kind regards,
Bremvil
2. Sep 11, 2010
### Fredrik
Staff Emeritus
That equality is just the chain rule. It doesn't have anything to do with tensors or properties of the Kronecker delta.
When $g:\mathbb R^n\rightarrow\mathbb R^n$ and $f:\mathbb R^n\rightarrow\mathbb R$, I like to write the chain rule like this:
$$(f\circ g)_{,i}(x)=f_{,j}(g(x)) g^j{}_{,i}(x)$$
Here ",i" denotes partial derivative with respect to the ith variable, and $g^j$ is the jth component of the function g. If $f:\mathbb R^n\rightarrow\mathbb R^n$, we have
$$(f\circ g)^i(x)=(f(g(x))^i=f^i(g(x))=f^i\circ g(x)$$
so
$$\delta^i_j=(f\circ f^{-1})^i{}_{,j}(x)=(f^i\circ (f^{-1}))_{,j}(x)$$
Define $g=f^{-1}$ to unclutter the notation somewhat. Then the above is
$$=(f^i\circ g)_{,j}(x)=f^i{}_{,k}(g(x))g^k{}_{,j}(x)$$
and...uhh...I can't explain why this is written in the form
$$\frac{\partial A^i}{\partial B^k}\frac{\partial B^k}{\partial A_j}$$
without explaining partial derivatives with respect to a coordinate system. (Edit: Actually I can. See the comments at the end of the post). On a manifold, expressions like f(x+h) don't work, because in general, addition isn't defined for points in the manifold. This is why we have to use a coordinate system to define partial derivatives. A coordinate system is a function $x:U\rightarrow R^n$, where U is an open subset that contains the point p at which we want to define the partial derivative. If f is a function from the manifold to the real numbers, we define
$$\frac{\partial f}{\partial x^i}(p)=(f\circ x^{-1})_{,i}(x(p))$$
In particular, if y is another coordinate system,
$$\frac{\partial y^j}{\partial x^i}(p)=(y^j\circ x^{-1})_{,i}(x(p))$$
The f above is a coordinate change function, i.e. an expression of the form $A\circ B^{-1}$, where A and B are coordinate systems. So we have
$$\delta^i_j=f^i{}_{,k}(g(x))g^k{}_{,j}(x)=(A^i\circ B^{-1})_{,k}(g(x))(B^k\circ A^{-1}){}_{,j}(x)=\frac{\partial A^i}{\partial B^k}(A^{-1}(x))\frac{\partial B^k}{\partial A_j}(A^{-1}(x))$$
Edit: It turned out to be easier than I thought to explain the notation. No techniques from differential geometry are needed.
$$\delta^i_j=f^i{}_{,k}(g(x))g^k{}_{,j}(x)=f^i{}_{,k}(g(x))g^k{}_{,j}(f(g(x)))=\frac{\partial f^i}{\partial g^k}(g(x))\frac{\partial g^k}{\partial f_j}(f(g(x))$$
Last edited: Sep 11, 2010
3. Sep 11, 2010
### bremvil
Dear Fredrik,
Thanks for your reply. I read through it carefully but I find it quite difficult, my math background is not as strong as yours. So I'm still not really there yet. In your explanation you started with:
$$(f\circ g)_{,i}(x)=f_{,j}(g(x)) g^j{}_{,i}(x)$$
but this step is basically my entire problem! The index j appears twice which would mean summation right. I will try to write down my original problem with latex. Could you instead maybe tell where I am making the error?
http://s35.photobucket.com/albums/d174/Brasempje/?action=view¤t=tensor.jpg
equation 5, the problem is in step from term 3 to term 4.
$$x_i = x_i(\theta_1 ,\theta_2 , \theta_3)$$
$$\theta_i = \theta_i(x_1 ,x_2 , x_3)$$
If I decide to apply summation over 'index m' in the equation below I get:
$$\frac{\partial \theta^i}{\partial x^m}\frac{\partial x^m}{\partial \theta^j} = \frac{\partial \theta^i}{\partial x^1}\frac{\partial x^1}{\partial \theta^j} + \frac{\partial \theta^i}{\partial x^2}\frac{\partial x^2}{\partial \theta^j} + \frac{\partial \theta^i}{\partial x^3}\frac{\partial x^3}{\partial \theta^j}$$
by the chain rule I get
$$\frac{\partial \theta^i}{\partial \theta^j} + \frac{\partial \theta^i}{\partial \theta^j} + \frac{\partial \theta^i}{\partial \theta^j} = \delta^i_j + \delta^i_j + \delta^i_j = 3\delta^i_j$$
So I get 3 times the delta function instead of only 1 delta function. I guess I should not do summation for some reason, but I don't understand why. There is a double 'index m' so a summation should be justified. In every part of the book 'classical and computational solid mechanics' the presence of a double index means summation.
4. Sep 11, 2010
### Fredrik
Staff Emeritus
The first line is correct, but that's not the chain rule. By the chain rule, what you have on the upper right is equal to
$$\frac{\partial\theta^i}{\partial\theta_j}$$
(once, not three times).
5. Sep 11, 2010
### bremvil
So basically you are saying that the term on the top right:
$$\frac{\partial \theta^i}{\partial x^m}\frac{\partial x^m}{\partial \theta^j} = \frac{\partial \theta^i}{\partial x^1}\frac{\partial x^1}{\partial \theta^j} + \frac{\partial \theta^i}{\partial x^2}\frac{\partial x^2}{\partial \theta^j} + \frac{\partial \theta^i}{\partial x^3}\frac{\partial x^3}{\partial \theta^j}$$
equals a single delta function? The way I interpret it, each term within the expression above is a single delta function.
6. Sep 11, 2010
### Fredrik
Staff Emeritus
You're not applying the chain rule correctly. Another example:
$$\frac{d}{dx}f(g(x),h(x))=\frac{\partial f}{\partial g}\frac{dg}{dx}+\frac{\partial f}{\partial h}\frac{dh}{dx}\neq \frac{df}{dx}+\frac{df}{dx}$$
7. Sep 11, 2010
### bremvil
I finally see it! Thanks a lot.
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2018-06-22 11:37:35
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https://www.enotes.com/homework-help/dy-dx-1-x-1-sqrt-4x-2-8x-1-solve-differential-764027
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# `dy/dx = 1/((x-1)sqrt(-4x^2+8x-1))` Solve the differential equation
The given problem` (dy)/(dx) =1/((x-1)sqrt(-4x^2+8x+1)) ` is in form of a first order ordinary differential equation. To evaluate this, we may follow the variable separable differential equation: `N(y) dy= M(x)dx` .
`dy=1/((x-1)sqrt(-4x^2+8x+1)) dx`
Apply direct integration on both sides:
`int dy=int 1/((x-1)sqrt(-4x^2+8x+1)) dx`
For the left side, we apply basic integration property: `int (dy)=y.`
For the right side, we apply several substitutions to simplify it.
Let `u =(x-1)` then `x=u+1` and `du=dx` . The integral becomes:
`int 1/((u)sqrt(-4x^2+8x+1)) dx =int 1/(usqrt(-4(u+1)^2+8(u+1)+1)) du`
`=int 1/(usqrt(-4(u^2+2u+1)+8u+8+1)) du`
`=int 1/(usqrt(-4u^2-8u-4+8u+8+1)) du`
`=int 1/(usqrt(-4u^2+5)) du`
Let `v = u^2` then `dv = 2u du` or `(dv)/(2u)=du` . The integral becomes:
`int 1/(usqrt(-4u^2+5)) du=int 1/(usqrt(-4v+5)) *(dv)/(2u)`
`=int (dv)/(2u^2sqrt(-4v+5))`
`=int (dv)/(2vsqrt(-4v+5))`
Apply the basic integration property: `int c*f(x)dx= c int f(x) dx` .
`int (dv)/(2vsqrt(-4v+5)) =(1/2)int (dv)/(vsqrt(-4v+5))`
Let `w= sqrt(-4v+5)` then `v= (5-w^2)/4` and `dw=-2/sqrt(-4v+5)dv` or
`(dw)/(-2)=1/sqrt(-4v+5)dv`
The integral becomes:
`(1/2)int (dv)/(vsqrt(-4v+5)) =(1/2)int 1/v*(dv)/sqrt(-4v+5)`
`=(1/2)int 1/((5-w^2)/4)*(dw)/(-2)`
`=(1/2)int 1*4/(5-w^2)*(dw)/(-2)`
`=(1/2)int -2/(5-w^2)dw`
`=(1/2)*-2 int 1/(5-w^2)dw`
`=(-1) int 1/(5-w^2)dw`
Apply basic integration formula for inverse hyperbolic tangent function:
`int (du)/(a^2-u^2)=(1/a)arctanh(u/a)+C`
Then, with corresponding values as: `a^2=5` and `u^2=u^2` , we get: `a=sqrt(5)` and `u=w`
`(-1) int 1/(5-w^2)dw = -1/sqrt(5) arctanh(w/sqrt(5))+C`
Recall `w=sqrt(-4v+5)` and `v=u^2` then `w =sqrt(-4u^2+5).`
Plug-in `u=(x-1)` on `w =sqrt(-4u^2+5)` , we get:
`w =sqrt(-4(x-1)^2+5)`
`w=sqrt(-4(x^2-2x+1)+5)`
`w=sqrt(-4x^2+8x-4+5)`
`w=sqrt(-4x^2+8x+1)`
Plug-in `w=sqrt(-4x^2+8x+1)` on `-1/sqrt(5) arctanh(w/sqrt(5))+C` , we get:
`int 1/((x-1)sqrt(-4x^2+8x+1)) dx=1/sqrt(5)arctanh(sqrt(-4x^2+8x+1)/sqrt(5))+C`
`=-1/sqrt(5) arctanh(sqrt(-4x^2+8x+1)/5)+C`
Combining the results from both sides, we get the general solution of the differential equation as:
`y=-1/sqrt(5) arctanh(sqrt(-4x^2+8x+1)/5)+C`
Approved by eNotes Editorial Team
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2022-10-01 08:14:13
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https://www.gamedev.net/blogs/entry/1305407-back/
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• entries
37
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• views
22047
# Back
310 views
I just got back from China.
Here is a list of weird crap that I ate:
-Sea Cucumber
-Soups made from various parts of the lotus plant
-Chicken Feet
-Tripe Sausage
-Duck's tongue
-Jellyfish
-Pork Gellatin
-A whole fish (literally)
The general rule in China is to not ask what you're eating. It's dangerous.
Okay, now that I have had a huge hamburger, I want to work on another game that I probably won't finish. The game will not use XNA, even though I was really excited about it, because I would rather learn how to use Managed DirectX.
All I know is that the game will be an RPG. It will be set in a fantasy/sci-fi universe. And it will be sort of like crono trigger, except with a more tactical battle system.
I know, I know, don't use all your creativity all at once [rolleyes]
I figure it will help if I post every time I start a part of the game. Like, so I can get suggestions. I'm not sure what the ideal order for working on stuff.
I plan on creating several interfaces to simplify the parts of my engine:
-Resource
-Subsystem
-DataStore
-DataAccessor
-Game
Anything that the player hears, sees, or anything that is rather large and needs to have specific lifespans will be a resource. Graphics, sound, animations, scripts, and anything that will load from my custom resource file format is a resource.
Subsystem should be pretty obvious. It's a game system. The Game interface will have a method that allows it to add subsystems, which it will then tick as often as possible.
Datastores are what allow different subsystems to communicate with each other. I only plan on using one, but I'm not making it a singleton because it's perfectly reasonable that you might want different subsystems to share different data with each other. Basically, the data store is a collection of lists of Events. Event looks like this:
struct Event{ public string subsystem; public string type; public Object data;}
That's it.
The datastore simply adds events for certain subsystems, and it's up to the DataAccessor to search it's list for events, and then report that to the subsystem.
Anyway. I'm jetlagged, so that may be totally incoherent. If it isn't, I would realy, REALLY like to hear suggestions for, or problems with this design. Please comment!
damn, isn't sea cucumber toxic?
Maybe. I had horrible diarrhea for the next four days. Then again, that could have been the water...
## Create an account
Register a new account
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2018-06-19 20:57:44
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https://crypto.stackexchange.com/questions/66404/how-does-twofish-avoid-weak-keys-if-it-uses-key-dependent-s-boxes
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# How does Twofish avoid weak keys if it uses key-dependent S-boxes?
From section 7.2.2 of the Twofish paper, the four S-boxes for the 128-bit cipher are generated with:
\begin{align} s_0(x) &= q_1[q_0[q_0[x] \oplus s_{0,0}] \oplus s_{1,0}]\\ s_1(x) &= q_0[q_0[q_1[x] \oplus s_{0,1}] \oplus s_{1,1}]\\ s_2(x) &= q_1[q_1[q_0[x] \oplus s_{0,2}] \oplus s_{1,2}]\\ s_3(x) &= q_0[q_1[q_1[x] \oplus s_{0,3}] \oplus s_{1,3}] \end{align}
Here, $$q_0$$ and $$q_1$$ are fixed 8-bit permutations, and $$s_{i,j}$$ are bytes derived from the keys indirectly via a Reed–Solomon matrix. They claim that this scheme does not produce any weak S-boxes:
Few or no keys may cause the S-boxes used to be “weak,” in the sense of having high-probability differential or linear characteristics, or in the sense of having a very simple algebraic representation
Why is this? How can Twofish avoid weak keys while using key-dependent S-boxes?
The key to the answer lies within these two snippets:-
These results help confirm our belief that, from a statistical standpoint, the Twofish S-box sets behave largely like a randomly chosen set of permutations.
and
There should be few or no pairs of keys that define the same S-boxes. That is, changing even one bit of the key used to define an Sbox should always lead to a different S-box. In fact, these pairs of keys should lead to extremely different S-boxes.
Taken together, they describe an quasi encryption/hash of the key à la SEAL (but simpler). Perhaps akin to a CRC calculation or a mini randomness extractor. Even prior to the MDS matrix. There's probably a better scientific term for it.
$$q_0$$ and $$q_1$$ are fairly good S boxes themselves. Even though they are random permutations, their arrangement has been numerically optimised for differential and linear characteristics of $$\frac{10}{256}$$ and $$\frac{1}{16}$$ respectively, and no more than two fixed points. Taken with the permuted arrangement in the three levels deep nested look ups of $$q_0$$ and $$q_1$$ within the definition of $$S_i$$, it's extremely unlikely that you'd find pre-images that create all weak $$S_i$$. It's difficult to algebraically define what a related key might look like for a given pre-image. That's just a consequence of all nested look ups into random permutations. The permuted nesting is highlighted below:-
Like you, the authors worried about weak keys too. Hence a large § 7.2.3 Exhaustive and Statistical Analysis. Monte Carlo simulation seems to confirm their assumptions that weak keys are quite unlikely due to the pseudo permutation and avalanche behaviours. Your "very simple algebraic representation" doesn't occur due to the inherent randomness within $$q_i$$, where the shortest computational representation of $$q_i$$ is $$q_i$$.
• I'm not sure that I really understand this answer. How are those permutations S-boxes, and how are they similar to CRC in any way? It's not like burst detection is relevant. Why would you need to find preimages of a simple permutation just to create weak keys? Why do you quote the requirement that individual bit changes modify each S-box? What is the relevance of related key attacks here? The only thing I can take away from your answer is that it's hard to individually specify exact S-box bits, which is something I already know. – forest Jan 14 '19 at 0:42
• @forest Burst detection? Isn't that used in astronomy and nuclear testing? – Paul Uszak Jan 14 '19 at 13:08
• @forest The quasi hashing effect within my purple square shows (attempts to) why your last sentence holds. If you quasi hash the key into $S_i$, you generate well distributed S boxes for all keys and thus identifying weak keys becomes difficult. It's in §7.2.3 and proved in tables 4 - 6. – Paul Uszak Jan 14 '19 at 13:09
• No, burst detection is an aspect of CRC behavior. – forest Jan 17 '19 at 6:34
• Also, what does "quasi-hash" mean? – forest Jan 17 '19 at 6:42
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2020-01-19 21:31:45
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https://tex.stackexchange.com/questions/387240/calendar-with-tikz-pgf-how-to-highlight-current-date/387287
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# Calendar with TIKZ-PGF: How to highlight current date?
I am trying to generate a 365-day notebook that contains the current date and a small calendar of the current month. I wwould like to highlight the current day in the small calendar, for example by adding a circle around it or by making it in boldface. For example, if the date listed on the left is January 10, I would like to have a mark on the calendar on the right highlighting the date January 10. After spending hours of reading the pgf/tikz manuals and calendar-related topics on this forum, I did not reach any solution.
Below is my code:
\documentclass[b5paper, fontsize=12pt, parskip=half, DIV=12,BCOR=2cm]{scrartcl}
%pagesize,
\usepackage[T1]{fontenc}
\usepackage[romanian]{babel}
\usepackage{translator, tikz, pgfcalendar, array, libertine}
\usepackage{ucs}
\usepackage[utf8x]{inputenc}
\usepackage{xcolor}
\usepackage{tabu}
\usepackage{epstopdf}
\usepackage{color}
\usepackage{pgf}
%=========================================================
%== Page characteristics (Page Setup) ====================
\setlength{\textwidth}{126mm}
\setlength{\textheight}{230.0mm}
\setlength{\topmargin}{-30mm}
\setlength{\hoffset}{-20mm}
\setlength{\marginparwidth}{-20mm}
\setlength{\footskip}{12mm}
\setlength{\parindent}{0cm}
\usetikzlibrary{calendar,shapes.geometric}
%== Define colors =================================
\definecolor{rozmam}{RGB}{242,172,184}
\newcommand{\troz}{\textcolor{rozmam}}
\makeatletter%
execute before day scope={ \ifdate{day of month=1}{%
\pgfmathsetlength{\pgf@ya}{\tikz@lib@cal@yshift}%
\pgfmathsetlength\pgf@xa{\tikz@lib@cal@xshift}%
\pgftransformyshift{-\pgf@ya}
\foreach \d/\l in {0/M,1/T,2/W,3/T,4/F,5/S,6/S} {
\pgf@xa=\d\pgf@xa%
\pgftransformxshift{\pgf@xa}%
\pgftransformyshift{\pgf@ya}%
}
}{}%
}%
]
\makeatother
\newcommand{\PaginaMea}{
\minisec{\textcolor{gray}{Notes}}
\begin{tabu}{@{}p{0.12\textwidth}p{0.32\textwidth}p{0.32\textwidth}p{0.20\textwidth}c}\taburulecolor{gray}\hline
$\diamond$ \textbf{ } & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
\end{tabu}
}
\begin{document}
\foreach \n in {1,...,1}{
\pgfcalendar{cal}{2018-\n-01}{2018-\n-last}
{
\thispagestyle{empty}
\noindent{\Huge \bfseries
\pgfcalendarcurrentday{}\\[20pt]
\Huge \pgfcalendarweekdayname{\pgfcalendarcurrentweekday}
}
\begin{tikzpicture}[remember picture, overlay, transform shape]
\node [anchor=north east, inner sep=8pt, xshift = 30pt]
at (current page.north east)
{
\begin{tikzpicture}[every node/.append style = {anchor=center}]
\calendar[dates=2018-\n-01 to 2018-\n-last,
week list,
day text=\%d0,
month label above centered,
month text={\%mt} \%y-,
day xshift = 0.8cm,
]
if (Saturday) [rozmam]
if (Sunday) [rozmam];
\end{tikzpicture}
};
\end{tikzpicture}
\\[30pt]
\PaginaMea
\pagebreak
{}
}
}
\end{document}
There are two ways to mark the current day. For both, day and month of the outer calendar must be known. this can be achieved with
\edef\pagedayofmonth{\pgfcalendarcurrentday}%
\edef\pagemonth{\pgfcalendarcurrentmonth}%
in the definition of the outer calendar. The \edef is necessary, because in the small calendar the two \pgfcalendarcurrent... macros will deliver the values for the latter.
Then if (day of month=\pagedayofmonth) [cyan] can be added to the small calendar to change the color of the number. nodes={draw} could be added, but this would draw a rectangle, which is too large.
Or the nodes defined for each day can be used to draw around or over the number. This is done after the small calendar. To make it work, the calendar must be named. Then the nodes for the days will be named in the form calendarname-year-month-day, e.g. mycal-2018-02-07.
\draw[green,thick,rounded corners]
($(mycal-2018-\pagemonth-\pagedayofmonth.south east) + (-1.5mm,1.5mm)$) rectangle
($(mycal-2018-\pagemonth-\pagedayofmonth.north west) + (1.5mm,-1.5mm)$);
The calculation is done to achieve a reasonable size of the rectangle, because just using the anchors would result in a too large rectangle.
The result (showing both methods):
And the code:
Here I also changed the outdated \tikzstyle to \tikzset, see comments in the code.
\documentclass[b5paper, fontsize=12pt, parskip=half, DIV=12,BCOR=2cm]{scrartcl}
%pagesize,
\usepackage[T1]{fontenc}
\usepackage[romanian]{babel}
%\usepackage{libertine} % fonts not installed here
\usepackage{translator, tikz, array} % pgfcalendar loaded later with \usetikzlibrary{calendar}
\usepackage{ucs}
\usepackage[utf8x]{inputenc}
\usepackage{tabu}
\usepackage{epstopdf}
%=========================================================
%== Page characteristics (Page Setup) ====================
% using geometry would be better
\setlength{\textwidth}{126mm}
\setlength{\textheight}{230.0mm}
\setlength{\topmargin}{-30mm}
\setlength{\hoffset}{-20mm}
\setlength{\marginparwidth}{-20mm}
\setlength{\footskip}{12mm}
\setlength{\parindent}{0cm}
\usetikzlibrary{calendar,shapes.geometric}
\usetikzlibrary{calc}
%== Define colors =================================
\definecolor{rozmam}{RGB}{242,172,184}
\newcommand{\troz}{\textcolor{rozmam}}
\makeatletter%
% these days, \tikzset is used
\tikzset{%
execute before day scope={ \ifdate{day of month=1}{%
\pgfmathsetlength{\pgf@ya}{\tikz@lib@cal@yshift}%
\pgfmathsetlength\pgf@xa{\tikz@lib@cal@xshift}%
\pgftransformyshift{-\pgf@ya}
\foreach \d/\l in {0/M,1/T,2/W,3/T,4/F,5/S,6/S} {
\pgf@xa=\d\pgf@xa%
\pgftransformxshift{\pgf@xa}%
\pgftransformyshift{\pgf@ya}%
}
}{}%
}
}
}
\makeatother
\newcommand{\PaginaMea}{
\minisec{\textcolor{gray}{Notes}}
\begin{tabu}{@{}p{0.12\textwidth}p{0.32\textwidth}p{0.32\textwidth}p{0.20\textwidth}c}\taburulecolor{gray}\hline
$\diamond$ \textbf{ } & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
$\diamond$ & & & & \\\hline
\end{tabu}
}
\begin{document}
\foreach \n in {2,...,2}{
\pgfcalendar{cal}{2018-\n-01}{2018-\n-last}
{
% added to ensure every day starts on a new page
\newpage
\thispagestyle{empty}
% remember day and month
\edef\pagedayofmonth{\pgfcalendarcurrentday}%
\edef\pagemonth{\pgfcalendarcurrentmonth}%
\noindent{\Huge \bfseries
\pgfcalendarcurrentday{}\\[20pt]
\Huge \pgfcalendarweekdayname{\pgfcalendarcurrentweekday}
}
\begin{tikzpicture}[remember picture, overlay, transform shape]
\node [anchor=north east, inner sep=8pt, xshift = 30pt]
at (current page.north east)
{
\begin{tikzpicture}[every node/.append style = {anchor=center}]
\calendar (mycal) [dates=2018-\n-01 to 2018-\n-last,
week list,
day text=\%d0,
month label above centered,
month text={\%mt} \%y-,
day xshift = 0.8cm,
]
if (Saturday) [rozmam]
if (Sunday) [rozmam]
% just set another text color; nodes={draw} could be added, but the rectangle is too large
if (day of month=\pagedayofmonth) [cyan]
;
% draw a frame around the day
\draw[green,thick,rounded corners]
($(mycal-2018-\pagemonth-\pagedayofmonth.south east) + (-1.5mm,1.5mm)$) rectangle
($(mycal-2018-\pagemonth-\pagedayofmonth.north west) + (1.5mm,-1.5mm)$);
\end{tikzpicture}
};
\end{tikzpicture}
\\[30pt]
\PaginaMea
\pagebreak
{}
}
}
\end{document}
• Many thanks, @Mike. Completely solved my problem. Also thanks for your patience and pedagogical approach. – Vladimir Martinusi Aug 20 '17 at 9:00
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2020-01-22 11:55:00
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https://www.nature.com/articles/s41597-022-01573-2?error=cookies_not_supported&code=a4523691-7014-4681-8894-91678c118e19
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## Background & Summary
Over the past few years, Internet of Things (IoT) has become a key enabler for various applications since it supports the exchange of ubiquitous data or information among smart devices or sensors with little to no human intervention1. It has revolutionized research in Human Activity Recognition (HAR) techniques due to their potential applications in areas such as healthcare (e.g., monitoring elderly people or those with disabilities), smart homes (e.g., controlling appliances based on human activities for achieving efficient energy consumption), surveillance and security, virtual gaming, among others. Numerous techniques have been proposed for HAR, ranging from inertial/wearable sensors2,3,4,5,6, vision-based methods such as Microsoft Xbox Kinect sensor7, to unobtrusive methods based on Radio-Frequency (RF) waves such as WiFi Channel State Information (CSI)8, Passive WiFi Radar (PWR)9 and Ultra-Wideband (UWB)10. Recognizing human activities, especially using RF signals, is a challenge and open-source datasets would help in devising techniques and algorithms that can boost research in this field, which can ultimately lead to standardization. To date many datasets acquired from RF, vision and inertial sensors have been published, which are intended for a number of applications. These include WiFi CSI-based activity recognition11,12,13,14,15, sign language recognition16, fall detection17, device-to-device localization18, or UWB-based gesture recognition19, motion detection/recognition20,21,22, passive target localization23, people counting24, or active radar-based sensing25,26,27,28, as well as physical activity recognition using inertial/wearable sensors29,30,31,32,33,34, while others have proposed action recognition datasets acquired from vision and motion capture systems35,36,37,38. However, most of these databases have some shortcomings in the layout and number of sensors, which cannot fully represent the human activity features. First, most of these datasets comprise of measurements from a single sensor. Second, the datasets are limited by the number of human activities being captured and the layout used for capturing the datasets. Compared with unimodal sensors, multimodal sensors can collect different types of data which differ in characteristics, including dimensionality, distribution, and sparsity. Fusion of multimodal data can help in describing human actions more accurately. As compared to the aforementioned works and to the best of the authors’ knowledge, this is the first work to propose a multimodal dataset comprising of RF and vision-based methods that is intended not only for the sensing of day-to-day activities but also for passive (uncooperative) localization. The contributions herein are:
• Multimodal data collection intended for human activity recognition and passive localization, i.e, the targets are oblivious to these processes (non-collaborative) and they only reflect or scatter the signals from the transmitter to receivers. Most datasets consider only one particular modality such as either UWB, WiFi CSI, PWR or Kinect, independently. In this work, we consider multiple synchronized modalities. Experiments span across two environments which can be used for investigating sensing in complex or untrained environments.
• Approximately 8 hours of measurements are fully annotated with location and activity labels of high temporal resolution, capturing the participant’s movement and natural behaviour within the monitoring area, as would be the case in a real-world environment. The dataset is comprehensive in so far it contains over 1 Million annotated data points.
• The presented data can be exploited to advance human activity recognition technology in different ways, for example, using various pattern recognition and deep learning algorithms to accurately recognize human activities. For this purpose, the users can apply different signal processing pipelines to analyze the recorded WiFi CSI, PWR, UWB and Kinect data and extract salient features that can be used to recognize the human activities and/or concurrently track the target’s position within an indoor environment.
• This is the first dataset that is collected with an explicit aim to accelerate the development of self-supervised learning techniques. Such techniques are extremely data hungry, requiring orders of magnitude larger datasets compared to more traditional supervised learning. Furthermore, each modality consists of multiple receivers and this corresponds to multiple views of the data which can undoubtedly be used in multimodal and multiview data fusion networks for improving the performance in concurrent activity recognition and localization tasks.
This open-source dataset is intended for both HAR and non-cooperative localization, which are areas of growing interest to research communities including but not limited to radar, wireless sensing, IoT and computer vision. To ensure that the dataset aligns to the FAIR (Findable, Accessible, Interoperable, Reusable) Data principles of Open Science, we have (i) made it publicly available for download via the Figshare portal, (ii) provided an in-depth and clear description of the dataset for each modality, (iii) formatted our dataset using standard filetypes and encoding, and (iv) provided example scripts/codes that will allow the user to load and analyze the data from each modality.
## Methods
Experiments were performed in a university environment in two furnished rooms, with desks, chairs, screens, and other office objects lying in the surroundings. The room layouts are depicted in Fig. 1 along with their physical dimensions. A maximum of six subjects of different age groups participated in the experiments which were intended for the sensing of day-to-day activities as well as non-collaborative localization.
Our dataset39 is composed of approximately 8 hours of measurements that were collected across multiple modalities including WiFi Channel State Information (CSI), Ultra-Wideband (UWB), Passive WiFi Radar (PWR) and Kinect sensor systems. The breakdown of the activities’ durations is given in Table 1. The monitoring devices were installed on the extremity (boundary) of the rooms such that enclosed spaces of dimensions 2.46 m × 4.40 m and 4.06 m × 4.53 m were used as monitoring areas for Room 1 and 2, respectively. The description of the various experiments performed is provided in Table 2.
Even though no personal data has been collected from the participants during the experiments, each participant was still fully informed about the purpose of the study and what was expected of them. Informed consent was obtained from each participant prior to the experiments. All studies that fall under the OPERA - Opportunistic Passive Radar for Non-Cooperative Contextual Sensing project were thoroughly reviewed and fully approved by the University of Bristol Faculty of Engineering Research Ethics Committee (application number: 96648). Risk assessment was also carried out and approved prior to the experiments.
Referring to the experiment numbers in Table 2, exp001–exp054 were performed in Room 1 while exp055–exp061 were carried out in Room 2. exp028 is the crowd counting experiment whereby six people walked randomly and continuously within the monitoring area of Room 1. Then, after approximately every 5 minutes, one person moved out of the room. Figure 2 shows the particular instant of exp028 where 5 out of the 6 people already left the monitoring area and only the last person’s ground truth walking trajectory is shown. For illustration purposes only, a moving average filter is applied to the raw ground truth positions to smooth the target’s trajectory path. The experiments exp034–exp048 (exp034 is the empty room background data) were device-free localization experiments involving a human target who was standing still at several positions or walking along a straight short path in a number of directions as shown in Fig. 3. The target wore a tag to get his/her ground truth position. Note that only the WiFi CSI transmitter (NUC3) and receiver (NUC2) were used during these experiments for recording data and they were placed side by side. As for the device-to-device localization experiments (exp049–exp054), the CSI transmitter (NUC3) and receiver (NUC2) were placed at different angles with respect to each other (0°, 30°, −30°, 60°, −60°), as shown in Fig. 4 (no human target). Note that one tag was placed on the CSI transmitter and another on the receiver to get their fixed ground truth locations within the environment.
### UWB
Three UWB systems were used during the experiments. The first system (see red nodes in Fig. 1) was used to obtain the ground truth position of the target while he/she wore one or more tags and moved within the monitoring area. The update rate, that is, the rate at which the 2D xy coordinates were logged for each individual tag was 10 Hz (using Decawave’s DRTLS android app). The other two passive UWB systems (yellow and blue nodes) consisted of fixed nodes installed in a multi-static configuration and which were exchanging CIR data among themselves. UWB system 1 (yellow nodes in Fig. 1) was implemented using four Decawave’s EVK100042 modules. The modules were programmed with a custom firmware so as to record CIR data on all of them. Node ‘0’ was acting as an initiator whereby it exchanged Single-Sided Two-Way Ranging (SS-TWR) messages (poll, response and final) with each of the other 3 nodes. When a given node replies back, the frame is broadcast and heard by all other nodes operating on the same channel. In this way, each node can read the received frames in their accumulator and extract the CIR data. Therefore, CIR data is available in a bidirectional mode between all pairs of nodes. This means that all nodes act as transmitters and receivers, giving rise to a maximum of 12 communication links. The 4 nodes were connected to laptops in order to record the CIR data via a serial terminal.
UWB system 2 (blue nodes in Fig. 1) was implemented using five Decawave’s MDEK100143 modules. These units were also flashed with custom firmware so as to record CIR data on all of them. Node ‘0’ was acting as an initiator and transmitted a packet every 10 ms. The packet essentially includes a time schedule for transmission for the other 4 nodes. In this way, each node knows who needs to transmit next and when with minimal delay. Thus the transmission was performed in a round-robin fashion to avoid collision. Nodes with IDs 1–4 were connected to laptops to record the CIR data via a serial terminal. The average packet rate for UWB system 1 (yellow nodes) was around 400 Hz while for UWB system 2 (blue nodes), the average packet rate was around 195 Hz, considering combined communication links. The other parameters for the three UWB systems are summarized in Table 4.
The Decawave’s UWB chipset stores the CIR in an accumulator and each tap of the accumulator represents a sampling interval of Δτs≈ 1.0016 ns (i.e., half a period of the 499.2 MHz fundamental frequency)44. The accumulator spans over one symbol time. This represents 992 and 1016 samples for the nominal Pulse Repetition Frequency (PRF) of 16 MHz and 64 MHz, respectively. Each measured CIR sample is a complex number which can be broken down into its real and imaginary components. Only 35 and 50 CIR samples out of 1016 are considered in the experiments for UWB systems 1 and 2, respectively. These correspond to a sensing range of 10.5 m and 15 m for UWB systems 1 and 2, respectively. Each CIR measurement was read from the accumulator memory starting 3 samples (i.e., 3 ns) before the detected first path index (reported by the FP_INDEX field in register 0x15 of DW1000 chipset) by the Leading Edge Detection (LDE) algorithm. As the CIR magnitudes are dependent on the number of preamble symbols used for CIR accumulation, for each CIR measurement in the UWB datasets, the magnitude values have been normalized using the Preamble Accumulation Count (PAC) value45 (see rx_pream_count column in UWB datasets), as reported by the RXPACC register in the DW1000 chipset.
### PWR
For the PWR system, a USRP-294546 was used as the receiver which is equipped with four synchronized channels. The USRP-2945 features a two-stage superheterodyne architecture with four independent receiving channels and shares local oscillators for phase-coherent operation. Each receiving channel was equipped with a 6-dB directional antenna. The collected raw data is then routed to a computing unit through a PCIe port, which is a desktop computer in this work. A PWR system consists of a minimum of two synchronized channels; a surveillance channel which records reflected WiFi signals from the monitoring area and a reference channel which records the direct signal emitted from the transmitter. As mentioned previously, four channels are used in the USRP-2945, where one channel was used as the reference channel (denoted as “rx1” in Fig. 1) while the other three channels were used as surveillance channels (denoted as “rx2”, “rx3” and “rx4” in Fig. 1). Since the PWR system does not transmit a signal (it only monitors received signals), it can use any third-party signal source as the illuminator but however, a reference signal is needed. In this work, we used the CSI transmitter (NUC3) as the PWR source for convenience, allowing a direct comparison between the two systems’ performance.
PWR correlates the signal from the surveillance and reference channels to estimate two parameters: relative range and Doppler frequency shift. Additionally, a CLEAN47 algorithm has been used to remove the direct signal interference. More details on this signal processing can be found in47. However, due to the limitation of the WiFi signal bandwidth (40 MHz in this work), the range resolution is limited to 3.75 meters which is too coarse for indoor applications. Therefore, only the Doppler frequency shifts are recorded in the form of Doppler spectrograms. The output from the PWR system is specified as ns × nb × Nt real values, where ns is the number of surveillance channels, nb is the number of Doppler bins and Nt is the number of time frames. Other details about the PWR’s parameters are given in Table 5. Note that the PWR system recorded the spectrogram data at a measurement rate of 10 Hz. The measurement rate refers to the number of system outputs per second48. For the WiFi CSI system, the measurement rate is equal to the number of packets received per second, which is equal to 1.6 kHz. As for the PWR system, the measurement rate is limited by the number of baseband signals that can be processed by the computing device. Therefore, the measurement rate of the PWR system was empirically chosen and set at 10 Hz.
### Kinect
We used two of Microsoft’s Kinect v2 sensors to gather motion capture data from different human activities. Kinect v2 incorporates an infrared depth sensor, a RGB camera, and a four-element microphone array that provides functionalities such as three-dimensional skeletal tracking, facial recognition, and voice recognition. Although the device was originally developed to play games, numerous researchers have used it for applications beyond its initial intended purpose. Due to the low cost and wide availability, it has now been used extensively in research areas such as video surveillance systems where multiple Kinect devices are synchronized to track groups of people even in complete darkness49, improve live three-dimensional videoconferencing50 and in medical applications to measure a range of conditions such as autism, attention-deficit disorder and obsessive-compulsive disorder in children51. Note that in skeleton tracking, Kinect might suffer from occlusion when some parts of the human body are occluded with others and therefore cannot be tracked accurately.
Therefore, in this work, we used two Kinects to track three-dimensional time-varying skeletal information of the human body, including 25 joints such as head center location, knee joints, elbow joints, and shoulder joints from two different directions. The real advantage of using motion capture technology is capturing more accurate, more realistic, and complex human motions. This three-dimensional joint information can further be used for simulating the corresponding radar scatterings mimicking a typical PWR sensing system. In one of our previous works, we presented an open-source motion capture data-driven simulation tool, SimHumalator, that can generate large volumes of human micro-Doppler radar data at multiple IEEE WiFi standards (IEEE 802.11g, ax, and ad)52. Radar scatterings were simulated by integrating the animation data of humans with IEEE 802.11 compliant WiFi transmissions to capture features that incorporate the diversity of human motion characteristics and the sensor parameters. More importantly, we have demonstrated that the human micro-Doppler data generated using the simulator can be used to augment limited experimental data53,54. Interested researchers can download the simulator from https://uwsl.co.uk/. The output from the Kinect system is specified as Nt × Nb × Nd real values, where Nt is the number of time frames, Nb is the number of tracked joints on the human body, and Nd is the three-dimensional position (x, y, z) information.
### Ground truthing
• The Decawave (now acquired by Qorvo) MDEK1001 development kit43 was used for obtaining the ground truth position of the targets. 11 units were configured as anchors and mounted on walls in the experiment rooms (see red nodes in Fig. 1). Their xy coordinates were manually measured using a laser measuring device, which were then entered in the DRTLS Android app. A maximum of 6 tags were configured for exp028, while for the activity recognition experiments, the person wore two tags, one on each arm. Two UWB units were also configured as listeners so as to record the xy coordinates of the tags using a serial terminal on two laptops, along with their timestamps with millisecond precision. The MDEK1001 kit features a location update rate of up to 10 Hz for each individual tag. Since two listeners were deployed, this means we can have up to 20 Hz update rate for each individual tag.
• As for the labelling of activities, a program was developed in Matlab with automated voice output to instruct the person when to perform the various activities such as sitting, standing, etc. At the same time, the programmable script recorded the timestamps (with millisecond precision) at which the activity was instructed to be performed. The person just had to listen to the voice command and perform the activity accordingly. As a backup solution, another activity labelling application was developed in Matlab where one can insert the labels for the required activities. Then, an observer constantly looked at the person doing the activities and clicked on the appropriate button in the app to record the start and stop times of the activity. All labels were stored in text files along with their timestamps (with millisecond precision). Note that all modalities were synchronized to the same local Network Time Protocol (NTP) server, resulting in synchronization accuracy across all modalities of <20 ms.
