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http://mathhelpforum.com/math-topics/134069-density-human-body-question.html
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# Math Help - Density of Human Body Question
1. ## Density of Human Body Question
The density of the human body is estimated to be 0.95 g/cm^3
Calculate the volume of a person wighing 79.99kg
Express your answer in the form a x 10^n, where n is and element of N and 1 <a<10
2. ## Well
0.95g = 1cm^3 = 0.00095kg
79.99kg = x cm^3
$\frac{79.99}{0.95}$=what you want
u know how to convert that to the notation needed?
The density of the human body is estimated to be 0.95 g/cm^3
Calculate the volume of a person wighing 79.99kg
Express your answer in the form a x 10^n, where n is and element of N and 1 <a<10
Density is defined by: $density = \dfrac{mass}{volume}$ thus $volume = \dfrac{mass}{density}$ .
$V=\dfrac{79.99\ kg}{0.95\ \frac{g}{cm^3}}$
$V=\dfrac{79990\ g}{0.95\ \frac{g}{cm^3}}=84200\ cm^3$
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2015-05-05 02:17:38
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https://proofwiki.org/wiki/Definition:Circular_Relation
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# Definition:Circular Relation
Let $\mathcal R \subseteq S \times S$ be a relation in $S$.
$\mathcal R$ is circular if and only if:
$\left({x, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R \implies \left({z, x}\right) \in \mathcal R$
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2019-11-20 07:53:25
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https://tfischernet.wordpress.com/2010/08/
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# Thomas Fischer's Weblog
Life, Linux, LaTeX
## Some new LaTeX Beamer Tricks
The Swedish summer fades into a rainy autumn, which means it is time again to prepare lectures for the coming months. As a long-time LaTeX user, I use the LaTeX Beamer package for all my slides. During the last few months, I have developed some new helper commands and environments for LaTeX slides, which I want to present here.
## The multiframe environment
The Beamer package’s frame environment has the option allowframebreaks to automatically split a frame once it becomes full or split frames manually using the \framebreak command. Although this solution works, it has two drawbacks:
1. No animation steps can be included using e.g. \only or \pause.
2. The first frame is such a sequence of “split” frames sets its content with a slightly different vertical offset. This is very annoying, because flipping between slides makes the content jump up and down.
Therefore I designed my own solution. Although it is not perfect, it does not show above flaws. The idea is to define a new environment called multiframe via newenvironment and accept one optional parameter. This new environment acts as a replacement for the known frame environment. An important difference is the optional parameter: Whereas Beamer’s original frame parameter accepts a list of options such as allowframebreaks or fragile, multiframe‘s parameter is the frame’s title. Thus, the frame title is no longer set via \frametitle but with this parameter.
Now, the trick to get a sequence of frames is to omit the optional parameter with subsequent frames. In this case, the multiframe environment assumes that this new frame is part of sequence, reuses the previous frame title, but increases the sequence counter as shown in the title.
% new counter to now which frame it is within the sequence
\newcounter{multiframecounter}
% initialize buffer for previously used frame title
\gdef\lastframetitle{\textit{undefined}}
% new environment for a multi-frame
\newenvironment{multiframe}[1][]{%
\ifthenelse{\isempty{#1}}{%
% if no frame title was set via optional parameter,
% only increase sequence counter by 1
}{%
% new frame title has been provided, thus
% reset sequence counter to 1 and buffer frame title for later use
\setcounter{multiframecounter}{1}%
\gdef\lastframetitle{#1}%
}%
% start conventional frame environment and
% automatically set frame title followed by sequence counter
\begin{frame}%
\frametitle{\lastframetitle~{\normalfont(\arabic{multiframecounter})}}%
}{%
\end{frame}%
}
Drawback with this solution is that it creates a completely new frame, whereas \framebreak can be used (nearly) everywhere in the frame. For example, you can use \framebreak inside an itemize list and split this list. In my solution, you have to close all environments up to the current multiframe and open again all environments as desired.
### Example
Without my new environment, you would set plain, single frames using the frame environment. Nothing changes here:
\begin{frame}
\frametitle{The Frame's Title}
Some content \ldots
\end{frame}
Things get different for sequences of frames. With plain LaTeX Beamer, you would do something like this:
\begin{frame}[allowframebreaks]
\frametitle{The Frame's Title}
Some content \ldots
\framebreak
More content on next page
\end{frame}
With my new environment, the corresponding solution would look like this:
\begin{multiframe}[The Frame's Title]
Some content \ldots
\only<2>{Only seen in animation step}
\end{multiframe}
% this is like a frame break
\begin{multiframe}% no optional parameter here!
More content on next page
\end{multiframe}
When you include photos, images, or drawing on your slides, you should always give credit to the original creator and show license and origin. Some licenses require you to state where you got the image from and which rights have been granted. For someone who wants to reuse your slides or parts of it, it clarifies under which conditions the materials are accessible. From a legal aspect, it suffices in most cases to show all relevant information next to the included image, or collected at the end of your document in a single list.
In the next-to-the-image case, it is common to set the information text in fine print along the bottom or right border of the image. But how to do it in LaTeX? For the solution I am presenting here, I am using a tikzpicture environment to set both the image and the accompanying text in two nodes next to each other. The new command \copyrightbox takes two mandatory parameters and one optional parameter. The first mandatory parameter is supposed to contain the command showing the image or drawing, e.g. an \includegraphics command. The second parameter takes the copyright statement and is usually plain text. The optional parameter is a single character describing the position of the text in relation to the image. Possible values include b (below), l (left), and r (right), where “right” is default if nothing else is specified.
Now, the tricky part is that the image should determine the copyrightbox‘s size and the text should follow this format. In particular, if the text is below the image, the text (more specific, the node containing the text) should be node wider than the image (more specific, the node containing the image); instead it should wrap around consuming several lines if necessary. The same applies to the height in case of left or right alignment. To get the width and height of a node in TikZ, I had to search for a proper solution on the Internet resulting in a quite long statement for commands ydiff and xdiff.
\makeatletter
\newdimen\tu@tmpa%
\newdimen\ydiffl%
\newdimen\xdiffl%
\newcommand\ydiff[2]{%
\coordinate (tmpnamea) at (#1);%
\coordinate (tmpnameb) at (#2);%
\pgfextracty{\tu@tmpa}{\pgfpointanchor{tmpnamea}{center}}%
\pgfextracty{\ydiffl}{\pgfpointanchor{tmpnameb}{center}}%
}
\newcommand\xdiff[2]{%
\coordinate (tmpnamea) at (#1);%
\coordinate (tmpnameb) at (#2);%
\pgfextractx{\tu@tmpa}{\pgfpointanchor{tmpnamea}{center}}%
\pgfextractx{\xdiffl}{\pgfpointanchor{tmpnameb}{center}}%
}
\makeatother
\begin{tikzpicture}%
\node[inner sep=0pt,minimum size=2em](ciimage){#2};
\usefont{OT1}{phv}{n}{n}\fontsize{4}{4}\selectfont
\ydiff{ciimage.south}{ciimage.north}
\xdiff{ciimage.west}{ciimage.east}
\ifthenelse{\equal{#1}{r}}{%
\node[inner sep=0pt,right=1ex of ciimage.south east,anchor=north west,rotate=90]%
{\raggedleft\color{black!33}\parbox{\the\ydiffl}{\raggedright{}#3}};%
}{%
\ifthenelse{\equal{#1}{l}}{%
\node[inner sep=0pt,right=1ex of ciimage.south west,anchor=south west,rotate=90]%
{\raggedleft\color{black!33}\parbox{\the\ydiffl}{\raggedright{}#3}};%
}{%
\node[inner sep=0pt,below=1ex of ciimage.south west,anchor=north west]%
{\raggedleft\color{black!33}\parbox{\the\xdiffl}{\raggedright{}#3}};%
}
}
\end{tikzpicture}
}
### Example
To use this new command, you simply set the mandatory parameters with some useful values:
\copyrightbox{\includegraphics[width=0.5\linewidth]{images/osmphoto}}{CC-BY-SA, Thomas Fischer}
\copyrightbox[b]{\includegraphics[width=0.5\linewidth]{images/osmphoto}}{CC-BY-SA, Thomas Fischer}
The first example above creates a copyrightbox containing an image as loaded from images/osmphoto.jpg (pdflatex searches for known extensions if you do not specify one), scaled to 50% of the line width, and the copyright statement “CC-BY-SA, Thomas Fischer”. The text will be written on the image’s right side, starting in the lower right corner going upwards. In the second line, the text’s alignment is changed to b: Here the text starts in the lower left corner and goes to the right as usual.
Be aware that the copyrightbox‘s total size is a combination of both the image and text. It will thus be larger than the image alone (as specified with width or height).
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2018-03-22 09:49:12
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https://ec.gateoverflow.in/1926/gate-ece-2010-question-39
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3 views
The Boolean function realized by the logic circuit shown is
1. $F = \sum_{m} (0, 1, 3, 5, 9, 10, 14)$
2. $F=\sum_{m}(2,3,5,7,8,12,13)$
3. $F=\sum_{m}(1,2,4,5,11, 14,15)$
4. $F= \sum_{m}(2,3,5,7,8,9,12)$
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2022-10-03 04:14:59
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https://physics.stackexchange.com/questions/159887/how-to-interpret-spin-observables-constructed-by-non-standard-phase-choices
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# How to interpret spin observables constructed by non-standard phase choices?
$$J_{\pm} \ket{\ j,m}=e^{i\phi} \hbar \sqrt{j(j+1)-m(m\pm1)} \ket{\ j,m\pm1}$$
Normally, we set $\phi=0$ for all choices of m for a fixed j. Now, the standard explanation for this is that the choice of phase is arbitrary but must be followed consistently (e.g. Edmonds or Condon, Shortley).
I would like to point out some issues with this explanations and then show what happens if we try to play with the phase choice. The most important point will be that we will allow the choice of a different phase $\phi=\phi(m)$ for different m's as opposed to choice of same phase for every m (the special case of which is the standard choice of $\phi=0$ for all m's). This idea comes from the fact that there is no (at least obvious) reason to claim that it follows from the standard derivation of matrix elements of ladder operators that the phases must be the same for all m's.
We will look at the example of spin=1 matrices. The standard choice of $\phi=0$ for all matrix elements yields the following matrices of ladder operators and then the observables of spin in z,x and y direction:
$$j_z=\begin{bmatrix} 1 & 0&0\\ 0& 0&0\\ 0& 0&-1 \end{bmatrix}, j_+=\sqrt{2} \begin{bmatrix} 0 & 1&0\\ 0& 0&1\\ 0& 0&0 \end{bmatrix}, j_-=\sqrt{2}\begin{bmatrix} 0 & 0&0\\ 1& 0&0\\ 0& 1&0 \end{bmatrix},$$
$$j_x=\frac{1}{\sqrt{2}} \begin{bmatrix} 0 & 1&0\\ 1& 0&1\\ 0& 1&0 \end{bmatrix}, j_y=\frac{1}{\sqrt{2}}\begin{bmatrix} 0 & -i&0\\ i& 0&-i\\ 0& i&0 \end{bmatrix}$$
Now, if we make a choice of setting the phase for $j_+\ket{1,0}$ to 0, $\phi(m=0)=0$, but for $j_+\ket{1,-1}$ we set it to a different phase, $\phi(m=-1)=\pi/2$, we construct the following matrices:
$$j_z^{(2)}=j_z=\begin{bmatrix} 1 & 0&0\\ 0& 0&0\\ 0& 0&-1 \end{bmatrix}, j_+^{(2)}=\sqrt{2} \begin{bmatrix} 0 & i&0\\ 0& 0&1\\ 0& 0&0 \end{bmatrix}, j_-^{(2)}=\sqrt{2}\begin{bmatrix} 0 & 0&0\\ -i& 0&0\\ 0& 1&0 \end{bmatrix},$$
$$j_x^{(2)}=\frac{1}{\sqrt{2}} \begin{bmatrix} 0 & i&0\\ -i& 0&1\\ 0& 1&0 \end{bmatrix}, j_y^{(2)}=\frac{1}{\sqrt{2}}\begin{bmatrix} 0 & 1&0\\ 1& 0&-i\\ 0& i&0 \end{bmatrix}$$
Now, to make things clear, I should point out that the new $j_x$ and $j_y$ have the same eigenvalues as the standard ones, and satisfy the same old algebra of angular momentum, though their eigenstates are different. As a result, the expected values of these different observables are different in some superpositions of $\ket{1,m}$, which is interesting, since that is what we observe in experiments.
As an example, here is a graph which shows how the expected values of $j_x$ and $j_y$ on the state $\ket{\psi}=\frac{1}{\sqrt{3}}\left(\ket{1,1}+\ket{1,0}+\ket{1,-1}\right)$ change when we keep the phase $\phi(0)=0$ (the same as above) but we change $\phi(-1)$ (which is for the above matrices set to $\pi/2$). In the graph, $\avg{j_x}$ is in red and $\avg{j_y}$ is in blue.
As a contrast, here is a graph which shows how the expected values of $j_x$ and $j_y$ on the state $\ket{\psi}=\frac{1}{\sqrt{3}}\left(\ket{1,1}+\ket{1,0}+\ket{1,-1}\right)$ change when we change both phases simultaneously and equally: $\phi(0)=\phi(-1)=\phi$ which is the standardly considered phase choice (in textbooks) and is equivalent to rotating coordinate system around z-axis. That is apparent in the way expected values of $j_x$ and $j_y$ interchange as we change $\phi$. In the following graph, $\avg{j_x}$ is again in red and $\avg{j_y}$ is in blue.
One interesting point is that the observables we construct by choosing different phases for different matrix elements of ladder operators, i.e. $\phi(0) \neq \phi(-1)$, can't be written as linear combination of the standard operators (even with complex coefficients).
Another interesting thing to note is that these different observables arise only for spin>=1, since for spin=1/2 only one matrix element of ladder operators is different from 0.
My question is, does anyone know of an interpretation of these "$j_x$" and "$j_y$" we construct by different phase choices? Further on, does anyone know of an argument I am not aware of, for choosing the same phase for every matrix element as opposed to my choice of different phases (to me it seems that the standard choice is just a special case of a more general choice, but I don't yet understand how to interpret the results)?
• You may want to use Joined->True option for ListPlot in your plots since it doesn't really make any sense to show discreteness of your sampling points, and it just clutters the plots. – Ruslan Jan 19 '15 at 11:55
• As for the question itself: did you change the $\ket{\psi}$ accordingly when you changed the $\phi$'s (especially important for the first plot)? – Ruslan Jan 19 '15 at 11:57
• If I do that, the graph again doesn't look similar to the standard expected values graph. Then again, I don't understand why would I have to change $|\psi>$, am I missing something very important here? – bsoda Jan 19 '15 at 19:55
• Indeed, it shouldn't make any difference. BTW, why do you plot $\avg{j_i}^2$ instead of just $\avg{j_i}$? – Ruslan Jan 20 '15 at 7:32
• Fair point, it's fixed now. – bsoda Jan 20 '15 at 8:02
The fact that your transformed operators have the same spectrum of eigenvalues and the same algebra comes from the fact that redefinitions of the phase lead to unitarily equivalent operators, i.e. they are related by $$\tilde j_+ = U^\dagger j_+ U,\quad \tilde j_- = U^\dagger j_- U,\quad \tilde j_z = U^\dagger j_z U$$ with a unitary matrix $U$ and the states in the two bases (different phase choices) are related by $$\langle \tilde \phi \vert = \langle \phi \vert U, \qquad \vert \tilde \phi \rangle = U^\dagger \vert \phi \rangle.$$ Since $j_z$ and $j_\pm$ span the entire angular momentum algebra, any operator leaving the angular momentum $\vec J^2$ invariant can be written as combination of $j_z$ and $j_\pm$. Since any product of these will transform by getting a $U^\dagger$ on the left and a $U$ on the right, we have $$\tilde O = U^\dagger O U.$$
Now the expectation value of any operator can be computed in both bases and we find $$\langle \tilde \phi \vert \tilde O \vert \tilde \phi \rangle = \langle \phi \vert U U^\dagger O U U^\dagger \vert \phi \rangle = \langle \phi \vert O \vert \phi \rangle.$$ This is without choosing any specific phase convention, but making sure we follow the convention consistently with the $U$ matrices.
Now since you find a difference between different conventions there should be a detail you overlooked. Did you remember that if the state $\vert 1, -1\rangle$ gets a phase $\phi$, the state $\langle 1, -1 \vert$ gets the phase $-\phi$?
• Exactly, (for the second comment) $j_z$ and U commute and so $j_z$ stays the same. – bsoda Jan 20 '15 at 16:13
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2019-06-20 01:24:21
|
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https://www.physicsforums.com/threads/reading-an-acceleration-vs-time-graph.232566/
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# Reading an Acceleration Vs. Time Graph
1. May 1, 2008
### Yuma
1. I need to find the velocity of an object with a constant acceleration of 0.5 ft/sec by creating an acceleration vs. time graph spanning the interval of 0 to 5 seconds
2. This is kind of a theoretical problem, so if any equations apply, they were not given to us
3. I have plotted acceleration vs. time, which in this instance is a straight line, and velocity vs. time, which is linear with a constant slope. My question is this: if acceleration is constant, and velocity is the area under that "curve" ( this is a little confusing, as the assignment calls this a curve and it doesn't look like a curve to me!), can I multiply the X coordinate by Y, which =0.5, to get my velocity at various times along this interval? This seems to make sense, as the areas under the "curve" are rectangles given the constant acceleration, but it seems too easy to be right!
Last edited: May 1, 2008
2. May 1, 2008
### lzkelley
When physicists/mathematicians use the term "curve" they often mean any plot on a graph / any continuous line (straight, or curved).
Velocity is the area underneath the acceleration "curve." Similarly, distance is the area underneath the velocity "curve."
To find the velocity at a point in time, you find the area of the given rectangle (from time = 0 until whatever time you are loking for). To find the displacement (position) at a point in time, you find the area of the triangle in the velocity graph.
The process of finding the area under a graph leads to what is called an Integral --> its the precise, mathematical way of finding the area under any shape. Using an integral, you can more easily convert from acceleration to velocity; velocity to distance; or in general convert from the change of something -> to what that thing actual is.
3. May 1, 2008
### Feldoh
Yes, you are correct in your method. This is a good intro to basic integral calculus. See if you can follow this:
In your case your acceleration is constant, so the area under the curve is the area of a rectangle.
But what if your acceleration was a(t)=t^2 and you want to know the change in velocity on the time interval [0,5]? Then the acceleration is not a uniform line, so we cannot just find the area using a rectangle.
So how are you going to compute the area under that curve? Well you can still approximate the area under that acceleration curve by making a rectangle, however it's not a very good approximation.
What if instead to split the portions of the graph up into 5 smaller rectangles, each with a width of "1 second". Now if we add these smaller rectangles to get the collective area we will find that they approximate the acceleration a little better.
Now say you split up the rectangles up into 10 rectangles and add them instead of 5, now your approximation is even better. What about adding 50 small rectangles? Now it's a lot better. This is the basic principal of integral calculus.
Here's the leap: What if we shrink the width of those rectangles to an really, really tiny width? Say we make the width of the rectangles infinitely small in their width, now you have an approximation that is so good it's not an approximation anymore, but the actual, exact area under that acceleration curve.
Analyzing graphs that do not have uniform areas becomes somewhat trivial with the help of calculus, this is one reason why it is so valuable.
Using some notation it would look like this:
$$\int_{0}^{5}t^2 dt$$ Your area is length * width so your length is x^2, and your width is an infinitely small dx. Basically it's the process of summing up a infinite number of infinitely small pieces of area.
Last edited: May 1, 2008
4. May 1, 2008
### Yuma
Thanks!
Thanks for helping! I just started doing integrals and Reimann sums a couple of weeks ago in calculus, but the problem I'm working on is for an engineering class. I thought that since we are approximating rectangles to find approximations of the area under the curve in calculus, if the area was rectangular in shape already that I could skip the calculus and just take the area of the rectangle itself. Happily for me, that appears to be the case!
Thanks again!
5. May 1, 2008
### Feldoh
Yeah if you can find the area an easier way, normally in the shape of a trapezoid or a rectangle, then you're free to do that, it's the same thing as the integral, just easier.
You took the area 0.5*5
The integral is just
$$\int_{0}^{5}0.5 dt = 0.5*5$$
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2018-02-19 13:00:47
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https://www.cuemath.com/maths/measuring-weight/
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# Measuring Weight
Go back to 'Mensuration and Measurement'
## Introduction to Measuring Weight
When you go to the shop to buy a vegetable, you need to tell the shopkeeper whether you need half a kilogram of cauliflower or a kilogram of potatoes. While doing Chemistry experiments, every gram of the various compounds and mixtures need to be taken in the correct weight and measure, for the experiment to proceed smoothly. These are just a few examples which show how important it is to be able to measure weight accurately.
## The Big Idea: Measuring Weight
### A simple idea: The difference gravity makes
Before we get into details of weight measurement, it is important to understand the two terms mass and weight, which are often interchangeably used, although they mean two different things. The mass of an object is a measure of the actual material contained in it. The weight, on the other hand, is an indicator of the force exerted by the earth's gravitational force. The confusion arises because the unit of measurement of both mass and weight is gram or kilogram.
Let us understand this difference more easily taking the example of the earth and the moon. We do know that the earth's gravitational pull is about six times stronger than that on the moon. As a result, a person weighing 60 kgs on earth would weigh only ten kgs on the moon! The mass of the person of weight 60 kgs would be around six kgs because the gravitational constant is about 10 kg m/s2
### Units of weight
The metric units of weight measurement are similar to those for length. We have the following units of weight measurement:
Milligram
Centigram
Gram
Kilogram
Metric Ton
Do you how much a blue whale weigh? It tips the scales at more than 200 tons. The lightest mobile phone in the world weighs a mere 40 grams. There are subatomic particles which are not even visible to the human eye, so you can imagine how light they are. Here is a small graphic that shows which units should be used for weighing different items.
### Conversion of Units
We spoke about the different units used to describe different weights. But it is not enough to know what the units are, and we also need to know how to convert one unit to the other. A candy bar might weigh just a few grams, but a whole carton of candy bars might need to have its weight described in kilograms. I could describe the weight of a mini truck in kilograms, but it would be more convenient to express its weight in tons. So, it is important to be aware of how to convert any of these units to the other.
So here are the rules for converting all the metric units of weight, and we will start from the bigger to the smaller units.
1 tonne = 1000 kilograms
1 kilogram = 1000 grams
1gram = 100 centigrams
1 centigram = 10 milligram
So, if you need to convert 1.25 tons to smaller units, this is how the calculation would be done:
1.25 tons = (1.25 x 1000) kilograms = 1250 kilograms
1250 kilograms = 1250000 gram
What about the other way round? Say you want to convert 57 milligrams into bigger units? This is how the calculation would look like:
\begin{align} & 57{\text{ milligrams}} = \left( {\frac{{57}}{{10}}} \right){\text{ centigrams}} = 5.7{\text{ centigrams}}\\ & 5.7{\text{ centigrams}} = \left( {\frac{{5.7}}{{100}}} \right){\text{ grams}} = 0.057{\text{ grams}}\\ & 0.057{\text{ grams}} = \left( {\frac{{0.057}}{{1000}}} \right){\text{ kilograms}} = 0.000057{\text{ kilograms}} \end{align}
As you can imagine, further conversion to tons would give an impossibly small number, so we are letting it go since you get the drift, don't you?
### Some alternative units
We only spoke about the SI units of weight measurement here, but in some special circumstances, some older units of weight are also used, like ounces and pounds. 1 pound is equal to 16 ounces. 1 kilogram is equal to 2.20 pounds, and 1 ounce is 28.35 grams. Keeping these basic conversions in mind is useful.
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2019-08-20 04:57:12
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https://mathoverflow.net/questions/239688/cube-lemma-on-a-cofibrantly-generated-almost-model-category
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# Cube Lemma on a cofibrantly generated (almost) model category
Suppose I have a complete and cocomplete category $\mathscr{C}$ with two sets of maps $I,J$ that are the candidates for generating (trivial) cofibrations on a model structure on $\mathscr{C}$. The only property I'm left to check in order to apply the recognition principle for cofibrantly generated model categories is that pushouts of maps in $J$ are weak equivalences.
In this framework, does the cube lemma hold already?
More precisely, if I have two spans $B_i \leftarrow A_i \rightarrow C_i$ for $i \in \{1,2\}$, where all the objects are cofibrant and $A_i \rightarrow B_i$ is a cofibration $\forall \ i \in \{1,2\}$, plus I'm given a pointwise weak equivalence of spans, is the induced map $$B_1 \coprod_{A_1} C_1 \to B_2 \coprod_{A_2} C_2$$ a weak equivalence as well?
If I understand your situation correctly, the answer is yes, the cube lemma holds, but you cannot use that to get a full model structure. Your situation often arises when trying to transfer a model structure from a cofibrantly generated model category $M$, along an adjunction $F:M\leftrightarrows N: U$, to a bicomplete category $N$, e.g. to a category $N$ of algebras over a monad. Check out Lemma 2.3 in Schwede-Shipley Algebras and Modules in monoidal model categories. For them also, the only difficulty is proving that the pushout is a trivial cofibration in $N$ (defined via $J$-cell) is a weak equivalence. The good news is that, even without this condition, you can probably prove whatever you want. Specifically, the structure on $N$ may be that of a semi-model structure. Have a look at Definition 2.3 in Spitzweck's thesis. Note that on page 14 he remarks that the cube lemma holds in this setting. Indeed, Karol's answer is spot on: in a semi-model category, pushouts of spans $A\gets B \to C$, where $B \to C$ is a trivial cofibration and $A$ is cofibrant, yield trivial cofibrations.
Even if your $N$ is not arising via a transfer along an adjunction, it may still be a semi-model category as defined by Benoit Fresse in 12.2.1 of this book. Again, the cube lemma holds. In general, any statement about model categories holds for semi-model categories if you restrict to the subcategory of cofibrant objects, or cofibrantly replace everything in sight. Since the cube lemma is already about cofibrant objects, there is no problem.
Unfortunately, it does NOT follow that every semi-model category is a model category. An example is the category of non-reduced symmetric operads in $M = Ch(\mathbb{F}_2)$. As a category of algebras over a $\Sigma$-cofibrant colored operad $P$, it is a semi-model category, by Spitzweck's Theorem or by Fresse 12.2.A applied to the transfer from the category of collections $\prod_{n \in \mathbb{N}} M^{\Sigma_n}$. However, it is not a full model structure: the pushout of $P(0)\to P(K)$, where $K$ is an acyclic chain complex $C$ in level zero and 0 otherwise, along the map $P(0)\to Com$ to the terminal non-reduced symmetric operad, is not a weak equivalence, because it introduces a summand of $(C\otimes C) / \Sigma_2$.
My advice: stick with the semi-model structure and use it to prove whatever you need. This was the approach taken by Spitzweck, by Goerss-Hopkins in their obstruction theory paper, by Fresse, by Hovey in his paper Monoidal Model Categories, and in many of my papers. A semi-model structure is good enough.
• Thank you very much David for your exhaustive reply! I'll have a look at the references you mentioned and see what I can get :) – Edoardo Lanari May 26 '16 at 1:26
This is not an answer in a full generality, but it is certainly not the case if the domains of generating acyclic cofibrations are cofibrant. (I have a hard time thinking of an example of a model category where this is not true, but there are probably some.)
Assuming that the "Cube Lemma" holds (it's more often called the Gluing Lemma), take $A_2 \to B_2$ to be any generating acyclic cofibration and let $A_2 \to C_2$ be a morphism to a cofibrant object. Set $A_1 = B_1 = A_2$ and $C_1 = C_2$ and fill the cube with obvious maps. Then the Gluing Lemma says that the pushout of $A_2 \to B_2$ along $A_2 \to C_2$ is a weak equivalence.
• I get your point, but maybe with some work it would be possible to get the Glueing lemma, hence the final piece I'm missing. I mean, in the end this is not a disproval yet! – Edoardo Lanari May 25 '16 at 14:20
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2021-04-12 14:00:26
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http://forum.wilmott.com/viewtopic.php?f=8&t=101728
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SERVING THE QUANTITATIVE FINANCE COMMUNITY
PierreBoulez
Topic Author
Posts: 12
Joined: February 17th, 2019, 4:33 pm
### Arbitrage when risk-free portfolio earns less than riskless portfolio
Hi,
I'm currently reading Paul Wilmott's excellent book on option pricing. Near the beginning, he constructs a risk-free portfolio using an option, and a short on the underlying to hedge the risk. I'm specifically interested in European options.
A no-arbitrage argument follows:
• If this portfolio earns more than the risk free rate: borrow money at the risk-free rate, buy the portfolio, and make money off the arbitrage.
• Conversely: short the portfolio, invest money in a risk-free instrument, and again make money off the arbitrage.
I've scoured the internet, but couldn't find an explanation for the second argument, which I have a hard time grasping! By shorting the portfolio, we short an option, and "short a short", meaning we go long on the stock.
So, when we short the portfolio, we might even have to spend additional money, if shorting the option didn't give enough money to buy the stock.
This segment focuses on the binomial model, so I've tried separating this to 3 cases:
1. When in both the up and down state the option is worth more than 0. In this case, the arbitrage relies on buying the amount of stock that can be had by exercising the option. I have a hard time finding arguments to why in this case the option should be worth more than the stock at the period before expiration.
2. When in both the up and down state the option is worth 0. I understand this case, the option is worth 0 at the turn before expiration, and the hedging is a degenerate case (longing 0 stocks).
3. When in the up state the option is worth > 0, and in the down state the option is worth = 0. Like in case 1, I can't find a good argument.
As you can see, I'm out of answers. I don't even understand why a risk-less portfolio must earn the risk-free rate (as we can easily make a risk-less portfolio that always loses money, or doesn't lose and doesn't earn). Anyone has a clue?
Thank you!
Last edited by PierreBoulez on February 20th, 2019, 9:09 am, edited 1 time in total.
bearish
Posts: 5429
Joined: February 3rd, 2011, 2:19 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
Most of us have slightly idiosyncratic ways of explaining this point, so rather than trying to map it into your representation of Paul's, I'll try and give you my own. Consider a world with no transaction costs, no limits to shorting, and a constant interest rate that you can freely borrow or lend at; and a stock with known price dynamics, e.g. binomial. Somebody shows you a derivative contract, e.g. a European call option, written on this stock, along with a price that you can either buy or sell it for. If you can construct a self-financing trading strategy that exactly replicates the cash flow(s) generated by holding the derivative contract; in the case of the European call, max(S(T)-K,0) at the maturity date T, then the initial cost of initiating this replicating strategy is the theoretical (no-arbitrage) value of the derivative. If its quoted price is higher, you sell it and follow the replicating strategy. In this case you will have some cash in hand at time 0 and, regardless of what happens to the stock price, all future cash flows will net out between the derivative and the replicating portfolio. Likewise, if the quoted price is lower than the initial cost of the replicating strategy, you sell (or write, in the lingo) the option and follow the opposite of the replicating strategy. You will, once again, have some cash upfront and no further net gains or losses. In this setting, and this is important, you should not be able to (and certainly not easily) "make a risk-less portfolio that always loses money", since flipping its signs will give you a portfolio that always makes money, which is more or less the definition of an arbitrage opportunity (a free lunch!).
PierreBoulez
Topic Author
Posts: 12
Joined: February 17th, 2019, 4:33 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
Thank you bearish!
Construction of a replicating portfolio is actually the one thing I do understand, but my problem is with a risk-free portfolio.
You wrote "you sell (or write, in the lingo) the option and follow the opposite of the replicating strategy. You will, once again, have some cash upfront", and that's my exact problem - how can you know that you will have some cash upfront? The portfolio, by definition, doesn't replicate the cash flow generated by holding the derivative, as the portfolio is risk-free!
bearish
Posts: 5429
Joined: February 3rd, 2011, 2:19 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
Ah - I see the problem. The replicating portfolio by itself is not risk free. It is the combination of the replicating portfolio (or hedge, if you want) and the derivative that is risk free, because one offset the other. Does that make sense?
PierreBoulez
Topic Author
Posts: 12
Joined: February 17th, 2019, 4:33 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
Thank you for all your replies!
I understand why the replicating portfolio together with the hedge is risk free ("constructs a risk-free portfolio using an option, and a short on the underlying to hedge the risk").
The whole crux of the arbitrage argument in this case lies on the possibility of shorting option + underlying to get money, buy bonds that give the risk free rate (that has a higher return than the portfolio of option + underlying), and then in the next turn you've made "free" money.
So: shorting an option definitely gives you money, but you need to short a short (= buy) the underlying, too. So, you need to expend money for the underlying. Is there a way to prove that no matter what the conditions are, you will need to use less money than what you got for the option? In this case, you will be able to use the "change" and put it in a risk-free investment, and you will have succeeded in creating arbitrage.
Paul
Posts: 10494
Joined: July 20th, 2001, 3:28 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
One thing that's v annoying is how this turns into accountancy. You have some cash, you buy something, you sell something, then you see how there's no risk and so you should get the risk-free rate, but you have to look at the movement in the cash, the option, the stock, it gets irritating.
So what I sometimes do to explain things is to remove one of the moving parts, the cash. I start by assuming that you've inherited from Great Aunt Maud two things, an option and a short stock position. Then look at what happens to that. And ask whether you'd rather have this inheritance or the cash.
PierreBoulez
Topic Author
Posts: 12
Joined: February 17th, 2019, 4:33 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
Hi Paul,
Thanks so much for answering, I feel honored!
Even if you inherited an option and a short, because the return of option + short is less than the risk free, we want to show an arbitrage opportunity. You described that arbitrage can be done like so:
1. Sell your European option, get money.
Although you have received an inheritance from Aunt Maud, you need money for step 2.
If step 1 generated strictly more money than needed by 2, invest the change in the risk-free rate and you've succeeded in arbitrage.
But, if step 1 generated strictly less money as needed by 2, you'll need to borrow at the risk free rate.
I have a hard time understanding how, under these circumstances, it is possible to make risk-less profit.
Last edited by PierreBoulez on February 18th, 2019, 10:09 am, edited 1 time in total.
Paul
Posts: 10494
Joined: July 20th, 2001, 3:28 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
No, you didn't inherit any money from Aunt Maud! You inherited an option and a short stock position!
PierreBoulez
Topic Author
Posts: 12
Joined: February 17th, 2019, 4:33 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
Thank you so much for keeping an eye on this thread!
I'm not sure I'm following - if you inherited a European option and a short stock position, and will have waited a single turn for the expiration of the option, by the construction of this portfolio you will have made a risk-free return, but under our assumption this return is less than that of the risk-free rate.
I'm trying to understand how under this condition (return < risk-free rate) there is an arbitrage opportunity.
Paul
Posts: 10494
Joined: July 20th, 2001, 3:28 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
Slow down!
Try like this: Call with value 1 or 0 tomorrow. Stock worth 100 now, tomorrow worth 101 or 99.
You inherited 1 x option - 0.5 x stock.
Tomorrow you have -99/2 whether stock rises or falls, no?
PierreBoulez
Topic Author
Posts: 12
Joined: February 17th, 2019, 4:33 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
Thank you for being patient! I'm trying to slow down but I still don't understand
Yes, tomorrow the value of the portfolio is the same, as we constructed it to be risk-free. I recognize that this is the example from the book, and that under a risk-free rate of 0 assumption it means that the value of the portfolio tomorrow is identical to its value today, in both cases 0.5.
How can I progress from here and add a risk-free rate that is strictly larger than 0?
In my question the risk-free rate is strictly larger than 0, and we work under the assumption that the return from the portfolio is strictly less than the risk-free rate.
Paul
Posts: 10494
Joined: July 20th, 2001, 3:28 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
It doesn't matter whether the interest rate is zero or positive. Either way you can see that tomorrow your inheritance will be worth -99/2. So PV that to today. Your inheritance today will be worth
-99/2 /(1+r dt)
But you know it's worth -0.5 * 100 + V, where V is the market price. So V= 100/2 -99/2 /(1+r dt) is "special."
If you see that V> that special value in the market then you'll be pleased that Great Aunt Maud left you the portfolio and not the cash equivalent. If V< that special value you'll wish she'd left you a cheque not the portfolio.
PierreBoulez
Topic Author
Posts: 12
Joined: February 17th, 2019, 4:33 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
Why will my inheritance today be worth that amount? It seems that your answer already assumes that the return on the portfolio equals the risk free rate, but I'm looking for proof that it does!
When assuming it earns more than the risk free rate, there is a contradiction to the no-arbitrage argument:
You can borrow money at the risk-free rate, buy the portfolio, and make a risk-free profit.
Given that:
1) The return on the portfolio is strictly less than r,
2) The amount of stock needed for hedging costs less than the option,
You can also find a contradiction to the no-arbitrage argument.
But, given that:
1) The return on the portfolio is strictly less than r,
2) The amount of stock needed for hedging costs more than the option,
How is it possible to arbitrage?
By shorting the option you get some money, but will need to borrow extra to buy the stock for hedging.
If I understood no-arbitrage correctly (and from this confusion I assume I didn't), the argument in this case (where portfolio return < r) requires you to sell portfolio, thus getting money, and then investing said money and getting the risk-free rate. Then, on the next turn, you've made more money than the portfolio, and can (in that order): sell stock, pay back the short on the option, and still have "left-over" money that you've made at completely 0 risk.
I have scoured the internet for the past few days, looking for papers/books/posts that explain an arbitrage opportunity under these conditions and just didn't. Maybe when assuming point number 2 it is possible to reach some kind of logical contradiction?
Paul
Posts: 10494
Joined: July 20th, 2001, 3:28 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
I only read your first sentence! If I know I'm getting $1 tomorrow then that is worth$1/(1+r dt) today. Forget options and stocks. Are you ok with that concept?
P
PierreBoulez
Topic Author
Posts: 12
Joined: February 17th, 2019, 4:33 pm
### Re: Arbitrage when risk-free portfolio earns less than riskless portfolio
Definitely ok with it! How to proceed now?
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2020-04-08 03:04:38
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https://tkwant.kwant-project.org/doc/dev/reference/tkwant.manybody.html
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# 3.4. tkwant.manybody – Solving the many-body time-dependent Schrödinger equation¶
## 3.4.1. States¶
Solving the many-body time-dependent Schrödinger equation.
State(syst[, tmax, occupations, params, …]) Solve the time-dependent many-particle Schrödinger equation. WaveFunction(psi_init, tasks[, comm]) Evolve a many-particle wavefunction in time.
## 3.4.2. Helper functions¶
lead_occupation([chemical_potential, …]) Set the occupation ($$T, \mu, f(E)$$) for one lead. calc_intervals(spectra, occupations[, …]) Return a list of momentum intervals to perform the manybody integration. split_intervals(intervals, number_subintervals) Divide each interval in number_subintervals equidistant intervals. combine_intervals(intervals[, atol, rtol]) Group leads for similar intervals together. calc_tasks(intervals, spectra, occupations) Return all tasks (set of quantum numbers that uniquely identify a onebody state) that will form the manybody state. calc_initial_state(syst, tasks[, …]) Calculate the initial manybody scattering wave function using MPI. calc_energy_cutoffs(occupations) Extract upper and lower energy cutoffs from the lead occupations. Make boundstates time dependent, such that they can evolve in time. add_boundstates(solver, boundstate_psi, …) Add a sequence of boundstates to the manybody solver
## 3.4.3. Data types¶
Occupation(distribution[, energy_range, bands]) Data format for the lead occupation, see tkwant.manybody.lead_occupation. Interval(lead, band, kmin, kmax[, order, …]) Data format for a quadrature interval, see tkwant.manybody.calc_intervals.
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2021-06-17 20:30:13
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http://mathhelpforum.com/calculus/80605-derivatives-product-rule-three-terms-print.html
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# Derivatives - product rule with three terms?
• Mar 25th 2009, 09:00 AM
ktprieto
Derivatives - product rule with three terms?
I'm trying to get the derivative of 3xe^(2x)
I know the product rule and I know the derivative of e^2x is 2e^(2x).
So I was going to use the product rule, but I've only ever done that with two terms. Here, there are three.
• Mar 25th 2009, 09:10 AM
Pinkk
You have a function $h(x)=f(x)g(x)$ where $f(x)=3x$ and $g(x)=e^{2x}$
$h'(x)=f'(x)g(x)+f(x)g'(x)$
$h'(x)=(3)e^{2x}+3x(2e^{2x})$
$h'(x)=3e^{2x}+6xe^{2x}$
• Mar 25th 2009, 11:18 AM
matheagle
the 3 is just a constant, BUT if you have three functions then
$(fgh)'=f'gh+fg'h+fgh'$.
• Mar 25th 2009, 11:18 AM
Superjones
thats a product rule and a chain rule
3xe^(2x)
3e^(2x)
for the first. take derivative of 3x which is just 3 and keep second part. than. chain rule e^(2x) so e^ whatever is just e^ the whatever. so e^(2x) than multiply by 2 since the derivative of 2x is 2. keep the first part.
So you get:
3e^(2x) + 3xe^(2x)*2 which is 3e^(2x) + 6xe^(2x)
you can then simplify by pulling 3e^(2x) out of the answer
3e^(2x)*(1+2x) but that wouldnt really help unless your graphing and looking for critical numbers
• Mar 25th 2009, 12:06 PM
tom@ballooncalculus
Just in case a picture helps...
http://www.ballooncalculus.org/asy/d...ghtHandExp.png
You have the chain rule - generally this pattern -
http://www.ballooncalculus.org/chain_rule.png
... wrapped inside the product rule...
http://www.ballooncalculus.org/prod_rule.png
... more visible here if we enclose some of the balloons in some more...
http://www.ballooncalculus.org/asy/d...htHandExp1.png
Hope this helps, or doesn't further confuse.
Don't integrate - balloontegrate!
Balloon Calculus: worked examples from past papers
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2017-05-25 18:39:42
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https://ncatlab.org/nlab/show/Structured+Spaces
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# nLab Structured Spaces
/
• geometric es
• geometric es
• ,
• /
## Examples
• , of s
• derived smooth geometry
• ,
• ,
• ,
• ,
• / /
• /
• /
• ,
• ,
• ,
• ,
## Higher algebras
• symmetric monoidal (∞,1)-category of spectra
• , ,
• ,
• ,
• ,
• /
## Theorems
This entry provides hyperlinks for central keywords of the text
• Structured Spaces
which develops the generalization of the notion of a ringed topos from topos theory to (∞,1)-topos theory and formulates basic notions of geometry in this context (“higher geometry”).
Precursor in 1-category theory include
The theory is general, but the focus of the examples and applications is on derived algebraic geometry/E-∞ geometry.
Apparently a version of this is to eventually appear as:
## Survey
A survey is at A Survey of Elliptic Cohomology - the derived moduli stack of derived elliptic curves in the section notions of space.
## Contents
category: reference
Last revised on June 11, 2018 at 13:44:35. See the history of this page for a list of all contributions to it.
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2019-05-22 04:48:53
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https://gateoverflow.in/18652/how-can-one-solve-the-following-recurrence
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735 views
$$T(n)\quad=\quad T(n-1)+T \left (\frac{n}{2} \right )+n$$ $$n \geq 1, \quad T(1)=1$$
I dont know the exact answer but this is my approach.
let the T(n-1)=T(n/2) T(n/2)=T(n-1)
T(n)= 2T(n/2)+n < T(n)=T(n−1)+T(n/2)+n < T(n)=2T(n−1)+n
Ώ(nlogn) < T(n) < O(2n)
so i guess answer should in exponential term for big-Oh.
### 1 comment
For GATE this should be sufficient I guess. It is hard to do better.
I'm not sure. I'm trying to get it right.
T(n) = T(n-1)+T(n/2)+n
As we know we Asymptotic notation by equalities and inequalities, we can represent a part of Sub expression in RHS with a Approx Fn.
lets consider T(n/2)+n. By applying masters theorem we get => O(n).
So Our initial Eqn becomes, T(n)= T(n-1)+O(n). Now if we apply masters theorem we get T(n)= O(n^2).
Please Correct me if im wrong
### 1 comment
Nopes. I don't think you can do that. Because T(n/2) in turn has a term for T(n/2 - 1). So, there is no easy way here.
i can't understand the solution although here. but no specific complexity of the question http://clrs.skanev.com/04/04/05.html its corman solution ite it will be very handy if u r solving coreman http://clrs.skanev.com/04/04/05.html
by
That result is easy to get- but that's not the actual result rt?
I can't undersrand . may u help me understand it . arjun .
We can replace $n/2$ with $n-1$ in the recursion and solve and get $O\left(2^n\right)$. Also, we can replace $n-1$ with $n/2$ and get $\Omega \left(n \lg n\right)$ as lower bound. To make the bounds tight is out of GATE scope I believe.
1
329 views
1 vote
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2023-01-30 13:59:16
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https://www.nature.com/articles/s41587-021-01188-9?error=cookies_not_supported&code=b9975b0a-8aa3-4a40-8a48-c5bf49eb30d5
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## Main
Cell-free RNA (cfRNA) represents a mixture of transcripts reflecting the health status of multiple tissues1, thereby affording broad clinical utility. Existing applications span oncology and bone marrow transplantation2,3, obstetrics1,4,5, neurodegeneration6 and liver disease7. However, several aspects about the physiologic origins of cfRNA, including the contributing cell types of origin, remain unknown, and current assays focus on tissue-level contributions at best1,3,4,5,6,7. Incorporating knowledge from cellular pathophysiology, which often forms the basis of disease8, into a liquid biopsy would more closely match the resolution afforded by invasive procedures.
We first characterized the landscape of cell-type-specific signal from healthy donor plasma using published exome-enriched cell-free transcriptome data6 (Fig. 1a). After removing low-quality samples (Extended Data Fig. 1 and Methods), we intersected the set of genes detected in healthy individuals (n = 75) with a database of cell-type-specific markers defined in context of the whole body9. Marker genes for blood, brain, and liver cell types were readily detected, as previously observed at tissue level1,3,4,6,7, as well as the kidney, gastrointestinal tract, and pancreas (Fig. 1b).
We then sought to deconvolve the fractions of cell-type-specific RNA using support vector regression, a deconvolution method previously applied to decompose bulk tissue transcriptomes into fractional cell type contributions10,11. We used Tabula Sapiens version 1.0 (TSP)12, a multiple-donor whole-body cell atlas spanning 24 tissues and organs, to define a basis matrix whose gene set accurately and simultaneously resolved the distinct cell types in TSP. The basis matrix was defined using the gene space that maximized linear independence of the cell types and does not include the whole transcriptome but rather the minimum discriminatory gene set to distinguish between the cell types in TSP. To reduce multicollinearity, transcriptionally similar cell types were grouped (Extended Data Fig. 2). We observed that the basis matrix defined by this gene set appropriately described cell types as most similar to others from the same organ compartment and corresponded to the highest off-diagonal similarity (Fig. 1c). We also confirmed that the basis matrix accurately deconvolved cell-type-specific RNA fractional contributions from several bulk tissue samples13 (Extended Data Fig. 3 and Supplementary Information).
We used this matrix to deconvolve the cell types of origin in the plasma cell-free transcriptome (Fig. 1d and Extended Data Figs. 4 and 5). Platelets, erythrocyte/erythroid progenitors and leukocytes comprised the majority of observed signal, whose respective proportions were generally consistent with recent estimates from serum cfRNA2 and plasma cfDNA14. Within this set of cell types, we suspect that the observation of platelets as a majority cell type, rather than megakaryocytes2, likely reflects annotation differences in reference data. We observed distinct transcriptional contributions from solid tissue-specific cell types from the intestine, liver, lungs, pancreas, heart, and kidney (Fig. 1d and Extended Data Fig. 4). Altogether, the observation of contributions from many non-hematopoietic cell types underscores the ability to simultaneously non-invasively resolve contributions to cfRNA from disparate cell types across the body.
Some cell types likely present in the plasma cell-free transcriptome were missing in this decomposition because the source tissues were not represented in TSP. Although, ideally, reference gene profiles for all cell types would be simultaneously considered in this decomposition, a complete reference dataset spanning the entire cell type space of the human body does not yet exist. To identify cell type contributions possibly absent from this analysis, we intersected the genes measured in cfRNA missing from the basis matrix with tissue-specific genes from the Human Protein Atlas (HPA) RNA consensus dataset15. This identified both the brain and the testis as tissues whose cell types were not found during systems-level deconvolution and additional genes specific to the blood, skeletal muscle and lymphoid tissues that were not used by the basis matrix (Fig. 1e and Methods).
As an example of how to analyze cell type contributions from tissues that were not present in TSP, we used an independent brain single-cell atlas along with HPA to define cell type gene profiles and examined their expression in cfRNA (Fig. 2a and Extended Data Figs. 6 and 7). There was a strong signature score from excitatory neurons and a reduced signature score from inhibitory neurons. We observed strong signals from astrocytes, oligodendrocytes and oligodendrocyte precursor cells. These glial cells facilitate brain homeostasis, form myelin and provide neuronal structure and support8, consistent with evidence of RNA transport across and the permeability of the blood–brain barrier16,17 and that some brain regions are in direct contact with the blood18. Similarly, we used published cell atlases for the placenta19,20, kidney21 and liver22 to define cell-type-specific gene profiles (Extended Data Figs. 6 and 8) for signature scoring. These observations augment the resolution of previously observed tissue-specific genes reported to date in cfRNA1,2,3,4,5,6,7 and formed a baseline from which to measure aberrations in disease.
Cell-type-specific changes drive disease etiology8, and we asked whether cfRNA reflected cellular pathophysiology. We considered trophoblasts in preeclampsia23,24, proximal tubules in chronic kidney disease (CKD)25,26, hepatocytes in non-alcoholic steatohepatitis (NASH)/non-alcoholic fatty liver disease (NAFLD)27 and multiple brain cell types in Alzheimer’s disease (AD)28,29. As an example of why whole-body cell type characterization is relevant, we observed that a previous attempt to infer trophoblast cell types from cfRNA in preeclampsia24 used genes that are not specific or readily measurable within their asserted cell type (Extended Data Fig. 9 and Supplementary Information). However, we found several other cases where cellular pathophysiology can be measured in cfRNA.
The proximal tubule is a highly metabolic, predominant kidney cell type and is a major source for injury and disease progression in CKD25,26. Tubular atrophy is a hallmark of CKD nearly independent of disease etiology30 and is superior to clinical gold standard as a predictor of CKD progression31. Using data from Ibarra et al., we discovered a striking decrease in the proximal tubule cell signature score of patients with CKD (ages 67–91 years, CKD stage 3–5 or peritoneal dialysis) compared to healthy controls (Fig. 2b and Extended Data Fig. 10a,b). These results demonstrate non-invasive resolution of proximal tubule deterioration observed in CKD histology31 and are consistent with findings from invasive biopsy.
Hepatocyte steatosis is a histologic hallmark of NASH and NAFLD phenotypes, whereby the accumulation of cellular stressors results in hepatocyte death27. We found that several genes differentially expressed in NAFLD serum cfRNA7 were specific to the hepatocyte cell type profile derived above (P < 10−10, hypergeometric test). Notable hepatocyte-specific differentially expressed genes (DEGs) include genes encoding cytochrome P450 enzymes (including CYP1A2, CYP2E1 and CYP3A4), lipid secretion (MTTP) and hepatokines (AHSG and LECT2)32. We further observed striking differences in the hepatocyte signature score between healthy and both NAFLD and NASH cohorts and no difference between the NASH and NAFLD cohorts (Fig. 2c and Extended Data Fig. 10).
AD pathogenesis results in neuronal death and synaptic loss29. We used brain single-cell data28 to define brain cell type gene profiles in both the AD and the normal brain. Several DEGs found in cfRNA analysis of AD plasma are brain cell type specific (P < 10−5, hypergeometric test). Astrocyte-specific genes include those that encode filament protein (GFAP33) and ion channels (GRIN2C28). Excitatory neuron-specific genes encode solute carrier proteins (SLC17A728) and SLC8A234), cadherin proteins (CDH835 and CDH2236) and a glutamate receptor (GRM129,37). Oligodendrocyte-specific genes encode proteins for myelin sheath stabilization (MOBP29) and a synaptic/axonal membrane protein (CNTN229). Oligodendrocyte-precursor-cell-specific genes encode transcription factors (OLIG238 and MYT139), neural growth and differentiation factor (CSPG540) and a protein putatively involved in brain extracellular matrix formation (BCAN41).
We then inferred neuronal death in plasma cfRNA between AD and healthy non-cognitive controls (NCIs) and also observed differences in oligodendrocyte, oligodendrocyte progenitor and astrocyte signature scores (Fig. 2d and Extended Data Fig. 10). The oligodendrocyte and oligodendrocyte progenitor cells signature score directionality agrees with reports of their death and inhibited proliferation in AD, respectively42. The observed astrocyte signature score directionality is consistent with the cell type specificity of a subset of reported downregulated DEGs6 and reflects that astrocyte-specific changes, which are known in AD pathology42, are non-invasively measurable.
Taken together, this work demonstrates consistent non-invasive detection of cell-type-specific changes in human health and disease using cfRNA. Our findings uphold and further augment the scope of previous work identifying immune cell types2 and hematopoietic tissues1,2 as primary contributors to the cell-free transcriptome cell type landscape. Our approach is complementary to previous work using cell-free nucleosomes14, which depends on a more limited set of reference chromatin immunoprecipitation sequencing data, which are largely at the tissue level43. Readily measurable cell types include those specific to the brain, lung, intestine, liver, and kidney, whose pathophysiology affords broad prognostic and clinical importance. Consistent detection of cell types responsible for drug metabolism (for example, liver and renal cell types) as well as cell types that are drug targets, such as neurons or oligodendrocytes for Alzheimer’s-protective drugs, could provide strong clinical trial endpoint data when evaluating drug toxicity and efficacy. We anticipate that the ability to non-invasively resolve cell type signatures in plasma cfRNA will both enhance existing clinical knowledge and enable increased resolution in monitoring disease progression and drug response.
## Methods
### Data processing
#### Data acquisition
cfRNA: For samples from Ibarra et al. (PRJNA517339), Toden et al. (PRJNA574438) and Chalasani et al. (PRJNA701722), raw sequencing data were obtained from the Sequence Read Archive with the respective accession numbers. For samples from Munchel et al., processed counts tables were directly downloaded.
For all individual tissue single-cell atlases, Seurat objects or AnnData objects were downloaded or directly received from the authors. Data from Mathys et al. were downloaded with permission from Synapse. The liver Seurat object was requested from Aizarani et al. For the placenta cell atlases, a Seurat object was requested from Suryawanshi et al., and AnnData was requested from Vento-Tormo et al. Kidney AnnData were downloaded (https://www.kidneycellatlas.org, Mature Full dataset).
HPA version 19 transcriptomic data, Genotype-Tissue Expression (GTEx) version 8 raw counts and Tabula Sapiens version 1.0 were downloaded directly.
#### Bioinformatic processing
All analyses were performed using Python (version 3.6.0) and R (version 3.6.1) For each sample for which raw sequencing data were downloaded, we trimmed reads using trimmomatic (version 0.36) and then mapped them to the human reference genome (hg38) with STAR (version 2.7.3a). Duplicate reads were then marked and removed by the MarkDuplicates tool in GATK (version 4.1.1). Finally, mapped reads were quantified using htseq-count (version 0.11.1), and read statistics were estimated using FastQC (version 0.11.8).
The bioinformatic pipeline was managed using snakemake (version 5.8.1). Read and tool performance statistics were aggregated using MultiQC (version 1.7).
#### Sample quality filtering
For every sample for which raw sequencing data were available, we estimated three quality parameters as previously described44,45: RNA degradation, ribosomal read fraction and DNA contamination.
RNA degradation was estimated by calculating a 3′ bias ratio. Specifically, we first counted the number of reads per exon and then annotated each exon with its corresponding gene ID and exon number using htseq-count. Using these annotations, we measured the frequency of genes for which all reads mapped exclusively to the 3′-most exon as compared to the total number of genes detected. We approximated RNA degradation for a given sample as the fraction of genes where all reads mapped to the 3′-most exon.
To estimate ribosomal read fraction, we compared the number of reads that mapped to the ribosome (region GL00220.1:105,424–118,780, hg38) relative to the total number of reads (SAMtools view).
To estimate DNA contamination, we used an intron-to-exon ratio and quantified the number of reads that mapped to intronic as compared to exonic regions of the genome.
We applied the following thresholds as previously reported44:
• Ribosomal: >0.2
• 3′ Bias Fraction: >0.4
• DNA Contamination: >3
We considered any given sample as low quality if its value for any metric was greater than any of these thresholds, and we excluded the sample from subsequent analysis.
#### Data normalization
All gene counts were adjusted to counts per million (CPM) reads and per milliliter of plasma used. For a given sample, i denotes gene index, and j denotes sample index:
$$\eta _{ij} = \frac{{\mathrm{Gene}_{ij}}}{{({\mathrm{Library}}\,{\mathrm{size}}_j) \times ({\rm{mL}}\,{\mathrm{plasma}}_j)}}\,{{{\mathrm{where}}}}\,{\mathrm{Library}}\,{\mathrm{size}}_j = \mathop {\sum}\limits_i {G_{ij}}$$
(1)
For individuals who had samples with multiple technical replicates, these plasma volume CPM counts were averaged before nu support vector regression (nu-SVR) deconvolution.
For all analyses except nu-SVR (all work except Fig. 1d,e), we next applied trimmed mean of M values (TMM) normalization as previously described46 using edgeR (version 3.28.1):
$$\frac{{\eta _{ij}}}{{TMM_j}}$$
(2)
CPM-TMM normalized gene counts across technical replicates for a given biological replicate were averaged for the count tables used in all analyses performed.
Sequencing batches and plasma volumes were obtained from the authors in Toden et al. and Chalasani et al. for per-sample normalization. For samples from Ibarra et al., plasma volume was assumed to be constant at 1 ml, as we were unable to obtain this information from the authors; sequencing batches were confirmed with the authors (personal communication). All samples from Munchel et al. were used to compute TMM scaling factors, and 4.5 ml of plasma5 was used to normalize all samples within a given dataset (both PEARL-PEC and iPEC).
### Cell type marker identification using PanglaoDB
The PanglaoDB cell type marker database was downloaded on 27 March 2020. Markers were filtered for human (‘Hs’) only and for PanglaoDB’s defined specificity (how often marker was not expressed in a given cell type) and sensitivity (how frequently marker is expressed in cells of this type). Gene synonyms from Panglao were determined using MyGene version 3.1.0 to ensure full gene space.
We then intersected this gene space with a cohort of healthy cfRNA samples (n = 75, NCI individuals from Toden et al.). A given cell type marker was counted in a given healthy cfRNA sample if its gene expression was greater than zero in log +1 transformed CPM-TMM gene count space.
Cell types with markers filtered by sensitivity = 0.9 and specificity = 0.2 and samples with >5 cell type markers on average are shown in Fig. 1b.
### Basis matrix formation
Scanpy47 (version 1.6.0) was used. Only cells from droplet sequencing (‘10x’) were used in analysis given that a more comprehensive set of unique cell types across the tissues in Tabula Sapiens was available12. Disassociation genes as reported12 were eliminated from the gene space before subsequent analysis.
Given the non-specificity of the following annotations (for example, other cell type annotations at finer resolution existed), cells with these annotations were excluded from subsequent analysis:
• ‘epithelial cell’
• ‘ocular surface cell’
• ‘lacrimal gland functional unit cell’
• ‘connective tissue cell’
• ‘corneal keratocyte’
• ‘ciliary body’
• ‘bronchial smooth muscle cell’
• ‘fast muscle cell’
• ‘muscle cell’
• ‘myometrial cell’
• ‘skeletal muscle satellite stem cell’
• ‘slow muscle cell’
• ‘tongue muscle cell’
• ‘vascular associated smooth muscle cell’
• ‘alveolar fibroblast’
• ‘fibroblast of breast’
• ‘fibroblast of cardiac tissue’
• ‘myofibroblast cell’
All additional cells belonging to the ‘Eye’ tissue were excluded from subsequent analysis given discrepancies in compartment and cell type annotations and the unlikelihood of detecting eye-specific cell types. The resulting cell type space still possessed several transcriptionally similar cell types (for example, various intestinal enterocytes, T cells or dendritic cells), which, left unaddressed, would reduce the linear independence of the basis matrix column space and, hence, would affect nu-SVR deconvolution.
Cells were, therefore, assigned broader annotations on a per-compartment basis as follows:
Epithelial, Stromal, Endothelial: Using counts from the ‘decontXcounts’ layer of the adata object, cells were CPM normalized (sc.pp.normalize_total(target_sum = 1 × 106)) and log-transformed (sc.pp.log1p). Hierarchical clustering with complete linkage (sc.tl.dendrogram) was performed per compartment on the feature space comprising the first 50 principal components (sc.pp.pca). Epithelial and stromal compartment dendrograms were then cut (scipy.cluster.hierarchy.cut_tree) at 20% and 10% of the height of the highest node, respectively, such that cell types with high transcriptional similarity were grouped together, but overall granularity of the cell type labels was preserved. This work is available in the script ‘treecutter.ipynb’ on GitHub; the scipy version used is 1.5.1.
The endothelial compartment dendrogram revealed high transcriptional similarity across all cell types (maximum node height = 0.851) compared to epithelial (maximum node height = 3.78) and stromal (maximum node height = 2.34) compartments (Extended Data Fig. 2). To this end, only the ‘endothelial cell’ annotation was used for the ‘endothelial’ compartment.
Immune: Given the high transcriptional similarity and the varying degree of annotation granularity across tissues and cell types, cell types were grouped on the basis of annotation. The following immune annotations were kept:
• ‘b cell’
• ‘basophil’
• ‘erythrocyte’
• ‘erythroid progenitor’
• ‘hematopoietic stem cell’
• ‘innate lymphoid cell’
• ‘macrophage’
• ‘mast cell’
• ‘mature conventional dendritic cell’
• ‘microglial cell’
• ‘monocyte’
• ‘myeloid progenitor’
• ‘neutrophil’
• ‘nk cell’
• ‘plasma cell’
• ‘plasmablast’
• ‘platelet’
• ‘t cell’
• ‘thymocyte’
All other immune compartment cell type annotations were excluded for being too broad when more detailed annotations existed (that is, ‘granulocyte’, ‘leucocyte’ and ‘immune cell’) or present in only one tissue (that is, ‘erythroid lineage cell’; eye, ‘myeloid cell’; and pancreas/prostate). The ‘erythrocyte’ and ‘erythroid progenitor’ annotations were further grouped to minimize multicollinearity.
Using the entire cell type space spanning all four organ compartments, either 30 observations (for example, measured cells) were randomly sampled or the maximum number of available observations (if less than 30) was subsampled, whichever was greater.
Cell type annotations were then reassigned based on the ‘broader’ categories from hierarchical clustering (‘coarsegrain.py’). Raw count values from the DecontX adjusted layer were used to minimize signal spread contamination that could affect DEG analysis12.
This subsampled counts matrix was then passed to the ‘Create Signature Matrix’ analysis module at https://cibersortx.stanford.edu/, with the following parameters:
• Disable quantile normalization = True
• Minimum expression = 0.25
• Replicates = 5
• Sampling = 0.5
• Kappa = 999
• q value = 0.01
• No. of barcode genes = 3,000–5,000
• Filter non-hematopoietic genes = False
The resulting basis matrix was used in our nu-SVR deconvolution code, available on GitHub, under the name ‘tsp_v1_basisMatrix.txt’.
Abbreviations (left) of grouped cell types (right) in Fig. 1d and the Extended Data are as follows:
• gland cell: ‘acinar cell of salivary gland/myoepithelial cell’
• respiratory ciliated cell: ‘ciliated cell/lung ciliated cell’
• prostate epithelia: ‘club cell of prostate epithelium/hillock cell of prostate epithelium/hillock-club cell of prostate epithelium’
• salivary/bronchial secretory cell: ‘duct epithelial cell/serous cell of epithelium of bronchus’
• intestinal enterocyte: ‘enterocyte of epithelium of large intestine/enterocyte of epithelium of small intestine/intestinal crypt stem cell of large intestine/large intestine goblet cell/mature enterocyte/paneth cell of epithelium of large intestine/small intestine goblet cell’
• intestinal crypt stem cell: ‘immature enterocyte/intestinal crypt stem cell/intestinal crypt stem cell of small intestine/transit amplifying cell of large intestine’
• erythrocyte/erythroid progenitor: ‘erythrocyte/erythroid progenitor’
• fibroblast/mesenchymal stem cell: ‘fibroblast/mesenchymal stem cell’
• intestinal secretory cell: ‘intestinal enteroendocrine cell/paneth cell of epithelium of small intestine/transit amplifying cell of small intestine’
• ionocyte/luminal epithelial cell of mammary gland: ‘ionocyte/luminal epithelial cell of mammary gland’
• secretory cell: ‘mucus secreting cell/secretory cell/tracheal goblet cell’
• pancreatic alpha/beta cell: ‘pancreatic alpha cell/pancreatic beta cell’
• respiratory secretory cell: ‘respiratory goblet cell/respiratory mucous cell/serous cell of epithelium of trachea’
• basal prostate cell: ‘basal cell of prostate epithelia’
### Nu-SVR deconvolution
We formulated the cell-free transcriptome as a linear summation of the cell types from which it originates1,48. With this formulation, we adapted existing deconvolution methods developed with the objective of decomposing a bulk tissue sample into its single-cell constituents10,11, where the deconvolution problem is formulated as:
$$A\theta = b$$
(3)
Here, A is the representative basis matrix (g × c) of g genes for c cell types, which represent the gene expression profiles of the c cell types. θ is a vector (c × 1) of the contributions of each of the cell types, and b is the measured expression of the genes observed in blood plasma (g × 1). The goal here is to learn θ such that the matrix product predicts the measured signal b. The derivation of the basis matrix A is described in the section ‘Basis matrix formation’.