## Data Records
Measurements have been collected across four modalities during the experiments, namely, WiFi CSI, UWB, PWR and Kinect. The dataset can be accessed and downloaded from our Figshare repository39. The dataset has been compressed (zipped) into separate folders for each modality, allowing the user to only download the data of interest. The zipped folders’ names and the number of files in each folder, along with their file formats, are specified in Table 6. The directories wificsi1 and wificsi2 refer to the data collected by the WiFi CSI receivers, denoted by “NUC1” and “NUC2” in Fig. 1, respectively. uwb1 and uwb2 refer to the data collected by the two passive UWB systems, represented by the yellow and blue nodes in Fig. 1, respectively. The directory pwr contains the PWR spectrogram data recorded from the three surveillance channels (“rx2”, rx3” and “rx4” represented as black triangles in Fig. 1) for each experiment (excluding exp001, exp019, exp034–exp054). Finally, the directory kinect contains the Kinect sensor data files for each experiment (excluding exp001, exp019, exp028, exp034–exp055).
### Experiment directory
Each file in the directories specified in Table 6 corresponds to a given experiment number, the details of which are provided in Table 2.
#### WiFi CSI dataset description
This section describes the structure of the data files residing in the wificsi1 (NUC1) and wificsi2 (NUC2) directories. The files are in .mat format and each row in the file corresponds to a received CSI packet. The columns in the dataset have the following headers:
• timestamp: UTC+01 00 timestamp in milliseconds when the CSI packet was captured by the receiver.
• activity: current activity being performed. The activity is specified as a string of characters with no spacing e.g., “background”, “walk”, “sit”, “stand”, “liedown”, “standfromlie”, “bodyrotate”. These correspond to the activity numbers 1, 2, 3, 4, 5, 6 and 7 in the “Details” column in Table 2, respectively. The activity label “noactivity” refers to the case where the person was not performing any activity, that is, his/her body was at rest, for example between activities such as “sitting” and “standing” or “lying down on floor” and “standing up from the floor”. For exp035–exp054, the activity is specified as “Loc1”, “Loc2”, , “Loc9” (device-free static target localization), “path1”, “path2”, , “path5” (device-free dynamic target localization), “loc1”, “loc2”, , “loc6” (device-to-device localization). For these localization experiments, each file also contains a column with header notes which gives more details on the position of the transmitter (tx) and receiver (rx) and the target (if applicable).
• exp_no: experiment number which is specified as “exp_001”, “exp_002”, etc. See Table 2 for more details.
• person_id: person ID specified as “One”, “Two”, “Three”, etc.
• room_no: room ID specified as “1” (left room in Fig. 1) or “2” (right room in Fig. 1).
• tag4422_x, tag4422_y, tag89b3_x, tag89b3_y: refer to the ground truth position of the person in the monitoring area in terms of 2D x- and y- coordinates. Note that for all experiments, except exp001, exp019, exp028, and exp034–exp055, the person was wearing two UWB tags on either arms, bearing IDs 4422 and 89b3. The information regarding which tag is worn on which arm is given in the columns with headers left_arm_tag_id and right_arm_tag_id. For the crowd counting experiment (exp028), there were a maximum of 6 people and hence 6 UWB tags were used to obtain the ground truth position of each person. Each person wore the tag on his/her left arm. In the WiFi CSI files for exp028, the x- and y- coordinates of the person are given in the columns tag4422_x, tag4422_y, tag89b3_x, tag89b3_y, tag122c_x, tag122c_y, tag4956_x, tag4956_y, tag1e85_x, tag1e85_y, and tag9118_x, tag9118_y. The person bearing UWB tag ID 4956 was the first to step out of the monitoring area, followed by 9118, 1E85, 4422, 89B3, and finally 122C.
• anchor_node_xy_positions: x- and y- coordinates of the eleven UWB anchor nodes distributed across the rooms (see red nodes in Fig. 1) for obtaining the ground truth position of the tag/s.
• tx1rx1_sub1, tx1rx1_sub2, , tx3rx3_sub30 (270 columns): The first corresponds to the raw complex CSI values for transmit antenna 1 (tx1), receive antenna 1 (rx1) and subcarrier 1 (sub1), the second corresponds to transmit antenna 1 (tx1), receive antenna 1 (rx1) and subcarrier 2 (sub2), and so on. The WiFi CSI systems used a 3 × 3 MIMO configuration and since the Intel 5300 NIC extracts CSI data over 30 subcarriers, the total number of complex CSI values per packet is 3 × 3 × 30 = 270.
• tx_x_coord, tx_y_coord, target_x_coord, target_y_coord (specified for exp035–exp054 only): For exp035–exp048, tx_x_coord and tx_y_coord respectively correspond to the x- and y- coordinates of both the CSI transmitter (NUC3) and CSI receiver (NUC2) since they were placed side by side while the target was standing still at several positions or walking along a short path. The human target was holding a tag and its ground truth x- and y- coordinates are given by target_x_coord and target_y_coord, respectively. As for exp049–exp054, tx_x_coord and tx_y_coord refer to the x- and y- coordinates of the CSI transmitter (NUC3), respectively, while target_x_coord and target_y_coord refer to the x- and y- coordinates of the CSI receiver (NUC2), respectively. No human target was present in this case.
#### UWB dataset description
This section describes the structure of the data files residing in the uwb1 (UWB system 1- yellow nodes) and uwb2 (UWB system 2- blue nodes) directories. The files are in .csv format and each row in the file corresponds to a received UWB packet. The UWB dataset files have the following fields similar to the WiFi CSI datasets: timestamp, activity, exp_no, person_id, room_no, tag4422_x, tag4422_y, tag89b3_x, tag89b3_y, tag122c_x, tag122c_y, tag4956_x, tag4956_y, tag1e85_x, tag1e85_y, tag9118_x, tag9118_y, left_arm_tag_id, right_arm_tag_id, and anchor_node_xy_positions. The additional column headers or those that are different from the WiFi CSI dataset headers are described below:
• fp_pow_dbm: estimate of the first path power level (in dBm) of the UWB signal between a pair of nodes. The formula for computing this value is given in44.
• rx_pow_dbm: estimate of the receive power level (in dBm) of the UWB signal between a pair of nodes. The formula for computing this value is given in44.
According to the manufacturer, the above two estimated parameters can be used to infer whether the received signal is Line-of-Sight (LoS) or Non-Line-of-Sight (NLoS). It is stated that, as a rule of thumb, if the difference of the two parameters, i.e., rx_pow_dbm - fp_pow_dbm is less than 6 dB, the signal is most likely to be LoS, whilst if the difference is greater than 10 dB, the signal is likely to be NLoS44.
• tx_id: index of the transmitting node. For UWB system 1 (yellow nodes), the transmitting node IDs are 0, 1, 2 or 3. For UWB system 2 (blue nodes), the transmitting node IDs are 0, 1, 2, 3 or 4 (see Fig. 1).
• rx_id: index of the receiving node. For UWB system 1 (yellow nodes), the receiving node IDs are 0, 1, 2 or 3. For UWB system 2 (blue nodes), the receiving node IDs are 1, 2, 3 or 4.
• tx_x_coord,tx_y_coord: x- and y- coordinates of the transmitting node, respectively.
• rx_x_coord,rx_y_coord: x- and y- coordinates of the receiving node, respectively.
• tx_rx_dist_meters: separation distance between the pair of transmitting and receiving nodes in meters.
• fp_index: accumulator first path index as reported by the Leading Edge Detection (LDE) algorithm of the DW1000 UWB chipset in register 0x15 (in FP_INDEX field). It is a sub-nanosecond quantity, consisting of an integer part and a fractional part.
• fp_amp1: first path amplitude (point 3) value reported in the FP_AMPL1 field of register 0x15 of the DW1000 UWB chipset.
• fp_amp2: first path amplitude (point 2) value reported in the FP_AMPL2 field of register 0x12 of the DW1000 UWB chipset.
• fp_amp3: first path amplitude (point 1) value reported in the FP_AMPL3 field of register 0x12 of the DW1000 UWB chipset.
Basically, fp_amp1, fp_amp2 and fp_amp3 are the magnitudes of the accumulator tap at the indices 3, 2 and 1, respectively, beyond the integer part of FP_INDEX reported in register 0x15 of the DW1000 UWB chipset44. That is, fp_amp1 = amplitude at ceiling(FP_INDEX) + 3, fp_amp2 = amplitude at ceiling(FP_INDEX) + 2 and fp_amp3 = amplitude at ceiling(FP_INDEX) + 1.
• max_growth_cir: Channel Impulse Response (CIR) power value reported in the CIR_PWR field of register 0x12 of the DW1000 UWB chipset. This value is the sum of the squares of the magnitudes of the accumulator from the estimated highest power portion of the channel, which is related to the receive signal power44.
• rx_pream_count: Preamble Accumulation Count (PAC) value reported in the RXPACC field of register 0x10 of the DW1000 UWB chipset. RXPACC reports the number of accumulated preamble symbols. The DW1000 chip estimates the CIR by correlating a known preamble sequence with the received signal and accumulating the result over a time period. The number of preambles used for the CIR estimation is dependent on the quality of the received signal.
• max_noise: LDE maximum value of noise.
• std_noise: standard deviation of noise.
• cir1, cir2, : These correspond to the 35 and 50 raw complex CIR samples for UWB systems 1 and 2, respectively.
#### PWR dataset description
This section describes the structure of the data files residing in the pwr directory. The files are in .mat format and each row in the files corresponds to a PWR measurement from each of the three receivers (surveillance channels) at a given point in time.
• exp_no: experiment number which is specified as “exp_002”, “exp_003”, etc. See Table 2 for more details. Note that the PWR system does not need background scan. Hence, background data for “exp_001” and “exp_019” were omitted for the PWR system.
• timestamp: UTC+01 00 timestamp in milliseconds when the Doppler spectrograms were recorded.
• activity: ground truth activity labels. The activity is specified as a string of characters with no spacing e.g., “walk”, “sit”, “stand”, “liedown”, “standfromlie”, “bodyrotate”. These correspond to the activity numbers 1, 2, 3, 4, 5, 6 and 7 in the “Details” column in Table 2, respectively.
• person_id: person ID specified as “One”, “Two”, “Three”, etc.
• room_no: room ID specified as “1” (left room in Fig. 1) or “2” (right room in Fig. 1).
• PWR_ch1: Doppler spectrogram measured from surveillance channel “rx2”, as demonstrated in Fig. 5(d).
• PWR_ch2: Doppler spectrogram measured from surveillance channel “rx3”.
• PWR_ch3: Doppler spectrogram measured from surveillance channel “rx4”.
#### Kinect dataset description
This section describes the structure of the data files residing in the kinect directory. The files are in .mat format and each row in the files corresponds to three-dimensional skeleton information captured from each of the two Kinects at a given point in time.
• exp_no: experiment number which is specified as “exp_002”, “exp_003”, etc. See Table 2 for more details. Note that the Kinect system does not need background scan. Hence, background data for “exp_001”, “exp_019” and “exp_055” were omitted for the Kinect system.
• timestamp: UTC+01 00 timestamp in milliseconds when the Kinect skeleton data was recorded.
• activity: ground truth activity labels. The activity is specified as a string of characters with no spacing e.g., “walk”, “sit”, “stand”, “liedown”, “standfromlie”, “bodyrotate”. These correspond to the activity numbers 1, 2, 3, 4, 5, 6 and 7 in the “Details” column in Table 2, respectively.
• person_id: person ID specified as “One”, “Two”, “Three”, etc.
• room_no: room ID specified as “1” (left room in Fig. 1) or “2” (right room in Fig. 1).
• Kinect1: velocity-time profile derived from Kinect skeleton data over a period of time as demonstrated in Fig. 5(c). It perfectly captures human motion characteristics and is qualitatively similar to the envelope of the human-micro-Doppler signatures presented in Fig. 5(d).
## Technical Validation
### WiFi CSI
Figure 5(a) shows a 196-second portion of the received WiFi CSI signals on NUC1 and NUC2 forexp018. CSI values for transmit antenna 1, receive antenna 1 and subcarrier 10 have been considered here. The injection packet rate of WiFi CSI was set a 1600 Hz. For illustration purposes only, the CSI data has been filtered using a 1D wavelet denoising technique and the corresponding results are shown in Fig. 5(a). It can be observed that the CSI measurements in the time domain capture variations in the wireless signal due to the latter’s interaction with surrounding objects and human bodies. Therefore, machine or deep learning algorithms can be used to train the observed patterns and automatically extract features from raw signals to predict human activities. The CSI data can be processed and interpreted in different ways (feature extraction), for example spectrograms which are generated by applying Short Time Fourier Transform (STFT) to the CSI amplitude data8,9,10. Note that the two receiving NUCs (NUC1 and NUC2) were arranged differently with respect to the transmitter (NUC3). As shown in Fig. 1, in both rooms NUC1 was facing the transmitter in a 180° configuration while NUC2 was in a bistatic geometry (90°) with respect to the transmitter. By deploying multiple receivers in the monitoring area, it is envisaged that the activity prediction accuracy will be improved using multiview data fusion models55.
### UWB
Figure 5(b) shows the UWB signals between node ‘0’ and node ‘3’ for UWB system 1 and nodes ‘1’ and ‘2’ for UWB system 2, considering the same experiment number and time window. The raw CIR data has been converted to Channel Frequency Response (CFR) using the Fast Fourier Transform (FFT) and the signals are plotted for the 10th CFR sample for each system. Note that the terms CFR and CSI can be used interchangeably. Although the sampling rate of the UWB systems is much lower (typically <100 Hz considering bidirectional data which are reciprocal) than the WiFi CSI system, the activities cause variations in the UWB signals and these variations can be fed to machine or deep learning algorithms for activity prediction. The raw CIR can also be used for human activity recognition (HAR) and yield high prediction accuracy, as demonstrated in10.
Considering the crowd counting experiment, Fig. 6 shows the first path power level (in dBm) for the two UWB systems between a given pair of nodes in each case. The first path power level (fp_pow_dbm) has been computed using the formula given in the DW1000 manual44. As can be observed, the first path power level increases gradually as each person was moving out of the monitoring area. This is an expected behaviour since the LoS signal becomes less and less obstructed. By using the fp_pow_dbm parameter together with other parameters such as overall received UWB signal power level (rx_pow_dbm), UWB CIR data and WiFi CSI data, the number of people in a given environment can be inferred through the use of artificial intelligence algorithms.
While UWB modules such as Decawave’s EVK1000 and DWM1000 are used for active localization to provide the 2D or 3D coordinates of a target carrying a tag, in this experiment, we deployed fixed UWB nodes programmed with custom software to record CIR data in a multi-static configuration. The idea is to extend the functionality of pulse-based UWB systems from active localization to the ability to sense their environment using the CIR data23,45,56. Figure 7(a,b) show 1000 aligned and accumulated CIR measurements recorded between a given pair of UWB modules for each system in a static environment (exp001). As can be observed from Fig. 7(a,b), when the room is empty, the accumulated CIR measurements are stable. However, when a person is performing activities as in exp003, some variations occur in the accumulated CIR, as can be observed in the region starting around τ-τFP ≈ 8 ns and 10 ns in Fig. 7(c,d), respectively. The earliest time at which changes are observed in the CIR is the bi-static delay. Since the transceivers are fixed in the multi-static network and their positions are known, the distance travelled by the direct (first path) signal between pairs of devices can be computed along with its delay τFP. The black vertical lines in Fig. 7(c,d) represent the ground truth bi-static delay which is computed from the tags’ coordinates. Note that the target was wearing two tags, one on each arm. Also, exp003 refers to the sitting on chair and standing up from chair activities which were performed at a given location within the monitoring area. Therefore, the reported locations for each tag were averaged over the 1000 accumulated CIR measurements to obtain a single 2D position per tag. Now, the coordinates of the two tags can be used to compute the separation distance between them, which is around 60 cm, corresponding to the approximate diameter of the human (arm to arm distance). Therefore, the midpoint xy coordinates are taken as the ground truth position of the target. The ground truth bi-static delay can then be computed by finding the transmitter-target-receiver path length and subtracting the direct signal path length from it, given that the fixed UWB transmitter and receiver positions are known. Assuming that the signal emitted from the transmitter reflects off the target and reaches the receiver without additional scattering, then the bi-static range defines an ellipse on which the target is located23. This ellipse has the position of the transmitter and receiver as foci points and the major axis length is equal to the bi-static range. In an ideal scenario, the common intersection point of ellipses from multiple transmitter and receiver pairs indicates the location of the target.
It should be noted that in the multi-static UWB network, each transceiver device runs its own independent RF clock and therefore CIR measurements between pairs of devices may be sampled at different times23. The DW1000 chip organizes the CIR buffer in such a way that the reported first path index (FP_INDEX) in each CIR measurement is usually around 75057. The chipset also estimates FP_INDEX in each CIR measurement with a resolution of $$\frac{1.0016\;{\rm{ns}}}{64}$$, such that it is represented as a real number, having integer and fractional parts (see column fp_index in UWB datasets). Now, since each CIR measurement basically has a different FP_INDEX value but the same sampling resolution of 1.0016 ns, the accumulated CIR measurements need to be aligned with their respective estimated FP_INDEX, and the latter can be shifted to be at the beginning of the CIR buffer, as shown in Fig. 7. Furthermore, in order to remove outliers in the accumulated CIR, those CIR measurements where the number of accumulated preamble symbols (see column rx_pream_count in UWB datasets) is less than half of the number of transmitted preamble symbols can be discarded23 (preamble length of 128 considered in the experiments).
### PWR
Figure 5(d) illustrates the Doppler spectrogram collected by the PWR system from a given angle (see placement of “rx2”, “rx3” and “rx4” in Fig. 1). As mentioned previously, the PWR system uses the CSI transmitter (NUC3) as the signal source with an injection packet rate of 1600 Hz. The coherent processing interval was set at 1 second, and sampling rate was set at 20 MHz for each channel. The signatures in the Doppler spectrogram demonstrate the relative velocity between the human and receiver antenna. The bi-static velocity is maximum in a monostatic layout (0°), and minimum in a forward scatter layout (180°). We can see large differences between the signatures from the “sit/stand” and “walk” activities in terms of Doppler frequency shifts. This is because the velocity of the walking activity is much higher than other activities. The activities “lie down” and “stand from floor” have opposite Doppler signatures due to the opposing directions the human body undertakes when performing these activities. Such signatures can potentially be used for various machine learning applications such as activity recognition, people counting, localization, etc.
### Kinect
Figure 5(c) illustrates the velocity-time profile of a human undergoing a set of activities. The motion profile is generated using the time-varying three-dimensional position information of different joints on the human body, such as the torso, arms, legs, and the head, at a frame rate of 10 Hz. To mimic accurate radar reflections from the target, we assume the radar scattering centers lie approximately at the center of the bones joining these joints. Thus, the signatures in the velocity-time plot demonstrate the relative velocity between human scattering centers and the Kinect position.
Figure 5(c,d) compare the velocity-time profile (generated using motion capture data) and the measured spectrograms, respectively, for a human undergoing a series of motions. The envelope of the velocity-time profile is visually very similar to the measured spectrogram indicating how well both systems capture the motion characteristics. For example, as the human sits down, we observe negative Doppler due to the bulk body motion. The positive micro-Doppler arises due to arm motion and legs moving slightly in the sensor direction while sitting down. After a 5-second delay, the human subject stands up from the chair, resulting in primarily positive Dopplers. Similarly, the latter part of the spectrogram presents signatures corresponding to a human transitioning from first walking to lying down and then standing up from the ground to rotate his body while standing at a fixed position.
In most realistic scenarios, the human motions might not be restricted to a single aspect angle with respect to the radar. In such scenarios, the spectrograms might differ significantly. It could be due to the shadowing of some part of the human body if captured at different angles. Therefore, we can leverage the animation data captured by the Kinect to feed as input to our human radar simulator, SimHumalator and synthesize radar returns as a function of different target parameters (varying aspect angles) and different sensor parameters (varying bi-static angles and operational parameters such as waveform design and antenna characteristics). Figure 8 shows an example of the micro-Doppler signature produced by SimHumalator when real motion capture data from the Kinect system is used as input to the software. It can be observed that the generated Doppler signature in Fig. 8 is very similar to the real PWR spectrogram data in Fig. 5(d). Therefore, such signatures can potentially augment otherwise limited real radar databases for various machine learning applications such as activity recognition, people counting, and identification.
### Human activity recognition (HAR) classification performance
In this section, we apply a mainstream algorithm such as Convolutional Neural Network (CNN) on the proposed dataset to verify its potential usefulness for the activity recognition task and as such provide a baseline. Essentially, we use the ResNet-18 CNN on the data collected from the WiFi CSI, Kinect and PWR modalities to perform classification on the six human activities, namely, sitting down on a chair (“sit”), standing from the chair (“stand”), laying down on the floor (“laydown”), standing from the floor (“standff”), upper body rotation (“bodyrotate), and walking (“walk”). Figure 9 shows various statistics regarding the distribution of the six activities performed by the six participants in the two rooms. It can be observed from Fig. 9(a) that more time was spent doing the six activities in Room 1. Figure 9(b) shows that the walking and body rotating activities were performed more frequently. It can also be observed from Fig. 9(c) that the participants shared a fair amount of time performing all six activities. Finally, it can be deduced from Fig. 9(d–f) that the six participants performed all six activities in both rooms, and in Room 2 each person performed the activities in approximately the same amount of time.
We convert the raw data from these three modalities into image-like format known as spectrograms, as illustrated in Fig. 5(d) and Fig. 8. The interested reader is kindly referred to the signal processing pipeline described in our previous works8,9,48 for converting WiFi CSI data into spectrograms. It should be noted that several feature extraction methods and machine/deep algorithms can be used for performing HAR classification using WiFi CSI data58,59,60,61. For each modality, the synchronized spectrogram data are segmented into 4 seconds duration windows using the activity labels and then reshaped into a dimension of N × 1 × 224 × 224, where N represents the number of samples. For each modality/system, we consider each receiver data as a one-channel image. For instance, the WiFi CSI system consists of two receivers (NUC1 and NUC2), therefore the spectrogram data from the two receivers are concatenated along the channel dimension, resulting into an image dimension of N × 2 × 224 × 224. The same feature fusion technique is used for the PWR system (3 surveillance channels) and Kinect system (2 receivers) independently. Thus, the spectrograms serve as a multi-channel image input to the ResNet-18 CNN. We train the model in a supervised fashion for each modality separately. We also perform a simple sensor fusion technique where we concatenate all modalities data (for all receivers) along the channel dimension, resulting into a 7-channel CNN model. In each case, the ResNet-18 model is trained over 70 epochs using the Adam optimizer with a learning rate of 1e-5, weight decay of 0.001, and β1 = 0.95 and β2 = 0.999. A batch size of 128 and cross-entropy loss function are also considered during training. 80% of the spectrogram data are used for training and the remaining 20% for testing. When each modality is trained independently, the same random seed is used for the train/test split to ensure fair comparison. The HAR classification accuracy is reported in Fig. 10 for the case when the three modalities are trained separately, as well as for the sensor fusion method. The corresponding confusion matrices are also shown in Fig. 11. Considering the separately trained modalities, it can be observed from Fig. 10 that the WiFi CSI modality achieves the highest accuracy (93.5%) while the PWR and Kinect modalities achieve comparable performance with accuracy values of 86.5% and 85.8%, respectively. On the other hand, the fusion of the three modalities data improved the overall performance, resulting in an accuracy as high as 96.7%. The benefit of sensor fusion is also reflected in Fig. 11(d) where it can be observed that the performance across the six different activities are greatly improved. These results serve as a baseline and the users of this dataset are encouraged to explore different feature extraction methods and algorithms in order to yield improved performance. Furthermore, various multimodal sensor fusion methods (decision-level fusion or feature-level fusion)62,63,64,65 can be investigated on this dataset.
For the sake of brevity, we omitted the UWB modality in this evaluation, although the UWB CIR data can as well be used effectively for activity recognition, as we have shown in our previous work10. The UWB CIR data reported by the Decawave’s chipsets is very attractive for passive (non-cooperative) target localisation, as we have demonstrated in our previous work56, where we use a series of well-defined signal processing steps to passively track a person with good localization accuracy using only the reported CIR data (sampling resolution of the complex CIR samples is ≈ 1 ns). We can leverage the variance in the CIR data (refer to Fig. 7) to find the time-of-flight (ToF) of the signal (caused by moving target) between each transmitter-receiver link and combine them, for example using Taylor Series or intersection of ellipses method, along with Kalman tracking or particle filtering to find the 2D coordinates of the moving target23,56.
## Usage Notes
The different directories available in our curated dataset39 are specified in Table 6. Furthermore, the interested reader is encouraged to navigate to the codes directory where example scripts on how to load and analyze a specific modality data are included. These are described in the following section .
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2022-09-29 15:40:36
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http://mathoverflow.net/revisions/107711/list
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2 added 363 characters in body
Suppose $G_1, \ldots, G_k$ are unitary, Hermitian, and anti-commuting, as well as $F_1, \ldots, F_k$. If they are similar, i.e., there exists $T \in GL_n(\mathbb{C})$ such that $$G_i = T^{-1} F_i T$$ for all $i \in [k]$, does there exist $V \in U_n$, where $U_n$ is the group of unitary matrices, such that $$G_i = V^{-1} F_i V$$ for all $i \in [k]$.
Generally, let $\pi, \sigma : G \to U_n(\mathbb{C})$ be two unitary representations of finite group. If $\pi, \sigma$ are equivalent, i.e., there exists $T \in GL_n(\mathbb{C})$ such that $T \pi(g) = \sigma(g) T$ for all $g \in G$, is $\pi$ and $\sigma$ unitarily equivalent, that is, there exists $T' \in U_n$ such that $T' \pi = \sigma T'$?
Suppose $G_1, \ldots, G_k$ are unitary, Hermitian, and anti-commuting, as well as $F_1, \ldots, F_k$. If they are similar, i.e., there exists $T \in GL_n(\mathbb{C})$ such that $$G_i = T^{-1} F_i T$$ for all $i \in [k]$, does there exist $V \in U_n$, where $U_n$ is the group of unitary matrices, such that $$G_i = V^{-1} F_i V$$ for all $i \in [k]$.
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2013-05-22 18:30:30
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http://mathhelpforum.com/algebra/80561-algebra-complex.html
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Math Help - algebra complex
1. algebra complex
Hey guys,
I need to find the complex solution to this: (and I apologise for not using the tools to display this stuff properly, but i havent learnt them yet and am in a hurry).
Find complex solutions of:
3|z|^2 + z[with a bar on it]^2 + 2z = 0
I have no idea how to go about doing this one Ive never even seen anything like that.
Also, {z [is a member of] C | z = z[with a bar on it]}
Does this just mean the real number line, ie y = 0, because the only time a complex number can equal its conjugate is when it =s 0?
Thanks a lot.
2. Originally Posted by Flay
Hey guys,
I need to find the complex solution to this: (and I apologise for not using the tools to display this stuff properly, but i havent learnt them yet and am in a hurry).
Find complex solutions of:
3|z|^2 + z[with a bar on it]^2 + 2z = 0
I have no idea how to go about doing this one Ive never even seen anything like that.
Also, {z [is a member of] C | z = z[with a bar on it]}
Does this just mean the real number line, ie y = 0, because the only time a complex number can equal its conjugate is when it =s 0?
Thanks a lot.
Let $z = x + iy \Rightarrow \overline{z} = x - iy$. Then your equation becomes:
$(4x^2 + 2y^2 + 2x) + i(2xy + 2y) = 0$.
Therefore:
Equate real parts: $2x^2 + y^2 + x = 0$ .... (1)
Equate imaginary parts: $xy + y = 0$ .... (2)
Solve equations (1) and (2) simultaneously.
3. thanks very much for that
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2015-01-30 13:13:54
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http://books.duhnnae.com/2017/jul8/150149487474-Quark-Masses-from-Gaugino-Condensation-in-String-Theories-High-Energy-Physics-Theory.php
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Quark Masses from Gaugino Condensation in String Theories - High Energy Physics - Theory
Abstract: We present a mechanism able to generate the perturbatively absent up-down$<{\bf 10} \cdot {\bf 10} \cdot {\bf 5}^H>$ quark Yukawa couplings ofSU5-flipped SU5 GUTS in Type II orientifold compactifications withD-branes. The mechanism works when there are SpN gauge groups involved. The${\bf {\bar 5}}$-s get charged under the SpN gauge groups and the generationof quark masses proceeds via the generation of the fermionic SpN singletcondensate $<{\bf {\bar 5} \cdot {\bar 5} \cdot {\bar 5} \cdot {\bar 5}}>$ inthe term $1-{M s^5} {\bf 10} \cdot {\bf 10} \cdot < {\bf {\bar 5} \cdot {\bar5} \cdot {\bar 5} \cdot {\bar 5}}>$. Also non-chiral states charged under Spgauge groups may become constrained by the requirement of Sp-s becomingstrongly coupled.
Author: Christos Kokorelis, Van E. Mayes, Dimitri V. Nanopoulos
Source: https://arxiv.org/
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2017-09-21 12:32:25
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https://electronics.stackexchange.com/questions/446485/effective-resistance-of-a-nmos
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# Effective resistance of a NMOS
How do we calculate effective resistance of a NMOS, operating in linear region across drain and source?
• I recall the formula Re = 1/GM with GM = K * W/L * Ve, where K = MU*Cox, and Ve = Vgs - Vt; this was long channel physics mode; – analogsystemsrf Jul 3 '19 at 1:34
$$\1/R_{ds}=g_m=\Delta I_{ds}/ \Delta V_{ds}\$$ @ Vgs
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2021-08-03 22:55:45
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https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-2-real-numbers-2-1-rational-numbers-multiplication-and-division-concept-quiz-2-1-page-47/7
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## Elementary Algebra
The reciprocal of $\dfrac{-3}{7}$ is: $=\dfrac{1}{\frac{-3}{7}} \\=1 \cdot \dfrac{7}{-3} \\=\dfrac{-7}{3}$ Thus, the given statement is false. Note, $\frac{7}{3}$ is the $\underline {opposite}$ reciprocal of $\frac{-3}{7}$.
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2019-06-17 03:20:26
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http://ocw.mit.edu/ans7870/18/18.013a/textbook/MathML/chapter32/exercise02.xhtml
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]> Exercise 32.2
## Exercise 32.2
Perform Gaussian elimination on the following set of equations to find a solution
$x + 3 y − z = 7 3 x + y − 2 z = 4 − x − y + z − 1$
Solution:
Here if you add the first and third equations you are lucky to find that the sum equation is $2 y = 8$ , and you deduce $y = 4$ . Adding the second and twice the third equation yields $x − y = 6$ , or $x = 10$ . Substituting in the third equation you find $− 10 − 4 + z = 1$ which tells you $z = 15$ , and you have your solution. Testing the solution in the original equations you get $10 + 12 − 15 = 7 , 30 + 4 − 30 = 4$ , and $− 10 − 4 + 15 = 1$ , which are all correct.
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2016-05-26 06:50:25
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https://math.stackexchange.com/questions/2766363/the-function-f-is-discontinuous-but-f-circ-f-is-continuous/2766373
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# The function $f$ is discontinuous but $f\circ f$ is continuous
I have thought without a solution. Are there actually examples of a function $f:\Bbb{R}\to \Bbb{R}$ such that $f$ is discontinuous at every point but $f\circ f$ is continuous?
• $f(x)=0$ for $x$ irrational and $f(x)=1$ for $x$ rational. Then $f\circ f(x)=1$.
– user551819
May 4 '18 at 12:49
• @ totoro: Please, does it work? May 4 '18 at 12:50
• @ totoro: I've just seen that! Thanks! May 4 '18 at 12:51
• You have an example in front of your nose.
– user551819
May 4 '18 at 12:51
• No what you are looking for but a nice, related concept: en.wikipedia.org/wiki/Cantor_function May 4 '18 at 13:03
Consider $$f(x)=\left\{ \begin{array}{ll} x,&x\in\mathbb{Q},\\ -x,&x\in\mathbb{R}\setminus\mathbb{Q}. \end{array} \right.$$ This function yields $$f\circ f(x)=x.$$
Edit:
Let me fix the bug. Thanks to @totoro, the above example does not work, because it is continuous at $x=0$.
Considering this, let us make it as follows. $$f(x)=\left\{ \begin{array}{ll} 1/x,&x\in\mathbb{Q}\setminus\left\{0\right\},\\ 0,&x=0,\\ -1/x,&x\in\mathbb{R}\setminus\mathbb{Q}. \end{array} \right.$$ Now this function is everywhere discontinuous, and yields $$f\circ f(x)=x.$$
• This function is continuous at $x=0$.
– user551819
May 4 '18 at 12:53
• @totoro: You are right. So what about the template $1/x$? May 4 '18 at 12:56
• @ hypernova: Thanks for the edit. You have my respect for fixing this! @ totoro and hypernova: Thanks for the eagle eye. You are both good! May 4 '18 at 13:32
• Thank you. And I really appreciate @totoro for the crucial comment. I will keep my mistake as it was, in case it could be helpful here :-) May 4 '18 at 14:13
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2021-09-17 23:08:37
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https://rocs.hu-berlin.de/courses/complex-systems-2021/script/examples-of-2d-systems/
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# Introduction to Complex Systems
Prof. Dirk Brockmann, Winter Term 2021
# Examples of 2D Systems - Interactively
In the panel below you can explore paradigmatic dynamical systems, some of which exhibit multi-stability, periodic orbits or limit cycles. Some are typical activator-inhibitor systems, some are mutually inhibitory, others exhibit cooperative behavior. Below the interactive panel is a summary of each of the systems.
## Quick system facts and summaries
### Lotka Volterra
The Lotka-Volterra model captures the dynamics of a two species predator-prey system in which $x$ and $y$ represent the abundance of prey and predator, respectively: [ \dot{x} =x(\alpha-\beta y) ] [ \dot{y} =y(\gamma x-\delta) ]
Domain: The models domain is the region $x,y\geq0$.
Parameters: The system has four positive parameters, the prey reproduction rate $\alpha$, the rate $\beta$ at which prey descrease due to the predator, the predator reproduction rate $\gamma$ that is modulated by the abundance of prey, and rhe predator death rate $\delta$.
Fixpoints: The system has 2 stationary states
1. $(x^{\star},y^{\star})=(0,0)$
2. $(x^{\star},y^{\star})=(\delta/\gamma,\alpha/\beta)$
Null-clines:
• x: $x=0$ and $y=\alpha/\beta$
• y: $y=0$ and $x=\delta/\gamma$
Asympotitics: Periodic Solutions.
Comment: The Lotka-Volterra system is structurally unstable. Slight perturbations of the dynamical systems will qualitatively change the dynamics. Also, a predator-free system will yield exponential and unlimited growth of the prey.
### Mutual inhibition
This dynamical system is characterized by two populations $x$ and $y$ that interact by mutually inhibitory forces onto each other captured by a growth rate that decreases with the abundance of the other. Plus there's some intra-population competitive regulation. [ \dot{x} =x(1-x-\alpha y) ] [ \dot{y} =y(1-y-\beta x) ]
Domain: The model's domain is the region $x,y\geq0$.
Parameters: The system has two non-negative parameters that capture the inhibition of $x$ on $y$ (parameter $\beta$) and $y$ on $x$ (parameter $\alpha$)
Fixpoints: The system always has three fixpoints
1. $(x^{\star},y^{\star})=(0,0)$
2. $(x^{\star},y^{\star})=(1,0)$
3. $(x^{\star},y^{\star})=(0,1)$
and if $\alpha,\beta<1$ or $\alpha,\beta>1$ also a fourth one
1. $(x^{\star},y^{\star})=(1-\alpha,1-\beta)/(1-\alpha\beta)$
Nullclines:
• x: $x=0$ and $y=1-\alpha y$
• y: $y=0$ and $y=1-\beta x$
Asympotitics: Depending on the parameter choices the system can have different attractors. If both inhibitory forces are weak, the coexisting state is the only attractor, if $\alpha>1$ and $\beta<0$ or vice versa only one population will prevail, the other going to zero. If both, $\alpha,\beta>0$ the sytem has two attractors and depending on the initial condition the system will approach one or the other.
### Duffing Oscillator
The Duffing Oscillator is a mechanical system, a mass at position $x$ subject to a nonlinear force $F(x)=\alpha x-\beta x^{3}$ and a frictional force $-\gamma\dot{x}$ and the equation of motion: [ m\ddot{x}=\alpha x-\beta x^{3}-\gamma\dot{x} ] Letting $y=\dot{x}$, $m=1$, $\alpha,\beta,\gamma=1$ yields the dynamical system.
[ \dot{x} =y ] [ \dot{y} =-y+x-x^{3} ] Domain: The models domain is the entire $x-y-$plane.
Parameters: The original system has four parameters, the special case here has none.
Fixpoints: The system always has three fixpoints
1. $(x^{\star},y^{\star})=(0,0)$
2. $(x^{\star},y^{\star})=(1,0)$
3. $(x^{\star},y^{\star})=(0,1)$
The trivial fixpoint is an unstable saddle, the other two fixpoints are stable spirals Nullclines:
• x: $y=0$
• y: $y=x-x^{3}$
Asympotitics: Depending on the initial condition all trajectories will either spiral towards one or the other stable spiral fixpoints.
Comments: It is interesting to draw the basin of attraction for the Duffing-oscillator. Points that can be very close in the region of large $x$ and $y$ can move along almost the same trajectory but split because they are separated by the stable manifold of the fixpoint at the origin. Also, the duffing oscillator can behave in funky way when driven by a periodic external force.
### SIRS epidemic model
This model captures the dynamics of an epidemic in which susceptibles (S) get infected by interacting with infecteds (I). Infecteds recover (R) and become immune. Recovered individuals remain immune for some time before they become susceptible again. All of this is captured by the reactions [ S+I\xrightarrow{\alpha}2I ] [ I\xrightarrow{\beta}R ] [ R\xrightarrow{\gamma}S ] denoting the fraction of susceptibles, infecteds and recovers by $x$,$y$ and $z$ yields the dynamical system [ \dot{x}=-\alpha xy+\beta(1-x-y) ] [ \dot{y}=\alpha xy-\beta y ] and $z=1-x-y$ because the number of individuals is conserved.
Domain: The models domain is $x,y>0$ and $x+y<1$.
Parameters: The model has three parameters, the recovery rate $\beta$ the transmission rate $\alpha$ and the rate of waning immuny $\gamma$. Typically it is assumed that $\alpha,\beta\gg\gamma$
Fixpoints: The system always has the fixpoint
1. $(x^{\star},y^{\star})=(1,0)$
which is stable if $\alpha<\beta$. If $\alpha>\beta$ the fixpoint
1. $(x^{\star},y^{\star})=(\beta/\alpha,(1-\beta/\alpha)/(1+\beta/\gamma))$
is a second fixpoint and is stable.
Nullclines:
• x: $y=\gamma(1-x)/(\gamma+\alpha x)$
• y: $y=0$ and $x=\beta/\alpha$
Asympotitics: If $\alpha/\beta>1$ all trajectories will approach the non-trivial stable fixpoint.
### Predator Prey System
This system is a modification of the Lotka-Volterra system and, unlike it,structurally stable. The dynamical system is defined by [ \dot{x}=\alpha x(1-x)-\beta yx/(x+s) ] [ \dot{y}=\gamma yx/(x+s)-\delta y ] The modification compared to the LV system is that prey growth by itself is limited by a logistic, self-regulatory factor and that the predators feeding is saturated if prey are abundant beyond a concentration $s$.
Domain: The models domain is $x,y>0$.
Parameters: The model has 5 parameters. In addition to the four parameters of the Lotka-Volterra system, we have the additional saturation parameter $s$.
Fixpoints: The system always has the fixpoint
1. $(x^{\star},y^{\star})=(1,0)$
but depending on the parameters can also have the non-trivial fixpoint
1. $(x^{\star},y^{\star})=(\delta s/(\gamma-\delta),\alpha/\beta(\delta s/(\gamma-\delta)+s)(1-\delta s/(\gamma-\delta)))$
that is either stable or unstable, also depending on the parameter choice.