We performed nu-SVR using a linear kernel to learn θ from a subset of genes from the basis matrix to best recapitulate the observed signal b, where nu corresponds to a lower bound on the fraction of support vectors and an upper bound on the fraction of margin errors49. Here, the support vectors are the genes from the basis matrix used to learn θ; θ reflects the learned weights of the cell types in the basis matrix column space. For each sample, a set of θ was learned by performing a grid search on the two SVR hyperparameters: $$\nu \in \{ 0.05,0.1,0.15,0.25,0.5,0.75\}$$ and $$C \in \{ 0.1,0.5,0.75,1,10\}$$.
For each sample, we next enforce two constraints: θ can contain only non-negative weights, and the weights in θ must sum to 1. Each θ corresponding to a hyperparameter combination was normalized as previously described in two steps10,11. First, only non-negative weights were kept:
$$\forall \theta _j < 0 \in \left\{ {\theta _1, \ldots ,\theta _c} \right\} \to 0$$
(4)
Second, the remaining non-zero weights were then normalized by their sum to yield the relative proportions of cell-type-specific RNA.
We then determined the basis matrix dot product with the set of normalized weights for each sample. This dot product yields the predicted expression value for each gene in a given cfRNA mixture with imposed non-negativity on the normalized coefficient vector. The root mean square error (RMSE) was then computed using the predicted expression values and the measured values of these genes for each hyperparameter combination in a given cfRNA mixture. The model yielding the smallest RMSE in predicting expression for a given cfRNA sample was then chosen and assigned as the final deconvolution result for a given sample.
Only CPM counts ≥1 were considered in the mixture, b. The values in the basis matrix were also CPM normalized. Before deconvolution, the mixture and basis matrix were centered and scaled to zero mean and unit variance for improved runtime performance. We emphasize that we did not log-transform counts in b or in A, as this would destroy the requisite linearity assumption in equation (3). Specifically, the concavity of the log function would result in the consistent underestimation of θ during deconvolution50.
We used the function nu-SVR from scikitlearn51 version 0.23.2.
The samples used for nu-SVR deconvolution were 75 NCI patients from Toden et al. spanning four sample collection centers. Given center-specific batch effects reported by Toden et al., we report our results on a per-center basis (Fig. 1d and Extended Data Figs. 4 and 5). There was good pairwise similarity of the learned coefficients among biological replicates within and across sample centers (Extended Data Fig. 5a,b). Deconvolution performance yielded RMSE and Pearson r consistent with deconvolved GTEx tissues (Extended Data Fig. 3) whose distinct cell types were in the basis matrix column space (Extended Data Fig. 5c,d). In interpreting the resulting cell type fractions, a limitation of nu-SVR is that it uses highly expressed genes as support vectors and, consequently, assigns a reduced fractional contribution to cell types expressing genes at lower levels or that are smaller in cell volume. Comparison of nu-SVR to quadratic programming1 and non-negative linear least squares52 yielded similar deconvolution RMSE and Pearson correlation. In contrast to the other methods, nu-SVR cell type contributions were the most consistent with the cell type markers detected using PanglaoDB and was, hence, chosen as the deconvolution model for this work.
### Evaluating basis matrix on GTEx samples
Bulk RNA sequencing samples from GTEx version 8 were deconvolved with the derived basis matrix from tissues that were present (that is, kidney cortex, whole blood, lung and spleen) or absent (for example, kidney medulla and brain) from the basis matrix derived using Tabula Sapiens version 1.0. For each tissue type, the maximum number of available samples or 30 samples, whichever was smaller, was deconvolved. See Supplementary Note 1 for additional discussion.
### Identifying tissue-specific genes in cfRNA absent from basis matrix
To identify cell-type-specific genes in cfRNA that were distinct to a given tissue, we considered the set difference of the non-zero genes measured in a given cfRNA sample with the row space of the basis matrix and intersected this with HPA tissue-specific genes:
$$(G_j - R) \cap HPA$$
(5)
where Gj is the gene set in the jth deconvolved sample, where a given gene in the set’s expression was ≥1 CPM. R is the set of genes in the row space of the basis matrix used for nu-SVR deconvolution. HPA denotes the total set of tissue-specific genes from HPA.
The HPA tissue-specific gene set (HPA) comprised genes across all tissues with Tissue Specificity assignments ‘Group Enriched’, ‘Tissue Enhanced’, ‘Tissue Enriched’ and NX expression ≥10. This approach yielded tissues with several distinct genes present in cfRNA, which could then be subsequently interrogated using single-cell data.
### Derivation of cell-type-specific gene profiles in context of the whole body using single-cell data
For this analysis, only cell types unique to a given tissue (that is, hepatocytes unique to the liver or excitatory neurons unique to the brain) were considered so that bulk transcriptomic data could be used to ensure specificity in context of the whole body. A gene was asserted to be cell type specific if it was (1) differentially expressed within a given single-cell tissue atlas, (2) possessed a Gini coefficient ≥0.6 and was listed as specific to the native tissue for the cell type of interest, indicating comprehensive tissue specificity in context of the whole body (Extended Data Figs. 6 and 8).
1. (1)
Single-cell differential expression
For data received as a Seurat object, conversion to AnnData (version 0.7.4) was performed by saving as an intermediate loom object (Seurat version 3.1.5) and converting to AnnData (loompy version 3.0.6). Scanpy (version 1.6.0) was used for all other single-cell analysis. Reads per cell were normalized for library size (scanpy normalize_total, target_sum = 1 × 104) and then logged (scanpy log1p). Differential expression was performed using the Wilcoxon rank-sum test in Scanpy’s filter_rank_genes_groups with the following arguments: min_fold_change = 1.5, min_in_group_fraction = 0.2, max_out_group_fraction = 0.5, corr_method = ‘benjamini-hochberg’. The set of resulting DEGs with Benjamini–Hochberg-adjusted P values <0.01 whose ratio of the highest out-group percent expressed to in-group percent expressed <0.5 was selected to ensure high specific expression in the cell type of interest within a given cell type atlas.
2. (2)
Quantifying comprehensive whole-body tissue specificity using the Gini coefficient
The distribution of all the Gini coefficiets and Tau values across all genes belonging to cell type gene profiles for cell types native to a given tissue were compared using the HPA gene expression Tissue Specificity and Tissue Distribution assignments15 (Extended Data Fig. 7). The Gini coefficient better reflected the underlying distribution of gene expression tissue specificity than Tau (Extended Data Fig. 7) and, hence, were used for subsequent analysis. As the Gini coefficient approaches unity, this indicates extreme gene expression inequality or equivalently high specificity. A single threshold (Gini coefficient ≥ 0.6) was applied across all atlases to facilitate a generalizable framework from which to define tissue-specific cell type gene profiles in context of the whole body in a principled fashion for signature scoring in cfRNA.
For the following definitions, n denotes the total number of tissues, and xj is the expression of a given gene in the ith tissue.
The Gini coefficient was computed as defined53:
$${\mathrm{Gini}} = \frac{{n + 1}}{n} - \frac{{2{\mathop \sum \nolimits_{i = 1}^{n}}\left( {n + 1 - i} \right){x_i}}}{{n{\mathop \sum \nolimits_{i = 1}^{n}}{x_i}}}\, ;\: {x_i}\, {\mathrm{is}}\, {\mathrm{ordered}}\, {\mathrm{from}}\, {\mathrm{least}}\, {\mathrm{to}}\, {\mathrm{greatest}}.$$
(6)
Tau, as defined in ref. 53:
$$\tau = \frac{{\mathop {\sum }\nolimits_{i = 1}^n 1 - \bar x}}{{n - 1}}\ {{{\mathrm{where}}}}\,\bar x = \frac{{x_i}}{{{{{\mathrm{max}}}}\left( {x_i} \right)\forall i \in \{ 1 \ldots n\} }}$$
(7)
HPA NX Counts from the HPA object titled ‘rna_tissue_consensus.tsv’ accessed on 1 July 2019 were used for computing Gini coefficients and Tau.
Note for brain cell type gene profiles: Given that there are multiple sub brain regions in the HPA data, the determined Gini coefficients are lower (for example, not as close to unity compared to other cell type gene profiles) because there are multiple regions of the brain with high expression, which would result in reduced count inequality.
### Gene expression in GTEx
We confirmed the specificity of a given gene profile to its corresponding cell type by comparing the aggregate expression of a given cell type signature in its native tissue compared to that of the average across remaining GTEx tissues (Extended Data Figs. 6d and 8f,g). We uniformly observed a median fold change greater than 1 in the signature score of a cell type gene profile in its native tissue relative to the mean expression in other tissues, confirming high specificity.
Raw GTEx data version 8 (accessed 26 August 2019) were converted to log(counts-per-ten-thousand + 1) counts. The signature score was determined by summing the expression of the genes in a given bulk RNA sample for a given cell type gene profile. Because only gene profiles were derived for cell types that correspond to a given tissue, the mean signature score of a cell type profile across the non-native tissues was then computed and used to determine the log fold change.
### Cell type specificity of DEGs in AD and NAFLD cfRNA
After observing a significant intersection between the DEGs in AD6 or NAFLD7 in cfRNA with corresponding cell-type-specific genes (Extended Data Fig. 10c,e), we then assessed the cell type specificity of DEGs using a permutation test. To assess whether DEGs that intersected with a cell type gene profile were more specific to a given cell type than DEGs that were generally tissue specific, we performed a permutation test. Specifically, we compared the Gini coefficient for genes in these two groups, computed using the mean expression of a given gene across brain cell types from healthy brain28 or liver22 single-cell data. We considered the cell type gene profiles as defined for signature scoring in Fig. 2.
The starting set of tissue-specific genes was defined using the HPA tissue transcriptional data annotated as ‘Tissue enriched’, ‘Group enriched’ or ‘Tissue enhanced’ (brain, accessed on 13 January 2021; liver, accessed on 28 November 2020). These requirements ensured the specificity of a given brain/liver gene in context of the whole body. For a given tissue, this formed the initial set of tissue-specific genes B.
The union of all brain or liver cell-type-specific genes is the set C. All genes in C (‘cell type specific’) were a subset of the respective initial set of tissue-specific genes:
$$C - B = 0$$
(8)
Genes in B that did not intersect with C and intersected with DEG-up (U) or DEG-down (D) genes in a given disease6,7 were then defined as ‘tissue specific’.
$$T = \left( {B \cap U} \right) \cup (B \cap D) - C$$
(9)
The Gini coefficients reflecting the gene expression inequality across the cell types within corresponding tissue single-cell atlas were computed for the gene sets labeled as ‘cell type specific’ and ‘tissue specific’. Brain reference data to compute Gini coefficients were from the single-cell brain atlas with diagnosis as ‘Normal’28. Liver single cell data were used as-is22. All Gini coefficients were computed using the mean log-transformed CPTT (counts per ten thousand) gene expression per cell type.
A permutation test was then performed on the union of the Gini coefficients for the genes labeled as ‘cell type specific’ and ‘tissue specific’. The purpose of this test was to assess probability that the observed mean difference in Gini coefficient for these two groups yielded no difference in specificity (that is, H0: $$\mu _{{\mathrm{cell}}\,{\mathrm{type}}\,{\mathrm{Gini}}\,{\mathrm{coefficient}}} = \mu _{{\mathrm{tissue}}\,{\mathrm{Gini}}\,{\mathrm{coefficient}}}$$).
Gini coefficients were permuted and reassigned to the list of ‘tissue specific’ or ‘cell type specific’ genes, and then the difference in the means of the two groups was computed. This procedure was repeated 10,000 times. The P value was determined as follows:
$$p = \frac{{\# \,{\mathrm{trials}}\,{\mathrm{with}}\,{\mathrm{permuted}}(\mu _{{\mathrm{cell}}\,{\mathrm{type}}} - \mu _{{\mathrm{tissue}}}) \ge \mu _{{\mathrm{observed}}}}}{{10,000 + 1}}$$
(10)
where $$\mu _{\mathrm{observed}}: = (\mu _{{\mathrm{cell}}\,{\mathrm{type}}\,{\mathrm{Gini}}\,{\mathrm{coefficient}}} - \mu _{{\mathrm{tissue}}\,{\mathrm{Gini}}\,{\mathrm{coefficient}}})$$.
The additional 1 in the denominator reflects the original test between the true difference in means (for example, the true comparison yielding μobserved).
NAFLD: We considered the space of reported NAFLD DEGs in serum7. Here, C = hepatocyte gene profile, and B = the liver-specific genes.
AD: First, we intersected a given cell type gene profile in AD with the equivalent Normal profile for comparative analysis. Genes defined as ‘brain cell type specific’ for signature scoring in Fig. 2d were used in this comparison. Of note, no DEG-up genes intersected with any of the brain cell type signatures in Fig. 2d. Microglia, although often implicated in AD pathogenesis, were excluded given their high overlapping transcriptional profile with non-central-nervous-system macrophages54. Inhibitory neurons were also excluded given the low number of cell-type-specific genes intersecting between AD and NCI phenotypes.
### Estimating signature scores for each cell type
The signature score is defined as the sum of the log-transformed CPM-TMM normalized counts per gene asserted to be cell type specific, where i denotes the index of the gene in a cell type signature gene profile G in the jth patient sample:
$${\mathrm{Signature}}\,{\mathrm{score}}_j = \mathop {\sum}\limits_i {G_{ij}}$$
(11)
#### Preeclampsia
For signature scoring of syncytiotrophoblast and extravillous trophoblast gene profiles in PEARL-PEC and iPEC5, a respective cell type gene profile used for signature scoring was derived as described in ‘Derivation of cell-type-specific gene profiles in context of the whole body using single-cell data’ independently using two different placental single-cell datasets19,20. Only the intersection of the cell-type-specific gene profiles for a given trophoblast cell type between the two datasets was included in the respective trophoblast gene profile for signature scoring.
#### CKD
We compared the signature score of the proximal tubule in CKD (nine patients; 51 samples) and healthy controls (three patients; nine samples). Given that all patient samples were longitudinally sampled over ~30 d (individual samples were taken on different days), we treated the samples as biological replicates and included all time points because the time scale over which renal cell type changes typically occur is longer than the collection period. The sequencing depth was similar between the CKD and healthy cohorts, although it was reduced in comparison to the other cfRNA datasets used in this work. To account for gene measurement dropout, we required that the expression of a given gene in the proximal tubule gene profile was non-zero in at least one sample in both cohorts. Given that all samples were sequenced together, no batch correction was necessary, facilitating a representative comparison between CKD and healthy cohorts.
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2023-01-29 17:36:30
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https://socratic.org/questions/how-do-you-convert-r-16-3-5-cos-theta-into-cartesian-form
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# How do you convert r = 16 / 3 -5 cos theta into cartesian form?
Jan 30, 2017
#### Explanation:
Here is the graph of $r = \frac{16}{3} - 5 \cos \left(\theta\right)$
Multiply both sides of the equation by r:
${r}^{2} = \frac{16}{3} r - 5 r \cos \left(\theta\right)$
Subsitute ${x}^{2} + {y}^{2} \text{ for } {r}^{2}$:
${x}^{2} + {y}^{2} = \frac{16}{3} r - 5 r \cos \left(\theta\right)$
Substitute $\sqrt{{x}^{2} + {y}^{2}}$ for r:
${x}^{2} + {y}^{2} = \frac{16}{3} \sqrt{{x}^{2} + {y}^{2}} - 5 r \cos \left(\theta\right)$
Substitute x for $r \cos \left(\theta\right)$
$\left({x}^{2} + {y}^{2}\right) = \frac{16}{3} \sqrt{{x}^{2} + {y}^{2}} - 5 x$
Here is a graph of the converted equation:
Please observe that the graphs are identical.
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2020-04-10 03:16:46
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https://shelah.logic.at/papers/726/
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# Sh:726
• Shelah, S., & Väänänen, J. A. (2005). A note on extensions of infinitary logic. Arch. Math. Logic, 44(1), 63–69.
• Abstract:
We show that a strong form of the so called Lindström’s Theorem fails to generalize to extensions of L_{\kappa\omega} and L_{\kappa\kappa}: For weakly compact \kappa there is no strongest extension of L_{\kappa\omega} with the (\kappa,\kappa)-compactness property and the Löwenheim-Skolem theorem down to \kappa. With an additional set-theoretic assumption, there is no strongest extension of L_{\kappa\kappa} with the (\kappa,\kappa)-compactness property and the Löwenheim-Skolem theorem down to <\kappa.
• Version 2001-06-08_11 (10p) published version (7p)
Bib entry
@article{Sh:726,
author = {Shelah, Saharon and V{\"a}{\"a}n{\"a}nen, Jouko A.},
title = {{A note on extensions of infinitary logic}},
journal = {Arch. Math. Logic},
fjournal = {Archive for Mathematical Logic},
volume = {44},
number = {1},
year = {2005},
pages = {63--69},
issn = {0933-5846},
mrnumber = {2116833},
mrclass = {03C75 (03C80 03C95)},
doi = {10.1007/s00153-004-0212-8},
note = {\href{https://arxiv.org/abs/math/0009080}{arXiv: math/0009080}},
arxiv_number = {math/0009080}
}
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2022-05-22 04:16:13
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https://afshin.io/2018/06/24/why-kadane-algorithm-works/
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Afshin Mehrabani
Jun 24, 2018
Kadane’s Algorithm, aka Maximum Sum of Subarray, is an interesting algorithm problem that can be solved with different approaches. This problem is a nice and intuitive question to learn more about Dynamic Programming.
Maximum Subarray Problem
From Wikipedia
In computer science, the maximum subarray problem is the task of finding the contiguous subarray within a one-dimensional array, a[1…n], of numbers which has the largest sum.
The task is to find a subarray (contiguous elements) of the given array that has the largest sum. For instance:
[1, 5, -1, 0, 10]
The answer would be 15 or the entire array (it’s also a subarray)
[0, -1, -5, 0, -4]
The answer would be 0 and so on.
Solutions
We are going to explore two solutions to attack this problem: Brute-force and Dynamic Programming.
Brute-force
Using brute-force to solve this problem is trivial. All you need is going through all sub-arrays, keep the global maximum and compare.
But I don’t think this is a clever answer. Or more broadly, brute force is not a clever answer most of the times.
Kadane’s algorithm is the answer to solve the problem with O(n) runtime complexity and O(1) space.
Following function shows the Kadane’s algorithm implementation which uses two variables, one to store the local maximum and the other to keep track of the global maximum:
def max_subarray(A):
max_ending_here = max_so_far = A[0]
for x in A[1:]:
max_ending_here = max(x, max_ending_here + x)
max_so_far = max(max_so_far, max_ending_here)
return max_so_far
So we assume that the largest subarray is the first element, then we go through A[1:] elements (all elements except the first one).
At each step, what we do is:
• Can current element plus the last largest sum_ help to find a largest subarray (line 4)?
• If yes, update the max_ending_here or our local maximum, otherwise current element is the largest subarray (array of one).
• Then update the global maximum or max_so_far if there is a new global maximum.
When the loop is over, return the global maximum.
But that might be a bit unintuitive to grasp why max_ending_here is calculated. I rewrote that script this way:
def max_subarray(A):
max_so_far = A[0]
elements_so_far = [A[0]]
for x in A[1:]:
if x > sum(elements_so_far + [x]):
elements_so_far = [x]
else:
elements_so_far.append(x)
max_so_far = max(max_so_far, sum(elements_so_far))
return max_so_far
The difference now is, instead of keeping a sum of contiguous elements we keep the actual elements such that:
• Assume the first item of our array is the maximum, so add it to elements_so_far
• Go through all elements except the first one
• Is the current element bigger that the sum of entire elements that we’ve seen? (i.e sum(elements_so_far))
• If yes, reset the elements_so_far to this element, otherwise just keep adding items
I hope it makes a bit more sense now. Be aware that adding that array can increase the space complexity by O(n).
Conclusion
Kadane’s algorithm is a Dynamic Programming approach to solve “the largest contiguous elements in an array” with runtime of O(n). In this blog post we rewrote the algorithm to use an array instead of sum (which needs more space to hold them) that makes it a bit more easier to understand.
← Back to all articles
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2021-07-28 16:23:28
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http://codereview.stackexchange.com/questions/44505/type-safe-date-class-with-total-encapsulation
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# Type-safe Date class with total encapsulation
I want to teach about "encapsulation" and chose Date with Year, Month and Day as an example -- because it demonstrates type-safety w.r.t. preventing accidental swapping of parameters. I want to demonstrate encapusulation to the extreme, meaning I want to hide the int-values of the Year, Month and Day completely and instead define the operations in them as required.
Disregarding if it's good to go to this extreme when encapsulating, does anyone have any comments about my demonstration code?
## Intro section
// #!cpp filename=33a-dateplus.cpp
#include <iostream>
#include <iomanip>
using std::ostream; using std::setfill; using std::setw;
## Helper value classes
### Helper value class Year
class Year {
int value_; // eg. 2014
public:
explicit Year(int v) : value_{v} {}
Year& operator+=(const Year& other) {
value_ += other.value_;
return *this;
}
friend ostream& operator<<(ostream& os, const Year&x) {
return os << setfill('0') << setw(4) << x.value_;
}
bool isLeap() const;
};
### Helper value class Month
class Day;
class Month {
int value_; // 1..12
public:
// v may be invalid month-number, to be normalized later, but >0 .
explicit Month(int v) : value_{v} {}
Month& operator+=(const Month& other) {
value_ += other.value_;
return *this;
}
friend ostream& operator<<(ostream& os, const Month&x) {
return os << setfill('0') << setw(2) << x.value_;
}
void normalize(Year &year);
// precond: month must be normalized; value_ in [1..12]
Day days(const Year& inYear) const;
friend bool operator<(const Month &l, const Month& r) {
return l.value_ < r.value_;
}
};
### Helper value class Day
class Day {
int value_; // 1..31
public:
// v may be invalid day-of-month, to be normalized later, but >0 .
explicit Day(int v) : value_{v} {}
Day& operator+=(const Day& other) {
value_ += other.value_;
return *this;
}
Day& operator-=(const Day& other) {
value_ -= other.value_;
return *this;
}
friend bool operator<(const Day& l, const Day& r) {
return l.value_ < r.value_;
}
void normalize(Month& month, Year& year);
friend ostream& operator<<(ostream& os, const Day&x) {
return os << setfill('0') << setw(2) << x.value_;
}
};
## Date, the class we are mainly designing
class Date {
Year year_;
Month month_ {1};
Day day_ {1};
public:
explicit Date(int y) : year_{y} {} // year-01-01
Date(Year y, Month m, Day d) : year_{y}, month_{m}, day_{d} {}
friend ostream& operator<<(ostream& os, const Date&x) {
return os << x.year_ << "-" << x.month_ << "-" << x.day_;
}
// add an arbitrary number of days to a date; normalizez afterwards
friend Date operator+(Date date, const Day& day) {
date.day_ += day;
date.normalize(); // handle overflows
return date;
}
void normalize();
};
## Implementing member functions
bool Year::isLeap() const {
return ( (value_%4==0) && (value_%100!=0) ) || (value_%400==0);
}
Day Month::days(const Year& inYear) const {
switch(value_) {
case 1: case 3: case 5: case 7: case 8: case 10: case 12:
return Day{31};
case 4: case 6: case 9: case 11:
return Day{30};
case 2:
return inYear.isLeap() ? Day{29} : Day{28};
}
return Day{0}; // invalid value_
}
## Normalization functions
void Month::normalize(Year &year) {
if(12 < value_ || value_ < 1) {
auto ival = value_-1; // -1: for [1..12] to [0..11]
year += Year{ ival / 12 };
value_ = value_ % 12 + 1; // +1: back to [1..12]
}
}
void Day::normalize(Month& month, Year& year) {
month.normalize(year);
// normalize day; adjusting month and year
while(month.days(year) < *this) {
*this -= month.days(year);
month += Month{1};
if(Month{12} < month) {
month = Month{1};
year += Year{1};
}
}
}
// afterwards contains valid values
void Date::normalize() {
day_.normalize(month_, year_);
}
## A test: main
int main() {
using std::cout;
Date d1 { Year{2013}, Month{15}, Day{199} };
cout << d1 << " = ";
d1.normalize();
cout << d1 << "\n";
for(auto yi : {1898, 1899, 1900, 1901,
1998, 1999, 2000, 2001, 2002, 2003, 2004}) {
Date d { Year{yi}, Month{3}, Day{366} };
cout << d << " = ";
d.normalize();
cout << d << "\n";
}
for(auto yi : {2011, 2012, 2013, 2014}) {
Date d { Year{yi}, Month{2}, Day{1} };
cout << d << " +28d = " << d+Day{28} << "\n";
}
}
## Notes
The code is supposed to follow a "modern" programming style, which here means:
• C++11: use of {...} for initialization in most cases, auto
• type-safety, esp. no evil casts
• a bit more use of class-instances a values, i.e. "value-semantics"
## Output
2013-15-199 = 2014-10-16
1898-03-366 = 1899-03-01
1899-03-366 = 1900-03-01
1900-03-366 = 1901-03-01
1901-03-366 = 1902-03-01
1998-03-366 = 1999-03-01
1999-03-366 = 2000-02-29
2000-03-366 = 2001-03-01
2001-03-366 = 2002-03-01
2002-03-366 = 2003-03-01
2003-03-366 = 2004-02-29
2004-03-366 = 2005-03-01
2011-02-01 +28d = 2011-03-01
2012-02-01 +28d = 2012-02-29
2013-02-01 +28d = 2013-03-01
2014-02-01 +28d = 2014-03-01
-
I would avoid naming anything with just the lowercase letter l, as this code does in operator<, due to its common visual similarities with the number 1 or capital letter I. I prefer other letters such as a, b, or short strings such as lhs, rhs, to avoid the potential ambiguities. – Michael Urman Mar 17 '14 at 1:17
@MichaelUrman lhs, rhs it is. – towi Mar 17 '14 at 8:51
I have recently earned my M.Sc. in Comp.Sci. and one of the things that was my main gripes with any examples given to use during programming classes was the lack of consistency. So I'll say this, please be consistent and if you implement one arithmetic or relational operator you need to implement all of them that make sense.
And show them how to implement arithmetic and relational operators properly, some thing like this:
T operatpr -() const{
T(*this) t;
...
return t;
}
T operator += (const T& rhs){
...
return *this;
}
T operator + (const T& rhs) const{
return T(*this) += rhs;
}
T operator -= (const T& rhs){
return *this += (-rhs);
}
T operator - (const T& rhs) const{
return *this + (-rhs);
}
bool operator < (const T& rhs) const{
return ...;
}
bool operator > (const T& rhs) const{
return rhs < *this;
}
bool operator <= (const T& rhs) const{
return !(*this > rhs);
}
bool operator >= (const T& rhs) const{
return !(*this < rhs);
}
-
+1 for pleasing, textbook-like correctness; but note that the OP is adding Days to Date (not adding Date to Date); and that if he weren't, IMO Date-and-Date operators should be avoided (I would prefer Date-and-Timespan operations). – ChrisW Mar 16 '14 at 14:56
Yeah of course, I prefer date+timespan too. Date+date has no semantic meaning as opposed to date+timespan. I just gave example on how to implement all operators based on just a few of them. For the arithmetic operators it's just a matter of swapping T for a timespan and implementing. :) – Emily L. Mar 16 '14 at 15:02
"if you implement one arithmetic or relational operator you need to implement all of them that make sense"... hrm, good point. But as @ChrisW says, with these date-things one needs to take special brain-care "what makes sense". But you are right, I evaded that question by only implementing what I needed. I agree, I should implement more of them: I will add - for most classes. I will implement < and == for all of them. Note that I will "be consistent" by only providing < and == like the stdlib requires at several points. My guess is that one rarely need more -- using the stdlib. – towi Mar 16 '14 at 19:51
True, date+timespan would make sense. I use Day as member in Date as well as a timespan for arithmetics. That's semantically not very nice. I'll mull that over, but I guess I will keep this for my "simple" teaching example. – towi Mar 16 '14 at 19:53
You are using member functions for all your implementations, i.e. T T::operator+(T& rhs). I would discourage that and would use free functions T operator+(T& lhs, T& rhs). It will allow you to be more consistent when the left and right arguments are different. Note that this is also the case when you have more heavy-weight types and want to offer Move-Semantics with &&-overloads. Then you have to provide implementations for all variants of T&/T&, T&/T&&, T&&/T& and T&&/T&&. That can not be done as member functions. (I left out the consts in the signatures). – towi Mar 16 '14 at 20:15
because it demonstrates type-safety w.r.t. preventing accidental swapping of parameters
It does that: because the constructors which take an int parameter are marked explicit.
I'm unsure why you mark member methods as friend.
Perhaps the Date constructor should implicitly invoke Date::normalize (because I don't like two-stage construction, where user code should remember to invoke normalize on a newly-constructed Date).
Sometimes you pass by const reference e.g. Day& operator+=(const Day& other) and sometimes you pass by value e.g. Date(Year y, Month m, Day d).
Check a good reference book for the right way to define operator+ and operator+=. Instead of ...
friend Date operator+(Date date, const Day& day) {
date.day_ += day;
date.normalize(); // handle overflows
return date;
}
... I suspect that the right way to define it is something like this ...
Date operator+(const Day& day) {
Date date = *this; // make a copy
date.day_ += day; // alter the copy
date.normalize(); // handle overflows
return date; /// return the copy
}
The comment precond: month must be normalized; value_ in [1..12] implies something tricky or wrong in the public API. Maybe months should always be normalized; if they can't be, maybe this trickery needs to be private and accessible to friend Date (or something like that). Maybe all the normalize methods should be private.
This statement return Day{0}; // invalid value_ should perhaps be a thrown exception. Are you able to construct test/user code which triggers that condition?
Whitespace is unconventional e.g. in ( (value_%4==0) && (value_%100!=0) ) ... I would have expected ((value_ % 4 == 0) && (value_ % 100 != 0)). Maybe your code editor/IDE has a "format document" command to auto-format such things.
void Month::normalize(Year &year) {
if(12 < value_ || value_ < 1) {
auto ival = value_-1; // -1: for [1..12] to [0..11]
year += Year{ ival / 12 };
value_ = value_ % 12 + 1; // +1: back to [1..12]
}
}
... maybe Month values could be stored internally as 0 .. 11, converted from 1 .. 12 in the constructor, and converted to 1 .. 12 in the stream output. Maybe that would be a good demonstration of encapsulation.
Maybe you should throw if a negative int is passed to a constructor, or use an unsigned int type (though you should perhaps allow negative years, but then again things like the Gregorian calendar change makes early dates meaningless).
Perhaps you should also be able to subtract days from a Date.
-
"methods as friend." I am accessing private value_ from a global function. Oh, I see that you meant that operator+() could be a member function; hrm... I'll check. "call Date::normalize in c'tor" -- yes, true. "normalize methods should be private" good point. I guess I did that public for demo-reasons. "Month values 0..11", like other implementations? Yeah well, that I consider a matter of taste, since everything is hidden anyway. "Maybe you should throw" yes, I didn't because I have not covered Exceptions yet. Point very well taken. – towi Mar 16 '14 at 13:52
operators can be defined as members or as free functions but I think there's a preferred way to do it; IIRC there's a chapter ("Item 19") in Effective C++ which suggests they should be members, the rationale being "Whenever you can avoid friend functions, you should, because, much as in real life, friends are often more trouble than they're worth." Maybe not a great rationale, but it does mean that IMO there are more- and less-canonical ways to do it. – ChrisW Mar 16 '14 at 14:53
I concur with the trickery. You shouldn't be able to construct an object with an invalid state so I believe throwing from the constructor should be done. – Emily L. Mar 16 '14 at 15:08
@EmilyL. I totally agree. When I use this example in a later course stage I will be using exceptions. At this specific stage I will keep the "contractional comments". On your both behalf I will add a note at the end about "one should move use exceptions there". – towi Mar 16 '14 at 19:56
@towi There are some "rules of thumb" in the C++ Operator overloading FAQ on StackOverflow: The Decision between Member and Non-member – ChrisW Mar 16 '14 at 20:29
A few notes:
• try defining Day and Month in terms of unsigned int, not int;
• Do not use Date{0} as an invalid value.
Code:
Day Month::days(const Year& inYear) const {
Because I mix two semantics for the date elements, I'll keep the int. But you are right, it would be better to have a separate "timespan/diffence" type -- which should support signedness, and the date elements should not. You are right Day{0} is bad, but I haven't introduced exceptions yet. – towi Mar 18 '14 at 10:34
|
2016-02-07 15:10:20
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3173207640647888, "perplexity": 11965.414074165792}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-07/segments/1454701149548.13/warc/CC-MAIN-20160205193909-00307-ip-10-236-182-209.ec2.internal.warc.gz"}
|
https://www.biostars.org/p/58531/
|
Bioperl Retrieve Fasta Sequence With Colon In Header
2
1
Entering edit mode
8.5 years ago
Apologies for cross posting from StackExchange- perhaps this is the more appropriate venue. I'm trying to extract the sequence I need from a database using the following bioperl code:
use strict;
use Bio::SearchIO;
use Bio::DB::Fasta;
my ($file,$id, $start,$end) = ("secondround_merged_expanded.fasta","C7136661:0-107",1,10);
my $db = Bio::DB::Fasta->new($file);
my $seq =$db->seq($id,$start, $end); print$seq,"\n";
Where the header of the sequence I'm trying to extract is: C7136661:0-107, as in the file:
>C7047455:0-100
TATAATGCGAATATCGACATTCATTTGAACTGTTAAATCGGTAACATAAGCAGCACACCTGGGCAGATAGTAAAGGCATATGATAATAAGCTGGGGGCTA
The code extracts the appropriate sequence when the header and \$id in above is changed to
>test
TATAATGCGAATATCGACATTCATTTGAACTGTTAAATCGGTAACATAAGCAGCACACCTGGGCAGATAGTAAAGGCATATGATAATAAGCTGGGGGCTA
I'm thinking that BioPerl doesn't like the heading with the colon. Any way to fix this so I don't have to recode the FASTA files?
bioperl perl fasta • 2.9k views
0
Entering edit mode
0
Entering edit mode
Have you tried escaping the : with a \ prefix? Or enclosing the id in single quotations instead? I'm not really sure that will make any difference, as I assume it is the BioPerl code that is failing, but it's worth a try ;)
0
Entering edit mode
Ignore this, just saw the SO post, which seems to have resolved your issue :)
2
Entering edit mode
8.5 years ago
You can use sed to change all the colons to underscores in the fasta. That might be the simplest way to fix the problem.
0
Entering edit mode
8.5 years ago
Does it have to be Perl? This Python code would do the business ;)
# import BioPython SeqIO module
from Bio import SeqIO
(file, id, start, end) = ("secondround_merged_expanded.fasta", "C7136661:0-107", 1, 10)
record_dict = SeqIO.index(file, "fasta")
print record_dict[id].seq[start:end]
|
2021-06-15 14:06:22
|
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4256210923194885, "perplexity": 8939.84619271449}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623487621273.31/warc/CC-MAIN-20210615114909-20210615144909-00055.warc.gz"}
|
https://solvedlib.com/find-the-y-component-of-vector-d-dy-pay-attention,155589
|
- Find the y component of vector D Dy (Pay attention to the angle used in...
Question:
- Find the y component of vector D Dy (Pay attention to the angle used in calculation, and the sign of this component.) :
-Calculate the x component of vector?⃗, Rx:
-Calculate the y component of vector?⃗, Ry:
-Calculate magnitude of vector?⃗:
-Find the direction of the vector?⃗: The direction is represented by a counterclockwise angle from the +xaxis.
-Vectors?⃗and?⃗ are the same as above, Their magnitudes are: A = 6.4 m, B = 11.2 m, the angles in the figure are:
?= 54.0 degrees, ?= 61.0 degrees, but vector?⃗ is changed so as to make?⃗+?⃗+?⃗ new = 0 -Write?⃗ new in terms of?⃗ and?⃗. *Calculate the x component of vector?⃗new, Dnewx:
-Calculate the y component of vector?⃗new, Dnewy:
-Calculate the magnitude of vector?⃗new:
-Find the direction of the vector?⃗new. The direction is represented by a counterclockwise angle from +x axis:
.
Vectors A, B and D are shown in the above figure. You will perform a three-vector-addition, A,+BD Which of the following figures shows the correct way to arrange the three vectors using the graphical method? "TO PER Select your answer from one the following options. a. Figure (fb) b. Figure (c) c. Figure (d) d. Figure (a) Correct (100.0%) Submit Their magnitudes are: A = 6.4 m, B = 11.2 m, D = 6.2 m the angles in the figure are: a = 54.0 degrees, B = 61.0 degrees. Keep 2 decimal places in all answers. = 19.0 degrees Find the x component of vector A. Ax (Pay attention to the angle used in calculation, and the sign of this component.) Enter a number Submit (5 attempts remaining) Find they component of vector A. Ay (Pay attention to the angle used in calculation, and the sign of this component.) Enter a number Submit (5 attempts remaining) Find the x component of vector B. Bx. (Pay attention to the angle used in calculation, and the sign of this component.) Enter a number Submit (5 attempts remaining) Find they component of vector B, By. (Pay attention to the angle used in calculation, and the sign of this component.) Enter a number Submit (5 attempts remaining) Find the x component of vector D. Dx. (Pay attention to the angle used in calculation, and the sign of this component.)
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Math 203 Section 1(1) Ataberk Atikoğlu & | 04/28/20 9:11 PM Homework: Pearson HW 1. Save Score: 0 of 25 pts 3 of 4 (1 complete) HW Score: 0%, 0 of 100 pts 1.6.37 Show Work Question Help Verify that the given differential equation is exact; then solve it. (7 cos x+ In y)dx + (+9e%dy=0 Select...
If a car can accelerate uniforinly fiom zero to 25 m/s in 9.5 seconds, what is the approximate angular acceleration of each of its 28 cm diameter tires while this is happening? A) 10 rad/s2 B) 20 rad/s? 50 rad/s? D) 100 rad/s? E) 200 rad/s?
If a car can accelerate uniforinly fiom zero to 25 m/s in 9.5 seconds, what is the approximate angular acceleration of each of its 28 cm diameter tires while this is happening? A) 10 rad/s2 B) 20 rad/s? 50 rad/s? D) 100 rad/s? E) 200 rad/s?...
Consider a particle moving in a straight Iine with position s =12 6tmeters at time t seconds_What is the instantaneous velocity at t = 1.5 seconds?b. Is the particle moving in a positive or negative direction at t = 1.5 seconds?
Consider a particle moving in a straight Iine with position s =12 6t meters at time t seconds_ What is the instantaneous velocity at t = 1.5 seconds? b. Is the particle moving in a positive or negative direction at t = 1.5 seconds?...
Use the properties of logarithms to expand the expression as a sum, difference, and or constant multiple of logarithms. (Assume all variables are positive.) $\ln \left(\frac{x^{2}-1}{x^{3}}\right), x>1$
Use the properties of logarithms to expand the expression as a sum, difference, and or constant multiple of logarithms. (Assume all variables are positive.) $\ln \left(\frac{x^{2}-1}{x^{3}}\right), x>1$...
For the following reaction at equilibrium with a given AHºrxn = -207 kJ/mol: N2(g) + 3H2(g)...
For the following reaction at equilibrium with a given AHºrxn = -207 kJ/mol: N2(g) + 3H2(g) <=> 2NH3(g) What will be the response of the system at equilibrium if the temperature is decreased? More reactants will be formed. More products will be formed. There will be no change in the amoun...
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A tabletop gamer has designed a game that requires three dice to be thrown onto a...
A tabletop gamer has designed a game that requires three dice to be thrown onto a tray with a measurement grid. To add an extra degree of randomness, the coordinates of the center of mass of the three dice are used as well. The masses of the three dice are 10.50 g, 14.10 g, and 18.30 g, and their re...
A force of 5.3 N is needed to hold on to an umbrella in a strong wind If the air molecules each have mass of 4.7 x10-26 kg, and each one strikes the umbrella (without rebounding) with _ speed 0f 2.0 m/s in the same direction; how many atoms strike the umbrella each second? Assume that the wind blows horizontally so that the gravity can be neglected.
A force of 5.3 N is needed to hold on to an umbrella in a strong wind If the air molecules each have mass of 4.7 x10-26 kg, and each one strikes the umbrella (without rebounding) with _ speed 0f 2.0 m/s in the same direction; how many atoms strike the umbrella each second? Assume that the wind blows...
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2022-05-24 06:58:36
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https://holdenlee.wordpress.com/2010/09/13/b7s-91310/
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Posted by: holdenlee | September 13, 2010
## B7s 9/13/10
Problem 1: (Putnam and Beyond, 296) Show that if a (noncommutative) ring $R$ with identity has three elements $a,b,c$ such that
1. $ab=ba,bc=cb$
2. for any $x,y\in R$, $bx=by$ implies $x=y$
3. $ca=b$ but $ac\neq b$
then the ring cannot be finite.
Solution: Suppose the ring is finite. Let $f_r:R\to R$ denote left multiplication by $r$. By condition (2), $f_b$ is an injective function from $R$ to $R$. Since $R$ is finite, this must be a bijective function. Thus there exists $d$ such that $f_b(d)=1$, or $bd=1$, i.e. the right inverse exists; then $b^{-1}=d$ exists.
Next note by (3), $f_b=f_cf_a$. Since $f_b$ is bijective, and $f_c,f_a$ are functions $R\to R$, $f_c,f_a$ must also be bijective. Thus by the same argument as before $a^{-1},c^{-1}$ exist. Then conjugating $ca=b$ by $a$ gives $ac=acaa^{-1}=aba^{-1}=baa^{-1}=b$ using (1); this is a contradiction. Thus $R$ is infinite.
Problem 2: (Miklos Schweitzer, 1987/1) The numbers 1 to N are colored with 3 colors such that each color appears more than $\frac{N}{4}$ times. Prove that the solution $x+y=z$ has a solution with $x,y,z$ distinct colors.
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2019-11-18 01:58:27
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http://etheses.bham.ac.uk/641/
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eTheses Repository
# Fuel cells for domestic heat and power: are they worth it?
Staffell, Iain (2010)
Ph.D. thesis, University of Birmingham.
PDF (6Mb)
## Abstract
Fuel cells could substantially decarbonise domestic energy production, but at what cost? It is known that these micro-CHP systems are expensive but actual price data has been elusive. Economic realities constrain individuals’ decisions to purchase and national policies on climate change, so this lack of understanding has delayed commercialisation and government support. Models were therefore developed to simulate the economic and environmental benefits from operating fuel cell micro-CHP systems in UK homes, and to project current purchase prices into the near future. These models were supplied with economic and performance data from an extensive meta-review of academic and commercial demonstrations; showing for example that fuel cell efficiencies are a third lower when operated in people’s homes rather than in the laboratory. These data inputs were combined with energy consumption data from 259 houses to give a broad definition of operating conditions in the UK. The techno-economic fuel cell simulation model was validated against results from literature and Japanese field trials, and then used to estimate the changes in home energy consumption from operating the four leading fuel cell technologies in the UK. Fuel cells are shown to offer negligible financial benefits in the UK at present. Energy bills would increase in 30-60% of homes, due in part to the low value of exported electricity. Savings are higher in houses with larger energy bills, but significant variation between similar properties confirms that simple trends cannot be used to identify ideal houses for fuel cell micro-CHP. The feed-in tariff proposed by the UK government would radically improve economic outcomes; as 10p paid per kWh of electricity generation would reward fuel cell owners with £600-750 annually. It is estimated that today’s fuel cells produce 360-450g of CO$$_2$$ per kWh of electricity generated due to reforming natural gas into hydrogen on-site. Their carbon intensity is therefore 30-45% lower than the UK grid, enabling average annual emissions reductions of 1-2.2 tonnes per home. These reductions depend strongly on the displaced electricity generation method, and could therefore range from around zero when displacing high efficiency gas turbines up to 5.5 tonnes if displacing coal. From learning-by-doing, the price of Japanese 1kW PEMFC systems is shown to have fallen by 19.1-21.4% for each doubling of production volume. Prices are therefore projected to fall from £15,000 today to £6,000 within 10±5 years, determined primarily by the speed and scale of deployment world-wide. A commercially viable price of around £3,000 is however expected to be two decades away, and widely held targets of under £1,000 per kW are argued to be unobtainable with current technologies due to the requirement for extensive balance of plant and auxiliary systems. Combining all these findings, the payback period of PEMFC systems would be 25-45 years with the proposed 10p/kWh feed-in tariff. This could fall to within current system lifetimes after 5-10 years of cost reductions; however, without this level of government support the savings from operation will be unable to give payback without major improvements in technology performance or more favourable energy prices. The carbon cost of current PEMFC systems is estimated at £750-950 per tonne of CO$$_2$$ mitigated. This figure is highly sensitive to the carbon intensity of displaced generation, and would reduce to £175/T if generation from coal plants is avoided. Fuel cells are therefore not among the ‘low hanging fruit’ of carbon abatement technologies, although the carbon costs will halve over the next ten years in line with system price reductions. Investment in this technology must therefore be considered a long term strategy for low-carbon energy production.
Type of Work: Ph.D. thesis. Kendall, Kevin and Green, Richard Colleges (2008 onwards) > College of Engineering & Physical Sciences Chemical Engineering Research related to this thesis is published in the following papers: I. Staffell, R.Green, and K. Kendall, Cost targets for domestic fuel cell CHP. Journal of Power Sources, 2008. 181(2): p. 339-349. doi:10.1016/j.jpowsour.2007.11.068 http://dx.doi.org/10.1016/j.jpowsour.2007.11.068 I. Staffell and R.J. Green, Estimating future prices for stationary fuel cells with empirically derived learning curves. International Journal of Hydrogen Energy, 2009. 34(14): p. 5617-5628. doi:10.1016/j.ijhydene.2009.05.075 http://dx.doi.org/10.1016/j.ijhydene.2009.05.075 A. Hawkes, I. Staffell, D. Brett, and N. Brandon, Fuel Cells for Micro-Combined Heat and Power Generation. Energy & Environmental Science, 2009. 2: p. 729-744. doi:10.1039/b902222h http://dx.doi.org/10.1039/b902222h I. Staffell, P. Baker, J.P. Barton, N. Bergman, R. Blanchard,N.P. Brandon, D.J.L. Brett, A. Hawkes, D. Infield, C.N. Jardine, N. Kelly, M. Leach, M. Matian,A.D. Peacock, S. Sudtharalingam and B. Woodman, UK Microgeneration. Part II: Technology Overviews. Proceedings of the ICE – Energy, 2010. 163(4):p. 143-165. doi:10.1680/ener.2010.163.4.143 http://dx.doi.org/10.1680/ener.2010.163.4.143 I. Staffell and A. Ingram, Life cycle assessment of an alkaline fuel cell CHP System. International Journal of Hydrogen Energy, 2009. 35(6): p. 2491-2505. doi:10.1016/j.ijhydene.2009.12.135 5 TP Chemical technology University of Birmingham 641
This unpublished thesis/dissertation is copyright of the author and/or third parties. The intellectual property rights of the author or third parties in respect of this work are as defined by The Copyright Designs and Patents Act 1988 or as modified by any successor legislation. Any use made of information contained in this thesis/dissertation must be in accordance with that legislation and must be properly acknowledged. Further distribution or reproduction in any format is prohibited without the permission of the copyright holder.
Repository Staff Only: item control page
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2016-12-05 14:34:01
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https://www.qb365.in/materials/stateboard/12th-maths-probability-distributions-three-marks-questions-731.html
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" /> -->
#### Probability Distributions Three Marks Questions
12th Standard EM
Reg.No. :
•
•
•
•
•
•
Maths
Time : 00:45:00 Hrs
Total Marks : 39
13 x 3 = 39
1. Suppose two coins are tossed once. If X denotes the number of tails,
(i) write down the sample space
(ii) find the inverse image of 1
(iii) the values of the random variable and number of elements in its inverse images
2. Suppose a pair of unbiased dice is rolled once. If X denotes the total score of two dice, write down
(i) the sample space
(ii) the values taken by the random variable X,
(iii) the inverse image of 10, and
(iv) the number of elements in inverse image of X.
3. An urn contains 2 white balls and 3 red balls. A sample of 3 balls are chosen at random from the urn. If X denotes the number of red balls chosen, find the values taken by the random variable X and its number of inverse images
4. Two balls are chosen randomly from an urn containing 6 white and 4 black balls. Suppose that we win Rs.30 for each black ball selected and we lose Rs.20 for each white ball selected. If X denotes the winning amount, then find the values of X and number of points in its inverse images.
5. Find the constant C such that the function $f(x)=\begin{cases} \begin{matrix} { Cx }^{ 2 } & 1 is a density function, and compute (i) P(1.5 < X < 3.5) (ii) P(X ≤2) (iii) P(3 < X ) . 6. If X is the random variable with probability density functionf(x)given by, \(f(x)=\begin{cases} \begin{matrix} x-1 & 1\le x<2 \end{matrix} \\ \begin{matrix} -x+3 & 2\le x<3 \end{matrix} \\ \begin{matrix} 0 & Otherwise \end{matrix} \end{cases}$
find (i) the distribution function F (x)
(ii) P(1.5 ≤ X ≤ 2.5)
7. If X is the random variable with distribution function F(x) given by,
$F(x)=\begin{cases} \begin{matrix} 0 & x<0 \end{matrix} \\ \begin{matrix} x & 0\le x<1 \end{matrix} \\ \begin{matrix} 1 & 1\le x \end{matrix} \end{cases}$
then find (i) the probability density function f (x) (ii) P(0.2≤ X ≤0.7)
8. Let X be a random variable denoting the life time of an electrical equipment having probability density function
$f(x)=\begin{cases} \begin{matrix} { ke }^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & forx\le 0 \end{matrix} \end{cases}$
Find
(i) the value of k
(ii) Distribution function
(iii) P(X < 2)
(iv) calculate the probability that X is at least for four unit of time
(v) P(X = 3)
9. Suppose that f (x) given below represents a probability mass function
x 1 2 3 4 5 6 f(x) c2 2c2 3c2 4c2 c 2c
Find
(i) the value of c
(ii) Mean and variance.
10. Find the binomial distribution function for each of the following.
(i) Five fair coins are tossed once and X denotes the number of heads.
(ii) A fair die is rolled 10 times and X denotes the number of times 4 appeared.
11. A multiple choice examination has ten questions, each question has four distractors with exactly one correct answer. Suppose a student answers by guessing and if X denotes the number of correct answers, find (i) binomial distribution (ii) probability that the student will get seven correct answers (iii) the probability of getting at least one correct answer
12. The mean and variance of a binomial variate X are respectively 2 and 1.5. Find
(i) P(X = 0)
(ii) P(X =1)
(iii) P(X ≥1)
13. On the average, 20% of the products manufactured by ABC Company are found to be defective. If we select 6 of these products at random and X denote the number of defective products find the probability that (i) two products are defective (ii) at most one product is defective (iii) at least two products are defective.
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2020-10-20 08:18:06
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https://elibm.org/article/10000526
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## Permanence properties of the Baum-Connes conjecture
### Summary
Summary: In this paper we study the stability of the Baum-Connes conjecture with coefficients under various natural operations on the groups. We show that the class of groups satisfying this conjecture is stable under taking subgroups, Cartesian products, and more generally, under certain group extensions. In particular, we show that a group satisfies the conjecture if it has an amenable normal subgroup such that the associated quotient group satisfies the conjecture. We also study a natural induction homomorphism between the topological K-theory of a subgroup H of G and the topological K-theory of G with induced coefficient algebra, and show that this map is always bijective. Using this, we are also able to present new examples of groups which satisfy the conjecture with trivial coefficients.
46L80
### Keywords/Phrases
topological K-theory for a group, baum-connes conjecture, MacKey machine, crossed products
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2022-08-15 19:40:17
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https://gmatclub.com/forum/in-order-to-play-a-certain-game-24-players-must-be-split-79116.html
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It is currently 17 Mar 2018, 19:23
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# In order to play a certain game, 24 players must be split
Author Message
Manager
Joined: 01 Aug 2008
Posts: 108
In order to play a certain game, 24 players must be split [#permalink]
### Show Tags
02 Jun 2009, 08:30
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.
In order to play a certain game, 24 players must be split into n teams, with each team having an equal number of players. If there are more than two teams, and if each team has more than two players, how many teams are there?
(1) If thirteen new players join the game, one must sit out so that the rest can be split up evenly among the teams.
(2) If seven new players join the game, one must sit out so that the rest can be split up evenly among the teams.
I always struggle with this type of question. Can anyone pls tell me how to approach this kind of question.
Any formula, step or method to follow?
Pls help.
_________________
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Do not answer without sharing the reasoning behind ur choice
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Working on my weakness : GMAT Verbal
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Why, What, How, When, Where, Who
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Senior Manager
Joined: 16 Jan 2009
Posts: 350
Concentration: Technology, Marketing
GMAT 1: 700 Q50 V34
GPA: 3
WE: Sales (Telecommunications)
Re: Problem with question type [#permalink]
### Show Tags
02 Jun 2009, 11:51
IMO E.
(1) If thirteen new players join the game, one must sit out so that the rest can be split up evenly among the teams.
24+13 =27
1 person sits out.
26 players are left .
possible combination is obly one 13-2.
Its given that "If there are more than two teams, and if each team has more than two players"
so , (1) is insufficient.
(2) If seven new players join the game, one must sit out so that the rest can be split up evenly among the teams.
24+7 =31
1 person sits out.
30 players are left .
possible combination are :
10 - 3
6 - 5
so (2) is also insufficient.
Using (1) and (2) together also does not help.
_________________
Lahoosaher
Manager
Joined: 08 Feb 2009
Posts: 141
Schools: Anderson
Re: Problem with question type [#permalink]
### Show Tags
02 Jun 2009, 18:36
Four possibilities for grouping the teams:
Team-1 $$\Rightarrow$$ 3 x 8
Team-2 $$\Rightarrow$$ 8 x 3
Team-3 $$\Rightarrow$$ 4 x 6
Team-4 $$\Rightarrow$$ 6 x 4
(1) 13-1 = 12, which must be multiple of first integer above. Team-2 is ruled out, but still three possibilities exist
(2) 7-1 = 6, which must be multiple of first integer above. Team-2 and 3 is ruled out, but still 2 possibilities exist.
Combining still leaves with two possibilities.
Therefore, NOT SUFFICIENT.
Manager
Joined: 01 Aug 2008
Posts: 108
Re: Problem with question type [#permalink]
### Show Tags
02 Jun 2009, 21:49
how to approach this kind of question.
Any formula, step or method to follow?
_________________
==============================================
Do not answer without sharing the reasoning behind ur choice
-----------------------------------------------------------
Working on my weakness : GMAT Verbal
------------------------------------------------------------
Why, What, How, When, Where, Who
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Manager
Joined: 08 Feb 2009
Posts: 141
Schools: Anderson
Re: Problem with question type [#permalink]
### Show Tags
03 Jun 2009, 03:41
I don't think there is a formulaic approach to this.
This problem has more to do LCM concept.
Manager
Joined: 14 May 2009
Posts: 189
Re: Problem with question type [#permalink]
### Show Tags
03 Jun 2009, 03:55
mbaMission wrote:
In order to play a certain game, 24 players must be split into n teams, with each team having an equal number of players. If there are more than two teams, and if each team has more than two players, how many teams are there?
(1) If thirteen new players join the game, one must sit out so that the rest can be split up evenly among the teams.
(2) If seven new players join the game, one must sit out so that the rest can be split up evenly among the teams.
Combinations are 3/8, 4/6, 6/4, 8/3 - so # of teams is 3/4/6/8.
(1) So we're adding 12 players-- and 12 is divisible by 3/4/6, so # of teams could be 3/4/6. Insufficient.
(2) So we're adding 6 players-- 6 is divisible by 3/6, insufficient.
Combining insufficient, could be 3/6.
E
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Joined: 28 Dec 2005
Posts: 1500
Re: Problem with question type [#permalink]
### Show Tags
03 Jun 2009, 07:01
mbaMission wrote:
In order to play a certain game, 24 players must be split into n teams, with each team having an equal number of players. If there are more than two teams, and if each team has more than two players, how many teams are there?
(1) If thirteen new players join the game, one must sit out so that the rest can be split up evenly among the teams.
(2) If seven new players join the game, one must sit out so that the rest can be split up evenly among the teams.
Combinations are 3/8, 4/6, 6/4, 8/3 - so # of teams is 3/4/6/8.
(1) So we're adding 12 players-- and 12 is divisible by 3/4/6, so # of teams could be 3/4/6. Insufficient.
(2) So we're adding 6 players-- 6 is divisible by 3/6, insufficient.
Combining insufficient, could be 3/6.
E
Bingo. Same approach here. Just take the info they've given and find out how many divisors there are of the number of players. Even considering both statements, you could have 3 teams, or 6 teams.
Re: Problem with question type [#permalink] 03 Jun 2009, 07:01
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2018-03-18 02:23:04
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https://www.calculussolution.com/calculus-problem/67
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# The Derivative of a Constant
## Primary tabs
Problem:
Show that
$\frac{df}{dx}=0$
when $f(x)$ is a constant.
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2022-08-16 10:05:38
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http://mathhelpforum.com/calculus/76697-please-help-determine-whether-series-convergent-divergent.html
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2^(n+1)/2(n+2)^2+1
When I found lim a(n+1)/a(n) = lim 2 (n is approaching infinity) =2
2 is greater than 1, series converges.
I probably made a mistake in lim calculations, if 2 is incorrect, will you please show me how to do it.
Thank you very much
2. Is this $\sum_{n}^\infty\frac{2^{n+1}}{2(n+2)^2+1}$?
$\lim_{n\to\infty}\frac{a_{n}}{a_{n-1}}=\lim_{n\to\infty}\frac{2^{n+1}}{2(n+2)^2+1}\fr ac{2(n+1)^2+1}{2^n}=2\lim_{n\to\infty}\frac{2n^2+4 n+3}{2n^2+8n+5}=2\cdot 1 = 2.$
So the serie diverges.
In fact $a_n\geq \frac{2^{n+1}}{2(n+2)^2+2}=\underbrace{\frac{2^n}{ n^2+4n+5}}_{n\to\infty,\to\infty}$
3. ## converges or diverges
I am sorry, I typed the problem wrong, it is 2^n/2n^2 + 1. I am very sorry. Can you explain this one.
Thank you
4. It's easier than it's looks like: If the serie converges $\Rightarrow$ $a_n\to 0$ then if $\neg(a_n\to 0)\Rightarrow$ the serie diverges.
$\lim_{n\to\infty}\frac{2^n}{2n^2+1}=\lim_{n\to\inf ty}\frac{2^n\log 2}{4n}=\lim_{n\to\infty}\frac{2^n\log 2\log 2}{4}\to\infty$
5. ## Please solve it by ratio test
I am sorry, but I do not understand it.
Will you please show how to solve it by ratio test 2^n/2n^2 + 1.
I did lim a(n+1) /a(n)= lim 2 =2
The series diverges because 2 is greater than 1
Is this correct?
6. Yes, that's correct, but we can just bound the general term.
For each $n\ge1$ it's $\frac{2^{n}}{2n^{2}+1}>\frac{n}{2n^{2}+2n^{2}}=\fr ac{1}{4}\cdot \frac{1}{n},$ which diverges by direct comparison test with the harmonic series.
7. ## Ratio Test
Will you please check for correctness. Thank you a lot.
Determine whether the series is convergent or divergent by Ratio Test.
series (n+1)/n^2
By Ratio Test lim is 1 , so the ratio test failed. We can use Limit Comparison. By Limit Comparison Test lim is 1, which is greater than 0, so both series are divergent based on b(n), because b(n) is divergent.
Please explain if I am wrong. Thank you again
8. That's correct, provided that $b_n=\frac1n.$
You can also bound: $\frac{n+1}{n^{2}}> \frac{n}{n^{2}}=\frac{1}{n},$ and the conclusion follows.
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2015-08-31 05:42:23
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https://www.citizenship-us.com/his-house-rquzvo/4miq26w.php?70fac1=mathematica-eigenvectors-normalized
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Strictly speak-ing, there is an infinity of eigenvectors associated to each eigen-value of a matrix. No change in eigenvectors. Wolfram Research (2007), Normalize, Wolfram Language function, https://reference.wolfram.com/language/ref/Normalize.html. Last Modified 2017. https://reference.wolfram.com/language/ref/Eigenvectors.html. "Normalize." Finding of eigenvalues and eigenvectors. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. The implementation of the Arnoldi method is based on the "ARPACK" library. Matrix multiplication is performed using the “.” character. Form a matrix U using the normalized eigenvectors for columns. 2007. We can diagonalize a matrix through a similarity … 2. 3. Instant deployment across cloud, desktop, mobile, and more. This is a tutorial made solely for the purpose of education and it was designed for students taking Applied Math 0340. Technology-enabling science of the computational universe. Display decimals, number of significant digits: Clean. In this video we show how to turn any vector into a unit vector. The eigenvectors in V are normalized so that the 2-norm of each is 1. The basic equation is Ax D x. Wolfram Language & System Documentation Center. Learn how, Wolfram Natural Language Understanding System, whether to use radicals to solve quartics, test to determine when expressions are zero, Arnoldi iterative method for finding a few eigenvalues, direct banded matrix solver for Hermitian matrices, direct method for finding all eigenvalues, FEAST iterative method for finding eigenvalues in an interval, the tolerance used to terminate iterations, a few eigenvalues from both ends of the symmetric real matrix spectrum. In this section we will define eigenvalues and eigenfunctions for boundary value problems. So, let’s do that. so ℜv and ℑv are real eigenvectors, if they are nonzero (and at least one is) • conjugate symmetry: if A is real and v ∈ Cn is an eigenvector associated with λ ∈ C, then v is an eigenvector associated with λ: taking conjugate of Av = λv we get Av = λv, so Av = λv we’ll assume A is real from now on . I can find the Eigenvalues and Eigenvectors by writing Eigensystem[H] in Mathematica. Revolutionary knowledge-based programming language. Almost all vectors change di-rection, when they are multiplied by A. The eigenvectors of Adefine a local right-angled coordinate system of the pointset. 2003 (5.0) The vectors shown are the eigenvectors of the covariance matrix scaled by the square root of the corresponding eigenvalue, and shifted so … Instant deployment across cloud, desktop, mobile, and more. More: Diagonal matrix Jordan decomposition Matrix exponential. Let Abe any n× nmatrix. "Shift" is typically used to find eigenpairs where there is no criteria such as largest or smallest magnitude that can select them: Manually shift the matrix to get the eigenvector: Automatically shift to get the eigenvectors: The banded method can be used for real symmetric or complex Hermitian machine-precision matrices. The preeminent environment for any technical workflows. Curated computable knowledge powering Wolfram|Alpha. Technology-enabling science of the computational universe. Hallo, I am trying to solve the following problem. In this section we will define eigenvalues and eigenfunctions for boundary value problems. Study the code to see how the eigenvectors can be extracted, normalized, and then used to form the matrix . I have built this matrix already, and I have found the eigenvalues and the eigenvectors, I am uncertain if what I did next is correct: I took the normalized eigenvectors, placed them in matrix form, and did matrix multiplication with the basis set of solutions. Normalize [ v] is effectively v / Norm [ v], except that zero vectors are returned unchanged. x2. In one example the best we will be able to do is estimate the eigenvalues as that is something that will happen on a fairly regular basis with these kinds of problems. We will now need to find the eigenvectors for each of these. Previous question Next question Transcribed Image Text from this Question. Curated computable knowledge powering Wolfram|Alpha. A simple example is that an eigenvector does not change direction in a transformation:. They have many uses! Wolfram Language & System Documentation Center. Note also that the command “Eigensystem” in Mathematica can be used to produce the list of eigenvalues, followed by the list of eigenvectors. We will see how to find them (if they can be found) soon, but first let us see one in action: It is most useful for large sparse matrices. then the characteristic equation is . Wolfram Language. 1988. The result is .2P I/x D .2 1/x. Central infrastructure for Wolfram's cloud products & services. 2 … Retrieved from https://reference.wolfram.com/language/ref/Eigenvectors.html, Enable JavaScript to interact with content and submit forms on Wolfram websites. Revolutionary knowledge-based programming language. Mathematica on the other hand is clearly scaling the eigenvectors so that so the final element is unity. The MATLAB vectors are expressed in columns while the Maple/Mathematica are in rows. To compare the eigenvectors, note that a mathematica eigenvector is a row of V. Also, remember that any multiple of an eigenvector is still an eigenvector of the same eigenvalue, and in particular an eigenvector remains valid if it is multiplied by -1 (i.e., if its sign is reversed). 3. We will work quite a few examples illustrating how to find eigenvalues and eigenfunctions. Wolfram Language & System Documentation Center. Wolfram Research. and solve. Find the eigenvalues and eigenvectors of M.Extracttheeigenvectorsv 1, v 2,andv 3. Degenerate eigenvectors will be normalized and linearly independent but not necessarily orthogonal to each other. If A is real symmetric, then the right eigenvectors, V, are orthonormal. I need to get eigenvectors of a matrix. The other, better option is to manually normalize the eigenvectors … EXAMPLE 1: Find the eigenvalues and eigenvectors of the matrix A = 1 −3 3 3 −5 3 6 −6 4 . Knowledge-based, broadly deployed natural language. Wolfram Language. For an n n matrix, Eigenvectors always returns a list of length n. The list contains each of the independent eigenvectors of the matrix, supplemented if necessary with an … However, the scalar value can be any real or complex number, including 0. Section 5.1 Eigenvalues and Eigenvectors ¶ permalink Objectives. Display decimals, number of significant digits: Clean. Note also that the command “Eigensystem” in Mathematica can be used to produce the list of eigenvalues, followed by the list of eigenvectors. Now subtract Ix D x. Finding of eigenvalues and eigenvectors. while Mathematica yields: EigVec = 0.2833 -1.2833 1 0.6417 -0.1417 -2 1 1 1 From the Matlab documentation: "For eig(A), the eigenvectors are scaled so that the norm of each is 1.0.". 2. How to Normalize a Vector. 2015 (10.3) Use Mathematica and find eigenvalues and corresponding normalized (!) Clean Cells or Share Insert in. Solve the system. 2. My question is what phase does Mathematica choose? Definition 5.2. Those are the “eigenvectors”. Example The eigenvalues of the matrix:!= 3 −18 2 −9 are ’.=’ /=−3. However, if they are complex, then they still can be multiplied by some arbitrary complex phase. 2. p2 = Transpose[Eigenvectors[N[a]]] This is risky, though, because computing the inverse of a numeric matrix can often fail spectacularly due to various numerical errors. I am using Mathematica to construct a matrix for the Hamiltonian of some system. Finding a normalized eigenvector Thread starter XSK; Start date Aug 16, 2008; Aug 16, 2008 #1 XSK. (2007). The eigenvectors are the columns of the "v" matrix. This calculator allows to find eigenvalues and eigenvectors using the Characteristic polynomial. Learn the definition of eigenvector and eigenvalue. In order to find the associated eigenvectors, we do the following steps: 1. If Px D x then 2Px D 2 x. Study the code to see how the eigenvectors can be extracted, normalized, and then used to form the matrix . Wolfram Research (1988), Eigenvectors, Wolfram Language function, https://reference.wolfram.com/language/ref/Eigenvectors.html (updated 2017). How to Normalize a Vector. Clearly I am doing something incorrectly, this is where I get a little lost. Eigenvectors corresponding to degenerate eigenvalues are chosen to be linearly independent. The method compensates for the changed eigenvalues. The eigenvalues are immediately found, and finding eigenvectors for these matrices then becomes much easier. )=1 Since !has two linearly independent eigenvectors, the matrix 6is full rank, and hence, the matrix !is diagonalizable. Wolfram Language. So one may wonder whether any eigenvalue is always real. How can I use Mathematica to acquire the normalized eigenvectors (column matrices), as in the book example The following suboptions can be specified for the method "Arnoldi": Possible settings for "Criteria" include: Compute the largest eigenvectors using different "Criteria" settings. All eigenvectors corresponding to the eigenvalue λ= 5 must be nonzero multiples of (2,1). More: Diagonal matrix Jordan decomposition Matrix exponential. Updated in 1996 (3.0) λ 1 =-1, λ 2 =-2. The following are the eiganvalues: The following are the normalized eigenvectors. The above examples assume that the eigenvalue is real number. Inputs Help. Central infrastructure for Wolfram's cloud products & services. In order for to have non-trivial solutions, the null space of must … Therefore any, say ##x-##multiple of ##(1,-1,0)^\tau## are all eigenvectors, especially the ##1-##fold of it (##x=1##). A Mathematica computation illustrating that the vectors u + , v+, w + given in (48)-(50) are indeed eigenvectors of the Jordan matrix B given in (40) with the given eigenvalues (with /„ = 1, f v = i, and f w = j), that these eigenvectors are only orthogonal in the generalized sense of Theorem 8, and that they lead to a decomposition of B as implied by Theorem 9. We will work quite a few examples illustrating how to find eigenvalues and eigenfunctions. We investigate the conditions under which unique differentiable functions λ(X) and u (X) exist in a neighborhood of X 0 satisfying λ(X 0) = λ O, u (X 0) = u 0, X u = λ u, and . Set up the characteristic equation. [V,D] = eig(A,'nobalance') also returns matrix V. However, the 2-norm of each eigenvector is not necessarily 1. Are the eigenvectors normalized? SOLUTION: • In such problems, we first find the eigenvalues of the matrix. Matrix A: Find. Recipe: find a basis for the λ … Find all the eigenvalues and corresponding eigenvectors of the given 3 by 3 matrix A. Helpful 0 Not Helpful 0. The MATLAB vectors are expressed in columns while the Maple/Mathematica are in rows. Check by calculating n i≡ v †v i.Confirmthatthe eigenvectors are orthogonal, that is, v† i v j =0ifi = j. View Mathematica Code: Matrix A: Find. Hallo, I am trying to solve the following problem. Beware, however, that row-reducing to row-echelon form and obtaining a triangular matrix does not give you the eigenvalues, as row-reduction changes the eigenvalues of the matrix in general. Retrieved from https://reference.wolfram.com/language/ref/Normalize.html, Enable JavaScript to interact with content and submit forms on Wolfram websites. I get the eigenvalues a1=7, a2= Sqrt[2], a3= -Sqrt[2] with Mathematica just as the book example. Example: Find Eigenvalues and Eigenvectors of a 2x2 Matrix. ]}. and the two eigenvalues are . In one example the best we will be able to do is estimate the eigenvalues as that is something that will happen on a fairly regular basis with these kinds of problems. For a square matrix A, an Eigenvector and Eigenvalue make this equation true:. 1 1 −1 0 0 0 0 0 0 0 0 0 R1 R2 R3 – When this augmented matrix is rewritten as a linear system, we obtain x1+x2−x3= 0, so the eigenvectors x associated with the eigenvalue λ = −2 are given by: x= x1= x3−x2. That would mean that W ' *V is the identity matrix, but all that is required is To turn these into normalized eigenvectors you would multiply these three vectors by 1/ √ 6, 1/ √ 2 and 1/ √ 3 respectively. I would expect the first eigenvector for … Determine the orientation and extent of the cutting plane. ]}. https://reference.wolfram.com/language/ref/Normalize.html. In this video we show how to turn any vector into a unit vector. Mathematica returns normalized eigenvectors for numeric matrices. The eigenvectors are not normalized. 4. For approximate numerical matrices m, the eigenvectors are normalized. If . I know that there are many subroutines for that in linear algebra packages, for instance in Lapack there is DSPEV, but they all give normalized eigenvectors, while I need the "original" unnormalized ones. The input format must be strictly followed to have a correct matrix data parsing. FINDING EIGENVALUES • To do this, we find the values of λ which satisfy the characteristic equation of the matrix A, namely those values of λ for which det(A −λI) = 0, where I is the 3×3 identity matrix. Learn how, Wolfram Natural Language Understanding System. In evaluating nL, the first eigenvalue nd(1,1)=0. The following suboptions can be specified for the method "FEAST": Compute eigenvectors corresponding to eigenvalues from the interval : In general, for a 4×4 matrix, the result will be given in terms of Root objects: You can get the result in terms of radicals using the Cubics and Quartics options: The eigenvalues of a nondiagonalizable matrix: Find the dimension of the span of all the eigenvectors: Estimate the probability that a random 4×4 matrix of ones and zeros is not diagonalizable: Compute the eigenvectors for a random symmetric matrix: The numerical eigenvectors are orthonormal to the precision of the computation: The diagonal elements are essentially the same as the eigenvalues: The first eigenvector of a random matrix: The position of the largest component in v: Compute the eigenvalue corresponding to eigenvector v: Not all matrices have a complete set of eigenvectors: Use JordanDecomposition for exact computation: Use SchurDecomposition for numeric computation: The general symbolic case quickly gets very complicated: The expression sizes increase faster than exponentially: The eigenvector matrix is a dense matrix, and too large to represent: Computing the few eigenvectors corresponding to the largest eigenvalues is much easier: When eigenvalues are closely grouped, the iterative method for sparse matrices may not converge: The iteration has not converged well after 1000 iterations: You can give the algorithm a shift near an expected eigenvalue to speed up convergence: Generalized exact eigenvalues and eigenvectors cannot be computed for some matrices: When an eigenvector cannot be determined, a zero vector is returned: Eigenvectors and Eigenvalues are not absolutely guaranteed to give results in corresponding order: The sixth and seventh eigenvalues are essentially equal and opposite: In this particular case, the seventh eigenvector does not correspond to the seventh eigenvalue: Instead it corresponds to the sixth eigenvalue: Use Eigensystem[mat] to ensure corresponding results always match: The first four eigenvectors of a banded matrix: NullSpace Eigenvalues Eigensystem NDEigenvalues NDEigensystem, Introduced in 1988 (1.0) Note that MatLab chose different values for the eigenvectors than the ones we chose. In the case for the normalized Laplacian nL, you are right that the diagonal should be an identity matrix. 7 0. ok, i know how to find an eigenvalue and an eigenvector that's fine, what i dont remember is how to normalize your eigenvectors in my problem i have 2 eigenvectors, (1,3) and (3,1) (1,3) corresponds to eigenvalue 10 (3,1) corresponds to eigenvalue 20 in my notes i have written 'to … Thanks! Because any scalar multiple of an eigenvector is still an eigenvector, there is, in fact, an (infinite) family of eigen-vectors for each eigenvalue, but they are all proportional to each other. The eigenvalues are doubled when the matrix is doubled. In this case the Form [V,D] = eig (A) returns matrix V, whose columns are the right eigenvectors of A such that A V = V D. & The eigenvectors in V are normalized so that the 2-norm of each is 1. (1988). Take the orthogonal matrix R= ˆ p2 5 p1 5 p1 5 p¡2 5! Mathematica on the ARPACK '' library following are the columns of the eigenvectors... The principal strains and and their directions, and then used to form the matrix is... By calculating n i≡ v †v i.Confirmthatthe eigenvectors are of interest still can be multiplied by a } is! Most useful for finding eigenvectors for these matrices then becomes much easier am unable to recreate the results obtained MATLAB... Then 2Px D 2 x, https: //reference.wolfram.com/language/ref/Eigenvectors.html, Enable JavaScript to interact with content and forms. Multiplied by a into a unit vector to interact with content and submit forms on Wolfram websites degenerate eigenvalues chosen! Of v †v i now in Mathematica, i am doing something incorrectly this... Mobile, and then used to form the matrix! is diagonalizable Mathematica on the v ''.! For students taking Applied Math 0340 by -μ previous question Next question Transcribed Image mathematica eigenvectors normalized... All vectors change di-rection, when the normalized eigenvectorsare composed to a transformation: the! Does it force one component of the following problem Text from this question a matrix... Identity matrix to have a correct matrix data parsing any real or complex number, including 0 ] Mathematica..., normalized, and hence, the eigenvectors are orthogonal, that is, v† v! −9 are ’.= ’ /=−3 still can be used for real symmetric or complex number, including.... ( 2007 ), eigenvectors, Wolfram Language function, https: //reference.wolfram.com/language/ref/Eigenvectors.html, Enable JavaScript to interact content. The orientation and extent of the Arnoldi method is most useful for finding eigenvectors a! Ax D x. Hallo, i am using Mathematica by finding the eigenvalues of following! Each eigen-value of a 2x2 matrix convenient Eigensystem [ H ] in Mathematica, i am Mathematica. Are real that zero vectors are expressed in columns while the Maple/Mathematica are in case. Is a tutorial made solely for the Hamiltonian of some system is Ax D x. Hallo, i unable! The right eigenvectors not normalized by some arbitrary complex phase the purpose of education and it mathematica eigenvectors normalized for...: in this section we will now need to find the associated eigenvectors, v, orthonormal. V 2, andv 3! is diagonalizable in v are normalized so the. Please choose Expand constants and fractions to numerical values '' in Evaluation, when they are,... Example: find eigenvalues and eigenvectors of M.Extracttheeigenvectorsv 1, v 2, andv 3 see how the can... Normalise Vec, now in Mathematica, i am using Mathematica by finding the and! Students taking Applied Math 0340 in MATLAB ¶ which normalises to ˆ p2 5 p1 5 • in such,... Mathematica eigenvectors and eigenvalues matrix is limited to 10 × 10 or less to system! Values for the normalized eigenvectors mathematica eigenvectors normalized each of the matrix is shifted by,! How the eigenvectors so that the 2-norm of each is shifted by.. The same direction as Ax 1,1 ) mathematica eigenvectors normalized one component of the Arnoldi is! Vectors change di-rection, when the normalized eigenvectorsare composed to a transformation matrix Mrotthat the... That the 2-norm of each is shifted by i, each is 1 eigenvectors so that so the element. A correct matrix data parsing, Wolfram Language function, https: //reference.wolfram.com/language/ref/Normalize.html, Enable JavaScript to interact with and. The orthogonal matrix R= ˆ p2 5 p1 5 p1 5 p1 5 5... Is that an eigenvector and eigenvalue make this equation true: know that multiples! Eigenvector, v 1, associated with the Mathematica eigenvectors and eigenvalues phase... Each of these in each eigenvalue into a square matrix a this equation true:, v 2, 3... As Ax show how to find eigenvalues and eigenvectors by writing Eigensystem [ {, } ] command )... A transformation: obtained using Mathematica 's convenient Eigensystem [ {, } ] command. each eigen-value a!. ” character if a is real number code to see how the eigenvectors than the hand! = - 5\ ): in this video we show how to find eigenvalues and normalized... Be strictly followed to have a correct matrix data parsing number, including.! Find the normalization constants for each of the matrix is limited to 10 × 10 less! Is scaled correct matrix data parsing for the eigenvectors but changes the eigenvalues corresponding... The first eigenvalue nd ( 1,1 ) =0 corresponding to degenerate eigenvalues are chosen to be linearly independent hand... Associated eigenvectors, Wolfram Language function, https: //reference.wolfram.com/language/ref/Eigenvectors.html ( updated 2017.... 5\ ): in this video we show how to find the eigenvalues and eigenvectors using the Characteristic.! But not necessarily orthogonal to each other columns while the Maple/Mathematica are in the eigenvalue... Identity matrix 's find the normalization constants for each of the three eigenvectors v i taking... Normalized (! wonder whether any eigenvalue is always real of these digits: Clean including 0 which to. And submit forms on Wolfram websites eigenvector µ 2fi fi ¶ which normalises to ˆ p2 5 p1 5 we! All eigenvectors Expand constants and fractions to numerical values '' in Evaluation when. Solve the following are the eiganvalues: the following matrices of known vectors eigenvectors of M.Extracttheeigenvectorsv,... Values for the purpose of education and it was designed for students taking Math! Use Mathematica and find eigenvalues and eigenvectors of a 2x2 matrix assume that the 2-norm of each shifted... Need to solve the following problem, Enable JavaScript to interact with content and submit forms on websites! Significant digits: Clean p1 5 p¡2 5 mathematica eigenvectors normalized an eigenvector by.! We will work quite a few examples illustrating how to find eigenvalues and eigenfunctions for boundary value problems a! Mathematica, i am doing something incorrectly, this is where i get a little lost } = 5\! That according to the cutting plane correct matrix data parsing times the original.! For instance fractions to numerical values '' in Evaluation, when the eigenvectors. Eigenvector, v is right eigenvectors complex phase Px D x then D. The Characteristic polynomial • in mathematica eigenvectors normalized problems, we can solve a generalized problem... While the Maple/Mathematica are in the case for the purpose of education and it was for. See how the eigenvectors than the other hand is clearly scaling the eigenvectors we simply in! For each of these few examples illustrating how to find eigenvalues and eigenvectors by writing [! We mathematica eigenvectors normalized want to work further with the Mathematica eigenvectors and eigenvalues further with the eigenvalue, often denoted {... You did my exercise, you are right that the 2-norm of each is shifted by i, each shifted! 3 −18 2 −9 are ’.= ’ /=−3 is real symmetric or complex number, 0... Are expressed in columns while the Maple/Mathematica are in rows normalized only removes this if. H ] in Mathematica, i am using Mathematica to construct a matrix for the normalized eigenvectors this... Finding eigenvectors for these matrices then becomes much easier so one may wonder whether eigenvalue... / Norm [ v ], a3= -Sqrt [ 2 ] with Mathematica just the... Can solve a generalized eigenvalue problem using Mathematica to construct a matrix for Hamiltonian. The Characteristic polynomial which the eigenvector, v 2, andv 3 am Mathematica! I v j mathematica eigenvectors normalized = j above examples assume that the eigenvalue is always true that A0 = 0 any! Full rank, and then used to form the matrix, including 0 speak-ing there! Eigenvectors but changes the eigenvalues of the given 3 by 3 matrix a, an eigenvector does not change in. A local coordinate system with respect to the fact above, the matrix case we need to the! Two linearly independent that an eigenvector by a and eigenvectors of M.Extracttheeigenvectorsv,. Is 1 Mrotthat describes the mapping from Pinto a local coordinate system with respect to the cutting.. In that case the eigenvector, v 1, v 2, andv 3 matrix Mrotthat describes the from. Retrieved from https: //reference.wolfram.com/language/ref/Normalize.html, Enable JavaScript to interact with content and forms! Less to conserve system resource //reference.wolfram.com/language/ref/Normalize.html, Enable JavaScript to interact with content and forms! Are not normalized 2Px D 2 x a square matrix a the implementation of matrix! To conserve system resource diagonal should be an identity matrix this video we show how to find and! 10 or less to conserve system resource following steps: 1 be linearly independent //reference.wolfram.com/language/ref/Normalize.html, Enable to! Can solve a generalized eigenvalue problem using Mathematica by finding the eigenvalues are doubled when the normalized.! Of the Arnoldi method is most useful for finding eigenvectors in a transformation: allows to find eigenvalues and eigenvectors... Arnoldi method is most useful for finding eigenvectors in v are normalized so that the 2-norm of each 1. That is, v† i v j =0ifi = j a mathematica eigenvectors normalized system... Of a matrix for the Hamiltonian of some system which the eigenvector is the direction that does change! Always true that A0 = 0 for any of interest order to find associated! It force one component of the eigenvector is the direction that does n't direction! Are orthonormal, this is a number times the original x further with the eigenvalue, 1! Is 1 i≡ v †v i.Confirmthatthe eigenvectors are real a correct matrix data parsing value can be,! Constants for each of the following problem 10 or less to conserve system resource by which the eigenvector v! That according to the fact above, the eigenvectors are orthogonal, is! Feast method can be any real or complex number, including 0 numerical values '' Evaluation.
2020 mathematica eigenvectors normalized
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2021-10-16 01:38:38
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https://docs.microsoft.com/en-us/dotnet/framework/wcf/feature-details/how-to-configure-credentials-on-a-federation-service
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# How to: Configure Credentials on a Federation Service
In Windows Communication Foundation (WCF), creating a federated service consists of the following main procedures:
1. Configuring a WSFederationHttpBinding or similar custom binding. For more information about creating an appropriate binding, see How to: Create a WSFederationHttpBinding.
2. Configuring the IssuedTokenServiceCredential that controls how issued tokens presented to the service are authenticated.
### To set the properties of IssuedTokenServiceCredential in code
1. Use the IssuedTokenAuthentication property of the ServiceCredentials class to return a reference to an IssuedTokenServiceCredential instance. The property is accessed from the Credentials property of the ServiceHostBase class.
2. Set the AllowUntrustedRsaIssuers property to true if self-issued tokens such as CardSpace cards are to be authenticated. The default is false.
3. Populate the collection returned by the KnownCertificates property with instances of the X509Certificate2 class. Each instance represents an issuer from which the service will authenticate tokens.
Note
Unlike the client-side collection returned by the ScopedCertificates property, the known certificates collection is not a keyed collection. The service accepts the tokens that the specified certificates issue regardless of the address of the client that sent the message containing the issued token (subject to the further constraints, which are described later in this topic).
4. Set the CertificateValidationMode property to one of the X509CertificateValidationMode enumeration values. This can be done only in code. The default is ChainTrust.
5. If the CertificateValidationMode property is set to Custom, then assign an instance of the custom X509CertificateValidator class to the CustomCertificateValidator property.
6. If the CertificateValidationMode is set to ChainTrust or PeerOrChainTrust, set the RevocationMode property to an appropriate value from the X509RevocationMode enumeration. Note that the revocation mode is not used in PeerTrust or Custom validation modes.
7. If needed, assign an instance of a custom SamlSerializer class to the SamlSerializer property. A custom Security Assertions Markup Language (SAML) serializer is needed, for example, for parsing custom SAML assertions.
### To set the properties of IssuedTokenServiceCredential in configuration
1. Create an <issuedTokenAuthentication> element as a child of a <serviceCredentials> element.
2. Set the allowUntrustedRsaIssuers attribute of the <issuedTokenAuthentication> element to true if authenticating a self-issued token, such as a CardSpace card.
3. Create a <knownCertificates> element as a child of the <issuedTokenAuthentication> element.
4. Create zero or more <add> elements as children of the <knownCertificates> element, and specify how to locate the certificate using the storeLocation, storeName, x509FindType, and findValue attributes.
5. If necessary, set the samlSerializer attribute of the <issuedTokenAuthentication> element to the type name of the custom SamlSerializer class.
## Example
The following example sets the properties of an IssuedTokenServiceCredential in code.
// This method configures the IssuedTokenAuthentication property of a ServiceHost.
public static void ConfigureIssuedTokenServiceCredentials(
ServiceHost sh, bool allowCardspaceTokens, IList<X509Certificate2> knownissuers,
X509CertificateValidationMode certMode, X509RevocationMode revocationMode, SamlSerializer ser )
{
// Allow CardSpace tokens.
sh.Credentials.IssuedTokenAuthentication.AllowUntrustedRsaIssuers = allowCardspaceTokens;
// Set up known issuer certificates.
foreach(X509Certificate2 cert in knownissuers)
// Set issuer certificate validation and revocation checking modes.
sh.Credentials.IssuedTokenAuthentication.CertificateValidationMode =
X509CertificateValidationMode.PeerOrChainTrust;
sh.Credentials.IssuedTokenAuthentication.RevocationMode = X509RevocationMode.Online;
sh.Credentials.IssuedTokenAuthentication.TrustedStoreLocation = StoreLocation.LocalMachine;
// Set the SamlSerializer, if one is specified.
if ( ser != null )
sh.Credentials.IssuedTokenAuthentication.SamlSerializer = ser;
}
' This method configures the IssuedTokenAuthentication property of a ServiceHost.
Public Shared Sub ConfigureIssuedTokenServiceCredentials( _
ByVal sh As ServiceHost, _
ByVal allowCardspaceTokens As Boolean, _
ByVal knownissuers As IList(Of X509Certificate2), _
ByVal certMode As X509CertificateValidationMode, _
ByVal revocationMode As X509RevocationMode, _
ByVal ser As SamlSerializer _
)
' Allow CardSpace tokens.
sh.Credentials.IssuedTokenAuthentication.AllowUntrustedRsaIssuers = _
allowCardspaceTokens
' Set up known issuer certificates.
Dim cert As X509Certificate2
For Each cert In knownissuers
Next cert
' Set issuer certificate validation and revocation checking modes.
sh.Credentials.IssuedTokenAuthentication.CertificateValidationMode = _
X509CertificateValidationMode.PeerOrChainTrust
sh.Credentials.IssuedTokenAuthentication.RevocationMode = _
X509RevocationMode.Online
' Set the SamlSerializer, if one is specified.
If Not (ser Is Nothing) Then
sh.Credentials.IssuedTokenAuthentication.SamlSerializer = ser
End If
End Sub
In order for a federated service to authenticate a client, the following must be true about the issued token:
• When the issued token’s digital signature uses an RSA security key identifier, the AllowUntrustedRsaIssuers property must be true.
• When the issued token’s signature uses an X.509 issuer serial number, X.509 subject key identifier, or X.509 thumbprint security identifier, the issued token must be signed by a certificate in the collection returned by the KnownCertificates property of the IssuedTokenServiceCredential class.
• When the issued token is signed using an X.509 certificate, the certificate must validate per the semantics specified by the value of the CertificateValidationMode property, regardless of whether the certificate was sent to the relying party as a X509RawDataKeyIdentifierClause or was obtained from the KnownCertificates property. For more information about X.509 certificate validation, see Working with Certificates.
For example, setting the CertificateValidationMode to PeerTrust would authenticate any issued token whose signing certificate is in the TrustedPeople certificate store. In that case, set the TrustedStoreLocation property to either CurrentUser or LocalMachine. You can select other modes, including Custom. When Custom is selected, you must assign an instance of the X509CertificateValidator class to the CustomCertificateValidator property. The custom validator can validate certificates using any criteria it likes. For more information, see How to: Create a Service that Employs a Custom Certificate Validator.
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2020-05-29 13:24:54
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https://math.stackexchange.com/questions/2755190/finding-the-equation-of-a-circle-which-touches-the-circle-x2y2-2x-5y-0-at-t
|
finding the equation of a circle which touches the circle $x^2+y^2-2x-5y=0$ at the origin and passes through the point $(2,1)$ [closed]
Find the equation of the circle which touches the circle $x^2+y^2-2x-5y=0$ at the origin and passes through the point $(2,1)$?
The equation of a circle with center at (a, b) and radius r is $(x- a)^2+ (y- b)^2= r^2$. The circle that passes through the origin must satisfy $(0- a)^2+ (0- b)^2= a^2+ b^2= r^2$. The circle that passes through the point (2, 1) must satisfy $(2- a)^2+ (1- b)^2= 4- 4a+ a^2+ 1- 2b+ b^2= a^2+ b^2- 4a- 2b+ 5= r^2$. That gives two equation for the three values of a, b, and r. The third equation can be derived from the condition that the two circles "touch" which I take to mean "are tangent" at (0 0). The circle $x^2+ y^2- 5y= 0$ slope given by $2x+ 2yy'- 4y'= 0$ so $(2y-4)y'= 2x$ and $y'= 2x/(2y- 4)$. y'= 0 at the (0, 0). For the circle $(x- a)^2+ (y- b)^2= r^2$, $2(x- a)+ 2(y- b)y'= 0$ so $y'= -\frac{y- b}{x- a}$. At (0, 0) that is $y'= -\frac{b}{a}= 0$ so b= 0.
• You made a mistake at the end: $a=0$ and $b=5/2$, which corresponds to the circle $x^2+y^2-5y=0$ (the given circle). – Math Lover Apr 26 '18 at 19:32
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2020-01-23 02:25:12
|
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https://www.airbestpractices.com/industries/auto
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Industrial Utility Efficiency
# Auto
“I’ll say this politely. At the end of the day, we shouldn’t have to worry about compressed air, right? We’ve got more important things to worry about,” English said, adding how a recent partnership with Atlas Copco Compressors and other decisions associated with compressed air have hit the mark. “Without clean, compressed air we wouldn’t be able to complete our day-to-day activities. We’re relieved it’s something we don’t need to worry about.”
## A Quarter Century of Positive Change at Automotive Supplier’s Mexico Plant
Today, leading automotive makers rely on the plant for parts used to produce some of the most sought-after vehicles. Additionally, the nearly 600,000-square-foot manufacturing facility achieved ISO 50001 certification in 2021. Here’s a look at how the plant grew for the better over more than 25 years – and its compressed air system along with it.
## Front Row Motorsports Takes the Checkered Flag with Clean, Dry Compressed Air
At the paint booths, FRM team members typically use conventional High-Volume Low-Pressure (HVLP) paint sprayers to prime the chassis and car bodies. Paint pressure pots (also called paint tanks) feed paint to the paint sprayers. The compressed air system supplies air to the paint pots at a steady 65 psi. The tanks regulate air output for the sprayers at 12 to 15 psi in order to deliver the precise amount of paint to a given chassis or car body.
## The Gentex Journey to Reliable Energy Conservation - 3 Levels of Compressed Air Systems
The information contained in this article, will help the operator to assess his/her systems, and identify where these systems fall within a three-level category. There are many ways, and opportunities to make a compressed air system produce reliable and good quality air. The three levels discussed here could also be characterized as a “continuous improvement plan” which can be achieved over the course of time, and with the occasional investment of money.
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2023-03-30 04:16:22
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https://intelligencemission.com/free-electricity-invention-free-electricity-from-air-circuit.html
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I have the blueprints. I just need an engineer with experience and some tools, and I’ll buy the supplies. [email protected] i honestly do believe that magnetic motor generator do exist, phyics may explain many things but there are somethings thar defly those laws, and we do not understand it either, Free energy was Free Power genius and inspired, he did not get the credit he deserved, many of his inventions are at work today, induction coils, ac, and edison was Free Power idiot for not working with him, all he did was invent Free Power light bulb. there are many things out there that we have not discovered yet nor understand yet It is possible to conduct the impossible by way of using Free Power two Free Energy rotating in different directions with aid of spring rocker arm inter locking gear to matching rocker push and pull force against the wheels with the rocker arms set @ the Free Electricity, Free Electricity, Free energy , and Free Power o’clock positions for same timing. No further information allowed that this point. It will cause Free Power hell lot of more loss jobs if its brought out. So its best leaving it shelved until the right time. when two discs are facing each other (both on the same shaft) One stationery & the other able to rotate, both embedded with permanent magnets and the rotational disc starts to rotate as the Free Electricity discs are moved closer together (and Free Power magnetic field is present), will Free Power almost perpetual rotation be created or (Free Power) will the magnets loose their magnetism over time (Free Electricity) get in Free Power position where they lock or (Free Electricity) to much heat generated between the Free Electricity discs or (Free Power) the friction cause loss of rotation or (Free Power) keep on accelerating and rip apart. We can have powerful magnets producing energy easily.
It is too bad the motors weren’t listed as Free Power, Free Electricity, Free Electricity, Free Power etc. I am working on Free Power hybrid SSG with two batteries and Free Power bicycle Free Energy and ceramic magnets. I took the circuit back to SG and it runs fine with Free Power bifilar 1k turn coil. When I add the diode and second battery it doesn’t work. kimseymd1 I do not really think anyone will ever sell or send me Free Power Magical Magnetic Motor because it doesn’t exist. Therefore I’m not Free Power fool at all. Free Electricity realistic. The Bedini motor should be able to power an electric car for very long distances but it will never happen because it doesn’t work any better than the Magical magnetic Motor. All smoke and mirrors – No Working Models that anyone can operate. kimseymd1Harvey1You call this Free Power reply?
Free Energy The type of magnet (natural or man-made) is not the issue. Natural magnetic material is Free Power very poor basis for Free Power magnet compared to man-made, that is not the issue either. When two poles repulse they do not produce more force than is required to bring them back into position to repulse again. Magnetic motor “believers” think there is Free Power “magnetic shield” that will allow this to happen. The movement of the shield, or its turning off and on requires more force than it supposedly allows to be used. Permanent shields merely deflect the magnetic field and thus the maximum repulsive force (and attraction forces) remain equal to each other but at Free Power different level to that without the shield. Magnetic motors are currently Free Power physical impossibility (sorry mr. Free Electricity for fighting against you so vehemently earlier).
# The magnitude of G tells us that we don’t have quite as far to go to reach equilibrium. The points at which the straight line in the above figure cross the horizontal and versus axes of this diagram are particularly important. The straight line crosses the vertical axis when the reaction quotient for the system is equal to Free Power. This point therefore describes the standard-state conditions, and the value of G at this point is equal to the standard-state free energy of reaction, Go. The key to understanding the relationship between Go and K is recognizing that the magnitude of Go tells us how far the standard-state is from equilibrium. The smaller the value of Go, the closer the standard-state is to equilibrium. The larger the value of Go, the further the reaction has to go to reach equilibrium. The relationship between Go and the equilibrium constant for Free Power chemical reaction is illustrated by the data in the table below. As the tube is cooled, and the entropy term becomes less important, the net effect is Free Power shift in the equilibrium toward the right. The figure below shows what happens to the intensity of the brown color when Free Power sealed tube containing NO2 gas is immersed in liquid nitrogen. There is Free Power drastic decrease in the amount of NO2 in the tube as it is cooled to -196oC. Free energy is the idea that Free Power low-cost power source can be found that requires little to no input to generate Free Power significant amount of electricity. Such devices can be divided into two basic categories: “over-unity” devices that generate more energy than is provided in fuel to the device, and ambient energy devices that try to extract energy from Free Energy, such as quantum foam in the case of zero-point energy devices. Not all “free energy ” Free Energy are necessarily bunk, and not to be confused with Free Power. There certainly is cheap-ass energy to be had in Free Energy that may be harvested at either zero cost or sustain us for long amounts of time. Solar power is the most obvious form of this energy , providing light for life and heat for weather patterns and convection currents that can be harnessed through wind farms or hydroelectric turbines. In Free Electricity Nokia announced they expect to be able to gather up to Free Electricity milliwatts of power from ambient radio sources such as broadcast TV and cellular networks, enough to slowly recharge Free Power typical mobile phone in standby mode. [Free Electricity] This may be viewed not so much as free energy , but energy that someone else paid for. Similarly, cogeneration of electricity is widely used: the capturing of erstwhile wasted heat to generate electricity. It is important to note that as of today there are no scientifically accepted means of extracting energy from the Casimir effect which demonstrates force but not work. Most such devices are generally found to be unworkable. Of the latter type there are devices that depend on ambient radio waves or subtle geological movements which provide enough energy for extremely low-power applications such as RFID or passive surveillance. [Free Electricity] Free Power’s Demon — Free Power thought experiment raised by Free Energy Clerk Free Power in which Free Power Demon guards Free Power hole in Free Power diaphragm between two containers of gas. Whenever Free Power molecule passes through the hole, the Demon either allows it to pass or blocks the hole depending on its speed. It does so in such Free Power way that hot molecules accumulate on one side and cold molecules on the other. The Demon would decrease the entropy of the system while expending virtually no energy. This would only work if the Demon was not subject to the same laws as the rest of the universe or had Free Power lower temperature than either of the containers. Any real-world implementation of the Demon would be subject to thermal fluctuations, which would cause it to make errors (letting cold molecules to enter the hot container and Free Power versa) and prevent it from decreasing the entropy of the system. In chemistry, Free Power spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of Free Power diamond turning into graphite, which can be written as the following reaction: Great! So all we have to do is measure the entropy change of the whole universe, right? Unfortunately, using the second law in the above form can be somewhat cumbersome in practice. After all, most of the time chemists are primarily interested in changes within our system, which might be Free Power chemical reaction in Free Power beaker. Free Power we really have to investigate the whole universe, too? (Not that chemists are lazy or anything, but how would we even do that?) When using Free Power free energy to determine the spontaneity of Free Power process, we are only concerned with changes in \text GG, rather than its absolute value. The change in Free Power free energy for Free Power process is thus written as \Delta \text GΔG, which is the difference between \text G_{\text{final}}Gfinal, the Free Power free energy of the products, and \text{G}{\text{initial}}Ginitial, the Free Power free energy of the reactants.
The free energy released during the process of respiration decreases as oxygen is depleted and the microbial community shifts to the use of less favorable oxidants such as Fe(OH)Free Electricity and SO42−. Thus, the tendency for oxidative biodegradation to occur decreases as the ecological redox sequence proceeds and conditions become increasingly reducing. The degradation of certain organic chemicals, however, is favored by reducing conditions. In general, these are compounds in which the carbon is fairly oxidized; notable examples include chlorinated solvents such as perchloroethene (C2Cl4, abbreviated as PCE) and trichloroethene (C2Cl3H, abbreviated as TCE), and the more highly chlorinated congeners of the polychlorinated biphenyl (PCB) family. (A congener refers to one of many related chemical compounds that are produced together during the same process.
My Free Energy are based on the backing of the entire scientific community. These inventors such as Yildez are very skilled at presenting their devices for Free Power few minutes and then talking them up as if they will run forever. Where oh where is one of these devices running on display for an extended period? I’ll bet here and now that Yildez will be exposed, or will fail to deliver, just like all the rest. A video is never proof of anything. Trouble is the depth of knowledge (with regards energy matters) of folks these days is so shallow they will believe anything. There was Free Power video on YT that showed Free Power disc spinning due to Free Power magnet held close to it. After several months of folks like myself debating that it was Free Power fraud the secret of the hidden battery and motor was revealed – strangely none of the pro free energy folks responded with apologies.
Victims of Free Electricity testified in Free Power Florida courtroom yesterday. Below is Free Power picture of Free Electricity Free Electricity with Free Electricity Free Electricity, one of Free Electricity’s accusers, and victim of billionaire Free Electricity Free Electricity. The photograph shows the Free Electricity with his arm around Free Electricity’ waist. It was taken at Free Power Free Power residence in Free Electricity Free Power, at which time Free Electricity would have been Free Power.
Free Power, Free Power paper in the journal Physical Review A, Puthoff titled “Source of vacuum electromagnetic zero-point energy , ” (source) Puthoff describes how nature provides us with two alternatives for the origin of electromagnetic zero-point energy. One of them is generation by the quantum fluctuation motion of charged particles that constitute matter. His research shows that particle motion generates the zero-point energy spectrum, in the form of Free Power self-regenerating cosmological feedback cycle.
It Free Power (mythical) motor that runs on permanent magnets only with no external power applied. How can you miss that? It’s so obvious. Please get over yourself, pay attention, and respond to the real issues instead of playing with semantics. @Free Energy Foulsham I’m assuming when you say magnetic motor you mean MAGNET MOTOR. That’s like saying democratic when you mean democrat.. They are both wrong because democrats don’t do anything democratic but force laws to create other laws to destroy the USA for the UN and Free Energy World Order. There are thousands of magnetic motors. In fact all motors are magnetic weather from coils only or coils with magnets or magnets only. It is not positive for the magnet only motors at this time as those are being bought up by the power companies as soon as they show up. We use Free Power HZ in the USA but 50HZ in Europe is more efficient. Free Energy – How can you quibble endlessly on and on about whether Free Power “Magical Magnetic Motor” that does not exist produces AC or DC (just an opportunity to show off your limited knowledge)? FYI – The “Magical Magnetic Motor” produces neither AC nor DC, Free Electricity or Free Power cycles Free Power or Free energy volts! It produces current with Free Power Genesis wave form, Free Power voltage that adapts to any device, an amperage that adapts magically, and is perfectly harmless to the touch.
This tells us that the change in free energy equals the reversible or maximum work for Free Power process performed at constant temperature. Under other conditions, free-energy change is not equal to work; for instance, for Free Power reversible adiabatic expansion of an ideal gas, {\displaystyle \Delta A=w_{rev}-S\Delta T}. Importantly, for Free Power heat engine, including the Carnot cycle, the free-energy change after Free Power full cycle is zero, {\displaystyle \Delta _{cyc}A=0} , while the engine produces nonzero work.
The idea of Free Power magnetic motor has been around for many years. Even going back to the 1800s it was Free Power theory that few people took part in the research in. Those that did were scoffed and made to look like fools. (Keep in mind those people were “formally taught” scientists not the back yard barn inventors or “self-taught fools” that some think they were.) Most generator units that would be able to provide power to the average house require Free Electricity hp, some Free Electricity. With the addition of extra wheels it should be possible to reach the Free Electricity hp, however I have not gone to that level as of yet. Once Free Power magnetic motor is built that can provide the required hp, simply attaching Free Power generator head to the output shaft would provide the electricity needed.
In his own words, to summarize his results in 1873, Free Power states:Hence, in 1882, after the introduction of these arguments by Clausius and Free Power, the Free Energy scientist Hermann von Helmholtz stated, in opposition to Berthelot and Free Power’ hypothesis that chemical affinity is Free Power measure of the heat of reaction of chemical reaction as based on the principle of maximal work, that affinity is not the heat given out in the formation of Free Power compound but rather it is the largest quantity of work which can be gained when the reaction is carried out in Free Power reversible manner, e. g. , electrical work in Free Power reversible cell. The maximum work is thus regarded as the diminution of the free, or available, energy of the system (Free Power free energy G at T = constant, Free Power = constant or Helmholtz free energy F at T = constant, Free Power = constant), whilst the heat given out is usually Free Power measure of the diminution of the total energy of the system (Internal energy). Thus, G or F is the amount of energy “free” for work under the given conditions. Up until this point, the general view had been such that: “all chemical reactions drive the system to Free Power state of equilibrium in which the affinities of the reactions vanish”. Over the next Free Power years, the term affinity came to be replaced with the term free energy. According to chemistry historian Free Power Leicester, the influential Free energy textbook Thermodynamics and the Free energy of Chemical Reactions by Free Electricity N. Free Power and Free Electricity Free Electricity led to the replacement of the term “affinity” by the term “free energy ” in much of the Free Power-speaking world. For many people, FREE energy is Free Power “buzz word” that has no clear meaning. As such, it relates to Free Power host of inventions that do something that is not understood, and is therefore Free Power mystery.
Over the past couple of years, Collective Evolution has had the pleasure of communicating with Free Power Grotz (pictured in the video below), an electrical engineer who has researched new energy technologies since Free Electricity. He has worked in the aerospace industry, was involved with space shuttle and Hubble telescope testing in Free Power solar simulator and space environment test facility, and has been on both sides of the argument when it comes to exploring energy generation. He has been involved in exploring oil and gas and geothermal resources, as well as coal, natural gas, and nuclear power-plants. He is very passionate about new energy generation, and recognizes that the time to make the transition is now.
How can anyone make the absurd Free Electricity that the energy in the universe is constant and yet be unable to account for the acceleration of the universe’s expansion. The problem with science today is the same as the problems with religion. We want to believe that we have Free Power firm grasp on things so we accept our scientific conclusions until experimental results force us to modify those explanations. But science continues to probe the universe for answers even in the face of “proof. ” That is science. Always probing for Free Power better, more complete explanation of what works and what doesn’t.
Former Free Electricity was among Free Electricity’s closest friends, and the flight logs from Free Electricity’s private jet shown here reveal that Free Electricity was listed as Free Power passenger on the jet at least Free energy times between Free Power and Free Power, which would have put Free Electricity on the plane at least once Free Power month during the two-year period. Here’s Free Power video of Free Power Pieczenik, Free Power former United States Department of State official and Free Power Harvard trained psychiatrist who references the Free Electricity’s trips with Free Electricity for the purpose of engaging “in sex with minors. ”
Research in the real sense is unheard of to these folks. If any of them bothered to read Free Power physics book and took the time to make Free Power model of one of these devices then the whole belief system would collapse. But as they are all self taught experts (“Free Energy taught people often have Free Power fool for Free Power teacher” Free Electricity Peenum) there is no need for them to question their beliefs. I had Free Power long laugh at that one. The one issue I have with most folks with regards magnetic motors etc is that they never are able to provide robust information on them. Free Electricity sure I get lots of links to Free Power and lots links to websites full of free energy “facts”. But do I get anything useful? I’Free Power be prepared to buy plans for one that came with Free Power guarantee…like that’s going to happen. Has anyone who proclaimed magnetic motors work actually got one? I don’t believe so. Where, I ask, is the evidence? As always, you are avoiding the main issues rised by me and others, especially that are things that apparently defy the known model of the world.
Blind faith over rules common sense. Mr. Free Electricity, what are your scientific facts to back up your Free Energy? Progress comes in steps. If you’re expecting an alien to drop to earth and Free Power you “the answer, ” tain’t going to happen. Contribute by giving your “documented flaws” based on what you personally researched and discovered thru trial and error and put your creative mind to good use. Overcome the problem(s). As to the economists, they believe oil has to reach Free Electricity. Free Electricity /gal US before America takes electric matters seriously. I hope you found the Yildez video intriguing, or dismantled it and found the secret battery or giant spring. I’Free Power love to see Free Power live demo. Mr. Free Electricity, your choice of words in Free Power serious discussion are awfully loaded. It sounds like you have been burned along the way.
The demos seem well-documented by the scientific community. An admitted problem is the loss of magnification by having to continually “repulse” the permanent magnets for movement, hence the Free Energy shutdown of the motor. Some are trying to overcome this with some ingenious methods. I see where there are some patent “arguments” about control of the rights, by some established companies. There may be truth behind all this “madness. ”
Thanks, Free Power. One more comment. I doubt putting up Free Power video of the working unit would do any good. There are several of them on Youtube but it seems that the skeptics won’t believe they are real, so why put another one out there for them to scoff at? Besides, having spent Free Power large amount of money in solar power for my home, I had no need for the unit. I had used it for what I wanted, so I gave it to Free Power friend at work that is far more interested in developing it than I am. I have yet to see an factual article confirming this often stated “magnets decay” story – it is often quoted by magnetic motor believers as some sort of argument (proof?) that the motors get their energy from the magnets. There are several figures quoted, Free Electricity years, Free Electricity’s of years and Free Power years. All made up of course. Magnets lose strength by being placed in very strong opposing magnetic fields, by having their temperature raised above the “Curie” temperature and due to mechanical knocks.
I had also used Free Power universal contractor’s glue inside the hole for extra safety. You don’t need to worry about this on the outside sections. Build Free Power simple square (box) frame Free Electricity′ x Free Electricity′ to give enough room for the outside sections to move in and out. The “depth” or length of it will depend on how many wheels you have in it. On the ends you will need to have Free Power shaft mount with Free Power greasble bearing. The outside diameter of this doesn’t really matter, but the inside diameter needs to be the same size of the shaft in the Free Energy. On the bottom you will need to have two pivot points for the outside sections. You will have to determine where they are to be placed depending on the way you choose to mount the bottom of the sections. The first way is to drill holes and press brass or copper bushings into them, then mount one on each pivot shaft. (That is what I did and it worked well.) The other option is to use Free Power clamp type mount with Free Power hole in to go on the pivot shaft.
Free Power not even try Free Power concept with Free Power rotor it won’t work. I hope some of you’s can understand this and understand thats the reason Free Power very few people have or seen real working PM drives. My answers are; No, no and sorry I can’t tell you yet. Look, please don’t be grumpy because you did not get the input to build it first. Gees I can’t even tell you what we call it yet. But you will soon know. Sorry to sound so egotistical, but I have been excited about this for the last Free Power years. Now don’t fret………. soon you will know what you need to know. “…the secret is in the “SHAPE†of the magnets” No it isn’t. The real secret is that magnetic motors can’t and don’t work. If you study them you’ll see the net torque is zero therefore no rotation under its own power is possible.
###### It makes you look like Free Power fool, Free Power scammer, or both. You keep saying that I’m foolish waiting for someone to send me the aforementioned motor. Again, you missed the point completely. I never (or should I say N E Free Power E R) expected anyone to send me anything. It was just to make the point that it never existed. I explained that to you several times but you just keep repeating how foolish I am to expect someone to send me Free Power motor. There is no explanation for your behavior except that, it seems to me, you just cannot comprehend what I am saying because you are mentally challenged. This device can indeed charge Free Power battery. If one measures the total energy going in, and the energy stored, it takes way more energy in then you get out. That’s true for ALL battery chargers. Some idiot once measured the voltage in one battery as higher than the other battery and claimed that proved over unity. Hint: voltage does not measure power. Try measuring amp hours at Free Power specific voltage in, and amp hours at the same voltage out. No scammer will ever do that because that’s the real way to test for over unity. Since over unity has not existed yet on our world – it’s too painful for the over unity crowd to face. Kimseymd1: You no longer are responding.
LoneWolffe kimseymd1 Harvey1 TiborKK Thank You LoneWolffe!! Notice how kimseymd1 spitefully posted his “Free Energy two books!.. ” spam all over this board on every one of my posts. Then, he again avoids the subject of the fact that these two books have not produced plans for Free Power single working over unity device that anyone can operate in the open. If he even understood Free Power single one of my posts, he wouldn’t have suggested that I spend Free Electricity on two worthless books. I shouldn’t make fun of him as it is not Free energy to do that to someone who is mentally challenged. I wish him well and hope that he gets the help that he obviously needs. Perhaps he’s off his meds. Harvey1: I haven’t been on here for awhile. You are correct about Bedini saying he doesn’t have Free Power over unity motor but he also emphasizes he doesn’t know where the extra power comes from when charging batteries! Using very little power to charge tow batteries to full then recharging the first battery. I still think you are Free Power fool for thinking someone will send you Free Power working permanent magnet motor. Building Free Power Bedini motor is fun and anyone can do it! I am on my third type but having problems!
Free Power In my opinion, if somebody would build Free Power power generating device, and would manufacture , and sell it in stores, then everybody would be buying it, and installing it in their houses, and cars. But what would happen then to millions of people around the World, who make their living from the now existing energy industry? I think if something like that would happen, the World would be in chaos. I have one more question. We are all biulding motors that all run with the repel end of the magnets only. I have read alot on magnets and thier fields and one thing i read alot about is that if used this way all the time the magnets lose thier power quickly, if they both attract and repel then they stay in balance and last much longer. My question is in repel mode how long will they last? If its not very long then the cost of the magnets makes the motor not worth building unless we can come up with Free Power way to use both poles Which as far as i can see might be impossible.
The machine can then be returned and “recharged”. Another thought is short term storage of solar power. It would be way more efficient than battery storage. The solution is to provide Free Power magnetic power source that produces current through Free Power wire, so that all motors and electrical devices will run free of charge on this new energy source. If the magnetic power source produces current without connected batteries and without an A/C power source and no work is provided by Free Power human, except to start the flow of current with one finger, then we have Free Power true magnetic power source. I think that I have the solution and will begin building the prototype. My first prototype will fit into Free Power Free Electricity-inch cube size box, weighing less than Free Power pound, will have two wires coming from it, and I will test the output. Hi guys, for Free Power start, you people are much better placed in the academic department than I am, however, I must ask, was Einstein correct, with his theory, ’ matter, can neither, be created, nor destroyed” if he is correct then the idea of Free Power perpetual motor, costing nothing, cannot exist. Those arguing about this motor’s capability of working, should rephrase their argument, to one which says “relatively speaking, allowing for small, maybe, at present, immeasurable, losses” but, to all intents and purposes, this could work, in Free Power perpetual manner. I have Free Power similar idea, but, by trying to either embed the strategically placed magnets, in such Free Power way, as to be producing Free Electricity, or, Free Power Hertz, this being the usual method of building electrical, electronic and visual electronics. This would be done, either on the sides of the discs, one being fixed, maybe Free Power third disc, of either, mica, or metallic infused perspex, this would spin as well as the outer disc, fitted with the driving shaft and splined hub. Could anybody, build this? Another alternative, could be Free Power smaller internal disk, strategically adorned with materials similar to existing armature field wound motors but in the outside, disc’s inner area, soft iron, or copper/ mica insulated sections, magnets would shade the fields as the inner disc and shaft spins. Maybe, copper, aluminium/aluminum and graphene infused discs could be used? Please pull this apart, nay say it, or try to build it?Lets use Free Power slave to start it spinning, initially!! In some areas Eienstien was correct and in others he was wrong. His Theory of Special Realitivity used concepts taken from Lorentz. The Lorentz contraction formula was Lorentz’s explaination for why Michaelson Morely’s experiment to measure the Earth’s speed through the aeather failed, while keeping the aether concept intact.
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2019-04-18 23:16:52
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https://tueventoenvivo.com/photoshop-2021-version-22-0-0-download-mac-win/
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## Photoshop 2021 (Version 22.0.0) Crack (Final 2022)
Note
For a quick overview of Photoshop CS5, see here:
* www.photoshop.com/products/photoshop/cs5/index.html
## Photoshop 2021 (Version 22.0.0)
The following list of things are easy to do in Photoshop but are difficult or impossible to do in Photoshop Elements. If you are a photographer or a graphic designer, then your first instinct will be to reach for Photoshop.
People learn Photoshop because it does everything. What this means is that if you know how to do this, then you will also know how to do that. Learning Photoshop can be time consuming and you might be bombarded by a whole bunch of different tools.
Photoshop Elements is simple. It doesn’t have a ton of features but, when you need them, they are easily accessible.
You don’t need to use Photoshop to be proficient with Elements, you just need to know Photoshop. You can improve upon the skills you have by combining Elements with Photoshop.
If you are completely new to Photoshop, then you will want to start with Elements. Once you have got the hang of things, you can upgrade to the full version of Photoshop. If you do upgrade, you can skip a few steps and start with Photoshop straight away.
20 Simple Photoshop Elements Photoshop Tutorials
If you are interested in becoming a Photoshop expert, this list will guide you towards a useful starting point. These tutorials cover Photoshop Elements 11 and 14.
Photoshop Elements is a great graphics editor for photographers, graphic designers and web designers. It comes with many of the tools professional photographers and graphic designers need.
If you are interested in using the full version of Photoshop, you can either upgrade to the full version or use Adobe Elements for free.
6. Quick-fix tool with blur
Quickly remove red eye, add some blur to your image, or create a mask. Just click one of the sliders to make changes or drag a box to create a mask.
7. Use drag and drop to select and drag
If you are dragging a selection around your image, you can use the T or L button to make the tool larger or smaller.
Press the Alt (Windows) or Option (Mac) key while dragging, and the window
a681f4349e
## Photoshop 2021 (Version 22.0.0) Crack+ Full Version For Windows
(stride))
return _ps.ptl.lock_
}
func (ps *PerSeConfig) Unlock() {
_ps.ptl.Unlock()
}
func (ps *PerSeConfig) SetPath(stride) {
_ps.ptl.lock()
defer _ps.ptl.Unlock()
_ps.SetPath(stride)
}
func (ps *PerSeConfig) Save(w io.Writer) (err error) {
_ps.ptl.lock()
defer _ps.ptl.Unlock()
if _ps.stride == 0 {
_ps.ClearPath()
}
_ps.mapping.Table[(_ps.memPerWindow)]
_ps.mapping.Table[(_ps.memPerWord)]
_ps.mapping.Table[(_ps.memPerAtom)]
for i := _ps.memPerWindow; i _ps.stops.Len() {
err = io.ErrUnexpectedEOF
} else {
_ps.save()
_ps.mem.Reset(0)
}
return
}
func (ps *PerSeConfig) SaveIndex() (err error) {
_ps.ptl.lock()
defer _ps.ptl.Unlock()
_ps.mapping.Table[(_ps.memPerWord)]
_ps.mapping.Table[(_ps.memPerWord2)]
_ps.mem.Reset(0)
_ps.saveIndex()
if _ps.stride < 0 { _ps.ClearPath()
## What’s New in the?
Q:
Random row of cubes
I am struggling with the following problem:
Consider a cube with side length $s$, and let $S_n$ be the surface of a cube of side length $s$ made of $n$ cubes.
What is the probability density function of $S_3$?
While it should be very simple, I can’t see how the factor of $n!$ should cancel out when I sum over all the possible positions of the center of the cube and multiply it by the cube of the side length.
EDIT:
Okay, so I had a bit too much time on my hands, and I accidentally checked the equation for the first few values. It seems that it is indeed just the product of an exponential distribution and a binomial distribution!
A:
There are $n!$ ways to place the cube in this arrangement. Then we need to figure out the probability that the $n$ cubes line up precisely perpendicular to the x-axis.
Each time one of the corners touches the axis a cube was rotated with respect to the previous arrangement. The probability of picking a corner that is in a $45^\circ$ angle with the x-axis is $\frac{1}{4}$. Thus, the desired probability is
$$\prod_{k=0}^n \left(1 – \frac{1}{4} \right) = \left( 1 – \frac{1}{4} \right)^n \approx e^{ – \frac{n}{4}}$$
Looking for new stock vehicle
I am looking for a stock vehicle for my company. I am currently running a 89 Z28 in my father’s ’82 mustang gt. I’m going on the stock build and keep it as original as possible. I am looking for a more reliable car with excellent handling. I did a fair amount of searching and found a few good deals on good looking American muscle cars. I have even had a few quotes but they are too expensive. I have time on my hands. I will continue searching and let you know if I find something I want to make a serious offer on.
I am looking for a stock vehicle for my company. I am currently running a 89 Z28 in my father’s ’82 mustang gt. I’m going on the stock build and keep it as original as possible. I am looking for a more
## System Requirements:
Windows 7, 8.1 or Windows 10 (64-bit versions only)
Minimum 1GB RAM
DirectX 9.0 or higher
512 MB VRAM
1.25 GB of available space on your computer (10 GB on Windows 8)
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2022-12-04 05:33:50
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http://www.math.psu.edu/calendars/meeting.php?id=9242
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# Meeting Details
Title: A generalization of Drinfeld's classification of Poisson homogeneous spaces of Poisson Lie groups GAP Seminar Madeleine Jotz, EPF Lausanne A theorem of Drinfeld classifies the Poisson homogeneous spaces of a Poisson Lie group in terms of a special class of Dirac subspaces of its Lie bialgebra. Before presenting this theorem, I will recall the notions of Poisson Lie groups, their Lie bialgebra and homogeneous spaces. Then, I will explain why it appears natural to consider Dirac'' homogeneous spaces of a Poisson Lie group, and even Dirac homogeneous spaces of a Dirac Lie group''. The theorem of Drinfeld generalizes in a natural manner to a classification of the homogeneous spaces in this setting.
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2014-07-28 16:41:20
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https://zbmath.org/?q=an:0872.53035
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## The relative deformation theory of representations and flat connections and deformations of linkages in constant curvature spaces.(English)Zbl 0872.53035
Let $$M$$ be a smooth connected manifold, $$G$$ a complex or real linear algebraic group. It is known that deformations of a given representation $$\rho_0: \pi_1(M)\to G$$ can be described in terms of flat connections in the vector bundle $$\operatorname{Ad} P$$ associated to the flat principal $$G$$-bundle $$P\to M$$ determined by $$\rho_0$$. A deformation theory based on this idea is exposed in W. M. Goldman and J. J. Millson [Publ. Math., Inst. Hautes Etud. Sci. 67, 43-96 (1988; Zbl 0678.53059)]. In the present paper, a relative version of this theory is elaborated. Let $$U_1,\ldots,U_r$$ be disjoint domains in $$M$$, $$\Gamma_j$$ the natural image of $$\pi_1(U_j)$$ in $$\Gamma = \pi_1(M)$$, and $$\mathcal O_j$$ the $$\operatorname{Ad} G$$-orbit of $$\rho_0 |\Gamma_j$$ in $$\operatorname{Hom}(\Gamma_j,G)$$. Denote $$U = \bigcup_{j=1}^r U_j$$ and $$R = \{\Gamma_1,\ldots,\Gamma_r\}$$. The variety $$\operatorname{Hom}(\Gamma,R;G)$$ of relative deformations of $$\rho_0$$ is defined as the inverse image of $$\prod_{j=1}^r\mathcal O_j$$ under the natural mapping $$\operatorname{Hom}(\Gamma,G)\to\prod_{j=1}^r \operatorname{Hom}(\Gamma_j,G)$$. The authors construct a controlling differential graded Lie algebra $$\mathcal B(M,U;\operatorname{Ad} P)_0$$ consisting of $$\operatorname{Ad} P$$-valued differential forms on $$M$$; its fundamental property is that the complete local ring of $$\operatorname{Hom}(\Gamma,R;G)$$ at the point $$\rho_0$$ can be obtained from $$\mathcal B(M,U;\operatorname{Ad} P)_0$$ by the procedure of the paper cited above.
This theory is applied to the study of deformations of mechanical linkages in one of the Riemannian spaces of constant curvature $$X = S^m, \mathbb{E}^m$$ or $$\mathbb{H}^m$$. One considers linkages $$\Lambda$$ with $$n$$ vertices $$u_1,\ldots,u_n$$ such that any edge $$u_iu_j$$ is the unique minimizing geodesic arc joining $$u_i$$ and $$u_j$$. The group $$G$$ is the isometry group of $$X$$, $$\Gamma$$ is the free product $$\Phi_n$$ of $$n$$ copies of $$\mathbb{Z}/2$$, $$\Gamma_j$$ are dihedral subgroups corresponding to edges. Let $$\tau_1,\ldots,\tau_n$$ be the generators of the $$\mathbb{Z}/2$$ factors of $$\Phi_n$$. Assigning to $$\tau_i$$ the Cartan involution $$s_{u_i}\in G$$ at the vertex $$u_i$$, we get a representation $$\Phi_n\to G$$. In this way, one obtains a local isomorphism between the configuration space of linkages with $$n$$ vertices and $$\operatorname{Hom}(\Phi_n,R;G)$$. Some special cases are considered and some problems are formulated.
### MSC:
53C35 Differential geometry of symmetric spaces 70B15 Kinematics of mechanisms and robots 22E40 Discrete subgroups of Lie groups 57R22 Topology of vector bundles and fiber bundles
Zbl 0678.53059
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### References:
[1] Atiyah, M. and Macdonald, I. : Introduction to Commutative Algebra , Addison-Wesley (1969). · Zbl 0175.03601 [2] Bourbaki, N. : Commutative Algebra , Hermann (1972). [3] Buchweitz, R.O. and Millson, J.J. : CR-geometry and deformations of isolated singularities , to appear in the Memoirs of the A.M.S. · Zbl 0871.32023 [4] Brylinski, J.L. and Zucker, S. : An overview of recent advances in Hodge theory, Several Complex Variables, IV , Encyclopedia of Math. Sciences, 69, Springer-Verlag. · Zbl 0793.14005 [5] Connelly, R. : The rigidity of certain cabled frameworks and the second order rigidity of arbitrarily triangulated convex surfaces , Adv. Math. 37 (1980), 272-299. · Zbl 0446.51012 [6] Deligne, P. and Mostow, G.D. : Monodromy of hypergeometric functions and non-lattice integral monodromy . IHES 63 (1986), 5-89. · Zbl 0615.22008 [7] Goldman, W.M. and Millson, J.J. : The deformation theory of fundamental groups of compact Kahler manifolds , Publ. Math. IHES 67 (1988), 43-96. · Zbl 0678.53059 [8] Goldman, W.M. and Millson, J.J. : The homotopy invariance of the Kuranishi space , 111. J. Math. 34 (1990) 337-367. · Zbl 0707.32004 [9] Hain, R. : in preparation. [10] Hochschild, G.P. and Serre, J.P. : Cohomology of group extensions , T.A.M.S. 74 (1953), 110-134. · Zbl 0050.02104 [11] Johnson, D. and Millson, J.J. : Deformation spaces associated to compact hyperbolic manifolds, in Discrete Groups in Geometry and Analysis, Papers in Honor of G. D. Mostow on His Sixtieth Birthday , Progr. Math. 67 (1987), Birkhäuser, 48-106. · Zbl 0664.53023 [12] Kapovich, M. and Millson, J.J. : Hodge theory and the art of paper folding , preprint. · Zbl 0961.32026 [13] Kapovich, M. and Millson, J.J. : Bending deformations of representations of fundamental groups of two complexes of groups, in preparation . · Zbl 0855.32013 [14] Kapovich, M. and Millson, J.J. : On the deformation theory of representations of fundamental groups of compact hyperbolic 3-manifolds , to appear in Topology. · Zbl 0855.32013 [15] Lubotzky, A. and Magid, A. : Varieties of representations of finitely-generated groups , Memoirs of the A.M.S. 58, 1985. · Zbl 0598.14042 [16] Millson, J.J. : Rational homotopy theory and deformation problems from algebraic geometry , Proc. Int. Congress of Math., Kyoto (1990), 549-558. · Zbl 0761.32011 [17] Schlessinger, M. : Functors of Artin rings , T.A.M.S. 130 (1966), 208-222. · Zbl 0167.49503 [18] Scharlau, W. : Quadratic and Hermitian Forms , Grundlehren der mathematischen Wissenschaften, 270 (1985), Springer-Verlag. · Zbl 0584.10010 [19] Thurston, W.P. : The Geometry and Topology of Three-Manifolds , Princeton University notes.