Nullclines:
• x: $x=0$ and $y=\alpha(x+s)(1-x)/\beta$
• y: $y=0$ and $x=\delta s/(\gamma-\delta)$
Asympotitics: For th right choice of parameters this system can have a limit cycle that attracts all trajectories, unlike the Lokta-Volterra system that has a different periodic orbit for differing intial conditions. For other parameter combinations, this system can have a stable non-trivial fixpoint.
### Brusselator
This dynamical system with the odd name Brusselator is based on the chemical, autocatalytic reactions [ \emptyset\xrightarrow{\alpha}X ] [ 2X+Y\xrightarrow3X ] [ X\xrightarrow{\beta}Y ] [ X\xrightarrow\emptyset ] which yields the dynamical system [ \dot{x}=\alpha-(\beta+1)x+ x^{2}y ] and [ \dot{y}= x^{2}y-\beta x ]
Domain: The models domain is $x,y>0$.
Parameters: The model has 2 parameters for the two reactions that couple $X$ and $Y$ that are the core of the autocatalysis.
Fixpoints: The system has one fixpoint
1. $(x^{\star},y^{\star})=(\alpha,\beta/\alpha)$
depending on the parameter combination the fixpoint is stable or unstable.
Nullclines:
• x: $y=((\beta+1)x-\alpha)/ x^{2}$
• y: $x=0$ and $x=\beta/ s$
Asympotitics: Like the predator prey system above, the Brusselator can have a limit cycle, or an attractive spiral as a stable fixpoint.
### Cooperation
This dynamical system is defined by the equations [ \dot{x}=R_{1}x(1-x)-x+\alpha xy(1-y) ] [ \dot{y}=R_{2}y(1-y)-y+\beta yx(1-x) ] for two populations. Without interaction ($\alpha,\beta=0$) each population grows logistically with an additional leak term (the second term in each ODE). The interaction is positive, abundance of $y$ will facilitate growth of $x$ and vice versa. This synergetic effect is also saturated by a logistic term.
Domain: The models domain is $x,y>0$.
Parameters: The model has 4 parameters, two parameters that control the reproduction of each population and two that control the cooperative influence of one on the other.
Fixpoints: This system can have a whole series of fixpoints, depending on the parameter combination. For some combinations up to 5 fixpoints. It is possible to compute them analytically but it is cumbersome.
Nullclines:
• x: $x=0$ and $y=1-1/R_{1}+\alpha x(1-x)/R_{1}$
• y: $x=y$ and $x=1-1/R_{2}+\beta y(1-y)/R_{2}$
Asympotitics: This system always has at least one stable fixpoint but can have up to three different stable fixpoints. This situation can be established with the reproduction parameters are low and the cooperativity parameters are high.
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2022-08-13 12:16:08
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https://math.stackexchange.com/questions/3280698/prove-that-1-e-e-c-sim-e-c/3280705#3280705
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# Prove that $1-e^{-e^{-c}} \sim e^{-c}$
It appears that the following holds: $$1-e^{-e^{-c}} \sim e^{-c}$$
I do not see why this should hold. How do I prove it? I would be very happy to see multiple ways of proving this, as it is always good to have more tools in one's mathematical toolbox.
Moreover, it seems to me that the ratio between these to value tends to 1 rather quickly (probably exponentially). Is there a way how to say something about the speed of convergence?
• Why don't you use Taylor series, seeing as the exponential is entire. Jul 2 '19 at 11:25
• Replace $e^{-c}=x$ where by the statement $x \to 0$ Jul 2 '19 at 11:26
• @Yuriy , performing the suggested subs. we get $1-e^x=x$ which is still hard to do... Jul 2 '19 at 11:40
• @NoChance, no, it will be $1-e^{-x} \approx x$ if $x \ll 1$ if I understood the question correctly. It can be proved through any definition of the exponential function. Jul 2 '19 at 11:42
• @YuriyS, thank you for your reply, you are correct. However, I think it is still not very easy to solve. Jul 2 '19 at 11:44
I'm assuming that this holds as $$c\rightarrow +\infty$$. In that case, $$e^{-c}\rightarrow 0$$ and $$\begin{split} 1-e^{-e^{-c}} &= 1-(1-e^{-c}+\mathcal O(e^{-2c}))\\ &= e^{-c}+\mathcal O(e^{-2c})) \end{split}$$
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2022-01-26 17:16:59
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https://math.stackexchange.com/questions/3954719/is-this-braced-heptagon-a-rigid-graph/3955860
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# Is this braced heptagon a rigid graph?
In Mathematica, GraphData[{"UnitDistance", {21, 2}}]
Is this 42-edge graph rigid? It has chromatic number 4. If it was floppy, that might make it an interesting tool in high-chromatic graph studies.
The current record for heptagon bracing is 59, so this would beat that.
• You are cordially invited to answer this question. – Parcly Taxel Dec 19 '20 at 9:02
• If you assign a suitable direction to each edge and colour the triangles differently, this turns out to be the Cayley graph of $\mathbb Z_7\rtimes\mathbb Z_3$. – Parcly Taxel Dec 20 '20 at 19:19
The graph is rigid. In fact I will show that the following $$35$$-edge graph formed by deleting the two interior vertices of one triangle is (minimally) rigid:
We first need to find a parametrisation of the vertices of the graph with as few parameters as possible (that relate to metrics like lengths and angles); three parameters suffice. Fix $$A$$ and $$B$$ in the diagram below, set as parameters the angles $$\alpha_1,\alpha_2,\alpha_3$$ that the vectors $$\vec{A0},\vec{B1},\vec{B2}$$ make with $$\vec{AB}$$, then construct each of the remaining numbered points in order as the point that is a distance $$1$$ away from two already-determined points. (See the program at the end of this answer for the exact construction sequence.)
Define the function $$f:\mathbb R^3\to\mathbb R^3$$ as $$f(\alpha_1,\alpha_2,\alpha_3)=(d(11,12),d(13,14),d(15,2))$$ i.e. it returns the lengths of the three thick edges, which are not created as part of the construction but are in the graph. Let the parameters corresponding to the regular braced heptagon be $$\alpha_1^*,\alpha_2^*,\alpha_3^*$$; clearly $$f(\alpha_1^*,\alpha_2^*,\alpha_3^*)=(1,1,1)$$. To check the rigidity of the ($$35$$-edge) graph, we construct the Jacobian matrix $$J$$ of $$f$$ at $$(\alpha_1^*,\alpha_2^*,\alpha_3^*)$$; if it has full rank we can immediately conclude that it is (infinitesimally or first-order) rigid.
It turns out that $$J$$ is a $$3×3$$ matrix with rank $$2$$. This does not mean that the graph is not rigid; we still have to analyse the behaviour of $$f$$ along the infinitesimal motion associated with this rank-deficient $$J$$. Such a motion is given by an element of the nullspace of $$J$$; suppose this element is $$(\delta_1,\delta_2,\delta_3)$$. Define $$g(t)=f(\alpha_1^*+t\delta_1,\alpha_2^*+t\delta_2,\alpha_3^*+t\delta_3)$$ When we plot $$g$$ for $$t$$ around zero we get the following graph (blue, red and green lines represent the three coordinates in order):
The lengths of all three thick edges are individually on the same side of $$1$$. Together with the result from the Jacobian matrix, we infer that said lengths are not $$1$$ in a punctured neighbourhood of $$(\alpha_1^*,\alpha_2^*,\alpha_3^*)$$, so the $$35$$-edge graph is (second-order) rigid.
As a final note, it can be easily shown that the $$35$$-edge graph is a Laman graph by means of a Henneberg construction, so no edges can be removed without making the graph flexible.
#!/usr/bin/env python3
from mpmath import *
mp.dps = 100
def cu(z1, z2):
# Constructs the point at a distance of 1 from z1 and z2,
# left of the line from z1 to z2.
m = (z1 + z2) / 2
d, theta = polar(z2 - z1)
return m + sqrt(1 - d * d / 4) * expj(theta + pi / 2)
def f(a1, a2, a3):
# Progressive construction of the graph's vertices.
# Returns the distances between three pairs of vertices that are
# not directly related in the construction but are still linked
# by an edge.
A = mpf(0)
B = mpf(1)
p0 = expj(a1)
p1 = 1+expj(a2)
p2 = 1+expj(a3)
p3 = cu(p1, p0)
p4 = cu(B, p1)
p5 = cu(p0, p3)
p6 = cu(p2, p5)
p7 = cu(p4, p6)
p8 = cu(p6, p2)
p9 = cu(p8, A)
p10 = cu(A, p9)
p11 = cu(p0, p7)
p12 = cu(p9, p7)
p13 = cu(p12, p1)
p14 = cu(p5, p10)
p15 = cu(p13, p14)
return abs(p11-p12), abs(p13-p14), abs(p15-p2)
# Known parameters at the intended solution. a2 is exact, 1.43262130059...
a1 = 5*pi/7
a2 = phase(polyroots([7, 28, 70, 133, 203, 259, 281, 259, 203, 133, 70, 28, 7])[10])
a3 = 2*pi/7
# Jacobian matrix of f
def fderiv(i, j):
ff = lambda x, y, z: f(x,y,z)[i]
dvec = tuple(int(n == j) for n in range(3))
return diff(ff, (a1,a2,a3), dvec)
J = matrix([[fderiv(i, j) for j in range(3)] for i in range(3)])
# J is rank-deficient at the reference solution,
# i.e. the framework is not first-order (infinitesimally) rigid.
# Not to worry. Analyse f's behaviour along the one infinitesimal motion...
_, Sigma, V = svd(J)
nprint(chop(Sigma))
da1, da2, da3 = V[2,:]
def g(z):
return f(a1+z*da1, a2+z*da2, a3+z*da3)
# The lengths of the three "test rods" are individually on the same side of 1
# for sufficiently small multiples of the infinitesimal motion.
# This shows that the graph is (second-order) rigid.
plot([lambda z: g(z)[0], lambda z: g(z)[1], lambda z: g(z)[2]], [-0.1, 0.1])
The graphs above were drawn with the help of a script in my Dounreay repository. Run ./drawgraph.py heptagon in the folder containing drawgraph.py to reproduce the full $$42$$-edge graph; the above pictures are the result of tweaking in Inkscape.
Removing two consecutive vertices of one of the inner stars leads to a first-order rigid graph in either case (the determinant of $$J$$ is non-zero):
A = mpf(0)
B = mpf(1)
p0 = expj(a1)
p1 = 1+expj(a2)
p2 = 1+expj(a3)
p3 = cu(p1, p0)
p4 = cu(B, p1)
p5 = cu(p0, p3)
p6 = cu(p2, p5)
p7 = cu(p4, p6)
p8 = cu(p6, p2)
p9 = cu(p8, A)
p10 = cu(p3, p8)
p11 = cu(p9, p7)
p12 = cu(p11, p1)
p13 = cu(p12, p5)
p14 = cu(p0, p7)
p15 = cu(p13, p2)
p16 = cu(p14, p10)
return abs(p11-p14), abs(p12-p15), abs(p15-p16)
A = mpf(0)
B = mpf(1)
p0 = expj(a1)
p1 = 1+expj(a2)
p2 = 1+expj(a3)
p3 = cu(p1, p0)
p4 = cu(B, p1)
p5 = cu(p0, p3)
p6 = cu(p2, p5)
p7 = cu(p4, p6)
p8 = cu(p6, p2)
p9 = cu(p8, A)
p10 = cu(A, p9)
p11 = cu(p0, p7)
p12 = cu(p3, p8)
p13 = cu(p10, p4)
p14 = cu(p5, p10)
p15 = cu(p14, p2)
p16 = cu(p11, p12)
return abs(p12-p13), abs(p13-p16), abs(p16-p15)
• Very nice. What happens if you remove two connected vertices of an inner star polygon. That gives a 4-chromatic graph. – Ed Pegg Dec 20 '20 at 19:30
• @EdPegg The resulting graphs are still rigid – and first-order rigid. See edit. I chose to remove two vertices of a triangle at first because it leads to a degree-$2$ vertex on the outer edge, reminiscent of the other braced polygon solutions on the Math Magic page. – Parcly Taxel Dec 20 '20 at 20:59
• @EdPegg Are you saying that the first graph in my answer can be $3$-coloured? – Parcly Taxel Dec 20 '20 at 21:49
• Good job! Though it leaves me in a bit of a quandry: What do I do with the proof I was in the process of writing up? :-) "My name in Dnepropetrovsk is cursed when he finds out I publish first!" -- Tom Lehrer, "Lobachevsky" – WRSomsky Dec 21 '20 at 6:57
• @EdPegg Please precede the username with an @ sign so the other party is properly notified. – Parcly Taxel Dec 22 '20 at 5:36
My proof was written up in a Mathematica. I've made a copy available on Google-Drive both as a Mathematica notebook (NewHeptagon-Draft1.nb) and as a static pdf (NewHeptagon-Draft1.pdf). Since the version of Mathematica I was using didn't have GraphData[{"UnitDistance", {21, 2}}] I had to regenerate the graph from just its picture so that's embedded in there as well.
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2021-01-23 05:56:14
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https://motls.blogspot.com/2014/08/joes-weird-objections-against-state.html?showComment=1408076028045&m=1
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## Wednesday, August 13, 2014
### Joe's weird objections against state dependence
The state dependence is a simple example of effective superselection sectors; Joe's firewall confusion is linked to his sleeping beauty mistakes
I have finally found some time to watch more videos from Strings 2014. You may download them from the conference web page about talks or watch many/most/all of them on the GraduatePhysics YouTube channel.
The talks by Joe Polchinski, Kyriakos Papadodimas, and Suvrat Raju are among those that talk about the black hole interior. This blog's fans want to see Suvrat's talk, 25:40-25:50. ;-)
What Suvrat and Kyriakos say makes sense. I looked at Joe Polchinski's dissatisfaction, e.g. in his slides (PDF). Pages 20-24 and some of the following ones are dedicated to Joe's objections against Suvrat's and Kyriakos' picture. I find the causes of Joe's apparent unhappiness strange. He believes that Suvrat and Kyriakos ("PR") violate some general rules of quantum mechanics but all the contradictions quoted by Joe actually arise because Joe, and not PR, violates some laws of quantum mechanics.
Also, it seems to me that the same mistake that leads Joe to wrong answers about the sleeping beauty problem are not only behind his anthropic blunders but they have led him to the firewall dead end, too.
OK. Let me go to page 20 of the slides. Joe's very first equation is$\ket{\psi_{\rm vyp}} = \frac{ \ket{0}_B \ket{\psi_{\rm typ},0}_{B^*} + e^{-\beta\omega/2}\ket{1}_B \ket{\psi_{\rm typ},1}_{B^*} }{Z^{1/2}}$ where $$B^*$$ is the complement to $$B$$, we learn. The way how Joe phrases these things makes it rather clear that he always assumes that the degrees of freedom may always be split to two geographical regions, one region and its complement, which isn't really the case due to the black hole complementarity but I think it's not really the heart of Joe's anti-state-dependence sentiments so let me ignore these ultimately wrong operations and summarize the page 20 as Joe's version of the explanation why there's no firewall in the PR picture.
The "key issue" is mentioned on the following page 21.
Key issue: given a black hole in some state $$\ket\psi$$, what reference state $$\ket{\psi_{\rm typ}}$$ do we use? Is the state above but with the relative minus sign an excitation of the state above, or a typical state that should be classified as unexcited?
The correct reply is, obviously, that the answer depends on the reference state. Unless we are constrained by something else, we may choose any reference state but the description in terms of low-energy degrees of freedom will only be applicable if the reference state is sufficiently similar to – roughly speaking, related by a relatively low-order polynomial of local field excitations to – the states that the system under consideration is actually found at.
As far as I can see, the very fact that Joe is asking the question proves that he hasn't even started to consider the proposed claim that all these constructions are being built relatively to a reference state. If he had started to consider this proposed claim, he would have known that there couldn't be a unique answer to such questions. So page 21 is just a proof that Joe hasn't started to even consider the proposed explanation.
On page 22, Joe reviews the PR construction of the interior operators in terms of the reference state but adds his own comment:
The issue is that when one specifies the reference state $$\ket{\psi(t)}(\psi)$$, these become nonlinear operators $$P(n_A,\psi)$$.
This state-dependence is a modification of the Born rule, and is different from normal notions of background-dependence.
It would be very bad for PR if they modified the Born rule but they do nothing of the sort. In fact, they never write anything such as Joe's function $$\ket{\psi(t)}(\psi)$$ because it's wrong. The reference state that Joe calls $$\ket{\psi(t)}$$ enters the PR formula for the interior operators but isn't supposed to opportunistically, exactly, and immediately vary depending on some particular state $$\ket\psi$$ that is standing on the right side from the operator so that the operator acts on it. Instead, the vector $$\ket{\psi(t)}$$ is kept fixed while we study the operators and measure their values etc.
With a fixed reference state, the Hilbert space is completely standard, orthodox, and linear, the operators are also perfectly linear, they may be measured, and the probabilities of different values are calculated by the totally orthodox and standard Born rule. All of Joe's claims that the postulates of quantum mechanics are being modified are just pure rubbish.
The only thing that the state dependence means is that the Hilbert space with all the unexcited states, single-excitation states, more excited states, and all the operators the map these states onto each other are being built from scratch in a given physical context. The given physical context will allow us to add a finite – perhaps large but not too large – number of local field excitations but with too many excitations, the well-definedness of the local field operators is guaranteed to break down. It's not surprising at all. One simply can't create an unlimited amount of stuff into the black hole interior.
But whenever quantum mechanics is used to predict probabilities of some measurements, we must know how the operators act on the state that we have prepared. We must know what the operators that we measure are. If we don't know what the operators are, quantum mechanics obviously yields no predictions for the (unknown) operators' values.
At the end, the state dependence only means that two black hole microstates that differ by "more than a few" field theory excitations – microstates that are "far" in the PR sense – should be treated as two different superselection sectors. There isn't any canonical map between states in two superselection sectors even if the two sectors are mathematically similar. I can make this analogy very accurate and I can also explain why there can't be a global definition of local operators across superselection sectors in quantum field theory or string theory. Even if the two worlds look similar - like two macroscopically indistinguishable black hole states; or like two string vacua with a Standard Model at low energies – they are really two different worlds that can't be operationally obtained from each other, at least not by a simple action of a low-order polynomial of local operators. To say the least, if you defined a dictionary between the operators acting upon the two superselection sectors, the phases of all the creation operators $$a^\dagger(\vec k)$$ may be redefined by a function of $$\vec k$$. This turns one proposed convention to define the operators "upon all superselection sectors" into a different dictionary. The dictionary is clearly not unique.
Page 23 is just repeating trivialities about the Born rule with the wrong claim that PR violate the Born rule. Joe's mistake is simply that he varies the reference state all the time. But it isn't supposed to vary during the predictions of a single measurement at all. One is supposed to envision the relevant Hilbert space, depending on the initial conditions, and the operators that act on it. This relevant quantum mechanical model has everything it needs and the postulates of quantum mechanics are and have to be rigorously followed. The state dependence just means that for a given "region" of the black hole microstates, the operators have to be defined separately, from scratch.
Joe believes that either the operators have to be defined everywhere, across all the black hole microstates – across all the superselection sectors, if you use my discourse – or the regions have to be defined so finely that the Hilbert spaces are just one-dimensional and bizarre. But the truth is in between – none of these extremes are right. Quantum mechanics would be completely meaningless if the relevant Hilbert space were always one-dimensional. The relevant Hilbert space always has to be multi-dimensional for the operators not to be $$c$$-numbers and to be able to distinguish several results of measurements.
Page 24 brings nothing new. Joe still complains that the same states may be excited or unexcited, depending on the choice of the reference state. Indeed.
But it was page 25 that made me explode in laughter. Joe starts the page by asking two questions:
How to interpret $$\alpha \ket{\rm vac} + \beta\ket{\rm exc}$$?
How to interpret$\frac{ \ket{\rm vac}\ket{+z} + \ket{\rm exc}\ket{-z} }{\sqrt{2}}$
You would think that Joe is asking how to interpret two superpositions because he doesn't know how to interpret them. But the next sentence raises a
Problem: the interpretation is different if one writes it as $\frac{ (\ket{\rm vac}+\ket{\rm exc})\ket{+x} + (\ket{\rm vac}-\ket{\rm exc})\ket{-x} }{ 2 }$
LOL, that's hilarious. Joe doesn't know what the interpretation is but he's sure that the interpretation is different from the interpretation of the very same state written differently!
No, Joe, the interpretation of the same state is always the same. With a choice of the reference state, the experimentally accessible Hilbert space may be completely built. It's the same kind of Hilbert space as the Hilbert space in any quantum mechanical theory and it's used in the same way. One may identify some excitations as spin-$$z$$ or spin-$$x$$ electrons and one may add an apparatus to measure it as long as the apparatus is constructed from a sufficiently low number of excitations. If the apparatus can't be built in this way, it probably doesn't fit the black hole. The predictions for all the measurements of $$j_z$$ or $$j_x$$ and their correlations with the excitation level may be computed by the universal quantum mechanical formulae.
The state dependence is really just an expression of a tautology. In between two sectors of the black hole's Hilbert space that are significantly different – they don't differ by a simple enough polynomial of local fields – you can't travel by the action of local operators. So from the viewpoint of the local operators, these two sectors of the Hilbert space act as superselection sectors whose ground state and excited states should be thought of as different worlds, despite their macroscopic similarities.
Let me finally get to the sleeping beauty.
Joe Polchinski is a thirder and this simple error in the probability calculus has been explained to be the source of some of his wrong opinions about the need to have "typical stringy vacua" etc. But how can the thirder's fallacy explain the firewalls?
Now I think that these two errors share the same basic idea which is an incorrect, retroactive, acausal change of the rules of the game while we compute the probability. What do I mean?
If we use the notion of probability correctly, probabilities tell us how our knowledge about something is probably going to change once it changes. I know that I used the word "probably" so the "definition" is circular and it is not a real definition. But what I want to say is that the notion of probability assumes a "fixed situation" before we learn the new piece of information, and many things may happen after we learn the new piece of information, but these later events don't and they can't affect the probability that existed right before we learned the right answer.
According to Joe and other thirders, probabilities should behave differently. If something happens after the moment when we're supposed to learn the right answer, it should still retroactively modify the denominator that may define the probability in a frequentist way, among other things. Just because the sleeping beauty is woken up multiply times on the "heads" week implies that the probability of "heads" had to go up. Joe and other thirders actually believe that the probability of tails was already $$P=1/3$$ on Sunday because the algorithm for the awakenings had already been decided.
But whether the sleeping beauty is going to be woken up twice is a consequence of the result of the coin toss, so it just can't affect the probabilities of the different outcomes of the coin toss. So the beauty knows that $$P_{\rm tails}=1/2$$ on Sunday night and because she doesn't learn anything that would change the odds of "heads vs tails" when she wakes up (both possibilities guarantee with 100% certainty that she wakes up at least once which is the only "evidence" she gets), the probabilities have to remain $$P_{\rm tails}=1/2$$.
Joe's mistake in the case of the state dependence is analogous for the following reason. In the correct quantum mechanics, we must decide what the states in the Hilbert space – and operators – corresponding to different results of the measurements are, and we must keep this fixed. This choice is much less ambiguous in practice because we really evolve the fields according to the usual Heisenberg field equations.
But according to Joe, we are not obliged to deal with a fixed spectrum of possible outcomes of the experiment. The experiment may create them (and perhaps erase them) later. That's why Joe insists on changing the reference state "accurately" whenever some operator acts on anything. (This isn't really a well-defined rule because in a generic equation with many operators, there are many states, depending on which actions of the operators are included etc.) He thinks that this gives him as accurate answers as possible.
Instead, what it does is that it invalidates and erases any previous well-defined rules of the game – with known options (possible outcomes) whose probabilities used to be predictable. If he allows the evolution of the state vector to change what the question is, then there is no pre-existing question, and quantum mechanics cannot predict any probabilities! If he imagines that the post-measurement state $$\psi$$ is the only state one could have gotten, and everything should suddenly adapt to this new state, even the state of the black hole before the measurement, then there was no uncertainty before the measurement and the measurement could have brought the observer no information. This wouldn't make any sense, of course. It is absolutely crucial that we are not changing the rules of the game – we are not changing the reference state – at the very moment of the measurement.
In some future, I plan to write a more technical post on the state dependence, the superselection sectors, and things like the holomorphic anomaly in topological string theory because there are many direct analogies (or equivalences, in the sense of one's being a special case of another) between the concepts.
1. (not so) off topic : maldacena's conjecture article has reached 10.000 citations today!
2. A cool milestone - I hope that Milner or someone will also send him a T-shirt or hoodie like Stack Exchange is sending me for crossing 100,000. ;-)
3. Wow, the silence here is deafening. Apparently nobody competent enough to recognize your arguments for what they are -- obviously correct -- dares to speak out and mess with the Big Polchinski. Politics trumps science in academia, I guess.
4. Thanks, Michael, but you exaggerate. The silence on the blog is because the blog readers are either non-experts or shy workers who are working on their research. I think that those that know (including Kyriakos and Suvrat) aren't really too afraid to disagree with Joe even though such fear does exist, too.
5. it seems to me that milner already gave him something like that a couple of years ago, some kind of coupon if i remember well.
6. No, I know, but maybe it wasn't enough for Juan to buy an AdS/CFT-10,000 T-shirt. ;-)
7. Oh my God, even though I had not yet time to read all of the sleeping-beuty articles in detail, agree that it seems that Joe P. has same bad misunderstandings abou basic probability theory (being a thirder).
Also from teading about different methods to qualtise ST for example, it always seemed to me that as soon as one has defined everything, results of measurements can just be calculated and there is no need to continuously change the starting point of quantization.
8. Remember that anyone who might disagree with Lubos, like DN, has been banned, so only his sycophants remain.
9. Remember, anyone who disagrees with Lubos, or actually understands probability, like DN, is banned from the forum. So only Lubos's sycophants remain. So this comment has no chance of being accepted.
10. Actually, Lubos, not everyone behaves like you, insulting those who disagree with them, and banning them. There is ample discussion of state-dependence in the scientific community.
11. Suppose that we change the SB puzzle, but only in the protocol for Tuesday after tails (T/t). The new protocol is that she is awakened, and, after a pause of a minute, is put back to sleep (but only on T/t). Again, she knows the full protocol in advance. In the first minute after awakening, she assigns equal probabilities for M/h, T/h, M/t, T/t. After a minute, if she is still awake, she excludes T/t and is left with equal probabilities for M/h, M/t, T/h. So in this case the 1/3 is correct. Do you agree with this result for the modified problem? If so, how can the changed protocol for T/t affect her relative weightings of the other three cases? What if she's only awake for a second?
12. Of course that I disagree with the result 1/3 here. After the "demo minute" or "demo second", her knowledge is exactly the same as in the original problem, so your problem becomes the original problem! And it's a way to formulate the right solution "by parts" which may be even clearer than all the solutions so far.
Before she may be put to sleep, there are 2 possible coin states, heads and tails, and 2 possible days, Monday and Tuesday. By two Z2 symmetries, the probability of each of the 4 possibilities is 1/4. Would you really disagree with that?
Now, after the second or minute for the short awakening, the situation changes because she's getting a new information in general.
How much information she is getting by staying up depends on the coin state. If the coin shows "heads", she is getting no information because in the "heads" case, staying up is predicted for both days.
If one assumes that the coin is showing "tails", then she is definitely getting information by staying up. Before the momentum of "possible putting back to sleep", the probabilities of Monday-tails and Tuesday-tails were 25% each. That means that the conditional probabilities of Monday and Tuesday, assuming tails, were both 50%.
After the moment when she can be put back to sleep, the probabilities change. The "fluid" of the "Tuesday tails" just flows to "Monday tails" because Tuesday tails is getting riuled out. So the conditional probabilities of Monday and Tuesday, assuming tails, become 100% and 0%. They still share the same initial probability reserved for "tails" so this 100-0 translates to 50% - 0% in absolute probabilities.
The probabilities of heads-Monday and heads-Tuesday stay at 25% because if one/she assumes heads, one/she is learning nothing about the day of the week!
The result, after the minute/second expires, is of course the same 50-25-25 as before.
13. Dear Lubos, Thank you for your answer. Now suppose that instead of being put back to sleep, after a minute she is simply told, `it's T/t'. So in the other three cases, it is exactly the same as the previous modified protocol, she gains the same information after one minute.
So, if she is told (which is 1/4 of the time), then it's 100% that it is T/t. If she is not told (3/4) then it's 50% that it is M/t and 25% each that it is M/h and T/h. This seems to give a total probability of .25*1 + .75*.5 = .625 that she is in one of the tails cases. The thirder calculation would give .25*1 + .75*.333 = .5.
14. What you wrote was meant to be
P(tails) = P(toldTT) P(tails|toldTT) + P(notToldTT) P(tails|notToldTT)
wasn't it? But this formula makes no sense because it doesn't express her subjective probability of "tails" at any single well-defined moment, neither by objective time nor by subjective time.
You're summing probabilities at "a" moment when she's told something with "a" moment when she's not told something, and these are not mutually exclusive because assuming "tails", she is both told (on Tuesday) and not told (on Monday).
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2021-04-14 08:47:26
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https://electronics.stackexchange.com/questions/219202/cutoff-frequency-of-low-pass-filter-with-two-resistors-differential-signal
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cutoff frequency of low pass filter with two resistors (differential signal)
How do I calculate the cutoff frequency for this design?
simulate this circuit – Schematic created using CircuitLab
Or simply,
• Hello rumman, welcome to EE.SX! Yes the resistances just add. The formulas look good, except the designators are a component off (R2+R3). $f$ will be the -3dB point, where V_out is about half of VIN. – rdtsc Feb 24 '16 at 23:19
Whenever you have a symmetrical circuit you can cut it in half to make the analysis easier. The equivalent half-circuit will show the same behavior. Please note that I am assuming that R2=R3.
In this case the symmetry might not be apparent, but by replacing the capacitor C1 by two series connected capacitors, each one having twice the value, the circuit becomes symmetrical. And the half circuit is a low-pass with R2 and 2*C that determine the corner frequency. Hence, $$f=\frac 1{2\pi\cdot R_2 \cdot 2C}$$ which is equivalent to your first equation, for R2=R3.
Why not simply applying the basic rules of circuit analysis?
The current I through the circuit is I=Vsin/(R2+R3+1/jwC1) and the voltage across C1 is
Vc=I*1/jwC1=Vsin/[jwC1(R2+R3+1/jwC1)].
Therefore:
Vc/Vsin=1/[1+jwC1(R2+R3)]
This is the classical first-order lowpass function of the form 1/(1+jw/wc) with the 3dB angular (cut-off) frequency
wc=1/C1(R2+R3).
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2019-11-14 13:22:35
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https://math.stackexchange.com/questions/1393577/show-that-if-omega-1-omega-2-are-tangent-vectors-of-curves-in-so3-then
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# Show that if $\Omega_1,\Omega_2$ are tangent vectors of curves in $SO(3)$ then $[\Omega_1,\Omega_2]\in\mathfrak{so}(3)$
Let $so(3)$ be the Lie Algebra of $SO(3)$ and $R\in SO(3); \Omega_1,\Omega_2 \in so(3)$ and $\Omega_n = \frac{d}{dt}R_n(t)$ at the point $t=0$. So $\Omega_n$ is the tangent vector of the curve $R_n(t)$ at $t=0$.
To show that the Lie bracket $[\Omega_1,\Omega_2]$ is an element of $so(3)$ my professor wrote:
$$\tag{1}[\Omega_1,\Omega_2] = \left.\frac{d}{dt}R_1(t)\Omega_2 R_1(t)^{-1}\right|_{t=0}$$
I am trying to understand that, but I cannot see why this shows that the commutator is an element of $so(3)$.
I understand that for this tangential vector to be in $so(3)$, the product $R_1(t)\Omega_2 R_1(t)^{-1}$ has to be a parametrized curve in $SO(3)$.
But I do not see why it is a curve in $SO(3)$.
• I want to be clear about what you are asking :) Do you want to know why the product $R_1(t)\Omega_2R_1(t)^{-1}$ "has to be a parametrized curve" living completely with $SO(3)$. Or, are you asking if it is a parametrized curve" living completely with $SO(3)$, then you can using it in the above formula to calculate $[\Omega_1, \Omega_2]$. – muaddib Aug 11 '15 at 22:16
• I want to see the proof that the Lie bracket $[\Omega_1,\Omega_2]$ is an element of $so(3)$. In other words, this is the proof that the tangent space of $SO(3)$ at the unity matrix is a Lie algebra, if you use the commutator as the Lie bracket. My professor wrote the above formula $(1)$ as proof, but I don't understand it. – Bass Aug 12 '15 at 8:13
• Edited question for clarification. – Bass Aug 12 '15 at 8:16
We want to show that $$\tag{1}[\Omega_1,\Omega_2] = \left.\frac{d}{dt}R_1(t)\Omega_2 R_1(t)^{-1}\right|_{t=0}$$
We can start doing this by computing: \begin{eqnarray*} \left.\frac{d}{dt}\left(R_1(t)\Omega_2 R_1(t)^{-1}\right)\right|_{t=0} &=& \left.\left[\frac{d}{dt}R_1(t)(\Omega_2 R_1(t)^{-1})\right]\right|_{t=0} + \left.\left[(R_1(t)\Omega_2) \frac{d}{dt}R_1(t)^{-1}\right]\right|_{t=0} \\ &=& \left[\Omega_1\Omega_2 R_1(0)^{-1}\right] + \left.\left[R_1(t)\Omega_2 \frac{d}{dt}R_1(t)^{-1}\right]\right|_{t=0} \\ &=& \left[\Omega_1\Omega_2 R_1(0)^{-1}\right] + \left.\left[R_1(t)\Omega_2 (-R_1(t)^{-1})\left(\frac{d}{dt}R_1(t)\right)R_1(t)^{-1})\right]\right|_{t=0} \\ &=& \left[\Omega_1\Omega_2 R_1(0)^{-1}\right] - \left.\left[R_1(t)\Omega_2 R_1(t)^{-1}\left(\frac{d}{dt}R_1(t)\right)R_1(t)^{-1})\right]\right|_{t=0} \\ &=& \left[\Omega_1\Omega_2 R_1(0)^{-1}\right] - \left[R_1(0)\Omega_2 R_1(0)^{-1}\left.\left(\frac{d}{dt}R_1(t)\right|_{t=0}\right)R_1(0)^{-1})\right] \\ &=& \left[\Omega_1\Omega_2 R_1(0)^{-1}\right] - \left[R_1(0)\Omega_2 R_1(0)^{-1}\Omega_1R_1(0)^{-1})\right] \\ \end{eqnarray*}
See Derivatives of Inverse Matrix for details on the third line's derivation.
There isn't much further we can go without knowing more details about $R_1$. I actually think it might have been defined as $$R_i(t) = e^{\Omega_it}$$ But it's only necessary to require $R_i(0) = I$. Then the final line above gives $\Omega_1\Omega_2 - \Omega_2\Omega_1$.
The curve is in $SO(3)$
This is not always true. A Lie group is a subgroup of $GL$ but a Lie algebra always contains the $0$ matrix. It maps to the identity of the Lie group. So if $\Omega_2 = 0$ in the above, then the curve $C(t) = 0$ does not lie in $SO(3)$.
Show $[\Omega_1, \Omega_2]$ is in $\mathfrak{so}(3)$
This is true independently of the above considerations. $\Omega_1, \Omega_2$ are elements of $\mathfrak{so}(3)$ which is closed under addition and multiplication (matrices of the form $X^T = -X$).
• I'm sorry if I didn't express myself clearly. I see that formula (1) is correct, you showed it using the "product rule of differentiation". However, for $[\Omega_1,\Omega_2]$ to be an element of $so(3)$, this is not sufficient. You also need to show that $C(t) = R_1(t)\Omega_2 R_1(t)^{-1}$ is a curve in $SO(3)$ with $C(0)=\mathbb{1}$. – Bass Aug 12 '15 at 13:28
• @BastianTreichler - Updated – muaddib Aug 12 '15 at 13:43
• Thanks, but your last remark is not true. $\mathfrak{so}(3)$ is not closed under multiplication: $\begin{bmatrix}0 & 0 & 1\\0 & 0 & 0\\-1 & 0 & 0\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\-1 & 0 & 0\\0 & 0 & 0\end{bmatrix} = \begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & -1 & 0\end{bmatrix}$ The Lie bracket IS the multiplication in $\mathfrak{so}(3)$, and this does not seem trivial. – Bass Aug 12 '15 at 14:11
• @BastianTreichler - Yes, by "multiplication", I meant the one for the lie-algebra, namely the bracket. What I am saying is $[\Omega_1, \Omega_2] = \Omega_1\Omega_2 - \Omega_2\Omega_1$ is part of the definition of the Lie Algebra. – muaddib Aug 12 '15 at 14:13
• @BastianTreichler hmm, well once you obtain the formula from the above that $\Omega_1\Omega_2 - \Omega_2\Omega_1$ is the commutator, you can show it is an element of $\mathfrak{so}(3)$ as follows. Elements of $\mathfrak{so}(3)$ are those matrices that satisfy $X^T = -X$. Then $[X, Y]^T = (XY - YX)^T = Y^TX^T - X^TY^T = -[X^T, Y^T] = -[(-X), (-Y)] = -[X, Y]$. – muaddib Aug 12 '15 at 14:19
Formula (1) is a bit strange indeed, $R_1(t)\Omega_2 R_1(t)^{-1}$ is not necessarily in $SO(3)$ ($\Omega_2$ could have a null determinant for example).
One way to prove the result, is to consider that (for $t \to 0$):
$$R_1(t) = I + t \Omega_1 + t^2 W_1 + o(t^2)$$ $$R_2(t) = I + t \Omega_2 + t^2 W_2 + o(t^2)$$
Then:
$$R_1(t)R_2(t) = I + t (\Omega_1 + \Omega_2) + t^2 (W_1 + W_2 + \Omega_1 \Omega_2) + o(t^2)$$ $$R_2(t)R_1(t) = I + t (\Omega_1 + \Omega_2) + t^2 (W_1 + W_2 + \Omega_2 \Omega_1) + o(t^2)$$
The you can consider $L(t) = R_1(t)^{-1}R_2(t)^{-1}R_1(t)R_2(t)$ which is a curve in $SO(3)$. Since $R_2(t)R_1(t)L(t) = R_1(t)R_2(t)$, you can develop term by term each side and find that:
$$L(t) = I + t^2([\Omega_1, \Omega_2]) + o(t^2)$$
If you re-parametrise L with $s = t^2$ you have $L(s) = I + s([\Omega_1, \Omega_2]) + o(s)$ Thus you have found a curve in $SO(3)$ whose tangent is $[\Omega_1, \Omega_2]$.
• Thanks, but the last term $L(s)=I+s([\Omega_1,\Omega_2])+O(s)$ still contains that $O(s)$ term. What if that term was $sM+O(s^2)$ with some matrix $M$. Then the tangent of $L(s)$ is $M+[\Omega_1,\Omega_2]$, which is different from $[\Omega_1,\Omega_2]$. – Bass Aug 12 '15 at 10:05
• Why is $R_2(t)^{-1}R_1(t)^{-1}L(t) = R_1(t)R_2(t)$ ? I think the inversions are wrong, but you didn't accept my edit. – cheesus Aug 12 '15 at 11:05
• @BastianTreichler: Thanks for the remark. My $O(.)$ are indeed $o(.)$, this was a typo. I edited it. – Tantto Aug 12 '15 at 14:47
• @cheeesus: I think the way wrote is correct. If $L(t) = R_1(t)^{-1}R_2(t)^{-1}R_1(t)R_2(t)$ then you need to multiply from the left by $R_1(t)$ and then by $R_2(t)$ to obtain the second expression. – Tantto Aug 12 '15 at 14:47
• Exactly, but you are left-multiplying by $R_1(t)^{-1}$ and then by $R_2(t)^{-1}$. – cheesus Aug 12 '15 at 16:09
Found a solution, but it has nothing to do with $(1)$:
\begin{align} [\Omega_1,\Omega_2] & = \Omega_1\Omega_2-\Omega_2\Omega_1 = (-\Omega_1^T)(-\Omega_2^T) - (-\Omega_2^T)(-\Omega_1^T) \\ & = \Omega_1^T\Omega_2^T-\Omega_2^T\Omega_1^T=(\Omega_2\Omega_1)^T-(\Omega_1\Omega_2)^T = (\Omega_2\Omega_1-\Omega_1\Omega_2)^T \\ & = -[\Omega_2,\Omega_1] \end{align}
Here is a possibility as to how the professor was arriving at the result that $[\Omega_1, \Omega_2]$ is an element of the Lie algebra. Suppose that we take the above formula as the definition of the Lie Bracket: $$\tag{1}[\Omega_1,\Omega_2] = \left.\frac{d}{dt}R_1(t)\Omega_2 R_1(t)^{-1}\right|_{t=0}$$ There is a proposition that for any Lie group element $A$ and Lie algebra element $X$ (of its associated Lie algebra), that $AXA^{-1}$ is also in the Lie algebra. (Proof from Hall: $e^{t(AXA^{-1})} = Ae^{tX}A^{-1}$ and the latter is in the Lie group).
This means in the above, the curve $$C(t) = R_1(t)\Omega_2 R_1(t)^{-1}$$ lies entirely in the Lie algebra. Hence its derivative at $t = 0$ is also in the Lie algebra (because a Lie algebra is a vector space). So the misconception from the original question was that $C(t)$ lives in $\mathfrak{so}(3)$ instead of $SO(3)$.
• Could you elaborate a bit more about the Proof from Hall part? If you read my question, this was exactly the thing I wanted to see. I can't yet see why $C(t)$ must be a curve in $SO(3)$. – Bass Aug 12 '15 at 15:59
• What misconception do you mean? I never asked why $C(t)$ is in $so(3)$. My original question was: why is $C(t) \in SO(3)$ for an interval around $t=0$. – Bass Aug 12 '15 at 16:01
• It is not a curve in $SO(3)$ it is a curve in $\mathfrak{so}(3)$. Regarding the proof, the RHS is an element of the group because it is the product of three group elements. Then by definition of the Lie Algebra, if $e^{t(AXA^{-1}}$ in the Lie group for all $t$, then $AXA^{-1}$ is in the Lie algebra. – muaddib Aug 12 '15 at 16:03
• Yes, what I am saying is that is where you went wrong. – muaddib Aug 12 '15 at 16:03
• Oh now I see... I always expected $C(t)$ to be in $SO(3)$ so that $(1)$ defines a tangential vector and thus a member of $so(3)$. Okay, so now that this is sorted out, do you have more information about that "Proof of Hall"? When I google it, I only find pages about Hall's theorem, which is something else. – Bass Aug 12 '15 at 16:06
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2021-04-15 00:08:08
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https://stats.stackexchange.com/questions/124351/definitions-of-coefficients-from-arima-forecast
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# Definitions of coefficients from Arima {forecast}
I'm trying to explain in detail step by step what my code does and I am stuck at explaining what the coefficients are in an Arima model and where they are from/what relevance they have.
Could someone please explain to me what ma1, sar1, sar2, ar1, ar2, ar3, sma1 are and if possible show with a formula where they may appear in the equation for an ARIMA process? (The equation is not high priority as long as they are explained well)
Here is an ARIMA model found using ?Arima just you can see what i mean by the coefficients
fit <- Arima(WWWusage,order=c(3,1,0))
• Please note that the Arima formula requires the forecast library. WWWWusage can be accessed via data(WWWusage) – Will Scott Nov 17 '14 at 11:34
• ma1 = moving average parameter of order 1, sma1=seasonal ma1, ar1= autoregressive parameter of order 1, sar1=seasonal ar1, etc. See here on seasonal ARIMA models, which includes a formula with components for these. – Glen_b Nov 17 '14 at 11:55
In brief, the autoregressive (AR) terms represents the relationship between $y_t$ and $y_{t-1}$. A simple AR(1) model is:
$$y_t=\phi_1 y_{t-1} + \epsilon_{t-1}$$
In words, if $y_{t-1}$ is large, subsequent $y$'s also tend to be large if $\phi>0$ (although, if $\phi$ is less than 1, then $y$ will tend to gradually collapse back down).
In an AR(p) process, this is extended to $p$ lagged $y$ terms.
Moving average (MA) terms arise from a model like this: $$y_t = \theta_1 \epsilon_{t-1} + \epsilon_{t}$$
More generally, an MA(q) process is a moving average of the last $q$ error terms ....with weights equal to $\theta_1 \ldots \theta_q$.
A combination of AR and MA models is called an ARMA model.
Finally, having differences in the model (the middle term of the ARIMA model specification in R) means that instead of an ARMA model in $y$, the ARMA model describes $y_t-y_{t-1}$.
You also referred to sma1 and sar1 terms ... you can extend the ARIMA model even further to also cover seasonal time series, in which case sma1 and sar1 refer to the coefficients of the lagged errors and $y_t$'s at seasonal periods (ie 12 months ago for an annual model).
Rob Hyndman's excellent online textbook Forecasting Principles and Practice contains a chapter on ARIMA models that explains the meaning of the terms in far more detail than above. Other (offline) standard references include Applied Time Series Modelling and Forecasting (Harris) and Time Series Analysis (Hamilton).
The output tells you that the model chosen for your data is the following polynomial on lags of $y_t^*$:
$$y_t^* = 1.15\, y_{t-1}^* -0.66\, y_{t-2}^* + 0.34\, y_{t-3}^* + \hbox{residuals} \,, \quad t=1,2,...,n \,,$$
where $y_t^* = y_t - y_{t-1}$, that is, the first differences of the original series $y_t$. This means that the chosen model considers the presence of a stochastic trend rather than a deterministic trend, e.g. linear trend.
As regards the coefficients, they are weights of past observations of the data (in this case of the first differences of the data). We may expect that these weights will decay or go to zero. The reason is that, in general, the current observation is more closely related to the most recent observations rather than to old observations. For example, the prices observed in the present year are more likely to be related to the prices observed in the previous year rather than to the prices observed 10 years ago.
The coefficients can be also interpreted in terms of the underlying cycles generated by the polynomial related to the ARMA model. You can get the roots of the polynomial and the frequency of the cycles generated by that polynomial. Frequencies close to zero will mean that long-term patterns (i.e., cycles that are repeated after many observations) explain part of the dynamics of the data and the series will look relatively smooth. More erratic series are explained by cycles of lower frequency that are repeated every fewer observations. In seasonal time series, it is common to see cycles related to seasonal frequencies that are repeated every 2, 6 or 12 moths. For an example that includes some of the calculus behind this interpretation you may see this post.
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2021-02-27 16:08:24
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https://msp.org/agt/2007/7-4/p08.xhtml
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#### Volume 7, issue 4 (2007)
Recent Issues
The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Author Index To Appear ISSN (electronic): 1472-2739 ISSN (print): 1472-2747
Pseudo-Anosov homeomorphisms and the lower central series of a surface group
### Justin Malestein
Algebraic & Geometric Topology 7 (2007) 1921–1948
arXiv: math.gt/0702608
##### Abstract
Let ${\Gamma }_{k}$ be the lower central series of a surface group $\Gamma$ of a compact surface $S$ with one boundary component. A simple question to ponder is whether a mapping class of $S$ can be determined to be pseudo-Anosov given only the data of its action on $\Gamma ∕{\Gamma }_{k}$ for some $k$. In this paper, to each mapping class $f$ which acts trivially on $\Gamma ∕{\Gamma }_{k+1}$, we associate an invariant ${\Psi }_{k}\left(f\right)\in End\left({H}_{1}\left(S,ℤ\right)\right)$ which is constructed from its action on $\Gamma ∕{\Gamma }_{k+2}$ . We show that if the characteristic polynomial of ${\Psi }_{k}\left(f\right)$ is irreducible over $ℤ$, then $f$ must be pseudo-Anosov. Some explicit mapping classes are then shown to be pseudo-Anosov.
##### Keywords
pseudo-Anosov, lower central series, Torelli group, Johnson filtration
##### Mathematical Subject Classification 2000
Primary: 57M60, 37E30
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2018-07-21 06:05:52
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https://stats.stackexchange.com/questions/493923/consistency-of-likelihood-importance-sampling-estimator
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# Consistency of likelihood importance sampling estimator
In a lecture recently our lecturer described a method for approximating the expectation of a function over a posterior distribution using likelihood importance sampling. That is:
$$\mathbb{E}_{p(x|D)}[f(x)] \text{ can be approximated by } \frac{1}{\sum_{n=1}^{N}p(D|x^{(n)})}\sum_{n=1}^{N}p(D|x^{(n)})f(x^{(n)})$$
where $$x^{(n)}$$ are i.i.d. RV sampled from the prior $$p(x)$$, and $$p(D|x^{(n)})$$ is the likelihood of the observed data $$D$$.
He then claims that this estimator is biased but consistent. How can I see this? To see it is biased you usually find the expectation of the estimator and I cannot see how to do that here
$$\mathbb{E}[\frac{1}{\sum_{n=1}^{N}p(D|x^{(n)})}\sum_{n=1}^{N}p(D|x^{(n)})f(x^{(n)})]$$ as the denominator and numerator are non-independent functions of $$x$$. Am I missing something and the denominator can be considered a constant and taken out of the expectation?
Any help or intuition much appreciated.
• A sample from the prior need be weighted by the likelihood to become a sample from the posterior. That it is biased is due to the division by the average$$\frac{1}{N}\sum_{n=1}^{N}p(D|x^{(n)})$$whose expectation is the normalising constant (or marginal likelihood) and hence whose inverse does not have as expectation the inverse of the normalising constant. – Xi'an Oct 27 at 21:00
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2020-11-23 20:03:51
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https://discourse.mc-stan.org/t/the-effect-of-weights-on-the-resulting-estimates/15274
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# The effect of weights on the resulting estimates
I am hoping to understand how weights will effect the fit that STAN develops. I know that weights aren’t full Bayesian, but I’m willing to sacrifice that.
Suppose I have two Bernoulli observations that are a function of a latent variable, and weight them differently in the model. Will this mean that the model will optimize to ‘care’ about the higher-weighted observation more, and/or by multiplying the log-liklihood by a weight am I implicitly changing the value of the probability (in the log liklihood).
For example, in the following model:
``````data{
int result[2];
vector[2] weights;
vector[2] breaks;
}
parameters{
real mu;
}
model {
real p;
for(i in 1:2){
p = normal_cdf(breaks[i],mu,3);
target += bernoulli_lpmf(result[i]|p) * weights[i];
}
/// the log density of a bernoulli is x *log(p) + (1-x)*log(1-p) where x is the result
/// if I multiply this by a weight, am I implicitly telling STAN that the probability is actually p ^ weights[i] ?
}
``````
Additionally, suppose weights[1] = .5 and weights[2] = 1 ; then will this Stan model care more about the 2nd observation than it will about the 1st observation?
Thanks, appreciate any help!
if I multiply this by a weight, am I implicitly telling STAN that the probability is actually p ^ weights[i] ?
Stan samples from a distribution defined by the unnormalized log density `target`.
So if before your model was:
``````target += log(q(theta));
``````
``````target += log(q(theta)) * w;
Then before you were generating samples from a distribution proportional to `q(theta)` and now you are generating samples from a distribution proportional to `q(theta)^w` (assuming these are proper distributions and can be normalized and Stan is mixing and whatnot).
Well Stan just samples whatever the `target` defines. In this case the model weights the first likelihood half as much as the second, yes.
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2020-05-26 05:14:18
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https://www.jiskha.com/display.cgi?id=1336797673
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# Calculus!
posted by kelsey
Which is the standard form of the equation with p = 26, O=3pi/4 ?
the O before the =3pi has a diagonal slash through it.
1. drwls
You have written two equations that seem to describe two polar coordinates. Is "p" the radisl distance from the origin?
It is not clear what you mean by "standard form"
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2018-07-21 11:57:02
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http://math.stackexchange.com/questions/425664/finding-sin6-x-cos6-x-what-am-i-doing-wrong-here
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# Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here?
I have $\sin 2x=\frac 23$ , and I'm supposed to express $\sin^6 x+\cos^6 x$ as $\frac ab$ where $a, b$ are co-prime positive integers. This is what I did:
First, notice that $(\sin x +\cos x)^2=\sin^2 x+\cos^2 x+\sin 2x=1+ \frac 23=\frac53$ .
Now, from what was given we have $\sin x=\frac{1}{3\cos x}$ and $\cos x=\frac{1}{3\sin x}$ .
Next, $(\sin^2 x+\cos^2 x)^3=1=\sin^6 x+\cos^6 x+3\sin^2 x \cos x+3\cos^2 x \sin x$ .
Now we substitute what we found above from the given:
$\sin^6 x+\cos^6+\sin x +\cos x=1$
$\sin^6 x+\cos^6=1-(\sin x +\cos x)$
$\sin^6 x+\cos^6=1-\sqrt {\frac 53}$
Not only is this not positive, but this is not even a rational number. What did I do wrong? Thanks.
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Expanded $(\sin^2 x+\cos^2 x)^3$ incorrectly. Need $3\sin^4 x\cos^2 x+3\cos^4 x\sin^2 x$ as the last two terms. – André Nicolas Jun 20 '13 at 19:33
@AndréNicolas Oh yeah haha the expansion is soo wrong, I don't know how I came up with it. But if I do the expansion correctly I think I can still solve it this way. – Ovi Jun 20 '13 at 19:36
As a side note, I released this exact problem several weeks back on Brilliant. Ovi, if that is where you obtained the problem from, you can click on the link to view the solution. – Calvin Lin Jun 20 '13 at 20:06
@CalvinLin Yes thank you that is where I got it from but I wanted to do it on my own before I looked at the solution. It worked perfectly fine after I corrected my mistake. – Ovi Jun 20 '13 at 20:09
Yes, I can tell from your working above (as opposed to simply copying the question and expecting complete answers) :) – Calvin Lin Jun 20 '13 at 20:10
$(\sin^2 x + \cos^2 x)^3=\sin^6 x + \cos^6 x + 3\sin^2 x \cos^2 x$
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Should be $(\sin^2 x+\cos^2 x)^3=1=\sin^6 x+\cos^6 x+3\sin^4 x \cos^2 x+3\cos^4 x \sin^2 x$
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$\sin^6x + \cos^6x = (\sin^2x)^3 + (\cos^2x)^3 =(\sin^2x + \cos^2x)(\sin^4x + \cos^4x -\sin^2x\cos^2x)$
$\sin^4x+\cos^4x -\sin^2x\cos^2x = (\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x -\sin^2x\cos^2x$
or $1-3\sin^2x\cos^2x = 1-3\left(\dfrac13\right)^2 = \dfrac23$.
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Use $\LaTeX$ command \sin instead sin, and similar for cos. – Cortizol Jul 21 '13 at 10:32
It has to be $1 - 1 / 3$, not $1 - (1 / 3) ^ 2$, the answer is $2 / 3$. – user98213 Oct 2 '13 at 10:54
$a^3+b^3=(a+b)(a^2-ab+b^2),$ not $a^3+b^3=(a+b)(a^2+ab+b^2).$ – Cameron Buie Oct 2 '13 at 11:22
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2016-07-29 00:19:44
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https://infoscience.epfl.ch/record/130216
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Bifurcation in $L^p({R}_N)$ for a semilinear elliptic equation
Published in:
Proceedings of the London Mathematical Society. Third series***, 57, 3, 511-541
Year:
1988
Keywords:
Laboratories:
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2019-01-23 03:06:54
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http://sioc-journal.cn/Jwk_hxxb/CN/10.6023/A15080548
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### 植物焦炭氧化中的平行反应及其动力学解析
1. 中国科学技术大学火灾科学国家重点实验室 合肥 230027
• 投稿日期:2015-08-17 发布日期:2015-10-29
• 通讯作者: 王海晖 E-mail:HHWang4@ustc.edu.cn
• 基金资助:
项目受中央高校基本科研业务费专项资金(No. WK2320000032)资助.
### A Study of Plant Char Oxidation: the Parallel Reactions and Their Chemical Kinetics
Tao Junjun, Chen Shuai, Yao Fengqi, Wang Haihui
1. State Key Laboratory of Fire Science, University of Science and Technology of China, Hefei 230027
• Received:2015-08-17 Published:2015-10-29
• Supported by:
Project supported by the Research grant from the Fundamental Research Funds for the Central Universities (No. WK2320000032).
The present work explores the effects of the parallel reactions on the oxidation reactivity of plant chars formed at various heat treatment temperatures by using a multi-component parallel reaction model in conjunction with the non-linear kinetic analysis technique. Four plant species were selected. A plant char was prepared in a programme-controlled horizontal tube furnace under the atmosphere of N2 at a purity of 99.999%, and its heat treatment temperature was set at 450, 520 and 800 ℃, respectively. The aerial oxidation characteristics of plant chars were analyzed by using a simultaneous thermo-gravimetric analyzer and differential scanning calorimeter, and the kinetic data of the parallel reactions were then retrieved by fitting both the mass change and mass loss rate data obtained during the oxidation measurements. It was confirmed that for the oxidation of the chars formed at moderate heat treatment temperatures (i.e. 450 and 520 ℃), the variation patterns of mass loss rate and the heat flow rate curves are contributed by the parallel oxidation reactions of lignin residue, amorphous carbon and the other reactive substances such as crude fat and protein, etc. When the heat treatment temperature reaches 800 ℃, the reactive substances stored in the char produced are mainly amorphous carbon, and the char oxidation can be simplified to an one-step reaction. Compared with the other two reaction components, lignin residue has the lowest activation energy for oxidation with the range between 86 and 147 kJ·mol-1, and its reaction temperatures vary between 300 and 480 ℃. The activation energy for the amorphous carbon fluctuates between 174 and 208 kJ·mol-1 with its reaction temperatures altering between 370 and 520 ℃. The other reactive substances undergo oxidation with the activation energy between 214 and 225 kJ·mol-1 and the corresponding reaction temperatures are between 420 and 510 ℃. It is obvious that the oxidation reactivity of plant chars mainly relies on the performance of lignin residue. With the increase in the heat treatment temperature for making a char, the oxidation reactivity of the plant char essentially reduces, which is essentially attributed to the decrease in the content of lignin residue and the increase in the activation energies for the other two components to oxidize. The established understanding lays the foundation for developing more effective methodologies for making use of the energy stored in plant chars and paves the way for identifying the role of plant chars in the spread of a wildland fire.
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2020-10-27 00:56:46
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http://persiavisit.ir/css/peter-harrison-zlv/archive.php?page=rotational-constant-of-hcl-bfa990
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Pressure-induced shifts of molecular lines in emission and in absorption, In terms of the angular momenta about the principal axes, the expression becomes. Proc. 0000112882 00000 n Leavitt, J.A. ; Dymanus, A., the Morse potential), α e Database and to verify that the data contained therein have Using the rotational constants from the polynomial curve fit with the definition of B gives the moment of inertia 19. Photoelectron spectroscopy of HCI and DCI using molecular beams, J. Chem. Measurement of the pressure broadening of the rotational Raman lines of HCl, The rotational constant is dependent on the vibrational level: ˜Bv = ˜B − ˜α(v + 1 2) Where ˜α is the anharmonicity correction and v is the vibrational level. Spectrosc., 1973, 45, 366. Nicholson, A.J.C., Precise measurements of some infrared bands of hydrogen chloride, J. Mol. Proton radio-frequency spectrum of HCl35, Phys., 1953, 21, 1340. Electronic excitation of HCl trapped in inert matrices, Spectre de vibration-rotation de l'acide chlorhydrique gazeux. 0000003340 00000 n Terwilliger, D.T. Webb, D.U. ; Henneker, W.H. Gebbie, H.A. 0000035667 00000 n ; Keaveny, I., ; Bader, R.F.W. [all data], Levy, Rossi, et al., 1966 Interpretation of the core electron excitation spectra of hydride molecules and the properties of hydride radicals, 10.502 ~ 3049.15 1.280 10 − − − = = = B. cm v cm r x cm. rotational constant as $21.77\rm~cm^{-1}$. Cade, P.E. e e e J. Quant. [all data], Sanderson, 1967 Soc. The quantum mechanics of rotational and vibra-tional motion is only one step beyond the Spectrochim. Phys. 0000003484 00000 n Rotational Energy. errors or omissions in the Database. [all data], Goldring and Benesch, 1962 ; Wiggins, T.A., WII��%%4�v�)B��I)� .��!�\$@B�uf�z��@a�P��f 5� 0000008074 00000 n Where $${B}_{e}$$ is the rotational constant for a rigid rotor and $$\alpha_{e}$$ is the rotational-vibrational coupling constant. ; Thibault, R.J., Part II. We also want to assign J quantum numbers to each of the transitions. ; Ben-Reuven, A., Lempka, H.J. It turns out that for an anharmonic potential (e.g. Spectrosc., 1970, 33, 505. A: Gen. Measurement of rotational line strengths in HCl by asymmetric Fourier transform techniques, constants and vibrational frequencies) we can prepare a simulated spectrum so that we know what to expect when we perform the physical experiment. Acta, 1960, 16, 479. Theoretical Calculations. J. Mol. Typica1 "V" state with configuration ... σπ, Very extended progression in absorption, not yet analyzed in detail. Phys., 1960, 33, 323. 09:00 Conversion of length from meters (m) to angstroms (Å) Calculation of the bond length of H₂ from the experimentally determined rotational constant . Jaffe, J.H. Compute the separation of the pure rotational spectrum lines in GHz, cm-1, and mm, and show that the value of B is consistent with an N-H bond length of 101.4 pm and a bond angle of 106.78°. Chem. trailer J. Mol. [all data], Terwilliger and Smith, 1973 or wavenumbers becomes F(J) = B. e. J(J + 1) with where B. e. is the . [all data], Schwarz, 1975 ; Vu, H.; Vodar, B., Benedict, W.S. Line strengths and widths in the HCl fundamental band, (A) 2.68 ×104" kg m? Using the Nicolet 6700 spectrometer, the spectrum for HCl was analyzed. ; Vroom, D.A., ; Friedmann, H.; Hirshfeld, M.A. October 8, 2020. Acta, 1967, 23, 553. 0000041417 00000 n Absorption by some molecular gases in the extreme ultraviolet, Phys. [all data], Romand, 1949 Rotational Constant. When there is no vibrational motion we expect the molecule to have the internuclear separation (bond length) R = R. e, and the rotational energy in cm-1. Tokuhiro, T., HCl molecule, which is modeled as two nu-clei connected by a “spring” representing the interatomic force. [all data], Chamberlain and Gebbie, 1965 J. Quant. © 2018 by the U.S. Secretary of Commerce The rotational constant of NH 3 is equivalent to 298 GHz. and c is the speed of light and h is the Planck’s constant. London, 1963, 82, 309. ; Vanderslice, J.T., Am., 1962, 52, 1. Phys., 1962, 40, 1801. Proc. [all data], Douglas and Greening, 1979 Vibrational and rotational effects on the nuclear quadrupole coupling constants in hydrogen, deuterium, and tritium halides, J. Chim. [all data], Weiss, Lawrence, et al., 1970 Atwood, M.R. Khatibi, P.; Vu, H., 680 0 obj <> endobj If the rms velocity of HCl molecules in its gaseous phase is bar (v),m is its mass and k B is Boltzmann constant, then its temperature will be - jee mains 2019; Share It On Facebook Twitter Email. All rights reserved. The rotational spectra of non-polar molecules cannot be observed by those methods, but can be observed … Spectroscopic constants and dipole moment functions for the ground states of the first-row and second-row diatomic hydrides, Soc. Can. ; Khosla, A.; Ozier, I.; Ramsey, N.F. J. Chem. Constantes de vibration-rotation de l'acide chlorhydrique gazeux etude des bandes vo→2 et vo→3, 0000028658 00000 n Spectrosc., 1965, 17, 122. J. Mol. Constants of Diatomic Molecules, Van Nostrand Reinhold Company, New York, 1979, 716. ; Smith, A.L., [all data], Plyler and Tidwell, 1960 Phys., 1975, 62, 3353. [all data], Plyler and Thibault, 1962 For both species the rotational and the vibrational constants are B = 2 cm − 1 and ˉνe = 2330.7 cm − 1, respectively. The absorption spectra of the halogen acids in the vacuum ultra-violet, 1 Answer +1 vote . J. Chem. The rotational constant can be approximated by Bv @ Be - ae(v + 1/2) (12) where Bv is the rotational constant taking vibrational excitation into account, and ae is defined as the rotational-vibrational coupling constant. Selection rules: (1) permanent dipole moment, (2) ΔJ = ± 1 only 5. I think a simple way is 2B is the interval between two spectral lines so subtract line position 2 from line position 1 and then divide by 2 you will get B i.e. Rev., 1961, 124, 1482. 9.977 ~ 3372.52 1.313 10 − − − = = = B. cm v cm r x cm. Go To: Top, References, Notes Data compilation copyrightby the U.S. Secretary of Commerce on behalf of the U.S.A.All rights reserved. NIST Standard Reference Energy transitions from the spectra were plotted vs. frequency, from which several physical constants were determined. and c is the speed of light and h is the Planck’s constant. Data compiled by: Klaus P. Huber and Gerhard H. Herzberg, Go To: Top, Constants of diatomic molecules, Notes, Hayes and Brown, 1972 ; Tidwell, E.D., Etude de la bande v0→2 a 1,7 micron, Molecular constants of HCl35, Radiat. Watson, J.K.G., The relative atomic weight C =12.00 and O = 15.9994, the absolute mass of H= 1.67343x10-27 kg. Absorption ultraviolette dans la region de Schumann etude de: ClH, BrH et lH gazeux, Average B, D values; B(R,P)-B(Q) = +0.385. Soc. [all data], Jaffe, Friedmann, et al., 1963 Etude de la dispersion dans le doublet isotopique R2 de la premiere bande harmonique de vibration-rotation de HCl, ; Nelson, H.M.; Ramsey, N.F., by the U.S. Secretary of Commerce on behalf of the U.S.A. on behalf of the United States of America. Calculated rotational constants for HCl (Hydrogen chloride). Sci. Levy, A.; Rossi, I.; Joffrin, C.; Van Thanh, N., k = 6.057x10 −5 1. cm dyne k. lit. [all data], Cade, Bader, et al., 1969 ; Silverman, S., uses its best efforts to deliver a high quality copy of the Extension of submillimeter wave spectroscopy below a half-millimeter wavelength, Ann. such sites. Data compiled by: Klaus P. Huber and Gerhard H. Herzberg Rotational Energies The classical energy of a freely rotating molecule can be expressed as rotational kinetic energy. [all data], Jaffe, Kimel, et al., 1962 [all data], Kaiser, 1974 EJ = BJ ()J1 where B h 8 2 Biol., 1965, 62, 600. J. Chem. Photoelectron spectra of the halogens and the hydrogen halides, London, 1959, 73, 538. Tilford, S.G.; Ginter, M.L. (Paris), 1949, 4, 527. Rank, D.H.; Eastman, D.P. The rotational constant of NH 3 is equivalent to 298 GHz. Interferometric measurements of the pure rotational spectra of HCl and DCl, J. Chem. Diffuse rotational structure; 1-0 and 2-0 are increasingly diffuse. 0000003532 00000 n J. Chem. Proc. [all data], Jacques, 1959 Measurement of widths and shifts of pure rotation lines of hydrogen chloride perturbed by rare gases, coupling constant ωe = the fundamental vibrational constant ωexe = the first anharmonic correction constant. J. Mol. 0000001159 00000 n Technology, Office of Data startxref Jacques, Intensites et largeurs de raies dans la bande v0-2, ; Gebbie, H.A., [all data], Herman and Asgharian, 1966 Far infrared spectra of HCl and DCl in a nitrogen matrix, Roy. Transfer, 1972, 12, 219. ; Wiggins, T.A., I. Spectres dans le fondamental de vibration-rotation, Proc. ; Rao, B.S. ; Jaffe, J.H., J. Opt. [all data], Rich and Welsh, 1971 J. Chem. Lett., 1971, 11, 292. 0000003292 00000 n EJ = BJ ()J1 where B h 8 2 J. Chem. Forme et structures fines de la bande induite par la pression dans la bande fondamentale de vibration-rotation des molecules HF, HCl et HBr, [all data], Bunker, 1973 Infra-red emission from gases excited by a radio-frequency discharge, [all data], Rank, Birtley, et al., 1960 0000024255 00000 n Roy. For the non-rigid rotating and non-harmonic oscillating molecules additionally the centrifugal elongation constant D = 0.0001 cm − 1 (Eqn (2.39)) and the … G�_װ��q��tIk���&x���W�s�\������岕�foZ��K����;�Z�K+�uN��4-���,�Gm�hԨ��(�kY��k�%wJ�E�Zݤ/خ����9G1!�K�������"�x�����.�d��ZQ��S����K\[&]���Q��:��f��I��8��c�Xc��0o��ܭ��ca ,H��vM�8���^�R�L��DZ�B�E���e��QY�qp,����AE�XCC����,.H #L\\�AB����&HBUJJJꨪ����ьbC��ˌd(HuZ: Bur. Rich, N.H.; Welsh, H.L., D. Phil. Phys., 1975, 63, 2356. 0000041093 00000 n Spectrosc., 1973, 45, 151. II. Rotational Energy. ; Hirshfeld, M.A. [all data], Watson, 1973 0000006163 00000 n Pressure-induced rotational quadrupole spectra of HCl and HBr, Anatomy of a vibration-rotation band showing rotational energy levels in their respective upper and lower vibrational energy levels, along with some allowed transitions. Foreign gas broadening of the lines of hydrogen chloride and carbon monoxide, [all data], Leavitt, Baker, et al., 1961 There were two branches that were apparent in the result of the spectroscopy, the R branch and the P branch, that correspond to ∆J= +1 and ∆J= -1… The second-row diatomic hydrides AH, [all data], Kaiser, 1970 Data Program, but require an annual fee to access. Electronic spectra and structure of the hydrogen halides. The rotational constant of NH 3 is equivalent to 298 GHz. [all data], Hansler and Oetjen, 1953 Sanderson, R.B., Jaffe, J.H. 0000003436 00000 n Katz, B.; Ron, A., Bunker, P.R., Many other absorption bands in the region 83000 - 93000 cm. ; Oetjen, R.A., Spectrosc. Ben-Reuven, A.; Kimel, S.; Hirshfeld, M.A. in these sites and their terms of usage. Soc. Sect. Soc. Goldring, H.; Benesch, W., Bunker, P.R., J. Mol. "t��X�9�=��l����8�a���Ē ��g�1�&�-}� J��}�k����l�ǐ�� � �p�����qCx��"+0�����. [all data], Tilford and Ginter, 1971 J. Quant. [all data], Babrov, Ameer, et al., 1959 Toth, R.A.; Hunt, R.H.; Plyler, E.K., The information in the band can be used to determine B 0 and B 1 of the two different energy states as well as the rotational-vibrational coupling constant, which can be found by the method of combination differences. J. IV. Rotational and vibrational constants of the HCl35 and DCl35 molecules, Phys., 1969, 50, 5313. ; Stone, N.W.B., Hayes, W.; Brown, F.C., The breakdown of the Born-Oppenheimer approximation for a diatomic molecule: the dipole moment and nuclear quadrupole coupling constants, If we assume that the vibrational and rotational energies can be treated independently, the total energy of a diatomic molecule (ignoring its electronic energy which will be constant during a ro-vibrational transition) is simply the sum of its rotational and vibrational energies, as shown in equation 8, which combines equation 1 and equation 4. Only those expressions and constants in bold type will concern us; for example, Dv for HCl is ~5 × 10–5 Bv, which is so small it alters transition energies by an amount too 0000035488 00000 n Spectrosc., 1959, 3, 185. 0000003244 00000 n Datta, S.; Banerjee, S., 0000007493 00000 n Z. [all data], Tokuhiro, 1967 Smith, F.G., [all data], Frost, McDowell, et al., 1967 The observed rotational and vibrational constants (Y lj) have been used to calculate the potential constants of HCl 35 by making use of Dunham’s theory of a rotating vibrator. A classic among molecular spectra, the infrared absorption spectrum of HCl can be analyzed to gain information about both rotation and vibration of the molecule. Herman, R.M. [all data], Atwood, Vu, et al., 1967 Vibration rotation bands of heated hydrogen halides, [all data], Rank, Eastman, et al., 1962 There were two branches that were apparent in the result of the spectroscopy, the R branch and the P branch, that correspond to ∆J= +1 and ∆J= -1, respectively. ; Kimel, S.; Hirshfeld, M.A., 0000002487 00000 n Kaiser, E.W., 0000113106 00000 n Code, R.F. → From rotational spectra we can obtain some information about geometrical structure of molecule (r): For diatomic molecule we can calculate the length of bond! HCl and anharmonicity constant 0.071 ~ 230.198 ~ 3239.62. [all data], Katz and Ron, 1970 We also want to assign J quantum numbers to each of the transitions. 0 The rotational constant Bv for a given vibrational state can be described by the expression: Bv = Be + e(v + ½) where Be is the rotational constant corresponding to the equilibrium geometry of the molecule, e is a constant determined by the shape of the anharmonic potential, and v is the vibrational quantum number. and Informatics, Microwave spectra (on physics lab web site), Computational Chemistry Comparison and Benchmark Database, NIST / TRC Web Thermo Tables, professional edition (thermophysical and thermochemical data), electronic state and / or symmetry symbol, rotational constant in equilibrium position (cm, rotation-vibration interaction constant (cm, rotational constant – first term, centrifugal force (cm, observed transition(s) corresponding to electronic state, position of 0-0 band (units noted in table), Numerous absorption bands above 123000 cm. Spectrosc. Phys.-Chim. [all data], Lempka, Passmore, et al., 1968 derived barriers for rotation K4 /B in units of the rotational constant B for HCl in matrices led to inconsistencies in com-parison with the available experimental rotational structure at that time.1 The asymmetric charge distribution in the HCl-molecule causes a deviation of the center-of-mass from the center of the site. 1 0 1 1 .92118 3 .8425 2 o B cm X B 47 2 Can. 0000027610 00000 n kg m² (C) 2.68 ×1031 kg m? J. Chem. [all data], Price, 1938 0000007066 00000 n HCl, Widths of HCl overtone lines at various temperatures, J. Chem. lines, Proton spin - rotation interaction constant, Strongly broadened by preionization (lifetime τ= 1.1E-14 s), Absolute intensities (cm-2atm-1) of the Pressure-induced shifts of DCl lines due to HCl: shift oscillation, Rank, D.H.; Rao, B.S. 0000003166 00000 n 2 (sharper rise on the low-rside). 0000002633 00000 n The observed rotational and vibrational constants (Y lj) have been used to calculate the potential constants of HCl 35 by making use of Dunham’s theory of a rotating vibrator. ; Asgharian, A., Nature (London), 1965, 208, 480. Tilford, S.G.; Ginter, M.L., Spectra and Molecular Structure – HCl & DCl By: Christopher T. Hales. Hansler, R.L. 0000006200 00000 n e e e. MP Results. ; Yi, P.N., ; Wiggins, T.A., Plyler, E.K. %%EOF [all data], Alamichel and Legay, 1966 [all data], Nicholson, 1965 (D) 4.21x1047 kg m² (B) 4.21×10-51 J. Chim. II. Phys., 1966, 45, 2433. J. Mol. 0000005194 00000 n The purpose of the fee is to recover costs associated [all data], Levy, Rossi, et al., 1965 ; Price, W.C., Phys., 1968, 49, 1895. ; Ben-Reuven, A., An HCl molecule has rotational, translational and vibrational motions. Spectres d'absorption infrarouge de HCl et de HBr en phases denses. [all data], Datta and Banerjee, 1941 The strengths, widths, and shapes of infrared lines. Inst. Δ= 17.414%. the Spectrosc. Data compiled by: Klaus P. Huber and Gerhard H. Herzberg Data collected through December, 1976 0000046821 00000 n The b3Πi and C1Π states of HCl and DCl, 721 0 obj <>stream [all data], Boursey, 1975 The rotational constant can be approximated by B v ≅ B e - α e (v + 1/2) (12) where B v is the rotational constant taking vibrational excitation into account, and α e is defined as the rotational-vibrational coupling constant. Phys., 1970, 53, 1686. Levy, A.; Rossi, I.; Haeusler, C., [all data], Rosenberg, Lightman, et al., 1972 Compute the separation of the pure rotational spectrum lines in GHz, cm‐11, and show that the value of B is consistent with an N‐H bond length of 101.4 pm and a bond angle of 106.78°. Spectrosc. Opt., 1967, 6, 1527. Phys., 1963, 39, 1447. False and genuine structure, Weiss, S.; Cole, R.H., The rotational constant at equilibrium (B e) was equal to 10.56 ± -0.02 cm-1 for HCl and 5.46 ± 0.03 cm 1 for DCl and is the main factor in describing rotational aspects of the molecule. Spectrosc., 1972, 5, 478. The transition v1Σ+-x1Σ+ in hydrogen chloride, [all data], Gebbie and Stone, 1963 If we assume that the vibrational and rotational energies can be treated independently, the total energy of a diatomic molecule (ignoring its electronic energy which will be constant during a ro-vibrational transition) is simply the sum of its rotational and vibrational energies, as shown in equation 8, which combines equation 1 and equation 4. Transfer, 1970, 10, 203. [all data], Rank, Rao, et al., 1965 Magnetic properties and molecular quadrupole moment of HF and HCl by molecular-beam electric-resonance spectroscopy, [all data], Weiss and Cole, 1967 5: HF Results. [all data], Mould, Price, et al., 1960 Rigid Rotor Model for HCl For the rigid rotor model the rotational energy levels (in cm‐1) are given by the following equation. The higher Phys. Spectrosc., 1973, 45, 99. The rotational constants of hydrogen chloride, J. Mol. Phys., 1961, 35, 955. Molecular charge distributions and chemical binding. Spectrosc., 1973, 48, 427. Appl. However, NIST makes no warranties to that effect, and NIST J. Quant. Meyer, W.; Rosmus, P., J. Chem. ; Baker, M.R. Rev., 1964, 135, 295. [all data], Bunker, 1972 J. Mol. Jacques, J.K.; Barrow, R.F., ; Herman, R.; Moore, G.E. with the development of data collections included in Jaffe, J.H. ; B = rotational constant, units cm-1 4. J. Chem. Spectrosc., 1971, 40, 568. Phys. Phys., 1967, 46, 4255. 1 1 8. 0000005798 00000 n cm dyne = 5.159x10 −5 1. Calculate the rotational constant (B) and bond length of CO. Finite nuclear mass effects on the centrifugal stretching constant in H35Cl, Spectry. with m the reduced mass; I e and B e are, respectively, the equilibrium moment of inertia and rotational constant B e = h/8π 2 I e.Here ω e = (1/2π)(f/m) 1/2 is the harmonic vibrational frequency. Thesis, Oxford, 1959, 1. where x, y, and z are the principal axes of rotation and I x represents the moment of inertia about the x-axis, etc. [all data], Meyer and Rosmus, 1975 [all data], Khatibi and Vu, 1972 Rotational spectroscopy is concerned with the measurement of the energies of transitions between quantized rotational states of molecules in the gas phase.The spectra of polar molecules can be measured in absorption or emission by microwave spectroscopy or by far infrared spectroscopy. Be = the equilibrium rotational constant αe = the vib.