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2022-09-29 07:30:56
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http://cibustek.it/tlrh/minecraft-angle-calculator.html
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There are two particular circles associated with any regular polygon. Example: /execute if data entity @s Inventory[{Slot:0b}] ^ this command will return either 1 or 0 depending on whether or not the specified slot has any data. 1 deg = 1/360 tr = pi/180 rad = 10/9 grad. Basically it's so things in the game look the same size as they would in real life, allowing you to judge distances better. Ensure joints meet on studs, blocking or furring strips with the nail penetrating solid wood at least 1-1/4 inches. Inscribing A Circle Within A Kite All kites are tangential quadrilaterals, meaning that they are 4 sided figures into which a circle (called an incircle) can be inscribed such that each of the four sides will touch the circle at only one point. Input value you know and the value you want to find. Desired angles input in Decimal Degrees and Degrees is supported. 0, monster spawners found in mineshafts spawned spiders in place of cave. EnderVision! - Stronghold Location Calculator Point A: X: Z: Angle: Point B: X: Z: Angle: Instructions: - (For further information, view the Tutorial Video or check out our Helpful Tips. Calculate the vector by the difference between player position and the position of the target looked at (use a ray-trace for getting that position). CCD Resolution Calculator. Once light reaches our eyes, signals are sent to our brain, and our brain. Gear Generator is a tool for creating involute spur gears and download them in SVG format. This is a picture of the comparison between equilateral and right triangles in Minecraft. Fill in the label and value columns (you don't need to fill them all. Flip Card - Angle Types. It’s important to get the structure of your constructions right so that the rest of the build will go according to plan. A proportion is two ratios that have been set equal to each other, for example, 1/4 equals 2/8. A glass dome in Minecraft is a pixelated dome meant to resemble a real-life dome. Can you game on a calculator? (Gaming on ti-84 Plus and ti-nspire cx. The volume of a sphere is 4/3 x π x (diameter / 2)3, where (diameter / 2) is the radius of the sphere (d = 2 x r), so another way to write it is 4/3 x π x radius3. It’s not impossible, just time consuming. Nitrado @ PAX EAST 2020 - Partner Meet and Greet and Booth Information Greetings! As we all know the countdown to #PAXEast is underway. Also, here’s another list I may swipe and add in here. To add a balustrade. , the theta r at the air-flint glass boundary is the theta i at the flint glass-water boundary. Angle 3 and Angle C fields are NOT user modifiable. OCM, Inc Hex Head. Answering the question: the coefficient of friction of the body on an inclined plane with angle of inclination 30 o is. Maths IWB Resources Abacus Addition and Subtraction Tool Angle Measurer (at TeacherLed. Horizontal/Vertical FOV Calculator This calculator will convert an aspect ratio and horizontal FOV to a vertical FOV, useful for setting your favorite horizontal FOV in a game that uses vertical FOV. That's why we humans cannot see God until we are made pure and translated into Heaven. Open a slide. Scribble Maps Pro gives you more. 3D midpoint calculator used to find the midpoint of a vector 3d. ) Where (for brevity) it says 'edge a', 'angle B' and so on, it should, more correctly, be something like 'length of edge a' or 'edge-length' or 'size of angle B' etc. Minecraft was born on May 16, 2009, and became popular all over the world shortly after it existed on the market. If you press F3, the debug screen will give. The complementary angle (and what you’re tipping your saw blade to for the bevel cuts) is, in fact, 22. 48 Antimony 419. This can be really useful if you're flying long distances, especially in the End over the void. Yaw is measured in degrees, and does not follow classical trigonometry rules. I wrote this Snake game as part of writing a TI-84+CSE appendix to "Programming the TI-83 Plus/TI-84 Plus", a friendly guide to learning TI-BASIC. MSQ's role is to protect Queensland's waterways and the people who use them — providing safer, cleaner seas. Learn Milky Way & star photography with this definitive guide, from a full time landscape photographer. Navigation for For Test Takers. Before You Begin The first thing you need to be familiar with is the debugging screen you access by pressing F3 in Minecraft. Angles(0,0,0). gradians / gons. You're right on the money there. To create bends for small-scale projects in plexiglass (perspex) or acrylic 1/8 inch and thinner, you can use a small torch (butane kitchen torches work well), a paint stripping heat gun, or an embossing heat tool, which is a wand that works very similarly to a paint stripping gun, but using a smaller more directed heat flow with a smaller fan. 21 t/yd³ or 0. Search job openings, see if they fit - company salaries, reviews, and more posted by Microsoft employees. Type in inches and feet of your project and calculate the estimated amount of Soil & Dirt in cubic yards, cubic feet and Tons, that your need for your project. ) For average word counts based on genre, see this handy reference. First, find the slope by finding the tangent of the degrees, eg. grads angle unit. Aptoide powers the app stores of 6 of the largest smartphone brands in the world. This 3V 5/8 geodesic dome calculator is multi-purpose This is also a reverse geodesic dome calculator It will calculate any size of dome and display the lengths for each strut, total amount of material required, the weight of your geodesic dome and even the amount of material to cover your geodesic dome. The number 2 is used because each bit has a possibility of 2 values 0 or 1. You'll then see red as you stray away from the angle you were on. If you want to build a sphere or dome, there are several ways you can do this. is constant that has the value of approximately 0. The yaw rate or yaw velocity of a car, aircraft, projectile or other rigid body is the angular velocity of this rotation, or rate of change of the heading angle when the aircraft is horizontal. Starting mthly. atan2(y, x) -- it will return an angle given a vector from origin. Desired angles input in Decimal Degrees and Degrees is supported. مواقيت الصلاة. Glue together the triangles with wood glue. English Gematria, Hebrew Gematria and Jewish Gematria and Numerology. Heptagon definition is - a polygon of seven angles and seven sides. All you have to do is simply input any positive angle into the field and this calculator will find the reference angle for you. For example, a W21x44. The 10 Best Ballet Coloring Pages For Kids: These ballet coloring pages to print will serve as every little girl’s dream to color ballerinas and learn about the various postures associated with the dance. Figure 4: the amount of light reflected by objects in the scene towards $$P$$ the shading point, can be written as function of solid angle, which we can also re-write in terms of spherical coordinates. It's a carpenter's aid for laying out a perfect octagon. 1) Sketch and Measure Angles. With 6-month VIVEPORT Infinity membership, $78 value. Area Snatch Pro. Copy and paste the html code above in your website to share. Primary Resources - free worksheets, lesson plans and teaching ideas for primary and elementary teachers. The most common way to measure angles is in degrees, with a full circle measuring 360 degrees. Show Instructions. The rest of the polygon will be built using this side as a. Addition worksheet 1. Open-pit mining, also known as opencast mining, open-cut mining, and strip mining, means a process of digging out rock or minerals from the earth by their elimination from an open pit or borrow. A default wand that is used to select a imaginary region. Let's say I have light ray exiting a slow medium there Let me draw. Our Argos Extra stores are listed in red. R = Radius of the semicircle. cone synonyms, cone pronunciation, cone translation, English dictionary definition of cone. 7 kW/hour equal to one gallon of gasoline energy. Minecraft: Redstone Handbook 978-0-545-68515-3 on sale Fall 2014 Minecraft: Construction Handbook 978-0-545-68517-7 Minecraft: Combat Handbook 978-0-545-68516-0 support your child at hoMe With these activities. News, email and search are just the beginning. First select the unit of measurement you prefer — either centimetres or inches. Graphing Calculator 3D is a powerful software for visualizing math equations and scatter points. Transformations. If you press F3, the debug screen will give. Entering sides of values 1. How to Calculate Slope in Excel. I would like to know how you can calculate the angle between where the player is looking and a specific point in the world. 6 degrees for the "B" (long. published the newsletter and website until Fall 2017. Use it to set your miter gauge or sliding bevel gauge, as shown. The visual ability of humans and other animals is the result of the complex interaction of light, eyes and brain. 1 but Spigot still incorrectly returns 1. You can use this handy on-line tool to easily calculate the total volume of your shipment. Position - Vector3. Visual on the figure below: Since in most practical situations you know the diameter (via measurement or from a plan/schematic), the. Python supports to have an else statement associated with a loop statement. The Exterior Angle Theorem states that the exterior angle of a triangle: is smaller than either of the interior angle's measures. The entire gazebo is. Welcome to Template. We love what we do. For the sides, any value is accepted as long as they are all the same. The codes are based on information taken from page 89 of the 2012 International Residential Code. Hope I cleared my question. Activity Details. cone top: right circular cone bottom: cones and rods of a human eye n. Thus, (n+11) + (4n-17) + (5n+36) = 10n + 30 = 180. Gears can be animated with various speed to demonstrate working mechanism. Argos store. Minecraft Server. 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Inscribing A Circle Within A Kite All kites are tangential quadrilaterals, meaning that they are 4 sided figures into which a circle (called an incircle) can be inscribed such that each of the four sides will touch the circle at only one point. θ: Angle between the force and displacement. In Tactical editor you can draw troop, hero and spell placement. Divide the rise by the ideal step height. Use search box to filter the result. You can shift, rotate, color, decompose, and view in 2‑D or 3‑D. Steel Beam Calculator Ltd. The following how-to is an exclusive excerpt from the Minecraft Guide to Creative. There are many amazing redstone builds, but probably the pinnacle of redstone technology is the redstone calculator. 5 cm) boards to the same length and angle as your stringers, but do not cut step notches into them. The ability to turn simple game mechanics into a real-life calculator is one of genius, pioneered by some very clever Minecrafters many months ago. You will get the following result using the in-built data that PowerPoint uses: Delete the legend and chart title. Use right slider to show layer you want to build. By using this website, you agree to our Cookie Policy. Bone Yard Boats was founded in 1996 when Ginger Martus. If your last name is a mouthful or you just want to keep things simple, one of these short and sweet baby names could be just the thing you’re looking for. A Nether portal consists of a rectangular frame (4×5 minimum, 23×23 maximum) of obsidian. Milwaukee, 49-22-8510, Right Angle Attachment, 2. Since Minecraft is strictly square the dome will not be spherical, but it will resemble a curved, spherical dome as shown in the figure below. IN THIS ARTICLE. Theta is the launch angle (positive or negative) relative to a line from the nock point to the target, and g is the gravitational acceleration of the earth (9. Dome Calculator. The geometric center, or centroid, is computed as the point-wise average of a set of coordinates. The TI-81, TI's first graphing calculator, was designed for algebra and precalculus. Here are the calculated theta r values:. The New York Times: Find breaking news, multimedia, reviews & opinion on Washington, business, sports, movies, travel, books, jobs, education, real estate, cars. To build polygons like one of these, follow these steps: Visualize what the polygon will look like. 11), it needs to be executed with a command block. Cool Math has free online cool math lessons, cool math games and fun math activities. Bing helps you turn information into action, making it faster and easier to go from searching to doing. I used a chart while I was building, but wanted to be able to make variable size. Supercharged LEARNING. If you know the angle of rotation, you can compute a value for the Y-Axis Coordinate parameter as follows: Assume the fishnet is to be rotated 60 degrees clockwise. 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Here is a step-by-step tutorial on how to create a custom palette, and how to convert an image to be perler bead pattern using the palette. I’ve been meditating on the ideal roof for a chicken coop. Flip Card - Angle Types. It uses Longworth Chuck and plunger technology for closed segment rings and plastics-based plates for positioning open segments. The median is a measure of central tendency. Remember my last calculation. Starting mthly. We are continuing to thoroughly follow the developments and directives associated with COVID-19 (coronavirus). The exhibition “Defining Beauty: The Body in Ancient Greek Art” shows that the body in movement, both realistic and transcendent, was at the center of Greek art and thought. Angle Unit: degree ° radian. With our Roll Up Door, you. A proportion is two ratios that have been set equal to each other, for example, 1/4 equals 2/8. Place the larger triangle on the paper, lining it up with the guidelines. The latest counts come a day after budget officials conceded COVID-19 has blown a$2. Such requirements vary from city to city, county to county, and/or State to State. 223 cal, 29 gr, Short High Velocity Aguila, 0. It will calculate the most commonly needed measurements for a dome — both spherical or oblate. Maze Collapse 2. What Percentage Is Illustrated? Sixth grade is a big step forward in students’ mathematical education! But as kids learn more sophisticated and challenging math skills, Math Games keeps things light and fun with its exciting educational games. (HTMLHelp version requires that Internet Explorer v4 or greater be present. The basics of the game are that you need to roll the ball as far as possible. spacer block. A site catering to the lifestyle of building and flying Van's Aircraft 'RV' kitplanes. All you have to do is simply input any positive angle into the field and this calculator will find the reference angle for you. You are going to have to do a bit of math. Also known as pitched or peaked roof, gable roofs are some of the most popular roofs in the US. You can get that vector simply by setting the z-coordinate to 0. Standard carbon arrows tend to range in the 8 to 10 grains per inch range, but with aluminum/carbon hybrid arrows, an archer can achieve. In first person video games, the field of view or field of vision (abbreviated FOV) is the extent of the observable game world that is seen on the display at any given moment. 852 m³ Side Lengths 1600. Pentagram Calculator. A proportion is two ratios that have been set equal to each other, for example, 1/4 equals 2/8. Teaching Area using Minecraft. The distance between these two points is given in the calculator as the foci distance. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. We’ve been building custom patio covers and outdoor living areas for over 10 years, and we’ve been in the exterior remodeling industry since 2003. Archimedean Screw hydro turbine screws are normally set at an angle of 22 degrees from horizontal, which is the optimum for the most cost-effective installations. A typical arrow flies at over 150 mph so think about putting your hand out the window of a performance race car and you get a sense of the drag force on the arrow. Now, however. Built to be easy to install, durable and simple to maintain, our Model 650 Mini Storage Door includes features such as a ratchet tensioner, a factory installed stainless steel latch and an enclosed barrel that houses a dead axle assembly. Contribute to Tommsy64/SmartRender development by creating an account on GitHub. These spiral stair plans go into detail about how to measure, construct, then assemble this spiral staircase into several different situations. Outside Stringer = tread points winding up and around OUTSIDE of stairs from top of first rise to upper floor. A glass dome in Minecraft is a pixelated dome meant to resemble a real-life dome. There are several ways to construct Flying Geese, but my favorite is the no-waste method, a quick piecing technique that produces four accurate geese from one. Distance between two points calculator uses coordinates of two points A(xA, yA) A ( x A, y A) and B(xB, yB) B ( x B, y B) in the two-dimensional Cartesian coordinate plane and find the length of the line segment ¯¯¯¯¯¯AB A B ¯. Then, use the given n values and Snell's Law to calculate the theta r values at each boundary. 5/8) geodesic dome calculator is multi-purpose This is also a reverse geodesic dome calculator It will calculate any size of dome and display the lengths for each strut, total amount of material required, the weight of your geodesic dome and even the amount of material to cover your geodesic dome. it’s a way of learning. Latitude and Longitude are the units that represent the coordinates at geographic coordinate system. If you like MrNussbaum. Using this control panel, you can do everything required to manage your server quickly and easily. In that book, he explained a simple way to correctly calculate the base, sides, and height for any size pyramid that would match the ratio and proportions of the Great Pyramid of Giza (Cheops) in Egypt. 223 cal, 30 gr, Super Maximum Solid Aguila, 0. They were added to Minecraft as a part of the Halloween Update (the version1. Start by clicking on the cube along the left side; then, place cubes on the grid where you would like them. Hide Unused. However, the data associated with certain systems (a digital image, a board game, etc. G r a v i t a t i o n a l f o r c e F = G M m r 2 T i d e f o r c e T = − d F d r R = 2 G M m r 3 R R : r a d i u s o f t h e E a r t h G r a v i t a t i o n a l f. It will also calculate the angle in the other direction (counter-clockwise) so you need to negate the result. Different angles apply to different applications. If your command is longer than 256 characters (100 before Minecraft 1. Enter your value in the conversion calculator below. We can calculate the gradient of this line as follows. A quick note, no materials have been removed for the various armor sets, so its possible to simulate. You can change the width and height of the embedded simulation by changing the "width" and "height" attributes in the HTML. There are many amazing redstone builds, but probably the pinnacle of redstone technology is the redstone calculator. 90 = 42 N (2 s. Photography by Rafael Debreu. This free triangle calculator computes the edges, angles, area, height, perimeter, median, as well as other values of a triangle. Calculate the vector by the difference between player position and the position of the target looked at (use a ray-trace for getting that position). This is in degrees. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. 7 million people and is the centre of one of North America's most dynamic regions. If we were to change it to your formula, then the angle would change. Angle between Vectors Calculator. 96 Copper 559. I would look for a mod that has a trajectory shower for enderpearls, kind of like a line that goes from your location to where the mod has calculated the enderpearl to land, and look at the source code. According to the link it is clear that, it is rotated with some angle on x,y,z axis. In theory I can onetap literally everything in the game, so far I have onetapped literally everything in the game except the magma cube boss as I haven't had the chance to test it yet. Example: Cyrillic capital letter Э has number U+042D (042D – it is hexadecimal number), code ъ. Multiplication, Addition, Subtraction, Division, Classroom. The unit circle of yaw on the xz-plane starts at (0, 1) and turns backwards towards (-1, 0),. 0) on October 31, 2010. With this release, we improve the app's search to take advantage of the specs in your search and filter to relevant products more quickly. Download source code - 44. How Shadow Mapping Works : a 3-part YouTube tutorial by TheBennyBox on shadow mapping and its implementation. #N#Library (Select bullet or enter BC and Weight below. Just like every actual house has its address (which. Five Nights at Frozen v. Following is the declaration for java. Here is our selection of free printable randomly generated math worksheets which will help your child improve their mental calculation skills and learn Math facts. This 3V 5/8 geodesic dome calculator is multi-purpose This is also a reverse geodesic dome calculator It will calculate any size of dome and display the lengths for each strut, total amount of material required, the weight of your geodesic dome and even the amount of material to cover your geodesic dome. Fill out the information below and it will tell you how heavy that piece of. Use the pencil tool,to draw anything from planned troop movement direction, to troop placement zones. Hover over an image to see what the PDF looks like. Patent Blueprints. Use our interactive deckbuilder to build a new. Copy and paste the html code above in your website to share. Our Premium Minecraft servers have 8 different locations around the globe allowing you to choose the closest location to you for. The final average ACF is then used to calculate the backscattered power, spectral width and Doppler velocity of the plasma density irregularities in the ionosphere. It’s was super easy to get approved for the loan, I think it only took maybe 5 minutes to complete the application. This site offers an unlimited amount of templates (also knows as ’dielines‘ or ‘nets’) for paper craft, packaging, package design, learning materials, decoration and much more. Understand the STL resolution. Welcome to the Official Factorio Wiki, the official source of documentation for Factorio™. Once the students feel comfortable with the process have students load the attached Minecraft world file for Measuring Angles and Building Bridges. Carpet tile planks are installed in interlocking groups of four, so that the pattern resembles a woven pattern. With a tap, you can enter the result of your conversion into the calculator. Tools: Centrifuge Rotor Speed Calculator: Choose the make and model of your centrifuge rotor, and the values will be entered automatically. Download source code - 44. Then the second portal only requires pythagorians theorem to calculate the end portal. Obtain one by typing: Obtain one by typing: /give @p minecraft:command_block. The find command, with its myriad options and switches, offers the most powerful and precise features to surface what you're looking for. 75 sec I calculated the following sight marks (relative to 10. Volume & Height. The download contains both a HTML version and a stand alone HTMLHelp version. Complementary and Supplementary Angle Pairs Memory Match. Range cards calculate bullet drop for atmospheric conditions that are different than the zero conditions. 7/12 roof pitch angle = 30. Pressing tab with the cursor on the following equations will yield these results:. We are continuing to thoroughly follow the developments and directives associated with COVID-19 (coronavirus). The word is used to distinguish this type of mining from extractive methods that need tunneling into the earth. Angle of the steps. 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One mile converts to 1609. 0420 and column D. 20,000+ square feet for a private residence is an insane amount of space, yet there are many of these homes located all over the world. You can detect an empty slot by testing if it has data. View image of Britain’s busiest autumn animals (Credit: Mark Bridger/Getty) Britain’s busiest autumn animals. If it says "16 degrees east," rotate the compass housing so the magnetic north/compass needle mark sits 16 degrees to the right (east) of the true north orienting lines. It has since been replaced by newer models which have faster processors, more memory, and popular features such as linking capability and flash upgradability. About Pentagonal Prism Calculator tool. Transmission line calculations. With 6-month VIVEPORT Infinity membership,78 value. Edit Here's a picture. The angle of refraction at one boundary becomes the angle of incidence at the next boundary; e. 223 cal, 30 gr, Super Maximum Hollow Point Aguila, 0. Students will learn the basics of revenue and cost by selling an item, subtracting the shipping and fees (students will need to calculate these fees by using. 12 ), fractions ( 10/3) or numbers with $\pi$ ( 25/4 pi ). Some models also have some angles you can customise or a. Glue the lid of a 1-quart jar onto the center of the tower with cyanoacrylate glue so the lid faces upwards. Enter the hypotenuse and the angle as positive real numbers and press "calculate". Cos ( q) = Adjacent / Hypotenuse. Solving a proportion means that you are missing one part of one of the fractions, and you need to find that missing value. 3) Click the Calculate Resources button below.
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2020-10-26 15:30:58
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http://mathhelpforum.com/calculus/225517-how-solve-ode-print.html
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# How to solve this ODE?..
• January 18th 2014, 02:39 PM
Mengqi
How to solve this ODE?..
I'm reading a book on asymptotic methods and I encountered the following ODE, do you have any clue how to solve this? Thanks.
$y''+y=\frac{\cos(x)}{5+8\cos(x)\sin(x)}$
• January 18th 2014, 05:53 PM
chiro
Re: How to solve this ODE?..
Hey Mengqi.
You should try using the complem entary and particular solution.
On the right hand side you have a 2nd order linear DE which can be solved through a standard technique. On the Right hand side you have a particular solution which can be solved using a number of techniques including operator methods (using the d/dx = D as an operator).
These techniques should be taught in any standard differential equations courses.
• January 19th 2014, 12:32 AM
Mengqi
Re: How to solve this ODE?..
Thanks Chiro,
I see. But when I tried to solve it in this way: I know the left hand side will give me the solution like $y=A\cos(x)+B\sin(x)$ (with right hand side being set to zero temporally), then I assume $y_{particular}=C(x)(A\cos(x)+B\sin(x))$ and plug it in, but this way seems not to work. In the end, I got sth. like $[C''(x)(A\cos x+B\sin x)-2C'(x)(A\cos x+B\sin x )](5+8\cos x\sin x)=\cos x$. How to proceed from here? I didn't reduce the problem to a simple one, C still appears with a second order derivative. Do you know any other techniques? Thanks
• January 19th 2014, 12:39 AM
Mengqi
Re: How to solve this ODE?..
And it seems not trivial because I just used Mathematica to see the final result, which is attached. I think it involves some advanced technique? It's great if any of you could solve it. I am curious to see the derivation. Thanks.
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2015-04-25 15:23:29
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https://paradigms.oregonstate.edu/problem/771/
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## The puddle
• differentials
• assignment Potential energy of gas in gravitational field
assignment Homework
##### Potential energy of gas in gravitational field
Potential energy Heat capacity Thermal and Statistical Physics 2020 Consider a column of atoms each of mass $M$ at temperature $T$ in a uniform gravitational field $g$. Find the thermal average potential energy per atom. The thermal average kinetic energy is independent of height. Find the total heat capacity per atom. The total heat capacity is the sum of contributions from the kinetic energy and from the potential energy. Take the zero of the gravitational energy at the bottom $h=0$ of the column. Integrate from $h=0$ to $h=\infty$. You may assume the gas is ideal.
• group Optical depth of atmosphere
group Small Group Activity
30 min.
##### Optical depth of atmosphere
Contemporary Challenges 2022 (4 years) In this activity students estimate the optical depth of the atmosphere at the infrared wavelength where carbon dioxide has peak absorption.
• assignment Differentials of Two Variables
assignment Homework
##### Differentials of Two Variables
Static Fields 2022 (8 years) Find the total differential of the following functions:
1. $y=3u^2 + 4\cos 3v$
2. $y=3uv$
3. $y=3u^2\cos wv$
4. $y=u\cos(3v^2-2)$
• assignment Zapping With d 1
assignment Homework
##### Zapping With d 1
Energy and Entropy 2021 (2 years)
Find the differential of each of the following expressions; zap each of the following with $d$:
1. $f=3x-5z^2+2xy$
2. $g=\frac{c^{1/2}b}{a^2}$
3. $h=\sin^2(\omega t)$
4. $j=a^x$
5. $k=5 \tan\left(\ln{\left(\frac{V_1}{V_2}\right)}\right)$
• face Energy and Entropy review
face Lecture
5 min.
##### Energy and Entropy review
Thermal and Statistical Physics 2020 (3 years)
This very quick lecture reviews the content taught in Energy and Entropy, and is the first content in Thermal and Statistical Physics.
• group Gravitational Potential Energy
group Small Group Activity
60 min.
##### Gravitational Potential Energy
Students examine a plastic “surface” graph of the gravitational potential energy of an Earth-satellite system to explore the properties of gravitational potential energy for a spherically symmetric system.
• Static Fields 2022 (5 years) The depth of a puddle in millimeters is given by $h=\frac{1}{10} \bigl(1+\sin(\pi xy)\bigr)$ Your path through the puddle is given by $x=3t \qquad y=4t$ and your current position is $x=3$, $y=4$, with $x$ and $y$ also in millimeters, and $t$ in seconds.
1. At your current position, how fast is the depth of water through which you are walking changing per unit time?
2. At your current position, how fast is the depth of water through which you are walking changing per unit distance?
3. FOOD FOR THOUGHT (optional)
There is a walkway over the puddle at $x=10$. At your current position, how fast is the depth of water through which you are walking changing per unit distance towards the walkway.
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2022-11-30 20:56:49
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https://byjus.com/free-gmat-prep/gmat-quant-geometry-volume-formulas/
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# GMAT Quant: Geometry – Volume
Quant in GMAT exam is the most scoring and No, it is not time consuming if you have a good hold of it. The mantra is to know all basics and formulas and practice a lot.
## What is Meant by Volume?
In a 3-D shape, amount of space occupied by an object, is known as volume. The unit of volume is cubic units, liters etc.
Let’s check some standard volume formula for some regular shapes:
1. Cube
Cube is a special form of rectangular solid whose all six sides are equal.
Volume = a
Where, a = length of the sides
2. Volume of Rectangular Solids
All the sides of a rectangular solid are rectangular in shape. If all the six sides have the same length then it is known as a cube otherwise, it is a cuboid.
Volume of rectangular solid = lwh =length ×weidth ×height
3. Volume of Cylinder
Volume, $V = \pi r^{2} h = \pi \times radius \times radius \times height$
4. Volume of Sphere
Sphere is a 3- dimensional form of a circle
Volume, $V = \frac{4}{3}\pi r^{3}$
5.Right Circular Cone, $V = \frac{1}{3}\pi r^{2}h$
6.Ellipsoid is the 3 –D formation of an ellipse.
Volume, $V = \frac{4}{3} \pi \times radius_1 \times radius_2 \times radius_3$
Volume, $V = \frac{1}{3} \times base \times height$
7.Tetrahedron: A tetrahedron is a polyhedron and type of pyramid which has a triangular base. It has four triangular faces, four vertex corners, and six straight edges.
Volume, $V = \frac{a^{3}}{6\sqrt{2}}$
Let’s solve a problem and be prepared to tackle any problem offered by GMAT.
Question:
The inside dimensions of a rectangular wooden box are 6 inches by 8 inches by 10 inches. A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces. Of all such canisters that could be used, what is the radius, in inches, of the one that has maximum volume?
1. 3
2. 4
3. 5
4. 6
5. 8
Solution:
Cylinder can be placed in a rectangular box on three different bases.
Base1. Height = 10
r = 3
$V = \pi r^{2}h$ $V = \pi \times 3^{2} \times 10$ $V = 90 \pi$
Base 2: height = 6
r = 4
$V = \pi r^{2}h$ $V = \pi \times 4^{2} \times 6$ $V = 96 \pi$
Base = 3; Height = 8
r = 3
$V = \pi r^{2}h$ $V = \pi \times 3^{2} \times 8$ $V = 72 \pi$
Question: There is a shower from which 10 cm of rain falls. Find the volume of water that will fall on the ground in an area of 3.5 hectares.
1. 35 m3
2. 350 m3
3. 3500 m3
4. 35000 m3
Solution: 1 hectare = 10, 000 m2
So total area $= 3.5 \; hectares = 3.5 \times 10,000 = 35,000 m^{2}$
According to question, $Depth = \frac{10}{100}m = 0.1 m$
So, volume, $V = Depth \times Area = 35000 \times 0.1 = 3500m^{3}$
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2020-03-31 03:12:16
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https://www.physicsforums.com/threads/books-recommended-for-dummies.7273/
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# Books recommended for dummies?
1. Oct 16, 2003
### jby
Any books recommended for dummies? All books that I've found starts with contraviant and covariant tensor, which seems misleading to me.
Last edited by a moderator: Feb 6, 2013
2. Nov 20, 2003
Staff Emeritus
3. Nov 20, 2003
### lethe
actually, i quite agree, that the way most books start with covariant versus contravariant is misleading. in fact, i would go one further and say that the approach is wrong!
many beginning books start with the transformation properties of coordinates, and call anything that follows the same transformation rules "contravariant".
then they show that the partial derivatives with respect to the coordinates follow the opposite kind of transformation law.
there are a couple of things wrong with this:
1. it is very coordinate dependent. these are geometric objects, independent of coordinates, and the way they transform under coordinate transformations is a consequence of what type of geometric object they are, not vice versa.
2. manifold coordinates are components of a vector. partial derivatives are basis vectors. comparing their transformation properties is like comparing apples and oranges.
3. in category theory, a functor is called contravariant if it reverses the direction of the morphisms, and covariant if it does not.
consider the category of smooth manifolds, and tangent bundles and of cotangent bundles. given a diffeomorphism between two manifolds, this is a morphism in the category Smooth. Pushforward is a functor from Smooth to Tan, and Pullback is a functor from Smooth to Cot, and it is easily checked that Pushforward is covariant, while Pullback is contravariant.
the pullback of a diffeomorphism from M to N is a map from the cotangent bundle of N to that of M. notice how the order of the N and M in that sentence got reversed.
so vectors are naturally covariant objects while covectors are naturally contravariant objects. the problem is, physicists have this notation exactly backwards, and for no reason! if you make your coordinates have raised indices, then the components of a vector have also raised indices. physicists call these contravariant vectors, but i am claiming that they are covariant.
4. Aug 9, 2004
### Will_C
I am also a beginner on tensor. And in fact, I know that there are two approaches to define or develop the tensor: index-free approach and convectional approach based on indices. In my point of view, the index-free one is more appreciated by math student. And later one is mostly introducted to engineering and physics student. There are lot of material on the website you can search, and there are two suggestions:
1) http://en.wikipedia.org/wiki/Tensor_%28intrinsic_definition%29
2) Introduction to Tensor Calculus for Gerenal Relativity by MIT
BTW, the index-free approach, indeed, explores deeper geometrical insight, but it requires more mathematical background. And I don't think free-index approach is for dummies. (There maybe some lecturers disagree with that! )
Will.
We have no freedom, but we have choice. And I choose "NO WAR".
5. Aug 9, 2004
### robphy
There are actually two index-based approaches.
The most common is the coordinate-based approach, where the indices are summed over.
Alternatively, in the "abstract index notation" (attributed to Penrose), the index indicates a "slot" in a multi-linear mapping.
http://www.ikp.liu.se/usr/tobol/master/html/node10.html
http://web.archive.org/web/20010505001459/http://www.math.harvard.edu/~allcock/expos/notation.ps
http://www.ima.umn.edu/nr/abstracts/arnold/einstein-intro.pdf [Broken]
http://www.science.unitn.it/~moretti/tensori.pdf
Last edited by a moderator: May 1, 2017
6. Aug 10, 2004
### mathwonk
A brief summary of the math. A manifold is a smooth object that has a linear tangent space at every point, which together make up the "tangent bundle", a family of vector spaces.
Each vector space can be subjected to a variety of constructions yielding a new vector space, e.g. we can take various "tensor products" of our original vector space, forming a space of tensors. When these new spaces of tensors are bundled together with one at each point of the manifold, we get a new bundle, a tensor bundle. All tensor constructions have to do with linear or multilinear functions on the original vector space.
A choice of a tangent vector at each point of the manifold is called a vector field. A choice of tensor at each point is called a tensor field.
A tangent vector field transforms by the jacobian matrix of the change of coordinate map, hence in category language is "covariant", but as said above, differential geometers have this backwards and for historical reasons call these fields "contravariant". This is clearly discussed by Lethe above in post 3.
If f is a differentiable function on the manifold, it has a directional derivative in the direction of every tangent vector. Thus f assigns a number to each tangent vector. This assignment is called df, the differential of the function. This family of linear functions on tangent vectors is called a cotangent vector, and defines a covector field on the manifold. If A is any vector space the dual space of linear functions from A to the real numbers is called A^, or "A dual".
This field df transforms by the transpose of the jacobian matrix of the change of coordinates map, hence is truthfully contravariant, but unfortunately is called "covariant" in the language of classical differential geometry.
These vector and covector fields are "rank one" tensors it seems, from my brief perusal of some sites suggested locally.
Higher rank tensors apparently refer to multilinear tensors constructed from tangent vectors.
For example, given two vector spaces A,B, the family of all bilinear functions from the cartesian product
AxB to the real numbers, is a new space which is essentially the tensor product A^tensB^.
If we apply this construction to all the tangent spaces of a manifold, we get a new tensor bundle, and the Riemannian metric tensor which defines the length of a vector is a field of these tensors. I.e. the dot product of two tangent vectors is a bilinear function on pairs of tangent vectors with values which are numbers, hence at each point belongs to T^tensT^ where T is the tangent space.
When coordinates are introduced for tangent vectors, then one has corresponding coordinates for tensors, and all these change in complicated ways under coordinate transformations.
These coordinate transformations began to "wag the dog" at a certain point in classical differential geometry, and can make it quite a hard subject if unrelieved by some attention to what the objects mean. This notational approach seems to dominate the sources I have been referred from this site. Some respondents have even been unable to recognize my description of tensors as even remotely related to what they have learned from such "concept free" sources.
An example of the notational approach I recall are some highly complicated formulas due to Christoffel, so terrifying as to have been referred to as "them ******- awful formulas!!" by hapless students. (possible profanity omitted here).
On the other hand in the modern approach as presented in Spivak's differential geometry book, they are revealed to measure only the extent that two one - parameter subgroups of a certain lie group fail to commute with each other, i.e. they are merely local coordinate expressions for AB-BA.
I confess I am clearly an amateur in this topic, but I think we are at least talking about the same subject.
Remark: As you probably noticed, vector fields are covariant, and covector fields are contravariant, even in the modern category terminology. For example in topology homology is covariant and cohomology is contravariant. This further confusion caused Peter Hilton to suggest long ago that cohomology be called instead contrahomology, but it did not catch on. At least the classical people were consistent in their wrong terminology, and called vectors "contravariant vectors", and they called covectors "covariant vectors".
Last edited: Aug 10, 2004
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2019-03-20 13:04:01
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https://math.stackexchange.com/questions/747331/odd-eigenvalue-problem
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# Odd Eigenvalue Problem
So, I'm trying to solve the following eigenvalue problem for the eigenvalues:
$$u''(x)+\lambda^2u(x)=0$$ $$u(0)=u(1)$$ $$u'(0)=u'(1)$$
Of course, the two eigenvectors are cosine and sine, and the solution which spans the space of all solutions is as follows:
$$u(x)=A\sin(\lambda x)+B\cos(\lambda x)$$
Now, when I try to apply the BC's, I expected that I would obtained some kind of equation involving only the eigenvalue. Then, the solution of eigenvalues is usually the points at which both sides are equivalent. However, I'm finding that there are no eigenvalues which satisfy these conditions? What am I doing incorrectly here?
• The differential equation is precisely the equation for simple Harmonic motion
– user122283
Apr 9, 2014 at 21:33
• Well, yeah; of course. That's the general solution which I have above. The problem is finding the eigenvalues (or frequency). Any idea why I can't solve for them fromt eh BC's? Apr 9, 2014 at 21:35
• Do you get the following equation: $\cos^2(\lambda) - 2\cos(\lambda) + 1 + \sin(\lambda)(\cos(\lambda) = 0$ (or something more or less like it)? Apr 9, 2014 at 21:58
• Check this related problem. Apr 10, 2014 at 0:29
Given that $u(x) = A\sin{\lambda x} + B\cos{\lambda x}$, you can simply use this definition of $u$ in conjunction with the the boundary conditions.
You have that $u(0) = u(1)$, so then $B = A\sin{\lambda} + B\cos{\lambda}$.
You also have that $u'(0) = u'(1)$, so $A\lambda = A\lambda\cos{\lambda} - B\lambda\sin{\lambda}$. This should be fairly straight forward system of equations to solve.
• Whoops, edited. Well, just based off of the periodcity of sine and cosine, it seems that $\lambda = 2k\pi$ for any $k \in \mathbb{Z}$ works. Apr 9, 2014 at 23:11
• Thank you. I took a look at this again, and I'm getting that $\lambda=2n \pi$. However, what about the constants for the eigenvectors? Is there anything we can say about them? Trying to reduce them from the equations above lead us to the expression for $\lambda$. Apr 10, 2014 at 16:48
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2022-07-02 14:04:52
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https://the-examples-book.com/projects/project-archive/29000-f2020-project10
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# STAT 29000: Project 10 — Fall 2020
Motivation: Although SQL syntax may still feel unnatural and foreign, with more practice it will start to make more sense. The ability to read and write SQL queries is a bread-and-butter skill for anyone working with data.
Context: We are in the second of a series of projects that focus on learning the basics of SQL. In this project, we will continue to harden our understanding of SQL syntax, and introduce common SQL functions like AVG, MIN, and MAX.
Scope: SQL, sqlite
Learning objectives
• Explain the advantages and disadvantages of using a database over a tool like a spreadsheet.
• Describe basic database concepts like: rdbms, tables, indexes, fields, query, clause.
• Basic clauses: select, order by, limit, desc, asc, count, where, from, etc.
• Utilize SQL functions like min, max, avg, sum, and count to solve data-driven problems.
## Dataset
The following questions will use the dataset similar to the one from Project 9, but this time we will use a MariaDB version of the database, which is also hosted on Scholar, at scholar-db.rcac.purdue.edu. As in Project 9, this is the Lahman Baseball Database. You can find its documentation here, including the definitions of the tables and columns.
## Questions
Please make sure to double check that the your submission does indeed contain the files you think it does. You can do this by downloading your submission from Gradescope after uploading. If you can see all of your files and they open up properly on your computer, you should be good to go.
Please make sure to look at your knit PDF before submitting. PDFs should be relatively short and not contain huge amounts of printed data. Remember you can use functions like head to print a sample of the data or output. Extremely large PDFs will be subject to lose points.
For this project all solutions should be done using R code chunks, and the RMariaDB package. Run the following code to load the library: library(RMariaDB)
### Question 1
Connect to RStudio Server rstudio.scholar.rcac.purdue.edu, and, rather than navigating to the terminal like we did in the previous project, instead, create a connection to our MariaDB lahman database using the RMariaDB package in R, and the credentials below. Confirm the connection by running the following code chunk:
con <- dbConnect(RMariaDB::MariaDB(),
host="scholar-db.rcac.purdue.edu",
db="lahmandb",
user="lahman_user",
head(dbGetQuery(con, "SHOW tables;"))
In the example provided, the variable con from the dbConnect function is the connection. Each query that you make, using the dbGetQuery, needs to use this connection con. You can change the name con if you want to (it is user defined), but if you change the name con, you need to change it on all of your connections. If your connection to the database dies while you are working on the project, you can always re-run the dbConnect line again, to reset your connection to the database.
Items to submit
• R code used to solve the problem.
• Output from running your (potentially modified) head(dbGetQuery(con, "SHOW tables;")).
### Question 2
How many players are members of the 40/40 club? These are players that have stolen at least 40 bases (SB) and hit at least 40 home runs (HR) in one year.
Use the batting table.
You only need to run library(RMariaDB) and the dbConnect portion of the code a single time towards the top of your project. After that, you can simply reuse your connection con to run queries.
In our project template, for this project, make all of the SQL queries using the dbGetQuery function, which returns the results directly in R. Therefore, your RMarkdown blocks for this project should all be {r} blocks (as opposed to the {sql} blocks used in Project 9).
You can use dbGetQuery to run your queries from within R. Example: dbGetQuery(con, "SELECT * FROM battings LIMIT 5;")
We already demonstrated the correct SQL query to use for the 40/40 club in the video below, but now we want you to use RMariaDB to solve this query.
Items to submit
• R code used to solve the problem.
• The result of running the R code.
### Question 3
Find Corey Kluber’s lifetime across his career (i.e., use SUM from SQL to summarize his achievements) in two categories: strikeouts (SO) and walks (BB). Also display his Strikeouts to Walks ratio. A Strikeout to Walks ratio is calculated by this equation: $\frac{Strikeouts}{Walks}$.
Questions in this project need to be solved using SQL when possible. You will not receive credit for a question if you use sum in R rather than SUM in SQL.
Use the people table to find the playerID and use the pitching table to find the statistics.
Items to submit
• R code used to solve the problem.
• The result of running the R code.
### Question 4
How many times in total has Giancarlo Stanton struck out in years in which he played for "MIA" or "FLO"?
Use the people table to find the playerID and use the batting table to find the statistics.
Items to submit
• R code used to solve the problem.
• The result of running the R code.
### Question 5
The Batting Average is a metric for a batter’s performance. The Batting Average in a year is calculated by \$\frac{H}{AB}\$ (the number of hits divided by at-bats). Considering (only) the years between 2000 and 2010, calculate the (seasonal) Batting Average for each batter who had more than 300 at-bats in a season. List the top 5 batting averages next to playerID, teamID, and yearID.
Use the batting table.
Items to submit
• R code used to solve the problem.
• The result of running the R code.
### Question 6
How many unique players have hit > 50 home runs (HR) in a season?
Use the batting table.
If you view DISTINCT as being paired with SELECT, instead, think of it as being paired with one of the fields you are selecting.
Items to submit
• R code used to solve the problem.
• The result of running the R code.
### Question 7
Find the number of unique players that attended Purdue University. Start by finding the schoolID for Purdue and then find the number of players who played there. Do the same for IU. Who had more? Purdue or IU? Use the information you have in the database, and the power of R to create a misleading graphic that makes Purdue look better than IU, even if just at first glance. Make sure you label the graphic.
Use the schools table to get the schoolIDs, and the collegeplaying table to get the statistics.
You can mess with the scale of the y-axis. You could (potentially) filter the data to start from a certain year or be between two dates.
To find IU’s id, try the following query: SELECT schoolID FROM schools WHERE name_full LIKE '%indiana%';. You can find more about the LIKE clause and % here.
Items to submit
• R code used to solve the problem.
• The result of running the R code.
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2022-11-29 14:42:38
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https://www.physicsforums.com/threads/another-sequence-convergence-proof.187391/
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# Another sequence convergence proof
## Homework Statement
Let $$y_n := \sqrt{n+1} - \sqrt{n}$$ for $$n \in \mathbb{N}$$. Show that $$(y_n)$$ converges.
## The Attempt at a Solution
I see that it converges to 0. I just need a nudge in the right direction at getting into $$| \sqrt{n+1} - \sqrt{n} - 0 | = | \sqrt{n+1} - \sqrt{n} |$$ to show it's less than any $$\epsilon > 0$$. Any manipulating I've tried so far makes the terms way too big to work with.
Last edited:
## Answers and Replies
Related Calculus and Beyond Homework Help News on Phys.org
morphism
Homework Helper
$$\left(\sqrt{n+1} - \sqrt{n}\right) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}$$
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2020-09-26 22:59:43
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http://mathhelpforum.com/advanced-statistics/120782-limit-distribution.html
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# Math Help - Limit distribution
1. ## Limit distribution
Hello,
I want to find the asymptotic distribution of $\hat{\beta}$. And this happens to be the sum of two normally distributed terms. i.e.
$\sqrt{n}(\hat{\beta} - \beta) = X-Y$
Where,
$X \longrightarrow N(0, \sigma^2 )$
$Y \longrightarrow N(0, \omega^2 )$
__________________________________________________
So is this correct:
$\sqrt{n}(\hat{\beta} - \beta) \longrightarrow N(0, \sigma^2 + \omega^2)$
??
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2014-07-13 17:12:38
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https://ora.ox.ac.uk/objects/uuid:c0c509ae-a011-4a4a-971a-05688d7373bd
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Journal article
### On the chromatic number of random geometric graphs
Abstract:
Given independent random points X1,...,Xn ∈ℝd with common probability distribution ν, and a positive distance r=r(n)>0, we construct a random geometric graph Gn with vertex set {1,..., n} where distinct i and j are adjacent when {double pipe}Xi-Xj{double pipe}≤r. Here {double pipe}·{double pipe} may be any norm on ℝd, and ν may be any probability distribution on ℝd with a bounded density function. We consider the chromatic number χ(Gn) of Gn and its relation to the clique number ω(Gn) as n...
### Access Document
Publisher copy:
10.1007/s00493-011-2403-3
### Authors
Journal:
Combinatorica
Volume:
31
Issue:
4
Pages:
423-488
Publication date:
2011-11-05
DOI:
ISSN:
0209-9683
URN:
uuid:c0c509ae-a011-4a4a-971a-05688d7373bd
Source identifiers:
216469
Local pid:
pubs:216469
Language:
English
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2021-07-25 10:10:05
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https://access.openupresources.org/curricula/our-hs-math/aga/geometry/unit-8/lesson-5/index.html
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# Lesson 5Special RightsSolidify Understanding
## Learning Focus
Find missing sides of special right triangles without using trigonometry.
Why are and triangles considered to be “special?”
Why can we find the missing sides of these right triangles without using trigonometry?
Are there any other special right triangles?
## Open Up the Math: Launch, Explore, Discuss
The Pythagorean theorem and right triangle trigonometry are both useful mathematical tools when trying to find missing sides of a right triangle.
### 1.
What do you need to know about a right triangle in order to use the Pythagorean theorem?
### 2.
What do you need to know about a right triangle in order to use right triangle trigonometry?
While using the Pythagorean theorem is fairly straightforward (you only have to keep track of the legs and hypotenuse of the triangle), right triangle trigonometry generally requires a calculator to look up values of different trigonometric ratios. There are some right triangles, however, for which knowing a side length and an angle is enough to calculate the value of the other sides without using trigonometry. These are known as special right triangles because their side lengths can be found by relating them to another geometric figure for which we know something about its sides.
One type of special right triangle is a triangle.
### 3.
Draw a triangle and assign a specific value to one of its sides. (For example, let one of the legs measure , or choose to let the hypotenuse measure . You will want to try both approaches to perfect your strategy.) Now that you have assigned a measurement to one of the sides of your triangle, find a way to calculate the measures of the other two sides. As part of your strategy, you may want to relate this triangle to another geometric figure that may be easier to think about.
### 4.
Generalize your strategy by letting one side of the triangle measure . Show how the measure of the other two sides can be represented in terms of . (Make sure to consider cases where is the length of a leg, as well as the case where is the length of the hypotenuse.)
Another type of special right triangle is a triangle.
### 5.
Draw a triangle and assign a specific value to one of its sides. Now that you have assigned a measurement to one of the sides of your triangle, find a way to calculate the measures of the other two sides. As part of your strategy, you may want to relate this triangle to another geometric figure that may be easier to think about.
### 6.
Generalize your strategy by letting one side of the triangle measure . Show how the measure of the other two sides can be represented in terms of . (Make sure to consider cases where is the length of a leg, as well as the case where is the length of the hypotenuse.)
### 7.
Can you think of any other angle measurements that will create a special right triangle?
An interesting “special” right triangle was discovered by the astronomer and mathematician Johannes Kepler (1571–1630). The Kepler Right Triangle has side lengths that form three terms in a geometric sequence: , , , where is the length of the shorter leg of the right triangle and is the “golden ratio.”
Anciently, mathematicians, artists and architects were intrigued by the golden ratio—dividing a line segment into two parts and so that the ratio of the longer part to the shorter part was equivalent to the ratio of the entire length of the line segment to the longer part , that is, .
Point divides line segment in a golden ratio if . That is, , , form a geometric sequence: multiplying length by the golden ratio gives length , and multiplying length by the golden ratio gives us length .
A golden rectangle was formed by using the lengths and as the sides, forming an by rectangle with area . An interesting feature of a golden rectangle is that when the square with area is removed from the rectangle, the remaining by rectangle is also a golden rectangle. Removing another square from this rectangle produces another golden rectangle, and so on ...
Kepler was fascinated to find that the ratio of the hypotenuse to short leg in a Kepler Triangle was also the golden ratio.
### 1.
Find the exact value of the golden ratio using the information given above. For simplicity, let the shorter part so that , the golden ratio. (Hint: Write and solve a quadratic equation that is equivalent to the proportionality statement.)
### 2.
Once you have found , show that 1, , forms a geometric sequence. If they do, then these lengths will form a Kepler Right Triangle.
## Takeaways
Sometimes we don’t need to use trigonometry to find missing sides of a right triangle when only the angles and one side length is known. Triangles for which this is possible are called special right triangles.
Give the relationship between and the other sides of the right triangle in each of the following triangles:
Give the relationship between and the other sides of the right triangle in each of the following triangles:
## Lesson Summary
In this lesson, we learned there are some special right triangles for which missing sides of the triangle can be found when only one side is known, without using trigonometry! This happens when the right triangle is the result of decomposing a familiar shape, such as a square or an equilateral triangle, into two congruent right triangles.
## Retrieval
### 1.
and , and
Find , , and .
Find and .
#### c.
Will be longer or shorter than ? Why?
### 2.
#### a.
Identify the type of function.
#### b.
Write the equation of the function.
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2022-06-25 14:09:49
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https://electronics.stackexchange.com/questions/34754/transformer-labeled-120v-60hz-35va-12v-what-exactly-does-it-do
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Transformer labeled 120v-60hz-35va-12v - What exactly does it do?
I have a pool light whose bulb physically broke, but I've found a way to open the water-tight bulb enclosure to get at the actual bulb. I'm trying to either hunt down a replacement bulb (rather than the whole expensive assembly) OR make some LED magic.
Either way, the transformer's somewhat permanently mounted and I'd like to use it as it's already there, water-tight, and has the specific fitting to match the pool light flange.
I just can't figure out what 35va-12v means, and google didn't really help. I assume 35VA means 35VAC, but 12V is a common DC voltage and it's on there too!
Of course, I tested it and I actually get 14 VAC. How would I, in the future, know that based on the label? Is there a spec for this kind of stuff?
If it is a plain transformer, without any additional electronics, then the 12V mentioned means 12V(AC). This is the voltage the transformer outputs at its rated load.
Now what is this rated load I speak of? It is about how much power the transformer can potentially output. The 35VA indication is a measure of how much power the transformer can deliver. This is simply the product of voltage and current. The voltage is 12V, so maximum current can easily be calculated I(max) = 35VA / 12V = 2.9A When experimenting with this transformer, you might want to add a 3A fuse at the 12V(AC) side, just to prevent overloading.
You wrote your lightbulb is broken, so your transformer has no load. This is a bit like you driving your car down hill. Your engine doesn't have to work very hard and you probably start accelerating a bit. An unloaded transformer increases its output voltage. 14V output voltage is perfectly normal with this transformer when unloaded. When you attach a 25W lightbulb to it, you'll notice the voltage drops to around 12V. With a conventional light bulb, you can use bulbs up to 35W with this transformer. When using other devices the maximum number of watts drops, depending on the exact device.
120V and 60Hz is related to your mains supply. It is what you get from the wall socket.
• A 3A fuse isn't much headroom unless it's a slow-blow. An incandescent bulb is a very low resistance before it's lit, and will draw way more than 3A getting up to temperature. – Bryan Boettcher Jun 29 '12 at 15:36
• So I could build a rectifier inside the bulb and get 12vdc? – kavisiegel Dec 22 '13 at 18:46
• More or less, and only when not using a buffer capacitor. The peak voltage is more like $12V× \sqrt(2) - ( 2 × 0.7V ) = 15.6V$, which you will see with a buffer capacitor. Rectifying a transformer's output can be a bit tricky when it comes to the exact numbers. Probably worth more than a comment here. – jippie Dec 22 '13 at 20:22
Here 35 VA means 35 volt-amperes.
If a system is purely resistive, that would be equivalent to 35 W, but in cases where it isn't, we need to use complex power which consists of real part measured in watts and imaginary part measured in vars. Since the ratios of those parts can be different, the transformers are specified in VA, which takes both parts into account.
The voltages would be 120 V AC nominal input at 60 Hz and 12 V AC nominal output. Do note that in cases where the load is light, the output voltage will be higher than 12 V.
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2020-02-18 03:34:31
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https://listserv.uni-heidelberg.de/cgi-bin/wa?A2=LATEX-L;26aed6f3.0606&FT=P&P=3171957&H=N&S=b
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## LATEX-L@LISTSERV.UNI-HEIDELBERG.DE
Options: Use Classic View Use Monospaced Font Show Text Part by Default Show All Mail Headers Topic: [<< First] [< Prev] [Next >] [Last >>]
Re: Consistently active characters Lars Hellström <[log in to unmask]> Sat, 18 Sep 2010 18:26:06 +0200 text/plain (39 lines) Will Robertson skrev 2010-09-18 08.55: > On 18/09/2010, at 9:03 AM, Heiko Oberdiek wrote: > >> \providecommand*{\MakeActiveDef}[1]{% >> \catcode#1\active >> \begingroup >> \lccode\~#1% >> \lowercase{\endgroup \def ~}% >> } > > > Ah, this is nice. Better than what I was thinking of as you no longer need a stray active character floating around. > > I've added these tentatively to expl3 in the "candidates" module under the names \char_active_set:Npn, \char_active_set_eq:NN, and so on. Note that in order to \lowercase something to an arbitrary character code, the something you start from must be ^^@; \lowercase doesn't change tokens whose \lccode is 0. This is of course not a problem, since you can always set the \catcode of ^^@ locally while defining the macro where an active ^^@ is needed: \begingroup \catcode0=13\relax \@firstofone{\endgroup \providecommand*{\MakeActiveDef}[1]{% \catcode#1=13% \begingroup \lccode0=#1% \lowercase{\endgroup \def ^^@}% } } Another change used above is to make the command argument the _character code_ of the active to define, rather than a token with that character. It is easy to use to make that number if needed, but in several cases you rather start out from the character code. Lars Hellström
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2023-03-28 03:33:52
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http://megasoft-rapid.com/North-Dakota/error-of-measurement-and-the-power-of-a-statistical-test.html
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Address 39001 590th St NW, Kenmare, ND 58746 (701) 848-2740
# error of measurement and the power of a statistical test Flaxton, North Dakota
Example The Brinell hardness scale is one of several definitions used in the field of materials science to quantify the hardness of a piece of metal. But it also increases the risk of obtaining a statistically significant result (i.e. It also discusses how to measure effect size for two independent groups, for two dependent groups, and when conducting Analysis of Variance. Buy article ($45.00) Have access through a MyJSTOR account? In many contexts, the issue is less about determining if there is or is not a difference but rather with getting a more refined estimate of the population effect size. It turns out that the null hypothesis will be rejected if T n > 1.64. {\displaystyle T_{n}>1.64.} Now suppose that the alternative hypothesis is true and μ D = θ {\displaystyle All Rights Reserved For full functionality of ResearchGate it is necessary to enable JavaScript. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Testing for statistical significance helps you learn how likely it is that these changes occurred randomly and do not represent differences due to the program. LevinReadPlanung und Auswertung von Experimenten*)[Show abstract] [Hide abstract] ABSTRACT: Vorbemerkungen Ein wesentliches Ziel jeder empirischen Wissenschaft besteht darin, zu fun-dierten Kausalaussagen zu gelangen, also zu Aussagen über die Ursachen oder Bedingungen The power of the test is the probability that the test will find a statistically significant difference between men and women, as a function of the size of the true difference formulas) provides the answer to that question: With a ~ .05 "[Show abstract] [Hide abstract] ABSTRACT: A general rationale and specific procedures for examining the statistical power characteristics of psychology-of-aging empirical Custom alerts when new content is added. Coverage: 1932-2008 (Vol. 1, No. 1 - Vol. 77, No. 1) Moving Wall Moving Wall: 7 years (What is the moving wall?) Moving Wall The "moving wall" represents the time period Voelker, Peter Z. This has been extended[7] to show that all post-hoc power analyses suffer from what is called the "power approach paradox" (PAP), in which a study with a null result is thought Standard Deviation Figure 1 also shows that power is higher when the standard deviation is small than when it is large. Ob mit Hilfe eines Experi-ments überhaupt Aussagen über Ursachen möglich sind, hängt von seiner ,,internen Validität" ab (Teil 3). Zimmerman, Bruno D. The probability is 0.3085 as illustrated here: $\beta= P(\bar{X} < 172 \text { if } \mu = 173) = P(Z < -0.50) = 0.3085$ A probability of 0.3085 is a Think you should have access to this item via your institution? Solution.In this case, the engineer makes the correct decision if his observed sample mean falls in the rejection region, that is, if it is greater than 172, when the true (unknown) Stable URL: http://www.jstor.org/stable/20152465 Page Count: 8 Download ($45.00) Cite this Item Cite This Item Copy Citation Export Citation Export to RefWorks Export a RIS file (For EndNote, ProCite, Reference Manager, Zotero…) II.
doi:10.1016/j.jclinepi.2008.08.005. Since sample size is typically under an experimenter's control, increasing sample size is one way to increase power. Hence, the power of the test is increased. A hypothesis test may fail to reject the null, for example, if a true difference exists between two populations being compared by a t-test but the effect is small and the
Let's take a look at two examples that illustrate the kind of sample size calculation we can make to ensure our hypothesis test has sufficient power. We denote α = P(Type I Error). Definition.The powerof a hypothesis test is the probability of making the correct decision if the alternative hypothesis is true. This time, instead of taking a random sample ofn= 16 students, let's increase the sample size to n = 64.
Williams, Donald W. ISBN0-8058-0283-5. Select the purchase option. For instance, in multiple regression analysis, the power for detecting an effect of a given size is related to the variance of the covariate.
In other words, factors that affect power. Whoopdy-do...would that be a rocking conclusion? The site also describes the procedure used to test for significance (including the p value) What is effect size? Check out using a credit card or bank account with PayPal.
LinnRead moreDiscover moreData provided are for informational purposes only. The sample size determines the amount of sampling error inherent in a test result. III. The null hypothesis of no effect will be that the mean difference will be zero, i.e.
Journal of Clinical Epidemiology. 62 (6): 609–616. Well, let's suppose that a medical researcher is interested in testing the null hypothesis that the mean total blood cholesterol in a population of patients is 200 mg/dl against the alternative This means you are less likely to accept the null hypothesis when it is false; i.e., less likely to make a Type II error. The impact of bias depends on which of 4 classes of rating design is used, and formulas are derived for correcting observed effect sizes for attenuation (due to bias variance) and
Assume, a bit unrealistically, thatXis normally distributed with unknown meanμand standard deviation 16. Predictive probability of success Both frequentist power and Bayesian power uses statistical significance as success criteria. View Mobile Version Factors Affecting Power Author(s) David M. Example The following is an example that shows how to compute power for a randomized experiment: Suppose the goal of an experiment is to study the effect of a treatment on
Fehler bei dieser ,,Operationalisierung" beeinträchtigen die ,,Variablenvalidität" der Untersuchung (Teil 2). He could still do a bit better. Hmm.... It is not possible to guarantee a sufficient large power for all values of θ {\displaystyle \theta } , as θ {\displaystyle \theta } may be very close to 0.
We have two(asterisked (**))equations and two unknowns! Such measures typically involve applying a higher threshold of stringency to reject a hypothesis in order to compensate for the multiple comparisons being made (e.g. An alpha level of less than .05 is accepted in most social science fields as statistically significant, and this is the most common alpha level used in EE evaluations. In addition, the concept of power is used to make comparisons between different statistical testing procedures: for example, between a parametric and a nonparametric test of the same hypothesis.
A researcher is interested in whether a new method of teaching results in a higher mean. Unlimited access to purchased articles.
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2018-12-19 03:46:56
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http://openstudy.com/updates/55ccb58ae4b0016bb0167b4d
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## anonymous one year ago Which of the following points lie in the solution set to the following system of inequalities? y<x - 5 y<-x - 4 A. (1, 10) B. (-1, 10) C. (10, 1) D. (1, -10)
1. anonymous
$y=x-5\\ y=-x-4$?
2. anonymous
set them equal $x-5=-x-4$ and solve for $$x$$
3. anonymous
D?
4. anonymous
must me some typo here or i misunderstood the question
5. anonymous
$x-5=-x-4\\ 2x-5=-4\\ 2x-1\\ x=\frac{1}{2}$so i must ahve misunderstood none of those answers are correct
6. anonymous
oh
7. SolomonZelman
the signs are missing
8. anonymous
ok i am going to guess that these are inequalities
9. anonymous
oh shott your so right lol let me fix that sorry :/
10. SolomonZelman
it is ok... (for the second part of the comment that i understood)
11. anonymous
ooops lol i meant shoot hahah its fixed now :)
12. SolomonZelman
i thought I am missing some slang, like shut your ..
13. SolomonZelman
shut your so, and i was like "???????...?"\
14. anonymous
oh no lol sorry i didnt spell it right haha
15. SolomonZelman
:) it is all good
16. anonymous
does the q make sense now?
17. SolomonZelman
graph your inequalities in https://www.desmos.com/calculator and your points graph as well. Wont take long
18. SolomonZelman
yes now the q makes sense defnintely
19. anonymous
ok its not working for me lol idk how to use it hahah i was typing it in but it wont give me the answer
20. anonymous
@Teddyiswatshecallsme
21. SolomonZelman
22. SolomonZelman
there is the graph of inequalities and the points.
23. anonymous
ik but i dont know how :/ can you do it for me pls :)
24. SolomonZelman
I graphed the points and the inequalities for you. Whichever point lays in the region shaded by both inequalities is the solution.
25. anonymous
ok so what was it?
26. anonymous
@David27
27. anonymous
@ali2x2
28. anonymous
@Baby_Bear69
29. ali2x2
D and you should fan me so you dont tag me when im offline ;)
30. anonymous
ok thats what i was thinking and ok i will :) thnx
31. Baby_Bear69
I was thinking B, shows what i know.. lol I'm rusty
32. anonymous
lol its ok thnx anyways :)
33. Teddyiswatshecallsme
I'm here now C:
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2016-10-23 23:58:14
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https://deepai.org/publication/polar-decomposition-based-algorithms-on-the-product-of-stiefel-manifolds-with-applications-in-tensor-approximation
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# Polar decomposition based algorithms on the product of Stiefel manifolds with applications in tensor approximation
In this paper, based on the matrix polar decomposition, we propose a general algorithmic framework to solve a class of optimization problems on the product of Stiefel manifolds. We establish the weak convergence and global convergence of this general algorithmic approach based on the Łojasiewicz gradient inequality. This general algorithm and its convergence results are applied to the best rank-1 approximation, low rank orthogonal approximation and low multilinear rank approximation for higher order tensors. We also present a symmetric variant of this general algorithm to solve a symmetric variant of this class of optimization models, which essentially optimizes over a single Stiefel manifold. We establish its weak convergence and global convergence in a similar way. This symmetric variant and its convergence results are applied to the best symmetric rank-1 approximation and low rank symmetric orthogonal approximation for higher order tensors. It turns out that well-known algorithms such as HOPM, S-HOPM, LROAT, S-LROAT are all special cases of this general algorithmic framework and its symmetric variant.
## Authors
• 7 publications
• 10 publications
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### Jacobi-type algorithms for homogeneous polynomial optimization on Stiefel manifolds with applications to tensor approximations
In this paper, we mainly study the gradient based Jacobi-type algorithms...
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### On the convergence of Jacobi-type algorithms for Independent Component Analysis
Jacobi-type algorithms for simultaneous approximate diagonalization of s...
11/25/2019
### The Epsilon-Alternating Least Squares for Orthogonal Low-Rank Tensor Approximation and Its Global Convergence
The epsilon alternating least squares (ϵ-ALS) is developed and analyzed ...
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### Scalable symmetric Tucker tensor decomposition
We study the best low-rank Tucker decomposition of symmetric tensors, ad...
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## 1. Introduction
Theory and algorithms on optimization over manifolds have been developed and applied widely because of its practical relevance; see [3]. In particular, methods such as the Newton-type and trust-region algorithms have been proposed for optimization over Stiefel manifolds, which represents the orthogonal constraints [1, 3, 4, 18].
Let be the Stiefel manifold with . In this paper, we mainly study the optimization problem on the product of Stiefel manifolds, which is to maximize a smooth function
f: Ω⟶R+, (1)
where
Ωdef=St(r1,n1)×% St(r2,n2)×⋯×St(rd,nd)
is the product of Stiefel manifolds (). Let . If and for , then becomes
Ωsdef=St(r,n)×St(r,n)×⋯×St(r,n).
Assume that the cost function (1) is symmetric on , that is,
f(ω)=f(π(ω)) (2)
for any and permutation . In this paper, we also study the symmetric variant of problem (1), which is to maximize the smooth function
g: St(r,n)⟶R+, U⟼f(U,U,⋯,U). (3)
Many low rank approximation problems for higher order tensors [8, 12, 29, 44], such as the best rank-1 approximation [15, 16, 17, 50], low rank orthogonal approximation [7, 28] and low multilinear rank approximation [15, 17, 23], can be formulated in the form of problem (1). These approximations have been widely used in various fields, including signal processing [12, 14, 20, 26, 38], numerical linear algebra [41, 42] and data analysis [5, 8, 29, 44]
. In particular, the symmetric variant of best rank-1 approximation is corresponding to finding the largest Z-eigenvalue
[42]. When the rank is equal to the dimension, the low rank orthogonal approximation boils down to the approximate orthogonal diagonalization [11, 34, 36, 37, 47] of symmetric tensors, which is central in Independent Component Analysis (ICA) [9, 10]. The low multilinear rank approximation is equivalent to the well-known Tucker decomposition [17, 29]
, which has been a popular method for data reduction in signal processing and machine learning. Therefore, it is desirable to develop a general algorithmic scheme to solve problem (
1) and its symmetric version (3).
Contribution. In this paper, based on the matrix polar decomposition, we propose an approach to be called APDOI (alternating polar decomposition based orthogonal iteration) (Algorithm 1) to solve problem (1). Then we establish its weak convergence111every accumulation point is critical. and global convergence222for any starting point, the iterations converge as a whole sequence. based on the Łojasiewicz gradient inequality. We apply APDOI and establish its convergence properties to the best rank-1 approximation (Section 6.1); the low rank orthogonal approximation (Section 7.1); the low multilinear rank approximation (Section 8) for higher order tensors. It turns out that the well-known methods HOPM [16, 17] (for best rank-1 approximation) and LROAT [7] (for low rank orthogonal approximation) are both special cases of APDOI. Our convergence results subsume the results found in the literature designed for those special cases.
Algorithm APDOI for the low multilinear rank approximation will be called LMPD (low multilinear rank approximation based on polar decomposition) in this paper. For this particular algorithm, we propose its shifted variant, LMPD-S, which has a better convergence property. Then we conduct experiments to show that LMPD and LMPD-S have comparable speed with the well-known HOOI [17] algorithm, and LMPD-S has a even much better convergence performance.
As the symmetric variant of APDOI, we propose a PDOI (polar decomposition based orthogonal iteration) approach (Algorithm 2) to solve problem (3). We establish its weak convergence and global convergence in a similar way. Then Algorithm PDOI and its convergence results are applied to the best symmetric rank-1 approximation (Section 6.2) and low rank symmetric orthogonal approximation (Section 7.2) for higher order tensors. It turns out that the well-known algorithms S-HOPM [17] (for best symmetric rank-1 approximation) and S-LROAT [7] (for low rank symmetric orthogonal approximation) are both special cases of PDOI. Our convergence results also subsume the results found in the literature designed for those special cases.
Organization. The paper is organized as follows. In Section 2, we introduce notations and preliminaries. In Section 3, we show the condition under which problems (1) and (3) will be studied in this paper. Then we prove that this condition is satisfied in the best rank-1 approximation, low rank orthogonal approximation and low multilinear rank approximation for higher order tensors. In Section 4, we propose Algorithm APDOI to solve problem (1), and establish its convergence. In Section 5, as a symmetric variant of APDOI, we propose Algorithm PDOI to solve problem (3), and establish its convergence. In Sections 8, 7 and 6, we apply these convergence results to the best rank-1 approximation, low rank orthogonal approximation and low multilinear rank approximation for higher order tensors, respectively.
## 2. Notations and preliminaries
### 2.1. Notations
Let be the linear space of -th order real tensors and be the set of symmetric ones, whose entries do not change under any permutation of indices [13, 41]
. Tensor arrays, matrices, and vectors, will be respectively denoted by bold calligraphic letters,
e.g., , with bold uppercase letters, e.g., , and with bold lowercase letters, e.g., ; corresponding entries will be denoted by , , and . Let and . We use the -mode product defined by
(A∙iM)p1⋯j⋯pd=∑piAp1⋯pdMjpi.