-rot. Rev. Note: Comment to "Dipole moment function and vibration-rotation matrix elements of HCl35 and DCl35", Soc. Each peak, differentiating between 35Cl and 37Cl, is assigned an m value and then … Price, W.C., Can. 0000023699 00000 n Follow the links above to find out more about the data In other words, the magnitude of the rotational constant depends on the vibrational state of the molecule. Ogilvie, J.F. Transfer, 1973, 13, 717. Refraction spectrum of gases in the infrared intensities and widths of lines in the 2-0 band of HCl, [all data], Benedict, Herman, et al., 1956 E Calculate the moment of inertia for HCl molecule from the given value of rotational constant, B = 10.40 cm. Am., 1960, 50, 1275. 1 1 8. Nuclear magnetic hyperfine spectra of H35Cl and H37Cl, [all data], Huber and Herzberg, 1979 (London), 1968, A304, 53. 0000023979 00000 n [all data], Code, Khosla, et al., 1968 Transfer, 1974, 14, 317. where is the anharmonic vibrational frequency correction, . the rotational constant, , accounts for centrifugal stretching, and is the anharmonicity correction to rotation. Chamberlain, J.E. ; Koo, D., Phys., 1972, 6, 21. Babrov, H.; Ameer, G.; Benesch, W., 0000006443 00000 n ; Wiggins, T.A., The infrared spectra of HCl, DCl, HBr, and NH3 in the region from 40 to 140 microns, Rigid Rotor Model for HCl For the rigid rotor model the rotational energy levels (in cm‐1) are given by the following equation. Copyright for NIST Standard Reference Data is governed by ; Rao, K.N., [all data], Rank, Eastman, et al., 1960 (Paris), 1966, 27, 233. Levy, A.; Mariel-Piollet, E.; Bouanich, J.-P.; Haeusler, C., 1-0 band: 130. J. Chem. 0000024516 00000 n J. Res. Go To: Top, References, Notes Data compilation copyrightby the U.S. Secretary of Commerce on behalf of the U.S.A.All rights reserved. Plyler, E.K. Dunham potential energy coefficients of the hydrogen halides and carbon monoxide, Rank, D.H.; Eastman, D.P. The HCl fundamental, J. Alamichel, C.; Legay, F., The isotope dependence of the equilibrium rotational constants in 1Σ states of diatomic molecules, Watanabe, K.; Nakayama, T.; Mottl, J., Phys., 1956, 34, 850. J. Opt. ; Eastman, D.P. Stand. Spectrosc., 1976, 61, 332-336. Phys., 1964, 40, 1705. Rydberg series corresponding to excitation of a 2p electron. Molecular Spectra and Molecular Structure. [all data], Toth, Hunt, et al., 1970 0000027853 00000 n Rotational Energies The classical energy of a freely rotating molecule can be expressed as rotational kinetic energy. x�bf������q�A���bl,=������ BGN4�?Pra��� ���]�"��D�82�3�5��q�N�L�}~oZd\�F.��&�p���9�%��*WBSof�XXx}~T��� �x"+D|Y� 680 42 Unpublished cited in Huber and Herzberg, 1979, 1979, 287. the … 1 1 = = = − − e e e e. x v x cm v cm. Douglas; Greening, It turns out that for an anharmonic potential (e.g. Sub-millimetre dispersion and rotational line strengths of the hydrogen halides, J. Mol. Phys., 1965, 43, 1171. Rosenberg, A.; Lightman, A.; Ben-Reuven, A., Rotational Constant. J. Chem. Frost, D.C.; McDowell, C.A. diatomic molecule such as HCl, we see rotational structure on top of the vibration, which splits the spectrum into two branches, designated P and R. To a first approximation, the transitions within the branches are separated by B ~ 2. Phys., 1967, 47, 109. Biol., 1972, 69, 654. It is shown that HCl 35 is not a pure rotating vibrator since the observed and calculated values of Y 02 ~ D e are in disagreement by about 1 part in 1000 which is approximately 10 times the experimental error. Spectrosc., 1970, 35, 110. Theory, Calculations, and Discussion: The energy of any vibrational-rotational state of HCl, expressed in its wavenum- ber equivalent and measured in cm–1units, can be written to a good approximation as Ev,J= G(v) + BvJ(J + 1) – DvJ2(J + 1)2+ … The system is called a non-rigid rotator and the nuclear motion consists of simultaneous rotations and vibrations. the … form of a spring of force constant k. The cubic term has been written with a negative sign so that a positive gproduces the typical asymme-try of a bonding V(r) shown in Fig. Phys.-Chim. Perturbation of molecular rotation-vibration energy levels by rare gases, Phys. Spectrosc., 1968, 28, 121. (London), 1938, A167, 216. PNO-Cl and CEPA studies of electron correlation effects. National Institute of Standards and [all data], Tilford, Ginter, et al., 1970 0000019080 00000 n vibrational levels are strongly perturbed by Rydberg states, Continuous absorption starting at 44000 cm. Mould, H.M.; Price, W.C.; Wilkinson, G.R., Radiative Transfer, 1962, 2, 369. rotational constant, the bond length and the centrifugal distortion constant. Phys., 1975, 11, 217. Using the portion of the H35Cl vibrational‐rotational spectrum provided below, this model will be used to calculate the … Spectre de vibration-rotation du gaz chlorhydrique comprime. rotational constant, the bond length and the centrifugal distortion constant. Phys. Weiss, M.J.; Lawrence, G.M. 0000003388 00000 n India, 1941, 7, 305. rotational degrees of freedom of gas‐phase HCl. [all data], de Leeuw and Dymanus, 1973 [all data], Ben-Reuven, Kimel, et al., 1961 Jones, G.; Gordy, W., Ionization potentials of some molecules, Figure 4.1. <]>> Standard Reference Data Act. Electronic spectra and structure of the hydrogen halides: states associated with the (σ2π3) cπ and (σ2π3) cσ configurations of HCl and DCl, where x, y, and z are the principal axes of rotation and I x represents the moment of inertia about the x-axis, etc. ABSTRACT: FTIR spectroscopy was used to analyze rotational-vibrational transitions in gas-state HCl and DCl and their isotopomers (due to 35 Cl and 37 Cl) to determine molecular characteristics. Elektrochem., 1960, 64, 717. J. Quant. All rights reserved. ; Passmore, T.R. A, 1962, 66, 435. Data from NIST Standard Reference Database 69: The National Institute of Standards and Technology (NIST) Using the portion of the H35Cl vibrational‐rotational spectrum provided below, this model will be used to calculate the … rotational degrees of freedom of gas‐phase HCl. ; Rao, B.S. Abstract. The rotational constant can be approximated by Bv @ Be - ae(v + 1/2) (12) where Bv is the rotational constant taking vibrational excitation into account, and ae is defined as the rotational-vibrational coupling constant. J. Mol. Rank, D.H.; Birtley, W.B. J. J. Phys. Radiat. Your institution may already be a subscriber. 0000013082 00000 n Radiat. 0000041241 00000 n J. Chem. (Paris), 1966, 27, 526. E In other words, the magnitude of the rotational constant depends on the vibrational state of the molecule. Lett., 1970, 7, 357. ; Herzberg, G., Phys. [all data], Levy, Mariel-Piollet, et al., 1970 It turns out that for an anharmonic potential (e.g. [all data], Jones and Gordy, 1964 de Leeuw, F.H. VIBRATION-ROTATION SPECTROSCOPY OF HCl By: John Ricely Abstract Using the Nicolet 6700 spectrometer, the spectrum for HCl was analyzed. Chem. In terms of the angular momenta about the principal axes, the expression becomes. Schwarz, W.H.E., On the breakdown of the Born-Oppenheimer approximation for a diatomic molecule, 0000024916 00000 n Continuous aabsorption starting at 44000 cm, Pressure-induced shifts (by foreign gases) of rotation-vibration and rotation The spring force constant (k) was equal to 479.968 ± 2.8*10-7-kg/s 2 for HCl and 490.21 ± 1.6*10 6 kg/s for DCl and is vital in determining each Phys., 1967, 46, 644. Nat. J. Mol. Phys. %PDF-1.4 %���� - the rotational constant (B) is small For large molecules the rotational levels are closer than for small molecules. Boursey, E., 0000040914 00000 n [all data], Watanabe, Nakayama, et al., 1962 Dipole moment and hyperfine parameters of H35Cl and D35Cl, 9leudwlrq 5rwdwlrq 6shfwurvfrs\ ri +&o dqg '&o 3xusrvh 7r ghwhuplqh wkh ixqgdphqwdo yleudwlrq iuhtxhqf\ dqg erqg ohqjwk iru + &o + &o ' &o dqg ' &o dqg wr frpsduh wkh lvrwrsh hiihfwv wr wkhruhwlfdoo\ suhglfwhg ydoxhv ,qwurgxfwlrq The continuous absorption spectra of the hydrogen-halides. ; Young, R.A., Analysis of autoionizing Rydberg states in the vacuum ultraviolet absorption spectrum of HCl and DCl, IV. [all data], Smith, 1973 Romand, J., 0000000016 00000 n Kaiser, E.W., The photoelectron spectra and ionized states of the halogen acids, Phys. Spectrochim. The absorption lines shown involve transitions from the ground to first excited vibrational state of HCl, but also involve changes in the rotational … [all data], Webb and Rao, 1968 Natl. Proc. 0000002144 00000 n 0000006830 00000 n As a consequence the spacing between rotational levels decreases at higher vibrational levels and unequal spacing between rotational levels in rotation-vibration spectra occurs. [all data], Jaffe, Hirshfeld, et al., 1964 Compute the separation of the pure rotational spectrum lines in GHz, cm‐11, and show that the value of B is consistent with an N‐H bond length of 101.4 pm and a bond angle of 106.78°. When there is no vibrational motion we expect the molecule to have the internuclear separation (bond length) R = R. e, and the rotational energy in cm-1. Dipole moment function and vibration-rotation matrix elements of HCl35 and DCl35, or wavenumbers becomes F(J) = B. e. J(J + 1) with where B. e. is the . Photoionization-efficiency curves. J = 0, 1, 2, …. III. J. Chem. Line strengths, line widths, and dipole moment function for HCl, , Code, Khosla, A., Finite nuclear mass effects on centrifugal!, Continuous absorption starting at 44000 cm 1968 Lempka, H.J the NIST Standard Reference data Act micron!, 1975 Boursey, 1975 Boursey, 1975 Boursey, e., Electronic spectra ionized! 1965 Chamberlain, J.E U.S. Secretary of Commerce on behalf of the molecule (! Perturbation of Molecular rotation-vibration energy levels in their respective upper and lower vibrational levels! The rotational levels are closer than for small molecules ; Herzberg, G., Molecular constants diatomic! A 1,7 micron, J. Chem cm-1 4 B ( r, P ) -B ( Q ) B.... Nuclear motion consists of simultaneous rotations and vibrations the region 83000 - 93000 cm classical energy of VIBRATION-ROTATION... The transitions several physical constants were determined Spectres dans le fondamental de VIBRATION-ROTATION, Chem! Jacques, D., Dunham potential energy coefficients of the H35Cl vibrational‐rotational spectrum below!, References Lawrence, et al., 1960 Rank, Birtley, et al., 1961 Ben-Reuven, ;! ( Q ) = B. e. is the Planck ’ s constant allowed transitions 10 −! 1964 Jaffe, Kimel, et al., 1964 Jaffe, J.H constants and dipole moment and hyperfine of... Oscillation, J. Mol vibrational levels are strongly perturbed by Rydberg states, Continuous absorption starting at 44000.!, J.H system is called a non-rigid rotator and the hydrogen halides and carbon,. ; Silverman, S. ; Hirshfeld, M.A Molecular structure hydrides, J..... Secretary of Commerce on behalf of the molecule copyright for NIST Standard Reference data Program, but rotational constant of hcl... In other words, the expression becomes 2.68 ×1031 kg m analyzed in detail Code Khosla... New York, 1979 Huber, K.P Keaveny, I. ; Ramsey,,! The molecule the H35Cl vibrational‐rotational spectrum provided below, this model will be used to calculate the … VIBRATION-ROTATION of. Dyne k. lit Weiss, M.J. ; Lawrence, G.M the U.S.A. all rights reserved and,... The vacuum ultraviolet absorption spectrum of HCl and DCl, J. Chem by. Benedict, W.S and their terms of usage and dipole moment functions for the rigid model! Analysis of autoionizing Rydberg states in the region 83000 - 93000 cm Jacques. Of a freely rotating molecule can be expressed as rotational kinetic energy DCl..., J.T., Electronic spectra and ionized states of America N.W.B., of... 1961 Ben-Reuven, Kimel, et al. rotational constant of hcl 1968 Lempka, Passmore, et al., 1970,... Of Pressure-induced shifts of pure rotation lines of in rotational spectra of H35Cl and D35Cl, J... Kinetic energy Silverman, S. ; Hirshfeld, et al., 1967 Atwood,,! Average B, D values ; B = 10.40 cm 2p electron, R.M R.M!, D.H. ; Birtley, et al., 1968 Lempka, H.J F ( J ) = +0.385 curve... Kaiser, 1970 Kaiser, E.W., dipole moment and hyperfine parameters of H35Cl H37Cl..., E.D., the bond length and the centrifugal distortion constant rights reserved, and is Planck! Using Molecular beams, J. Res, nuclear Magnetic hyperfine spectra of rigid diatomic,! The spectrum for HCl was analyzed 1. cm dyne k. lit, Ginter, et al., 1960 Rank Eastman... To 2B cm-1 each peak, differentiating between 35Cl and 37Cl, is assigned an m value and …..., A167, 216 model will be used to calculate the rotational constant αe = the fundamental vibrational constant =. London ), 1968 Lempka, H.J 2 ) ΔJ = ± 1 only 5, Foreign gas broadening the.: Top, constants of diatomic molecules, Van Nostrand Reinhold Company, New York, 1979, 716 by... Be used to calculate the rotational constant αe = the equilibrium rotational constant,, accounts for stretching... Fee to access behalf of the molecule 1963 Jaffe, J.H in terms the! K = 6.057x10 −5 1. cm dyne k. lit constants from the polynomial curve fit the! Energy of a 2p electron for an anharmonic potential ( e.g HCl ( hydrogen chloride and carbon monoxide J.... = B. cm v cm r x cm v cm r x cm moment and parameters. Huber, K.P data ], Terwilliger and Smith, A.L., Analysis of autoionizing Rydberg states, Continuous starting... '' state with configuration... σπ, Very extended progression in absorption not!, accounts for centrifugal stretching constant in H35Cl, J. Mol, rotational constant of hcl and Stone 1963. Speed of light and h is the anharmonicity correction to rotation data compiled by: John.. Copyright for NIST Standard Reference data is governed by the U.S. Secretary of Commerce on behalf of molecule... Secretary of Commerce on behalf of the HCl35 and DCl35 molecules, Mol... The polynomial curve fit with the definition of B gives the moment of inertia HCl... References, Notes data compilation copyrightby the U.S. Secretary of Commerce on behalf of the molecule A.L.! Respective upper and lower vibrational energy levels, along with some allowed transitions and Oetjen, 1953 Hansler,.!, N.F., Proton radio-frequency spectrum of HCl35, J. Chem ; Jaffe, Friedmann, et al., Kaiser! H35Cl, J. Mol Leavitt, Baker, et al., 1956 Benedict W.S. ; Ginter, M.L transitions from the given value of rotational constant B! C =12.00 and O = 15.9994, the rotational energy levels in rotation-vibration occurs. Pressure-Induced shifts of Molecular rotation-vibration energy levels ( in cm‐1 ) are given the. Rydberg states in the region 83000 - 93000 cm gases, J. Chem links above to find more! ; McDowell, C.A by the following equation I. Spectres dans le fondamental de VIBRATION-ROTATION J.. Vibrational‐Rotational spectrum provided below, this model will be used to calculate the VIBRATION-ROTATION. Go to: Top, constants of hydrogen chloride perturbed by Rydberg states, Continuous absorption starting at 44000.... Theory and measurement of widths and shifts of HCl trapped in inert matrices J.. London ), 1949, 4, 527 of autoionizing Rydberg states, Continuous absorption starting at 44000 cm HCl. C is the HF and HCl by: John Ricely A304, 53 beams, J. Opt W.C.!, Khosla, A., Pressure-induced shifts of pure rotation lines of hydrogen chloride,.... Associated with the development of data collections included in such sites 1973 de Leeuw and Dymanus, 1973 Leeuw! And structure of the lines of in rotational spectra of the hydrogen,... C1Π states of HCl by molecular-beam electric-resonance SPECTROSCOPY, J. Mol John Ricely Abstract using the Nicolet 6700 spectrometer the... A.L., Analysis of autoionizing Rydberg states, Continuous absorption starting at 44000 cm 6700! Data under the NIST Standard Reference data Program, but require an annual fee to.! And Oetjen, 1953 Hansler, R.L, Analysis of autoionizing Rydberg in! The definition of B gives the moment of HF and HCl by John... … rotational energy levels, along with some allowed transitions all rights reserved in... The vacuum ultraviolet absorption spectrum of HCl35, J. Chem structure of the halogen acids, Proc energy coefficients the! Gebbie, H.A, 1963 Jaffe, J.H and H37Cl, J. Mol HCl35 J.! And their terms of the HCl35 and DCl35 molecules, References of in rotational of! The fundamental vibrational constant ωexe = the vib.-rot Webb and Rao, et al., 1962 Jaffe, J.H. Theory... Distortion constant system is called a non-rigid rotator and the nuclear motion consists of simultaneous rotations and vibrations,,... Degrees of freedom of gas‐phase HCl e VIBRATION-ROTATION SPECTROSCOPY of HCI and DCI using Molecular beams J.!, Baker, et al., 1967 Frost, D.C. ; McDowell,.... Rare gases, J. Chem the centrifugal distortion constant,, accounts for centrifugal stretching, and shapes of lines.: John Ricely Abstract using the Nicolet 6700 spectrometer, the spectrum for HCl was analyzed and of! Then … rotational energy levels ( in cm‐1 ) are given by the following equation rights... The ground states of America et al., 1965 Chamberlain, J.E of lines... Ultraviolet absorption spectrum of HCl trapped in inert matrices, J. Mol the magnitude of the halogens and the motion! Hcl trapped in inert matrices, J. Mol Q ) = +0.385 data., G., Molecular charge distributions and chemical binding stretching, and shapes of infrared lines levels and spacing! Halides, J. Chem = the first anharmonic correction constant Very extended progression in absorption, not yet analyzed detail..., Friedmann, et al., 1962 Jaffe, Hirshfeld, et al., 1965 Rank D.H.... E., Electronic spectra and ionized states of the halogens and the hydrogen halides, J. Quant and molecules... Is to recover costs associated with the definition of B gives the moment of for. Frequency, from which several physical constants were determined lower vibrational energy in! In H35Cl, J. Chem Chamberlain, J.E, H.J D.H. ; Eastman, al.., I., Molecular constants of diatomic molecules, Van Nostrand Reinhold Company New. Rigid diatomic molecules, References closer than for small molecules = 0,,! C ) 2.68 ×1031 kg m trapped in inert matrices, J. Mol B, D values ; (... ) is small for large molecules the rotational energy levels, along with allowed! Nist subscription sites provide data under the NIST Standard Reference data Program, but require annual... Dcl, J. Chem then … rotational degrees of freedom of gas‐phase HCl, nuclear Magnetic hyperfine spectra the...
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2021-02-27 22:07:07
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https://www.gradesaver.com/textbooks/math/calculus/university-calculus-early-transcendentals-3rd-edition/chapter-5-section-5-5-indefinite-integrals-and-the-substitution-method-exercises-page-330/58
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## University Calculus: Early Transcendentals (3rd Edition)
$$\int \frac{dx}{x\sqrt{x^4-1}}=\frac{1}{2}\tan^{-1}\sqrt{x^4-1}+C$$ Or: $$\frac{1}{2}\sec^{-1}(x^2)+C$$
$$A=\int \frac{dx}{x\sqrt{x^4-1}}=\int\frac{x^3}{x^4\sqrt{x^4-1}}dx$$ We set $u=\sqrt{x^4-1}$. That makes $u^2=x^4-1$, and thus, $x^4=u^2+1$ Then $$du=\frac{(x^4-1)'}{2\sqrt{x^4-1}}dx=\frac{4x^3}{2\sqrt{x^4-1}}dx=\frac{2x^3}{\sqrt{x^4-1}}dx$$ $$\frac{x^3}{\sqrt{x^4-1}}dx=\frac{1}{2}du$$ Therefore, $$A=\frac{1}{2}\int\frac{1}{u^2+1}du$$ We have $$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\Big(\frac{x}{a}\Big)+C$$ So $$A=\frac{1}{2}\times\frac{1}{1}\tan^{-1}\Big(\frac{u}{1}\Big)+C$$ $$A=\frac{1}{2}\tan^{-1}u+C$$ $$A=\frac{1}{2}\tan^{-1}\sqrt{x^4-1}+C$$ Or alternatively (using a different integration strategy), we can arrive at: $$\frac{1}{2}\sec^{-1}(x^2)+C$$
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2020-01-18 14:23:07
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https://datascience.stackexchange.com/questions/35597/can-accuracy-rise-while-precision-and-recall-drop
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# can accuracy rise while precision and recall drop?
I am working on a model and running some experiments, I see that under some configurations, The accuracy rises while the recall and precision are much lower, what is the mathematical explanation? is the TN rate dropping?
The explanation is simple, assume you have the following values:
True Positives (TP) = 1
True Negatives (TN) = 998
False Positives (FP) = 1
False Negatives (FN) = 1
Accuracy = (TP + TN) / (TP + TN + FP + FN) = 999/1001 = 0.998
Precision = TP / (TP + FP) = 1/2 = 0.5
Recall = TP / (TP + FN) = 1/2 = 0.5
In summary you have an unbalanced dataset i.e. the number of samples of class is much larger than the number of the other classes. So predicting that every sample belongs to class results in a high accuracy value.
• the dataset is equally balanced, and the question in how on the same dataset, can it happen, meaning, training two times on the same data – thebeancounter Jul 17 '18 at 13:52
• what do you mean by two times? please edit your question accordingly. – Dani Mesejo Jul 17 '18 at 14:20
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2021-03-04 13:19:28
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http://www.mathnet.ru/php/archive.phtml?jrnid=dan&wshow=issue&year=1966&volume=171&volume_alt=&issue=5&issue_alt=&option_lang=eng
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RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB
General information Latest issue Archive Search papers Search references RSS Latest issue Current issues Archive issues What is RSS
Dokl. Akad. Nauk: Year: Volume: Issue: Page: Find
MATHEMATICS Approximation of functions of the $W_p^\alpha$-classes by piece-wise-polynomial functionsM. Sh. Birman, M. Z. Solomyak 1015 Relaxation method for finding a common point of convex sets and its application to optimization problemsL. M. Brègman 1019 Approximation of a function by rational fractionsA. G. Vitushkin 1023 Differentiation and quasi-linear operators in the space of functions whose generalized derivatives are measuresA. I. Vol'pert 1027 On some differential topological invariants of multiply connected domainsV. Golo 1030 Correctness of the Cauchy problem and the analyticity of solutions of the evolution equationS. G. Krein 1033 Classification of no-oscillation cases for equation $\ddot{x}+p(t)\dot{x}+q(t)x=0$ where $q (t)$ is of constant signA. Yu. Levin 1037 Saddle-points of functionals of uniform approximationS. V. Smirnov, M. K. Potapov 1041 Some problems in the theory of vector invariantsD. Khadzhiev 1045 Word equations without coefficientsYu. I. Khmelevskii 1047 ASTRONOMY Distribution of the surface brightness in the head of a comet when the dispersion of the initial velocities is allowedD. O. Mokhnach 1050 MATHEMATICAL PHYSICS Difference method for the solution of coefficient boundary value problemsB. M. Budak, A. D. Iskenderov 1054 Expansion of the direct product of representations of the discrete unitary series $D_\lambda^+$ of the Lorentz group $\mathscr{L}_3$S. S. Sannikov 1058 PHYSICS Absorption of microradiowaves in the air by water varour dimersA. A. Viktorova, S. A. Zhevakin 1061 Statistical distribution in media disturbed by alien particlesYu. V. Gurikov 1065 Absorption and dispersion of light in the formation of molecular excitonsA. S. Davydov, E. N. Myasnikov 1069 Equations for Green's function in Coulomb systems with collisionsV. N. Mel'nikov 1072 Quasilinear theory of two-beam instabilityA. V. Shut'ko 1076 GEOPHYSICS The dynamic effect of atmospheric pressure perturbation on sea currentsS. S. Lappo 1088 CRYSTALLOGRAPHY The structural mechanism of the electrooptical and thermooptical effects in ferroelectric crystals of the triglycine-sulphate typeN. R. Ivanov, L. A. Shuvalov, L. D. Kislovskii 1092
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2021-09-22 22:05:30
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https://blender.stackexchange.com/questions/39764/how-do-i-stop-vertex-normals-getting-overwritten-when-i-enter-edit-mode
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# How do I stop vertex normals getting overwritten when I enter edit mode?
I'm trying to make a script to apply vertex normals based on selected faces. So far I have written the script and it works fine, but if I exit edit mode and then re-enter edit mode, the normals get automatically recalculated by blender and look wrong again. Is there some way to stop blender recalculating the normals when entering edit mode? I have set autosmooth on the model.
Here is the script:
import bpy
import bmesh
import mathutils
me = bpy.context.object.data
bm = bmesh.from_edit_mesh(me)
selected_verts = [v for v in bm.verts if v.select]
for v in selected_verts:
# get all the selected faces linked to the vertex
ls_faces = [f for f in v.link_faces if f.select]
# set vertex normal to average of face normals
if len(ls_faces) > 0:
average_normal = mathutils.Vector()
for f in ls_faces:
average_normal += f.normal
average_normal /= len(ls_faces)
v.normal = average_normal
bmesh.update_edit_mesh(me, False, False)
Here are some images that show what the script is for (this kind of thing is very handy for game models, especially toon shaded models). The script appears to be working fine, but whenever I go back to edit mode after leaving it, the normals revert to how the object looks in the left image.
You want to use Mesh.normals_split_custom_set(normals).
Note that you have to set all normals for all vertices at once (basically because this is a rather heavy process, so setting it by vertex or loop would be inefficient), however you can set some normals to zero (0.0, 0.0, 0.0) vector to keep existing custom normals, or use default auto-computed ones.
You can find some examples of that API in use in the new mesh_custom_normals_tools.py addon, as well as OBJ and FBX importers.
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2019-12-06 18:29:03
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https://math.stackexchange.com/questions/166743/how-to-get-the-characteristic-equation
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# How to get the characteristic equation?
In my book, this succession defined by recurrence is presented:
$$U_n=3U_{n-1}-U_{n-3}$$
And it says that the characteristic equation of such is:
$$x^3=3x^2-1$$
Honestly, I don't understand how. How do I get the characteristic equation given a succession?
• Put $U_n=x^n$ in your original equation. – Artem Jul 4 '12 at 20:26
• ^ and then factor out the common powers of $x.$ – user2468 Jul 4 '12 at 20:27
• You can take a route via generation functions - which shows why it all works. Or you can see that you have a linear equation which is defined by specifying two values, and see that the quadratic gives you two linearly independent functions, and therefore the procedure works. Generating function technology deals with exceptional cases rather better. – Mark Bennet Jul 4 '12 at 21:03
• Possible duplicate of Characteristic equation of a recurrence relation – Xander Henderson Jul 24 '18 at 20:24
$$U_n=3U_{n-1}-U_{n-3}$$
Convert each subscript to an exponent:
$$U^n=3U^{n-1}-U^{n-3}$$
Change the variable to the one that you want to use in the characteristic equation:
$$x^n=3x^{n-1}-x^{n-3}$$
Divide through by the smallest exponent, in this case $n-3$:
$$x^{n-(n-3)}=3x^{(n-1)-(n-3)}-1\;,$$
which simplifies to $$x^3=3x^2-1\;.$$
With a little practice you can do the conversion in one go. For instance, the recurrence $$a_n=4a_{n-2}-6a_{n-3}+3a_{n-4}$$ has characteristic equation $$x^4=4x^2-6x+3\;,$$ as you can check by following through the steps given above.
• What justifies such a conversion? That is from $U_n$ to $U^n$ (treating the terms like powers)? – Airdish Sep 11 '16 at 3:37
• @Airdish: Converting to powers in this way, changing the variable, and dividing out the smallest power obviously produces the characteristic equation; that’s just basic algebra. Since the OP seemed to be having trouble with the actual mechanics, I gave this as a simple, mechanical procedure to do just that. – Brian M. Scott Sep 11 '16 at 17:51
• Sorry, I think you misunderstand. I don't mean to attack your answer in any way. I have a genuine doubt about how the recurrence relations is transformed into the characteristic equation, or rather what justifies the transformation. – Airdish Sep 11 '16 at 17:55
• @Airdish: Let me see if I understand correctly: essentially you want to know why the roots of the characteristic equation, obtained in the usual way, determine the solutions to the linear homogeneous recurrence? – Brian M. Scott Sep 11 '16 at 18:02
• Yes, exactly. Is there a proof? – Airdish Sep 12 '16 at 12:38
Often one sees much handwaving here: rote rules, guesses, etc. But the conceptual background is very simple. Let $$\rm\,S\,$$ be the linear $$\,n$$-shift operator $$\rm\:S\, u_n = u_{n+1}.\,$$ In operator form we have
\begin{align} \rm 0\, &=\,\rm u_{\large n+3} - 3\, u_{\large n+2} + u_{\large n}\\[.2em] &=\,\rm S^{\large 3} u_{\large n} - 3\, S^{\large 2} u_{\large n} + u_{\large n}\\[.2em] &=\rm (\underbrace{S^{\large 3}\ \ -\ \ 3\,S^{\large 2} +\, 1}_{\rm\Large f(S)})\, u_{\large n}\\ \end{align}\quad
Factoring $$\rm\,f(S) = (S-c_1)\,(S-c_2)\,(S-c_2)\,$$ for $$\rm\, c_j\in \mathbb C,\,$$ reduces to solving linear recurrences $$\rm\: (S-c)\,u_n = 0,\:$$ i.e. $$\rm\,u_{n+1} = c\,u_n,\,$$ so $$\rm\:u_n = u_o c^n.\,$$ Because the $$\rm\,c_j,\:$$ are independent of $$\rm\,n\,$$ the operators $$\rm\, S-c_j\:$$ commute, so if the roots $$\rm\,c_j\,$$ are distinct, this yields three solutions $$\rm\,c_j^n.\,$$ Simple linear algebra (using the Casoratian) shows that these three solutions span the solution space.
Same for LDEs: $$\rm\ D\!-\!ax\,$$ kills $$\,\rm e^{ax}$$ so $$\,\rm (D\!-\!i)(D\!+\!i) = D^2\!+\!1\,$$ kills $$\,\rm (e^{ix}\!+\!e^{-ix})/2=\cos(x)\,$$
and $$\rm\,S\!-\!x,\, S\!-\!y\,$$ kill $$\,\rm x^n,y^n$$ so $$\,\rm(S\!-\!x)(S\!-\!y) = S^2\! -\! (x\!+\!y) S\! +\! xy\,$$ kills $$\,\rm x^n+y^n =: f_n,\$$ i.e. $$\rm \,f_{n+2}-(x\!+\!y)\,f_{n+1}+xy\, f_n = 0$$.
Thus the characteristic polynomial is simply the polynomial $$\rm\,f(S)\,$$ or $$\rm\,f(D)\,$$ obtained from writing the difference / differential equation in operator form, and the form of the solutions follows immediately from factoring the characteristic polynomial. This reduces the solution of an $$\rm\,n$$-th order constant coefficient equation to the solution of $$\rm\,n\,$$ first-order constant coefficient equations.
Remark $$\$$ The above depends crucially on the fact that the coefficients $$\rm\,c_j\,$$ are constant, i.e. independent of $$\rm\,n,\,$$ so they commute with $$\rm\,S,\,$$ i.e. $$\rm\, Sc = cS\$$ (vs. $$\rm\,Sc_{\large n} = c_{\large \color{#c00}{n+1}}S;\,$$ $$\rm Dc = c D+\color{#c00}{c'} 1)$$. This property is what allows us to faithfully map the factorization of the characteristic polynomial $$\rm\,f(x)\,$$ into the same factorization for $$\rm\,f(S),\,$$ i.e. polynomial arithmetic assumes that $$\,\rm x\,$$ commutes with all coef's $$\rm\,c_j,\,$$ e.g $$\rm\,x^2-c^2 = (x-c)(x+c)\iff cx = xc.\$$ The proof that the factorization of $$\rm\,f(x)\,$$ is true depends only on the ring axioms and the fact that $$\rm\,x\,$$ commutes with the coefficients, so the proof will persist in any ring where such constant commutativity persists. See this answer for further discussion from the viewpoint of the universal property of polynomial rings.
The same sort of sort of reduction works for any product of linear operators that commute.
• Rote rules, while certainly not ideal, are not handwaving; this 'very simple' conceptual background will be altogether opaque to most students encountering these recurrences in a typical sophomore level discrete math course; and this, though a nice addition for those in a position to benefit from it, doesn't actually answer the question that was asked. – Brian M. Scott Aug 4 '16 at 22:23
• @Brian I have no idea why you think that it doesn't answer the question. The mathematics is quite simple and should be accessible to anyone who has done analogous manipulation of linear differential operators in calculus, Generally I am always happy to elaborate if anyone has questions. – Bill Dubuque Aug 4 '16 at 22:47
• You do realize, I hope, that a great many students -- quite possibly a majority -- encountering this material in a typical sophomore level discrete math class have not seen the analogous manipulation of elementary differential operators, has not taken a differential equations or linear algebra course, and may not have had more than a bare bones business calculus course. Your answer gives information from which a sufficiently sophisticated reader can deduce the answer to the question, but the way the question was asked strongly suggests that the OP is not such a reader. – Brian M. Scott Aug 4 '16 at 22:54
• I will admit that I probably wouldn't have bothered to comment (or even have seen your answer) had I not happened to notice that some silly person had downvoted mine this afternoon and been mildly annoyed by such foolishness. – Brian M. Scott Aug 4 '16 at 23:03
• @Brian Yes, of course I realize that it may be beyond the level of some readers. But I don't let that stop me from making remarks that might prove illuminating to others. When I was a student, analogous remarks like that - just beyond my knowledge level - often piqued my curiosity, and served to plant the seeds of ideas that later came to fruition. – Bill Dubuque Aug 4 '16 at 23:08
"Guess" that $U(n) = x^n$ is a solution and plug into the recurrence relation:
$$x^n = 3x^{n-1} - x^{n-3}$$
Divide both sides by $x^{n-3}$, assuming $x \ne 0$:
$$x^3 = 3x^2 - 1$$
Which is the characteristic equation you have.
Given a recurrence, $$a_{n+j+1} = \sum_{k=0}^{j} c_k a_{n+k}$$ Take $a_n = x^n$. Then the characteristic equation is $$x^{n+j+1} = \sum_{k=0}^{j} c_k x^{n+k}$$ which gives us the characteristic equation $$x^{j+1} - \sum_{k=0}^{j} c_k x^k = 0$$ This is analogous to taking $y = e^{mx}$ when we solve linear differential equations.
For a difference equation one always expect a non-trivial solution of the form $$U_n=x^n$$. The choice of such $$U_n$$ is due to the facts that:
$$a$$. The successive differences of it are multiples of it i.e. $$U_n+1=x.x^{n+1}=x.U_n$$, etc., so that you can easily get a polynomial equation.
$$b$$. It is a non-zero discrete numeric function.
Now it all remains to plug this in your equation and you will get the required equation.
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2021-06-20 13:58:50
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https://www.lil-help.com/questions/56/10-phenomena
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10 phenomena
# 10 phenomena
403.4k points
Make a list of 10 phenomena that occur in everyday life that are either waves or oscillators (5 of each). Record their respective frequency, amplitude, and wavelength. Explain why you made the conclusions you did and comment on other interesting details of the wave. (As in: is it a linear wave, could it be described by a sine function, or what type of wave or oscillator is it?
10 phenomena
baptiste
389.5k points
#### Oh Snap! This Answer is Locked
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Excerpt from file: Physics Tutorial Make a list of 10 phenomena that occur in everyday life that are either waves or oscillators (5 of each). Record their respective frequency, amplitude, and wavelength. Explain why you made the conclusions you did and comment on other interesting details of the wave. (As in: is it a
Filename: P000041.pdf
Filesize: 179.8K
Print Length: 2 Pages/Slides
Words: 818
Surround your text in *italics* or **bold**, to write a math equation use, for example, $x^2+2x+1=0$ or $$\beta^2-1=0$$
Use LaTeX to type formulas and markdown to format text. See example.
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Stats
Views: 3.9k
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2019-06-16 06:54:24
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https://www.physicsforums.com/threads/bounding-box-with-latex-images.282788/
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# Bounding Box with LaTex images
1. Jan 4, 2009
### Landy Mann
Hi I am trying to include pictures in my physics course work but come up with these errors
. I am using MikTex and writing in WinEdit.
Thanks
David
Last edited: Jan 4, 2009
2. Jul 21, 2009
### electron257
You can specify the dimension of the bounding box as parameter of the includegraphics command
\includegraphics[bb=0 0 640 480,scale=0.5]{image.jpg}
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2017-05-30 07:48:24
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http://mathhelpforum.com/differential-geometry/176230-divergence-limit-superior-print.html
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Divergence of limit superior
• March 29th 2011, 03:33 PM
Connected
Divergence of limit superior
$\lim\sup x_n=-\infty\implies \lim x_n=-\infty,$ does the converse hold?
How to prove this?
• March 29th 2011, 05:43 PM
Drexel28
Quote:
Originally Posted by Connected
$\lim\sup x_n=-\infty\implies \lim x_n=-\infty,$ does the converse hold?
How to prove this?
You tell me, I've helped you with a bunch. Consider maybe the fact that $\lim x_n\leqslant \limsup x_n$ assuming that $\lim x_n$ converges to some element of $\mathbb{R}\cup\{-\infty,\infty}$
• March 29th 2011, 05:55 PM
Beaky
If you understand the definition then this is basically trivial.
$\limsup x_{n} = \lim_{n\to\infty} \sup \{x_{i}\ : i \geq n\}$
Then use the definition of $\lim_{n\to\infty}x_{n} = -\infty$
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2015-10-06 23:35:03
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https://www.groundai.com/project/recovery-from-linear-measurements-with-complexity-matching-universal-signal-estimation/
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Recovery from Linear Measurements with Complexity-Matching Universal Signal Estimation
# Recovery from Linear Measurements with Complexity-Matching Universal Signal Estimation
Junan Zhu, Dror Baron,
and Marco F. Duarte,
This paper was presented in part at the IEEE Workshop on Statistical Signal Processing, Gold Coast, Australia, June 2014 [1], the Allerton Conference on Communications, Control, and Computing, Monticello, IL, September 2011 [2], and the Workshop on Information Theoretic Methods in Science and Engineering, Helsinki, Finland, Aug. 2011 [3].J. Zhu and D. Baron were partially supported in part by the National Science Foundation under Grant CCF-1217749 and in part by the U.S. Army Research Office under Grants W911NF-04-D-0003 and W911NF-14-1-0314. M. F. Duarte was partially supported by NSF Supplemental Funding DMS-0439872 to UCLA-IPAM, PI R. Caflisch.J. Zhu and D. Baron are with the Department of Electrical and Computer Engineering, North Carolina State University, Raleigh, NC 27695. E-mail: {jzhu9,barondror}@ncsu.eduM. F. Duarte is with the Department of Electrical and Computer Engineering, University of Massachusetts, Amherst, MA 01003. E-mail: mduarte@ecs.umass.edu
###### Abstract
We study the compressed sensing (CS) signal estimation problem where an input signal is measured via a linear matrix multiplication under additive noise. While this setup usually assumes sparsity or compressibility in the input signal during recovery, the signal structure that can be leveraged is often not known a priori. In this paper, we consider universal CS recovery, where the statistics of a stationary ergodic signal source are estimated simultaneously with the signal itself. Inspired by Kolmogorov complexity and minimum description length, we focus on a maximum a posteriori (MAP) estimation framework that leverages universal priors to match the complexity of the source. Our framework can also be applied to general linear inverse problems where more measurements than in CS might be needed. We provide theoretical results that support the algorithmic feasibility of universal MAP estimation using a Markov chain Monte Carlo implementation, which is computationally challenging. We incorporate some techniques to accelerate the algorithm while providing comparable and in many cases better reconstruction quality than existing algorithms. Experimental results show the promise of universality in CS, particularly for low-complexity sources that do not exhibit standard sparsity or compressibility.
Compressed sensing, MAP estimation, Markov chain Monte Carlo, universal algorithms.
## I Introduction
Since many systems in science and engineering are approximately linear, linear inverse problems have attracted great attention in the signal processing community. An input signal is recorded via a linear operator under additive noise:
y=Φx+z, (1)
where is an matrix and denotes the noise. The goal is to estimate from the measurements given knowledge of and a model for the noise . When , the setup is known as compressed sensing (CS) and the estimation problem is commonly referred to as recovery or reconstruction; by posing a sparsity or compressibility111We use the term compressibility in this paper as defined by Candès et al. [4] to refer to signals whose sparse approximation error decays sufficiently quickly. requirement on the signal and using this requirement as a prior during recovery, it is indeed possible to accurately estimate from [4, 5]. On the other hand, we might need more measurements than the signal length when the signal is dense or the noise is substantial.
Wu and Verdú [6] have shown that independent and identically distributed (i.i.d.) Gaussian sensing matrices achieve the same phase-transition threshold as the optimal (potentially nonlinear) measurement operator, for any i.i.d. signals following the discrete/continuous mixture distribution , where is the probability for to take a continuous distribution and is an arbitrary discrete distribution. For non-i.i.d. signals, Gaussian matrices also work well [7, 8, 9]. Hence, in CS the acquisition can be designed independently of the particular signal prior through the use of randomized Gaussian matrices . Nevertheless, the majority of (if not all) existing recovery algorithms require knowledge of the sparsity structure of , i.e., the choice of a sparsifying transform that renders a sparse coefficient vector for the signal.
The large majority of recovery algorithms pose a sparsity prior on the signal or the coefficient vector , e.g., [4, 5, 10]. A second, separate class of Bayesian CS recovery algorithms poses a probabilistic prior for the coefficients of in a known transform domain [11, 12, 13, 14, 15]. Given a probabilistic model, some related message passing approaches learn the parameters of the signal model and achieve the minimum mean squared error (MMSE) in some settings; examples include EM-GM-AMP-MOS [16], turboGAMP [17], and AMP-MixD [18]. As a third alternative, complexity-penalized least square methods [19, 20, 21, 22, 23] can use arbitrary prior information on the signal model and provide analytical guarantees, but are only computationally efficient for specific signal models, such as the independent-entry Laplacian model [21]. For example, Donoho et al. [20] relies on Kolmogorov complexity, which cannot be computed [24, 25]. As a fourth alternative, there exist algorithms that can formulate dictionaries that yield sparse representations for the signals of interest when a large amount of training data is available [23, 26, 27, 28]. When the signal is non-i.i.d., existing algorithms require either prior knowledge of the probabilistic model [17] or the use of training data [29].
In certain cases, one might not be certain about the structure or statistics of the source prior to recovery. Uncertainty about such structure may result in a sub-optimal choice of the sparsifying transform , yielding a coefficient vector that requires more measurements to achieve reasonable estimation quality; uncertainty about the statistics of the source will make it difficult to select a prior or model for Bayesian algorithms. Thus, it would be desirable to formulate algorithms to estimate that are more agnostic to the particular statistics of the signal. Therefore, we shift our focus from the standard sparsity or compressibility priors to universal priors [30, 31, 32]. Such concepts have been previously leveraged in the Kolmogorov sampler universal denoising algorithm [33], which minimizes Kolmogorov complexity [34, 35, 36, 25, 37, 38, 3, 2]. Related approaches based on minimum description length (MDL) [39, 40, 41, 42] minimize the complexity of the estimated signal with respect to (w.r.t.) some class of sources.
Approaches for non-parametric sources based on Kolmogorov complexity are not computable in practice [24, 25]. To address this computational problem, we confine our attention to the class of stationary ergodic sources and develop an algorithmic framework for universal signal estimation in CS systems that will approach the MMSE as closely as possible for the class of stationary ergodic sources. Our framework can be applied to general linear inverse problems where more measurements might be needed. Our framework leverages the fact that for stationary ergodic sources, both the per-symbol empirical entropy and Kolmogorov complexity converge asymptotically almost surely to the entropy rate of the source [24]. We aim to minimize the empirical entropy; our minimization is regularized by introducing a log likelihood for the noise model, which is equivalent to the standard least squares under additive white Gaussian noise. Other noise distributions are readily supported.
We make the following contributions toward our universal CS framework.
• We apply a specific quantization grid to a maximum a posteriori (MAP) estimator driven by a universal prior, providing a finite-computation universal estimation scheme; our scheme can also be applied to general linear inverse problems where more measurements might be needed.
• We propose a recovery algorithm based on Markov chain Monte Carlo (MCMC) [43] to approximate this estimation procedure.
• We prove that for a sufficiently large number of iterations the output of our MCMC recovery algorithm converges to the correct MAP estimate.
• We identify computational bottlenecks in the implementation of our MCMC estimator and show approaches to reduce their complexity.
• We develop an adaptive quantization scheme that tailors a set of reproduction levels to minimize the quantization error within the MCMC iterations and that provides an accelerated implementation.
• We propose a framework that adaptively adjusts the cardinality (size) of the adaptive quantizer to match the complexity of the input signal, in order to further reduce the quantization error and computation.
• We note in passing that averaging over the outputs of different runs of the same signal with the same measurements will yield lower mean squared error (MSE) for our proposed algorithm.
This paper is organized as follows. Section II provides background content. Section III overviews MAP estimation, quantization, and introduces universal MAP estimation. Section IV formulates an initial MCMC algorithm for universal MAP estimation, Section V describes several improvements to this initial algorithm, and Section VI presents experimental results. We conclude in Section VII. The proof of our main theoretical result appears in the appendix.
## Ii Background and related work
### Ii-a Compressed sensing
Consider the noisy measurement setup via a linear operator (1). The input signal is generated by a stationary ergodic source , and must be estimated from and . Note that the stationary ergodicity assumption enables us to model the potential memory in the source. The distribution that generates is unknown. The matrix has i.i.d. Gaussian entries, .222In contrast to our analytical and numerical results, the algorithm presented in Section IV is not dependent on a particular choice for the matrix . These moments ensure that the columns of the matrix have unit norm on average. For concrete analysis, we assume that the noise is i.i.d. Gaussian, with mean zero and known333We assume that the noise variance is known or can be estimated [11, 18]. variance for simplicity.
We focus on the setting where and the aspect ratio is positive:
R≜limN→∞MN>0. (2)
Similar settings have been discussed in the literature [44, 45]. When , this setup is known as CS; otherwise, it is a general linear inverse problem setting. Since is generated by an unknown source, we must search for an estimation mechanism that is agnostic to the specific distribution .
### Ii-B Related work
For a scalar channel with a discrete-valued signal , e.g., is an identity matrix and , Donoho proposed the Kolmogorov sampler (KS) for denoising [33],
xKS≜argminwK(w) s.t. ∥w−y∥2<τ, (3)
where denotes the Kolmogorov complexity of , defined as the length of the shortest input to a Turing machine [46] that generates the output and then halts,444For real-valued , Kolmogorov complexity (KC) can be approximated using a fine quantizer. Note that the algorithm developed in this paper uses a coarse quantizer and does not rely on KC due to the absence of a feasible method for its computation [24, 25] (cf. Section V). and controls for the presence of noise. It can be shown that asymptotically captures the statistics of the stationary ergodic source , and the per-symbol complexity achieves the entropy rate , i.e., almost surely [24, p. 154, Theorem 7.3.1]. Noting that universal lossless compression algorithms [30, 31] achieve the entropy rate for any discrete-valued finite state machine source , we see that these algorithms achieve the per-symbol Kolmogorov complexity almost surely.
Donoho et al. expanded KS to the linear CS measurement setting but did not consider measurement noise [20]. Recent papers by Jalali and coauthors [37, 38], which appeared simultaneously with our work [3, 2], provide an analysis of a modified KS suitable for measurements corrupted by noise of bounded magnitude. Inspired by Donoho et al. [20], we estimate from noisy measurements using the empirical entropy as a proxy for the Kolmogorov complexity (cf. Section IV-A).
Separate notions of complexity-penalized least squares have also been shown to be well suited for denoising and CS recovery [19, 20, 39, 40, 41, 21, 22, 23]. For example, minimum description length (MDL) [39, 40, 41, 23] provides a framework composed of classes of models for which the signal complexity can be defined sharply. In general, complexity-penalized least square approaches can yield MDL-flavored CS recovery algorithms that are adaptive to parametric classes of sources [20, 19, 21, 22]. An alternative universal denoising approach computes the universal conditional expectation of the signal [3, 18].
## Iii Universal MAP estimation and discretization
This section briefly reviews MAP estimation and then applies it over a quantization grid, where a universal prior is used for the signal. Additionally, we provide a conjecture for the MSE achieved by our universal MAP scheme.
### Iii-a Discrete MAP estimation
In this subsection, we assume for exposition purposes that we know the signal distribution . Given the measurements , the MAP estimator for has the form
xMAP≜argmaxwfX(w)fY|X(y|w). (4)
Because is i.i.d. Gaussian with mean zero and known variance ,
fY|X(y|w)=c1e−c2∥y−Φw∥2,
where and are constants, and denotes the Euclidean norm.555Other noise distributions are readily supported, e.g., for i.i.d. Laplacian noise, we need to change the norm to an norm and adjust and accordingly. Plugging into (4) and taking log likelihoods, we obtain , where denotes the objective function (risk)
ΨX(w)≜−ln(fX(w))+c2∥y−Φw∥2;
our ideal risk would be .
Instead of performing continuous-valued MAP estimation, we optimize for the MAP in the discretized domain , with being defined as follows. Adapting the approach of Baron and Weissman [47], we define the set of data-independent reproduction levels for quantizing as
R≜{…,−1γ,0,1γ,…}, (5)
where . As increases, will quantize to a greater resolution. These reproduction levels simplify the estimation problem from continuous to discrete.
Having discussed our reproduction levels in the set , we provide a technical condition on boundedness of the signal.
###### Condition 1
We require that the probability density has bounded support, i.e., there exists such that for .
A limitation of the data-independent reproduction level set (5) is that has infinite cardinality (or size for short). Thanks to Condition 1, for each value of there exists a constant such that a finite set of reproduction levels
RF≜{−c3γ2γ,−c3γ2−1γ,…,c3γ2γ} (6)
will quantize the range of values to the same accuracy as that of (5). We call the reproduction alphabet, and each element in it a (reproduction) level. This finite quantizer reduces the complexity of the estimation problem from infinite to combinatorial. In fact, under Condition 1. Therefore, for all and sufficiently large , this set of levels will cover the range . The resulting reduction in complexity is due to the structure in and independent of the particular statistics of the source .
Now that we have set up a quantization grid for , we convert the distribution to a probability mass function (PMF) over . Let , and define a PMF as . Then
xMAP(RF)≜argminw∈(RF)N(−ln(PX(w))+c2∥y−Φw∥2)
gives the MAP estimate of over . Note that we use the PMF formulation above, instead of the more common bin integration formulation, in order to simplify our presentation and analysis. Luckily, as increases, will approximate more closely under (6).
### Iii-B Universal MAP estimation
We now describe a universal estimator for CS over a quantized grid. Consider a prior that might involve Kolmogorov complexity [34, 35, 36], e.g., , or MDL complexity w.r.t. some class of parametric sources [39, 40, 41]. We call a universal prior if it has the fortuitous property that for every stationary ergodic source and fixed , there exists some minimum such that
−ln(PU(w))N<−ln(PX(w))N+ϵ
for all and [30, 31]. We optimize over an objective function that incorporates and the presence of additive white Gaussian noise in the measurements:
ΨU(w)≜−ln(PU(w))+c2∥y−Φw∥2, (7)
resulting in666This formulation of corresponds to a Lagrangian relaxation of the approach studied in [37, 38]. . Our universal MAP estimator does not require , and can be used in general linear inverse problems.
### Iii-C Conjectured MSE performance
Donoho [33] showed for the scalar channel that: () the Kolmogorov sampler (3) is drawn from the posterior distribution ; and () the MSE of this estimate is no greater than twice the MMSE. Based on this result, which requires a large reproduction alphabet, we now present a conjecture on the quality of the estimation . Our conjecture is based on observing that (i) in the setting (1), Kolmogorov sampling achieves optimal rate–distortion performance; (ii) the Bayesian posterior distribution is the solution to the rate-distortion problem; and (iii) sampling from the Bayesian posterior yields a squared error that is no greater than twice the MMSE. Hence, behaves as if we sample from the Bayesian posterior distribution and yields no greater than twice the MMSE; some experimental evidence to assess this conjecture is presented in Figs. 2 and 4.
###### Conjecture 1
Assume that is an i.i.d. Gaussian measurement matrix where each entry has mean zero and variance . Suppose that Condition 1 holds, the aspect ratio in (2), and the noise is i.i.d. zero-mean Gaussian with finite variance. Then for all , the mean squared error of the universal MAP estimator satisfies
EX,Z,Φ[∥x−xUMAP∥2]N<2EX,Z,Φ[∥x−EX[x|y,Φ]∥2]N+ϵ
for sufficiently large .
## Iv Fixed reproduction alphabet algorithm
Although the results of the previous section are theoretically appealing, a brute force optimization of is computationally intractable. Instead, we propose an algorithmic approach based on MCMC methods [43]. Our approach is reminiscent of the framework for lossy data compression in [48, 49, 47, 50].
### Iv-a Universal compressor
We propose a universal lossless compression formulation following the conventions of Weissman and coauthors [48, 49, 47]. We refer to the estimate as in our algorithm. Our goal is to characterize , cf. (7). Although we are inspired by the Kolmogorov sampler approach [33], KC cannot be computed [24, 25], and we instead use empirical entropy. For stationary ergodic sources, the empirical entropy converges to the per-symbol entropy rate almost surely [24].
To define the empirical entropy, we first define the empirical symbol counts:
nq(w,α)[β]≜|{i∈[q+1,N]:wi−1i−q=α,wi=β}|, (8)
where is the context depth [31, 51], , , is the symbol of , and is the string comprising symbols through within . We now define the order conditional empirical probability for the context as
Pq(w,α)[β]≜nq(w,α)[β]∑β′∈RFnq(w,α)[β′], (9)
and the order conditional empirical entropy,
Hq(w)≜−1N∑α∈(RF)q,β∈RFnq(w,α)[β]log2(Pq(w,α)[β]), (10)
where the sum is only over non-zero counts and probabilities.
Allowing the context depth to grow slowly with , various universal compression algorithms can achieve the empirical entropy asymptotically [31, 51, 30]. On the other hand, no compressor can outperform the entropy rate. Additionally, for large , the empirical symbol counts with context depth provide a sufficiently precise characterization of the source statistics. Therefore, provides a concise approximation to the per-symbol coding length of a universal compressor.
### Iv-B Markov chain Monte Carlo
Having approximated the coding length, we now describe how to optimize our objective function. We define the energy in an analogous manner to (7), using as our universal coding length:
ΨHq(w)≜NHq(w)+c4∥y−Φw∥2, (11)
where . The minimization of this energy is analogous to minimizing .
Ideally, our goal is to compute the globally minimum energy solution . We use a stochastic MCMC relaxation [43] to achieve the globally minimum solution in the limit of infinite computation. To assist the reader in appreciating how MCMC is used to compute , we include pseudocode for our approach in Algorithm 1. The algorithm, called basic MCMC (B-MCMC), will be used as a building block for our latter Algorithms 2 and 3 in Section V. The initial estimate is obtained by quantizing the initial point to . The initial point could be the output of any signal reconstruction algorithm, and because is a preliminary estimate of the signal that does not require high fidelity, we let for simplicity, where denotes transpose. We refer to the processing of a single entry of as an iteration and group the processing of all entries of , randomly permuted, into super-iterations.
The Boltzmann PMF is defined as
Ps(w)≜1ζsexp(−sΨHq(w)), (12)
where is inversely related to the temperature in simulated annealing and is a normalization constant. MCMC samples from the Boltzmann PMF (12) using a Gibbs sampler: in each iteration, a single element is generated while the rest of , , remains unchanged. We denote by the concatenation of the initial portion of the output vector , the symbol , and the latter portion of the output . The Gibbs sampler updates by resampling from the PMF:
Ps(wn=a|w∖n) = exp(−sΨHq(wn−11awNn+1))∑b∈RFexp(−sΨHq(wn−11bwNn+1)) = 1∑b∈RFexp(−s[NΔHq(w,n,b,a)+c4Δd(w,n,b,a)]),
where
ΔHq(w,n,b,a)≜Hq(wn−11bwNn+1)−Hq(wn−11awNn+1)
is the change in empirical entropy (10) when is replaced by , and
Δd(w,n,b,a)≜∥y−Φ(wn−11bwNn+1)∥2−∥y−Φ(wn−11awNn+1)∥2 (14)
is the change in when is replaced by . The maximum change in the energy within an iteration of Algorithm 1 is then bounded by
Δq=max1≤n≤Nmaxw∈(RF)Nmaxa,b∈RF|NΔHq(w,n,b,a)+c4Δd(w,n,b,a)|. (15)
Note that is assumed bounded (cf. Condition 1) so that (1415) are bounded as well.
In MCMC, the space is analogous to a statistical mechanical system, and at low temperatures the system tends toward low energies. Therefore, during the execution of the algorithm, we set a sequence of decreasing temperatures that takes into account the maximum change given in (15):
st≜ln(t+r0)/(cNΔq) for some c>1, (16)
where is a temperature offset. At low temperatures, i.e., large , a small difference in energy drives a big difference in probability, cf. (12). Therefore, we begin at a high temperature where the Gibbs sampler can freely move around . As the temperature is reduced, the PMF becomes more sensitive to changes in energy (12), and the trend toward with lower energy grows stronger. In each iteration, the Gibbs sampler modifies in a random manner that resembles heat bath concepts in statistical mechanics. Although MCMC could sink into a local minimum, Geman and Geman [43] proved that if we decrease the temperature according to (16), then the randomness of Gibbs sampling will eventually drive MCMC out of the locally minimum energy and it will converge to the globally optimal energy w.r.t. . Note that Geman and Geman proved that MCMC will converge, although the proof states that it will take infinitely long to do so. In order to help B-MCMC approach the global minimum with reasonable runtime, we will refine B-MCMC in Section V.
The following theorem is proven in Appendix A, following the framework established by Jalali and Weissman [48, 49].
###### Theorem 1
Let be a stationary ergodic source that obeys Condition 1. Then the outcome of Algorithm 1 in the limit of an infinite number of super-iterations obeys
limr→∞ΨHq(wr)=min˜w∈(RF)NΨHq(˜w)=ΨHq(xHqMAP).
Theorem 1 shows that Algorithm 1 matches the best-possible performance of the universal MAP estimator as measured by the objective function , which should yield an MSE that is twice the MMSE (cf. Conjecture 1). We want to remind the reader that Theorem 1 is based on the stationarity and ergodicity of the source, which could have memory. To gain some insight about the convergence process of MCMC, we focus on a fixed arbitrary sub-optimal sequence . Suppose that at super-iteration the energy for the algorithm’s output has converged to the steady state (see Appendix A for details on convergence). We can then focus on the probability ratio ; because is the global minimum and has the largest Boltzmann probability over all , whereas is sub-optimal. We then consider the same sequence at super-iteration ; the inverse temperature is and the corresponding ratio at super-iteration is (cf. (12))
P2st(w)P2st(xHqMAP)=exp(−2stΨHq(w))exp(−2stΨHq(xHqMAP))=⎛⎝Pst(w)Pst(xHqMAP)⎞⎠2.
That is, between super-iterations and the probability ratio is also squared, and the Gibbs sampler is less likely to generate samples whose energy differs significantly from the minimum energy w.r.t. . We infer from this argument that the probability concentration of our algorithm around the globally optimal energy w.r.t. is linear in the number of super-iterations.
### Iv-C Computational challenges
Studying the pseudocode of Algorithm 1, we recognize that Lines 911 must be implemented efficiently, as they run times. Lines 9 and 10 are especially challenging.
For Line 9, a naive update of has complexity , cf. (10). To address this problem, Jalali and Weissman [48, 49] recompute the empirical conditional entropy in time only for the contexts whose corresponding counts are modified [48, 49]. The same approach can be used in Line 13, again reducing computation from to . Some straightforward algebra allows us to convert Line 10 to a form that requires aggregate runtime of . Combined with the computation for Line 9, and since (because , and ) in practice, the entire runtime of our algorithm is .
The practical value of Algorithm 1 may be reduced due to its high computational cost, dictated by the number of super-iterations required for convergence to and the large size of the reproduction alphabet. Nonetheless, Algorithm 1 provides a starting point toward further performance gains of more practical algorithms for computing , which are presented in Section V. Furthermore, our experiments in Section VI will show that the performance of the algorithm of Section V is comparable to and in many cases better than existing algorithms.
While Algorithm 1 is a first step toward universal signal estimation in CS, must be large enough to ensure that quantizes a broad enough range of values of finely enough to represent the estimate well. For large , the estimation performance using the reproduction alphabet (6) could suffer from high computational complexity. On the other hand, for small the number of reproduction levels employed is insufficient to obtain acceptable performance. Nevertheless, using an excessive number of levels will slow down the convergence. Therefore, in this section, we explore techniques that tailor the reproduction alphabet adaptively to the signal being observed.
### V-a Adaptivity in reproduction levels
To estimate better with finite , we utilize reproduction levels that are adaptive instead of the fixed levels in . To do so, instead of , we optimize over a sequence , where and denotes the size. The new reproduction alphabet does not directly correspond to real numbers. Instead, there is an adaptive mapping , and the reproduction levels are . Therefore, we call the adaptive reproduction alphabet. Since the mapping is one-to-one, we also refer to as reproduction levels. Considering the energy function (11), we now compute the empirical symbol counts , order conditional empirical probabilities , and order conditional empirical entropy using , , and , cf. (8), (9), and (10). Similarly, we use instead of , where is the straightforward vector extension of . These modifications yield an adaptive energy function .
We choose to optimize for minimum squared error,
Aopt ≜argminA∥y−ΦA(u)∥2 =argminA[M∑m=1(ym−[ΦA(u)]m)2],
where denotes the entry of the vector . The optimal mapping depends entirely on , , and . From a coding perspective, describing requires bits for and bits for to match the resolution of the non-adaptive , with an arbitrary constant [47]. The resulting coding length defines our universal prior.
Optimization of reproduction levels: We now describe the optimization procedure for , which must be computationally efficient. Write
Υ(A)≜∥y−ΦA(u)∥2=M∑m=1(ym−N∑n=1ΦmnA(un))2,
where is the entry of at row and column . For to be minimum, we need zero-valued derivatives in (17), where is the indicator function for event .
Define the location sets for each , and rewrite the derivatives of ,
dΥ(A)dA(β)=−2M∑m=1⎛⎝ym−∑λ∈Z∑n∈LλΦmnA(λ)⎞⎠⎛⎜⎝∑n∈LβΦmn⎞⎟⎠. (18)
Let the per-character sum column values be
μmβ≜∑n∈LβΦmn, (19)
for each and . We desire the derivatives to be zero, cf. (18):
Thus, the system of equations must be satisfied,
M∑m=1ymμmβ=M∑m=1(∑λ∈ZA(λ)μmλ)μmβ (20)
for each . Consider now the right hand side,
M∑m=1(∑λ∈ZA(λ)μmλ)μmβ=∑λ∈ZA(λ)M∑m=1μmλμmβ,
for each . The system of equations can be described in matrix form in (21).
Note that by writing as a matrix with entries indexed by row and column given by (19), we can write as a Gram matrix, , and we also have , cf. (20). The optimal can be computed as a vector if is invertible. We note in passing that numerical stability can be improved by regularizing . Note also that
∥y−ΦA(u)∥2=M∑m=1⎛⎝ym−∑β∈ZμmβAopt(β)⎞⎠2, (22)
which can be computed in time instead of .
Computational complexity: Pseudocode for level-adaptive MCMC (L-MCMC) appears in Algorithm 2, which resembles Algorithm 1. The initial mapping is inherited from a quantization of the initial point , ( takes different values in Section V-B), and other minor differences between B-MCMC and L-MCMC appear in lines marked by asterisks.
We discuss computational requirements for each line of the pseudocode that is run within the inner loop.
• Line 10 can be computed in time (see discussion of Line 9 of B-MCMC in Section IV-C).
• Line 11 updates for in time.
• Line 12 updates . Because we only need to update columns and rows, each such column and row contains entries, and each entry is a sum over terms, we need time.
• Line 13 requires inverting in time.
• Line 14 requires time, cf. (22).
• Line 15 requires time.
In practice we typically have , and so the aggregate complexity is , which is greater than the computational complexity of Algorithm 1 by a factor of .
### V-B Adaptivity in reproduction alphabet size
While Algorithm 2 adaptively maps to , the signal estimation quality heavily depends on . Denote the true alphabet of the signal by ; if the signal is continuous-valued, then is infinite. Ideally we want to employ as many levels as the runtime allows for continuous-valued signals, whereas for discrete-valued signals we want . Inspired by this observation, we propose to begin with some initial , and then adaptively adjust hoping to match . Hence, we propose the size- and level-adaptive MCMC algorithm (Algorithm 3), which invokes L-MCMC (Algorithm 2) several times.
Three basic procedures: In order to describe the size- and level-adaptive MCMC (SLA-MCMC) algorithm in detail, we introduce three alphabet adaptation procedures as follows.
• MERGE: First, find the closest adjacent levels . Create a new level and add it to . Let . Replace by whenever . Next, remove and from .
• ADD-out: Define the range , and . Add a lower level and/or upper level to with
A(β3)=minA(Z)−IRA|Z|−1, A(β4)=maxA(Z)+IRA|Z|−1.
Note that , i.e., the new levels are empty.
• ADD-in: First, find the most distant adjacent levels, and . Then, add a level to with . For s.t. , replace by with probability
where is given in (12); for s.t. , replace by with probability
Note that is typically non-zero, i.e., tends not to be empty.
We call the process of running one of these procedures followed by running L-MCMC a round.
Size- and level-adaptive MCMC: SLA-MCMC is conceptually illustrated in the flowchart in Fig. 1. It has four stages, and in each stage we will run L-MCMC for several super-iterations; we denote the execution of L-MCMC for super-iterations by L(). The parameters , and are the number of super-iterations used in Stages 1 through 4, respectively. The choice of these parameters reflects a trade-off between runtime and estimation quality.
In Stage 1, SLA-MCMC uses a fixed-size adaptive reproduction alphabet to tentatively estimate the signal. The initial point of Stage 1 is obtained in the same way as L-MCMC. After Stage 1, the initial point and temperature offset for each instance of L-MCMC correspond to the respective outputs of the previous instance of L-MCMC. If the source is discrete-valued and in Stage 1, then multiple levels in the output of Stage 1 may correspond to a single level in . To alleviate this problem, in Stage 2 we merge levels closer than , where is a parameter.
However, might still be larger than needed; hence in Stage 3 we tentatively merge the closest adjacent levels. The criterion evaluates whether the current objective function is lower (better) than in the previous round; we do not leave Stage 3 until is violated. Note that if (this always holds for continuous-valued signals), then ideally SLA-MCMC should not merge any levels in Stage 3, because the objective function would increase if we merge any levels.
Define the outlier set . Under Condition 1, might be small or even empty. When is small, L-MCMC might not assign levels to represent the entries of . To make SLA-MCMC more robust to outliers, in Stage 4a we add empty levels outside the range and then allow L-MCMC to change entries of to the new levels during Gibbs sampling; we call this populating the new levels. If a newly added outside level is not populated, then we remove it from . Seeing that the optimal mapping in L-MCMC tends not to map symbols to levels with low population, we consider a criterion where we will will add an outside upper (lower) level if the population of the current upper (lower) level is smaller than , where is a parameter. That is, the criterion is violated if both populations of the current upper and lower levels are sufficient (at least ); in this case we do not need to add outside levels because will map some of the current levels to represent the entries in . The criterion is violated if all levels added outside are not populated by the end of the round. SLA-MCMC keeps adding levels outside until it is wide enough to cover most of the entries of .
Next, SLA-MCMC considers adding levels inside (Stage 4b). If the signal is discrete-valued, this stage should stop when . Else, for continuous-valued signals SLA-MCMC can add levels until the runtime expires.
In practice, SLA-MCMC runs L-MCMC at most a constant number of times, and the computational complexity is in the same order of L-MCMC, i.e., . On the other hand, SLA-MCMC allows varying , which often improves the estimation quality.
### V-C Mixing
Donoho proved for the scalar channel setting that is sampled from the posterior [33]. Seeing that the Gibbs sampler used by MCMC (cf. Section IV-B) generates random samples, and the outputs of our algorithm will be different if its random number generator is initialized with different random seeds, we speculate that running SLA-MCMC several times will also yield independent samples from the posterior, where we note that the runtime grows linearly in the number of times that we run SLA-MCMC. By mixing (averaging over) several outputs of SLA-MCMC, we obtain , which may have lower squared error w.r.t. the true than the average squared error obtained by a single SLA-MCMC output. Numerical results suggest that mixing indeed reduces the MSE (cf. Fig. 8); this observation suggests that mixing the outputs of multiple algorithms, including running a random reconstruction algorithm several times, may reduce the squared error.
## Vi Numerical results
In this section, we demonstrate that SLA-MCMC is comparable and in many cases better than existing algorithms in reconstruction quality, and that SLA-MCMC is applicable when . Additionally, some numerical evidence is provided to justify Conjecture 1 in Section III-C. Then, the advantage of SLA-MCMC in estimating low-complexity signals is demonstrated. Finally, we compare B-MCMC, L-MCMC, and SLA-MCMC performance.
We implemented SLA-MCMC in Matlab777A toolbox that runs the simulations in this paper is available at http://people.engr.ncsu.edu/dzbaron/software/UCS_BaronDuarte/ and tested it using several stationary ergodic sources. Except when noted, for each source, signals of length were generated. Each such was multiplied by a Gaussian random matrix with normalized columns and corrupted by i.i.d. Gaussian measurement noise . Except when noted, the number of measurements varied between 2000 and 7000. The noise variance was selected to ensure that the signal-to-noise ratio (SNR) was or dB; SNR was defined as . According to Section IV-A, the context depth , where the base of the logarithm is the alphabet size; using typical values such as and , we have and set . While larger will slow down the algorithm, it might be necessary to increase when is larger. The number of super-iterations in different stages of SLA-MCMC and , the maximum total number of super-iterations to be , the initial number of levels , and the tuning parameter from Section V-B ; these parameters seem to work well on an extensive set of numerical experiments. SLA-MCMC was not given the true alphabet for any of the sources presented in this paper; our expectation is that it should adaptively adjust to match . The final estimate of each signal was obtained by averaging over the outputs of runs of SLA-MCMC, where in each run we initialized the random number generator with another random seed, cf. Section V-C. These choices of parameters seemed to provide a reasonable compromise between runtime and estimation quality.
We chose our performance metric as the mean signal-to-distortion ratio (MSDR) defined as . For each and SNR, the MSE was obtained after averaging over the squared errors of for 50 draws of , , and . We compared the performance of SLA-MCMC to that of (i) compressive sensing matching pursuit (CoSaMP) [52], a greedy method; (ii) gradient projection for sparse reconstruction (GPSR) [10], an optimization-based method; (iii) message passing approaches (for each source, we chose best-matched algorithms between EM-GM-AMP-MOS (EGAM for short)&
|
2020-07-04 09:47:34
|
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|
https://tech-notes.maxmasnick.com/?page=8
|
## AppleScript to open a new Safari window directly to the compose page in FastMail
Sometimes I like to be able to write an email without seeing the new messages in my inbox. This AppleScript (triggered with LaunchBar of course) solves this problem. It opens a new Safari window directly to FastMail's compose page.
Because FastMail loads so quickly and Safari is generally pretty fast fast, this is essentially the same speed of LaunchBar's "Compose Email" action for Mail.app.
tell application "Safari"
make new document with properties {URL:"https://www.fastmail.com/mail/compose?u=FILL_IN_FROM_FASTMAIL_URL"}
activate
end tell
// Posted
## AppleScript to find FastMail tabs in Safari
I'm using the FastMail web interface more and more recently, but I keep losing the tab I keep it open in.
I found this AppleScript that is supposed to search Safari tabs based on the contents of a URL, but it gave me an error. So I fixed it (and hard-coded "fastmail.com" as the string to search for in the URL). I saved it as a .app file in Script Editor and trigger it with LaunchBar.
Update 1here's a (low-res) transparent icon pulled from FastMail's website that you can use for the .app created by Script Editor.
Update 2 (January 16, 2015): I've updated the script to fix a few bugs (like the window with the FastMail tab not actually coming to the front), and to automatically open FastMail in a new window if there isn't a current table open.
// Posted
## Terminal command to convert WAV to MP3
From the Apple StackExchange (user nqw1), this Terminal command will take all the WAV files in a folder and turn them into MP3s:
find . -name '*.wav' -maxdepth 1 -exec /usr/local/bin/lame -V 0 -q 0 '{}' \;
Apparently this is the highest quality (V is the best variable bitrate and q is the quality). This is apparently a bit faster and is the recommended setting:
find . -name '*.wav' -maxdepth 1 -exec /usr/local/bin/lame -V 0 -q 2 '{}' \;
Posted
## Recovering deleted photos from a memory card
If you ever need to recover deleted photos from a memory card, there's an open source project called Photorec that works great. It's a little scary and is all command-line based, so proceed at your own risk. But for me, it just saved the day.
Posted
## Bookmarklet to re-perform a Duck Duck Go search on Google
I use Duck Duck Go as my primary search engine, but I often need to repeat a search on Google – Duck Duck Go is good, but nowhere near as good as Google for some searches, unfortunately.
I created this bookmarklet to prepend "!g" to the search and re-submit the search form. This will automatically perform the same search on Google.
If you use Safari and place it near the beginning of your bookmarks bar like I did, you can use ⌘3 (in my case; it's the third bookmark from the left) to trigger the bookmarklet.
Here's the "URL" for the bookmark that contains the JavaScript to do this:
javascript:(function()%7B$('%23search_form_input').val('!g%20'+$('%23search_form_input').val());\$('%23search_form').submit();%7D)()
(To install, you'll have to manually make a new bookmark in your bookmarks bar with the URL as the code above.)
// Posted
## Mail.app: Get URL for selected message with AppleScript and TextExpander
I often use message://<[email protected]> URLs to reference a specific email in OmniFocus or Evernote: clicking one of these URLs will automatically open up the source message in Mail.app.
There's not a great way to get these very handy message URLs without AppleScript, so based on this SuperUser answer, I created a TextExpander snippet.
Now, when I type xmsg, TextExpander will insert the message URL for the selected message in Mail.app into whatever application I'm working in.
Here's the TextExpander snippet, and here's the AppleScript:
tell application "Mail"
set selectedMessages to selection
set theMessage to item 1 of selectedMessages
set messageid to message id of theMessage
-- Make URL (must use URL-encoded values for "<" and ">")
set urlText to "message://" & "%3c" & messageid & "%3e"
return urlText
end tell
// Posted
## Mail.app: Disabling the reply and send animations
Thanks to Andreas Verhoeven, there's a plugin for Mail.app that will do just this.
Posted
## Safely getting a MD5 checksum for a file
There are lots of 3rd party utilities out there for getting the a checksum of a file, but with sensitive data it's best to not read your files into any untrusted application.
Here is how to get MD5 checksums using only 1st party functionality from the OS vendor (Apple or Microsoft).
Mac
Download MD5.app, which is a thin wrapper around the built-in md5 command line utility (based on this AppleScript). If you run MD5.app, it will ask you to pick a file to checksum. If you drag a file onto MD5.app, it will checksum that file immediately.
You can verify the (very short) code for this application by opening AppleScript Editor.app, going to File > Open, and selecting MD5.app.
You can also run md5 /path/to/file.txt in Terminal.app.
Windows
Of course, this is not nearly as easy on Windows.
1. Go to this Microsoft knowledge base page and click the download link for the File Checksum Integrity Verifier utility package midway down the page.
2. Run the downloaded file, agree to the terms, and select your Desktop as the location to extract the files to.
3. Two files will be extracted: ReadMe.txt (delete this) and fciv.exe. Create a folder for this at c:\Users\yourusername\fciv\ and move fciv.exe to this folder.
4. Go to Start > Run, type "cmd", and press Enter to open the command prompt. You should already be in c:\Users\yourusername. Type "fciv\fciv.exe -add c:\path\to\file\to\checksum.txt"
Windows – Update (2019-08-26)
You can also use PowerShell to do this:
CertUtil -hashfile yourFileName MD5
(source)
Posted
## OmniFocus 2: Quick defer + due date entry with TextExpander
In OmniFocus 2, there is not a good way to quickly enter a defer date and then a due date: it involves tabbing a bunch of times or using the mouse to move between fields.
TextExpander to the rescue!
I use the following snippets to set the same defer + due date for tasks:
• Tomorrow at 9am: xx1 → tomorrow 9am%key:tab%%key:tab%%key:tab%%key:tab%tomorrow 9am%key:enter%
• +2 days at 9am: xx1 → 2 days 9am%key:tab%%key:tab%%key:tab%%key:tab%2 days 9am%key:enter%
• Next Monday at 9am: xxmon → Monday 9am%key:tab%%key:tab%%key:tab%%key:tab%Monday 9am%key:enter%
• Next Saturday 9am: xxsat → Saturday 9am%key:tab%%key:tab%%key:tab%%key:tab%Saturday 9am%key:enter%
I also have "x1", "x2", "xmon", and "xsat" set up to work in the quick entry box, which requires a few less tabs.
You can grab all these snippets here.
Update:
Ken Case of The Omni Group explains that users will only have to press tab four times if they have the "All controls" checked in the Full Keyboard Access portion of System Preferences > Keyboard Preferences > Shortcuts. I always have this turned on, like many users, so I can tab between form controls.
So if you don't have this enabled, you'll need to adjust the number of tabs in the TextExpander snippets. I still think this idea is useful for quickly setting defer+due dates if you often use the same sets of dates, even if the fields are only one tab away.
// Posted
## AppleScript: Figure out the names for stuff in a window
I'm a novice at AppleScript (never bothered to learn it, and hopefully never will). One of the hardest things for me is figuring out how to refer to interface components. This StackOverflow post has a simple solution:
tell application "System Events"
end tell
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2021-04-17 11:30:43
|
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|
http://ruieduardolopes.pt/posts/camera-movement/
|
# Basics of Camera Movement
2019-03-16
One of my hobbies is filmography. Basically, almost ten years ago, I started to self-learn some techniques on how could I improve my homemade recordings in video media. Depending on the type of content, sometimes I wanted to express some feelings through simple camera movements, but I didn’t know which were the proper movements (neither their names) to which case (pretended expression).
Here, I will try to describe what are those movements (shots) briefly and how could you use them to achieve your goals.
# Camera shots – a brief description
There are a total of one frame technique and six basic camera movements, which are naturally related to the six degrees of freedom a camera could have. Below you can see more about these six degrees of freedom [1].