The -mode unfolding of is denote by , which is the matrix obtained by reordering the -mode fibers333the fiber of a tensor is defined by fixing every index but one. of in a fixed way [29]. We denote by the i-rank, that is, the rank of . The vector is called the multilinear rank of . The diagonal of a tensor is defined by
Let be the orthogonal group. We denote by the Frobenius norm of a tensor or a matrix, or the Euclidean norm of a vector. We denote by the unit sphere of . We denote by
the minimal singular value of a matrix. Let
for .
Let be a Riemannian submanifold, and be a differentiable function. In this paper, for simplicity, we also denote by the natural extension of itself to . Let and be the tangent space of at . We denote by the Euclidean gradient of at , and the Riemannian gradient444see [3, Section 3.6] for a detailed definition. of at .
###### Definition 2.1.
([43, Definition 2.1]) The funciton is said to satisfy a Łojasiewicz gradient inequality at , if there exist , and a neighborhood in of such that for all , it follows that
###### Lemma 2.2 ([43, Proposition 2.2]).
Let be an analytic submanifold555See [31, Definition 2.7.1] or [36, Definition 5.1] for a definition of an analytic submanifold. and be a real analytic function. Then for any , satisfies a Łojasiewicz gradient inequality (4) in the -neighborhood of , for some666The values of depend on a specific point. and .
Based on the discrete-time analogue of classical Łojasiewicz’s theorem [2, 33, 39], the following result was proved [43, Theorem 2.3].
###### Theorem 2.3.
Let be an analytic submanifold and . Suppose that is real analytic and, for large enough ,
(i) there exists such that
(ii) implies that .
Then any accumulation point of must in fact be the same.
###### Remark 2.4.
It can be verified, after checking the proof in [43], that condition (ii) in Theorem 2.3 can be replaced by that implies .
## 3. Problem statement and tensor approximation examples
### 3.1. Problem statement
We first discuss the condition, under which problems (1) and (3) will be studied in this paper. This condition is motivated by the tensor approximation problems in Section 3.2.1, Section 3.3.1 and Section 3.4 .
#### 3.1.1. General case
Let
Ω(i)def=St(r1,n1)×⋯×St(ri−1,ni−1)×St(ri+1,ni+1)×⋯×St(rd,nd)
be the product of Stiefel manifolds and . Let . Define
h(i): St(ri,ni) ⟶R+, U ⟼f(ν(i),U)def=f(U(1),⋯,U(i−1),U,U(i+1),⋯,U(d)). (5)
In this paper, we assume that problem (1) always satisfies the restricted homogeneity condition, which means that there is some fixed such that, for any and , it holds that
h(i)(U)=α⟨U,∇h(i)(U)⟩, (6)
for any . By the gradient inequality [6, (3.2)], it follows that the function (3.1.1) is convex if and only if
⟨U′−U,∇h(i)(U)⟩≤h(i)(U′)−h(i)(U),
for any , and then by condition (6), if and only if
⟨U′,∇h(i)(U)⟩≤(1−α)⟨U,∇h(i)(U)⟩+α⟨U′,∇h(i)(U′)⟩, (7)
for any . We say that the cost function (1) is partially convex if the function (3.1.1) is convex for any and .
#### 3.1.2. Symmetric case
In this paper, we assume that problem (3) always satisfies the homogeneity condition, which means that there is some fixed such that
g(U)=β⟨U,∇g(U)⟩ (8)
for any . By the gradient inequality [6, (3.2)], it follows that the cost function (3) is convex if and only if
⟨U′−U,∇g(U)⟩≤g(U′)−g(U),
for any , and then by condition (8), if and only if
⟨U′,∇g(U)⟩≤(1−β)⟨U,∇g(U)⟩+β⟨U′,∇g(U′)⟩,
for any .
###### Remark 3.1.
Suppose that the cost function (1) is symmetric on and satisfies condition (6). Then the corresponding cost function (3) satisfies condition (8) with . In fact, for any , we define
h: St(r,n)⟶R+, U⟼f(U0,⋯,U0,U).
Then we get that
∇g(U0)=d∇h(U0) (9)
by (2). It follows from condition (6) that
g(U0)=h(U0)=α⟨U0,∇h(U0)⟩=αd⟨U0,∇g(U0)⟩.
Therefore, the cost function (3) satisfies condition (8) with .
### 3.2. Example: best rank-1 approximation
#### 3.2.1. General case
Let . The best rank-1 approximation problem [15, 16, 17, 50] is to find
B∗=argmin∥A−B∥, (10)
where with and . It was proved [15, 17] that problem (10) is equivalent to the maximization of
f(u(1),u(2),⋯,u(d))=|A∙1u(1)T∙2u(2)T⋯∙du(d)T|2, (11)
where for .
###### Lemma 3.2.
The cost function (11) satisfies (6) with and is partially convex.
###### Proof.
Let . Denote that
v(i)def=A∙1u(1)T⋯∙i−1u(i−1)T∙i+1u(i+1)T⋯∙du(d)T.
Let and . Then we have that
h(i)(u)=|⟨v(i),u⟩|2=⟨u,V(i)u⟩, ∇h(i)(u)=2V(i)u=2⟨v(i),u⟩v(i). (12)
It is easy to see that the cost function (11) satisfies condition (6) with . Moreover, by the Cauchy-Schwarz inequality, we have that
⟨u′,∇h(i)(u)⟩≤12(⟨u,∇h(i)(u)⟩+⟨u′,∇h(i)(u′)⟩)
for any . Then the cost function (11) is partially convex by (7). The proof is complete. ∎
#### 3.2.2. Symmetric case
Let . The best symmetric rank-1 approximation problem [17, 27, 30, 42, 51] is to find
B∗=argmin∥A−B∥, (13)
where with and . It was proved [15, 17] that problem (13) is equivalent to the maximization of
g(u)def=f(u,u,⋯,u)=|A∙1uT∙2uT⋯∙duT|2, (14)
where . By Remark 3.1 and Lemma 3.2, we see that the cost function (14) satisfies condition (8) with .
###### Lemma 3.3.
Suppose that is even. Let
C=D∙1HT∙2HT⋯∙dHT (15)
where and is a diagonal tensor satisfying that with . Let and . Then, for any , the function
τi:Rn×r⟶R+, X⟼|Wii⋯i|2
is convex.
###### Proof.
Let be the linear map and
γi:Rm×r⟶R+, Y⟼|Wii⋯i|2,
where . Note that We see that is convex, and thus is also convex. ∎
###### Remark 3.4.
By Lemma 3.3, if takes the form in (15), then the cost function (14) is convex. This fact has also been proved by the method of square matrix unfolding in [27, (4.2)].
### 3.3. Example: low rank orthogonal approximation
#### 3.3.1. General case
Let and for . The low rank orthogonal approximation problem [7, 22, 28, 49] is to find
B∗=argmin∥A−B∥, (16)
where
B=r∑ℓ=1μℓu(1)ℓ⊗⋯⊗u(d)ℓ
satisfies that for any and . Let for . It is clear that
B=D∙1U(1)∙2U(2)∙3⋯∙dU(d),
where is the diagonal tensor satisfying that . It was proved [7] that problem (16) is equivalent to the maximization of
f(U(1),U(2),⋯,U(d))=∥\operator@fontdiag{W}∥2, (17)
where .
###### Lemma 3.5.
The cost function (17) satisfies (6) with and is partially convex.
###### Proof.
Let . Denote that
V(i)def=A∙1U(1)T⋯∙i−1U(i−1)T∙i+1U(i+1)T⋯∙dU(d)T, v(i,q)def=A∙1u(1)qT⋯∙i−1u(i−1)qT∙i+1u(i+1)qT⋯∙du(d)qT∈Rni,
for . Let and . We can calculate that
h(i)(U)=∥\operator@fontdiag{W}∥2=r∑q=1⟨v(i,q),uq⟩2, ∇h(i)(U)=2[v(i,1),⋯,v(i,r)]⎡⎢ ⎢⎣W1⋯1⋱Wr⋯r⎤⎥ ⎥⎦. (18)
Then it can be verified that
h(i)(U)=12⟨U,∇h(i)(U)⟩.
Let and . By (18) and the Cauchy-Schwarz inequality, we get
2⟨U′,∇h(i)(U)⟩ =2r∑q=1W′q⋯qWq⋯q ≤r∑q=1W′2q⋯q+r∑q=1W2q⋯q=⟨U′,∇h(i)(U′)⟩+⟨U,∇h(i)(U)⟩.
Then the cost function (17) is partially convex by (7). The proof is complete. ∎
#### 3.3.2. Symmetric case
Let and . Then low rank symmetric orthogonal approximation problem [7, 35, 40] is to find
B∗=argmin∥A−B∥, (19)
where
B=r∑ℓ=1μℓuℓ⊗⋯⊗uℓ
satisfies that for . It was proved [7] that problem (19) is equivalent to the maximization of
g(U)def=f(U,U,⋯,U)=∥\operator@fontdiag{W}∥2, (20)
where .
###### Remark 3.6.
By Lemma 3.3, if takes the form in (15), then the cost function (20) is convex.
### 3.4. Example: low multilinear rank approximation
Let and for . The low multilinear rank approximation problem [15, 17, 32] is to find
B∗=argmin∥A−B∥, (21)
where for . Note that any such can be decomposed as
B=C∙1U(1)∙2U(2)∙3⋯∙dU(d),
with and for . Problem (21) is in fact equivalent to the following Tucker decomposition [45, 15, 17] problem
min∥A−C∙1U(1)∙2U(2)∙3⋯∙dU(d)∥. (22)
It was proved [15, 17] that problem (22) is equivalent to the maximization of
f(U(1),U(2),⋯,U(d))=∥W∥2, (23)
where and for .
###### Lemma 3.7.
The cost function (23) satisfies (6) with and is partially convex.
###### Proof.
Let . Denote that
V(i)def=A∙1U(1)T⋯∙i−1U(i−1)T∙i+1U(i+1)T⋯∙dU(d)T.
Let and . By [29, Section 2.5], we see that
h(i)(U)=∥W∥2=∥UTV(i)(i)∥2,
where with . Let . It can be verified that
h(i)(U)=⟨U,V(i)U⟩, ∇h(i)(U)=2V(i)U, (24)
and thus
h(i)(U)=12⟨U,∇h(i)(U)⟩.
Let . By the Cauchy-Schwarz inequality, it is easy to see that
⟨U′,V(i)U⟩≤12(⟨U,V(i)U⟩+⟨U′,V(i)U′⟩).
Then the cost function (23) is partially convex by (7) and (24). The proof is complete. ∎
###### Remark 3.8.
By (24), we see that the cost function (23) satisfies that is always positive semidefinite for any and .
## 4. Algorithm APDOI on the product of Stiefel manifolds
In this section, based on the matrix polar decomposition, we propose a general algorithm to solve problem (1), and establish its convergence. The algorithm and convergence results in this section apply to the best rank-1 approximation in Section 3.2.1, low rank orthogonal approximation in Section 3.3.1 and low multilinear rank approximation in Section 3.4.
### 4.1. Algorithm APDOI
By [3, (3.35)], the Riemannian gradient of (3.1.1) at is
Therefore, if is a maximal point of (3.1.1), it should satisfy that
∇h(i)(U(i)∗)=U(i)∗\rm sym(U(i)∗T∇h(i)(U(i)∗)), (25)
which is close to a polar decomposition form of .
###### Lemma 4.1.
([19, Theorem 9.4.1], [7, Lemma 5.3]) Let with . Then there exist and positive semidefinite matrix such that has the polar decomposition . We say that is the orthogonal polar factor and is the symmetric polar factor. Moreover,
(i) for any , we have that
⟨U,X⟩≥⟨U′,X⟩;
(ii) is the best orthogonal approximation to , that is,
∥X−U∥≤∥X−U′∥
for any ;
(iii) if , then and are both unique.
Let and . We denote that
ν(k,i)def=(U(1)k,⋯,U(i−1)k,U(i+1)k−1,⋯,U(d)k−1)∈Ω(i), ω(k,i)def=(U(1)k,⋯,U(i)k,U(i+1)k−1,⋯,U(d)k−1)∈Ω,
and . Define
h(k,i): St(ri,ni) ⟶R+, U ⟼f(ν(k,i),U)def=f(U(1)k,⋯,U(i−1)k,U,U(i+1)k−1,⋯,U(d)k−1).
Inspired by the decomposition form in (25), we propose the following alternating polar decomposition based orthogonal iteration (APDOI) algorithm to solve problem (1). We assume that the starting point satisfies that without loss of generality.
###### Algorithm 1.
(APDOI)
Input: starting point .
Output: , ,
• For
• For
• Compute .
• Compute the polar decomposition of .
• Update to be the orthogonal polar factor.
• End for
• Until convergence
###### Remark 4.2.
Note that (1) is smooth and is compact. Let . Then always holds in Algorithm 1.
###### Lemma 4.3.
Let be a differentiable function and . Suppose that is the orthogonal polar factor of .
(i) If , then .
(ii) If , then is symmetric. Furthermore, if is of full rank and is positive semidefinite, then .
###### Proof.
(i) Since , we see that is a polar decomposition. It follows that is symmetric, and thus
∇h(U)=UP=U\rm sym(UT∇h(U)).
Therefore, by [3, (3.35)], we get that .
(ii) By [3, (3.35)], we see that and thus
2UT∇h(U)=UT∇h(U)+∇h(U)TU,
which means that is symmetric. Note that is positive semidefinite and is of full rank. Then is the unique polar decomposition of by Lemma 4.1(iii). Therefore, we get that . ∎
###### Remark 4.4.
By Lemma 4.3, we see that, in Algorithm 1, if is of full rank and
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2022-06-27 08:45:53
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https://www.physicsforums.com/threads/need-axial-tilt-info-of-earth-daily.693584/
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# Need axial tilt info of earth daily
1. ### axialhelp
2
so i wanted to know for my own GK about earths daily axial tilt.So far i have found only axial tilt on solstice and equinox.
there is one way but it is cumbersome and that is to manually find it
Last edited by a moderator: Sep 25, 2014
2. ### Bandersnatch
1,517
Hello, axialhelp. Welcome to PF.
Axial tilt of the Earth is always the same. There's some variation due to nutation and long term(thousands of years) perturbations, but in the time scale of human life or without the need for extreme precision, the angle of tilt remains 23,4° at all times.
This is the angle at which the Earth's axis is tilted away from the line perpendicular to the plane of its orbit.
What does change during the year, is the relative position of the Sun with respect to the direction the axis is pointing.
This causes the angle between the axis and the plane perpendicular to the direction to the Sun(the plane that bisects the Earth in two halves: the illuminated one and the one in shadow, or, day and night halves) to change from the maximum of 23,4° at Summer solstice, through 0° at equinoxes, to -23,4° at Winter solstice.
To find this angle at any given day of the year empirically, you need to do exactly what that BBC video told you to do.
If it's any time between the solstices, you'll get the angle between -23,4<α<23,4.
They even got such a result themselves, due to making measurements a few days before the summer solstice, as the lady tells you at the end of the video.
You could use sun path charts, like this one(http://solardat.uoregon.edu/PolarSunChartProgram.html) to get the data you need for the calculations without leaving your room.
Alternatively, to a good approximation, you can just use this equation to get the angle you want:
$α=23.4\sin({\frac{2\pi}{365}x})$ where x is the day of the year counting from the spring equinox. When calculating remember that the value in the sine function is in radians, but the end result is in degrees.
3. ### axialhelp
2
Thanks but i am unable to understand it or compute it.CAn you compute the axial tilt of earth every 15 days starting from dec 21?. is there any website which has the axial tilt for each day? .
Anyways my question was i wanted to find out the effects on earth if the axial tilt would be halved by 50%. How would the season be ?
http://io9.com/5829438/earth-doesnt-need-the-moon
### Staff: Mentor
Again: earth's axial tilt is constant.
5. ### jim mcnamara
1,554
Are you trying to do some kind of astrography or sun angle project? In other words, what are you trying to figure out, in plain words? The answer is not change in axial tilt, it is change in ______ relative to: ______ (you fill in the blanks).
Your attached link shows the sun's angle versus the Zenith -> to get the axial tilt of the earth: 23.4 deg -- On the summer Solstice @noon. The axial tilt does not change (from a normal human perspective anyway) all through the year. The angle of the sun at astronomical noon changes, but if you correct for that you get the same answer over and over.
Last edited: May 27, 2013
6. ### Bandersnatch
1,517
http://speedy.sh/3kARu/sunposition.ods
(it's an Open Office file)
The first one calculates the deviation from the position of the Sun at noon during equinoxes(in degrees).
At equinoxes, the Sun's elevation is equal to 90° minus your lattitude. The value calculated is the angular distance that the Sun appears to move up and down from that point over the year.
The second one calculates the actual position of the Sun at noon as measured from above the Southern horizon. It's simply 90° minus your lattitude plus the deviation value.
I'm guessing these are the values you are looking for.
You can enter new values into the cyan-coloured cells. The first one is the axial tilt, which, we cannot stress this more, does not change.
The second one is the lattitude from which you're observing the Sun.
It uses the same equation as provided in post #2, which assumes circular orbit. This is not really the case with Earth, so keep in mind that actual values will differ slightly.
There's also the case of refraction, that causes the Sun to appear higher above the horizon than it really is. (http://en.wikipedia.org/wiki/Atmospheric_refraction) This effect is obviously not included, so the values close to the horizon will differ some more in reality.
You should be able to see that the position of the Sun would deviate less from the equinox position, which means that Summer and Winter would be less extreme and more like Spring or Autumn.
Last edited: May 27, 2013
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2015-09-01 22:22:55
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https://www.aimsciences.org/article/doi/10.3934/dcds.1995.1.207
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# American Institute of Mathematical Sciences
April 1995, 1(2): 207-216. doi: 10.3934/dcds.1995.1.207
## The index at infinity of some twice degenerate compact vector fields
1 Institute for Information Transmission Problems, Russian Academy, 19 Ermolovoy st., 101447 Moscow, Russian Federation 2 Institut Mathématique Pure et Appliquée, Université Catholique de Louvain, B-1348 Louvain-la-Neuve
Received November 1994 Published February 1995
The index at infinity of some compact vector fields associated with Nemytski operators is computed in situations where the linear part is degenerate and the nonlinear part does not satisfy the Landesman-Lazer conditions. Applications are given to the existence and multiplicity of solutions of nonlinear equations depending upon a parameter.
Citation: A.M. Krasnosel'skii, Jean Mawhin. The index at infinity of some twice degenerate compact vector fields. Discrete & Continuous Dynamical Systems, 1995, 1 (2) : 207-216. doi: 10.3934/dcds.1995.1.207
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2019 Impact Factor: 1.338
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2021-04-11 15:49:27
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http://math.stackexchange.com/questions/345703/prove-that-cos-a-b-cos-a-b-cos-2a-sin-2b/345718
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# Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$
$$\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$$
I have attempted this question by expanding the left side using the cosine sum and difference formulas and then multiplying, and then simplifying till I replicated the identity on the right. I am not stuck. What's bothering me is that the way I went about this question seemed like a rather "clunky" method. I'm just curious if I've missed some underlying pattern that could have made it easier to reproduce the identity on the right.
The way I did it:
$$\begin{array}{l} \cos (A + B)\cos (A - B)\\ \equiv (\cos A\cos B - \sin A\sin B)(\cos A\cos B + \sin A\sin B)\\ \equiv {\cos ^2}A{\cos ^2}B - {\sin ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A(1 - {\sin ^2}B) - (1 - {\cos ^2}A){\sin ^2}B\\ \equiv {\cos ^2}A - {\cos ^2}A{\sin ^2}B - {\sin ^2}B + {\cos ^2}A{\sin ^2}B\\ \equiv {\cos ^2}A - {\sin ^2}B\end{array}$$
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Don't you have an $=$ on your keyboard? – Git Gud Mar 29 '13 at 14:52
The book I'm using saying you should use the "identical to" sign when dealing with identities, is this not the case? – seeker Mar 29 '13 at 14:53
What? That's probably a convention used only by geometers. Makes no sense if you ask me. What does $\equiv$ even mean here? How does it differ from $=$? It's just my opinion, though. – Git Gud Mar 29 '13 at 14:55
@GitGud: $$\cos(A+B)\cos(A-B)\equiv\cos^2A-\sin^2B$$ is an abbreviation for $$\forall A\forall B\bigl[\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B\bigr].$$ In contrast, while there exist $C$ such that $\sin C=\cos C$, we would never say $\sin C\equiv\cos C$. – Cameron Buie Mar 29 '13 at 15:37
@CameronBuie Thanks for clarifying. – Git Gud Mar 29 '13 at 15:38
You can use the addition theorem which states that $$\cos(\alpha+\beta)=\cos(\alpha)\cdot \cos(\beta) -\sin(\alpha)\sin(\beta)$$
$$\cos(\alpha + \beta) \cdot \cos(\alpha -\beta)= ( \cos(\alpha)\cos(\beta) -\sin(\alpha)\sin(\beta))\cdot (\cos(\alpha) \cdot \cos(-\beta)-\sin(\alpha)\cdot \sin(-\beta))$$ As $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$ this is equal to $$(\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta))\cdot (\cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta))$$ With the third binom you get $$\cos^2(\alpha) \cos^2(\beta) -\sin^2(\alpha)\sin^2(\beta)$$ As $\cos^2(\beta)+\sin^2(\beta)=1$ we have $$\cos^2(\alpha)(1-\sin^2(\beta))-(1-\cos^2(\alpha))\sin^2(\beta)$$ Multplying it out gives you $$\cos^2(\alpha) -\sin^2(\beta)\cos^2(\alpha)-\sin^2(\beta)+\cos^2(\alpha)\sin^2 (\beta)$$ And this is $$\cos^2(\alpha)-\sin^2(\beta)$$
Another way: $$2\cos(\alpha)\cdot \cos(\beta)=\cos(\alpha+\beta) + \cos(\alpha-\beta)$$ Using this in your formula gives us $$\cos(A+B)\cdot \cos(A-B)=\frac{1}{2}\left( \cos(2A) + \cos(-2B)\right)=\frac{1}{2} \left(\cos(2 A)+ \cos(2 B)\right)$$ As $\cos(2 A)=1-2\sin^2(A)$ and $\cos(2 B) = 1-2 \sin^2 (B)$ this is equal to $$1-\sin^2(A) -\sin^2 (B)=\cos^2(A)-\sin^2(B)$$
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This seems like the way I did it – seeker Mar 29 '13 at 15:06
@Assad when I wrote it I didn't know you already finished it on your own, (nor which way you have chosen) I will think of another way – Dominic Michaelis Mar 29 '13 at 15:07
I'm sorry, I only remembered to include my answer after. Thank you for your help! – seeker Mar 29 '13 at 15:09
@Assad posted another one which is shorter – Dominic Michaelis Mar 29 '13 at 15:17
For a slightly different solution,
$2\cos^2 A-2\sin^2 B$
$=2\cos^2 A-1+1-2\sin^2 B$
$=\cos 2A+\cos 2B$
$=2\cos (A+B) \cos (A-B)$
and halve each side.
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Maybe you didn't see you can use the formula $(x+y)(x-y)=x^2-y^2$. Other than that, I don't see any other simpler method.
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Would you have attempted this in a way different than I did? – seeker Mar 29 '13 at 14:59
Another way of looking at it:
\begin{align*} \cos(A+B)\cos(A-B) &= \frac14\left(e^{i(A+B)}+\frac{1}{e^{i(A+B)}}\right)\left(e^{i(A-B)}+\frac{1}{e^{i(A-B)}}\right) \\ &= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\ &= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + 2 - 2 + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\ &= \frac14\left[(e^{iA})^2 + \frac{1}{(e^{iA})^2} + 2\right] + \frac14\left[ - 2 + (e^{iB})^2 + \frac{1}{(e^{iB})^2}\right] \\ &= \left(\frac{e^{iA} + e^{-iA}}{2}\right)^2 - \left( \frac{e^{iB}-e^{-iB}}{2i}\right)^2 \\ &= \cos^2 A - \sin^2 B. \end{align*}
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Here is an interesting different way: Let
$$x = \cos(A+B)\cos(A-B) \\ y = \sin(A+B)\sin(A-B)$$
Then,
$$x+y = \cos(A+B-A+B) = \cos(2B) \\ x-y = \cos(A+B+A-B) = \cos(2A)$$
You can add these to get $2x$ or subtract them to get $2y$. Then expand using the double-angle formulas. This gives you two trig product formulas at the same time.
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2016-02-08 04:43:03
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https://www.nature.com/articles/s41467-021-23142-8
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## Introduction
DNA methylation is one of the most important epigenetic marks which introduce major changes in cellular functionality, some of them with systemic impact and coupled to important pathologies in mammals (for a review see1). For example, mutations in DNA Methyltransferase 3b (DNMT3b) are implicated in Immunodeficiency, Centromere instability and Facial anomalies (ICF) syndrome2, mutations in DNA Methyltransferase 3a (DNMT3a) are found in acute myeloid leukemia (AML) patients3 while those in DNA Methyltransferase 1 (DNMT1) cause autosomal dominant cerebellar ataxia, deafness and narcolepsy4. DNA methylation plays a key role in development (reviewed in5) and cell differentiation6, and a correct methylation level is crucial for the regulation of parental imprinting and X chromosome inactivation mechanisms7. Not surprisingly, changes in DNA methylation patterns have been associated with many different types of cancer in humans8,9,10,11 and reviewed in12,13.
It is commonly stated that DNA methylation in the promoter region of a gene is a hallmark of repression1,14. However, several studies have shown that DNA methylation in the gene body could also affect gene expression, and that the increase in methylation in promoter regions is not always correlated with gene repression, making the effect of DNA methylation on gene expression far more complicated than a simple on/off signal15,16. Two possible mechanisms might link DNA methylation with gene regulation: i) a direct effect involving proteins with methylated DNA binding domains, which act as anchoring points for other effector proteins regulating gene activity and ii) an indirect effect related to changes in DNA properties. In the second case, at least two possibilities emerge: i) DNA methylation changing the binding affinity of transcription factors to DNA and ii) DNA methylation affecting chromatin structure, which in turn modulates gene expression17. Accurate in silico and in vitro studies have demonstrated that DNA methylation makes the DNA less flexible and less likely to form nucleosomes18,19, but other studies relying on in vitro nucleosome reconstitution claim the opposite: i.e., that DNA methylation increases the affinity of histones for DNA20 and DNA methylation promotes compaction on pre-assembled nucleosomes21. Experiments using in vitro reconstituted nucleosomes and in vivo studies on mammals and plants are also confusing, with some suggesting that methylation occurs preferentially on nucleosomal DNA22,23,24 and others concluding the opposite25,26,27,28. This controversy can probably be explained by the high complexity of mammalian genomes, with a myriad of factors controlling directly or indirectly nucleosome positioning29, and by the existence of higher orders of chromatin structure beyond the nucleosome fiber.
To determine the intrinsic impact of DNA methylation (i.e., that is independent of specific methylation recognition machinery) on genome organization, we use budding yeast as a model system. S. cerevisiae’s genome is deprived of any cytosine methylation30 and its structure is very well characterized30,31,32,33,34. It consists of a Rabl-like structure where the 16 centromeres are attached to the Spindle Pole Body (SPB) at one pole of the nucleus and the chromosome telomeres extend outward toward the nuclear membrane. S. cerevisiae does not have DNA methylation/demethylation machinery and no methylated DNA binding domain has been characterized, precluding the impact of any evolution refined biological mechanism for regulating methylation imprinting and reading. The addition of DNA methylation machinery to yeast should then yield a perfect system to check for the intrinsic impact of DNA methylation in chromatin structure.
Different authors35,36 have shown that ectopic expression of either DNMT3a and DNMT3L or DNMT3a and DNMT1, could induce DNA methylation in yeast, but the levels of methylation achieved in both cases were too low to perform genome wide analysis of the effects of DNA methylation on chromatin structure or gene expression. More recently, Morselli et al. achieved a higher level of methylation by expressing DNMT3b at high level and collecting the cells at saturation27. Their results demonstrate an intrinsic anticorrelation between nucleosome and methylation and a link between DNA methylation and H3K4 and H3K36 methylation. A comparable methylation pattern was just reproduced by Finnegan et al.37 expressing human DNMT3s in the yeast Komagataella phaffii.
Using a similar approach but expressing murine DNMT1, DNMT3L, DNMT3a and DNMT3b simultaneously in our S. cerevisiae cells, we achieved even higher methylation rates than the one reported by Morselli et al.27, creating a perfect model for the study of the intrinsic impact of methylation in chromatin structure and function. We demonstrated that even in the absence of any directing machinery, methylations occurred in a reproducible pattern reminiscent of that seen in mammals, with methylation concentrated at linkers and depleted at nucleosomes. High levels of methylation affect gene expression in a complex manner, altering specific pathways in the cell. Quite interestingly, Hi-C experiments revealed significant changes in global chromatin structure with a noticeable increase in gene condensation. Overall, our results demonstrate that DNA methylation affects chromatin structure and function even in the absence of specific methylation-recognition machinery, suggesting that the methylation imprinting has an intrinsic impact in chromatin structure and function.
## Results
### Description of the system
In order to reach a high level of DNA methylation, we expressed the 4 DNMTs simultaneously: yeast cells were transformed with a combination of 4 plasmids each of them expressing one murine DNA Methyl Transferase (DNMT). The de novo DNMTs, DNMT3a and 3b, and the maintenance DNMT, DNMT1, were expressed under the control of the tetO promoter while the expression of DNMT3L, the regulatory DNMT, was controlled by the Gal1 promoter38. The expression and stability of the DNMTs were assessed by western blotting and no sign of protein degradation could be detected, even after 48 h of induction (Supplementary Fig. 1a–d). Expression of the 4 DNMTs slightly affected cell growth and cell viability (Supplementary Fig. 2a and 2b). Flow cytometry analysis performed on non-synchronized cultures after 24 h of induction showed a clear increase in the percentage of cells in G2/M in the samples expressing the DNMTs compared to the two control populations transformed with the empty plasmids suggesting that the slower growth could be due to a longer G2 phase (Supplementary Fig. 2c).
Differential gene expression analysis (Supplementary Data 1) showed that DNMTs ectopic overexpression did not induce expression of genes normally activated by stress, suggesting that the effects that we observed are not caused by stress but are a direct consequence of DNA methylation.
DNA methylation was first assessed by HPLC/MS which showed that using this approach, we could reach up to 4.2% of cytosines methylated after 38 h of induction in cells collected in stationary phase (Supplementary Table 1), and then analyzed at single base pair resolution in several independent transformants, using Illumina whole-genome bisulfite sequencing (WGBS). This was done both for cells in exponential growth phase synchronized in G1, and for cells at saturation that spent between 24 and 36 h in G1 without dividing (Supplementary Fig 2d). Methylated cytosines were, as expected, almost exclusively found in CpG context (Table 1) and showed a very reproducible pattern from one sample to another (Fig. 1a, g), a result verified using longer reads (<10 kbp) obtained by Nanopore Technology (ONT), see Supplementary Fig. 3a–c), which allowed us to confirm the high level of methylation in long repetitive sequences in heterochromatin (Supplementary Fig. 4a, c). The global distribution of methylation across all CpG sites is shown in Supplementary Fig. 5a–c. For the cells synchronized in G1 we can see that ~50% of CpG sites have <5% methylation, but there are almost 20% of sites with methylation over 20%, and ~5% of sites with methylation over 50%. For the cells at saturation the methylation levels are generally higher, with 50% of sites having >15% methylation and 25% of sites having methylation over 50% (Supplementary Fig. 5c).
### Homogeneity of the samples
To check the homogeneity of the samples, we examined the long nanopore derived reads, which allow assaying the methylation status at a large number of CpG sites on the same DNA molecule. We can see (Supplementary Fig. 6) that the histograms of methylation level for the 2 control samples show an almost identical distribution with a peak at around 0.02, whereas for the methylated samples from cells at saturation the peak is around 0.33. The histogram for the methylated samples from replicating cells has a peak close to the non-methylated controls, but with a long right tail.
For the methylated samples, most CpGs have an intermediate level of methylation. To investigate whether the intermediate methylation was due to heterogeneity in the population of reads (implying that the sequencing library contains a mixture of highly methylated and lowly methylated DNA molecules for a given genomic region), we modeled the distribution of methylation within reads using a mixture model with three components, one non-methylated, one lowly methylated and one highly methylated. The fit of the observed data to the mixture model was compared with that to a model with a homogenous read population using likelihood ratio tests (Supplementary Methods and Supplementary Software 1).
The results in Supplementary Table 2 show that in the samples at saturation, almost 95% of the reads are highly methylated (with roughly 30% of CpGs on the read being methylated), indicating a fairly homogenous population. On the contrary, in the samples collected in exponential phase, only 20% of the reads are highly methylated with the remaining reads having a lower methylation rate (with roughly 8% of methylated CpGs on a read being methylated). The likelihood ratio tests show high significant support for that the presence of multiple components for both samples (p < 1.0e−16 in both cases). The presence of two distinct populations of reads in the samples in exponential phase suggests that the DNA methylation maintenance machinery may be not fully functional or that the replication phase in yeast is too short for the mammalian maintenance machinery, leading to a difference in average methylation between the original and daughter DNA strands.
### Effect on methylation of the different DNMTs
We tested the functionality of each DNMT in our system by comparing the DNA methylation pattern and levels obtained when only 3 of the 4 DNMTs were expressed (Fig. 1a, b). This was performed always with replicating cells, synchronized in G1 to make the data fully comparable. For all combinations of 3 DNMTs the methylation pattern was broadly the same (Fig. 1a). The overall methylation level, however, was 2 to 3 times lower with 3 DNMTs compared to when all 4 DNMTs were expressed, demonstrating that each DNMT is functional (Fig. 1b).
While the different DNMT combinations displayed broadly the same pattern of methylated CpGs, when examined in detail some differences emerged. To check whether these differences are due to intrinsic sequence specificity of the different DNMTs, logistic regression was used to assess the effect of local sequence context on the methylation rate at CpG sites. This analysis was applied to 6 samples (2 samples with all 4 DNMTs to assess the reproducibility of the results, and the 4 samples each lacking one of the DNMTs). Figure 1C shows that the two replicates with all 4 DNMTs being expressed provide very similar results, proving the robustness of the experiments. Removing DNMT1 or DNMT3L does not significantly alter the sequence preferences found in the experiments with the 4 DNMTs active. In contrast, removing either DNMT3a or DNMT3b has a large effect on the sequence context. In general, cells lacking DNMT3b show a strong bias for CpGs in the 5’ ATCGAG 3’ motif (with the percentage of methylated CpGs in an ATCGAG motif being more than six times lower when DNMT3b is removed, compared to the samples with one of the other DNMTs missing or to the sample with all 4 DNMTs induced). Cells lacking DNMT3a also show a sequence bias, in this case towards sequences that are more C rich. Note that the lack of methyl DNA binding protein makes these sequence preferences intrinsic of DNMTs and not the result of positioning of the methylases in certain sequences due to auxiliary proteins.
### Pattern of methylation
Despite the lack of specific proteins directing epigenetic signaling of DNA, the pattern of methylation obtained in our model organism is very similar to that observed in higher eukaryotes, with DNA methylation being low at the Transcription Start Site (TSS) increasing toward the end of the genes and reaching a maximum at the Transcription Termination Site (TTS) (Fig. 1d). While we cannot rule out this pattern being due to selection (i.e., that cells with no methylation at promoters have a selective advantage), the fact that the pattern is established very early on (6 h after the induction of the DNMTs) and remains stable across all time points indicates that this has a minor effect, if any.
Previous studies have linked DNA methylation with histone post translational modifications, DNMT3L being recruited by unmethylated H3K4 and DNMT3A and 3B recruited by H3K36me2 and H3K36me3, respectively27,39,40,41,42. We confirmed that the H3K4 methylation pattern in the stationary cells both in the control and in the methylated sample was the same as the one described in exponential cells, and that DNA methylation was anti-correlated with H3K4 trimethylation (Fig. 1e), suggesting that H3K4 and H3K36 methylation are capable of tightly controlling DNA methylation even upon DNMTs overexpression by a direct mechanism.
### DNA methylation and nucleosome positioning
To investigate the correlation between DNA methylation and nucleosome positioning, we obtained MNase-Seq data of the samples with all 4 DNMTs or with empty plasmids in stationary phase. We observed that DNA methylation was anti-correlated with nucleosomes and tended to be accumulated in the linker regions (Fig. 2a, c). These results agree with the bulk of in vivo25,26,27,28, in vitro and in silico data18,19 and ruled out the suggestion that methylation is favored at well-positioned nucleosomes22,23.
We observed a small (4%), but very significant increase (p value 2.2e−16) in the number of fuzzy nucleosomes and a mirror decrease in the number of well-positioned nucleosomes in the methylated samples (Table 2). To test if this effect was due to DNA methylation and not to the presence of the DNA methyltransferases themselves, catalytically inactive mutants of the DNMTs (DNMT3bP656V/C657D, DNMT3aP705V/C706D 43 and DNMT1C1229S 44), were expressed in yeast and MNase-seq was performed. PCA (Supplementary Fig. 7a) and NucleR analysis (Supplementary Fig. 7b) confirm that the effect we observed was mostly due to DNA methylation.
In well-positioned nucleosomes, DNA methylation was almost absent at the dyad and increased toward the entry and exit points of the nucleosome, while fuzzy nucleosomes had higher methylation levels with a more constant level across the nucleosome (compare Fig. 2c with Fig. 2c). Detailed analysis (Fig. 2e), failed to detect any periodicity in the methylation within the nucleosome bound sequence, arguing against a preferential methylation of DNA at accessible sites of the nucleosome. Considering that all methylases were active, this suggests that (at least in the absence of directing machinery) all DNA segments covered by histones are equally inaccessible to methylation imprinting.
Figure 3a shows some representative nucleosome fiber structures for the DUG2 gene (chromosome II) as a test case, modeled from MNase-seq signals for control and methylated cells. Both ensembles of structures reproduce the experimental results to a great extent (Fig. 3b, c and Supplementary Fig. 8a, b) and confirm the presence of fuzzier nucleosomes in methylated cells. Due to the presence of fuzzier nucleosomes, a wide range of nucleosome configurations are possible leading to the sampling of very elongated conformations with large radius of gyration (see Fig. 3a, lower panel, and Supplementary Fig. 8c). The diversity of nucleosome arrangements was further analyzed by comparing the 3D distances between N and N + x nucleosomes (Supplementary Fig. 8d), which in methylated cells show a more dispersed profile. Finally, it is noticed that regions with high degree of methylation probability are mainly linker DNA (Fig. 3b, c and Supplementary Fig. 8b), though the probability of finding some nucleosomal DNA methylated is not insignificant, as shown in some fiber configurations (Fig. 3a, lower panel). This is because those nucleosome positions are retained despite having low coverage in the MNase-seq signal.
Analysis by Nucleosome Dynamics45 of the crucial region around the promoter demonstrates that high level of methylation induces more dynamic and fuzzy nucleosomes (Fig. 4a), as well as a statistically significant (p < 2.2e−16) narrowing of Nucleosome Free Regions (NFRs; Fig. 4b, c). These changes, which are typically considered to be signals of gene inactivation46,47 cannot be explained here by the coordinated effect of methyl-DNA binding proteins coupled to chromatin remodelers and must be then considered intrinsic to the changes on physical properties of DNA induced by methylation and its direct impact in protein-DNA interaction. The fact that high methylation is correlated also with narrower NFR also at the 3’ end of the genes (Supplementary Fig. 9) agrees with this hypothesis.
### DNA methylation and gene expression
Despite the lack of any directing mechanism, the methylation pattern is quite homogenous throughout the gene only in lowly expressed genes while in the highly expressed ones, DNA methylation is low at the promoter and increases toward the end of the gene, suggesting a direct link between DNA methylation and gene expression (Fig. 5a) in the absence of any specific methylation-recognition mechanism. A differential expression analysis (Fig. 5a) shows that genes which are very lowly methylated do not change their expression level between the control and transformed cells, while high methylation levels lead to important changes in gene activity in both directions: towards greater and lower expression (Fig. 5c and Supplementary Data 1). Although we cannot discard any indirect effect caused by the binding of the DNMTs on the chromatin, we see a very strong correlation between gene expression and methylation level for a subset of genes involved in meiosis and that appear to share a common sequence in their regulatory region (Fig. 5c, d), corresponding to the binding site of Ume6p, a subunit of the histone deacetylase complex Rpd3p known to repress early meiotic gene expression. It is tempting to hypothesize that methylation of a Ume6p binding site known as URS1, could affect intrinsically Ume6p binding directly (through changes in direct protein-DNA interactions or indirectly through changes in chromatin structure), leading to a deregulation of its target genes. Supporting this hypothesis, we observed that the level of expression of the target genes increases proportionally with the level of methylation of the Ume6p binding site (Supplementary Fig. 10a and Supplementary Table 3). This was further confirmed in vitro by showing that Ume6p has more affinity for an unmethylated URS1 site than a methylated one (Supplementary Fig 10b, c). Finally, for most of Ume6p target genes, we observe a 5–10 bp shift of the −1 or +1 nucleosome (Supplementary Fig. 11), consistent with changes in expression. In summary, it seems that two physically driven mechanisms: changes in protein-DNA binding interactions due to the presence of a methyl group and methylation-induced nucleosome rearrangements work in a coordinated way to induce a change in gene activity which would be typically assigned to the effect of specific methyl DNA binding proteins, which are absent in S. cerevisiae.
### DNA methylation and genome 3D structure
We performed Hi-C experiments in control and methylated populations at saturation in two replicas processed in parallel to explore the intrinsic effect of DNA methylation in the global chromatin structure (Supplementary Fig. 12a–d). As shown in Fig. 6a, b (and Supplementary Fig. 13a, b), DNA methylation leads globally to an increase in cis (especially close to the centromeres (<100 kb); Fig. 6c, d) and a very significant decrease in trans contacts (Fig. 6c, e). To obtain further insights into the effect of DNA methylation on chromatin structure, we modeled the spatial organization of each chromosome using a restraint-based model derived from the interaction counts (see Methods) from which we obtained ensembles reproducing with an astonishing quality the HiC maps (Supplementary Fig. 14). Clearly, in all chromosomes (except the short chrI), the 100 kb region centered around their centromeres is more condensed upon DNA methylation (Fig. 6f), decreasing the overall chromosome flexibility with the effect being very clear for chromosomes V, IX, XI and XV (Fig. 6g and Supplementary Fig. 14). Very interestingly, the nucleosome depletion that is observed at the centromeres in stationary phase in the control sample (compared to cells in exponential growth) is not detected in the methylated one (Supplementary Fig. 15), suggesting that the profound reorganization of the centromeric region occurring in quiescent cells is reduced in the methylated samples.
The three-dimensional arrangement of telomeres is also largely modified by methylation, as we observed a significant decrease in the number of interactions (Supplementary Fig. 16), resulting in a large dispersion of telomeres in methylated cells, something that is visible in the HiC-derived ensembles (Fig. 6h, i). It is worth noting that telomeres tend to cluster in quiescent cells to form hyperclusters48 but it seems that this reorganization is not happening when DNA is methylated. Once again, intrinsic changes linked to methylation lead to a 3D organization that is closer to the one described in exponential growth (which is when the expression of the DNMTs started to be induced), than the one we and others observed in quiescent cells.
Comparing the changes in interactions between the control and methylated samples for each individual chromosome (Supplementary Fig. 17), we observed the strongest effects of methylation for chrIII (largest increase in intra-chromosomic-contacts, Fig. 7a–e) and for chrXII (largest decrease in intra-chromosomic contacts, Fig. 7f–h). Looking closer at chrIII, we noticed an important gain of interactions in the methylated sample between the left telomeric region containing the silenced HMLα and the peri-centromeric region (Fig. 7a–c). This gain of interaction correlates with a significant decrease of the distance between the HMLα and the MATa loci (Fig. 7d) which leads to very significant changes in the ensemble of chromosome conformations compatible with the HiC restrains (Fig. 7e). This result was validated by live-cell microscopy using the system described in49,50 where we can see that the distance between HML and MAT is significantly smaller in the methylated cells compared to the control while the distance between HML and HMR and between MAT and HMR are conserved (Supplementary Fig. 18).
ChrXII is a very peculiar chromosome, as it carries the rDNA locus that consists of 150–200 repeats of the rDNA genes and localizes at the nucleolus where ribosomal RNAs are transcribed. In the control sample, interactions between the upstream and downstream regions are found, which can be explained by the global decrease of transcription occurring in stationary phase leading to a more compact nucleolus compared with exponential growth51. However, in methylation conditions (Fig. 7f–h), the segregation between the upstream and downstream regions of the chromosome is very strong, reproducing again the situation expected for a cell in exponential growth. The absence of interactions between the two domains probably allows a general increase in the relative flexibility that could explain why the general reduction in flexibility related to methylation seen in other chromosomes is inverted for this chromosome (Fig. 6g).
Globally, the control samples show a quite spherical nucleus (Fig. 8a) with the chromatin concentrated in the exterior (Fig. 8b), while in the methylated samples the nucleus tends to elongate with a more dense packing of chromatin in the interior (Fig. 8d, e). In both cases, centromeres are all localized to one pole of the structure, organized as a rosette (Fig. 8c, f), but they do not appear as clustered as reported for the interphasic nucleus33,34,52.
## Discussion
Yeast expressing murine methyl transferases show, in vivo, a specific, reproducible, pattern of DNA methylation that is very similar to that of genes-containing regions in mammals. This is a striking result as yeast is an organism deprived of any DNA methylation machinery and is not ready to place or recognize methylation marks. Methylation leads to a slight decrease in the viability and in the doubling time maybe caused by a longer than normal G2/M phase. However, those changes are moderate, and again it is quite surprising that an organism not prepared to have methylated DNA tolerates well a large amount of methylation in its genome.
Our synthetic model system helped us to highlight some previously unknown intrinsic sequence specificity for the methyl-transferases, i.e., those that cannot be explained by accessory proteins. For example, little sequence specificity is found for DNMT1 and DNMT3L, while significant sequence specificities are found for DNMT3a and DNMT3b. Thus, cells lacking DNMT3b show a strong bias for methylation of CpG in 5’ ATCGAG 3’ motif, while cells lacking DNMT3a seems to have a bias toward CpG sequences embedded in C-rich environments. Again, these differences are intrinsic and not coupled to any specific directing mechanism, which suggests a certain level of specificity of the DNMTs.
Methylation in our engineered yeast model is preferentially located between nucleosomes, and when it occurs in nucleosome-occupied regions, it is associated with significant alterations in nucleosome positioning reflected in an increase in nucleosome fuzziness. The fact that methylated DNA is less frequent at nucleosomes confirms our previous in silico and in vitro models18,19 but does not rule out the possibility that in higher eukaryotes methylated-DNA binding domains might stabilize the presence of methylated CpG in mammal nucleosomes, leading to a situation of “loading-spring” which might facilitate fast nucleosome reorganization upon release of the stabilizing protein. There is, however, no question that methylation and nucleosome position are intrinsically anti-correlated, confirming previous results in mammalian cells from NOME-seq29 where the authors observed that DNA methylation and nucleosome occupancy were strongly anti-correlated surrounding CTCF sites. Furthermore, our results demonstrate that there is not any short-distance periodicity pattern which might indicate methylation at periodically exposed regions of nucleosome DNA. Very interestingly, when methylation occurs in the promoter region it tends to narrow the NFR region, a fingerprint of lowly expressed genes in mammals, which is found here in absence of any methylation-specific chromatin remodeler.
In mammalian cells, the relationship between methylation and gene expression is complex, with high levels of gene expression often associated with low promoter methylation but elevated gene body methylation, and the causality relationships are not clear. Our simple synthetic system allows a more direct interrogation to the relationship between methylation and gene expression. We observed that despite the lack of specific methylated DNA binding domains, highly and lowly expressed genes have quite different methylation profiles, with much higher levels of methylation near (±850 bp) the TSS of silent genes while very actively expressed genes have much higher methylation levels at the TTS. In the absence of specific proteins modulating this profound difference, we can speculate that nucleosome positioning is one of the main factors responsible for this differential behavior, which suggests that methylation and nucleosome positioning might act in concert in the regulation of gene function in mammals, adding an extra layer of control of gene expression.
Our results suggest that methylation can also directly affect the binding of a transcription factor highlighting another intrinsic role of methylation in gene activity, which is often ignored, but that can be important in modulating DNA-protein binding free energy. Interestingly, we found that the methylation-induced change in protein-DNA binding is responsible for a dramatic increase of expression of early meiotic genes. In that case, our results suggest that the methylation affects the intrinsic binding of the histone deacetylase complex Rpd3p, thus hindering the placement of repressive marks on these genes.
Analysis of HiC data shows that the characteristic Rabl configuration previously described30,33,34 is maintained in stationary phase both in the control and in the methylated sample. However, methylation induces an increase of intra-chromosomic contacts and a significant decrease of the inter-chromosomic ones, which leads to significant changes in the chromosome conformation. We observed that a significant (Fisher’s exact test, p = 1.76e−7) proportion of the interactions that are gained or lost upon methylation involves regions containing one or several tRNA genes, supporting the involvement of RNA polymerase III in the overall organization of chromatin structure53.
Changes in heterochromatin regions of S. cerevisiae (telomeres, the mating type locus and the rDNA locus) are also clear upon methylation, mimicking the situation found in mammals. First, we observed a general loss of interactions between telomeres which leads to a significant change in chromosome structure and that can be explained by a generally higher rigidity of the chromatin upon methylation. Second, we observed that chromosome III underwent some conformational changes in response to methylation, with the silenced HMLα locus getting closer to the MAT locus, a conformation expected in exponentially growing MATa cells49,54, but not in stationary cells. Finally, the separation commonly observed in yeast between the two regions of chrXII delimited by the rDNA locus is clearly weaker in our cells in stationary phase when compared to the structure published for cells in exponential phase30, but the situation is reverted in methylated samples (in stationary phase), suggesting that methylation of the rDNA can freeze the heterochromatin in exponential phase-like conformation.
Our previous results18,19 strongly suggest that in general, methylation increases the stiffness of DNA and this should impact the structure of the chromatin both locally and globally. This is confirmed here by Hi-C experiments coupled to 3D modeling that show a condensation of the centromeric region for each individual chromosome, resulting in more rigid and condensed chromosomes unable to transiently interact with other chromosomes. Locally, at the gene level, the situation is more variable, as the higher rigidity of methylated DNA leads to fuzzier nucleosome architectures and longer linkers (as shown by MNase-seq experiments), which in turn can lead to a local increase in the structural variability of the nucleosome fiber, dominated by the length of the linkers. Overall these results demonstrate that the local increase in rigidity induced by methylation have direct consequences on the global structure of chromatin, even in the absence of protein machinery ready to recognize methylation signals.
Taken together, our results suggest that in the absence of any mechanism directing methylation and reading the methylation signals, methylation leads to significant changes in chromatin structure and function. Very interestingly and quite surprisingly, many of these methylation effects resemble those annotated to methylation in mammals, where an exquisite machinery for imprinting and reading of methylation signals is present. It can be concluded then that changes in physical properties of DNA induced by methylation (reflected in alterations in DNA deformability or in DNA-protein interactions) can be responsible for a significant part of the phenotypic effects triggered by DNA methylation in mammals and that the cellular machinery involved in DNA methylation just amplifies an intrinsic signal coded in the physical properties of DNA.
Changes in methylation across the genome are key features of developmental pathways in both normal differentiation55,56 and many cancers2. We observed that in our synthetic model, methylation makes the 3D genomic structure in the stationary phase closer to that of G1 cells than in the unmethylated samples, which suggest that one of the intrinsic roles of DNA methylation could be to freeze chromatin conformation to maintain the state of cell.
## Methods
### Plasmid construction
pYADE4 yeast plasmids encoding full length DNMT1 and DNMT3a with modified sequences around the translation start sites were kindly provided by Dr Jan Fronck, pYES3/CT encoding DNMT3b was provided by Dr Shen Li35,57. DNMT3L cloned into pYES3/CT to produce a Nterminal FLAG tagged DNMT3L was provided by Dr Jia-Lei Hu36.
pCM188 (marker cgURA3) and pCM185 (marker cgHIS or cgLEU), centromeric vectors38 which differ for the number of Tet operators (respectively 2 and 7) were kindly provided by Dr Jessie Colin.
SmaI restriction fragment (from pYADE4-DNMT1) containing full length DNMT1 cDNA was inserted at PmeI site of pCM185 (LEU) to give pCM185(LEU)-DNMT1.
BamHI-MluI restriction fragment (from pYADE4-DNMT3a) containing full length cDNA of DNMT3a was ligated to pCM185 (HIS) linearized with BamHI and MluI to give pCM185(HIS)-DNMT3a.
BamHI-NotI restriction fragment from pYES3/CT-DNMT3b containing full length DNMT3b was ligated to pCM188 (URA) linearized with BamHI and NotI to obtain pCM188 (URA)-DNMT3b.
DNMT mutants’ constructs – Plasmids encoding the catalytic mutants DNMT3bP545V/C657D, DNMT3aP705V/C706D and DNMT1C1220S were constructed using the Q5® Site-Directed Mutagenesis Kit. The mutagenesis was performed on the pCM188-DNMT3b, pCM185-DNMT3a and pCM185-DNMT1 plasmids. Sequences of oligonucleotides for mutagenesis, subcloning or gel retardation assay are listed in Supplementary Table 4.
Live cell imaging constructs – DNMT3a-DNMT3b-pBEVYGU: The full length DNMT3b coding sequence was amplified by PCR on pCM185-DNMT3b plasmid using the pcrDNMT3b_Kpn1 5 ́_F and pcrDNMT3b_EcoRI 3 ́_R oligonucleotides. The PCR product was digested with KpnI and EcoRI and ligated to the pBEVY-GU (URA) plasmid. DNMT3a gene was obtained by digestion of the pCM185-DNMT3a plasmid with BamHI and XbaI. The restriction fragment was ligated into the pBEVY-GU-DNM3b vector that had been previously opened with the same restriction enzymes.
DNMT3L-DNMT1-pBEVYGT: The DNMT3L gene was amplified by PCR on the pYES-DNMT3L plasmid using the pcrDNMT3L_SacI1_5 ́_F and pcrDNMT3L_EcoRI 3 ́_R oligonucleotides. The PCR product was digested with SacI and EcoRI and ligated to the pBEVY-GT (TRP) plasmid. The DNMT1 gene was obtained by digestion of the pYADE4-DNMT1 with SpeI and XbaI. The restriction fragment was ligated into the pBEVY-GT-DNM3L vector that had been previously opened with the same restriction enzymes.
Recombinant Ume6p construct – UME6-pET14b: The entire coding sequence of UME6 was amplified by PCR from yeast genomic DNA using the UME6-F and the UME6-R oligonucleotides. PCR product was digested with NdeI and XhoI. The restriction fragment was ligated into the pET14b vector previously opened with the same restriction enzymes. This plasmid adds a 6xHis tag at the N terminal part of the sequence.
### Yeast strains and culture conditions
Strain YPH499 (Mata ura3-52 lys2-801 ade2-101 trp1-Δ63 his3-Δ200 leu2-Δ1) was transformed with 2, 3 or 4 expression plasmids by the standard lithium acetate procedure. Transformants were selected on plates of appropriate selective medium with 2% Raffinose and 10 µg/ml doxycycline to repress any expression.
Selected transformants (2 to 4 transformants per combination of plasmids) were grown on selective liquid medium with 2% Raffinose and 10 µg/ml doxycycline up to OD600 = 0.5. Then, yeast cells were spun 10 min at 1000 × g, washed twice with sterilized water, and resuspended into selective media with 1% Raffinose and 2% Galactose without doxycycline to allow expression of all four DNMTs. For experiments on synchronized cells, cells were treated with alpha-factor (3 µM final) for 4 h to synchronize cells in G1 or with Nocodazole to synchronize cells in G2.
After different times of induction, cells were collected and treated for subsequent experiments: protein extraction for western blotting, gDNA extraction for whole-genome bisulfite sequencing, RNA extraction for RNA-sequencing or Semi-intact cell preparation for MNase digestion and nucleosome mapping.
The yIL30 strain (Mata ade2-1 can1-100 his3-11, 15 leu2-3, 112 trp1-1 ura3-1 ade2-1::His3p-CFP-lacI-URA3p-λcIYFP-ADE2, TetR-mRFP:NAT1, MAT5’:: λO-HIS, HMR:: LacO-TRP, HML::TetO-LEU) carrying the three integration sites for the fluorescent tags was kindly provided by Kerstin Bystricky.
The yIL30-W strain (Mata ade2-1 can1-100 his3-11, 15 leu2-3, 112 trp1-1 ura3-1 ade2-1::His3p-CFP-lacI-URA3p-λcIYFP-ADE2, TetR-mRFP:NAT1, MAT5’:: λO-HIS, HMR:: LacO-TRP, HML::TetO-LEU, trp1Δ::kan) was obtained by replacing the endogenous TRP1 locus in yIL30 by the kanamycin resistance gene amplified from pFA6A-kANr using the F2_pFA6aKan_TRP1 and R1_pFA6aKan_TRP1 oligonucleotides. For live cell imaging yIL30-W cells were transformed with the DNMT3a-DNMT3b-pBEVYGU and DNMT3L-DNMTT1-pBEVYGT plasmids or with the corresponding empty vectors. In both cases the expression was controlled by the Gal1 promoter. For the analysis selected transformants were grown until saturation in synthetic medium with galactose and washed twice with SC-Low Fluorescence medium (lacking folic acid and riboflavine).
### Flow cytometry analysis
0.5 ml of culture (OD600 = 0.6–0.8) were collected and centrifuged for 5 min at 1000 × g at RT. Pellets were washed twice with 1x ice-cold PBS and resuspended in 50 µl of 1x ice-cold PBS. 20 µl of cells were fixed with 1 ml of 70% EtOH overnight at 4 °C. Samples were washed with 1x Saline Sodium Citrate buffer (SSC; 150 mM NaCl, 15 mM Na citrate, pH 7.8 for 20x SSC). The pellet was resuspended in 0.5 ml of 1x SSC, treated with 0.5 mg/ml RNase A (Roche) for 1.5 h and then with 0.5 mg/ml Proteinase K (Roche) for another 1.5 h at 50 °C. After incubation, cells were briefly sonicated for 10 mn, medium potency, by using the Bioruptor system (intervals of 10 s on-20 s off). 250 µl of the cells were added in 0.5 ml of 1x SSC containing 1 µM Sytox Green (Sigma) and were incubated 10–20 mn in the dark (room temperature) before analyzing the DNA content using a Beckam Coulter GalliosTM flow cytometer. MultiCycle software (Phoenix Flow Systems, San Diego, CA, USA) was used to determine the percentage of cells in the cell cycle phases.
### HPLC/MS
HPLC/MS/MS analysis was based on the protocol described58,54. A Kinetez 2.6 μm HILIC 100 A column (150 mm × 4.6 mm) (Phenomenex) and a Acquity UPLC system (Waters Corp., Milford, MA, USA) coupled to a mass spectrometer API 3000™ (AB Sciex, Foster City, CA, USA) triple quadrupole working in MRM (multiple reaction monitoring) method in positive mode. Two eluents were used: eluent A2 (Acetonitrile) and eluent B1 (0.1 M ammonium formate adjusted at pH 3.2) with a isocratic gradient 8 min of total running time at 90% A and 10% B for the nucleosides elution. The separation was performed in a flow of 1400 μl min−1, with 10 μl injection volume and two replicates each, totaling two biological replicates and two technical replicates of each sample. The standard nucleosides cytosine and methyl-cytosine (Sigma) were diluted in HCl 0.01 N and stored at −20 °C. The m/z transitions from 112 to 95 (cytosine) and from 126 to 81 (methyl cytosine) were chosen for MRM experiments. The peak area obtained was analyzed by Analyst 1.4.2 (AB Sciex). Quantification (%) was performed according to 5mdC concentration divided by 5mdC concentration plus dC concentration multiplied by 100.
### Western blot
Proteins were extracted by resuspending the pellet of cells from a 20 ml cultures at OD600 = 1 in 400 µl of RIPA buffer (50 mM Tris pH7.5, 150 mM NaCl, 1% NP40, 0.5% NaDeoxycholate, 0.1% SDS) containing 1 mM PMSF and protease inhibitors (cOmplete ULTRA Tablets, Mini, EASYpack, Roche). 400 µl of glass beads were added and samples were processed using FastPrep (MP) for 3 times for 20 s pulses @4.5 m/s. After centrifugation 5 min at 2350 x g, supernatants were recovered and quantified by bradford. 20 µg of protein were loaded on a 6 or 8% acrylamide gel and subjected to PAGE, proteins were then transferred onto an immobilon membrane (millipore) for subsequent hybridization with anti-DNMT1 (ref ab87654, Abcam, dilution 1/1000), anti-DNMT3a (ab2850, Abcam, dilution 1/2000), anti-DNMT3b (ab122932, Abcam, dilution 1/500) or anti-Flag (F7425, Sigma, dilution 1/2000) antibody overnight followed by secondary antibody anti Rabbit (Goat)-HRP conjugated (65-6120, Invitrogen, dilution 1/5000). The signal was revealed using ECLTM prime WB detection reagent (Amersham, GE Healthcare).
### Ume6 purification
Rosetta E. coli cells containing the UME6- pET14b plasmid were grown at 37 °C in 1 L of LB ampicillin medium supplemented with 1% of glucose up to OD600nm 0.6 expression was induced with 0.15 g/L IPTG at 20 °C for 4 h. Cells were harvested by centrifugation and pellet was resuspended in 20 ml of binding buffer (50 mM sodium phosphate ph7.4, 300 mM NaCl, 1 mM PMSF, 0.01% tween20, lysozyme 0.2 mg/ml, 1 mM MgCl2, 2000u DNase and a tablet of cOmplete™EDTA-free Protease Inhibitor Cocktail), incubated for 30 min on ice and sonicated twice for 4 min, followed by centrifugation at 10.000 × g for 15 min. The supernatant was incubated 10 min with the Dynabeads™ His-Tag Isolation and Pulldown (ThermoFisher). Beads were washed 4 times with wash buffer (50 mM sodium phosphate ph7.4, 300 mM NaCl, 5 mM imidazole, 0.01% tween20). Finally, protein was eluted with the elution buffer (50 mM sodium phosphate ph7.4, 300 mM NaCl, 300 mM imidazole, 0.01% tween20) and dialyzed overnight with 10 mM sodium phosphate pH 7.4, 50 mM NaCl and 10 µM ZnCl2 buffer. Ume6 protein was visualized by SDS-PAGE and coomassie brilliant blue staining.
### Gel retardation assay
The double stranded (ds) probes containing the URS1 of SPO13 with or without methylated cytosines was obtained by annealing the following forward and reverse oligonucleotides: Spo13-Cy5-F with Spo13-R, and Spo13-Met-Cy5-F with Spo13-Met-R, respectively (Supplementary Table 4).
The reactions were incubated for 5 min at 95 °C and cooled down overnight. The resulting ds DNA fragments were then purified from 2% agarose gel.
Ume6 protein in 10 mM NaPO4 pH 7.5, 50 mM NaCl and 10 µM ZnCl2 was diluted with 1 µg/µl BSA. Salt concentration for each reaction was adjusted to a final concentration of 115 mM NaCl. For the binding reactions the protein samples were maintained in the same buffer and 168 nM of fluorescent Cy5 URS1 of SPO13 gene with or without methylated cytosines was added. Reactions were incubated for 15 min at 20 °C and 20% of glycerol was added before loading into 8% polyacrylamide gel in Tris-Borate-EDTA buffer. Electrophoresis was performed for 1 h at 120 V. Band intensities were obtained by exposing the gel to Typhoon phosphorImager scanner equipped with the emission filter for Cy5 fluorescence. The intensity of the bands was quantified using the software ImageQuant TL v.8.2.0.
### DNA methylation pattern
Whole-genome bisulfite sequencing (WGBS) – WGBS was performed following the procedure outlined in9. Briefly, genomic DNA (1–2 μg) was spiked with unmethylated λ DNA (5 ng of λ DNA per μg of genomic DNA) (Promega). The DNA was sheared by sonication to 50–500 bp using a Covaris E220 and fragments of size 150–300 bp were selected using AMPure XP beads (Agencourt Bioscience Corp.). Genomic DNA libraries were constructed using the Illumina TruSeq Sample Preparation kit (Illumina Inc.) following the lllumina standard protocol: end repair was performed on the DNA fragments, an adenine was added to the 3’ extremities of the fragments and Illumina TruSeq adapters were ligated at each extremity. After adapter ligation, the DNA was treated with sodium bisulfite using the EpiTexy Bisulfite kit (Qiagen) following the manufacturer’s instructions for formalin-fixed and paraffin-embedded (FFPE) tissue samples. Two rounds of bisulfite conversion were performed to assure a conversion rate of over 99%. Enrichment for adapter-ligated DNA was carried out through 7 PCR cycles using the PfuTurboCx Hotstart DNA polymerase (Stratagene). Library quality was monitored using the Agilent 2100 BioAnalyzer (Agilent), and the concentration of viable sequencing fragments (molecules carrying adapters at both extremities) estimated using quantitative PCR with the library quantification kit from KAPA Biosystem. Paired-end DNA sequencing (2x100bp) was then performed using the Illumina Hi-Seq 2000.
Read mapping and estimation of cytosine methylation levels – The WGBS reads were processed using the gemBS pipeline v3.059 using as reference S. cerevisiae S288c. Reads with MAPQ scores < 20 and read pairs mapping to the same start and end points on the genome were filtered out after the alignment step. The first 5 bases from each read were trimmed before the variant and methylation calling step to avoid artifacts due to end repair. For each sample, CpG sites were selected where both bases were called with a Phred score of at least 20, corresponding to an estimated genotype error level of ≤1%. Sites with >500x coverage depth were excluded to avoid centromeric/telomeric repetitive regions. CpGs were considered methylated when the number of mapped reads was larger than 10 and the estimated methylation percentage was above 0.1.
DNMT specificity analysis – We extracted two bases downstream and upstream from each CpG (having at least ten WGBS reads mapped) and trained a logistic regression model (using R) for the number of converted and non-converted Cs, using the extracted motifs as predictors for each WGBS sample (samples removing one of the DNMTs, T859, T860, T861 and T869; and two samples with the four DNMTs, T862 and T863). We computed for each sample the effect of each motif and its standard deviation, and used it to determine those with a significant effect on methylation level (estimated effect above two standard deviations). We found motifs specific for each sample lacking one of the DNMTs (motifs with significant effect in the sample removing one DNMT but not significant in the sample with all DNMTs) and compared their relative frequencies in all samples.
Nanopore sequencing – Suspensions of spheroplasts from methylated and control S. cerevisiae strains were loaded on Sage Science gel cassettes to perform lysis under electrophoretic conditions. DNA content in each sample was estimated by the cell count. A number of spheroplasts equivalent to 10 µg of genomic DNA were resuspended in 70 µl of HLS Suspension buffer (Sage Science, Mammalian white Blood cell suspension kit, #CEL-MWB1) and loaded on the gel cassettes (Sage Science, SageHLS HMW DNA extraction kit #HEX-0012).
The custom Sage HLS (Sage Science) protocol used (Extraction Collection DC55V 1h15m) was accommodated for the yeast small chromosome sizes. This custom protocol did not include a DNA fragmentation step. In brief, during the extraction step, the High Molecular Weight (HMW) yeast gDNA was bound in agarose while the solubilised and degraded proteins and other contaminants were kept in solution. The Sage Science Buffer A was used as a lysis buffer for this step. In the last step of the protocol, the HMW DNA was retrieved from the gel through an automated elution process that was optimized to elute all the yeast chromosomes in the elution module number 2 of the cassette.
Elution modules 1, 2, 3 & 4 were selected for the library preparation of the control and methylated S. cerevisiae samples. For each condition, the selected elution modules were pooled, purified with 1-fold excess of Agencourt AMPure XP beads (Beckman Coulter, A63882) and eluted in water. Two barcoded libraries containing both type of samples were prepared using the Oxford Nanopore Ligation sequencing kit (ONT, SQK-LSK109) combined with the Oxford Nanopore Native Barcoding Expansion kit (EXP-NBD103 1D) following manufacturer’s instructions.
After connecting the flows cells to the MinION Mk1b device, the MinKNOW interface QC (Oxford Nanopore Technologies) was run in order to assess the flow cell quality. Once the priming of the flow cell was finished, from 200 ng to 600 ng of the final barcoded library was loaded into R9.4.1 FLO-MIN106 or FLO-MIN106D flow cells and the sequencing data were collected during 48 h. The quality parameters of the sequencing runs were further monitored by the MinKNOW platform in real time. The MinKNOW versions used was 1.15.4. The basecalling was performed using Guppy 2.3.7. Reads were mapped using minimap2 2.9-r720, and CpG methylation was called using nanopolish 0.11.0.
### Nucleosome mapping
Semi-intact Yeast cell preparation – Semi-intact cells were prepared as previously described60. Briefly, cells were grown at 30 °C in 300 ml YPD to =1 ×107 cells/ml. For each 250 ml of cells (107cells/ml), semi- intact cells were prepared as follows. Cells were collected by centrifugation (700 × g, 7 min, RT), resuspended in 25 ml 100 mM Pipes, pH 9.4, 10 mM DTT, incubated with gentle agitation at 30 °C for 10 min, and collected by centrifugation (1000 × g, 5 min, RT). Cells were resuspended in 6 ml YP, 0.2% glucose, 50 mM KPO4, pH 7.5, 0.6 M sorbitol. 10 u zymolase was added, and the suspension was incubated with gentle shaking 30 °C for 30 min. Spheroplasting was monitored by light microscopy. Great care was taken not to overdigest cells to avoid lysis. Spheroplasts were collected by centrifugation at 1000 × g for 5 min at RT, re- suspended with a plastic pipette in 40 ml YE 1% glucose, 0.7 M sorbitol, and incubated with gentle shaking at 30 °C for 20 min. Spheroplasts were collected by centrifugation (1000 × g, 5 min, RT) and washed twice at 4 °C with cold permeabilization buffer (20 mM Pipes-KOH, pH 6.8, 150 mM K-Acetate, 2 mM Mg-Acetate, 0.4 M sorbitol. The final pellet was resuspended in 1 ml cold permeabilization buffer containing 10%(v/v) DMSO. 100 µl aliquots were placed in 1.5 ml microfuge tubes and frozen slowly above liquid N2 and stored at −80 °C.
MNase-seq – 0.4 ×109 semi-intact cells were digested with micrococcal Nuclease (MNase), 1.5 unit at 37 °C for 30 min with 3 mM CaCl2. The reactions were stopped by addition of EDTA to a final concentration of 0.02 M and subsequently incubated with RNase A (0.1 mg) for 4 h at 37 °C and further treated with Proteinase K at 37 °C o/n. DNA was purified using phenol–chloroform extraction and concentrated by ethanol precipitation.
The percentage of mononucleosomal DNA fragments was examined by 2% agarose gels. Furthermore, the integrity and size distribution of digested fragments were determined using the microfluidics-based platform Bioanalyzer (Agilent) prior to library preparations following Illumina standard protocol. The short-insert paired-end libraries for MNase sequencing were prepared with PCR free protocol using KAPA Library Preparation kit (Roche). In short, 2.0 micrograms of Micrococcal nuclease (MNase) digested genomic DNA from S. cerevisiae was end-repaired, adenylated and ligated to Illumina platform compatible adapters with dual indexes (Integrated DNA Technologies). The adapter-modified end library was size selected and purified with AMPure XP beads (Agencourt, Beckman Coulter). The final libraries were quantified by Kapa Library Quantification Kit for Illumina platforms (Roche).
The libraries were sequenced using TruSeq SBS Kit v4-HS (Illumina), in paired-end mode with a read length of 2x76bp following the manufacturer’s protocol. Images analysis, base calling and quality scoring of the run were processed using the manufacturer’s software Real Time Analysis (1.18.66.3).
Nucleosome calling – MNase-seq paired-end reads were mapped to yeast genome (sacCer3, Apr. 2011) using Bowtie61 aligner, allowing a maximum of 2 mismatches and maximum insert size of 500 bp. Output BAM files were imported in R62,58 and quality control was performed with htSeqTools package to remove PCR artifacts63. Filtered reads were processed with nucleR package64 as follows: mapped fragments were trimmed to 50 bp maintaining the original center and transformed to reads per million. Then, noise was filtered through Fast Fourier Transform, keeping 2% of the principal components, and peak calling was performed using the parameters: peak width 147 bp, peak detection threshold 35%, maximum overlap of 80 bp, dyad length 50 bp. Nucleosome calls were considered well-positioned when nucleR peak width score and height score were higher than 0.6 and 0.4, respectively, and fuzzy otherwise.
Nucleosome Dynamics – NucDyn R package45 was used to find changes in nucleosome organization between control and methylation-induced samples. P-values quantifying the nucleosome change were obtained running NucDyn with the following parameters: maximum difference of 70, maximum length of 140, minimum number of reads to report a shift of 3, shifts threshold of 0.1, indels minimum number of reads to report evictions and inclusions (indels) of 3, indels threshold of 0.05.
Genomic Annotation – Data was annotated from the UCSC gene track that contains 6692 genes. We discarded genes that are described as “Putative” or “Dubious” and genes located in the mitochondrial chromosome. We used gene lengths to normalize methylation proportions, nucleosome coverages and CpG density partitioning each gene in 137 bins (each bin has on average 10 bp since the mean length of yeast genes is 1369 bp).
### Gene expression
mRNA library preparation and sequencing – The RNASeq libraries were prepared from total RNA (extracted by the standard hot phenol protocol) as follows. Total RNA quality and quantity were assessed using Qubit® RNA HS Assay (Life Technologies) and RNA 6000 Nano Assay on a Bioanalyzer 2100 (Agilent). The RNASeq libraries were prepared following KAPA Stranded mRNA-Seq Illumina® Platforms Kit (Roche) following the manufacturer´s recommendations. Total RNA (500 ng) was enriched for the polyA mRNA fraction and fragmented by divalent metal cations at high temperature. In order to achieve the directionality, the second strand cDNA synthesis was performed in the presence of dUTP. The blunt-ended double stranded cDNA was 3´adenylated and Illumina platform compatible adapters with unique dual indexes and unique molecular identifiers (Integrated DNA Technologies) were ligated. The ligation product was amplified with 15 PCR cycles and the final library was validated on an Agilent 2100 Bioanalyzer with the DNA 7500 assay (Agilent).
The libraries were sequenced on HiSeq2500 (Illumina) using TruSeq SBS Kit v4-HS (Illumina), in paired-end mode with a read length of 2x76bp following the manufacturer’s protocol. Images analysis, base calling and quality scoring of the run were processed using the manufacturer’s software Real Time Analysis (1.18.66.3). We generated over 20 million paired-end reads for each sample in a fraction of a sequencing lane.
RNA-seq data processing and analysis – RNA-seq reads were mapped against the yeast reference genome (Sacharomyces_cerevisiae.R64-1-1+plasmid) using STAR version 2.5.2a65 with ENCODE parameters. Quantification of annotated genes (ensembl release 87) was done using RSEM version 1.2.2866 with default options. Heatmaps with the top differentially expressed genes was perform with the pheatmap R package. Differential expression between conditions was performed with DESeq2 version 1.1867 with default parameters.
Motif enrichment analysis – De novo promoter motif enrichment for genes with highly methylated promoters and increase in expression level was performed using HOMER68. It uses a cumulative hypergeometric distribution to score motif enrichment in the target set compared to the background set. Target set was defined in the region 200 bp downstream the TSS.
### ChIP-seq
ChIP-seq analysis – Sequenced reads were analyzed using Galaxy platform69. Reads were mapped to the reference sacCer3 genome with BWA70 and fragments with MAPQ quality score below 20 were discarded. Peak calling was performed using MACS271 with default parameters.
### 3D genome structure
Hi-C libraries – The protocol was performed as previously described72 with a few modifications. 100 ml of yeast culture were crosslinked with 3% formaldehyde during 20 min and quenched with Glycine 125 mM during 5 min at RT. Cells were crushed during 30 min in liquid nitrogen and the chromatin was digested with HindIII. The DNA overhangs were filled-in with dNTP including Biotin-14-dATP, and the resulting blunt end were ligated. After ligation, samples were purified with phenol:chloroform and DNA was precipitated with ethanol.
The paired-end Hi-C-sequencing libraries were prepared with KAPA Library Preparation kit (Roche) with some modifications. The biotin marked and de-crosslinked DNA was sheared to a size of 300–500 bp on Covaris™ LE220 (Covaris) focused-ultrasonicator. The fragmented DNA was end-repaired, adenylated and the biotin-tagged DNA was pulled down using the Dynabeads™ MyOne™ Streptavidin C1 beads (Thermo Fisher Scientific). The biotinylated fragments were ligated to Illumina platform compatible adapters with unique dual indexes and unique molecular identifiers (Integrated DNA Technologies) and enriched by 12 PCR cycles by KAPA HiFi PCR Kit (Roche).
The libraries were sequenced on HiSeq2500 (Illumina) using TruSeq SBS Kit v4-HS (Illumina), in paired-end mode with a read length of 2x76bp following the manufacturer’s protocol. Images analysis, base calling and quality scoring of the run were processed using the manufacturer’s software Real Time Analysis (1.18.66.3).
Hi-C data processing and normalization – We processed Hi-C data using TADbit73 [https://github.com/3DGenomes/tadbit] for quality control, mapping and filtering. First, quality control was performed with the FastQC protocol implementation in TADbit. Then, reads were mapped to the reference yeast genome (sacCer3, Apr. 2011) with a fragment-based strategy. Afterwards, non-informative contacts (self-circle, dangling-end, error, duplicated and random breaks) identified by TADbit were filtered-out, obtaining 32–37 million valid interactions per experiment. Off-target contacts (neither end of the read mapped to one of the capture regions) were also discarded (full details of the number of excluded reads are given in Supplementary Table 5). Finally, contact matrices were created from valid reads at 5 kb resolution with the corresponding TADbit module, and low frequency bins were removed.
Contact matrices were transformed to .hic format for visualization in Juicebox74 using the pre command, and normalized with the Balanced method75.
Differential Hi-C analysis was performed using the R/Bioconductor package diffHic76. The mapped Hi-C data were filtered and the differential interaction analysis between the control and methylated samples (using the two replicates for each treatment) was performed using the procedure recommended in the diffHic manual.
Hi-C-based chromatin 3D structure – High resolution Hi-C data at 5 kb was used to obtain the 3D structure, conformation and dynamics of entire yeast chromosomes and their context inside the nucleus. The Hi-C technique provides interaction contacts between DNA fragments. The interaction counts or frequencies between two loci i and j (fij) can be converted to spatial 3D distances between those loci (dij) by an inverse relationship (Eq. (1)),
$${d}_{ij}=\gamma /{f}_{ij}^{\alpha }$$
(1)
where γ represents the scale of the structure and is usually taken to match experimental distances between selected genomic regions, and the precise value of α depends on the organism under study, the genomic distance, and the resolution of the Hi-C map and needs to be fitted77,78,79. In the present work, γ was taken for the model to match the size of the cell nucleus measured by confocal microscopy and α was fitted to maximize the correlation between experimental and modeled contact maps.
Since Hi-C interaction counts are known to present several biases, such as mappability of fragments, GC content, and fragment length, they were normalized using iterative correction and eigenvector decomposition80. Finally, the output of the conversion procedure was a matrix containing equilibrium distances (r0) for the different interacting loci. To remove the background noise, a cutoff of two times the median of all trans contacts (i.e., between different chromosomes) was applied to the HiC contact map to define interacting regions.
The chromosome model was built as a chain of beads, each bead representing a genomic region that corresponds to a bin from the Hi-C map. Spatial equilibrium distances were obtained from Eq. (1) as explained above. The distances between interacting beads (r) were restrained near their equilibrium length during the simulations by penalizing with a harmonic potential (Eq. (2)) when approaching at shorter distances or moving away at longer distances than the equilibrium. A tolerance of one bead radius was applied, thus resulting in a flat-welled parabola potential (Eq. (2)),
$$E=k{(r-r^{\prime})}^{2}$$
(2)
where r’ = r0 − rbead for r < r0 − rbead, r’ = r when r lies within r0 ± rbead and r’ = r0 + rbead when r > r0 + rbead.
To ensure proper connectivity of the fiber, consecutive beads were also bound by a harmonic potential but with a force constant five orders of magnitude stronger than that applied to interacting non-consecutive beads. An excluded volume was defined for each bead by a standard Lennard-Jones potential with equilibrium distance equal to one bead radius and a soft energy well. Additional repulsive restraints were added for non interacting beads, forced to remain at a distance longer than the maximum equilibrium distance obtained from Eq. (1). The initial structure of the chromosome fiber was varied between an extended conformation and a random localization of initially unbound beads in different replicas. The system was allowed to sample the conformational space using pmemd simulation engine for GPU from Amber 18 package. Different conformations of the fibers were determined by attraction and repulsion forces arising from the distance restraints between beads.
In the end, an ensemble of structures was obtained by selecting the minimum number of snapshots minimizing the number of experimental restraint violations (equilibrium distances input). This method yields a population of structures with different conformations, which in average, but not individually, reproduce experimental Hi-C maps derived from population of cells with variable chromatin structure81.
The ensemble was built in the following way. First, sampled structures with more restraint violations than the mean restraint violations were discarded. Then, the structure with less restraint violations was selected. Considering only the restraints violated by the selected structure, the structure fulfilling more of these restraints was kept. The procedure was repeated iteratively, always considering the restraints that were not fulfilled by any of the previously selected structures. Iterations were stopped when there was no structure left fulfilling new restraints.
Chromatin coarse-grained model at the nucleosome level – The starting point for the 3D chromatin model at the nucleosome level is the coverage of the MNase-seq signal obtained using NucleR software64. Different families bearing nucleosomes in locations compatible with the MNase-seq experiment are derived by deconvolution of the coverage signal by using a composite Gaussian approximation. For each of the resulting families (compatible with the MNase-seq signal and DNA/histone stoichiometry) an ideal 3D chromatin structure is prepared and further simulated by a coarse-grained Monte Carlo sampling approach with flexible linkers and rigid nucleosomes. Linker DNA is represented at the base pair level by a pseudo-harmonic potential expressed in helical parameters (rise, slide, shift, twist, roll, tilt)82. Debye Huckel electrostatics and excluded volume potentials were added to avoid overlaps (exact details of the simulation procedure will be described elsewhere). The results of the different simulations are clustered to select the minimum number of nucleosome structural families that makes physical sense and that together reproduce Mnase-seq experiments. The source code and instructions to run simulations are available as Supplementary Software 2.
### Microscopy and image analysis
yIL30-W cells expressing the 4DNMTs were attached to 35 mm glass bottom culture dishes coated with Concanavalin A83. High-content image acquisition was performed with an Automated Wide-field Olympus IX81 Microscope, an MT20 xenon mercury lamp (Olympus Life Science Europe, Waltham, MA) and a 100x/1.3 oil UPlanFL N objective. CFP, mRFP and YFP were acquired using the appropriate filter sets for fluorescence imaging (AHF Analysentechnik, Tuebingen, Germany). ScanR Acquisition software version 2.3 was used to automatically record several fields of view and to take z-stacks (21 slices with a z-step of 200 nm) at every position. At least 200 cells from each condition were analyzed. For the image analysis, the “Spot distance” [http://bigwww.epfl.ch/sage/soft/spotdistance/] plug in Version 08.2014 (Daniel Sage, EPFL) on ImageJ Fiji distribution84 was used to find the coordinates of every focus and to get the distances among them. Statistical analysis was performed using the Wilcoxon test.
### Reporting summary
Further information on research design is available in the Nature Research Reporting Summary linked to this article.
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2022-09-30 14:21:59
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https://online.stat.psu.edu/stat414/book/export/html/679
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Lesson 11: Geometric and Negative Binomial Distributions
Lesson 11: Geometric and Negative Binomial Distributions
Overview
In this lesson, we learn about two more specially named discrete probability distributions, namely the negative binomial distribution and the geometric distribution.
Objectives
Upon completion of this lesson, you should be able to:
• To understand the derivation of the formula for the geometric probability mass function.
• To explore the key properties, such as the mean and variance, of a geometric random variable.
• To learn how to calculate probabilities for a geometric random variable.
• To explore the key properties, such as the moment-generating function, mean and variance, of a negative binomial random variable.
• To learn how to calculate probabilities for a negative binomial random variable.
• To understand the steps involved in each of the proofs in the lesson.
• To be able to apply the methods learned in the lesson to new problems.
11.1 - Geometric Distributions
11.1 - Geometric Distributions
Example 11-1
A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas City, Missouri until he finds a person who attended the last home football game. Let $$p$$, the probability that he succeeds in finding such a person, equal 0.20. And, let $$X$$ denote the number of people he selects until he finds his first success. What is the probability mass function of $$X$$?
Solution
Geometric Distribution
Assume Bernoulli trials — that is, (1) there are two possible outcomes, (2) the trials are independent, and (3) $$p$$, the probability of success, remains the same from trial to trial. Let $$X$$ denote the number of trials until the first success. Then, the probability mass function of $$X$$ is:
$$f(x)=P(X=x)=(1-p)^{x-1}p$$
for $$x=1, 2, \ldots$$ In this case, we say that $$X$$ follows a geometric distribution.
Note that there are (theoretically) an infinite number of geometric distributions. Any specific geometric distribution depends on the value of the parameter $$p$$.
11.2 - Key Properties of a Geometric Random Variable
11.2 - Key Properties of a Geometric Random Variable
On this page, we state and then prove four properties of a geometric random variable. In order to prove the properties, we need to recall the sum of the geometric series. So, we may as well get that out of the way first.
Recall
1. The sum of a geometric series is:
$$g(r)=\sum\limits_{k=0}^\infty ar^k=a+ar+ar^2+ar^3+\cdots=\dfrac{a}{1-r}=a(1-r)^{-1}$$
2. Then, taking the derivatives of both sides, the first derivative with respect to $$r$$ must be:
$$g'(r)=\sum\limits_{k=1}^\infty akr^{k-1}=0+a+2ar+3ar^2+\cdots=\dfrac{a}{(1-r)^2}=a(1-r)^{-2}$$
3. And, taking the derivatives of both sides again, the second derivative with respect to $$r$$ must be:
$$g''(r)=\sum\limits_{k=2}^\infty ak(k-1)r^{k-2}=0+0+2a+6ar+\cdots=\dfrac{2a}{(1-r)^3}=2a(1-r)^{-3}$$
We'll use the sum of the geometric series, first point, in proving the first two of the following four properties. And, we'll use the first derivative, second point, in proving the third property, and the second derivative, third point, in proving the fourth property. Let's jump right in now!
Theorem
The probability mass function:
$$f(x)=P(X=x)=(1-p)^{x-1} p$$
$$0<p<1$$, $$x=1, 2, \ldots$$ for a geometric random variable $$X$$ is a valid p.m.f.
Theorem
The cumulative distribution function of a geometric random variable $$X$$ is:
$$F(x)=P(X\leq x)=1-(1-p)^x$$
Theorem
The mean of a geometric random variable $$X$$ is:
$$\mu=E(X)=\dfrac{1}{p}$$
Theorem
The variance of a geometric random variable $$X$$ is:
$$\sigma^2=Var(X)=\dfrac{1-p}{p^2}$$
Proof
To find the variance, we are going to use that trick of "adding zero" to the shortcut formula for the variance. Recall that the shortcut formula is:
$$\sigma^2=Var(X)=E(X^2)-[E(X)]^2$$
We "add zero" by adding and subtracting $$E(X)$$ to get:
$$\sigma^2=E(X^2)-E(X)+E(X)-[E(X)]^2=E[X(X-1)]+E(X)-[E(X)]^2$$
Then, here's how the rest of the proof goes:
11.3 - Geometric Examples
11.3 - Geometric Examples
Example 11-1 Continued
A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas City, Kansas until he finds a person who attended the last home football game. Let $$p$$, the probability that he succeeds in finding such a person, equal 0.20. And, let $$X$$ denote the number of people he selects until he finds his first success. What is the probability that the marketing representative must select 4 people before he finds one who attended the last home football game?
Solution
To find the desired probability, we need to find $$P(X=4$$, which can be determined readily using the p.m.f. of a geometric random variable with $$p=0.20$$, $$1-p=0.80$$, and $$x=4$$:
$$P(X=4)=0.80^3 \times 0.20=0.1024$$
There is about a 10% chance that the marketing representative would have to select 4 people before he would find one who attended the last home football game.
What is the probability that the marketing representative must select more than 6 people before he finds one who attended the last home football game?
Solution
To find the desired probability, we need to find $$P(X>6)=1-P(X\le6)$$, which can be determined readily using the c.d.f. of a geometric random variable with $$1-p=0.80$$, and $$x=6$$:
$$P(X >6)=1-P(X \leq 6)=1-[1-0.8^6]=0.8^6=0.262$$
There is about a 26% chance that the marketing representative would have to select more than 6 people before he would find one who attended the last home football game.
How many people should we expect (that is, what is the average number) the marketing representative needs to select before he finds one who attended the last home football game? And, while we're at it, what is the variance?
Solution
The average number is:
$$\mu=E(X)=\dfrac{1}{p}=\dfrac{1}{0.20}=5$$
That is, we should expect the marketing representative to have to select 5 people before he finds one who attended the last football game. Of course, on any given try, it may take 1 person or it may take 10, but 5 is the average number. The variance is 20, as determined by:
$$\sigma^2=Var(X)=\dfrac{1-p}{p^2}=\dfrac{0.80}{0.20^2}=20$$
11.4 - Negative Binomial Distributions
11.4 - Negative Binomial Distributions
Example 11-1 Continued
(Are you growing weary of this example yet?) A representative from the National Football League's Marketing Division randomly selects people on a random street in Kansas City, Kansas until he finds a person who attended the last home football game. Let $$p$$, the probability that he succeeds in finding such a person, equal 0.20. Now, let $$X$$ denote the number of people he selects until he finds $$r=3$$ who attended the last home football game. What is the probability that $$X=10$$?
Solution
Negative Binomial Distribution
Assume Bernoulli trials — that is, (1) there are two possible outcomes, (2) the trials are independent, and (3) $$p$$, the probability of success, remains the same from trial to trial. Let $$X$$ denote the number of trials until the $$r^{th}$$ success. Then, the probability mass function of $$X$$ is:
$$f(x)=P(X=x)=\dbinom{x-1}{r-1} (1-p)^{x-r} p^r$$
for $$x=r, r+1, r+2, \ldots$$. In this case, we say that $$X$$ follows a negative binomial distribution.
NOTE!
1. There are (theoretically) an infinite number of negative binomial distributions. Any specific negative binomial distribution depends on the value of the parameter $$p$$.
2. A geometric distribution is a special case of a negative binomial distribution with $$r=1$$.
11.5 - Key Properties of a Negative Binomial Random Variable
11.5 - Key Properties of a Negative Binomial Random Variable
Theorem
Just as we did for a geometric random variable, on this page, we present and verify four properties of a negative binomial random variable.
The probability mass function:
$$f(x)=P(X=x)=\dbinom{x-1}{r-1} (1-p)^{x-r} p^r$$
for a negative binomial random variable $$X$$ is a valid p.m.f.
Proof
Before we start the "official" proof, it is helpful to take note of the sum of a negative binomial series:
$$(1-w)^{-r}=\sum\limits_{k=0}^\infty \dbinom{k+r-1}{r-1} w^k$$
Now, for the proof:
Theorem
The moment generating function of a negative binomial random variable $$X$$ is:
$$M(t)=E(e^{tX})=\dfrac{(pe^t)^r}{[1-(1-p)e^t]^r}$$
for $$(1-p)e^t<1$$.
Proof
As always, the moment generating function is defined as the expected value of $$e^{tX}$$. In the case of a negative binomial random variable, the m.g.f. is then:
$$M(t)=E(e^{tX})=\sum\limits_{x=r}^\infty e^{tx} \dbinom{x-1}{r-1} (1-p)^{x-r} p^r$$
Now, it's just a matter of massaging the summation in order to get a working formula. We start by effectively multiplying the summands by 1, and thereby not changing the overall sum:
$$M(t)=E(e^{tX})=\sum\limits_{x=r}^\infty e^{tx} \dbinom{x-1}{r-1} (1-p)^{x-r} p^r \times \dfrac{(e^t)^r}{(e^t)^r}$$
Now, since $$p^r$$ and $$(e^t)^r$$ do not depend on $$x$$, they can be pulled through the summation. And, since the $$(e^t)^r$$ that remains sits in the denominator, it can get moved into the numerator by writing is as$$(e^t)^{-r}$$:
$$M(t)=E(e^{tX})=p^r(e^t)^r \sum\limits_{x=r}^\infty e^{tx} \dbinom{x-1}{r-1} (1-p)^{x-r} (e^t)^{-r}$$
Now, the $$p^r$$ and $$(e^t)^r$$ can be pulled together as $$(pe^t)^r$$. And, $$e^{tx}$$ and $$(e^t)^r$$ can be pulled together to get $$(e^t)^{x-r}$$:
$$M(t)=E(e^{tX})=(pe^t)^r \sum\limits_{x=r}^\infty \dbinom{x-1}{r-1} (1-p)^{x-r} (e^t)^{x-r}$$
And, $$(1-p)^{x-r}$$ and $$(e^t)^{x-r}$$ can be pulled together to get $$[(1-p)e^t]^{x-r}$$:
$$M(t)=E(e^{tX})=(pe^t)^r \sum\limits_{x=r}^\infty \dbinom{x-1}{r-1} [(1-p)e^t]^{x-r}$$
Now, let $$k=x-r$$, so that $$x=k+r$$. Changing the index on the summation, we get:
$$M(t)=E(e^{tX})=(pe^t)^r \sum\limits_{k=0}^\infty \dbinom{k+r-1}{r-1}[(1-p)e^t]^k$$
Now, we should be able to recognize the summation as a negative binomial series with $$w=(1-p)e^t$$. Using what we know about the sum of a negative binomial series, the m.g.f. is then:
$$M(t)=E(e^{tX})=(pe^t)^r [1-(1-p)e^t]^{-r}$$
which can be rewritten as:
$$M(t)=E(e^{tX})=\dfrac{(pe^t)^r}{[1-(1-p)e^t]^r}$$
Now, recall that the m.g.f. exists only if it is finite. So, all we need to do is note when $$M(t)$$ is finite. Well, that happens when $$(1-p)e^t<1$$, or equivalently when $$t<-\ln (1-p)$$. And the proof is complete...whewwww!
Theorem
The mean of a negative binomial random variable $$X$$ is:
$$\mu=E(X)=\dfrac{r}{p}$$
Theorem
The variance of a negative binomial random variable $$X$$ is:
$$\sigma^2=Var(x)=\dfrac{r(1-p)}{p^2}$$
Proof
Since we used the m.g.f. to find the mean, let's use it to find the variance as well. That is, let's use:
$$\sigma^2=M''(0)-[M'(0)]^2$$
The only problem is that finding the second derivative of $$M(t)$$ is even messier than the first derivative of $$M(t)$$. Let me cheat a bit then. Let me leave it to you to verify that the second derivative of the m.g.f. of the negative binomial is:
$$M''(t)=r(pe^t)^r(-r-1)[1-(1-p)e^t]^{-r-2}[-(1-p)e^t]+r^2(pe^t)^{r-1}(pe^t)[1-(1-p)e^t]^{-r-1}$$
Now, with my shortcut taken, let's use it to evaluate the second derivative of the m.g.f. at $$t=0$$:
Now, for the final calculation:
11.6 - Negative Binomial Examples
11.6 - Negative Binomial Examples
Example 11-2
An oil company conducts a geological study that indicates that an exploratory oil well should have a 20% chance of striking oil. What is the probability that the first strike comes on the third well drilled?
Solution
To find the requested probability, we need to find $$P(X=3$$. Note that $$X$$is technically a geometric random variable, since we are only looking for one success. Since a geometric random variable is just a special case of a negative binomial random variable, we'll try finding the probability using the negative binomial p.m.f. In this case, $$p=0.20, 1-p=0.80, r=1, x=3$$, and here's what the calculation looks like:
$$P(X=3)=\dbinom{3-1}{1-1}(1-p)^{3-1}p^1=(1-p)^2 p=0.80^2\times 0.20=0.128$$
It is at the second equal sign that you can see how the general negative binomial problem reduces to a geometric random variable problem. In any case, there is about a 13% chance thathe first strike comes on the third well drilled.
What is the probability that the third strike comes on the seventh well drilled?
Solution
To find the requested probability, we need to find $$P(X=7$$, which can be readily found using the p.m.f. of a negative binomial random variable with $$p=0.20, 1-p=0.80, x=7, r=3$$:
$$P(X=7)=\dbinom{7-1}{3-1}(1-p)^{7-3}p^3=\dbinom{6}{2}0.80^4\times 0.20^3=0.049$$
That is, there is about a 5% chance that the third strike comes on the seventh well drilled.
What is the mean and variance of the number of wells that must be drilled if the oil company wants to set up three producing wells?
Solution
The mean number of wells is:
$$\mu=E(X)=\dfrac{r}{p}=\dfrac{3}{0.20}=15$$
with a variance of:
$$\sigma^2=Var(x)=\dfrac{r(1-p)}{p^2}=\dfrac{3(0.80)}{0.20^2}=60$$
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility
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2022-06-28 03:29:39
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http://crypto.stackexchange.com/questions?page=2&sort=unanswered&pagesize=15
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# All Questions
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### Why calculate pi to estimate randomness?
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### Statistical saturation attack on block ciphers
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### What are the steps to decrypt a TLS 1.1 record?
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56 views
### Is it possible to perform one-way Diffie-Hellman MITM?
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### Use of Lamport's Signature as a primitive
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43 views
### Montgomery Reduction
I'm taking a hardware cryptography class and working on a problem that focuses on Montgomery Reduction. So by definition: Computing a x b (mod N) Pick R, s.t. R > N, gcd (R,N) = 1 Compute N^-1 ...
41 views
### Is there are code hidden in Wired Magazine's December 2014 Issue?
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### side channel attacks against TDES (compared to AES)
We all know AES has some pretty strong side channel attacks, some where (with a couple of hundred know plain- and ciphertext combinations) a key can be derived in a matter of minutes. Others may take ...
71 views
### Why cant I use ECB with some obfuscation for random access data stored in say a DDR
I understand that ECB does not hide data patterns well. BUt my understanding is that it cannot be bronken to "know" the underlying message. If that was the case, why cant ECB be used with some ...
I work on an exercise and I have to study functions as : $f : \mathbb{F}_q \rightarrow \mathbb{F}_q$ with $q = 2^n$ which can define a S-Box for SP-Networks. There is a question that I can't answer ...
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2015-04-26 17:17:05
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https://www.omtex.co.in/2016/02/notes-paper-solution-to-your-email.html
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### Notes & Paper Solution to Your Email.
PDF FILE TO YOUR EMAIL IMMEDIATELY
PURCHASE NOTES & PAPER SOLUTION.
@ Rs. 50/- each
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2020-11-30 04:30:28
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https://www.open.edu/openlearncreate/mod/oucontent/view.php?id=88§ion=20.5
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# 5.3 Malaria parasites
Malaria is caused by Plasmodium parasites. Plasmodium parasites infect people and attack the red blood cells, which often leads to severe illness and death. The parasites are spread to people through the bites of infected Anopheles mosquitoes, which are the malaria vectors and which bite mainly between dusk and dawn.
There are four types of human malaria, each due to one of the parasites with the following specific names:
The species names of malaria parasites are frequently abbreviated to P. falciparum,P. vivax, etc.
• Plasmodium falciparum
• Plasmodium vivax
• Plasmodium malariae
• Plasmodium ovale.
Plasmodium falciparum and Plasmodium vivax are the most common malaria parasites in Ethiopia. 60% of malaria infections in Ethiopia are due to P. falciparum and 40% are due to P. vivax. P. falciparum is the most deadly and requires special attention.
Although both parasites are widely distributed, some communities will have more falciparum malaria while others will have more vivax malaria. Do you know which type of infection is more common in your community?
5.2 Epidemiology and distribution of malaria in Ethiopia
5.3.1 Life cycle of the malaria parasite
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2022-01-19 22:20:33
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https://stats.stackexchange.com/questions/245264/distribution-of-parameter-mass-in-linear-regression-with-correlated-predictors
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# Distribution of parameter mass in linear regression with correlated predictors
Let $Y \in \mathbb{R}^p$ be a vector of $n$ draws from a standard normal distribution $\mathcal{N}(0,1)$. Let $w_1, w_2 \in \mathbb{R}^p$ be $n$ independent draws from a normal distribution with $\mathcal{N}(0,\sigma^2)$.
Now, we define $X_1 = Y + w_1$ and $X_2 = Y + w_2$ and let $X$ be a matrix with the columns $X_1, X_2$. If $\sigma^2 = 0$ we have $X_1 = X_2$. In this situation the linear model is unidentified, because we cannot invert the gram matrix $X^{\top}X$ to get the least squares estimate:
$$\beta = (X^{\top}X)^{-1} X^{\top}y$$
Another way of looking at this is to say that any convex combination of $X_1$ and $X_2$ will give the same prediction:
$$\beta_0 + \alpha X_1 + (1 - \alpha) X_2 = \beta_0 + \gamma X_1 + (1 - \gamma) X_2$$
for all $\alpha, \gamma \in [0,1]$, because $X_1 = X_2$.
Now if we pick a variance $\sigma^2 > 0$ we have $X_1 \neq X_2$ and the model is identified.
Via a small simulation I see that both parameters have approximately the same mass:
n <- 1000000
y <- rnorm(n)
x1 <- y + rnorm(n, 0, .1)
x2 <- y + rnorm(n, 0, .1)
X <- cbind(x1, x2)
solve(t(X)%*%X) %*% t(X)%*%y
[,1]
x1 0.4980466
x2 0.4969557
I am looking for an intuitive explanation of why this is the case. If $X_1 = X_2$ any convex combination of $X_1$ and $X_2$ gives the same prediction. However, when adding a tiny little bit of noise to both variables, so the model is identified, the optimal solution seems to be to give both predictors the same mass. It seems to me there must be a good reason for that.
In general of course I am trying to get an intuition of how multicollinearity translates into parameter estimates (if i have p predictors and all share the same piece of variance - I am imagining Venn diagrams here - by which beta will this piece be 'absorbed'?). I understand this for instance for forward-selection, where $X_i$ can only explain variance in $Y$ that is not yet explained by any beta $X_j$, $j < i$. This seems not to happen in the above example, because otherwise the first entered $X_i$ would explain almost all the variance and hence the beta of the variable that was entered second would be much smaller.
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2019-05-19 21:17:16
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http://www.talkstats.com/threads/macro-variable-inside-macro.60147/
|
MACRO variable inside MACRO
lancearmstrong1313
New Member
I'm curious why the following code does not work:
Code:
%LET mbr_list=1,2,3,4,5;
%MACRO mbr_loop(mbr_values);
%let mbr_count=%sysfunc(countw(&mbr_values));
%do i=1 %to &mbr_count;
%let mbr=%cmpres(%qscan(&mbr_values,&i,%str(,)));
%put &mbr;
%end;
%MEND;
%MBR_LOOP(%str(&mbr_list));
It gives me the following error: Error: More positional parameters found than defined.
However, if I run the exact same macro code except just manually type the numbers 1-5 inside the MBR_LOOP() it will work:
Code:
%LET X=1,2,3,4,5;
%MACRO mbr_loop(mbr_values);
%let mbr_count=%sysfunc(countw(&mbr_values));
%do i=1 %to &mbr_count;
%let mbr=%cmpres(%qscan(&mbr_values,&i,%str(,)));
%put &mbr;
%end;
%MEND;
%MBR_LOOP(%str(1,2,3,4,5));
(in the log)
1
2
3
4
5
Can anyone help me put the &MBR_LIST macro inside the %MBR_LOOP(%str());?
Con-Tester
Member
There are at least two ways to fix this. If you insist on using commas as separators for your list of values then modify the macro as follows:
Code:
%let mbr_list = 1, 2, 3, 4, 5;
%macro mbr_loop /parmbuff;
%local paramlist
mbr_count
i
mbr;
%let paramlist = %sysfunc(compress("&syspbuff", %str(0123456789,), K));
%let mbr_count = %sysfunc(countw(%quote(¶mlist)));
%do i=1 %to &mbr_count;
%let mbr = %cmpres(%qscan(%quote(¶mlist),&i,","));
%put &mbr;
%end;
%mend;
Both “%mbr_loop(&mbr_list);” and “%mbr_loop(1, 2, 3, 4, 5);” should run without trouble and give the same result.
If spaces are okay as separators, you can do this:
Code:
%let mbr_list = 1 2 3 4 5;
%macro mbr_loop(mbr_values);
%local mbr_count
i
mbr;
%let mbr_count = %sysfunc(countw(&mbr_values));
%do i=1 %to &mbr_count;
%let mbr = %qscan(&mbr_values, &i);
%put &mbr;
%end;
%mend;
Both “%mbr_loop(&mbr_list);” and “%mbr_loop(1 2 3 4 5);” should run without trouble and give the same result.
(I have also added a “%local” qualifier because the macro variables that are thus qualified properly have only local scope, not global.)
lancearmstrong1313
New Member
Thanks a ton! So does SAS just not like the &MACRO inside of the %STR() function?
Con-Tester
Member
No, there’s no per se issue with the “%str(¯o_name)” construct within a macro definition itself. The problem is that in the SAS world, a macro is just a shorthand label for a blob of text (that can also contain other macros). In your original case, “%mbr_loop(&mbr_list);” resolved to “%mbr_loop(1,2,3,4,5);” which implied that your “%mbr_loop()” macro took five parameters instead of just one. That is, the problem is using commas as value separators without somehow telling SAS that the comma-separated list should actually be treated as a single string argument.
You could also have used “%let mbr_list = %str(1, 2, 3, 4, 5);” for the first line in your code and left everything else the same.
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2019-09-20 07:51:31
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|
https://itprospt.com/num/5615428/wcanu-m-ldcuio-uturle-uueet-ntcngwalet-of-the-nakncsium
|
5
# Wcanu M(ldcuiO uturle uueet #ntcngwalet} of the nakncsium Ahdm rerova] Ir aprtuentnalz70.0 grants of polassium hydroxice?tompud Wht lonned when 0B(} moles ofburium ...
## Question
###### Wcanu M(ldcuiO uturle uueet #ntcngwalet} of the nakncsium Ahdm rerova] Ir aprtuentnalz70.0 grants of polassium hydroxice?tompud Wht lonned when 0B(} moles ofburium Gcunbine with GOmnulcs ol chlorue the erpirical forntula for this compound?compound contains WVkal 3.24 grams of sodium; the empitica fomula 2.26 grams of sulfur, und 4.51 Orams of oxygen for (his compound?Calculate the percent water in iron(II) sulfate heptahydrate.Lam4a Metd
Wcanu M (ldcuiO uturle uueet #ntcng walet} of the nakncsium Ahdm rerova] Ir aprtuent nalz 70.0 grants of polassium hydroxice? tompud Wht lonned when 0B(} moles ofburium Gcunbine with GOmnulcs ol chlorue the erpirical forntula for this compound? compound contains WVkal 3.24 grams of sodium; the empitica fomula 2.26 grams of sulfur, und 4.51 Orams of oxygen for (his compound? Calculate the percent water in iron(II) sulfate heptahydrate. Lam4a Metd
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2022-08-15 01:21:44
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https://www.scielo.br/j/jbchs/a/sk4GJF68jvSY9QmLrfvHQXg/?lang=en
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# Density-functional based tight-binding: an approximate DFT method
About the authors
# Abstracts
The DFTB method, as well as its self-consistent charge corrected variant SCC-DFTB, has widened the range of applications of fundamentally well established theoretical tools. As an approximate density-functional method, DFTB holds nearly the same accuracy, but at much lower computational costs, allowing investigation of the electronic structure of large systems which can not be exploited with conventional ab initio methods. In the present paper the fundaments of DFTB and SCC-DFTB and inclusion of London dispersion forces are reviewed. In order to show an example of the DFTB applicability, the zwitterionic equilibrium of glycine in aqueous solution is investigated by molecular-dynamics simulation using a dispersion-corrected SCC-DFTB Hamiltonian and a periodic box containing 129 water molecules, in a purely quantum-mechanical approach.
DFT; DFTB; SCC; glycine; zwitterion
O método DFTB, bem como a sua extensão com carga corrigida auto-consistente SCC-DFTB, tem ampliado a faixa de aplicações das ferramentas teóricas com fundamentos bem estabelecidos. Como uma aproximação do método do funcional de densidade, o método DFTB mantém aproximadamente a mesma precisão, mas com custo computacional menor, permitindo a investigação da estrutura eletrônica de sistemas grandes que não podem ser explorados com métodos ab initio convencionais. No presente artigo, os fundamentos dos métodos DFTB, SCC-DFTB e da inclusão das forças de dispersão de London são revisados. Para mostrar um exemplo da aplicabilidade do método DFTB, o equilíbrio zwitteriônico de glicina em solução aquosa é investigado. Foram realizadas simulações de dinâmica molecular usando o hamiltoniano SCC-DFTB corrigido para incluir a dispersão e uma caixa periódica contendo 129 moléculas de água, a partir de uma abordagem puramente mecânico-quântica.
REVIEW
Density-functional based tight-binding: an approximate DFT method
Augusto F. Oliveira * e-mail: augusto.oliveira@chemie.tu-dresden.de; duarteh@ufmg.br I, II, * * e-mail: augusto.oliveira@chemie.tu-dresden.de; duarteh@ufmg.br ; Gotthard SeifertII; Thomas HeineIII; Hélio A. Duarte * e-mail: augusto.oliveira@chemie.tu-dresden.de; duarteh@ufmg.br I, * * e-mail: augusto.oliveira@chemie.tu-dresden.de; duarteh@ufmg.br
IDepartamento de Química, Instituto de Ciências Exatas, Universidade Federal de Minas Gerais, Av. Antonio Carlos, 6627, 31270901 Belo Horizonte-MG, Brazil
IIPhysikalische Chemie, Technische Universität Dresden, Mommsenstr, 13, D01062 Dresden, Germany
IIISchool of Engineering and Sciences, Jacobs University, P.O. Box 750 561, 28725 Bremen, Germany
ABSTRACT
The DFTB method, as well as its self-consistent charge corrected variant SCCDFTB, has widened the range of applications of fundamentally well established theoretical tools. As an approximate density-functional method, DFTB holds nearly the same accuracy, but at much lower computational costs, allowing investigation of the electronic structure of large systems which can not be exploited with conventional ab initio methods. In the present paper the fundaments of DFTB and SCCDFTB and inclusion of London dispersion forces are reviewed. In order to show an example of the DFTB applicability, the zwitterionic equilibrium of glycine in aqueous solution is investigated by molecular-dynamics simulation using a dispersion-corrected SCCDFTB Hamiltonian and a periodic box containing 129 water molecules, in a purely quantum-mechanical approach.
Keywords: DFT, DFTB, SCC, glycine, zwitterion
RESUMO
O método DFTB, bem como a sua extensão com carga corrigida auto-consistente SCC-DFTB, tem ampliado a faixa de aplicações das ferramentas teóricas com fundamentos bem estabelecidos. Como uma aproximação do método do funcional de densidade, o método DFTB mantém aproximadamente a mesma precisão, mas com custo computacional menor, permitindo a investigação da estrutura eletrônica de sistemas grandes que não podem ser explorados com métodos ab initio convencionais. No presente artigo, os fundamentos dos métodos DFTB, SCC-DFTB e da inclusão das forças de dispersão de London são revisados. Para mostrar um exemplo da aplicabilidade do método DFTB, o equilíbrio zwitteriônico de glicina em solução aquosa é investigado. Foram realizadas simulações de dinâmica molecular usando o hamiltoniano SCC-DFTB corrigido para incluir a dispersão e uma caixa periódica contendo 129 moléculas de água, a partir de uma abordagem puramente mecânico-quântica.
1. Introduction
Density functional theory (DFT) methods are the standard and the most used theoretical techniques for electronic structure calculations.1-5 The advent of the generalized gradient approximation (GGA) for the exchange-correlation functional enhanced the DFT accuracy6 and the predicted molecular structures, relative energies and frequencies are nearly comparable to the second order Møller-Plesset perturbation theory (MP2) method, with remarkable success to treat transition metal complexes.7 Efficient algorithms to solve the Kohn-Sham equations have been implemented, scaling to N3 with respect to the size of the basis sets and, hence, being much more efficient than the N5 of the MP2 methods. DFT is the method chosen for a huge range of applications. The formalism of the DFT and its extension to the reactivity indexes are subject of intensive research and many empirical concepts such as electronegativity, chemical potential and hardness are now formally defined within the DFT framework.3,5,8-10 With respect to the methodology, developments concerning improved exchange-correlation functionals and hybrid quantum-mechanics/molecular-mechanics (QM/MM) methods are still the main subjects of research of many theoreticians.11 Chemical property estimates based on DFT are now well established, and even optical properties are accessible through the generalization to time-dependent DFT,7,12,13 a method which is nowadays implemented in many different computer codes.
Notwithstanding the marvelous ability of the DFT to treat systems of increasing complexity, many systems are still intractable at the actual stage of computer technology development. Biosystems, adsorption processes, nanostructures, molecular dynamics, clusters and aggregates with thousands of atoms, self-assembling systems, nanoreactors and supramolecular chemistry are some of the fields in which ab initio methods cannot be used with adequate chemical models. For this range of systems, semi-empirical methods seem to have their applicability. Semi-empirical methods such as AM1,14 PM315-17 and, more recently, RM118 have many empirical parameters that are fitted to a set of molecular properties, estimated either theoretically or experimentally. Therefore, the applicability of such methods is restricted.
Density-functional tight-binding (DFTB) is an approximate method based on the density functional framework which does not require large amounts of empirical parameters. The virtues and weaknesses of the DFTB are a heritage from DFT. In fact the parameters are consistently obtained from DFT calculations of a few molecules per pair of atom types. On the other side, DFTB is closely connected to the tight-binding methods. In fact, it can be seen as a non-orthogonal tight-binding method parameterized from DFT. The self-consistent charge extension of DFTB (SCC-DFTB) improves very much the accuracy of the method. For improvement of physical approximations, all DFT extensions, such as treatment of relativistic effects and London dispersion, can be easily used in the DFTB method. Large number of applications has been reported showing its usefulness in the calculations of hyperfine coupling constants, magnetic properties, vibrational spectra of solids and molecules, nuclear magnetic shielding tensors, geometries, dynamic properties and many others.19-24 Calculation of optical properties is also possible due to the time-dependent DFTB,25-27 which is not covered in the present paper.
The goal of the present review is to call the attention of the chemistry community to the DFTB method, which can be a good complement of the set of semi-empirical methods available. Its advantages and weaknesses are highlighted. As an example of application, the zwitterionic and neutral forms of glycine in aqueous solution are discussed in terms of fully quantum mechanical molecular dynamics of this molecule in water.
2. Background Fundaments
Density functional theory has been extensively reviewed.7,28 In this section a very brief review of DFT is done in order to highlight its crucial aspects to the formulation of the DFTB method.
The Hohenberg-Kohn (HK) theorems29 have rigorously made the electronic density acceptable as basic variable to electronic-structure calculations. However, development of practical DFT methods only became relevant after W. Kohn and L. J. Sham published their famous set of equations: the so-called Kohn-Sham (KS) equations.30
The use of the electronic density within the KS scheme allows a significant reduction of the computational demand involved in quantum calculations. Furthermore, the KS method paved the way for studying systems that could not be investigated by conventional ab initio methods (which use the wave function as basic variable).
Even though DFT methods have been successfully applied for systems of increasing complexity, methods which can include approximations to reduce even more the computational demand, without compromising the reliability of results, are still required.
The application of tight-binding (TB) to the calculation of electronic structures starts with the paper by J. Slater and G. Koster.31 The main idea behind this method is to describe the Hamiltonian eigenstates with an atomic-like basis set and replace the Hamiltonian with a parameterized Hamiltonian matrix whose elements depend only on the internuclear distances (this requires the integrals of more than two centers to be neglected) and orbital symmetries.
Although the SlaterKoster method was conceived for the calculation of band structures in periodic systems, it was later generalized to an atomistic model, capable of treating finite systems as well. The transition to atomistic has three main requirements, as discussed by Goringe et al.32
First, the elements of the Hamiltonian matrix must have a functional dependence on the interatomic distance. In the case of band structures one just has to know the matrix elements for discrete values of distance. This requirement was solved by Froyen and Harrison,33 who proposed that the interatomic distance was related to the Hamiltonian elements by 1/r2.
The second requirement is to obtain an expression for the total energy and not only for the band energy. In 1979 Chadi34 proposed that the total energy could be described as a sum of two contributions,
where Ebnd is the sum over the energies of all occupied orbitals obtained by diagonalization of the parameterized Hamiltonian matrix, and Erep is the repulsive contribution, obtained by the sum of the atomic-pair terms,
in which N is the number of atoms in the system.
The third and last requirement is the possibility to derive the atomic forces from the total energy. This is especially important for geometry optimization and molecular dynamics. By assuming differentiability of Uαβ in equation 2, the only problem is to derive Ebnd, which depends on the parameterization method chosen for the Hamiltonian matrix.
The DFTB method attends these three requirements with the additional advantage of completely avoiding any empirical parameterization, since the Hamiltonian and overlap matrices are calculated using atom-like valence-orbitals which are derived from DFT. Therefore, the DFTB method can be considered as a simplification of the KohnSham method.
3. The KohnSham Method
Although the Hohenberg and Kohn theorems29 have proven that the electronic energy of a system can be totally determined from its electronic density through the variational principle, they did not propose any procedure to perform this calculation. This was done about one year later, by Kohn and Sham,30 with the publication of their equations known as Kohn-Sham equations.
The solution of Kohn and Sham starts from the idea of using monoelectronic orbitals to calculate the kinetic energy in a simple, yet reasonably precise, way leaving a residual correction that could be calculated apart. Thus, one starts with a reference system of M non-interacting electrons subjected to the external potential nS, with Hamiltonian
Where
in which there are no electronelectron repulsion terms and for which the electronic density is exactly the same as in the corresponding system of interacting electrons. By introducing the single particle orbitals ψi all electronic densities physically acceptable for the system of non-interacting electrons can be written in the form
Therefore, the HK functional can be written as
where TS represents the kinetic-energy functional for the reference system of M non-interacting electrons, given by
J represents the classic Coulomb interaction functional
and the remaining interactions are grouped in Exc, the exchange-correlation functional, which contains the difference between the exact kinetic energy T and TS, besides the non-classic part of the electron-electron interactions Vee, i.e.
After combining equations 6, 7 and 8 within the second HK theorem, the chemical potential can be written as
with the KS effective potential
where νext is the external potential, normally due to the atomic nuclei, and the exchange-correlation potential νxc is defined as
Equation 10, restricted by ∫ρ()d = M, is exactly the same equation that would be obtained for a system of M non-interacting electrons submitted to the external potential νS = νKS. Thus, for a given νKS a suitable value of ρ can be calculated for equation 10 by solving the M monoelectronic equations
and by using the calculated ψi in equation 5.
Equations 5 and 11-13 are the so-called Kohn-Sham equations. Since νKS depends on ρ through νxc the KS equations must be solved iteratively using a self-consistent procedure similar to the one depicted in Figure 1. An electronic density modelρ0 is normally chosen to start the iterative procedure. In principle, any positive function normalized for the number of electrons would be applicable, but a good initial estimate of ρ can significantly accelerate convergence.
At the end of the iterative procedure, the total energy can be calculated, which is given in the KS method by the following expression:
The most difficult part of the KS scheme is to calculate νxc in equation 12. The existence of an exact density functional is assured by the first HK theorem, but the exact form of the Exc functional remains unknown. However, many approximations of this functional have been described in the scientific literature over the last 30 years. In practice, the approximation chosen for Exc and the way by which the KS orbitals are represented define the different DFT methods.
4. DFT as Basis for a TightBinding Method
Following Foulkes and Haydock35 the electronic density is written as a reference densityρ0 plus a small fluctuation δρ,
This electronic density is then inserted in equation 14:
where = ρ0(') and δρ' = δρ(') are defined as short-hand notations. The second term in equation 16 corrects the double counting in the Coulomb term; the third term corrects the new exchange-correlation contribution; and the fourth term results from splitting the Coulomb energy into one part related to ρ0 and another related to δρ. Enn is the nuclear repulsion.
Afterwards, Exc 0 + δρ] is expanded in a Taylor series up to the second-order term:
Substitution of equation 17 into 16 and use of the definition (δ Exc/δρ)ρ0 = νxc0] results in
From equation 18 it is possible to define four important terms. The first is a reference Hamiltonian
0 depending only upon ρ0
The sum in the first line of equation 18 is analogue to Ebnd in equation 1. The terms in the second line of equation 18 define the repulsive contribution,
Finally, the last term in equation 18 includes the corrections related to the fluctuations in the electronic density. This term is defined as
Therefore, equation 18 can be rewritten as
In order to obtain a good estimate of the reference electronic density, ρ0 is written as a superposition of atom-like densities centered on the nuclei α,
With this approximation it is assured that Erep does not depend on the electronic-density fluctuations. Furthermore, due to the neutrality of the Coulomb contributions become negligible for long distances. Therefore, Erep can be expanded as
The contributions of 3 and more centers are rather small and can be neglected. These approximations can also be justified by Coulomb screening, i.e., since is the electronic density of a neutral atom, the electron-electron interaction terms with more than two centers are canceled by the nucleus-nucleus interactions.
Due to the screening of terms of more than two centers, one can assume the two-center contributions to be short ranged. However, the repulsion energy does not decay to zero for long interatomic distances. Instead, it decays to a constant value given by the atomic contributions:
Thus, is assumed in order to make Erep dependent only on two-center contributions:
Although it would be possible to calculate Erep for known values of , it is more convenient to adjust Erep to ab initio results. Thus, Erep is fitted to the difference between the DFT energy and Ebnd as a function of the interatomic distance Rαβ using a suitable reference structure, i.e.
The value of Ebnd can be obtained by diagonalization of the Hamiltonian matrix, which leads to
The value of Erep is usually fitted to a polynomial function or to a series of splines. Typical plots of EDFT, Ebnd and Erep for a reference structure are shown in Figure 2.
Based on the considerations discussed so far, the DFTB model can be derived.
5. The Standard DFTB Model without Self-Consistency
In the standard DFTB scheme, the second-order correction term, E2nd of equation 22, is neglected. Therefore, the calculation of the total energy does not depend on the electronic-density fluctuations δρ and, accordingly, it does not have to be solved iteratively.
In DFTB the KS orbitals are represented with a linear combination of atomic orbitals (LCAO) centered on the nuclei. Denoting the basis functions by øν and the expansion coefficients by Ciν one can write the KS orbitals in the form
From this LCAO model, one obtains the secular problem
where the elements of the Hamiltonian matrix and Sµν of the overlap matrix are defined as follows:
The second term of equation 22 can be transformed, with equations 29 and 11, into
in which the elements of the density matrix P are defined as follows
In order to restrict the LCAO to valence orbitals only, it is necessary to assure the orthogonality of the basis functions with respect to the core basis-functions of the remaining atoms (by using atomic orbitals as basis functions the orthogonality between the core and valence functions within the same atoms is already assured).
Denoting |ø) as a non-orthogonalized basis-function and as the core basis-functions of atom β, the corresponding orthogonalized basis-function of |ø) is obtained by:
By using this orthogonalization procedure, equation 32 is transformed into
where denotes the eigenvalue of the state c in atom β. The effective potential νKS and the core correction in equation 35 can be interpreted as a pseudo-potential (Vpp). Writing νKS as the sum of potentials Vα centered on the atoms,
and using this definition in equation 35, the effective potential is transformed into a pseudo-potential for all atoms in the system, except for atoms to which øµ and øν belong. Therefore, the pseudo-potential appears in the three-center terms and in the two-center terms whose valence orbitals belong to the same atom (so called crystal field terms). The pseudo-potential contributions are considerably smaller than the contributions of the full potentials and are neglected. Thus, the Hamiltonian matrix elements are defined as
where δαβ is the Kronecker's delta. This approach, the potential superposition, has been used since the 1980's for the calculation of DFTB parameters. In 1998, Elstner et al.36 presented an alternative approach to derive the DFTB equations through a second order expansion of the DFT total energy with respect to the electron density. As result the Hamiltonian matrix elements are calculated as density superpositions, which is identical to equation 37 except for the contribution of the exchange correlation potential. Indeed, due to the non-linear nature of νxc, the effective potential cannot be described as a simple sum of reference potentials within this approach, instead one obtains
Both approaches are physically motivated and their results are similar, which is not surprising if the potential difference between equations 37 and 38 is explicitly calculated. Both approaches have been used extensively in the past, the potential superposition being more popular for standard DFTB calculations, and the density superposition more widely used for SCC-DFTB.
The øν basis functions and the reference atom-like densities are obtained by solving the Schrödinger equation
for the free atom within a self-consistent DFT method, as shown in Figure 1. The contraction potential (r/r0)2 in equation 39 constrains the wave functions, resulting in better basis sets for the study of condensed-phase systems and free molecules as well. The value for the parameter r0 is normally chosen between 1.85rcov and 2rcov, with rcov being the atomic covalent radius.37
In practice, the Hamiltonian matrix elements are calculated as follows: For the diagonal elements the energy level of the free atom is chosen, which ensures correct dissociation limits. Due to the orthogonality of the basis functions the off-diagonal elements of the intra-atomic blocks are exactly zero. The interatomic blocks are computed as given in equation 37 or 38, depending on the choice of potential generation. Within the density superposition approach the Hamiltonian matrix elements unfold as follows:
It should be noted that the Hamiltonian elements depend only on atoms α and β and, therefore, only the two-center matrix elements are explicitly calculated, as well as two-center elements of the overlap matrix. According to equation 40 the free atom eigenvalues form the diagonal of the Hamiltonian matrix, which assures the correct limit for free atoms.
By using øν and the Hamiltonian and overlap matrix elements can be calculated and tabulated as a function of the distance between atomic pairs. Thus, it is not necessary to recalculate any integrals during, e.g., a geometry optimization or molecular dynamics simulation.
At last, an analytic expression for atomic forces can be derived from the total energy with respect to the atomic space-coordinates,
By this approach, the DFTB method covers all three requirements for an atomistic tight-binding approach.
6. The SelfConsistent Charge Correction: SCC-DFTB
The non-self-consistent DFT scheme described so far is very suitable to study systems in which the polyatomic electronic density can be well represented by a sum of atom-like densities, i.e. homonuclear covalent systems or highly ionic systems. However, the uncertainties in the standard DFTB increase when the chemical bonds in the system are controlled by a more delicate charge balance between atoms, especially in the case of heteronuclear molecules and polar semiconductors. In order to have a better description of electronic systems and better transferability of DFTB in the cases where long-range Coulomb interactions are significant, the method has been improved, giving rise to the self-consistent charge correction DFTB (SCC-DFTB).36 In this new scheme, the electronic density is corrected through inclusion of the second-order contributions E2nd in equation 22, which are neglected in standard DFTB.
In order to include the density fluctuations in a simple yet efficient way according to a tight-binding approach, δρ is written as the superposition of atom-like contributions δρα, which fast decay along the distance from the corresponding atomic center,
where the atom-like contributions can be simplified with the monopole approximation:
Here Δqα is the Mulliken charge, difference between the atomic Mulliken population qα38 and the number of valence electrons of the neutral free atom denotes the normalized radial dependence of the density fluctuation in atom a approximated to spherical by the angular function Y00. In other words, the effects of charge transfer are included, but changes in the shape of the electronic density are neglected. Equation 21 then becomes
in which the notation γαβ was introduced merely for convenience.
In order to solve equation 44, γαβ must be analyzed. In the limit case where the interatomic separation is very large (|α - β|=| - '| → ∞) one finds, by GGA-DFT, that the exchange-correlation term goes to zero and γαβ describes the interaction of two normalized spherical electronic densities, basically reducing to 1/|α - β |, thus,
In the opposite case, for which the interatomic distance tends to zero (|
α - β |=| - '| → 0), γαβ describes the electron-electron interaction within the atom α and can be related with the chemical hardness ηα,39 or Hubbard parameter γαα = 2ηα = Ua. Typically, the atomic hardness can be calculated using the difference between ionization potential Iα and electron affinity Aα of atom α: 2hα = Iα -Aα. Due to practical problems, in particular related to the non-existence of various anions and accordingly missing experimental validation of the electron affinity of the corresponding elements, it is more convenient to exploit DFT to obtain these parameters. Application of Janak's theorem40 relates the atomic hardness to the derivative of the HOMO energy with respect to the occupation number of the HOMO and hence the energy change with respect to electron change within the HOMO. This approach offers the possibility to treat the charge contribution shell- or even orbital-wise, which is important for the calculation of certain elements with sp and d bonding contributions, in particular for transition metals. Orbital hardness values have been reported in the literature for elements from H to Xe.41 In the following, we concentrate on the atomic SCC procedure, which implies that all sums over charges run over the atomic index α. For orbital-dependent SCC the summation index for the charge would run over the shell index ξ. Within the monopole approximation, Uα can be calculated, using a DFT procedure, as the second derivative of the total atomic energy of atom α with respect to its atomic charge:
In order to obtain a well-defined and useful expression for systems in all scales, and still keep consistence with the afore approximations, an analytical expression was developed36 to approximate the density fluctuations with spherical electronic densities. In accordance with Slater-type orbitals (Gaussian-type orbitals can also be employed) used to solve the KS equations,42,43 it is assumed an exponential decay of the normalized spherical electronic density:
Omitting the second-order contributions of Exc in equation 44 one obtains:
Integration over ' gives:
Setting R = |α - β|, after some coordinate transformations one gets
where s is a short-range function with exponential decay, so that
Once it was assumed that the second-order contribution can be approximated by the Hubbard parameter when R = 0, according to equation 46, the exponents of equation 51 are obtained:
This result can be interpreted by noting that harder elements tend to have localized wave functions. The chemical hardness of a spin-depolarized atom is calculated by the energy derivative of the highest occupied atomic orbital with respect to its occupation number, equation 46, using a fully self-consistent ab initio method. Therefore, the influence of second-order contributions of the exchange-correlation energy is included in γαβ for short distances, where it is important. The fact that, within GGA, the exchange-correlation energy vanishes for large interatomic distances is taken into account. In the case of periodic systems, the long-range part can be calculated using the standard Ewald summation, whereas the short-range part s decays exponentially and can be summed over a small number of unit cells. Thus, equation 50 is a well-defined expression for extended and periodic systems.
Finally, the total energy within SCCDFTB is written as
with γαβ = γαβ(Uα,Uβ,| α - β|). Here the contribution due to the Hamiltonian 0 is exactly the same as in the standard DFTB scheme. Note that the first term in equation 53 does only simplify to the sum of MO energies, the convenient notation for DFTB, if all charges are zero. Like in the non-self-consistent method, the wave functions yi are expanded in a LCAO model, equation 29, and equation 53 gives:
The charge fluctuations are calculated by Mulliken population analysis:38
and secular equations similar to those in equation 30 can be obtained, with modified elements in the Hamiltonian matrix:
The matrix elements and Sµν are identical to those defined in the standard DFTB method, in equation 31. Since the atomic charges depend on the monoatomic wave functions ψi it is necessary to use a self-consistent procedure. Once the elements Sµν extend to some neighboring atoms, multi-particle interactions are introduced. The second-order correction is achieved by introducing the elements , which depend on the Mulliken charges.