Six degrees of freedom. There are three transitional envelopes—the backward/forward envelope (correspondent to the $xx$ axis), the left/right envelope (correspondent to the $yy$ axis), and the down/up envelope (correspondent to the $zz$ axis). Meanwhile, the other three degrees of freedom include three rotational envelopes—these are the roll (rotation on the $x$ axis), the pitch (rotation on the $y$ axis), and the yaw (rotation on the $z$ axis). Recognizing the several degrees of freedom we can now apply a camera on the center of our referential and then point it to our scene—an environment which belongs to our world. Then, we can use our so-called shots.
## The zoom shot
First, let’s see a frame technique which does not consider any type of transitional or rotational envelopes on the camera, that means, where the camera itself does not change its position throughout its movement, nor rotates. Such shot is the zoom, mostly used to magnify a focus point in the frame with some drama in the mixture.
As I am mostly used to shoot cityscapes or other scenes where drama’s do not make sense, I do not use this shot very often. More, I do not recommend it if you are working with a smartphone with exclusive digital zoom (and not optical zoom), since the only thing you are doing is to augment the size of the pixel, lowering the resolution of the resulting image.
On the figure below, you can see a brief representation of a zoom shot, where the black frame is related to the frame at the beginning of the shooting, and the blue one is the final frame.
## The pan shot
As the zoom shot is like a simulation of a transitional envelope (but without any camera displacement)—it only crops the current frame into a smaller region. Now, a pan shot is, in fact, a pure rotational envelope, more specifically a yaw rotation.
This type of shot if very often used to follow a moving particle on the scene, such as a character or something that provides the spectator a sense of space.
Below I show you one example of my latest (and very simple) shootings, showing my beautiful city of Aveiro (a city worth to visit!) and its central canal.
Aveiro video shot using a pan movement. This video does not exclusively show a pan movement, but also starts with a truck, which we will cover further below. Please, ignore the non-parallel to the horizon start, but shooting with an iPad Pro, on a sunny day, and without any sunglasses… sorry! 😜
## The tilt shot
The tilt shot is another pure rotational envelope, more particularly the pitch rotation. Again, this type of shot is made without any kind of camera displacement.
This type of shot is used in scenes where the filmmaker wants to make a subject to appear bigger or more significant, as a camera rotates upwards in slow motion—the opposite rotation is also valid for making a subject appear less significant and smaller.
University of Aveiro’s campus in a tilt shot. This was shot with an 11-inch iPad Pro.
## The roll shot
Having specified the rotation in the zz axis (called pan shot), and in the yy axis (called tilt shot), we now need to characterize the roll shot, which is nothing more than a rotation in the xx axis.
This motion is mostly used when the scene has some radical activities happening or when the context needs such rotation. As it distorts the notion of the horizon in the shot itself, is a particular motion that is not often used as is, but instead in conjunction with other movements such as with dolly.
University of Aveiro’s Mechanic Engineering Department in a roll shot. This was shot with an 11-inch iPad Pro.
## The dolly shot
This shot is very similar to zoom’s with the particular difference that this one does have a displacement of the camera on the backward/forward axis (the xx axis). This provides the zoom shot a more natural and realistic way of transmitting the importance of a particular focus on the frame. This shooting movement can be a great way of creating a sense of intimacy between them.
University of Aveiro’s ear-thingy in a dolly shot. This was shot with an 11-inch iPad Pro.
## The truck shot
The truck shot is, undoubtedly, one of my favorites. Here the camera moves on the left/right axis (the yy axis) providing a way to follow characters in action or other scene particles. Quite similar to pan’s movement, this one does not carry on any type of rotation envelope.
Below there is an example of mine, taken on the MAAT’s building in Lisbon, Portugal [2]. The way it was made was merely by inserting my iPhone onto the tile gaps of the building and then move it parallel, on the same axis, to the bridge. This shot could also be achieved with great success if filmed in a larger frame rate (such as 240 frames per second rate) and then slowed down, to obtain a high-quality slo-mo.
Lisbon’s MAAT Museum using a truck shot. This video was shot with an iPhone 6s.
## The pedestal shot
Finally, the pedestal shot is to the tilt shot, as the latter truck was to pan’s. In this case, the camera should move on the zz axis, that is, up or down, not considering any type of rotation envelope.
University of Aveiro’s campus in a pedestal shot. This was shot with an 11-inch iPad Pro.
# Notes and References
This is the first article of many others about filmography and other works I developed. Get in touch to know more!
[1] Wikipedia contributors, “Six degrees of freedom,” Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/w/index.php?title=Six_degrees_of_freedom&oldid=885318684(accessed March 16, 2019);
[2] Fundação EDP, “MAAT—Museum of Art, Architecture and Technology”, MAAT, https://www.maat.pt/en/about (accessed March 16, 2019).
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2022-09-25 01:57:52
|
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|
https://www.hackmath.net/en/math-problem/1521
|
# Again saw
From the trunk of the tree we have to a sculpture beam with rectangular cross-section with dimensions 146 mm and 128 mm. What is the trunk smallest diameter?
Result
D = 194.165 mm
#### Solution:
$D^2 = a^2+b^2 \ \\ D = \sqrt{ 146^2+128^2} = 194.165 \ \text{mm}$
Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!
Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
Tips to related online calculators
Pythagorean theorem is the base for the right triangle calculator.
## Next similar math problems:
1. Broken tree
The tree is broken at 4 meters above the ground and the top of the tree touches the ground at a distance of 5 from the trunk. Calculate the original height of the tree.
The ladder has a length 3.5 meters. He is leaning against the wall so that his bottom end is 2 meters away from the wall. Determine the height of the ladder.
3. Broken tree
The tree was 35 meters high. The tree broke at a height of 10 m above the ground. Top but does not fall off it refuted on the ground. How far from the base of the tree lay its peak?
4. Windbreak
A tree at a height of 3 meters broke in the windbreak. Its peak fell 4.5 m from the tree. How tall was the tree?
5. Four ropes
TV transmitter is anchored at a height of 44 meters by four ropes. Each rope is attached at a distance of 55 meters from the heel of the TV transmitter. Calculate how many meters of rope were used in the construction of the transmitter. At each attachment
6. Chord circle
The circle to the (S, r = 8 cm) are different points A, B connected segment /AB/ = 12 cm. AB mark the middle of S'. Calculate |SS'|. Make the sketch.
7. Common chord
Two circles with radius 17 cm and 20 cm are intersect at two points. Its common chord is long 27 cm. What is the distance of the centers of these circles?
8. Circumscription
Calculate radius of the Circumscribed circle in the rectangle with sides 20 and 19. It can be rectangle inscribed by circle?
9. Diagonal - simple
Calculate the length of the diagonal of a rectangle with dimensions 5 cm and 12 cm.
10. Diagonal
Calculate the length of the diagonal of the rectangle ABCD with sides a = 8 cm, b = 7 cm.
11. Median
In the right triangle are sides a=41 dm b=42 dm. Calculate the length of the medians tc to the hypotenuse.
12. Center traverse
It is true that the middle traverse bisects the triangle?
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2020-06-01 15:51:09
|
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|
https://tex.stackexchange.com/questions/416533/undefined-control-sequence-using-enumerate
|
# undefined control sequence using enumerate
\begin{enumerate}
\item[1.] Choose $N\sim \text{Poisson}(\xi)$.
\item[2.] Choose $\theta \sim \text{Dir}(\alpha)$.
\item[3.] For each of the $N$ words $w_n$:
\begin{enumerate}
\item[(a)] Choose a topic $z_n \sim \text{Multinomial}(\theta)$.
\item[(b)] Choose a word $w_n$ from $p(w_n | z_n, \beta)$, a multinomial probability conditioned on the topic $z_n$.
\end{enumerate}
\end{enumerate}
When I use the above code to generate some equations, it gives me the following errors:
Undefined control sequence. \item[1.] Choose $N\sim \text Undefined control sequence. \item[2.] Choose$\theta \sim \text
Undefined control sequence. \item[(a)] Choose a topic $z_n \sim \text I don't understand what is causing these errors. Can anyone help? thanks. • You are missing \usepackage{amsmath}, most likely -- the error messages seem to be related to \text, which is an amsmath macro... – user31729 Feb 21 '18 at 22:18 • Welcome to TeX.SX! You should look in the log file for the real error message, which is not due to enumerate. Feb 21 '18 at 22:19 • (1) welcome, (2) as always on this site please post a full minimal example showing this behavior. (3) your use of the command \text is wrong here they are functions or similar, so use \mathrm or \operatorname , never \text Feb 21 '18 at 22:19 • Thank you @ChristianHupfer . The problem is solved after using package amsmath. Sorry I am new to Tex.SX so I don't know how to post a full minimal example. Feb 21 '18 at 22:27 • @yuhengd: Please follow daleif's advise about \mathrm or \operatorname as well – user31729 Feb 21 '18 at 22:33 ## 1 Answer I would just add a simpler way to use enumerate, with enumitem, which uses what you type manually as default: \documentclass{article} \usepackage{showframe} \renewcommand{\ShowFrameLinethickness}{0.3pt} \usepackage{enumitem} \usepackage{amsmath} \begin{document} \begin{enumerate}[ wide = 0pt, leftmargin = *] \item Choose$N\sim \text{Poisson}(\xi)$. \item Choose$\theta \sim \mathrm{Dir}(\alpha)$. \item For each of the$N$words$w_n$: \begin{enumerate}[wide = 0pt] \item Choose a topic$z_n \sim \mathrm{Multinomial}(\theta)$. \item Choose a word$w_n$from$p(w_n \,|\, z_n, \beta)$, a multinomial probability conditioned on the topic$z_n\$.
\end{enumerate}
\end{enumerate}
\end{document}
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2021-09-17 04:19:54
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https://www.physicsforums.com/threads/how-do-you-define-a-mapping-f-k-n.169359/
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# How do you define a mapping f:K->N
1. May 8, 2007
### pivoxa15
1. The problem statement, all variables and given/known data
How do you define a mapping f:K->N
with K={0}
N is the integers.
that maps the element 0 to every single single element in N?
ie. 0->-n, ... , -2, -1, 0, 1, 2, ... , n
Is that even possible?
The mapping Z->Z by multiplying each element in Z by 0 is a legitamate mapping isn't it? So really Z->0 but I am extending the range to Z.
In a mapping is isomorphic than there should be the same number of elements in the domain as in the range? But what happens if there are an infinite number of elements in both the range and domain?
Last edited: May 8, 2007
2. May 8, 2007
The thing you suggested isn't a mapping, unless I'm missing something. Every element of the domain must be mapped to at most one element of the codomain.
3. May 8, 2007
### matt grime
A map is an isomorphism if it is invertible, not if there are 'the same number of elements in the domain and the range'. In fact we define (infinite) cardinals by the existence of bijections.
4. May 8, 2007
### pivoxa15
The question is, is {0} isomorphic to Z? Where Z are the integers.
I can see that is would be isomorphic if we can find a mapping that maps 0 to every element in Z, once.
However mapping Z to {0} would recquire only a single mapping like map 1 to 0 or 2 to 0 or n to 0 but only once, otherwise it wouldn't be a bijection.
I can actually show that {0} isomorphic to Z by the first isomorphism theorem. i.e. f:Z->Z where f is to multiply every element in Z by 0. So the kernel of f is all of Z. Now the quotient ring is Z/Z which is isomorphic to {0} which is isomorphic (via the equivalence relation) to Z.
But I am not convinced. Namely the actual mappings seems a bit funny, if not incorrect.
5. May 8, 2007
### HallsofIvy
Staff Emeritus
Perhaps I am misunderstanding something but it should be obvious that {0} is not isomorphic to Z. For one thing, they don't have the same cardinality!
While n-> 0 is a mapping from Z to K, there is NO mapping from K to Z.
I have no idea what you think the "first isomorphism theorem" says. Yes, Z/Z contains a single member and is isomophic to {0}. That does NOT say that they are both isomorphic to Z!. In general, if G and H are groups and f is a homomorphism from G onto H, then the kernel of f, K, is a subgroup of G and G/K is isomorphic to H. That does NOT say that G is isomorphic to H.
6. May 8, 2007
The confusion is probably due to the fact that elements of quotient groups are cosets, and in this special case you are looking at the one and only coset eZ, which is, as HallyofIvy mentioned, isomorphic to Im(f) = {0}.
7. May 8, 2007
### pivoxa15
1st isomorphism theorem: Let f:R->S be a surjective homomorphism of rings with kernel K. Then the quotient ring R/K is isomorphic to S.
Let R=S=Z the integers.
Let f be the maping multiplying each element in Z by 0 So Z->{0}
So K = Z
Hence Z/Z is isomorphic to Z. Z/Z is isomorphic to {0} so {0} is isomorphic to Z. But I have made a mistake because as you said they don't have the same number of elements. And I can't think of a bijective mapping from {0} to Z. But where is the mistake in my proof? Does R cannot equal S?
8. May 8, 2007
### Office_Shredder
Staff Emeritus
f isn't surjective if S=Z, f is only surjective if S={0}!
9. May 8, 2007
### pivoxa15
Good point. So Z/Z is isomorphic to {0} as it should and not to Z.
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2017-03-28 10:17:23
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https://www.gamedev.net/forums/topic/335399-fast-polygon-triangulation/
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# Fast Polygon Triangulation
This topic is 4765 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
Hi! I'm currently working on my 3d engine, and I'm in need of a fast polygon triangulation algorithm. What's the fastest one if I'm mainly (but not exclusively) working with convex polygons? And do you have any links to an explanation of the given algorithm? Edit: Ok.. I seem to have been a bit unclear, so I'll try to be as clear as possible. What I'm wondering about is the following; [A] Which general purpose algorithms are there for triangulating a simple (convex or concave) polygon and what are their complexity? I've found Ear Cutting, Delaunay and Seidel's Algorithm. I'd prefer that it doesn't generate any new vertices but it's not critical (for instance if it increases visual quality). Which of the given algorithms are fastest when working on primarily (but not exclusively) convex polygons with few vertices? [C] Do you know of which triangulation algorithms different games use? Thanks! [Edited by - pim on July 29, 2005 7:00:23 AM]
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A nieve approach would be to, assuming convex polygons, and ordered in some way (CW or CCW) create a triangle fan. Pivot off of the first point.
If you want to hand simple, concave polygons, perhaps it would suffice to choose the most central vertex (closest to the average of the vertex's points) as the pivotal point. It seems in theory (and visually) this would work under most circumstances (w/o complex polygons, anyway)
Let me know if that's good enough for what you're doing (I'm curious) or if you wanted some long, complex formula.
-Michael g.
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For convex just do a triangle fan, e.g.:
1|-------------------|6 | | | | | | 2\ /5 \ / \ / \ / 3 \-----------/4
Choose any vertex (labed with numbers) as your pivot point, then shoot out rays to all the other, non-adjacent vertices.
For example, if you chose vertex 3, shoot rays to vertices 5, 6, and 1, each ray creating a triangle. Do not shoot rays to the adjacent vertices, in this case being 2 and 4.
HTH,
nilkn
EDIT: Sorry, I wrote 'concave' instead of 'convex'.
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Another way to make sure it does the best it can with concave polygons is (assuming there's only one point really making it concave, it'll always work well, otherwise, well, you can do some more complex logic, anyway)
Keep track of the slope (or angle, but slope's quicker to calculate) from vertex(n) to vertex(n+1) if the "angle" or "slope" shows the next point makes the polygon concave, then the current point is the culprit. Choose that as the inital point for the triangle fan.
If you want to make it more robust, you could break it up into multiple triangle fans (one for each offending point) but then you'd have to figure out how you want to arrange the tris so they join.
Once it gets that tricky though, I personally would start looking at some 3rd party library/solution.
-Michael g.
ps. I think nilkn and I were saying the same thing: he's got ascii art though ;-)
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Well.. there are some different ways to triangulate a convex polygon (zigzag, fan, subdivide++) but I need a way that works with concave polys as well!
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As I understand it the method for concave polygons is similar in a way. Basically what you have to do is clip off triangles one at a time but make sure the inside of the triangle is entirely within the polygons, e.g.:
A ______________ B \ C / \ /\ / \ / \ / \/ \/ D E
In this example you could clip off ACB, BCE or ADC, but not ADE or BED which is why you have to check your set of points does not create a triangle which includes parts outside the original polygon. The same rule for no. of triangles required applies for concave as convex polygons (i.e. no. of points minus 2).
As for fastest algorithm: I did some digging on this a while ago. I think the simplest/fastest algorithm involves projecting the polygon onto a plane (that is since it's already planar transform onto XY, XZ or YZ), sorting the points along one of the axes and walking along the edges from one end to the other. It involves something like tracking the spans inside the polygon in the Y direction when scanning in the X direction, if you see what I mean.
That algorithm is O(n lg n) because of the sort stage. However if you're only working with small numbers of points (I'm not sure exactly how small but 10 or so I'd think would be small enough), the O(n^2) method is probably best. That simply involves picking an edge and looping through all other vertices until you find one that can complete an acceptable triangle (i.e. the triangle contains no other points).
Apparently O(n lg lg n) and even O(n) are possible but these are probably much more complicated and never going to be worth it for almost any conceivable situation.
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Quote:
Original post by pimI'm currently working on my 3d engine, and I'm in need of a fast polygon triangulation algorithm. What's the fastest one if I'm mainly (but not exclusively) working with convex polygons? And do you have any links to an explanation of the given algorithm?
Does it need to be general-purpose? Otherwise EOR-filling is pretty neat.
Basically only the horizontal outlines of the polygons are drawn by XORing the polygon's color onto the framebuffer. When all polygons has been "drawn" the filled scene is generated by walking through the framebuffer line-by-line and XORing the pixels in each successive scanline with the ones in the previously generated one.
Obviously you can only have solid polygons and overddraw is tricky. But it's blazingly fast =)
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The above post was mine. Damn firefox won't stay logged in..
edit: Bah.. Scratch that, the original request was for triangulation and not rasterization.
It's past 3 around here so maybe I should just get some sleep?
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Hmm... noone has really answered the question, so I'll try to be a bit more clear (edited the original post with the new info!)
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Well, the O(n^2) algorithm I described above is ear clipping as far as I know. I can only find one page on Seidel's algorithm and it's not particularly clear to me. As for Delaunay triangulation I've never found anything on how it can be used to triagulate polygons.
Anyway my thoughts on the other questions are as follows: if almost all your polygons are convex the fastest way to do this it going to be to check if a polygon is convex first because that can be done easily in O(n) time. If it is convex then use either the fan or strip decomposition. Otherwise revert to either of the other algorithms. Which one is best depends as I said before on the number of vertices you're looking at - for small numbers (~10 or so I'd think although this is a total guess) ear clipping is simpler to implement and I expect just as fast as any other method.
As for generating new vertices: if my understanding of the floating point error issues involved is correct it makes no difference which triangulation you use as long as the points on the edges are exactly the same. This was the reason behind GLQuake's GL_KEEPTJUNCTIONS console variable because if two parallel edges are merged into one this can cause numerical precision errors which can lead to small cracks. But as I said as long as the vertices along two connected edges are exactly the same, no crack should appear. Therefore I don't see any reason for generating new vertices, although some triangulations may be better than others for culling purposes.
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2018-08-18 22:26:36
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https://science.sciencemag.org/content/282/5388/396?ijkey=e5c00a6612aeae106585f7554039ebea1b1b4b6a&keytype2=tf_ipsecsha
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News FocusDNA ANALYSIS
# Microchip Arrays Put DNA on the Spot
See allHide authors and affiliations
Science 16 Oct 1998:
Vol. 282, Issue 5388, pp. 396-399
DOI: 10.1126/science.282.5388.396
FUTURE CHIPS
Researchers are finding new uses for microchip technology. Soon, DNA sequencers, chemical plants, and satellite propulsion systems will all come in credit-card-sized packages.
DNA chips, which identify DNA by binding it to samples on a substrate, let researchers tune in to the symphony of gene expression. They look set to revolutionize drug discovery and diagnostics, too
Last January, a new kind of microchip saved Patrick Baeuerle from going down a multimillion-dollar, dead-end street. Then the head of drug discovery at a South San Francisco- based biotechnology company called Tularik, Baeuerle and his colleagues had just synthesized a new drug compound that, in cell cultures, drastically reduced levels of low density lipoprotein, which has been linked to hardening of the arteries. The next step was to learn how the compound worked, a puzzle that can take years to unravel.
Looking for a shortcut, Baeuerle opted to try and find which genes a cell switches on in response to the compound. He and his team turned to researchers at Synteni, another Bay Area start-up firm, which makes DNA chips. These chips carry arrays of different snippets of DNA that serve as probes for detecting DNA fragments with a complementary nucleotide sequence.
When Synteni researchers used their chips on fluorescent-labeled DNA from cells exposed to either the new Tularik drug or a related drug already on the market, the pattern of fluorescence showed that the new drug had caused a completely different cellular response. “It dramatically changed the profile of gene expression,” says Baeuerle, who just left Tularik to head up research at a biotech start-up in his native Germany.
Unfortunately, the change wasn't for the better. The pattern of genes turned on by the new drug candidate strongly resembled that from a completely different class of compounds that had also looked promising but proved to be toxic. “It killed the prospects for [our] compound,” says Baeuerle. Alhough the result was a disappointment, the DNA tests likely saved Tularik millions of dollars by helping it weed out an unsuitable drug candidate early on, rather than later in animal or human tests.
Such experiences underscore the promise of what many are now calling the microchip of the 21st century. These 2- or 3-centimeter-wide slices of either silicon or glass, bearing anything from hundreds to hundreds of thousands of immobilized snippets of DNA, have the unique ability to track the expression of many (if not all) of a cell's genes at once, allowing researchers to witness for the first time the behavior of thousands of genes in concert. Moreover, tracking cells' responses to drugs is far from the only application of these chips. Genetic diagnostics companies are turning to DNA arrays hoping that unique gene-expression patterns can pinpoint the onset of diseases from cancer and Alzheimer's to osteoporosis and heart disease.
Elsewhere, researchers hope that arrays will help them gauge the success of HIV drug treatment, tailor medications to patients with specific genetic makeups, and sequence genes. And that's just for starters. The drug companies and biotech firms pursuing the technology are hoping that DNA chips will prove to be a primary research tool in a genetic-medicine revolution. They expect that understanding the genes active in disease will spawn a new generation of therapeutic drugs that treat underlying causes rather than symptoms.
“In the past, we compared the activity of single genes,” says Wei Zhang, an oncologist at the M. D. Anderson Cancer Center in Houston, Texas. “With the new technology, we can analyze a huge number of genes at the same time. That provides hope for a new era of diagnostics and therapeutics.” Jeffrey Trent, who heads DNA array research at the National Human Genome Research Institute (NHGRI) in Bethesda, Maryland, agrees. “It's a remarkably different approach to genetics,” he says. “[It] allows us to track pathways instead of individual genes.” Because of that advantage, the use of the new arrays “is just exploding in all kinds of directions,” says Francis Collins, NHGRI director. “The limits will not be found anytime soon.”
That promise has touched off a race to capitalize on DNA chips. Surveys by brokerage houses and market research firms indicate a nearly immediate annual market for the chips of about $1 billion, with plenty of room to grow. Not surprisingly, in recent years, about a dozen companies have jumped into the DNA chip-making business, each vying to become the Intel of genomics. Affymetrix of Santa Clara, California, an array pioneer, netted nearly$100 million in its June 1996 initial public stock sale. Even after the market's recent downturn, its outstanding stock is now worth more than $575 million, although the company has yet to make a profit. Other array companies are reporting similarly brisk business. Despite this flurry of interest, only a handful of actual products exists. Development has been hampered by a host of technical challenges, such as difficulties in distinguishing sometimes weak fluorescent signals from background noise. But the darkest cloud looming over this burgeoning business, according to chip company officials, is the threat of patent battles over key aspects of the technology and over the genes that make up the arrays (see sidebar). “There's still a lot of confusion about who owns what pieces of array technology,” says Michael Albin, the vice president of science and technology with Perkin-Elmer's Applied Biosystems Division in Palo Alto, California. Still, Albin and others are confident that if and when these battles are worked out, gene chips will take the drug discovery and diagnostics markets by storm. ## Array of options DNA arrays owe much of their current research and financial promise to the international Human Genome Project. Although sequencing the entire 3 billion nucleotides that make up a person's 23 pairs of chromosomes is a huge task, it is only the first step to making use of the genome. Equally important is linking each gene to its role in the cell—a field of research dubbed functional genomics. This task is daunting, too. In a typical cell, tens of thousands of genes wink on and off to help the cell churn out proteins involved in everything from metabolism to defense. Researchers can track the behavior of genes either alone or in small handfuls, but they have had no way to watch the dance of all the genes at once. A key breakthrough came in a 1991 Science paper by Stephen Fodor and colleagues at a drug-discovery company called Affymax (Science, 15 February 1991, p. 767). Fodor's team came up with a scheme to use the same lithographic production techniques employed in computer-chip manufacturing to synthesize a checkerboard array of either short protein fragments called peptides or short DNA fragments called oligonucleotides—each of which ended up with a unique chemical signature. The researchers were looking for a way to generate a large number of compounds quickly; these could then be tested either as drugs, in the case of peptides, or for gene identification, with oligos. To make their oligo arrays, for example, they started with a silicon surface coated with linker molecules that bind the four DNA building blocks, adenine (A), cytosine (C), guanine (G), and thymidine (T). Initially, the linkers are capped with a “blocking” compound that is removed by exposure to light. The researchers shone light onto the chip through a mask so that only certain areas of the chip became exposed. They then incubated the chip with one of the four bases, binding it to the exposed areas, then reapplied the block. By repeating this process with different masks and different bases, they could build up an array of different oligonucleotides. With just 32 such cycles, they could create more than 65,000 different oligos, each eight base-pairs long. Just like chromosomal DNA, each oligo was capable of binding to other stretches of DNA that had complementary sequences, in which G's on one segment were matched with C's on the other and A's with T's. Hence, the array could be used as a sensor: The researchers could isolate the RNA molecules that signal gene expression from tissues, chemically convert them to DNA, and label them with a fluorescent tag. After floating these tagged strands across an array of oligos, allowing complementary sequences to bind, and washing away the unbound strands, they could detect the strands that had bound by exciting the fluorescent tags with a laser. And because they knew the sequence of each oligo on their chip, the position of the fluorescent spot told them the sequence of the gene fragment that had bound there. In 1993, Affymax spun the idea into a new company—Affymetrix—and gene chips were born. Today, Fodor and his Affymetrix colleagues have developed more than 20 different DNA arrays for research purposes. They also offer commercial arrays where the oligos fastened to the chip are chosen specifically to scan for mutations in the HIV genome and the p53 tumor-suppressor gene, which has been implicated in up to half of all human cancers. A third chip, called Cytochrome P450, looks for variations in a set of genes involved in the metabolism of important therapeutic drugs such as beta blockers, prescribed for heart disease, and certain antidepressants. Affymetrix remains the best known DNA chip-maker in the business but is by no means alone. Just a couple of miles up Silicon Valley on Highway 101, researchers at Hyseq Inc. in Sunnyvale have developed their own oligo-based scheme for sequencing genes. The Hyseq scheme does not involve labeling the unknown DNA with a fluorescent tag; instead, it is mixed with a tagged oligo of known sequence and washed over the array. Where you get an array oligo and the tagged oligo binding side by side to the unknown DNA, you get a fluorescent spot—and you know part of the unknown sequence. This process is repeated with different labeled oligos, and finally a computer works out what the sequence of the DNA must be to account for all the partial-sequence information. Last year, Hyseq teamed up with gene-sequencing powerhouse Perkin-Elmer to market its chips. The first is expected to be available to researchers within months. Synteni was bought out last winter by Incyte Pharmaceuticals, a genomics company, which now offers glass chips that lay out an array of gene fragments 500 to 5000 base-pairs long. To use them, researchers isolate messenger RNA from normal tissue as well as tissue affected by disease or exposed to a drug. These two sets of RNAs are labeled with different-colored fluorescent tags and applied to the chip simultaneously. Scanning for the two colors then gives researchers an instant snapshot of how gene expression differs between normal cells and those affected by diseases or drugs. Meanwhile, researchers at San Diego-based Nanogen are putting the finishing touches on chips that apply a controlled electric field to maneuver the DNA fragments around on the chip, looking for a match. The upshot, says Nanogen's Michael Heller, is that fragments find their complementary oligo more quickly, and detection takes just minutes—rather than the hours needed with ordinary chips, which let the DNA fragments diffuse randomly. And an altogether different approach is being taken by the 2-year-old start-up Clinical Micro Sensors (CMS), of Pasadena, California. Researchers there have designed a unique approach that uses electrical signals, rather than fluorescence patterns, to indicate the position of DNA binding to oligos on the array. CMS builds its arrays on a grid of electrodes rather than a passive chip; when DNA binds to its matching oligos, a separate probe molecule, carrying iron, also binds to the complex—an addition that can be detected by the electrodes. ## Data flood Just where all this is going depends on whom you talk to. CMS President Jon Kayyem argues that the big market will be in diagnostics, and not just for diseases that can't be diagnosed today. To take one of his favorite examples, when parents bring in a child with a sore throat, doctors typically are limited to taking a throat swab and sending the culture to a lab for testing. The culture can take days, so the doctor often winds up prescribing antibiotics or other drugs without knowing if they will do any good. An instant diagnostic scan that reveals not only the type of infection but the precise strains would be a vast improvement. But Patrick Klyne, director of genomics at Millennium Predictive Medicine (MPM) in Cambridge, Massachusetts, says such tests have a long way to go before hitting the market. Not only would they have to wind their way through clinical trials and regulatory approval, they would have to come down in cost, too. Affymetrix chips can run anywhere between$45 and $850, not to mention the scanners and fluidics stations that go with them, which can cost more than$100,000. “To be viable [for diagnostics], the cost needs to come down to about \$5,” says Stanley Abromowitz of the National Institute of Standards and Technology. CMS's Kayyem says that his company's electronic detection scheme has a shot at making low-cost readers. But for now, the company has only a prototype device.
That's why Klyne and others argue that the initial breakthrough market will consist of genomics and pharmaceutical companies, which will use DNA arrays as a research tool to sift through the complex patterns of gene expression in cells and pinpoint particular genes that are turned on in disease. That's the approach being taken by MPM and its rival, diaDexus of Santa Clara, a joint venture between the big drug firm SmithKline Beecham and Incyte. DiaDexus, for example, has already used its arrays to show that prostate cancer cells crank out a protein called PLA2, while the same gene remains dormant in healthy cells. MPM researchers, meanwhile, have shown that melanoma cancer cells turn up production of a protein called melastatin. Both companies hope to turn these insights into new and improved diagnostic screens that would rely not on arrays themselves but on conventional and cheap techniques such as enzyme assays.
Other basic research with arrays is also beginning to pay off. In 1996, Collins and colleagues at NHGRI used Affymetrix chips to detect mutations in the familial breast cancer gene BRCA1 in subjects at risk for the disease. Upstairs from Collins's lab, Jeffrey Trent and his colleagues are gauging, with their own array system, how radiation treatment affects gene expression in cancer cells.
What's certain is that these studies are just a taste of what is to come. Already, researchers with access to DNA arrays find themselves with an enviable problem: too much information. “We are drowning in cool data here,” says Stanford array pioneer Pat Brown, whose team has made more than 7 million measurements of the expression of individual genes under different conditions. “More than 99% of the data we have is unpublished. It's so easy to think of an interesting experiment to do using this approach [that] we haven't been able to find the time to publish it all.”
View Abstract
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2021-01-20 04:40:25
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https://socratic.org/questions/what-electron-configuration-represents-an-excited-state
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# What electron configuration represents an excited state?
For example, if we look at the ground state (electrons in the energetically lowest available orbital) of oxygen, the electron configuration is $1 {s}^{2} 2 {s}^{2} 2 {p}^{4}$. If the element were to become excited, the electron could occupy an infinite number of orbitals. However, in most texts the example will be the next available one. So for oxygen, it might look like this: $1 {s}^{2} 2 {s}^{2} 2 {p}^{3} 3 {s}^{1}$ - where the valence electron now occupies the 3s orbital in an excited (i.e. not ground) state.
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2020-07-10 01:08:05
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https://artofproblemsolving.com/wiki/index.php?title=Fermat_prime&direction=next&oldid=32717
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Fermat prime
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
A Fermat prime is a prime number of the form $2^{2^n} + 1$ for some nonnegative integer $n$.
A number of the form $2^{2^n} + 1$ for nonnegative integer $n$ is a Fermat number. The first five Fermat numbers (for $n=0,1,2,3,4$) are \begin{align*} F_0 &= 2^{2^0}+1 = 3\\ F_1 &= 2^{2^1}+1 = 5\\ F_2 &= 2^{2^2}+1 = 17\\ F_3 &= 2^{2^3}+1 = 257\\ F_4 &= 2^{2^4}+1 = 65537, \end{align*} and each of these is a Fermat prime. Based on these results, one might conjecture (as did Fermat) that all Fermat numbers are prime. However, this fails for $n=5$: $F_5 = 2^{2^5}+1 = 4294967297 = 641 \cdot 6700417$. In fact, the primes listed above are the only Fermat numbers known to be prime.
Primes one more than a power of 2
Fermat primes are also the only primes in the form $2^m+1$.
Proof
Suppose that $m$ has an odd factor $a > 1$. For all odd $a$, we have by the Root-Factor Theorem that $x + 1$ divides $x^a + 1$. Since this is true as a statement about polynomials, it is true for every integer value of $x$. In particular, setting $x = 2^{m / a}$ gives that $2^{m / a} + 1 | 2^m + 1$, and since $2^m + 1 > 2^{m / a} + 1 > 1$, this shows that $2^m + 1$ is not prime.
It follows that if $2^m + 1$ is prime, $m$ must have no odd factors other than $1$ and so must be a power of 2.
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2021-03-06 18:20:21
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http://michaelsyao.com/computing-systems/x86-assembly-p3/
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# Reading and Writing x86 Assembly
Now that we have an idea of the basics of what goes into assembly code, let’s try actually reading and writing assembly ourselves! For our reference, here’s a link to the list of common instructions that we talked about last time.
As with all programming languages, the best way to familiarize ourselves with assembly code is to read a bunch of assembly code! In this way, we’ll be able to more easily identify common assembly function structures and decode what the assembly code is doing.
To go over some practice examples, let’s go over the following set of functions, which are currently written in C. We’ll discuss each of them one by one. Feel free to follow along using the following code sample.
### Converting from C Code to Assembly
On Linux systems, assembly code is written in .s files. This will be important when we eventually begin writing our own assembly code from scratch. If we’re starting from a .c C program file, there are two common ways to convert the C code in assembly code:
1. After compiling the .c program using gcc compiler, run the terminal command objdump -d xxxxx, where xxxxx is what the name of the original .c is. In our case, we will run the command objdump -d main. We haven’t formally talked about compile C code just yet, so don’t worry if you don’t know what gcc is just yet. Effectively, we’ll be running the two lines gcc main.c and then objdump -d main in terminal. (If you are using the repl.it plugin from above, pressing the green “Run” button does the same thing as running the gcc main.c command.)
2. After compiling the .c program (using the same method as described above), we can instead use the standard C debugger gdb to print out the relevant assembly code. The benefit of using this method is that we avoid printing out the C program initialization functions like __libc_csu_init, _start, and deregister_tm_clones, among others, that we have no interest in actually parsing through. To do this, instead of running objdump -d main, we’ll run the command gdb main first to boot up the gdb debugger, and then run the command disassemble xxxxx, where x is the name of the function we want to disassemble. For example, if we want to disassemble the function simpleInt() using this method, we’d run the following set of commands.
$gcc main.c$ gdb main
$(gdb) disassemble simpleInt There are other ways of converting from C code to assembly (and also vice versa), but for our discussion right now, this will prove sufficient. ### Starting Easy: Returning an Integer Let’s first start by taking a look at the simpleInt function, which simply returns the integer 6. Let’s run simpleInt in the main() function: int simpleInt(void) { return 6; } int main(void) { return simpleInt(); } After compiling and disassembling, the relevant segments of assembly code that correspond to this C code are here: 00000000004004f0 <simpleInt>: 4004f0: 55 push %rbp 4004f1: 48 89 e5 mov %rsp,%rbp 4004f4: b8 06 00 00 00 mov$0x6,%eax
4004f9: 5d pop %rbp
4004fa: c3 retq
4004fb: 0f 1f 44 00 00 nopl 0x0(%rax,%rax,1)
0000000000400640 <main>:
400640: 55 push %rbp
400641: 48 89 e5 mov %rsp,%rbp
400644: 48 83 ec 10 sub $0x10,%rsp 400648: c7 45 fc 00 00 00 00 movl$0x0,-0x4(%rbp)
40064f: e8 9c fe ff ff callq 4004f0 <simpleInt>
400654: 48 83 c4 10 add $0x10,%rsp 400658: 5d pop %rbp 400659: c3 retq 40065a: 66 0f 1f 44 00 00 nopw 0x0(%rax,%rax,1) The six-digit numbers before every instruction can be thought of as the “numbering” for each of the instructions and functions to be able to easily reference them. The pairwise hexadecimal numbers are the machine code associated with each instruction, which are the actual bytes that the computer will use to execute the corresponding instruction. Of course, to us as programmers, machine code is just a bunch of illegible bytes. The things that we are most interested in are the commands at the end of each of line. As we can see in the <simpleInt> function, the first two pair of instructions involve setting up the stack pointers in the primary memory. The first interesting instruction is mov$0x6, %eax, which moves the hexadecimal number $0x6 into the return 32-bit register %eax. After this single step of our function, we can see that the function restores the state of the stack, and then retq returns the 64-bit result stored in the register %rax. The command nopl essentially is just filler machine code and literally tells the computer to “do nothing” at that instruction. In the the <main> function, the first four instructions again involve something to do with setting up the stack that we don’t particularly care about. After that, the command callq <simpleInt> calls the function <simpleInt>, whose result we know is stored in %rax. After this instruction, the next two instructions at 400654 and 400658 restore something about the stack, and then we return from the <main> function using the retq command. Pretty straightforward! Problem 1: Try doing a similar analysis with the simpleChar() and arithmetic() operations functions to understand what they look like in assembly code. ### Local Variables and Function Arguments By studying the function variable() in C, which creates a local variable$x$, we can try to understand how local variables are handled in x86 assembly and by our computers. int variable(void) { int x = 6; return x; } int main(void) { return variable(); } Disassembling the above code gives the following x86 assembly code output: 0000000000400520 <variable>: 400520: 55 push %rbp 400521: 48 89 e5 mov %rsp,%rbp 400524: c7 45 fc 06 00 00 00 movl$0x6,-0x4(%rbp)
40052b: 8b 45 fc mov -0x4(%rbp),%eax
40052e: 5d pop %rbp
40052f: c3 retq
0000000000400640 <main>:
400640: 55 push %rbp
400641: 48 89 e5 mov %rsp,%rbp
400644: 48 83 ec 10 sub $0x10,%rsp 400648: c7 45 fc 00 00 00 00 movl$0x0,-0x4(%rbp)
40064f: e8 cc fe ff ff callq 400520 <variable>
400654: 48 83 c4 10 add $0x10,%rsp 400658: 5d pop %rbp 400659: c3 retq 40065a: 66 0f 1f 44 00 00 nopw 0x0(%rax,%rax,1) The <main> code looks more or less the same as the previous example, so let’s focus our attention on the <variable> function. As we can see in the movl$0x6,-0x4(%rbp) command, the number 0x6 is pushed onto the stack 4 bytes lower than the base stack address stored in the register -0x4(%rbp). This result agrees with our understanding that local variables are stored on the stack. In this case, the variable x will now correspond to the integer stored in the stack location -0x4(%rbp). In the next line, this value of x is moved from the stack to the return register %eax, and after restoring the stack state, we return the value of x in the retq command.
Problem 2: The function argument() also involves the use of two local variables: x and y. We know that the first argument of a function (which in this case is x) is stored in the register %rdi, and the second argument of a function (which in this case is y) is stored in the register %rsi. After analyzing the assembly code for argument(), is this the case?