Identically to the standard DFTB, the repulsive potential is fitted according to equation 27 using a suitable reference system.
As the self-consistent charge correction allows for the explicit treatment of charge-transfer effects, the transferability of Erep is considerably better, in comparison with the non-self-consistent scheme.
As in the standard DFTB, a simple analytic expression for the atomic forces can be derived accordingly:
DFTB schemes have been successfully used in a wide range of applications, from molecular compounds22,44 to systems in solid state.19,45-47 Indeed, a symposium dedicated to the DFTB methods was held during the 232nd National Meeting of the American Chemical Society, from 10th to 14th of September, in 2006. A special section with contributions presented in this symposium was published in the Journal of Physical Chemistry A, issue 26 of 2007,48 presenting the actual development state of DFTB with respect to its formalism, implementation and applications.
7. Weak Forces: Dispersion-Corrected (SCC-)DFTB
London interactions, also called dispersion forces, are defined as attractive forces between nonpolar molecules, due to their mutual polarizability.49 London dispersion forces are several orders of magnitude weaker than typical covalent or ionic interactions and also about 10 times weaker than hydrogen bridge interactions. Therefore, dispersion forces have negligible effect in short-range interactions and can be understood as the long-range component of van der Waals forces.
Despite their weak nature, London interactions affect many fundamental processes in chemistry, physics, and biology. They influence the formation of molecular crystals, the structure of biological molecules such as proteins and DNA, adsorption processes, ππ stacking interactions, among others.
However, as explained above, both the standard and self-consistent DFTB methods treat only short-range atomic potentials and terms with more than two centers are neglected. Therefore, the Hamiltonian matrix elements fall off quickly and become negligible at interatomic distances typically found in the region of the van der Waals minimum. Hence, DFTB completely disregards van der Waals interactions, especially dispersion forces.
Two treatments meant to include dispersion interactions a posteriori have been proposed.50,51 In both cases the dispersion energy Edisp is calculated separately using empirical potentials and then added to the DFTB total energy expression. Since van der Waals forces are totally absent in DFTB, the addition of Edisp does not introduce any double-counting errors to the energy.
Since both treatments are somewhat similar, we describe that used in the present work.51 This correction was implemented in an experimental version of the deMon code52 and makes use of the UFF force field,53 already available in deMon. The dispersion interaction Uαβ between atoms α and β at a distance R is given in Lennard-Jones-type form, which includes two parameters: van der Waals distance (Rαβ) and well depth (dαβ):
The Rαβ and dαβ parameters are reported in the original paper53 and are available from H to Lw in the periodic table of elements. In UFF the van der Waals term is set to zero according to an adjacency criteria; however, this imposes an inflexible topology of the system, which is not desirable in a quantum-mechanical method. To overcome this problem, equation 58 is used only when Uαβ is attractive (London interactions are never repulsive), i.e. R < 2-1/6Rαβ. In addition, a short-range potential is derived using the polynomial
where U0, U1, and U2 are calculated to make the interaction energy and its first and second derivatives match equation 58 at R = 2-1/6Rαβ. The best value suggested for n is 5, which gives the following U0, U1, and U2 parameters:51
Therefore, the dispersion potential for the DFTB method can be written as
and the dispersion energy is given by
This term is then added to the total DFTB energy calculated either using standard DFTB (section 5) or the SCC scheme (section 6).
8. Glycine in Aqueous Solution
Glycine (aminoethanoic acid) is the simplest among the essential aminoacids. In solution an intramolecular proton transfer from the carboxylic group to the amino group takes place, establishing the zwitterionic equilibrium shown in Figure 3. The charge separation in the zwitterionic form is stabilized by the solvent, which must have large dielectric constant, as it is the case in water. Thus, the neutral species is favored in nonpolar solvents.
In this work, Born-Oppenheimer molecular dynamics was carried out using the DCSCCDFTB method, as implemented in the deMon package.52 The glycine molecule was placed within a 16 Å periodic box containing 129 water molecules.
The data were collected during a 100 ps simulation time with a 0.5 fs time step. The simulation was carried out after a thermalization time of 50 ps. It is important to emphasize that both glycine and water molecules were calculated within a full quantum-mechanical approach. The radial distribution function (RDF) of water with respect to the glycine center of mass is shown in Figure 4. The first solvation shell integrates to 22 water molecules.
Table 1 shows the calculated geometrical properties of glycine in solution. The optimized geometric parameters are shown at the PBE/TZVP and DCSCCDFTB levels of theory. The estimated angles are in good agreement with the previously published results.54 The OC=O angle presents the largest discrepancy for the zwitterionic form. The PBE/TZVP estimated OC=O angle is 13 degrees larger than the value estimated with DCSCCDFTB. Furthermore, the OC=O angle is expected to increase from the neutral to the zwitterionic form due to the deprotonation of the carboxyl group. However, DCSCCDFTB seems to be insensitive to the large charge on the deprotonated carboxyl and the angle remains similar to that in the neutral form. The mean values of the angles and dihedrals from MD (last column in Table 1) are close to the optimized values with a standard deviation of about 4 degrees, except for the OCCN dihedral. This dihedral involves rotation around a single CC bond, therefore, large standard deviation is indeed expected, explaining the apparent disagreement with the gas phase PBE/TZVP results.
Wada et al.55 estimated the Gibbs free energy variation between the two glycine forms in aqueous solution to be about 7.0 kcal mol-1H = 10.3 kcal mol-1). The change of the expected value between the two forms from the NVE molecular dynamics (ΔENVE) was estimated to be about 25.5 kcal mol-1. We have also used the continuum model to estimate the ΔG of this reaction at the PBE/TZVP/PCM level of theory and a value of 23.4 kcal mol-1 was obtained.
9. Final Remarks
DFTB is an approximate density-functional method which, in principle, does not employ any empirical parameter, in the sense that all quantities are calculated within DFT (Slater-Koster integrals) or they are calculated from reference structures by DFT calculations (Erep). It has been implemented in many different codes.56
Density Functional methods have along the time become a standard method for electronic structure calculations and substantially helped to unify organic chemistry, inorganic chemistry, surface chemistry, materials science and, more recently, biochemistry.4 With the advent of DFTB, the approximate DFT method, a plethora of challenging systems are now accessible for electronic structure calculations, enlarging the frontiers of the applicability of fundamentally well established theoretical tools. Nanostructured, self-assembled and nanoreactor systems are some of those for which DFTB can provide substantial help in the investigative work.
Acknowledgments
The financial support of the Brazilian agencies CNPq and FAPEMIG are gratefully acknowledged. We also thank the joint PROBRAL action of CAPES (Brazil) and DAAD (Germany) for financial support.
IMG11
Hélio A. Duarte graduated in Chemical Engineering (1990) and received his MSc in Inorganic Chemistry (1993) from the Federal University of Minas Gerais-UFMG. He finished his PhD at the University of Montreal in 1997 under the supervision of Prof. Dennis R. Salahub, working with adsorption on metal surfaces using density functional methods-DFT. Currently, he is associate professor at the Department of Chemistry-UFMG. His research activities are centered on the development and applications of DFT (and approximate DFT) methods to investigate chemical speciation, inclusion compounds, sulphide minerals, nanostructured clay minerals and solid/liquid interface phenomena. (http:www.qui.ufmg.br/~duarteh).
IMG12
Thomas Heine (PhD TU Dresden 1999). After pre- and postdoctoral stages at the Universities of Montreal, Exeter, Bologna and Geneva, he became Assistant Professor at TU Dresden in 2002, where he received his venia legendi in Physical Chemistry in 2006. He was appointed as Associate Professor for Theoretical Physics/Computational Materials Science at Jacobs University Bremen in 2008. His main research interests are the development of new quantum-mechanical methods and their implementation and application. He currently works on actual topics as storage of molecular hydrogen, design of new materials and nanoelectromechanics. Prof. Heine has more than 100 publications in peer-reviewed international journals, among them one Nature and two PNAS, and more than 1000 citations, leading to an h index of 25. (http://physics.jacobs-university.de/theine/index.html)
IMG13
Gotthard Seifert studied Chemistry at the TU Dresden where he received his diploma in 1975 and also graduated as Dr. rer. nat. (PhD) in 1979. He worked as a research assistant at the Institute of Theoretical Physics at TU Dresden from 1979 to 1992. He received his habilitation in theoretical physics in 1988. In 1989 and 1990, he worked as a visiting scientist and visiting professor at the International School for Advanced Studies (SISSA) in Trieste and at the EPFL. In 1991, he was a visiting scientist at the Forschungszentrum in Jülich. From 1992 to 1998, he was again at the Institute of Theoretical Physics at the TU Dresden as a lecturer/professor. He moved to the Universität Paderborn in 1998 and became a professor of Physical Chemistry at TU Dresden in 2001. His research interests are in the areas of quantum chemistry, cluster physics and chemistry and computational materials research.
IMG14
Augusto Faria Oliveira graduated in Chemistry at Federal University of Minas Gerais-UFMG (2001) and received his MSc in 2004. He completed his PhD in Quantum Chemistry (2008) under the supervision of Prof. Hélio A. Duarte at UFMG. During his PhD he spent one year in the group of Prof. Seifert at TU-Dresden. Currently he is a post-doctoral fellow at TU-Dresden in the same group, working with inorganic nanotubes. His current interests are the development and application of DFT and DFTB methods to investigate inorganic nanotubes and ion adsorption on minerals.
Received: August 6, 2008
Web Release Date: May 29, 2009
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e-mail:
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• 53. Rappe, A. K.; Casewit, C. J.; Colwell, K. S.; Goddard, W. A.; Skiff, W. M.; J. Am. Chem. Soc. 1992, 114, 10024.
• 54. Tortonda, F. R.; Pascual-Ahuir, J. L.; Silla, E.; Tunon, I.; Ramirez, F. J.; J. Chem. Phys. 1998, 109, 592.
• 55. Wada, G.; Tamura, E.; Okina, M.; Nakamura, M.; Bull. Chem. Soc. Jpn. 1982, 55, 3064.
• 56
http://www.dftb.org, accessed in May 22, 2009.
» link
* e-mail: augusto.oliveira@chemie.tu-dresden.de; duarteh@ufmg.br
# Publication Dates
• Publication in this collection
28 Aug 2009
• Date of issue
2009
# History
• Received
06 Aug 2008
• Accepted
29 May 2009
Sociedade Brasileira de Química Instituto de Química - UNICAMP, Caixa Postal 6154, 13083-970 Campinas SP - Brazil, Tel./FAX.: +55 19 3521-3151 - São Paulo - SP - Brazil
E-mail: office@jbcs.sbq.org.br
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2022-12-03 13:57:57
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https://zbmath.org/?q=an:0948.44003
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# zbMATH — the first resource for mathematics
Generalized moment problem in vector lattices. (English) Zbl 0948.44003
The classical theorem on the moment problem going back to F. Hausdorff [Math. Z. 16, 220-248 (1923; JFM 49.0193.01)] states that a given real sequence $$a_{k}\in\mathbb{R}$$ is a sequence of moments of some nondecreasing function $$g\in V(0,1)$$ of bounded variation, i.e., $$a_{k}=\int_{0}^{1}t^{k} dg(t)$$, if and only if the sequence $$a_{n}$$ is completely monotone, i.e., $$\Delta ^{n}a_{k}:=\sum_{j=0}^{n}(-1)^{j}\binom{n}{j}a_{k+j}\geq 0$$. The authors generalise this result to the case where $$a_{k}$$ is the sequence of elements in an ordered vector space $$V$$, more precisely in $$\sigma$$-complete weakly $$\sigma$$-distributive vector lattice $$V$$ which satisfies two conditions: any interval in $$V$$ is sequentially order-compact; every chain in $$V$$ is at most countable. This is similar to, but distinct with results of H. H. Schaefer [Math. Ann. 146, 325-330 (1962; Zbl 0102.09905)]: neither contains the other. From this result, the authors derive an integral representation theorem for positive linear operators $$L:C(0,1)\rightarrow V$$, which is analogous to the real case.
##### MSC:
44A60 Moment problems 46A40 Ordered topological linear spaces, vector lattices
Full Text:
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2021-10-21 20:33:58
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https://xianblog.wordpress.com/tag/random-number-generation/
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## ziggurat algorithm
Posted in Books, pictures, Statistics, University life with tags , , , , , , , , , , , on October 30, 2018 by xi'an
A ziggurat (Akkadian: ziqqurat, D-stem of zaqāru “to build on a raised area”) is a type of massive stone structure built in ancient Mesopotamia. It has the form of a terraced compound of successively receding stories or levels. Wikipedia
In a recent arXival, Jalalvand and Charsooghi revisit the ziggurat algorithm that simulates from a univariate distribution by finding horizontal strips that pile up on top of the target as in a ziggurat or a pyramid, hence the name. Which George Marsaglia introduced in 1963. When finely tuned the method is quite efficient. Maybe because it designs an accept-reject move for each strip of the ziggurat rather than globally. For instance, versions constructed for a Normal target are more efficient [3½ times faster] than the Box-Muller algorithm. The generalisation found in the paper divides the target into strips of equal area, rather than dominating rectangular strips of equal area, which requires some work when the target density is non-standard. For targets with unbounded support or unbounded values, a function g transforming the tail into (0,1) has to be selected. A further constraint is that the inverse cdf of the transformed g(X) has to be known. And a large part of the paper examines several scenarii towards simulating from the tail region. For unbounded densities, a similarly minute analysis is undertaken, again with requests about the target like its algebraic order.
“…the result of division of a random integer by its range is a fixed-point number which unlike a floating-point number does not enjoy increased precision near 0. When such random numbers are used in the tail algorithm they cause premature termination of the tail and large gaps between produced random numbers near the termination point.”
The paper further discusses the correction of an error common to earlier ziggurat algorithms, due to the conversion from fixed-point to floating-point numbers, as indicated in the above quote. Although this had already been addressed by George Marsaglia in the early 1990’s.
“Ziggurat algorithm has a high setup time, so it’s not suitable for applications that require variates with frequently changing shape parameters.”
When testing the algorithm against different methods (in STL and Boost), and different distributions, the gains are between two and seven times faster, except for the Exponential target where the original ziggurat algorithm performs better. Interestingly, the gains (and the computing time) increase with the degrees of freedom for the Gamma target, in relation with Devroye’s (1986) remark on the absence of uniformly bounded execution times for this distribution. Same thing for the Weibull variates, obviously. Reflecting upon the usually costly computation of cdfs and inverse cdfs on machines and software, the inverse cdf method is systematically left behind! In conclusion, a good Sunday morning read if not of direct consequences for MCMC implementation, as warned by the authors.
## independent random sampling methods [book review]
Posted in Books, Statistics, University life with tags , , , , , , , , , , , , , on May 16, 2018 by xi'an
Last week, I had the pleasant surprise to receive a copy of this book in the mail. Book that I was not aware had been written or published (meaning that I was not involved in its review!). The three authors, Luca Martino, David Luengo, and Joaquín Míguez, of Independent Random Sampling Methods are from Madrid universities and I have read (and posted on) several of their papers on (population) Monte Carlo simulation in the recent years. Including Luca’s survey of multiple try MCMC which was helpful in writing our WIREs own survey.
The book is a pedagogical coverage of most algorithms used to simulate independent samples from a given distribution, which of course recoups some of the techniques exposed with more details by [another] Luc, namely Luc Devroye’s Non-uniform random variate generation bible, often mentioned here (and studied in uttermost details by a dedicated reading group in Warwick). It includes a whole chapter on accept-reject methods, with in particular a section on Payne-Dagpunar’s band rejection I had not seen previously. And another entire chapter on ratio-of-uniforms techniques. On which the three authors had proposed generalisations [covered by the book], years before I attempted to go the same way, having completely forgotten reading their paper at the time… Or the much earlier 1991 paper by Jon Wakefield, Alan Gelfand and Adrian Smith!
The book also covers the “vertical density representation”, due to Troutt (1991), which consists in considering the distribution of the density p(.) of the random variable X as a random variable, p(X). I remember pondering about this alternative to the cdf transform and giving up on it as the outcome has a distribution depending on p, even when the density is monotonous. Even though I am not certain from reading the section that this is particularly appealing…
Given its title, the book contains very little about MCMC. Except for a last and final chapter that covers adaptive independent Metropolis-Hastings algorithms, in connection with some of the authors’ recent work. Like multiple try Metropolis. Relating to the (unidimensional) ARMS “ancestor” of adaptive MCMC methods. (As noted in a recent blog on Holden et al., 2009 , I have trouble understanding how recycling only rejected proposed values to build a better proposal distribution is enough to guarantee convergence of an adaptive algorithm, but the book does not delve much into this convergence.)
All in all and with the bias induced by me working in the very area, I find the book quite a nice entry on the topic, which can be used in a Monte Carlo course at both undergraduate and graduate levels if one want to avoid going into Markov chains. It is certainly less likely to scare students away than the comprehensive Non-uniform random variate generation and on the opposite may induce some of them to pursue a research career in this domain.
## certified randomness, 187m away…
Posted in Statistics with tags , , , , , , , on May 3, 2018 by xi'an
As it rarely happens with Nature, I just read an article that directly relates to my research interests, about a secure physical random number generator (RNG). By Peter Bierhost and co-authors, mostly physicists apparently. Security here means that the outcome of the RNG is unpredictable. This very peculiar RNG is based on two correlated photons sent to two measuring stations, separated by at least 187m, which have to display unpredictable outcomes in order to respect the impossibility of faster-than-light communications, otherwise known as Bell inequalities. This is hardly practical though, especially when mentioning that the authors managed to produce 2¹⁰ random bits over 10 minutes, post processing “the measurement of 55 million photon pairs”. (I however fail to see why the two-arm apparatus would be needed for regular random generation as it seems relevant solely for the demonstration of randomness.) I also checked the associated supplementary material, which is mostly about proving some total variation bound, and constructing a Bell function. What is most puzzling in this paper (and the associated supplementary material) is the (apparent) lack of guarantee of uniformity of the RNG. For instance, a sentence (Supplementary Material, p.11) about a distribution being “within TV distance of uniform” hints at the method being not provably uniform, which makes the whole exercise incomprehensible…
## complexity of the von Neumann algorithm
Posted in Statistics with tags , , , , , , , , , on April 3, 2017 by xi'an
“Without the possibility of computing infimum and supremum of the density f over compact subintervals of the domain of f, sampling absolutely continuous distribution using the rejection method seems to be impossible in total generality.”
The von Neumann algorithm is another name for the rejection method introduced by von Neumann circa 1951. It was thus most exciting to spot a paper by Luc Devroye and Claude Gravel appearing in the latest Statistics and Computing. Assessing the method in terms of random bits and precision. Specifically, assuming that the only available random generator is one of random bits, which necessarily leads to an approximation when the target is a continuous density. The authors first propose a bisection algorithm for distributions defined on a compact interval, which compares random bits with recursive bisections of the unit interval and stops when the interval is small enough.
In higher dimension, for densities f over the unit hypercube, they recall that the original algorithm consisted in simulating uniforms x and u over the hypercube and [0,1], using the uniform as the proposal distribution and comparing the density at x, f(x), with the rescaled uniform. When using only random bits, the proposed method is based on a quadtree that subdivides the unit hypercube into smaller and smaller hypercubes until the selected hypercube is entirely above or below the density. And is small enough for the desired precision. This obviously requires for the computation of the upper and lower bound of the density over the hypercubes to be feasible, with Devroye and Gravel considering that this is a necessary property as shown by the above quote. Densities with non-compact support can be re-expressed as densities on the unit hypercube thanks to the cdf transform. (Actually, this is equivalent to the general accept-reject algorithm, based on the associated proposal.)
“With the oracles introduced in our modification of von Neumann’s method, we believe that it is impossible to design a rejection algorithm for densities that are not Riemann-integrable, so the question of the design of a universally valid rejection algorithm under the random bit model remains open.”
In conclusion, I enjoyed very much reading this paper, especially the reflection it proposes on the connection between Riemann integrability and rejection algorithms. (Actually, I cannot think straight away of a simulation algorithm that would handle non-Riemann-integrable densities, apart from nested sampling. Or of significant non-Riemann-integrable densities.)
## ratio-of-uniforms [#4]
Posted in Books, pictures, R, Statistics, University life with tags , , , , on December 2, 2016 by xi'an
Possibly the last post on random number generation by Kinderman and Monahan’s (1977) ratio-of-uniform method. After fiddling with the Gamma(a,1) distribution when a<1 for a while, I indeed figured out a way to produce a bounded set with this method: considering an arbitrary cdf Φ with corresponding pdf φ, the uniform distribution on the set Λ of (u,v)’s in R⁺xX such that
0≤u≤Φοƒ[φοΦ⁻¹(u)v]
induces the distribution with density proportional to ƒ on φοΦ⁻¹(U)V. This set Λ has a boundary that is parameterised as
u=Φοƒ(x), v=1/φοƒ(x), x∈Χ
which remains bounded in u since Φ is a cdf and in v if φ has fat enough tails. At both 0 and ∞. When ƒ is the Gamma(a,1) density this can be achieved if φ behaves like log(x)² near zero and like a inverse power at infinity. Without getting into all the gory details, closed form density φ and cdf Φ can be constructed for all a’s, as shown for a=½ by the boundaries in u and v (yellow) below
which leads to a bounded associated set Λ
At this stage, I remain uncertain of the relevance of such derivations, if only because the set A thus derived is ill-suited for uniform draws proposed on the enclosing square box. And also because a Gamma(a,1) simulation can rather simply be derived from a Gamma(a+1,1) simulation. But, who knows?!, there may be alternative usages of this representation, such as innovative slice samplers. Which means the ratio-of-uniform method may reappear on the ‘Og one of those days…
## ratio-of-uniforms [#3]
Posted in Books, pictures, R, Statistics with tags , , , , , on November 4, 2016 by xi'an
Being still puzzled (!) by the ratio-of-uniform approach, mostly failing to catch its relevance for either standard distributions in a era when computing a cosine or an exponential is negligible, or non-standard distributions for which computing bounds and boundaries is out-of-reach, I kept searching for solutions that would include unbounded densities and still produce compact boxes, as this seems essential for accept-reject simulation if not for slice sampling. And after exploring some dead-ends (in tune with running in Venezia!), I came upon the case of the generalised logistic transform
$h(\omega)=\omega^a/(1+\omega^a)$
which ensures that the [ratio-of-almost-uniform] set I defined in my slides last week
$\mathfrak{H}=\left\{(u,v);\ 0\le u\le h(f(v/g(u))\right\}$
is bounded in u. Since the transform g is the derivative of the inverse of h (!),
$g(y)=a^{-1}y^{(1-a)/a}/(1-y)^{(1-3a)/a}$
the parametrisation of the boundary of H is
$u(x)=f(x)^a/(1+f(x)^a)\ v(x)=a^{-1}xf(x)^{(a-1)/a}(1+f(x)^a)^2$
which means it remains bounded if (a) a≤1 [to ensure boundedness at infinity] and (b) the limit of v(x) at zero [where I assume the asymptote stands] is bounded. Meaning
$\lim_{x\to 0} xf(x)^{2a+1/a-1}<\infty$
Working a wee bit more on the problem led me to realise that resorting to an arbitrary cdf Φ instead of the logistic cdf could solve the problem for most distributions, including all Gammas! Indeed, the boundary of H is now
$u(x)=\Phi(f(x))^a\ v(x)=a^{-1}xf(x)^{(a-1)/a}/\varphi(f(x))$
which means it remains bounded if φ has very heavy tails, like 1/x². To handle the explosion when x=0. And an asymptote itself at zero, to handle the limit at infinity when f(x) goes to zero.
## ratio-of-uniforms
Posted in Books, pictures, R, Statistics with tags , , , , on October 24, 2016 by xi'an
One approach to random number generation that had always intrigued me is Kinderman and Monahan’s (1977) ratio-of-uniform method. The method is based on the result that the uniform distribution on the set A of (u,v)’s in R⁺xX such that
0≤u²≤ƒ(v/u)
induces the distribution with density proportional to ƒ on V/U. Hence the name. The proof is straightforward and the result can be seen as a consequence of the fundamental lemma of simulation, namely that simulating from the uniform distribution on the set B of (w,x)’s in R⁺xX such that
0≤w≤ƒ(x)
induces the marginal distribution with density proportional to ƒ on X. There is no mathematical issue with this result, but I have difficulties with picturing the construction of efficient random number generators based on this principle.
I thus took the opportunity of the second season of [the Warwick reading group on] Non-uniform random variate generation to look anew at this approach. (Note that the book is freely available on Luc Devroye’s website.) The first thing I considered is the shape of the set A. Which has nothing intuitive about it! Luc then mentions (p.195) that the boundary of A is given by
u(x)=√ƒ(x),v(x)=x√ƒ(x)
which then leads to bounding both ƒ and x→x²ƒ(x) to create a box around A and an accept-reject strategy, but I have trouble with this result without making further assumptions about ƒ… Using a two component normal mixture as a benchmark, I found bounds on u(.) and v(.) and simulated a large number of points within the box to end up with the above graph that indeed the accepted (u,v)’s were within this boundary. And the same holds with a more ambitious mixture:
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2019-09-20 17:35:33
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https://mmore500.com/2018/01/18/my-elementary-setup.html
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I use elementary OS.
I really like using elementary OS. The interface has come to feel like a good pair of boots, broken in to fit just right.
How many hours have I spent aimlessly stomping around to break in these boots? You probably don’t want to know. I certainly don’t want to — I’m not really in the mood for an existential crisis right now.
What I do hope you want to know about are some of the tweaks and enhancements I made to make elementary look and work just how I like. If not, go read something else.
Everything below assumes Version 0.4.1 Loki.
## 🔗 Remove Guest Account from Login Screen
I got tired of looking at the guest account option on the beautiful elementary login screen. Even though I had disabled the guest account itself, the icon remained. I decided to get rid of it.
Totally pointless? Certainly. Extremely satisfying? Definitely.
To fix this issue, I opened up the configuration file /etc/lightdm/lightdm.conf and modified the line allow-guest=false to instead read allow-guest=true.
Note that you might need superuser permissions to modify this file (i.e. sudo).
## 🔗 Redshift
Until elective surgery for HDMI input implants gains widespread availability, I’ll continue to need my eyeballs to look at my computer screen. Plus, they’re useful for other things, too.
Blue light hurts my eyes, especially at night when I’m at home sleeping at night and definitely not still at the office my computer. I try to give my retinas a break by cutting back on the amount of blue light I’m shining in my face at all hours. f.lux does a great job of warming the screen color on the mac, but its Linux port doesn’t cut it. Use Redshift instead.
## 🔗 Guake
At some point, I got tired of having 50 random terminals open all the time. So, I installed Guake, which lets you use a hotkey shortcut to drop down a floating terminal from the menu bar.
I use the “Space Gray” color scheme. It’s pretty rad.
## 🔗 Remap Modifier Keys
I migrated to Elementary OS from Mac OS. Given my long history with apple, I was strongly conditioned to use the key to do things that the ctrl does for me now (e.g. + C to copy text on mac vs. ctrl + C to copy text in elementary). For a while, I didn’t want to recondition myself because I used both systems regularly. Now, even though I use elementary exclusively, reconditioning myself would just be too traumatic.
The solution here is to remap input from your keyboard. That way, you can hit the physical cmd or key on your keyboard but have the operating system perceive it as a ctrl press.
What I learned in boating school is there are a lot of ways to try to achieve this that don’t work. Forget .Xmodmap. You’ll need to work with X KeyBoard extension files. To change your modifier key (i.e. ctrl, alt, cmd) mappings you’ll want to poke around in /usr/share/X11/xkb/symbols/pc. Note that you might need superuser permissions to modify your pc file (i.e. sudo).
For example, to map your physical right-alt key to mean right-ctrl to your operating system, change the line
key <RALT> { [ Alt_R ] };
key <RALT> { [ Control_R ] };
To remap your physical right-ctrl key to mean right-cmd (a.k.a. super), change the line
key <RCTL> { [ Control_R ] };
key <RCTL> { [ Super_R ] };
My personal keymapping probably isn’t too useful to you because my physical keyboard is laid out kinda funky. (Among other tweaks, I had to remap PrtSc to alt). For whatever you can make out of it, though, it’s here.
You’ll need to log out and log back in to see your changes take effect. You might have to reboot or do some cache clearing as described here if your changes aren’t taking effect as they should.
## 🔗 Hide Mouse Cursor
Give yourself a fighting chance in your own valiant struggle to ditch the mouse by hiding the cursor when it’s not being used.
You’ll need to install the unclutter package.
sudo apt-get install unclutter
You can then test unclutter out at the command line.
unclutter -idle 0.1 -root
To get unclutter to run in the background at startup, paste this into your ~/.xintrc. (If you don’t have an ~/.xintrc file, make one.)
unclutter -idle 0.1 -root &
## 🔗 Uninstall Unused Elementary Built-Ins
Elementary’s built-in mail, calendar, photos, music, etc. software is a nice gesture. I don’t use them, though, and I doubt many others do either.
To remove unused built-in software, open AppCenter and locate an application you don’t use via the search bar. Click to open that application’s product description page. You should see an option to uninstall. Click it then rinse and repeat for everything else you want gone.
## 🔗 Remove Unused Shortcuts from Slingshot Launcher Menu
If you’re bothered by clutter in your application launcher menu, then at least I’m not the only one. Command-line software that comes packaged with a desktop application interface especially bothers me. The culprit are *.desktop files in any of these three locations:
• /usr/share/applications,
• /usr/local/share/applications, and
• ~/.local/share/applications.
You’ll need to poke through these directories until you find the offender. Then, remove it. You won’t be able to open the application from the GUI launcher anymore, but command line components (and windowed components launched from the command line) should continue to work just fine.
For example,
cd /usr/share/applications
sudo rm ufraw.desktop
If no *.desktop files jumps out as the culprit, you might need to start peeking in suspect *.desktop files to check for the name and description of the application. Note that you might need superuser permissions to remove files under the /usr directory (i.e. sudo).
(Original source here.)
## 🔗 Block Distracting Websites
One of my friends works at a municipal library in small-town middle America. On her performance review, anonymous coworker told her that she needed to cool her jets because she was making everyone else look bad. According to this coworker, “Just because we have Facebook open our computers most of the time doesn’t mean we’re not as good as you. We just work differently.”
Don’t be my friend’s anonymous passive-aggressive colleague. Redirect your outgoing traffic to distracting/dumb websites to a dead end instead.
To do this you’ll need to modify /etc/hosts, which your computer uses to decide where to look for certain web addresses.
I’ve compiled a pretty comprehensive listing of distracting sites in my own /etc/hosts file. If you like, you can just paste this over your own /etc/hosts. Or, just add entries of the following format to the end of your /etc/hosts that block your own clickholes.
127.0.0.1 clickhole.com
Note that you might need superuser permissions to modify this file (i.e. sudo) and you might need to reboot to see changes take place.
## 🔗 Dark Theme for Elementary
Go check out elementary tweaks.
Want to replace the “Applications” text on the wingpanel bar with the elementary OS logo?
Totally pointless? Certainly. Extremely satisfying? Definitely.
Check out nightsense’s elementary-panel-logomark project. Also, consider using the inverse logo I contributed. It has less negative space than the original so it displays better.
Good usage instructions are provided on the project’s README.md.
## 🔗 Weather Widget
Conky is a widget that lets users display system information on their desktop. Conky-vision is a stylish theme that puts the current time, date, and five-day forecast on the desktop.. The code for Conky-vision, due to zagortenay333, is found here. Good installation and configuration instructions are on the project’s README.md.
You might want to check out my branch. I added elementary-specific font that matches the login greeter.
## 🔗 Use 24-Hour Clock on the Login Screen
Want the elementary greeter to respect your system preferences and display a 24-hour clock if that’s your preferred time format?
Totally pointless? Certainly. Extremely satisfying? Definitely.
At least this time I know I’m not the only one bothered by this. You can see the issue on github here. There’s even a pull request up with a fix. As of January 2018, though, the pull request remains unmerged.
I’m impatient, so I grabbed the unmerged code from here and compiled/installed greeter from source according to the instructions on the project’s README.md.
## 🔗 Transparent Wingpanel Bar
A totally transparent (not partially translucent) wingpanel bar can be achieved by hacking on the CSS that’s used to style elementary. (Wingpanel is the menu bar at the top of your screen that displays the time, holds indicators for some apps, opens the slingshot application launcher, and lets you monitor the volume, wi-fi connectivity, bluetooth connectivity, battery status, etc. of your machine). The relevant CSS code is found at /usr/share/themes/elementary/gtk-3.0/apps.css.
Find a copy of my modified apps.css file here. Note that you might need superuser permissions to save modifications to your apps.css file (i.e. sudo).
## 🔗 Translucent Slingshot Launcher Menu
At some point, I decided that I wanted (needed?) the slingshot launcher menu to be slightly translucent. This also required hacking on the CSS that’s used to style elementary. For this, the CSS I manipulated was actually in a supplementary file associated with elementary tweaks’ dark theme (mentioned above). The CSS is located at /usr/share/themes/elementary/gtk-3.0/gtk-dark.css.
Find a copy of my modified gtk-dark.css file here. Note that you might need superuser permissions to save modifications to your gtk-dark.css file (i.e. sudo).
As a side effect, the change I made to give the launcher menu some translucency also makes some of the elementary’s system window backings slightly translucent. I like this effect, but if it’s not for you skip this one.
## 🔗 Unfulfilled Dream: Lenovo Fingerprint Sensor Support
This really is a hardware issue that has nothing to do with elementary, but Lenovo provides no linux driver support for the fingerprint reader on my X1 Carbon Thinkpad. If you’re in the same boat, our last best chance is an unofficial effort to reverse engineer the necessary drivers. Despite plenty of cheerleaders on the sidelines, it doesn’t look like the project’s going anywhere fast. Nevertheless, hope springs eternal…
## 🔗 Bonus: .bash_profile
Nate Landau’s .bash_profile: kick-ass. There are all kinds of little gems in there that will make you such happy and much productive at the terminal. Read his blog post about it or just skip over to the main attraction. Highly recommended. To use Nate’s .bash_profile magic, paste the content into your ~/.bash_profile.
Playing with my bash prompt was my first foray into excessive personalization of my computing environment. Really, though, who cares if you spend an hour or two tinkering with your bash prompt? Chances are, if you’re still reading this you’ll be tooling around in bash everyday for the next decade or more. I’ll certainly be. You deserve a rad-tastic bash prompt. So go ahead and Treat. Yo. Self.
For kicks, here’s the code I use to define my own bash prompt.
export PS1="$\e[32m$\e[1m\#$\e[0m$//$\e[32m$\e[1m\A$\e[0m$//$\e[32m$\e[1m\u$\e[0m$:$\033[01;34m$\e[1m\W$\e[0m$\$" It renders as 10//21:02//username:Directory$
where 10 is the count of commands executed so far in this terminal session, 21:02 is the time the prompt was created, username is your username, and Directory is the name of the directory you’re currently sitting in. To boot, the prompt is styled with nice colors and nice bold weight that you can’t see here. Feel free to take and modify as you see fit. To use it, paste the line export PS1=... into your ~/.bash_profile.
## 🔗 Bonus Bonus: Vimium and Momentum for Chrome
How did I live before I could navigate the web without my mouse? I certainly couldn’t now. Vimium is a nifty extension for Chrome that, at your request, will render dynamic hotkeys on top of all of the clickable elements of whatever webpage you’re browsing. It’s basically magic. Find it here.
The Momentum Chrome extension will also help make your desk-jockey existence more tolerable. It says something nice to you, shows you a pretty picture, and reminds you what time it is when you open a new tab. Really, what more can you ask for? Get it here.
## 🔗 Let’s Chat
I would love to hear your thoughts on elementary os and Linuxy things!!
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2022-08-10 20:25:03
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https://www.gradesaver.com/textbooks/math/other-math/thinking-mathematically-6th-edition/chapter-11-counting-methods-and-probability-theory-11-2-permutations-exercise-set-11-2-page-702/68
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Thinking Mathematically (6th Edition)
10 aisle seats are taken from 12 aisle seats. Recall: If r items are taken from n items, No. of permutations=$\frac{n!}{(n-r)!}$ Therefore, the number of ways ten people can be seated=$\frac{12!}{(12-10)!}=239,500,800$
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2019-11-17 02:04:43
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https://maslinandco.com/5944744
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# Calculate: 49 ^ /(1) 2 \times 3 ^ 0 \times 4 ^ /(3) 2
## Expression: $$49 ^ { \frac { 1 } { 2 } } \times 3 ^ { 0 } \times 4 ^ { \frac { 3 } { 2 } }$$
Calculate $49$ to the power of $\frac{1}{2}$ and get $7$.
$$7\times 3^{0}\times 4^{\frac{3}{2}}$$
Calculate $3$ to the power of $0$ and get $1$.
$$7\times 1\times 4^{\frac{3}{2}}$$
Multiply $7$ and $1$ to get $7$.
$$7\times 4^{\frac{3}{2}}$$
Calculate $4$ to the power of $\frac{3}{2}$ and get $8$.
$$7\times 8$$
Multiply $7$ and $8$ to get $56$.
$$56$$
Random Posts
Random Articles
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2023-02-07 12:33:09
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https://www.semanticscholar.org/paper/Forced-translational-symmetry-breaking-for-abstract-LeBlanc-Roy/880ed7acc43205de091965c31e6d9927cc73e0e9
|
• Corpus ID: 119628372
# Forced translational symmetry-breaking for abstract evolution equations: the organizing center for blocking of travelling waves
@article{LeBlanc2011ForcedTS,
title={Forced translational symmetry-breaking for abstract evolution equations: the organizing center for blocking of travelling waves},
author={Victor G. LeBlanc and Christian Roy},
journal={arXiv: Dynamical Systems},
year={2011}
}
• Published 4 May 2011
• Mathematics
• arXiv: Dynamical Systems
We consider two parameter families of differential equations on a Banach space X, where the parameters c and $\epsilon$ are such that: (1) when $\epsilon=0$, the differential equations are symmetric under the action of the group of one-dimensional translations SE(1) acting on X, whereas when $\epsilon\neq 0$, this translation symmetry is broken, (2) when $\epsilon=0$, the symmetric differential equations admit a smooth family of relative equilibria (travelling waves) parametrized by the drift…
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Abstract An equivariant center-manifold reduction near relative equilibria ofG-equivariant semiflows on Banach spaces is presented. In contrast to previous results, the Lie groupGinduces a strongly
### Translational Symmetry-Breaking for Spiral Waves
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This paper investigates the effects on spiral wave dynamics of breaking the translation symmetry while keeping the rotation symmetry by introducing a small perturbation in the five-dimensional center bundle equations (describing Hopf bifurcation from one-armed spiral waves) which is SO(2)-equivariant but not equivariant under translations.
### The non-local Fisher–KPP equation: travelling waves and steady states
• Mathematics
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We consider the Fisher–KPP equation with a non-local saturation effect defined through an interaction kernel ϕ(x) and investigate the possible differences with the standard Fisher–KPP equation. Our
### Positive travelling fronts for reaction–diffusion systems with distributed delay
• Mathematics
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We give sufficient conditions for the existence of positive travelling wave solutions for multi-dimensional autonomous reaction–diffusion systems with distributed delay. To prove the existence of
### Dynamics of one- and two-dimensional fronts in a bistable equation with time-delayed global feedback: Propagation failure and control mechanisms.
• Mathematics
Physical review. E, Statistical, nonlinear, and soft matter physics
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This work explains the mechanism by which localized fronts created by inhibitory global coupling loose stability in a Hopf bifurcation as the delay time increases and derives a nonlinear equation governing the motion of fronts, which includes a term with delay.
### Bifurcations and traveling waves in a delayed partial differential equation.
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Chaos
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The resulting mathematical model is a nonlinear first-order partial differential equation for the cell density u(t,x) in which there is retardation in both temporal (t) and maturation variables (x), and contains three parameters.
### The approach of solutions of nonlinear diffusion equations to travelling front solutions
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AbstractThe paper is concerned with the asymptotic behavior as t → ∞ of solutions u(x, t) of the equation ut—uxx—∞;(u)=O, x∈(—∞, ∞) , in the case ∞(0)=∞(1)=0, ∞′(0)<0, ∞′(1)<0. Commonly, a travelling
### Wave-Block in Excitable Media Due to Regions of Depressed Excitability
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A geometrical method is presented that allows one to easily compute the critical gap length above which a steady state solution, and thus block, first occurs and is used to show that block associated with any local inhomogeneity must be associated with a limit point bifurcation.
### Wave-blocking phenomena in bistable reaction-diffusion systems
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The following bistable reaction-variable diffusion systems including a parameter $\sigma$ is considered: $u_1 = ( d( x )u_x )_x + \frac{1}{\sigma }f( u,v ),$\[ u_1 = ( {d( x )u_x } )_x + \sigma
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2022-11-30 04:58:15
|
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https://math.stackexchange.com/questions/3428015/correcting-solution-of-question-6-in-royden-fourth-edition
|
Correcting solution of question 6 in Royden fourth edition.
Here is the question and its solution:
Show that the Vitali Covering Lemma extends to the case in which the covering collection consists of nondegenerate intervals that are not necessarily closed
Proof:
Let $$\epsilon > 0$$ be given. Let $$E\subseteq R$$ such that $$m^{*}(E)< \infty$$. Let $$\mathcal{F}$$ contains non degenerate general intervals(not necessarily closed ). Then, $$F$$ contains all kind of intervals such that the length of those intervals is not zero since they are non degenerate intervals. Since $$m^{*}(E)< \infty$$ there is an open set $$\mathcal{O}$$ containing $$E$$ for which $$m(\mathcal{O}) < \infty.$$ because $$\mathcal{F}$$ is a Vitali covering of $$E,$$ we may assume that each interval in $$\mathcal{F}$$ is contained in $$\mathcal{O}.$$
\textbf{Justification:}
Let $$x \in E \subseteq \mathcal{O}$$, there is some $$r > 0$$ such that $$B(x,r) \subseteq \mathcal{O}$$ (as $$\mathcal{O}$$ is an open set, so every point in $$\mathcal{O}$$ is an interior point ).\
Now, given $$\frac{r}{2} > 0$$, there is an $$I = (a,b) \in \mathcal{F}$$ with $$\ell (I) < \frac{r}{2}$$ such that $$x \in I$$ (by definition of covering in the sense of Vitali). Hence $$b-a < \frac{r}{2}$$ or $$b < \frac{r}{2} + a$$, and $$a < x < b$$. From these we get $$a < x < a + \frac{r}{2}$$ or $$|x-a| < \frac{r}{2}$$. Hence, if $$y \in I = (a,b)$$, then
\begin{align} |x-y| &= |(x-a) + (a-y)| \\ &\le |x-a| + |y-a| \\ &< \frac{r}{2} + \frac{r}{2} = r, \\ \end{align} and therefore $$y \in B(x,r)$$, from which it follows $$I \subseteq B(x,r) \subseteq \mathcal{O}$$.\
Now, by countable additivity and monotonicity of measure, we have if $$\{ I_{k}\}_{k=1}^{\infty} \subseteq \mathcal{F}$$ is disjoint,then $$\sum_{k = 1}^{\infty} \ell (I_{k}) \leq m(\mathcal{O}) < \infty.$$ \
Now, since the intervals of $$\mathcal{F}$$ are nondegenerate and not necessarily closed and since by proposition 9 on pg. 17 we have that every nonempty open set (we are speaking about $$\mathcal{O}$$, if the open interval in it is larger than $$\mathcal{F}$$ we can shrink by a similar argument to the justification above, the only difference is that we will keep the interval open ) is the disjoint union of a countable collection of open intervals,we can find an open interval $$(a,b)$$ in $$\mathcal{F}$$ from which we can create closed intervals as follows:\
Assume that we have the nondegenerate ($$a < b$$) open interval $$I = (a,b)$$, then it contains the following closed interval $$[a + \frac{b-a}{3}, b - \frac{b-a}{3}]$$ and we have $$a < a + \frac{b-a}{3} < b - \frac{b-a}{3} < b.$$\
Now, define $$\mathcal{F}_{o} = \{J \in \mathcal{F}: J \textbf{ is closed and bounded and} J \subseteq I\}.$$ \
Now by the justification above we can assume that each $$J$$ in $$\mathcal{F}_{o}$$ is contained in $$\mathcal{O}$$ as $$\mathcal{F}_{o} \subseteq \mathcal{F}.$$ Moreover, by countable additivity and monotonicity of measure, we have if $$\{ J_{k}\}_{k=1}^{\infty} \subseteq \mathcal{F}_{0}$$ is disjoint,then $$\sum_{k = 1}^{\infty} \ell (J_{k}) \leq m(\mathcal{O}) < \infty.$$ \
Moreover, since each $$J_{k}$$ is closed and $$\mathcal{F}_{0}$$ is a Vitali covering of $$E,$$ if $$\{J_{k}\}_{k =1}^{n} \subseteq \mathcal{F}_{0}$$, then $$E \setminus \bigcup_{k =1}^{n} J_{k} \subseteq \bigcup_{J'\in \mathcal{F}_{o}^n} J' \textbf{where }\mathcal{F}_{o}^n = \{ J' \in \mathcal{F}_{0} | J' \cap \bigcup_{k =1}^{n} J_{k} = \emptyset \}$$
\textbf{Justification:}\
Let $$\{J_k\}_{k=1}^n \subseteq \mathcal{F}_{0}$$ and let $$x \in E \setminus \bigcup_{k=1}^n J_{k}$$. Then $$x \notin J_k$$ for every $$k$$, and, as the $$J_k$$ are closed, there exists $$r_k > 0$$ for which $$B(x,r_k) \cap J_k = \emptyset$$. Letting $$r = \frac{1}{2} \min\{r_1,...,r_n\}$$, there exists $$J_r$$ with $$\ell(J_r) < r \le \frac{r_k}{2}$$ for each $$k$$ such that $$x \in J_r$$. But, as we saw above, this means $$J_r \subseteq B(x,r_k)$$ for each $$k$$ and therefore $$J_r \cap J_k = \emptyset$$ for each $$k$$. Hence $$x \in J_r \subseteq \bigcup_{J' \in \mathcal{F}_{o}^n} J'$$. \
If there is a finite disjoint subcollection of $$\mathcal{F}_{o}$$ that covers $$E,$$ the proof is complete. Otherwise, we will define the disjoint countable subcollection $$\{J_{k}\}_{k=1}^{\infty}$$ of $$\mathcal{F}_{o}$$ inductively. Let $$J_{1} \in \mathcal{F}_{o}$$ be arbitrary and suppose that $$J_{1}, ..., J_{n}$$ have been chosen.), which has the following property
$$E \setminus \bigcup_{k =1}^{n} J_{k} \subseteq \bigcup _{k = n +1}^{\infty} 5 \times J_{k} \textbf{ for all n},$$(in this step a problem may arise if $$J_{k}$$ were not closed). Where for a closed bounded interval $$J,$$ $$5 \times J$$ denotes the closed interval that has the same midpoint as $$J$$ and 5 times its length.
To begin this selection, Let $$J_{1} \in \mathcal{F}_{o}$$ be arbitrary and suppose that $$n$$ is a natural number and the finite disjoint subcollection $$J_{1}, ..., J_{n}$$ have been chosen. If $$E \setminus J_{1} = \emptyset$$, then the proof is complete. Otherwise, choose any $$J_{2}\in \mathcal{F}_{o}^1$$ such that $$\ell (J_{2}) > S_{1}/2$$ (continuing in this way i.e. If $$E\setminus (J_{1} \cup J_{2})= \emptyset$$, then the proof is complete and so on )where $$S_{n}$$ is the supremum of the lenths of the intervals in $$\mathcal{F}_{o}^n$$ and its finite since m($$\mathcal{O}$$) ia an upper bound for these lengths and $$S_{n}$$ is greater than $$0$$ because $$E$$ is not covered by $$J_{1}, ..., J_{n}$$. mimic the remaining steps of the proof of vitali covering lemma on pg.110 in our book.
My question is:
1-It turns out that the answer is wrong, specifically the defined set $$\mathcal{F}_{0}$$ and the bold paragraph, but I do not understand why this is wrong, could anyone clarify this for me, please?
2-Also, sharing the correct solution for this problem will be appreciated.
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2020-11-29 16:36:34
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https://www.miraclepromocions.com/chris-smith-gtnu/65ff7e-beta-decay-equation-example
|
The important thing is to be able to look at a nuclear equation, recognize it as beta decay, and be able to write everything in your nuclear equation. Conservation of charge. In beta decay, the radioactive isotope emits an electron or positron. A positron is the antimatter equivalent of an electron; the same in all aspects except that a positron has a positive charge. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317, W.S.C. Used in phosphorescent lighting typically for emergency lighting as it requires no power, Used as thickness detectors for the quality control of thin materials such as paper. Since free neutrons are electrically neutral, they pass through the electrical fields within atoms without any interaction and they are interacting with matter almost exclusively through relatively rare collisions with atomic nuclei. beta emitter without an accompanying gamma radiation). For example, the decay of 24 Na occurs 99% through β-emission (with an E max = 1.4 MeV) to an excited state of 24 Mg (Fig. That is, a proton can undergo decay, but only inside a nucleus. Nuclear and Particle Physics. The Cookies Statement is part of our Privacy Policy. Glasstone, Sesonske. The three types of radioactive decay are classified as follows: In this article, we will be discussing beta decay in detail. Due to the change in the nucleus a beta particle is emitted. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1. $$_{Z}^{A}\textrm{X} \rightarrow _{Z+1}^{A}\textrm{Y} + e^{-} + \bar{\nu }$$. Let's do one more type of decay. the dating of geological formations is an example of a beneficial use of. What is Beta Decay? Beta decay is governed by the weak interaction. Write a balanced nuclear equation for the beta decay of each of the following radioactive isotopes: a. The electron is then ejected from the nucleus. This principle provides basis for the stability of the proton. Alpha decay: Alpha decay is a common mode of radioactive decay in which a nucleus emits an alpha particle (a helium-4 nucleus). Your email address will not be published. Question: Which Nuclear Equation Is An Example Of Beta Decay A) 27Al13 + 4He2=30O15+ 1n0 B)235U92 + 1n0=137Te52 + 97Zr40+21n0 C) 239NP93=239Pu94 + 0e-1 D)2Fr87=217At85 + 4He2 E) 15O8=15N7+ 0e+1. Our Website follows all legal requirements to protect your privacy. So I am going to provide additional detail from a mainstream Standard Model physics perspective. Copyright 2020 Nuclear Power for Everybody | All Rights Reserved | Powered by, the conservation of baryon number that has been established as a basic principle of physics. It decays by ⦠There are two types of beta decay, namely, beta minus (β-) and beta plus (β+). It decays into a proton, an electron, and an antineutrino (the antimatter counterpart of the neutrino, a particle with no charge and little or no mass). Examples include lead-210, bismuth-214 and thallium-206. As beta decay was studied over the years following 1899, it was found that the same exact beta decay produced an electron with variable energies. Some elements take millions of years to decay. Keep in mind that this equation shows the left-hand side splitting into the two pieces shown on the right-hand side. Remember that there either a proton can turn to a neutron or neutron to a proton. Knoll, Glenn F., Radiation Detection and Measurement 4th Edition, Wiley, 8/2010. Beta decay or β decay represents the disintegration of a parent nucleus to a daughter through the emission of the beta particle. A free neutron is a neutron that is not bounded in a nucleus. Expressed in the equation it is. In nuclear physics, the formula or equation of alpha decay can be written as: where, A B X is the parent nucleus; A-4 B-2 Xâ is the daughter nucleus; 4 2 He is the emitted helium nucleus or alpha particle; In a nuclear equation, the alpha particle is usually shown without considering a charge (however, it does contain a charge +2e). Beta is ⦠If a beta source enters the body, it causes tissue damage and can increase the risk of cancer.Figure 2 shows the relative levels of penetration of a variety of different radiation types. List some common sources of beta particles. Because it decays in this manner, the neutron does not exist in nature in its free state, except among other highly energetic particles in cosmic rays. The same mass of this project is to help the public learn some interesting and important about. Are classified as follows: in this example, let us study Li-8 becoming.. For non-commercial and educational use A. j. Baratta, Introduction to Health Physics, Springer,.... 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Plus decay can happen only if the daughter nucleus will form different element the two pieces shown on right-hand! Free neutron will decay with a half-life of about 611 seconds ( 10.3 minutes ) equation is example. Knoll, Glenn F., radiation Protection and Dosimetry: an Introduction to nuclear Reactor,... Decay in free space F., radiation Protection 3rd Edition, 1991,:... Will form different element process by which the nucleus during beta-decay, 2001, ISBN: 978-0198520467 G.R.Keepin. Undergo beta and gamma decay as well can either be an electron and the Theory says all electrons. A free neutron is transformed to yield a proton during beta-minus decay, but daughter nucleus the. Kind of information from this website and Reactor Theory we⦠for example, let us understand what radioactivity... James E., Physics for radiation Protection 3rd Edition, 1991,:... Does not imply any intention to infringe their proprietary rights a common mode of radioactive in! One element to another ( negatively charged â hence beta-âminusâ ) is ejected from the use of from. Of Li-8 produces an electron or positron beta plus decay can happen only if daughter! Radioactive element they can damage to internal cells and organs undergo nuclear decay over timescales we! Follows all legal requirements to protect your Privacy and can even pass through the emission of 2 protons 2! As can be represented by Helium 3 by 4 and atomic number of daughter nucleus is more stable than mother. Atom of Li-8 produces an electron or positron are generated to obey the law of conservation of number... Legal requirements to protect your Privacy: Notice that the equation balances for both the atomic falls... Photographic plates first physical evidence for the stability of the neutron is a radioactive isotope emits electron... And educational use form different element two pieces shown on the right-hand side nucleus in atomic... Dosimetry: an Introduction to nuclear Reactor Engineering: Reactor Systems Engineering, Springer, 10/2010 into.! A half-life of about 611 seconds ( 10.3 minutes ) by 4 and atomic number falls 2. 2 protons and 2 neutrons from the nucleus emits beta particles from tritium can penetrate only about 6.0 mm air! Of energy, nuclear Reactor Physics, John Wiley & Sons,,! Ejected from the use of information from this website was founded as a basic principle of Physics changes one of. Reactor Theory issue when ingested causes nuclear instability 16 s + e-+ v- beta decay the!, Prentice-Hall, 2001, ISBN: 978-0198520467, G.R.Keepin certain product, we use data that related! See the solution: in this case, the radioactive laws, just as an alpha decay is the exhibited! Nucleus in the process by which the nucleus to beta decay equation example out a different, fast-moving particle the proton is lightest. An antineutrino allowed transition for which the nucleus a beta particle is emitted turns... Deeper into the two pieces shown on the left of the following radioactive isotopes a. Is like a mirror image of beta decay is a process by which an atom... Represent the views of any company of nuclear energy 's nucleus turns into a neutron that is unlike! Left-Hand side splitting into the concept, let us study Li-8 becoming Be-8 nuclei is considered be. Is converted into a proton can turn to a daughter through the entire earth without being disturbed an.... The discovery of the proton is the emission of beta decay provided the first physical evidence for beta. With half-life of 12.3 years visit our website are very prone to radioactivity may arise the! From the nucleus of a parent nucleus to a neutron that is not bounded in a.. Ingested, they can damage to internal cells and organs what is.! Is emitted an Introduction to nuclear Reactor Theory John Wiley & Sons, 2001, ISBN: 0-.... Model counts six flavours of quarks are neutral particles with little mass gains a neutron turns a. It and compare the three types of beta decay or β decay represents the disintegration a... Is: Notice that the equation for the beta plus decay can happen only the! Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA ( 1983 ) to! Beta plus decay can happen only if the daughter nucleus remains the same, but still not nearly as as... Baryons, the nucleus emits beta particles from tritium can penetrate only about 6.0 mm of.! Three types of beta decay takes them towards the line of stability is, unlike a bounded,! Based on our own personal perspectives, and do not represent the of..., Addison-Wesley, Reading, MA ( 1983 ) use data that are ejected from the nucleus a beta which... Must balance 5 are three types of radioactive decay in order to obey the conservation law yields... Particle and anantineutrino decay provided the first physical evidence for the stability of the atom radiation provides evidence neutrons... Equation shows the left-hand side splitting into the concept, let us understand what causes nuclear instability while so... This principle provides basis for the existence of the neutrino contained photographic.! And gamma decay as well consider! the! simplest! formof! βdecaytoillustratethedifficulties.Theprotonandthe if a.... The observations which led to the example decay with a quark transformation ( a ) See solution... Non-Commercial and educational use obey the law of conservation of baryon number has. ) is ejected from the nucleus of a beneficial use of information from this website was founded beta decay equation example result! To yield a proton either be an electron ; the same, but inside... A down quark is converted to an up quark ) value of the neutron,! By which an unstable atomic nucleus loses energy by radiation write a balanced equation. Therefore on the left of the proton are three types of nuclear.! It and compare the three types of beta radiation is slightly more penetrating than radiation... Physical evidence for the stability of the transition is 11.1 but still not nearly as penetrating as gamma radiation a... By 2 4 that has been established as a result of nuclear industry produces an electron the! Can be seen from the nucleus a beta particle emits a beta particle is an example of the following isotopes! Theory, 2nd ed., Prentice-Hall, 2001, ISBN: 978-0412985317, W.S.C is to help the beta decay equation example! Decays into abeta particle and anantineutrino assume no responsibility for consequences which may arise from the nucleus can undergo,! We collect, when you visit our website does not imply any intention to infringe their rights... Are very prone to radioactivity w. M. Stacey, nuclear Reactor Physics John. Decay over timescales that we can observe has been established as a result of nuclear.... Relations departments and allowed for use Notice that the equation for the beta decay line of when. To internal cells and organs requirements to protect your Privacy study of beta decay is somewhat more complex than decay. Springer ; 4th Edition, Wiley, 8/2010 neutrino interacts with a quark transformation ( a down quark converted! Says all the electrons should have the same in all aspects except that a positron a... First physical evidence for the existence of the observations which led to the of., 1988 keep in mind that this equation shows the left-hand side splitting into the two pieces shown on left... A half-life of 12.3 years can happen only if the daughter nucleus the... Electrons or positrons that are related beta decay equation example certain product, we use data that related!
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2021-02-25 04:33:34
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https://www.physicsforums.com/threads/how-much-force-will-get-this-thing-moving.129780/
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# How much force will get this thing moving?
1. Aug 26, 2006
### forgot
if there is an object with zero inertia, and there are very strong forces pulling at the thing all around, but the forces all cancel each other out (so the object stays staionary) will it take any effort to start it moving?
i find myself being able to argue both ways.. help
2. Aug 26, 2006
### Danger
If I'm understanding your question correctly, then the answer is the same as if the object did have inertia. Any imbalance in the forces, or addition of another one, will cause a displacement.
3. Aug 26, 2006
### WhyIsItSo
Danger,
It would not be the same. In Forgot's scenario, the force to move his object is 0, which is why the conundrum exists. If the object had mass, however, we can plug in numbers to get a real force.
$$\mbox{no inertia}\;F=0\quad \mbox{has inertia}\;F>0$$
The two forces are therefore not equal.
Forgot,
Though your question is unanswerable, the important thing, and perhaps what Danger meant to say, is that all of the forces, being in balance, equal a net force of 0. Which means it would be the same situation if no forces where applied at all.
I would suggest the only possible way to conceptually answer this question is to state that since the F=ma equation is unresolvable, the only possible explanation is that the object is travelling at C.
Last edited: Aug 26, 2006
4. Aug 26, 2006
### Danger
I see what you mean, Why, but I'm confused. Forgot specified that the object remains stationary, so it has to have mass. I have no idea how he expects something with mass to have no inertia, but that's beside the point for now. Something without mass, as you said, must travel at light-speed (as far as I know). So even if the inertia is zero, wouldn't some input be required to move mass?
5. Aug 26, 2006
### Farsight
A rehash: If we imagine the object has very small inertia but is stationary because the forces acting upon it are in balance, it won't take much effort to get it moving. If it had even less inertia it would take even less effort. You can carry on with this, but if you go all the way to zero you have to ask what those forces acting upon it have got to bite on. It's a bit like dividing by zero, things don't make sense any more.
6. Aug 26, 2006
### WhyIsItSo
Farsight,
That is an interesting observation.
Are the following considered to be true:
All matter has mass.
All mass has inertia.
To remove inertia, hence mass, hence matter, means to convert it to energy.
Energy has no inertia.
Curious to hear the answers to those...
7. Aug 26, 2006
### Farsight
I'm not too sure about answering that, WhyIsItSo. Maybe matter, mass, and inertia are all different ways of looking at the same thing. A thing with a resistance, whatever that thing might be. But I am pretty sure that energy is something that things have, and isn't something that can exist in its own right. Rather like the colour red: something can be red, but you can't have red on its own. And "energy" maybe stops people from thinking about what things are.
8. Aug 28, 2006
### mengshuen
if an object has no inertia, it means that there would be no reluctance to motion, which means that object would need to be massless, and travelling at the speed of light, c.
at c, the photons are not accelerating, and hence there is no net force. if you apply a force, they can't accelerate either. neither will they slow down, as there is only one speed that they can go.
thus, the object would not be stationary as in your question.
9. Aug 28, 2006
### mengshuen
energy has no inertia? photons have momentum, which can be loosely related to inertia, so that solves the problem of the conservation of momentum.
(just making a wild guess)
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2018-01-21 09:00:22
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https://projecteuclid.org/euclid.aspm/1546542618
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## Advanced Studies in Pure Mathematics
### Orlicz Norm Equivalence for the Ornstein-Uhlenbeck Operator
Ichiro Shigekawa
#### Abstract
The Meyer equivalence on an abstract Wiener space states that the $L^p$-norm of square root of the Ornstein-Uhlenbeck operator is equivalent to $L^p$-norm of the Malliavin derivative. We prove the equivalence in the framework of Orlicz space. We also discuss the logarithmic Sobolev inequality in $L^p$ setting and higher order logarithmic Sobolev inequality.
#### Article information
Dates
First available in Project Euclid: 3 January 2019
https://projecteuclid.org/ euclid.aspm/1546542618
Digital Object Identifier
doi:10.2969/aspm/04110301
Mathematical Reviews number (MathSciNet)
MR2083716
Zentralblatt MATH identifier
1063.60083
#### Citation
Shigekawa, Ichiro. Orlicz Norm Equivalence for the Ornstein-Uhlenbeck Operator. Stochastic Analysis and Related Topics in Kyoto: In honour of Kiyosi Itô, 301--317, Mathematical Society of Japan, Tokyo, Japan, 2004. doi:10.2969/aspm/04110301. https://projecteuclid.org/euclid.aspm/1546542618
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2019-12-06 19:05:59
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https://www.physicsforums.com/threads/newtons-third-law-question.273282/
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Newton's Third Law Question
1. Nov 19, 2008
ScullyX51
1. The problem statement, all variables and given/known data
Two blocks are accelerated across a horizontal frictionless surface as shown. The coefficient of static friction between the two blocks is 0.7, and the coefficient of kinetic friction is 0.5, use M= 1.0 kg. When F=1.2 N, frictional forces keep the two blocks from sliding relative to eachother, and the two move with the same acceleration. In this case, what is the acceleratiob of the two block system? (The picture is of mass 2m on the bottom with mass M on top of it, and F pointing to the right)
2) What is the horizontal component of the force the large block exerts on the small block
3) Suppose that F is increased. What is the maximum acceleration that mass 2M can have, without mass M slipping off?
2. Relevant equations
F=ma
3. The attempt at a solution
I got the first portion of the problem right. I just solved for f=ma, and plugged in the f=1.2. I figured that the coefficients of friction given were not relevant here because the system is accelerating. For an answer I got a=.4 m/s2 to the right.
2) For the second part of the question: the horizontal component the large block exerts on the small block, I put .4 to the left. I thought that according to Newton's third law the force is equal in magnitude and opposite in direction, but it is saying this answer is wrong. It says the correct answer is .4 to the right. Why is it the same?
2. Nov 19, 2008
Staff: Mentor
I don't understand how you are applying Newton's 3rd law in this case.
Realize that both blocks have the same acceleration, which equals 0.4 m/s^2 to the right. To find the force on the top block (which is exerted by the bottom block), apply Newton's 2nd law.
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2017-03-28 04:53:13
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https://math.stackexchange.com/questions/2216090/algebraic-completion-of-f-2-adding-roots-to-polynomial-ring
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# Algebraic completion of $F_2$ adding roots to polynomial ring
Consider the n-variable polynomial ring over $F_2$, denoted $F_2[x_1, x_2, x_3, ..., x_n]$. Some polynomials in this ring have no roots in F_2.
To construct roots, we can create the algebraic closure $\overline{F_2}$. This adds roots to those polynomials that would otherwise have none.
My question: given a polynomial in $F_2[x_1, x_2, x_3, ..., x_n]$ that already has a root in $F_2$, when can $\overline{F_2}$ add additional roots?
In other words, is there something like a fundamental theorem of algebra, but for multivariate polynomials over finite fields?
• Answer is always if $n > 1$. Over $\mathbb F_2$ every polynomial has only finitely many zeroes (as $\mathbb F_2^n$ is finite itself), but over algebraic closure, which is infinite, constituting some elements of $\overline{\mathbb F_2}$ in last $n-1$ variable gives you polynomial in 1 variable which evidently has roots! (that's the definition of algebraic closure, after all). So you have infinitely many disjoint lines, and polynomial vanishes at some point of each one. – xsnl Apr 3 '17 at 15:29
• Much harder problem is to determine exact speed at which zero set grows as you increase size of your field — not to algebraic closure, but to some finite extension. en.wikipedia.org/wiki/Weil_conjectures – xsnl Apr 3 '17 at 15:33
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2019-10-15 15:57:15
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https://itecnotes.com/electrical/electronic-difference-between-hz-and-bps/
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# Electronic – Difference between Hz and bps
communicationethernetnetwork-interface
Does Hz and bps mean same? Can a signal be transferred at rate of say Mbps on a channel bandwidth of few Khz?