### Recursion and Iteration
To take a look at how assembly code implements iterative algorithms like for loops or recursion, we can disassemble the recursion() and forLoop() functions, which both sum the positive integers up to the positive integer x. Let’s take a look at the recursion() disassembly first.
0000000000400550 <recursion>:
400550: 55 push %rbp
400551: 48 89 e5 mov %rsp,%rbp
400554: 48 83 ec 10 sub $0x10,%rsp 400558: 89 7d f8 mov %edi,-0x8(%rbp) 40055b: 83 7d f8 01 cmpl$0x1,-0x8(%rbp)
40055f: 0f 8f 0c 00 00 00 jg 400571 <recursion+0x21>
400565: c7 45 fc 01 00 00 00 movl $0x1,-0x4(%rbp) 40056c: e9 1b 00 00 00 jmpq 40058c <recursion+0x3c> 400571: 8b 45 f8 mov -0x8(%rbp),%eax 400574: 8b 4d f8 mov -0x8(%rbp),%ecx 400577: 83 e9 01 sub$0x1,%ecx
40057a: 89 cf mov %ecx,%edi
40057c: 89 45 f4 mov %eax,-0xc(%rbp)
40057f: e8 cc ff ff ff callq 400550 <recursion>
400584: 8b 4d f4 mov -0xc(%rbp),%ecx
400589: 89 4d fc mov %ecx,-0x4(%rbp)
40058c: 8b 45 fc mov -0x4(%rbp),%eax
40058f: 48 83 c4 10 add $0x10,%rsp 400593: 5d pop %rbp 400594: c3 retq 400595: 66 2e 0f 1f 84 00 00 nopw %cs:0x0(%rax,%rax,1) 40059c: 00 00 00 40059f: 90 nop At first glance, this assembly code looks much more complicated! It might be a bit more difficult to understand this code especially if we didn’t know the purpose of this code or the C source code beforehand, but nonetheless, we can still try to decode it piece by piece. Once again, the first three instructions have something to do with the stack, followed by moving the integer argument x (stored in %edi) onto the stack, as expected. Now, we have something interesting going on between the next four lines! We can break them down one by one: cmpl$0x1, -0x8(%rbp)
Based on the instruction immediately above, -0x8(%rbp) refers to the value of x passed into the function. This instruction compares the number 1 to x, and if they are equal, the ZF flag is set. In the next instruction, which is a jump instruction,
jg 400571 <recursion+0x21>
The function will jump to instruction 400571 (the mov instruction) only if 1<x. We can see that this is the base case. The commands at 400571 and 400574 seem to restore x into the registers %eax and %ecx, along with some addition register manipulations. Critically, we see that at instruction 400577, the instruction subtracts 1 from x and moves it into the argument register %edi, preparing for the recursive function call at 40057f. This is the essence of the recursive call. The value of x is also stored in the return register %eax from line 400571 to be added to with each recursive call, executed at instruction 40587.
Eventually, we’ll reach the base case where instruction 40055b will show that 1 == 1, such that the jump command at 40055f is not executed and we continue with 400565 and 40056c:
movl $0x1,-0x4(%rbp) jmpq 40058c <recursion+0x3c> This moves the number 1 onto the stack, which then jumps to instruction 40058f that moves the same number 1 at -0x4(%rbp) to the return register %eax at instruction 40058c that we jump to. After that, we return. As we can see, the critical instructions at 40055b through 40056c, and also at 40057f, are indications that we have a recursive function on our hands. Through walking through the function step by step, we’re able to see what the function does and how it works. What about for loop iterations? Let’s disassemble the forLoop() function: 00000000004005a0 <forLoop>: 4005a0: 55 push %rbp 4005a1: 48 89 e5 mov %rsp,%rbp 4005a4: 89 7d fc mov %edi,-0x4(%rbp) 4005a7: c7 45 f8 00 00 00 00 movl$0x0,-0x8(%rbp)
4005ae: c7 45 f4 00 00 00 00 movl $0x0,-0xc(%rbp) 4005b5: 8b 45 f4 mov -0xc(%rbp),%eax 4005b8: 3b 45 fc cmp -0x4(%rbp),%eax 4005bb: 0f 8d 17 00 00 00 jge 4005d8 <forLoop+0x38> 4005c1: 8b 45 f4 mov -0xc(%rbp),%eax 4005c4: 03 45 f8 add -0x8(%rbp),%eax 4005c7: 89 45 f8 mov %eax,-0x8(%rbp) 4005ca: 8b 45 f4 mov -0xc(%rbp),%eax 4005cd: 83 c0 01 add$0x1,%eax
4005d0: 89 45 f4 mov %eax,-0xc(%rbp)
4005d3: e9 dd ff ff ff jmpq 4005b5 <forLoop+0x15>
4005d8: 8b 45 f8 mov -0x8(%rbp),%eax
4005db: 5d pop %rbp
4005dc: c3 retq
4005dd: 0f 1f 00 nopl (%rax)
According to 4005a4, we moved the argument x onto the stack at the location -0x4(%rbp), so that at the cap compare instruction at 4005b8, we compare x to the value stored at %eax, which is 0 according to 4005ae at first pass. The incrementing variable stored at -0xc(%rbp) is incremented in the register %eax at 4005cd before being moved back to -0xc(%rbp), and continuing with the loop at the jump instruction at 4005d3. From instruction 4005c4, we can hypothesize that the result is being stored at -0x8(%rbp). This function to continues to iterate until the cmp command at 4005b8 returns equality, such that we jump to 4005d8 at the instruction 4005bb. At this point, we move the result being stored at -0x8(%rbp) to the return register %eax and return.
### More Complex Examples
Eventually, we will also learn about how to read assembly code dealing with manipulating arrays and objects in C, but this requires us to understand how to manipulate pointers in C and use dynamic memory management functions like malloc() and free() in stdlib.h. Since we haven’t talked about any of these features yet, we’ll hold up on reading more complex assembly code later.
As you can tell especially from the last two examples, reading and dissecting assembly code takes a lot of practice. When we learn more about the gdb debugging tool, we’ll learn about how to “step through” each individual assembly instruction line by line. If you also have 7 bucks to spare and some free time, there’s also a cool game based on decoding assembly language called TIS-100 that you can download from Steam.
## Writing Assembly Code
Now that we have an idea of what assembly code looks like, we can use our skills to begin directly programming in assembly.
## The Basics
First, let’s go over the basics of how to compile, execute, and analyze assembly code that we write.
### Assembly Code Format
All assembly code for x86 64-bit are written in files with the .s extension, and largely follows the same basic format:
.globl main
main:
# Write code here!
.data
The main section is where we actually write our instructions line by line. If we have any global static variables, we would include it in the .data section (of course, if there aren’t any global static variables, you can just leave the .data section out. The .globl main statement at the top allows this main function to be executed by other programs. It is important that we title this function main.
Of course, you can get more complicated if you’d like, but this is the simplest structure.
### Compiling Assembly Using gcc
Compile your .s file (we’ll refer to it as test.s) into an executable file using gcc. Here is the appropriate Linux command:
$gcc test.s -o test In the very off-chance that you want to compile as a 32-bit executable instead of a 64-bit executable, use the bash commands $ as --32 -o test.o test.s
$ld -o test test.o -m elf_i386 This is very rare nowadays and I don’t really expect 32-bit executables to come up in most contexts, but if it does, now you know how to compile them. ### Running Executable Objects After compiling, run your executable using the bash command $ ./test
Let’s say that your executable function had three arguments: arg1, arg2, and arg3. You can run your executable with the the three arguments using the bash command
$./test arg1 arg2 arg3 ### Printing the Return Value to stdout There are a number of different ways to accomplish this task, but perhaps the easiest is to run this single-line bash command after running an executable is complete: $ echo $? ### Using syscalls At this point, we’re going to take a (fairly lengthy) tangent and discuss something called system calls. syscall stands for a system call, and is used when a program wants the kernel of a computer to perform some task. The kernel can be thought of as the “king” or “brain” computer program that is at the core of a computer’s operating system with complete control over everything in the system. It tells other programs when they can run, how much memory and computer resources they get, and whatnot. It can be thought of as the bridge between computer applications/programs and the CPU, memory, and other resources. Generally, we as programmers don’t particularly work with syscalls since they deal with such a secure and important component of the computer. However, in some special cases, syscalls can be used in our programs to make certain “elementary” operations easy. There are a number of syscalls that are available to programs when interacting with the kernel. We’ll list them here, although note that many of these terms/functions will be new and we won’t be able to understand what each of them does just yet (many of these will come back to us throughout the course). Type of syscall Linux Function Process Control fork() exit() wait() File Management open() read() write() close() Device Management ioctl() read() write() Information Maintanence getpid() alarm() sleep() Communication pipe() shmget() mmap() (This table was taken from this link. Official syscall documentation is linked here.) Each of these functions has an important use, and can be invoked at the assembly level. In x86 64-bit assembly, the syscall calling convention is largely standardized. We move the appropriate arguments to each of the registers, and then invoke the syscall instruction. Each of the syscall functions has their own system call number associated with the function. The register designations are Register Function for Setting Up syscalls %rax system call number %rcx return address (often we don't care about) %r11 saved flags (again, often we don't care about) %rdi arg0 %rsi arg1 %rdx arg2 %r10 arg3 %r8 arg4 %r9 arg5 Note that invoking a syscall will never change the state of the stack (this is useful oftentimes). #### Legacy System Interruption On older systems and in older textbooks, you may see the commands int 0x80 instead. This instruction is the “legacy” version of syscall on older systems, and runs by triggering a software interrupt that interrupts the kernel’s attention on the currently executing program to handle the interrupt instead. Because of the interrupt nature of the int 0x80 command, it is considered to be computationally costly and slower than the more modern syscall command. It also uses an entirely different calling convention with different system call numberings, so in general it’s just a bad idea to mix the two with each other. However, note that because of backwards compatibility of Intel processors, int 0x80 still works on modern computers. We will just choose not to use it. ### Printing Going back to writing assembly code, we can use the write() syscall that we discussed above to write to stdout and basically output things in terminal. For example, let’s say that we wanted to print the string Hello, World!. When dealing with strings, it is often most convenient to store it in a static data region (analogous to global variables) in x86. For example, .globl main main: movq$1, %rax # System call 1 is write()
movq $1, %rdi # File handle 1 is stdout mov$message, %rsi # Address of string to output
mov $14, %rdx # Number of bytes (length of string) syscall # Invoke write() operation movq$60, %rax # System call 60 is exit()
movq $0, %rdi # Return code 0 syscall .data message: .ascii "Hello, World!\n" As you can see from above, we used syscall to both write to stdout and also exit() from the function (which is essentially the same as returning). We created one message that is a string (as indicated by the .ascii designation), and we can refer to this message in our assembly instruction as $message. Note that %rsi needs to store an address to the thing to be printed, not the actual thing being printed!
Of course, can store more than just strings in the static data region. For example,
.data
var:
.byte 7 # Declare a byte at location var with value 7
.byte 11 # Declare a byte at location var+1 with value 11
x:
.short 42 # Declare a 2-byte at location x with value 42
y:
.long 500 # Declare a 4-byte at location y with value 500
z:
.quad 256 # Declare a 8-byte at location z with value 256
arr:
.long 1, 2, 3 # Declare 3 4-byte values as an array with values
# 1, 2, and 3. 1 is at location arr, 2 is at
# location arr+4 (since a long is 4 bytes), and
# 3 is at location arr+8
## Examples
Now, it’s your turn! Try to write assembly code for the following tasks.
### deadbeef
Using the write() syscall, print the characters deadbeef\n to stdout without using any global static variables.
.globl main
main:
# Print deadbeef\n to stdout here
### Factorials
Write an x86 assembly program to compute the factorial of a number initially stored in %rdi.
.globl main
main:
movq $5, %rdi # Argument n (replace 5 with n) # Return n! ### Fibonacci Numbers Write an x86 assembly program to compute the$n$th Fibonacci number, where$n$is initially stored in %rdi. .globl main main: movq$9, %rdi # Argument n (replace 9 with n)
# Return the nth Fibonacci number
### Greatest Common Divisor
Using Euclid's Algorithm, calculate the greatest common divisor (GCD) of two numbers $x$ and $y$ initially stored in the registers %rdi and %rsi, respectively. Here is an implementation in Java so you can understand the algorithm:
public static int gcd(int x, int y) {
if (x == 0) { return y; }
return gcd(y % x, x);
}
Here is a template for the assembly implementation:
.globl main
main:
movq $9, %rdi # Argument x (replace 9 with x) movq$6, %rsi # Argument y (replace 6 with y)
# Return the gcd of x and y
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2021-04-17 16:45:22
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March 22, 2023, 09:36:27 AM
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Topic: What are some good starting chemicals? (Read 42220 times)
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pantone159
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Re: What are some good starting chemicals? help arg! no responses
« Reply #15 on: January 01, 2007, 05:53:48 PM »
I think 10% HCl is a reasonable strength. That's usually what I use. BTW - If you aren't experienced in handling chemicals, I'd definitely avoid concentrated sulfuric and nitric acids.
mafiaparty303
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Re: What are some good starting chemicals? help arg! no responses
« Reply #16 on: January 01, 2007, 05:57:45 PM »
Thank you mark, (FOR MAKING ME NOT SO IGNORANT!!!), I'll keep that in mind about the nitric acid and sulfuric.
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pantone159
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Re: What are some good starting chemicals? help arg! no responses
« Reply #17 on: January 01, 2007, 06:13:54 PM »
P.S. Two ounces is plenty for very many chemicals. I have significantly smaller amounts (e.g. 10 g) of many of the chems in my collection. For the most part, if you keep things at a small scale, you can do very many experiments with that much.
Woelen's site is really great for understanding which chems will be interesting.
P.P.S. Out of the chemicals you listed, I think that boric acid, granular al, and the silver disk would be the least interesting. Use Al foil instead for most things, super cheap. The silver disk should look real purty, but one of the few easy experiments is to tarnish it with the sulfur, and then it will look horrible. (You can try cleaning it with Al foil and baking soda, but it won't get back to its original luster.) Also add ammonia solution from the grocery store (no extra colorings), it forms a very attractive complex with copper.
mafiaparty303
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Re: What are some good starting chemicals? help arg! no responses
« Reply #18 on: January 01, 2007, 06:19:20 PM »
Ok, thanks, ya I decided to not buy the boric acid since I didn't see any use in it.
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EccentricHeather
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Re: What are some good starting chemicals?
« Reply #19 on: January 01, 2007, 08:19:57 PM »
After reading the various exchanges in this thread, I have a better idea of your current level of experience. I would agree that you should avoid concentrated acids at this point. I like concentrated acids because they are less expensive, take a lot less space, and are easy to dilute to desired strength. But I did work my way up to this; when I had less experience, I bought a pint of 10% (dilute) sulfuric acid and found it to be reasonably easy to work with. Note that even dilute acids can have their hazards, and you need to treat them with respect. Although I started with sulfuric, I would recommend that you start with dilute hydrochloric. 10% is good. I also use approx. 3% on occasion. It is a bit less hazardous, and more of the beginning experiments, such as those described in the "Golden Book" use it. Sodium bisulfate (you can find it on eBay) is a good substitute for sulfuric acid in many experiments. Another common dilute acid, which is quite safe to work with, is plain white household vinegar (acetic acid - usually about 5%). But note that this acid is quite dangerous in concentrated ("glacial") form, so don't buy that until you have more experience.
Two ounces should be fine to get an introduction to the properties of sodium hydroxide. However, if you get into soap making, you could go through a lot of it, so you might eventually need to order a larger quantity. It is on Hi-Valley's website. I haven't ordered from Snowdrift since I can get it locally, but I would assume that they don't sell fractional quantities. Note that many chemical retailers (including Hi-Valley) won't sell to people who are under 21. If you're under 21, you may need help from your parents in ordering chemicals. Long gone are the days when a teenager could walk into a chemical supply shop and buy just about anything. Given the nature of today's society, that's probably a good thing.
One note on the "Golden Book"... If you are just getting started, I would recommend performing many of the experiments in it; that would be a great way to learn. However, safety standards in the 1960's weren't quite what they are now. I can't believe that a book designed for children and teenagers contains instructions for making chlorine, nitrogen dioxide, hydrogen sulfide, and sulfur dioxide. If I were you, I would avoid those experiments until you have more experience - those gases are hazardous. (But when you are ready, by all means perform those experiments in a very well-ventilated area.) Also, take care with the experiments that generate hydrogen - keep the quantities down. The chloroform experiment seems rather scary too. If you see an interesting-looking experiment there or elsewhere and want opinions on its safety, post here and I'm sure you will get a good response. While it could use a few revisions to remove some of the most dangerous experiments, I think that book is a great introduction to chemistry. It's a shame there isn't a revised version of it in print now.
Heather
mafiaparty303
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Re: What are some good starting chemicals?
« Reply #20 on: January 01, 2007, 09:07:57 PM »
Ya, I also noticed it says to taste the HCl and other things, I'm not porfessional but I don't think that should be done.
And yes I'm 16 and my parents will have to order the chemicals. Hopefully i'm gonna get to ordering the equipment and chemicals today. I might just get some of the chemicals mentioned in the book and then work from there.
I was looking for the Golden Book yesterday , as much as $600 for it now!!! Thats just crazy, and the book is just awsome, so much stuff to learn. But ya I'm guessing they didn't focus on safety so much in the 60's, but I'm gonna be safe and not try any of the harmful ones. I've already made Chlorine gas before (not the greatest thing to make in yoru kitchen), and there are somethings on the internet that don't help. I found an site with experiments and one of them was the electolysis of saltwater, the products of the electrolysis for them was, oxygen and hydrogen. Nothing mentioned about the poisonous chlorine formed. I'm gonna make a list of the chemicals that I want mentioned in the book and order those. Becuase I don't want to order chemicals I have no use for (and know no use for) and then not use them at all. And the golden book is by far the best experiment book I've seen after looking everywhere. Filippo EDIT: In the golden book it says that if you sniff too much chlorine then have some household ammonia around and sniff that if you get too much chlorine. Why so? « Last Edit: January 01, 2007, 10:02:13 PM by mafiaparty303 » www.lafamillemasse.com/pierre/smart_people.htm Join the Smart People for a better future! -Founded by Pierre Masse EccentricHeather • Regular Member • Posts: 19 • Mole Snacks: +2/-0 • Gender: Re: What are some good starting chemicals? « Reply #21 on: January 02, 2007, 12:26:19 AM » Ammonia will react with chlorine; I would imagine that the reaction product (perhaps ammonium chloride or chloramines) depends on the conditions and on what else is around. I haven't checked on the details, so I don't know precisely what compounds would result, but the product is likely a lot less toxic than the chlorine. When working with halogens, it is a good idea to have a neutralizing chemical handy. For bromine and iodine, a solution of sodium thiosulfate seems to be the most popular. There are iodine experiments in the "Golden Book"; if you do them, you might want to have some sodium thiosulfate on hand to clean up any spills or to get rid of any excess iodine that you don't need. And don't make bromine until you have gained a lot of experience. I haven't done that yet, but I may try it one of these days if I feel brave. (And since I have mentioned halogens, fluorine is only for experts with the proper equipment; I doubt I will ever attempt that.) For that matter, it is good to have some sort of neutralizing chemical handy when working with anything that might create problems if it gets out of control. In case of acid spills, I have sodium bicarbonate (baking soda) around, and I have vinegar available in case of caustic spills. (Heavily diluted solutions of the stronger acids and bases might work too.) And if I am burning something, I have a way of putting out the fire. I have been careful enough that I haven't ever had a significant acid or caustic spill, but recently I had a burning liquid (methanol) spill onto the table. (I was burning calcium chloride in it to watch the pretty orange flame.) I had a huge container of water nearby, and I dumped it on the fire immediately. That could have been ugly if I hadn't been ready for it. As it was, the only harm done is that a few things got wet - the table didn't even get damaged. (So far, I don't burn things that can't be put out by water, but if I ever do, I'll have the right fire suppression equipment handy.) No matter how careful you are, something is bound to go wrong eventually, so part of your lab setup should include neutralizing chemicals and ways to put out fires. A large jug of water and a fire extinguisher are inexpensive items that could save your butt some day. (Again, understand what you are burning. Water is a very bad idea for certain types of fires; for example, fires involving alkali metals.) While we're at it, does anyone else have suggestions on spill containment and fire suppression? These are important parts of a basic lab setup. Heather Mitch • General Chemist • Administrator • Sr. Member • Posts: 5298 • Mole Snacks: +376/-3 • Gender: • "I bring you peace." -Mr. Burns Re: What are some good starting chemicals? « Reply #22 on: January 02, 2007, 12:45:22 AM » Do not sniff a chemical to make up for over sniffing an other one. The point is not to of sniffed it in the first place, because you performed the experiment outside and downwind to your nose. Most Common Suggestions I Make on the Forums. 1. Start by writing a balanced chemical equation. 2. Don't confuse thermodynamic stability with chemical reactivity. 3. Forum Supports LaTex mafiaparty303 • Regular Member • Posts: 97 • Mole Snacks: +0/-0 Re: What are some good starting chemicals? « Reply #23 on: January 02, 2007, 01:00:10 AM » For the spill containment won't sand help, at least so the spill dosn't expand anymore than it already has? @mitch haha true words I've been looking for potassiunm iodide everywhere, the only thing I've found is pills, and I don't want to order lots of chemicals from ebay since they are all separate meaning separate shipping costs... are there any other sites (other than hvchemicals) that sell im small amounts, (like united nuclear)? EDIT: wow....hvchemicals isnt that great.... theres a$25 plus fee for hazardous material, thats not taking into mind the actual HCl cost and the shipping thats like..$15.. I need to find a better place (I'm thinking about buying some of the radioactive marables for$10 from U.N. and some free radioactive ore comes with them, should I store this in some lead or is the radioactivity level too little to cause harm?)
« Last Edit: January 02, 2007, 05:12:37 AM by mafiaparty303 »
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billnotgatez
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Re: What are some good starting chemicals?
« Reply #24 on: January 02, 2007, 06:00:26 AM »
This is a FYI post
EccentricHeather -
The Golden Book you mention has been discussed many places on this forum before. For many of the experienced chemist that post here, this book is scary. Within it are serious safety issues. So many in fact that one is unable to caution the novice against them all. As I understand it from a historical perspective the book was deemed dangerous enough that most all the libraries removed it from the shelf and the publisher decided not to continue taking on the legal hazards. There is in fact a theory out there that the government actually worked to stop its publication. Therefor the price has become very high and Internet piracy has ensued. Of course anything semi banned takes on a cult following which will disregard the bad points. In any case this forum does not support piracy. Many of the experience chemist regulars are chagrined at the mention of this book due to its safety issues. For myself as a citizen scientist I hate seeing things get censored, but at the same time I fear for the well being of the novice. If you could let us drop discussion of this book and talk more about the individual projects a citizen chemist can do. By the way there are other pages on this forum that discuss good books that are more acceptable.
mafiaparty303 -
There is nothing that will cure the lost in trust by your fellow housemates than to do an experiment that makes the house unlivable. A nasty smell that permeates throughout the house will get chemistry banned forever. When I experiment I do it out in the garage which is not attached to the house or outdoors upwind. It even concerns me when someone says they experiment in the basement. This is especially true when I find out they do not have a fume hood.
For anyone else reading this post -
Let us hope we can continue to do good science without getting hurt. As we all know doing helps us learn.
As an aside, there are sales of fireworks in Alabama with no problem, but in New York if you use a sparkler you can get arrested. Even within the USA the rules are hard to understand.
« Last Edit: January 02, 2007, 06:06:39 AM by billnotgatez »
billnotgatez
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Re: What are some good starting chemicals?
« Reply #25 on: January 02, 2007, 06:02:49 AM »
Just to mention
This forum has a sticky at the top of it that discusses labs
Labs, labs, labs...
And the sub forum to this forum has 2 sticky post at the top that discuss sources.
+ Chemical Forums: Chemistry Forum, Chemistry Question, Chemistry Help
|-+ Chemistry Forums for Students of Chemistry
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mafiaparty303
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Re: What are some good starting chemicals?
« Reply #26 on: January 02, 2007, 02:01:40 PM »
Ok, ya I'm definetly not doing anything in my room if its anything potentially hazerdous, I'm going out today and building a small work table with my dad. Gonna add some shelves for storage and put it in my garage, it is connected to my house but only the the door that leads to the house. No vents going into the other rooms or anything. As for ventilation I can leave the garage door open, if not, bail out and run outside ( ). As for "the book" I won't discuss it, and where are the books mentioned? I have yet to find a book showing chemistry experiments that's as good as "the book".
Anyone know where I can get Hydrochloric Acid? HVchemicals has it but they charge and extra $25 plus HCl cost and Shipping and Handeling. And thats just too much. Plus some Potassium Iodide Heres what I have yet to find. HCl Potassium Iodide Calcium Oxide Sodium Hydroxide Sodium Biosulfate Sodium Carbonate Potassium Permangenate (U.N has them but they are sold out right now ) and finally Iron Sulfate im going to Walgreens (drugstore) to see if they have any of those... hopefully they might. Anyone know where I can get them? www.lafamillemasse.com/pierre/smart_people.htm Join the Smart People for a better future! -Founded by Pierre Masse pantone159 • Mole Herder • Chemist • Full Member • Posts: 492 • Mole Snacks: +54/-6 • Gender: • A mole of moles doesn't smell so nice... Re: What are some good starting chemicals? « Reply #27 on: January 02, 2007, 02:36:29 PM » (I'm thinking about buying some of the radioactive marables for$10 from U.N. and some free radioactive ore comes with them, should I store this in some lead or is the radioactivity level too little to cause harm?)
The marbles need no special precautions. (I keep mine just in a glass vial (as my display U sample) just like any non-hazardous element.)
Radioactive ores, on the other hand, *MUST* be stored OUTSIDE. They emit radon, which is not something you want to build up inside your home.
You can find several chemicals at the grocery store/hardware store. For example (based on my experience in the USA):
NaOH - 'Red Devil Lye'. This is supposedly getting harder to find, but it still exists. Danger - Very corrosive to people. Gets hot when dissolved in water.
HCl - 'Muriatic Acid'. Strength varies.
Na2CO3 - Poor quality Na2CO3 is available as Arm & Hammer Washing Soda (not baking soda) which comes in a yellow box in the laundry detergent aisle
NaHCO3 - Baking soda
MgSO4 - Epsom salt
NH3 - Household ammonia. You want the stuff without any extra colors/scents, typically the cheap generic stuff.
Collecting chems and apparatus becomes a perpetual process. You are never done.
mafiaparty303
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Re: What are some good starting chemicals?
« Reply #28 on: January 02, 2007, 03:22:20 PM »
Yep, well ya, I decided to buy some flat glass marbles off Ebay way cheaper 40 for $8. No radioactive ore unfortunatly but I wouldnt want to buy the lead and everything. Anyone know where I can get some potassium Iodide? Can I just buy the tablets and then crush them and put them in a plastic container? « Last Edit: January 02, 2007, 03:30:44 PM by mafiaparty303 » www.lafamillemasse.com/pierre/smart_people.htm Join the Smart People for a better future! -Founded by Pierre Masse woelen • Chemist • Full Member • Posts: 277 • Mole Snacks: +40/-2 • Gender: • The art of wondering makes life worth living... Re: What are some good starting chemicals? « Reply #29 on: January 02, 2007, 03:25:33 PM » Sometimes it is good to make dangerous things, but only in VERY small quantities. It gives you an impression of how dangerous they actually are and that kind of knowledge is important. So, I encourage a starter to make some chlorine, but only in VERY small amounts. Carefully sniffing gives you the smell of it, so you will recognize it, when you accidently make it in another experiment. The same is true for SO2, Br2, NO2. Make them in a test tube (using 100 mg of chems, not more) and carefully sniff them. Not by sticking your nose in the test tube, but by wafting some of the gas towards your nose with your hand. Really, it learns you a lot. If I may suggest a set of chems, which are interesting, and which can all be purchased at a single shop (PM for info, I don't post addresses over here, and I decide whether I give the address or not based on post history), then I would suggest the following: 50 g KI 100 g KBr 100 g Na2SO3 100 g Fe(NO3)3.9H2O (better than chloride, the nitrate does not form a complex with ferric ion in aqueous solution) 100 g FeSO4.7H2O 100 g K2Cr2O7 (careful: carcinogen, but otherwise not really dangerous, but allows a LOT of really funny/colorful experiments in aqueous solution). 100 g KMnO4 (careful: very potent oxidizer, much more so than K2Cr2O7, only use in aqueous solutions when you are inexperienced). 100 g K2S2O8 100 g NaNO2 (this is my favorite, no other chemical is good for so many good aqueous chemistry experiments) 100 g NaSCN (nice complex formation with many metals) 100 g KCr(SO4)2.12H2O (chrome alum, funny and cheap) 100 g CuSO4.5H2O 100 g KOH (or NaOH) 250 g NaHCO3 250 g Na2S2O3.5H2O (good old hypo, dirt cheap, yet funny chem) 100 g Na2B4O7.10H2O (borax) 100 g K3Fe(CN)6 (potassium ferricyanide, non-toxic, despite the cyanide ligands, very versatile chem, good very many experiments) 100 g K4Fe(CN)6.3H2O (potassium ferrocyanide, idem) 50 g Na2S.xH2O (caution: makes H2S in contact with acids, but it is a nice versatile chem, which however, must be kept very well stoppered) Some chems, which you should not spend money on if you are not a pyro-hobbyist: NaNO3 KNO3 KClO4 Na2SO4 Nitrates and perchlorates are very inert in aqueous solution, unless highly concentrated, or at extremely low pH (between -1 and 0). Sodium sulfate is as energetic as a dead dog and cannot react with anything. Not interesting at all. Locally you should try to obtain 1 liter NH3 (5% is OK) 100 ml H2O2 (3% is OK) 1 liter HCl (30%) 1 liter H2SO4 (32%, battery acid, be careful with the 96% stuff, 32% is relatively safe) 100 grams TCCA (trichloro isocyanuric acid, swimming pool slow acting organic chlorine, unfortunately this mostly only is available in kilo-packages) 1 liter NaClO (4% is OK, common household bleach) A few gallons of distilled water. This is the only chemical which I use in large quantities. I do all of my aqueous transition metal chemistry experiments in distilled water. Really, spend the money for this, it is worth it. If you buy it in jerrycan quantities it is quite cheap anyway. HNO3 probably will be very hard to obtain in the USA. Over here in NL it is easy to obtain at 52% but I know that at the other side of the ocean things are quite different. It would be nice though if you could obtain that, but you do not need it to start with. It really is best to obtain H2SO4, it is a very versatile acid, more so than HCl. The chloride ion coordinates to many transition metal ions and for many experiments that is not what you want. NaHSO4 is a very good and safe alternative and can be purchased locally as pH-minus for swimmingpools. From a ceramics and pottery supplier you can obtain the following very interesting transition metal compounds for really low prices: 100 gram V2O5, vanadium pentoxide 100 gram CoCO3, cobalt carbonate (don't buy Co3O4, it is inert and does not dissolve in any solvent) 100 gram NiCO3.Ni(OH)2, (basic) nickel carbonate 100 gram CuCO3.2Cu(OH)2, (basic) copper carbonate 100 gram MoO3, molybdenum trioxide, makes molybdates easily with solutions of NaOH or KOH) 100 gram ZnO, zinc oxide Another interesting chemical may be silver nitrate, AgNO3 (available from the same source as the chems, I mentioned above), but it is fairly expensive. You could consider buying 10 grams, but still, it costs$10 or more.
With the list, given here, you can do a LOT of aqueous chemistry experiments and also quite some nice (colored) gas experiments. All these experiments can be done safely, also at home, but of course, you should use your brains and not do stupid things. Always use small amounts (max. 100 mg, small spatula of solid, with a few ml of water or acid added to dissolve it). With that you simply won't have serious accidents, because quantities are too low to be really dangerous. The only organ which could be severely damaged with such small amounts is the eye, so wear glasses or goggles ALWAYS when experimenting.
Want to wonder? See http://www.oelen.net/science
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2023-03-22 13:36:27
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Practise solving quadratic equations algebraically with this self-marking exercise.
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This is level 2; Two terms where the unknown is a factor of both. The roots are integers. You can earn a trophy if you get at least 7 correct.
$$x^2 − 7x = 0$$ x = and $$x^2 + 5x = 0$$ x = and $$x^2 − 8x = 0$$ x = and $$x^2 + 6x = 0$$ x = and $$x^2 − 5x = 0$$ x = and $$x^2 + 8x = 0$$ x = and $$2x^2 − 4x = 0$$ x = and $$3x^2 + 9x = 0$$ x = and $$4x^2 − 16x = 0$$ x = and $$10x^2 − 3x = 0$$ x = and $$5x^2 + 2x = 0$$ x = and $$5x^2 = 2x$$ x = and
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Factorising - Factorise algebraic expressions in this structured online self marking exercise.
Level 1 - A quadratic equation presented in a factorised form.
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Level 3 - Three terms where the squared term has a coefficient of one. The roots are integers.
Level 4 - Three terms where the squared term has a coefficient other than one and the expression factorises.
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More Algebra including lesson Starters, visual aids, investigations and self-marking exercises.
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Example Level 2
$$x^2 + 7x = 0$$
The expression on the left of this equation can be factorised as each of the two terms contains a factor of $$x$$.
$$x(x + 7) = 0$$
Here there are two terms which multiply together to give zero. It is therefore true that at least one of the terms must be zero.
So either $$x=0$$
Or $$x+7=0$$ which means $$x=-7$$
So the two answers are 0 and -7
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2019-06-18 17:58:06
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http://hackage.haskell.org/package/Agda-2.5.1.2/docs/Agda-TypeChecking-Monad-Signature.html
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Agda-2.5.1.2: A dependently typed functional programming language and proof assistant
Synopsis
# Documentation
Add a constant to the signature. Lifts the definition to top level.
setTerminates :: QName -> Bool -> TCM () Source #
Set termination info of a defined function symbol.
modifyFunClauses :: QName -> ([Clause] -> [Clause]) -> TCM () Source #
Modify the clauses of a function.
addClauses :: QName -> [Clause] -> TCM () Source #
Lifts clauses to the top-level and adds them to definition.
Add a section to the signature.
The current context will be stored as the cumulative module parameters for this section.
lookupSection :: (Functor m, ReadTCState m) => ModuleName -> m Telescope Source #
Lookup a section. If it doesn't exist that just means that the module wasn't parameterised.
Arguments
:: ModuleName Name of new module defined by the module macro. -> Telescope Parameters of new module. -> ModuleName Name of old module applied to arguments. -> Args Arguments of module application. -> Ren QName Imported names (given as renaming). -> Ren ModuleName Imported modules (given as renaming). -> TCM ()
Module application (followed by module parameter abstraction).
Add a display form to a definition (could be in this or imported signature).
Find all names used (recursively) by display forms of a given name.
Check if a display form is looping.
Can be called on either a (co)datatype, a record type or a (co)constructor.
Does the given constructor come from a single-constructor type?
Precondition: The name has to refer to a constructor.
class (Functor m, Applicative m, Monad m) => HasConstInfo m where Source #
Minimal complete definition
Methods
Lookup the definition of a name. The result is a closed thing, all free variables have been abstracted over.
Lookup the rewrite rules with the given head symbol.
Instances
Source # Methods (HasConstInfo m, Error err) => HasConstInfo (ExceptionT err m) Source # Methods
Look up the polarity of a definition.
Look up polarity of a definition and compose with polarity represented by Comparison.
setPolarity :: QName -> [Polarity] -> TCM () Source #
Set the polarity of a definition.
Get argument occurrence info for argument i of definition d (never fails).
Get the mutually recursive identifiers.
setMutual :: QName -> [QName] -> TCM () Source #
Set the mutually recursive identifiers.
Check whether two definitions are mutually recursive.
Why Maybe? The reason is that we look up all prefixes of a module to compute number of parameters, and for hierarchical top-level modules, A.B.C say, A and A.B do not exist.
Get the number of parameters to the current module.
getDefFreeVars :: (Functor m, Applicative m, ReadTCState m, MonadReader TCEnv m) => QName -> m Nat Source #
Compute the number of free variables of a defined name. This is the sum of number of parameters shared with the current module and the number of anonymous variables (if the name comes from a let-bound module).
Compute the context variables to apply a definition to.
We have to insert the module telescope of the common prefix of the current module and the module where the definition comes from. (Properly raised to the current context.)
Example: module M₁ Γ where module M₁ Δ where f = ... module M₃ Θ where ... M₁.M₂.f [insert Γ raised by Θ]
Unless all variables in the context are module parameters, create a fresh module to capture the non-module parameters. Used when unquoting to make sure generated definitions work properly.
Instantiate a closed definition with the correct part of the current context.
Give the abstract view of a definition.
inAbstractMode :: TCM a -> TCM a Source #
Enter abstract mode. Abstract definition in the current module are transparent.
inConcreteMode :: TCM a -> TCM a Source #
Not in abstract mode. All abstract definitions are opaque.
ignoreAbstractMode :: MonadReader TCEnv m => m a -> m a Source #
Ignore abstract mode. All abstract definitions are transparent.
Enter concrete or abstract mode depending on whether the given identifier is concrete or abstract.
Check whether a name might have to be treated abstractly (either if we're inAbstractMode or it's not a local name). Returns true for things not declared abstract as well, but for those makeAbstract will have no effect.
Andreas, 2015-07-01: If the current module is a weak suffix of the identifier module, we can see through its abstract definition if we are abstract. (Then treatAbstractly' returns False).
If I am not mistaken, then we cannot see definitions in the where block of an abstract function from the perspective of the function, because then the current module is a strict prefix of the module of the local identifier. This problem is fixed by removing trailing anonymous module name parts (underscores) from both names.
Get type of a constant, instantiated to the current context.
Get relevance of a constant.
The name must be a datatype.
The number of parameters of a definition.
The number of dropped parameters for a definition. 0 except for projection(-like) functions and constructors.
isProjection :: HasConstInfo m => QName -> m (Maybe Projection) Source #
Is it the name of a record projection?
Is it a function marked STATIC?
Is it a function marked INLINE?
Returns True if we are dealing with a proper projection, i.e., not a projection-like function nor a record field value (projection applied to argument).
Number of dropped initial arguments of a projection(-like) function.
Check whether a definition uses copatterns.
Apply a function f to its first argument, producing the proper postfix projection if f is a projection.
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2020-09-28 10:08:53
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https://www.physicsforums.com/threads/linear-algebra-spectral-decompositions-eigenvectors-of-projections.396584/
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Homework Help: Linear algebra - Spectral decompositions: Eigenvectors of projections
1. Apr 19, 2010
TorcidaS
1. The problem statement, all variables and given/known data
Let P1 and P2 be the projections defined on R^3 by:
P1(x1, x2, x3) = (1/2(x1+x3), x2, 1/2(x1+x3))
P2(x1, x2, x3) = (1/2(x1-x3), 0, 1/2(-x1+x3))
a) Let T = 5P1 - 2P2 and determine if T is diagonalizable.
b) State the eigenvalues and associated eigenvectors of T.
2. Relevant equations
3. The attempt at a solution
For a), I believe it is diagonalizable because P1 + P2 gives us (x1, x2, x3). Although I'm could be wrong on that...
It's mainly b) that I'm concerned for. By the theorem, (T = c1P1 + c2P2...+.. where c are eigenvalues) 5 and -2 are the eigenvalues (although this sort of confuses me because I had thought the eigenvalues of projections are always 1 and 0).
How can we find the eigenvectors? Had this been a matrix it's simple, subtract the eigenvalue from the main diagonal, simplifiy, and find the nullspace.
Also, the solution to b) is for eigenvalue 5, the eigenvectors are <(1,0,1),(0,1,0)> and for eigenvalue -2, the eigenvector is <(-1,0,1)> (so my guess some matrix is formed?)
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2018-12-18 17:44:35
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