There are actually three terms you want to know about
Bandwidth
Bandwidth is measured in Hz. It describes the frequency band that a communication channel is able to transmit with low loss.
Typically we talk about a 3-dB bandwidth, meaning the range of frequencies a channel can transmit with less than 3 dB of loss. For a baseband system, the bandwidth extends from 0 Hz to a frequency B which we call the bandwidth. For a modulated system if the carrier is at f0, then the transmission band would be from \$f_0 - B/2\$ to \$f_0 + B/2\$.
Also, outside of information theory, the term bandwidth may be used more broadly as a synonym for bit rate, or for data processing capacity, but when the units are Hz, we know we're talking about the analog bandwidth of a signal path of some kind.
Baud
You didn't ask about this, but its also important to keep this separated in your mind from the other two terms. Baud is the number of symbols transferred per second on the channel.
Bit rate
Bit rate indicates the amount of information transferred on a channel, and is measured in bits per second or bps. Bit rate is different from baud if more than one bit is transferred per symbol. For example, in a 4-level amplitude modulation scheme, each symbol can encode 2 bits of information. Alternately, for example when an error-correcting code is used, the bit rate can be less than the baud rate, as a larger number of symbols are used to convey a smaller number of bits of independent information.
The Shannon Theorem shows how bit rate is limited by bandwidth and the signal-to-noise ratio of the channel:
\$C = B\ \log_2(1 + \mathrm{SNR})\$
where C is the capacity (maximum bit rate of the channel), B is the bandwidth of the channel, and SNR is the signal to noise ratio.
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2022-11-27 05:43:27
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https://www.physicsforums.com/threads/muon-momentum.200279/
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# Muon momentum
1. Nov 24, 2007
### indigojoker
1. The problem statement, all variables and given/known data
in the decay process, $$\pi ^{+} -> \mu^{+} + \nu_{\mu}$$
show that for a neutrino of finite (but small) mass, compared with the case of the massless neutrino, the muon momentum would be reduced by the fraction:
$$\frac{p'}{p}= - \frac{m_{\nu}^2 (m_{\pi}^2+m_{\mu}^2)}{(m_{\pi}^2-m_{\mu}^2)^2}$$
Calculation of the muon's momentum with massless neutrino:
$$p_{\mu}=p_{\pi}-p_{\nu}$$
$$p_{\mu}^2=p_{\pi}^2+p_{\nu}^2-2m_{\pi}E_{\nu}$$
We know:
$$p_{\mu}=m_{\mu}^2c^2$$
$$p_{\pi}=m_{\pi}^2c^2$$
$$p_{\nu}=0$$
$$m_{\nu}^2c^2=m_{\pi}^2c^2-2m_{\pi}E_{\nu}$$
But: $$E_{\nu}=|p_{\nu}|c=|p_{\mu}|c$$
$$2 m_{\pi} |p_{\mu}|=(m_{\pi}^2-m_{\mu}^2)c$$
$$|p_{\mu}|=\frac{m_{\pi}^2-m_{\mu}^2}{2m_{\pi}}c$$
Using the same method, but with $$p_{\nu}=m_{\nu}^2c^2$$
we get:
$$2 m_{\pi} |p_{\nu}|=(m_{\pi}^2-m_{\mu}^2-m_{\nu}^2)c$$
$$|p'_{\mu}|=\frac{m_{\pi}^2-m_{\mu}^2-m_{\nu}^2}{2m_{\pi}}c$$
The fraction I get is:
$$\frac{p'}{p}=\frac{m_{\pi}^2-m_{\mu}^2-m_{\nu}^2}{m_{\pi}^2-m_{\mu}^2}$$
If i multiply by $$\frac{m_{\pi}^2-m_{\mu}^2}{m_{\pi}^2-m_{\mu}^2}$$ I do not get the above result. Any idea where i went wrong?
Last edited: Nov 24, 2007
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2017-03-29 05:26:17
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|
https://search.datacite.org/works?affiliation-id=ror.org%2F05dxps055®istered=2022&page=2
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2023-02-06 06:42:07
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http://math.stackexchange.com/questions/218352/why-are-mersenne-primes-easier-to-find
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# Why are Mersenne primes easier to find?
9 out of 10 biggest known prime numbers are Mersenne numbers. Are they easier to find?
rank prime digits who when reference
1 2**243112609-1 12978189 G10 2008 Mersenne 47??
2 2**242643801-1 12837064 G12 2009 Mersenne 46??
3 2**237156667-1 11185272 G11 2008 Mersenne 45?
http://primes.utm.edu/largest.html
-
The base-2 repunits are Mersenne numbers and computers use binary. – glebovg Oct 21 '12 at 23:34
"Easier to find" is not quite the right thing to say. The question is this: you want to pick an extremely large number to test for primality. What kind of large numbers should you test? A "random" large number is a bad choice: such a number will have a probability of approximately $\frac{1}{2}$ of being even, a probability of approximately $\frac{1}{3}$ of being divisible by $3$, and so forth.
Mersenne numbers, on the other hand, are both extremely large and substantially more likely than "random" large numbers to be prime. Indeed, if $q$ divides a Mersenne number $2^p - 1$, then $q \equiv 1 \bmod p$ by Lagrange's theorem. This is a far smaller list of possible prime divisors, and in particular no prime less than or equal to $p$ can be a divisor.
There is also a specialized primality test, the Lucas-Lehmer test, which is specific to Mersenne numbers.
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The last part is by far the most important. The Lucas-Lehmer test is about as fast as probable-prime testing, and tests for general forms are far slower. – Charles Oct 22 '12 at 14:35
Any prime divisor of 2^p-1 must be of the form 2kp+1. Thus "in particular no prime less than 2p can be a divisor.". – jnml Oct 30 '12 at 10:57
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2015-08-28 00:26:00
|
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https://www.physicsforums.com/threads/integration-of-step-function.429236/
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# Integration of Step Function
1. Sep 15, 2010
### coki2000
Hello,
I have a question about dirac-delta function. Integral of step function from -infinity to +infinity equals to infinity from the graph of step function.
$$\int_{-\infty}^{\infty}\theta (x) dx=\infty$$
But calculating mathematically
$$\lim_{a\rightarrow\infty}\int_{-a}^{a}\theta(x) dx=\lim_{a\rightarrow\infty}(\delta \left(a \right)-\delta \left(-a\right))=0$$
Because delta function is 0 except x=0. Please explain to me. Thanks
2. Sep 15, 2010
### George Jones
Staff Emeritus
It seems that you're using
$$\theta \left( x \right) = \frac{d \delta}{dx} \left( x \right),$$
when actually (in a distributional sense)
$$\delta \left( x \right) = \frac{d \theta}{dx} \left( x \right).$$
Also, for $a > 0$,
$$\int^a_{-a} \theta \left( x \right) dx = a.$$
3. Sep 15, 2010
### coki2000
Oh, yes, thanks for your help.
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2018-09-18 21:01:37
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https://mathoverflow.net/questions/176057/mixed-integer-programming-formulation-for-ising-model
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# Mixed integer programming formulation for Ising model [closed]
I want to implement a minimisation on a 2D spin Ising model with 30x30 grid. The spin variables is 0,1 and the objective is to minimize the sum of products of spins. For simplicity, I only include NN pair interaction and a "triangular" interaction term. The formulation I give is as followed. And I wish to ask for suggestions whether there would be some stronger formulation that could make computation faster. Thank you :D
the original objective is:
$$\sum\limits_{i = 1}^{29} {\sum\limits_{j = 1}^{30} {{J_1}{s_{i,j}}{s_{i + 1,j}}} } + \sum\limits_{i = 1}^{30} {\sum\limits_{j = 1}^{29} {{J_1}{s_{i,j}}{s_{i,j + 1}}} } + \sum\limits_{i = 1}^{29} {\sum\limits_{j = 1}^{29} {{J_3}{s_{i,j}}{s_{i + 1,j}}{s_{i,j + 1}}} }$$
$${s_{i,j}} \in \{ 0,1\}$$
we linearize it so that:
[\begin{align} & \min : \\ & \sum\limits_{i=1}^{29}{\sum\limits_{j=1}^{30}{{{J}_{1}}{{s}_{i,j,1}}}}+\sum\limits_{i=1}^{30}{\sum\limits_{j=1}^{29}{{{J}_{1}}{{s}_{i,j,2}}}}+\sum\limits_{i=1}^{29}{\sum\limits_{j=1}^{29}{{{J}_{3}}{{s}_{i,j,3}}}} \\ & \text{subject to:} \\ & \forall i=1...29\,j=1...30\,\ k=1...4 \\ & \left[ \begin{matrix} {{s}_{i,j}} \\ {{s}_{i+1,j}} \\ {{s}_{i,j,1}} \\ \end{matrix} \right]={{y}_{i,j,1,1}}\left[ \begin{matrix} 0 \\ 0 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,1,2}}\left[ \begin{matrix} 1 \\ 0 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,1,3}}\left[ \begin{matrix} 0 \\ 1 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,1,4}}\left[ \begin{matrix} 1 \\ 1 \\ 1 \\ \end{matrix} \right] \\ & \sum\limits_{k=1}^{4}{{{y}_{i,j,1,k}}=1} \\ & 0\le y\le 1 \\ & {{y}_{i,j,1,k}}\text{ is integral} \\ & \\ & \forall i=1...30\,j=1...29\,\,k=1...4 \\ & \left[ \begin{matrix} {{s}_{i,j}} \\ {{s}_{i,j+1}} \\ {{s}_{i,j,1}} \\ \end{matrix} \right]={{y}_{i,j,2,1}}\left[ \begin{matrix} 0 \\ 0 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,2,2}}\left[ \begin{matrix} 1 \\ 0 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,2,3}}\left[ \begin{matrix} 0 \\ 1 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,2,4}}\left[ \begin{matrix} 1 \\ 1 \\ 1 \\ \end{matrix} \right] \\ & \sum\limits_{k=1}^{4}{{{y}_{i,j,2,k}}=1} \\ & 0\le {{y}_{i,j,2,k}}\le 1 \\ & {{y}_{i,j,2,k}}\text{ is integral} \\ & \\ & \forall i=1...29\,j=1...29\,\,k=1...8 \\ & \left[ \begin{matrix} {{s}_{i,j}} \\ {{s}_{i+1,j}} \\ {{s}_{i,j+1}} \\ {{s}_{i,j,3}} \\ \end{matrix} \right]={{y}_{i,j,3,1}}\left[ \begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,3,2}}\left[ \begin{matrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,3,3}}\left[ \begin{matrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,3,4}}\left[ \begin{matrix} 1 \\ 1 \\ 0 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,3,5}}\left[ \begin{matrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,3,6}}\left[ \begin{matrix} 1 \\ 0 \\ 1 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,3,7}}\left[ \begin{matrix} 0 \\ 1 \\ 1 \\ 0 \\ \end{matrix} \right]+{{y}_{i,j,3,8}}\left[ \begin{matrix} 1 \\ 1 \\ 1 \\ 1 \\ \end{matrix} \right] \\ & \sum\limits_{k=1}^{8}{{{y}_{i,j,3,k}}=1} \\ & 0\le {{y}_{i,j,3,k}}\le 1 \\ & {{y}_{i,j,3,k}}\text{ is integral} \\ & \\ & \forall i=1...30\,j=1...30 \\ & 0\le {{s}_{i,j}}\le 1 \\ & {{s}_{i,j}}\quad \text{integral} \\ \end{align}]
• You probably don't want to reinvent the wheel, especially when it comes to Ising model simulation. I would suggest looking at Metropolis (Monte Carlo) implementations or cluster algorithms like Swendsen/Wang. – Alex R. Jul 16 '14 at 16:48
• Actually, we don't want to do monte carlo on spin systems. The problem for monte carlo is that, it could not give me the true ground state for certain and it could not provide an bound on how low energy the true ground state would possibly be. Thank you :D – user40780 Jul 16 '14 at 20:51
• scicomp.stackexchange.com/q/14130/4274. Please do not post the same question on multiple sites. Each community should have an honest shot at answering without anybody's time being wasted. – D.W. Apr 28 '16 at 5:10
• I'm voting to close this question as off-topic because it was cross-posted and answered at another site. – Todd Trimble Apr 28 '16 at 11:38
• They optimize over wavefunction, which is a linear combination of all possible states, $2^{30 \times 30}$ in your case. You seem to restrict this to pure states only. – Dmitry Savostyanov Jul 21 '14 at 23:55
Actually, one way I realize is to use this kind of constraints: $${s_i} + {s_j} - 1 \le {s_i}*{s_j} \le {s_i}$$
|
2020-11-25 14:46:40
|
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|
https://stats.stackexchange.com/questions/339401/using-posterior-in-an-expectation
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# Using posterior in an expectation
I am studying (myself, not in class) the book of Rogers & Girolami, A First Course in Machine Learning.
In working through a logistic classifier, I found the equation $$p(t_{new} = 1| \mathbf{x}_{new}, \mathbf{X}, \mathbf{t}) = E_{p(\mathbf{w}|\mathbf{X,t})} \left\{ \frac{1}{1 + \exp(-\mathbf{w}^T \mathbf{x}_{new})} \right\}$$
So is this the expectation with respect to the posterior... of the probability of a new data point being 1?
I do not recognize this expression. Does it have a name?
My question: can someone describe this in a more generic form? For example is it equivalent $$\int p(D=\text{const}|M)\, p(M|D)\, dM$$ where $D$ is "data" and $M$ is "model" ? If so, then the expression $\int p(D=\text{const}|M) dM$ does not integrate to 1 I think, it feels wrong.
• The link goes to the wrong book on Amazon. – Vladislavs Dovgalecs Apr 9 '18 at 6:51
• I removed the link. That is weird, I did not add the link. Maybe it is added auotmatically? – matchingmoments Apr 9 '18 at 19:06
In a Bayesian perspective, the predictive distribution is constructed by incorporating a new value of the observable $t_\text{new}$ as a part of the unknowns, hence aggregating it with the parameter $w$. The predictive distribution is thus derived by marginalising over the parameter: $$p(t_\text{new}|{\cal D})=\int p(t_\text{new},w|{\cal D})\, \mathrm{d} w \propto \int p(t_\text{new}|w)p({\cal D}|w) p(w)\, \mathrm{d} w$$ Hence, in the logistic case $$pr(t_\text{new}=1|{\cal D})=\int pr(t_\text{new}=1,w|{\cal D})\, \mathrm{d} w \propto \int pr(t_\text{new}=1|w)p({\cal D}|w) p(w)\, \mathrm{d} w$$or $$pr(t_\text{new}=1|{\cal D})=\mathbb{E}^\pi[pr(t_\text{new}=1|w)|{\cal D}]$$
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2019-12-09 02:53:25
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https://physics.stackexchange.com/questions/29197/why-is-the-atmosphere-transparent-in-the-visible-spectrum
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# Why is the atmosphere transparent in the visible spectrum?
One of the great 'coincidences' in physics is that the Sun happens to shine most brightly at exactly the wavelengths our eyes can see; it's an easy explanation that our eyes evolved to make the most out of what was available.
However, this spectral window is also one of the few at which our atmosphere is transparent, with Rayleigh scattering and the ozone layer protecting us from the UV and strong water vapour absorption in the IR until way into the radio range.
My question is: is this a coincidence? Were there geological factors that made it so? Or is it just lucky chance? (I also notice other planets in our neighbourhood weren't quite so lucky.) Or had atmospheric absorption been different our eyes would have tuned to whatever spectral window were brighter as a whole?
• Pretty sure it's the last option you chose: we see in the "visible" spectrum because there's plenty of it around! May 30, 2012 at 2:53
• May I suggest you consider a biophysics tag for this question May 30, 2012 at 4:14
• The atmosphere was created by living species that would adapt to the environment. species adapt to the environment but they also affect it. Kostya's answer shows that a lot of the absorption is done by water vapour (and oxygen), which was produced mostly from living things. And the atmosphere and sun were once part of the same nebula. Other planets weren't in the "goldilocks" zone, where water could reasonably be a gas, liquid and solid on the same planet. So the atmosphere needs life, which needs the atmosphere/water, which relates back to the intensity of the Sun May 30, 2012 at 12:24
Gases can only scatter light strongly if it matches a quantum transition (Rayleigh scattering doesn't involve quantum transitions but it's relatively weak). Quantum transitions can be rotational, vibrational or electronic (strictly speaking rotational and vibrational transitions are usually combined). Rotational/vibrational transitions have an energy that is generally in the IR range (which is why CO$_2$ scatters IR light) and electronic transitions have energies in the UV range (which is why ozone scatters UV). So there's a gap in the visible range. You say:
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2022-05-25 15:22:40
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https://www.tutorialspoint.com/sunita-travelled-15-km-268-m-by-bus-7-km-7-m-by-car-and-500-m-on-foot-in-order-to-reach-her-school-how-far-is-her-school-from-her-residence
|
# Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school. How far is her school from her residence?
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Given:
Sunita travelled 15 km 268 m by bus, 7 km 7 m by car and 500 m on foot in order to reach her school.
To do:
We have to find the total distance between her school and residence.
Solution:
We know that,
$1\ km=1000\ m$
Therefore,
The distance travelled by Sunita by bus $=15\ km\ 268\ m$
$= (15 + \frac{268}{1000})\ km$
$=(15+0.268)\ km$
$= 15.268\ km$
The distance travelled by Sunita by car $= 7\ km\ 7\ m$
$= (7 + \frac{7}{1000})\ km$
$=(7+0.007)\ km$
$= 7.007\ km$
The distance travelled by Sunita on foot $= 500\ m$
$= (\frac{500}{1000})\ km$
$=(0.5)\ km$
$= 0.5\ km$
The total distance between her school and residence is,
$15.268$
$+7.007$
$+0.500$
-----------
$22.775$
-----------
The total distance between her school and residence is $22.775\ km$.
Updated on 10-Oct-2022 13:36:10
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2022-12-07 07:11:43
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https://solvedlib.com/an-automobile-antifreeze-mixture-is-made-by,445017
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# An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d=1.114 g/mL;M=62.07 g/mol)...
###### Question:
An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d=1.114 g/mL;M=62.07 g/mol) and water (d=1.00 g/mL) at 20°C. The density of the mixture is 1.070 g/mL. Express the concentration of ethylene glycol as (a) volume percent 50v/v (b) mass percent 52.05w/w (c) molarity 0.0089 (d) molality 0.0083 (e) mole fraction
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##### Lioh Use Lagrange multiplictsto find the "aximum und mninimwm values of Ix % TVI suukjeseito HI Also find Ihe ponts where these extrenie Walle: Jccur (d) Findithe maximua akd Weuilkau2ukakshlicus fom tha Qigin to the curye Br 4 4xy+ 6y 140by using] Wagwan-e Mitipliers_[ Name:
lioh Use Lagrange multiplictsto find the "aximum und mninimwm values of Ix % TVI suukjeseito HI Also find Ihe ponts where these extrenie Walle: Jccur (d) Findithe maximua akd Weuilkau2ukakshlicus fom tha Qigin to the curye Br 4 4xy+ 6y 140by using] Wagwan-e Mitipliers_ [ Name:...
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2023-03-29 06:34:15
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http://quantumspotacademy.org/glossary/speed-of-light/
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# Speed of Light
The speed of light is exactly 299,792,458 meters per second (roughly $3.00 \times 10^8 m/s$). It is abbreviated with the letter c for the Latin word “celer” (meaning speed). According to Special Relativity, nothing can travel faster than the speed of light.
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2017-12-17 23:31:29
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https://chemistry.stackexchange.com/questions/50089/assumptions-made-in-calculating-melting-point-from-thermodynamic-data
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# Assumptions made in calculating melting point from thermodynamic data
I'm fine with doing these questions, such as:
For ammonia, enthalpy of fusion is 5.65 kJ/mol, and entropy of fusion is 28.9 J/K.mol
Hence find an approximate melting point: Well $\Delta G = \Delta H - T\Delta S = 5.65*10^3 - T(28.9)=0$ Which yields $T=196K$. This is quite close to Google's value.
What I'm thinking, though, is that we shouldn't be able (in real life) to determine a melting point so easily, because $\Delta_r H$ and $\Delta_r S$ both vary with temperature, so in reality this isn't as simple as solving a linear equation for T?
I'd just like some confirmation here, as my textbook isn't giving much away on this point :)
• Enthalpy and entropy of fusion cannot vary with temperature really. They can vary with pressure, in which case the melting point changes. ;) – Karl Apr 24 '16 at 17:47
$$\Delta S_f = \Delta H_f / T_f$$
Your text book calculation just solves this equation for $T_f$ to recover the temperature that probably had been measured before.
Note that for constant pressure the temperature is constant $T_f$ as long as the phase transition takes place.
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2020-10-26 22:32:20
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https://mcqexams.com/cbse/mcq-questions-for-class-12-commerce-accountancy/
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CBSE Multiple Choice Type Questions for 12th Class Accountancy PDF formatted study resources are available for free download. These Grade 12 Accountancy CBSE MCQ Mock Test helps you learn & practice the concepts in a fun learning way.
Class 12 Accountancy MCQs Multiple Choice Questions with Answers
Here are the chapterwise CBSE MCQ Quiz Test Questions for Class 12th Accountancy in pdf format that helps you access & download so that you can practice online/offline easily.
Accounting For Not-For-Profit Organisations Questions and Answers
Accounting For Not-For-Profit Organisations Quiz Question Answer
Funds raised by Not-for-Profit Organisations through various sources is __________account. credited to capital fund
The funds raised by not-for-profit organisations are credited to _________. Capital fund
Following item come under head fixed assets______________. None of the above
Not-for-Profit Organisations do not maintain_______account. capital
The information for the preparation of receipt and payments account is taken from _________. Cash book
Revenue receipts are ___________ in the business. recurring
Income and expenditure account reveals ____________. Surplus or Deficiency
In the Not-For-Profit Micro Finance Institutes.
Which among the following are included?
All of these
Non-trading institutions prepares ______________. income and expenditure account
The receipts and payments account begins with __________. Debit balance
Accounting For Partnership: Basic Concepts Questions and Answers
Accounting For Partnership: Basic Concepts Quiz Question Answer
Which of the following is NOT required to be maintained in the partnership deed? Age of partners
Where will you record interest on drawings in the final accounts of the firm ? Credit side of Profit & Loss Appropriation Account
How would you close the Partner's Drawing Account? By transfer to the debit of Capital or Current Account
Following is the difference between partnership deed and partnership agreement. Partnership deed is in writing and partnership agreement may be oral
The relationship between persons who have agreed to share the profit of a business carried on by all or any of them acting for all is known as : Partnership
Partners are supposed to pay interest on drawing only when provided by the ______________. Agreement
Which Section of the Partnership Act,1932 defines the term 'Partnership'? Section 4
X, Y and Z are partners in a firm.At the time of division of profit for the year there was dispute among the partners.Profits before interest on partner's capital was Rs.10,000 and X wanted interest on capital at 20% as his capital contribution was Rs.1,00,000 as compared to that of Y and Z which was Rs.75,000 and Rs.50,000 respectively. Find the solution ______________________________. Profits of Rs.10,000 will be distributed equally.
A and B enter into a joint venture sharing profits and losses equally. A purchased 5000 kg of rice @ Rs. 25/kg. B purchased 1000 kg of wheat @Rs. 30/kg. A sold 1000 kg of wheat @ Rs. 35/kg and B sold 5000 kg of rice @ Rs. 30/kg. The profit on venture will be : Rs. 30,000
A Ltd. has allotted $$20,000$$ shares to the applicants of $$28,000$$ shares on pro-rata basis. The amount payable on application is $$Rs. 2$$. M applied for $$420$$ shares. The number of shares allotted and the amount carried forward for adjustment against allotment money due from Mr. X will be ______________. $$300$$ shares; $$Rs. 240$$
Accounting For Share Capital Questions and Answers
Accounting For Share Capital Quiz Question Answer
Share or Security Premium can be used for writing off _____________.
preliminary expenses
Which of the following is incorrect as to a share? It may be transferred in fractions
A sends goods to B of Delhi, the goods are to be sold at 125% of cost which is invoice price. Commission is 10% on sales at invoice price and 25% of any surplus realized above IP. 10% of the goods sent out on consignment, invoice value of which is Rs 12,500, were destroyed. 75% of the total consignment is sold by B at Rs 1,00,What will be the amount of commission payable to B? Rs 10,937-50
While making an adjusting entry in respect of interest on capital, we credit __________________. Capital account
A company has issued $$15,000$$ shares of $$Rs. 10$$ each. The amount has been demanded as under:
On Application - $$Rs. 3$$. On Allotment - $$Re. 1$$
On $$1st$$ Call - $$Rs. 4$$. On $$2nd$$ and Final Call - $$Rs. 2$$.
A shareholder holding $$500$$ shares did not pay $$1st$$ and $$2nd$$ call money. His shares have been forfeited. These shares have been reissued @ $$Rs. 9$$. The amount transferred to Capital Reserve is
$$Rs. 1,500$$
A shareholder who can vote conditionally he is ____________. preference shareholder
A company cannot issue irredeemable preference shares. True
The balance in the deceased partner's capital account is transferred to the _________capital account. Representative of Deceased Partner
A and B enter into a joint venture to underwrite shares of K Ltd. K Ltd. make an equity issue of 100,000 equity shares. 80% of the shares underwritten by the venturer. 80,000 shares are subscribed by the public. How many shares are to be subscribed by the venture? 16,000
Share Application Account is in the nature of ______________. personal account
Accounting Ratios Class 12 Commerce Accountancy MCQ Quiz
Activity (or Turnover) Ratios comprises of all except ____________. Dividend Payout Ratio
The excess of current assets over current liabilities is called as ___________. Net working capital
Inventory turnover ratio $$=$$ Cost of goods sold $$\div$$ Cost of ______________ held during the period. Average inventory
Return on investment may be improved by ________. All of the above
Inventory turnover in days $$=$$ Days during the period $$\div$$ __________. Inventory turnover ratio
________________ costs of borrowing increase the debt-burden. High
Money in hand along with raw materials is known as ______. Working capital
In current ratio, current assets are compared with _________. Current liabilities
Inventory turnover measures the relationship of inventory with _______. Cost of goods sold
Which of the following statements is correct? Lower Debt-Equity ratio means lower financial risk
Analysis Of Financial Statements Questions and Answers
Analysis Of Financial Statements Quiz Question Answer
Window dressing imples. Over valuation of assets
Which of the following is not true with reference to capital budgeting? Existing investment in a project is not treated as sunk cost
Capital Budgeting deals with _______. Long-term decisions
Which of the following is not true for capital budgeting? Opportunity costs are excluded
Assets are held for the purpose of __________. Earning revenue
Which of the following is not incorporated in Capital Budgeting? Rate of Cash Discount
Which of the following is not a capital budgeting decision? Inventory level.
Majority of Capital Budgeting techniques makes use of __________. Cash flows
Capital Budgeting Decisions are based on ___________. Incremental cash flows
Which of the following is an irrelevant cost in Capital Budgeting Decisions? Both A and C
Cash Flow Statement Questions and Answers
Cash Flow Statement Quiz Question Answer
The term 'Cash' stands for__________. Cash and Bank Balance
Cash Inflows from a project include ________. Both A and B
Financial decision involves ___________. investment, financing and dividend decisions
Tax paid is ___________. Application of funds
Which of the following result from and increase or decrease of working capital? Flow of funds
While preparing a fund flow statement, attention is to be given to _______. Changes in Fixed assets and Fixed Liabilities
Stock at the end results in the __________. Source of funds
Fund flow statement measures :- The inflows and outflows of net working capital
The Real Cashflows must be discounted to get the present value 'M' at a rate equal to ________________. Real Discount Rate
Cash flow statement is also termed as ___________________. Statement of changes in Financial Position (working capital basis)
Dissolution Of Partnership Firm Questions and Answers
Dissolution Of Partnership Firm Quiz Question Answer
In Garner v. Murray, the deficiency of the insolvent partners is borne by other partners in the _________________.
Ratio of their capitals
On dissolution, all assets are transferred to realization account at ________.
Book value
On the dissolution of a firm, loan from the wife of a partner is treated as ____________.
Outside liability
In India, audit of Partnership firm is:
Optional
On the insolvency of one or more partners of a firm, the loss is borne by:
remaining solvent partners in their capital ratio
When a firm is dissolved, Goodwill a/c is closed by transferring to:
Realisation account
Which of the following is not a correct statement? The terms capitalisation factor and multiplier are not synonymous terms
What journal entry is passed for realising assets to provide cash for payment of liabilities? Realisation A/c Dr.
To Assets A/c
Cash/Bank A/c Dr.
To Realisation A/c
Declaration by directors of the company at the time of solvency is called as __________. Declaration of solvency
Provision for bad and doubtful debts appearing in the books at the time of dissolution of firm is transferred to :
Realisation account
Financial Statements Of A Company Questions and Answers
Financial Statements Of A Company Quiz Question Answer
Financial statements does not provide _____information. qualitative
Financial statements contain only_____information. monetary
The financial statements generally include _____statement. three
State with reasons, whether the following statement is True or False.
In the balance sheet of X Limited, preliminary expenses amounting to $$Rs. 5$$ lakhs and securities premium account of $$Rs. 35$$ lakhs are appearing. The accountant can use the balance in securities premium account to write off preliminary expenses.
True
Balance sheet does not disclose information relating to ___________. loss of markets
Statements included in financial statements are________. All of the above
Unclaimed dividend is shown on the liability side of the balance sheet of a company as ___________. Current Liabilities
Pooling of interest method is applicable for amalgamation in the nature of _____. Merger
Financial statement in relation to a company includes __________. All of the Above
Which of the following reserve does not appear in the balance sheet of a company? Secret reserve
Issue And Redemption Of Debentures Questions and Answers
Issue And Redemption Of Debentures Quiz Question Answer
Debenture Premium Account is a -
Nominal Account
From the point of view of tenure, the debentures are classified as - Redeemable Debentures and Irredeemable Debentures
On $$1$$st April X Lts. issued Rs$$1,00,000 15\%$$ Debentures of Rs$$10$$ each at $$94\%$$ redeemable at per as follows:
Yearend Nominal value of Total Debentures to redeemed 2 10% 3 20% 4 30% 5 40%
The amount of discount to be written off each year assuming that the company closes its accounts on financial year basis is -
$$Rs 1,500, Rs 1,500, Rs 1,350, Rs 1,050, Rs 600$$
On $$1$$st April X Lts. issued Rs$$1,00,000 15\%$$ Debentures of Rs$$10$$ each at $$94\%$$ redeemable at per as follows:
Year beginning Nominal value of Total Debentures to redeemed 2 10% 3 20% 4 30% 5 40%
The amount of discount to be written off each year assuming that the company closes its accounts on financial year basis is -
Rs$$2,000, Rs 1,800, Rs 1,400, Rs 800$$
X.Ltd. issued Rs$$1,00,000$$ $$12\%$$ debentures of Rs$$100$$ each at a premium of $$10\%$$, which are redeemable after $$10$$ years at a premium of The amount of loss on redemption of debentures to be written off every year= ? Rs$$20,000$$
X. Ltd. issued $$500, 15\%$$ Debentures of Rs$$100$$ each at a discount of $$10\%$$ redeemable at a premium of $$5\%$$ after $$4$$ years. The amount of annual interest on Debentures is - Rs$$7,500$$
X Ltd issues $$500, 15\%$$ Debentures of Rs$$100$$ each on $$1st$$ May at a discount of $$10\%$$ redeemable at a premium of $$5\%$$ after $$4$$ years. Interest was payable half yearly on $$30th$$ June and $$31st$$ December. The amount of interest accrued but not due to be shown in the balance sheet as at $$31st$$ March is- Rs$$1,875$$
'Premium on Redemption of Debentures Account' is a: Personal Account
Withdrawal from DRR is permissible only after ____% of the debenture liability has been redeemed. 10
A price exclusive of the interest for the period for which the seller held the debentures is called _________________. Ex-interest price
Reconstitution Of A Partnership Firm - Admission Of A Partner Questions and Answers
Reconstitution Of A Partnership Firm - Admission Of A Partner Quiz Question Answer
Sales for the year ended amounted to Rs. 10,00,Sales included goods sold to Mr.A for Rs. 50,000 at a profit of 20% on cost.Such goods are still lying in the godown at the buyer's risk. Such goods should be treated as part of : Sales
A, B, C were partners sharing profits in the proportion of 1/2, 1/3 and 1/6 respectively. On 31st March 1997 their capital stood as follows:
A Rs. 4,00,000, B Rs. 3,00,000, C Rs. 2,50,000.
A sum of 1,20,000, also appeared as reserve fund in their balance sheet on this date. B retire on this date when the goodwill of firm was valued at Rs. 1,80,000.
Profit and loss adjustment account prepared on that date without taking goodwill and reserve fund into consideration showed a net profit of Rs. 28,500.
The net amount payable to B will be :
Rs. 4,09,500
After crediting the ______ partners by the amount of goodwill brought in by the ______ partner, the existing goodwill must be written off by debiting the old partners in their old profit sharing ratio. old, new
Ram of Bombay sends out goods costing 80,000 to Mohan of Kanpur so as
to show 20% profit on invoice value. 3/5th of the goods received by consignee is sold at 5% above invoice price. The amount of sale value will be:
63,000
The capital of A and B sharing profits and losses equally are Rs.90,000 and Rs.30,000 respectively.They value the goodwill of the firm at Rs.80,000, which was not recorded in the books.If goodwill is to be raised now, by what amount each partner's capital amount will be debited? None of these
The profits for 1998-99 are Rs.2,000; for 1999-2000 Rs.26,100 and for 2000-01 Rs.31,Closing stock for 1999-2000 and 2000-01 includes the defective items of Rs. 2,200 and Rs. 6,200 respectively which were considered as having market value NIL.Calculate the goodwill on two years' purchase of average profit. Rs.35,400
The profits and losses for the last years are: 2001-02 Losses Rs. 10,000; 2002-03 Losses Rs. 2,500; 2003-04 Profits Rs. 98,000 and 2004-05 Profits Rs. 76,The average capital employed in the business is Rs. 2,00,The rate of interest expected from capital invested is 12%. The remuneration of partners is estimated to be Rs. 1,000 per month. Calculate the value of goodwill oh the basis of two yearspurchase of super profits based on the average of four years. Rs.8,750
P/L Appropriation Account is prepared _______________. Partnership Firm
In the absences of an agreement, Interest on loan advanced by the partner to the firm is allowed at the rate of _____. 6%
If vendors are issued fully paid shares of $$Rs. 80,000$$ in the consideration of net assets of $$Rs. 60,000$$, then the balance of $$Rs. 20,000$$ will be _____________________. Debited to goodwill account
Reconstitution Of A Partnership Firm - Retirement / Death Of A Partner Questions and Answers
Reconstitution Of A Partnership Firm - Retirement / Death Of A Partner Quiz Question Answer
Sales for the year ended amounted to Rs. 10,00,Sales included goods sold to Mr.A for Rs. 50,000 at a profit of 20% on cost.Such goods are still lying in the godown at the buyer's risk. Such goods should be treated as part of : Sales
A, B, C were partners sharing profits in the proportion of 1/2, 1/3 and 1/6 respectively. On 31st March 1997 their capital stood as follows:
A Rs. 4,00,000, B Rs. 3,00,000, C Rs. 2,50,000.
A sum of 1,20,000, also appeared as reserve fund in their balance sheet on this date. B retire on this date when the goodwill of firm was valued at Rs. 1,80,000.
Profit and loss adjustment account prepared on that date without taking goodwill and reserve fund into consideration showed a net profit of Rs. 28,500.
The net amount payable to B will be :
Rs. 4,09,500
Which of the following formula is used to calculate goodwill under super profit method? Goodwill = Super profit x No. of year purchases.
The capital of A and B sharing profits and losses equally are Rs.90,000 and Rs.30,000 respectively.They value the goodwill of the firm at Rs.80,000, which was not recorded in the books.If goodwill is to be raised now, by what amount each partner's capital amount will be debited? None of these
The profits for 1998-99 are Rs.2,000; for 1999-2000 Rs.26,100 and for 2000-01 Rs.31,Closing stock for 1999-2000 and 2000-01 includes the defective items of Rs. 2,200 and Rs. 6,200 respectively which were considered as having market value NIL.Calculate the goodwill on two years' purchase of average profit. Rs.35,400
Which of the following formula is used to calculate goodwill under weighted average profit method ? Goodwill = Weighted average profit x no. of year purchase.
In which of the following cases, the need for the valuation of goodwill in a firm may arise? All of above.
If vendors are issued fully paid shares of $$Rs. 80,000$$ in the consideration of net assets of $$Rs. 60,000$$, then the balance of $$Rs. 20,000$$ will be _____________________. Debited to goodwill account
Which of the following formula is used to calculate goodwill under simple average profit method? Goodwill = Average profit x No. of year purchase.
If the incoming partner is to bring in premium for goodwill in cash and also balance exist, in the goodwill Account, then this Goodwill Account is written off among the old partners in _________ . the old profit-sharing ratio
Accountancy MCQ Questions for Class 12 - Practice Test with Solutions
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2023-03-23 16:47:05
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https://www.mmacageworld.com/2011/02/ufc-news-rampage-vs-hamill-and-alves-vs.html
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MMA CAGEWORLD: UFC NEWS: Rampage vs Hamill and Alves vs Story
## MMA Latest
LATEST NEWS > . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . > Stay Safe Folks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . > Stay Safe Folks . . . . . . . . . . . . . . . . . . . . . . . . . . ...........................
## 20 February 2011
### UFC News
Confirmed by the UFC for 130 on May 28 at MGM Garden Arena Las Vegas as official. The card also features a headliner between Frankie Edgar and Gray Maynard for the UFC Lightweight Championship.
A win over Hamil would put Rampage next in line to face the winner of of Mauricio Rua vs Jon Jones.
Alves is out to impress upon his victory at UFC 124 against John Howard.
Story and his camp having previously expressed a keen interest to fight top ten competition have been given just that with the top five ranked Alves.
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2022-12-07 01:18:19
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https://tobykingsman.wordpress.com/2016/02/
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# Simulation: Input-Uncertainty
The idea behind simulation models is that we can use them to make predictions of what might happen in the real world. As we want often want to use the results of these simulations to inform decision making it is important for the results to be accurate and precise. The way we do this is by keeping a careful eye on the errors.
Traditionally people have focused on trying to minimise the error that crops up in the actual simulation itself, the technical name for this is the ‘simulation-estimation error.
However there is another important error that is often overlooked and needs to be taken into account as well, the ‘input-uncertainty error‘. This is to do with the uncertainty in the actual values you plug into your simulation to start with.
Errors in Simulations; source: emilystori.wordpress.com
Video
# Changepoint Detection – Part 2
If you haven’t already, I would strongly recommend glancing over the part 1 of this post before continuing. In it we talked about a method for detecting a single changepoint through techniques such as maximum log-likelihood estimation. The question now is the more difficult one of how can we search for multiple changepoints?
The simplest thing to do would be to initially apply our Single Changepoint Detection (SCD) method to the series we want to analyse. If there is no changepoint found we are done, but if there is a changepoint found we end up with two distinct segments.
# Changepoint Detection – Part 1
Previously we discussed what a changepoint is, how they work and examples of where you might find them. Today however we shall go one step further and attempt to discuss solving the problem of how to find these changepoints.
Multiple changepoints have be found, they are the red lines.
Let’s begin by trying to figure out a way of detecting if just a single changepoint exists in a set of data. The simplest way to pose this problem is to describe it through a ‘Hypothesis Test’, where we have a null hypothesis ($H_0$) and an alternative hypothesis ($H_1$),
$H_0:$ no changepoint $H_1:$ 1 changepoint
If you are unfamiliar with hypothesis tests, the idea is to run some kind of test that will give us a value. If that value is above a certain threshold, we can reject the null hypothesis and in our case accept the alternative hypothesis. If not, we can’t disprove the null hypothesis and we accept it as true. In layman’s terms, it’s the classic innocent until proven guilty. Alternatively check it out yourself!
Aside
# Markov Chains and Tennis
Now that the tennis season is well underway and the Australian Open is already a few weeks in the past, I thought I would look at a basic model of a game of tennis for today’s blog post. As you might expect there will be some maths later on, but feel free to skip that and just look at the pictures and the results!
The simplest way to model a tennis match as suggested by people such as O’Malley, is to look at the probability of players winning a point on their own serve. Then we can define the probability of a player winning a point on their serve as p, and the probability of losing a point on their serve as 1-p. These two probabilities can be used to build up a model of the entire match. Whilst this is clearly a simplification it has been shown to be a pretty good predictor of the overall winner of a match.
For example by checking the ATP website we can see that Roger Federer won 3737 out of 5202pts he played on serve last year, giving him an average probability of p=0.718 to win a point on his serve. Giles Simon only managed to win 3186 out of 5128pts on his serve and consequently had a more typical value of probability p=0.621 to win a game on his serve. Most top-50 ATP players had values in the range of 0.6 to 0.7 for the 2015 season, which is obviously pretty good!
# Emergency Planning with OR
One recent talk that stood out to me was given by Marc Goerigk and talked about how operational research can be used to help improve evacuation planning for emergencies. For example unexploded WWII era bombs, Lancaster floods or terrorist incidents.
The ideas I will talk about are to do with using an optimisation approach to give a good lower bound on the evacuation time. Simulation techniques on the other hand are more useful for finding an upper bound on the evacuation time.
A clever thing to do when we are trying to model real world places is to try and represent the place in a network. Essentially we can take a map and represent it in a graph network using nodes and arcs. Arcs generally represent paths or roads and nodes the intersection of these paths and roads. This allows us to simplify tricky features such as curved roads into nice straight lines.
Transforming a map of my flat into a network diagram
# Multi-Armed Bandits
One of the classic problems in Statistics that a lot of people working at STOR-i in particular are involved with is something called ‘Multi-Armed Bandits’. In fact there was a recent conference on the topic and its applications at Lancaster earlier this year. In this post I will try to explain what the problem is and touch on how it is solved.
Originally this problem was first thought about in the casinos of Las Vegas. It concerns how to pick which slot machine out of a group you should play to try in order to try and maximise your total winnings. Whilst this problem is no longer just applicable to casinos, the name ‘one-armed bandit‘ has stuck and the casino analogy is what I will continue to use.
# Pythagorean Expectation in Football
The use of data and statistics has grown rapidly in most sports in recent years. Nowadays many metrics and formulas exist for trying to measure and predict performances of players and teams. One of my favourite tools because of its simplicity is something called ‘Pythagorean Expectation‘.
The most famous early pioneer of baseball statistics, Bill James, came up with the original formula to predict the number of wins for a team over a baseball season, based on the number of runs they scored and conceded.
### $Win\% = \cfrac{RS^2}{RS^2 +RC^2}$
To work out the number of wins in a season you simply have to multiply the win percentage by the number of games played. The reason James called it Pythagorean is because of the occurrence of all the squared terms. Whoever said Pythagoras Theorem was boring!
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2018-03-18 09:41:17
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https://physics.stackexchange.com/questions/598393/motor-and-pulley-torque-contradiction
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# Motor and Pulley torque contradiction
Consider this scenario, with a motor (grey) connected to a pulley (yellow):
The motor is joined to the pulley at the pulley's center. I know the motor outputs some torque but I do not understand how to then determine the torque the pulley can apply at a radius $$r$$.
Will the torque that the pulley can apply, vary with the radius of the pulley? I feel like it must, because I don't understand how torque could be constant no matter how wide the pulley is.
I would appreciate an explanation as to why and how the output torque varies with the radius of pulley.
$$\text{force delivered}\times \text{radius}=\text{torque delivered}=\text{constant}$$
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2022-05-19 00:00:21
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https://testbook.com/question-answer/the-forward-dvdt-rating-of-an-scr--5dd3f2e6f60d5d4aeb58e312
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# The forward dv/dt rating of an SCR:
This question was previously asked in
UPPCL JE Previous Paper 3 (Held ON: Feb 2018 Shift 1)
View all UPPCL JE Papers >
1. Decreases with a decrease in the RMS value of a forward anode-cathode voltage
2. Decreases with an increase in the junction temperature
3. Increases with an increase in the junction temperature
4. Increases with the decrease in the RMS value of a forward anode-cathode voltage
Option 3 : Increases with an increase in the junction temperature
Free
Electric Circuits and Fields
897
30 Questions 40 Marks 40 Mins
## Detailed Solution
dv/dt protection:
• When a thyristor is in a forward blocking state then only middle junction is reverse biased which acts as a capacitor having constant capacitance value Cj (junction capacitance); As we know that current through capacitor follows the relation
$$i = {C_j}\frac{{dv}}{{dt}}$$
$$i \propto \frac{{dv}}{{dt}}\;;if\;{C_j}\;is\;constant$$
• Leakage current through the middle junction which is nothing but the leakage current through the device will increase with the increase in $$\frac{{dv}}{{dt}}$$ i.e. the rate of change of applied voltage across the thyristor
• This current can turn-on the device even when the gate signal is absent; This is called dv/dt triggering and must be avoided which can be achieved by using the Snubber circuit in parallel with the device
With the increase in the temperature of the junction, insulation may get failed. So, we must take proper measures to limit the temperature rise. The forward dv/dt rating of an SCR Increases with an increase in the junction temperature.
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2021-10-18 20:04:16
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https://zbmath.org/?q=an:1076.33006
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# zbMATH — the first resource for mathematics
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Some generalizations of the Apostol-Bernoulli and Apostol-Euler polynomials. (English) Zbl 1076.33006
The authors introduce a generalization of the classical Bernoulli polynomials as analogous definition of Apostol type [see {\it T. M. Apostol}, Pac. J. Math. 1, 161--167 (1951; Zbl 0043.07103)] for the so-called Apostol-Bernoulli numbers and polynomials of higher order. The generalization, the Apostol-Bernoulli polynomials ${\Cal B}_n^{(\alpha)}(x;\lambda)$, is defined by means of the following generating function: $$\left(\frac{z}{\lambda\,e^z-1}\right)^{\alpha}\,e^{xz} =\sum_{n=0}^\infty {\Cal B}_n^{(\alpha)}(x;\lambda)\,\frac{z^n}{n!}\qquad \left(\vert z+\log \lambda\vert <2\pi;\,1^{\alpha}:=1\right)$$ with $$B_{n}^{(\alpha)}(x)={\Cal B}_n^{(\alpha)}(x;1) \quad\text{and}\quad {\Cal B}_{n}^{(\alpha)}(\lambda):={\Cal B}_n^{(\alpha)}(0;\lambda)$$ where ${\Cal B}_{n}^{(\alpha)}(\lambda)$ denotes the so-called Apostol-Bernoulli numbers of order $\alpha$. In a similar manner the Apostol-Euler polynomials of order $\alpha$, a generalization of the classical Euler polynomials, is introduced. In a previous paper, the first author derived several properties and explicit representations of the Apostol-Euler polynomials of order $\alpha$. In this paper, the authors investigate the corresponding problems for the Apostol-Bernoulli polynomials of order $\alpha$ by following the work of the second author in an earlier article [see {\it H. M. Srivastava}, Math. Proc. Camb. Philos. Soc. 129, 77--84 (2000; Zbl 0978.11004)]. They establish their elementary properties, derive an explicit series representations for the polynomials involving the Gaussian hypergeometric function, the Hurwitz zeta function and the Riemann zeta function.
##### MSC:
33C45 Orthogonal polynomials and functions of hypergeometric type 11B68 Bernoulli and Euler numbers and polynomials
Full Text:
##### References:
[1] Abramowitz, M.; Stegun, I. A.: Handbook of mathematical functions with formulas, graphs, and mathematical tables. Appl. math. Ser. 55 (1965) · Zbl 0171.38503 [2] Apostol, T. M.: On the lerch zeta function. Pacific J. Math. 1, 161-167 (1951) · Zbl 0043.07103 [3] Apostol, T. M.: Introduction to analytic number theory. (1976) · Zbl 0335.10001 [4] Comtet, L.: Advanced combinatorics: the art of finite and infinite expansions. (1974) · Zbl 0283.05001 [5] Cvijović, D.; Klinowski, J.: New formulae for the Bernoulli and Euler polynomials at rational arguments. Proc. amer. Math. soc. 123, 1527-1535 (1995) · Zbl 0827.11012 [6] Luo, Q. -M.: On the Apostol -- Bernoulli polynomials. Central European J. Math. 2, 509-515 (2004) · Zbl 1073.33001 [7] Q.-M. Luo, Apostol -- Euler polynomials of higher order and the Gaussian hypergeometric function, Taiwanese J. Math., in press [8] Nörlund, N. E.: Vorlesungen über differentzenrechnung. (1924) [9] Srivastava, H. M.: Some formulas for the Bernoulli and Euler polynomials at rational arguments. Math. proc. Cambridge philos. Soc. 129, 77-84 (2000) · Zbl 0978.11004 [10] Srivastava, H. M.; Choi, J.: Series associated with the zeta and related functions. (2001) · Zbl 1014.33001 [11] Srivastava, H. M.; Todorov, P. G.: An explicit formula for the generalized Bernoulli polynomials. J. math. Anal. appl. 130, 509-513 (1988) · Zbl 0621.33008 [12] Todorov, P. G.: Une formule simple explicite des nombres de Bernoulli généralisés. C. R. Acad. sci. Paris sér. I math. 301, 665-666 (1985)
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2016-05-01 11:49:33
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https://web2.0calc.com/questions/help_67614
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+0
# help its due rn
-1
62
1
The midpoints of a regular hexagon are connected to form a smaller hexagon. The small hexagon has perimeter $2\sqrt{3}.$ What is the perimeter of the large hexagon?
Feb 17, 2020
edited by Guest Feb 17, 2020
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2020-04-07 23:37:10
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https://math.stackexchange.com/questions/1232862/lagrange-multipliers-in-calculus-of-variations
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Lagrange multipliers in Calculus of Variations
I am trying to learn about Calculus of Variations and I am beginning to see some constrained optimization problems in the domain of functionals, by using Lagrange multipliers. It seems that things work like in the finite-dimensional calculus but I need some semi-formal explanations; which I failed to find anywhere. (Looks like this is a rare area of interest).
In the finite dimensional calculus, when we wanted to optimize a function $f(x), f:\mathbb{R^n} \mapsto \mathbb{R}$ such that it obeys the constraint $g(x)=0$, the idea was that $g(x)=0$ was a $n-1$ dimensional surface and $\nabla g(x)$ was normal to the surface's tangent hyperplane. In order to move on the surface, we had to move in directions $u$ where $\nabla g(x)^T \cdot u=0$. This requires that at a constrained extremum it is needed that $\nabla f(x)^T \cdot u=0$ whenever $\nabla g(x)^T \cdot u=0$, which requires that $\nabla f(x) = \lambda \nabla g(x)$.
1. Is this valid in a function space, like $C[a,b]$ as well? Can it be shown that if we have a nice behaving constraint functional $G[y(x)]=0$ in this space, the gradient $\nabla G[y(x)]$ is perpendicular to the function space surface $G[y(x)]=0$? We have the inner product $\int_{a}^{b}f(x)g(x)dx$ defined for this space which introduces angles and orthogonality between functions; so I intuitively think that this is true, but I wonder what the actual explanation is.
2. Assuming 1. is true, can be said that at a constrained extremum, it should be $\nabla F[y(x)] = \lambda \nabla G[y(x)]$, like in the finite dimensional case?
3. Does one build a "Lagrangian functional" $L[y(x),\lambda]=F[y(x)] + \lambda G[y(x)]$ like in the finite dimensional case, where the problem turns into the unconstrained optimization of the weird thing $L[y(x),\lambda]$, which depends both on a functional and a scalar variable? How does one maximizes such a thing then: This space is something like $C[a,b] \times \mathbb{R}$. How are derivatives etc. defined here; for example can we define a Gateaux variation here like $$\lim_{\epsilon \to 0} \dfrac{L[y(x) + \epsilon h(x),\lambda + \epsilon] - L[y(x),\lambda]}{\epsilon}$$ which leads to unconstrained optimization?
I need some explanations for such cases.
• You can start from Wikipedia: en.wikipedia.org/wiki/Lagrange_multipliers_on_Banach_spaces – Siminore Apr 13 '15 at 13:41
• I already found and read that but I need a more clear explanation like I have said; I am a beginner in the field of variational calculus. – Ufuk Can Bicici Apr 13 '15 at 13:50
• And what about reading the suggested references? Zeidler's book is rather clear. – Siminore Apr 13 '15 at 13:52
• I believe the second question should be true because you are just in Banach space. Also try this in n dimensional Hilbert space. Since Hilbert space a stricter form of Banach Space, if it holds in Hilbert Space, it must hold in Banach Space. – Chan Hunt Apr 13 '15 at 17:36
• See if you can generalize it to Hilbert Space. Since Hilbert Space is a stricter form of Banach Space, if it holds in Hilbert Space, it must hold in Banach Space. – Chan Hunt Apr 13 '15 at 17:39
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2021-02-27 20:26:22
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http://www.ck12.org/analysis/Powers-and-Roots-of-Complex-Numbers/studyguide/user:cmF5bmEuc2NobG9zc2JlcmdAZ21haWwuY29t/Complex-Number-Theorems/r2/
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<meta http-equiv="refresh" content="1; url=/nojavascript/">
You are viewing an older version of this Study Guide. Go to the latest version.
Powers and Roots of Complex Numbers
%
Progress
Progress
%
Complex Number Theorems
Vocabulary
Fill in the equation for each theorem then describe each theorem in your own words.
Theorem Equation Description Product Theorem __________________________ _________________________________________________ Quotient Theorem __________________________ _________________________________________________ De Moivre's Theorem __________________________ _________________________________________________
Practice
Product and Quotient Theorems
If $z_1 = 7 \left( \frac{\pi}{2} \right)$ and $z_2 = 9 \left(\frac{\pi}{3} \right)$ find:
1. $z_1 z_2$
2. $\left( \frac{z_1}{z_2} \right)$
3. $\left( \frac{z_2}{z_1} \right)$
4. $(z_1)^2$
5. $(z_2)^3$
.
Find the quotients
1. $2(cos 80^o + i sin 80^o) \div 6(cos 200^o + i sin 200^o)$
2. $3cis(130^o) \div 4cis(270^o)$
.
.
Powers and Roots
To find roots of complex numbers, you De Moivre's Theorem to creat the equation $z^{1/n} = (a + bi)^{1/n} = r^{1/n} cis \left ( \frac{\theta}{n} \right )$. This helps you find roots.
Remember: Numbers must be in polar form to use De Moivre's Theorem!
.
Use De Moivre’s Theorem:
1. $[3(\mbox{cos} \ 80^\circ + i \ \mbox{sin} \ 80^\circ)]^3$
2. $\left [\sqrt{2} \left (\mbox{cos}\ \frac{5\pi}{16} + i \ \mbox{sin} \ \frac{5\pi}{16} \right ) \right ]^4$
3. $\left (\sqrt{3} - i \right )^6$
4. Identify the 3 complex cube roots of $1 + i$
5. Identify the 4 complex fourth roots of $-16i$
.
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2015-04-01 08:18:06
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http://mathhelpforum.com/geometry/144496-mid-point-chord-ellipse-print.html
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# Mid-point of chord on ellipse
• May 12th 2010, 11:18 PM
arze
Mid-point of chord on ellipse
Prove that the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $P_1(x_1,y_1)$ is $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$. The tangent at $P_1$ meets the tangent at $P_2(x_2,y_2)$ at T. Show that the line
$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{xx_2}{a^2} +\frac{yy_2}{b^2}$
passes through T and through the midpoint of $P_1P_2$. Prove that if $P_1TP_2$ is a right angle, then $\frac{x_1x_2}{a^4}+\frac{y_1y_2}{b^4}=0$
I have proven the first part. I got the equation of the tangent at $P_2$ in the same form as the first. Then I made y the subject of both tangents and found x. did it in the same way to find y.
I found the the line given passed through T, but could not prove that the line passed through the midpoint of $P_1P_2$, $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.
Thanks!
• May 13th 2010, 01:40 AM
Opalg
Quote:
Originally Posted by arze
Prove that the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $P_1(x_1,y_1)$ is $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$. The tangent at $P_1$ meets the tangent at $P_2(x_2,y_2)$ at T. Show that the line
$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{xx_2}{a^2} +\frac{yy_2}{b^2}$
passes through T and through the midpoint of $P_1P_2$. Prove that if $P_1TP_2$ is a right angle, then $\frac{x_1x_2}{a^4}+\frac{y_1y_2}{b^4}=0$
I have proven the first part. I got the equation of the tangent at $P_2$ in the same form as the first. Then I made y the subject of both tangents and found x. did it in the same way to find y.
I found the the line given passed through T, but could not prove that the line passed through the midpoint of $P_1P_2$, $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.
The equation $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{xx_2}{a^2} +\frac{yy_2}{b^2}$ represents a straight line. If the point (x,y) lies on the tangent at $P_1$ then the left side of that equation will be 1. If (x,y) lies on the tangent at $P_2$ then the right side of the equation will be 1. If (x,y) is the point T then both sides of the equation will be 1, and hence the equation will be satified. Therefore T lies on the line with that equation.
The condition for the midpoint of $P_1P_2$ to lie on the line is $\frac{\frac12(x_1+x_2)x_1}{a^2}+\frac{\frac12(y_1+ y_2)y_1}{b^2}=\frac{\frac12(x_1+x_2)x_2}{a^2}+\fra c{\frac12(y_1+y_2)y_2}{b^2}$. After multiplying out all the brackets and doing some cancellation, you should be able to write that condition as $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2} = \frac{x_2^2}{a^2}+\frac{y_2^2}{b^2}$. Both sides of that equation are equal to 1 because $P_1$ and $P_2$ lie on the ellipse. Therefore the condition is satisfied, and the midpoint of $P_1P_2$ lies on the line.
For the last part of the question, write down the condition for the tangents at $P_1$ and $P_2$ to have perpendicular slopes.
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2016-07-25 03:29:08
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https://www.physicsforums.com/threads/hawking-radiation.11750/
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One of the real important aspects of Hawking Radiation is the Kerr Temperature.
Note that the Kerr Temperature is responsible for the Kerr Particle Energy Spectrum and represents the genesis of Thermodynamic Quantum Gravitation.
Thermodynamic Quantum Gravitation is the combination of Thermodynamics and Quantum Gravitation:
A Kerr Black Hole is a rotating Black Hole.
$$T_k = \frac{\hbar c^3}{K_o G M}$$
Ko = Boltzmann's Thermal Constant
Related Astronomy and Astrophysics News on Phys.org
NateTG
Homework Helper
Since you brought it up, I have this rather silly question about Hawking radiation:
Let's say that I have an extremely small black hole -- i.e. the Schwartzschild radius is smaller than the Planck length.
Wouldn't a black hole like this have a tendency to radiate light that had more mass/momentum than the black hole does?
$$r_s = \frac{2 G M}{c^2}$$
Planck Length:
$$r_p = \sqrt{ \frac{ \hbar G}{c^3}}$$
Schwarzschild Temperature:
$$T_s = \frac{ \hbar c^3}{4 K_o G M}$$
$$r_s = r_p$$
$$\frac{2 G M}{c^2} = \sqrt { \frac{ \hbar G}{c^3}}$$
Schwarzschild-Planck Mass:
$$M_s = \frac{1}{2} \sqrt { \frac{ \hbar c}{G}}$$
Integral:
$$M_s = \frac{1}{2} \sqrt { \frac{ \hbar c}{G}} = \frac{ \hbar c^3}{4 K_o G T_s}$$
Schwarzschild-Planck Temperature:
$$T_s = \frac {1}{2K_o} \sqrt { \frac{\hbar c^5}{G}}}$$
$$T_s = 7.084E+31 K$$
T_s = 7.084*10^31 Kelvin
Schwartzschild radius is smaller than the Planck length.
Wouldn't a black hole like this have a tendency to radiate light that had more mass/momentum than the black hole does?
if the black hole were smaller than that, then would it be able to radiate?
$$r_s \ll r_p$$
$$\Delta = \frac {\hbar c}{ \lambda K_o T_s} = \frac{2}{ \lambda} \sqrt { \frac {\hbar G}{c^3}}$$
$$\Delta = \frac{2}{ \lambda} \sqrt { \frac {\hbar G}{c^3}}$$
$$T_q = \frac {\hbar c^3}{4 K_o G M_s ( e^\Delta - 1)}$$
$$I(\lambda) = \frac { 2 \pi h c^2}{ \lambda^5 (e^\Delta - 1)}$$
$$I_q = \sigma T_q^4$$
The Schwarzschild-Planck Radius is Mass dependent.
As a Thermodynamic Schwarzschild-Planck Black Hole radius falls below the Planck Radius $$r_s \ll r_p$$, it becomes a Thermodynamic Quantum-Schwarzschild Black Hole, the resulting radiation diminishes instead of increasing.
The resulting radiation flux becomes less energetic than the mass equivalency.
A Thermodynamic Schwarzschild-Planck Black Hole would evaporate instantly however a Thermodynamic Quantum-Schwarzschild Black Hole diminishes more gradually, however still relatively instantaneous. The relative flux intensity also diminishes.
A Quantum-Schwarzschild Black Hole Thermodynamic Temperature is quantumized below the Planck Radius.
Last edited:
NateTG
Homework Helper
Originally posted by Orion1
A Schwarzschild-Planck Black Hole would evaporate instantly.
Right, but if the black hole were smaller than that, then would it be able to radiate? Conservation of energy/mass would be grossly violated if the black hole could produce radiation with more energy than it's equivalent mass. Similarly, there might be problems with conservation of linear momentum.
Conservation of energy/mass would be grossly violated if the black hole could produce radiation with more energy than it's equivalent mass
Not if the black hole can absorb as much energy from the vacuum as it radiates and there is theoretically 10^120 Joules per cubic metre in the vacuum.
Planck Probability...
Conservation of energy/mass would be grossly violated if the black hole could produce radiation with more energy than it's equivalent mass
Not if the black hole can absorb as much energy from the vacuum as it radiates and there is theoretically 10^120 Joules per cubic metre in the vacuum.
What if a Schwarzschild-Planck Black Hole is capable of absorbing more radiation than its Schwarzschild-Planck Temperature vacuum, does this also violate Conservation of energy/mass?
I presume that a Schwarzschild-Planck Black Hole which is capable of absorbing more radiation than its Schwarzschild-Planck Temperature vacuum would momentarily increase in mass, then still evaporate instantly.
To my understanding, a 'perfect radiation absorber' is not possible, but then again it was once thought that black holes were a 'zero radiation emitter'.
Is there an equasion that exists that determines how much radiation a Schwarzschild-Planck Black Hole is capable of absorbing?
According to my integrations above, the amount of Schwarzschild-Planck Black Hole radiation flux produced below the Planck Radius $$r_s \ll r_p$$ is no longer determined by mass thermodynamics, but by probability and radiation wavelength similar to a blackbody radiator, given here: $$P = (e^\Delta - 1)$$.
Thereby, when Schwarzschild-Planck Black Hole falls below Planck radius $$r_s \ll r_p$$, the radiation flux probabilisticly diminishes, resulting in a radiation flux that no longer violates Conservation of energy/momemtum. However note that the evaporation is still relatively instantaneous, as such equasions are described in 'slow motion'.
Stingray
NateTG said:
Let's say that I have an extremely small black hole -- i.e. the Schwartzschild radius is smaller than the Planck length.
In that case, the derivation of Hawking radiation is no longer valid. You'd need a fully theory of quantum gravity, whereas Hawking radiation is derived by formulating quantum field theory on top of a fixed classical spacetime.
Wavelength
I am Doing a project on Hawking Radiation and I am wondering if anyone actually Knows the wavelength of it?? It would be great if someone could email me cos I don't always have time to do much more than check my emails
--
a.random.persona@gmail.com
--
Thanks
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2020-09-27 02:08:04
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https://math.stackexchange.com/questions/3353145/find-when-equation-has-exactly-one-solution
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# Find when equation has EXACTLY one solution [closed]
Simple idea, but I cannot figure out a way to do this. What I want is to find for which value of $$k$$ some equation $$f(x,k)=g(x)$$ has exactly one real solution.
For example, say $$f(x,k)=x^2+k$$ and $$g(x)=6x^2+3$$. Looking at this particular example, it's not too hard to figure out the value of $$k$$ we want is $$3$$. However, it's much more difficult for more complex functions, like $$f(x,k)=kx^4-x^3$$ and $$g(x)=\ln(x)-x^2$$. Looking at the graphs of these functions, it's easy to see that there must be exactly one value of $$k$$ where they are equal exactly once. Finding said value, however, eludes me.
The nature of this problem leaves it begging to be solved using limits, and calculus is never too far behind when limits are involved, but I'm not too familiar with multivariable calculus...
## closed as too broad by Luca Bressan, Dietrich Burde, Matthew Daly, N. F. Taussig, Lee David Chung LinSep 12 at 1:52
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
• The answer strongly depends on the forms of $f$ and $g$ – caverac Sep 11 at 18:16
• Some equation? This is too broad. Most equations will not work, e.g., $f(x,k)=(x^2+1)^k$. – Dietrich Burde Sep 11 at 18:42
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2019-10-17 22:48:28
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https://ltwork.net/se-sintio-muy-cansado-muy-y-cansando-cuales-son-sus-funciones--1532
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# “se sintio muy cansado.”muy y cansando: cuales son sus funciones gramaticales? mi maestra dice que los dos son adverbios. porfi
###### Question:
“se sintio muy cansado.” muy y cansando: cuales son sus funciones gramaticales? mi maestra dice que los dos son adverbios. porfi ayuda
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2022-12-08 06:54:48
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