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# Open Access System for Information Sharing
Thesis
Cited 0 time in Cited 0 time in
곱함수 연산자의 르베그 공간 계측
Title
곱함수 연산자의 르베그 공간 계측
Authors
Date Issued
2013
Publisher
포항공과대학교
Abstract
The generalized Riesz mean of order $\alpha$ associated with a function $\rho$ is defined by \begin{equation*} \widehat{S^{\alpha}_tf}(\xi)=\Big(1-\frac{\rho(\xi)}t\Big)_+^\alpha \widehat{f}(\xi), \quad \xi = (\xi ', \xid)\in \mathbb R^d\times \mathbb R. \end{equation*} When $\rho(\xi)= \xi$, $S^\alpha_t$ is the standard Bochner-Riesz operator in $\mathbb R^{d+1}$ (referred to later as spherical means, as the set $=1$ is a sphere). When $\rho(\xi)=\frac{ \xi' ^2}{\xi_{d+1}^2}$ with a cutoff in $\xi_{d+1}$, it is called a cone multiplier operator in $\mathbb R^{d+1}$. There has been a lot of work devoted to the problem of $L^p$ and $L^p-L^q$ boundedness of the Bochner-Riesz and cone multiplier operators. They are generally known as difficult problems in this field and there are still many open problems. In fact, for the cone multiplier there are some additional difficulties as compared with Bochner-Riesz problem. In this thesis, we consider $L^p$ and maximal $L^p$ estimates for the generalized Riesz means which are associated with the cylindrical distance function $\rho(\xi) = \max \{ \xxi , \xid \}$, $(\xxi,\xid)\in\mathbb R^{d}\times\mathbb R$. We prove these estimates up to the currently known range of the spherical Bochner-Riesz and its maximal operators. This is done by establishing implications between the corresponding estimates for the spherical Bochner-Riesz and the cylindrical multiplier operators. Secondly, we obtain some sharp $L^{p}-L^{q}$ estimates and the restricted weak-type endpoint estimates for the multiplier operator of negative order associated with conic surfaces in $\mathbb{R}^{3}$ which have finite type degeneracy.
URI
http://postech.dcollection.net/jsp/common/DcLoOrgPer.jsp?sItemId=000001554465
http://oasis.postech.ac.kr/handle/2014.oak/1731
Article Type
Thesis
Files in This Item:
There are no files associated with this item.
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# A question about sigma-algebras and generators
Suppose you are given a set $\Omega$ and a collection $\mathcal{G}$ of subsets of $\Omega$. Assume further that $A \subset \Omega$. Now let $\sigma_{\Omega} (\mathcal{G})$ denote the smallest sigma-algebra on $\Omega$ containing $\mathcal{G}$, and let $\sigma_{A}(\mathcal{G} \ \cap A)$ denote the smallest sigma-algebra on A containing the collection $\mathcal{G} \ \cap A$.
Is it true that $\sigma_{A}(\mathcal{G} \ \cap A) = \sigma_{\Omega} (\mathcal{G}) \cap A$ ?
The inclusion " $\subset$ " is clear, since if $\mathcal{H}$ is a sigma-algebra on $\Omega$ containing $\mathcal{G}$, then $\mathcal{H} \cap A$ is a sigma-algebra on $A$ containing $\sigma_{A}(\mathcal{G} \ \cap A)$. But what about the other inclusion?
Thanks for your help! Regards, Si
-
I don't understand the notation $\mathcal{G}\cap A$, since $\mathcal{G}$ is the set of subsets of $\Omega$ while $A$ is the subset of $\Omega$. – Jack Nov 17 '11 at 15:15
@Jack: I think that's why Si wrote "the family $\mathcal{G} \ \cap A$", to make clear that every set in $\mathcal{G}$ is being intersected with $A$, much like when we write $gH$ for a coset of a subgroup. – joriki Nov 17 '11 at 15:21
Let $$\mathcal{B} = \left\{B \subset X | B \cap A \in \sigma_A(\mathcal{G} \cap A)\right\}.$$ Notice that $\mathcal{G} \subset \mathcal{B}$. It is easy to see that $\mathcal{B}$ is a $\sigma$-algebra. For example, if $B_j \in \mathcal{B}$, then $$\left(\bigcup B_j\right) \cap A = \bigcup (B_j \cap A) \in \sigma_A(\mathcal{G} \cap A),$$ because $B_j \cap A \in \sigma_A(\mathcal{G} \cap A)$.
Therefore, since $\mathcal{B}$ is a $\sigma$-algebra containing $\mathcal{G}$, we can conclude that $\sigma(\mathcal{G}) \subset \mathcal{B}$. So, for every $B \in \sigma(\mathcal{G})$, $B \cap A \in \sigma_A(\mathcal{G} \cap A)$. In other words, $$\sigma(\mathcal{G}) \cap A \subset \sigma_A(\mathcal{G} \cap A).$$
+1, nice! To put it in words: It doesn't matter what happens outside $A$; the elements of $\mathcal G$ happen to have certain parts outside $A$, and these may interact in complicated ways to determine $\sigma_\Omega(\mathcal{G})$, but when we cut everything down to $A$ none of that matters; we might as well consider all possible extensions outside $A$ of the elements of $\sigma_A(\mathcal G\cap A)$ (as you did in $\mathcal B$); then the result is clearly a sigma algebra, and yet we don't get more than $\sigma_A(\mathcal G\cap A)$ when we cut it down to $A$. – joriki Nov 17 '11 at 15:57
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# Solar panel maximum efficiency at specific frequency and temperature?
I talked to a professor about solar panels and their efficiency. It seemed that the main reason solar panels aren't that efficient is because it can only accept a single energy input size. Anything below is ignored and anything above only gives the energy of that single limit.
That is, say the limit is 1W. A photon with 0.999W energy wouldn't give any voltage. A photon with 10W energy would give the same voltage as a 1W photon.
Does this mean I can get very high efficiency from a light source with a specific frequency? For example a solar-powered calculator in a room with only sodium lights. How high?
How does the efficiency change depending on the amount of light or the temperature of the solar cell? How about an extremely small amount of light, like 1nW, close to 0K?
This could be terribly wrong as there was quite a language barrier between us. I want to learn, so please point out any misstakes.
In theory, you could get almost $100\%$ efficiency from a solar cell exposed to light with the photon energy just above the band gap. Each absorbed photon generates an electron of almost the same energy. The problem is that there are only so many band gaps available, so you have to find a light source at the correct wavelength to match one. The intensity of the light doesn't change the efficiency, just how much power you can get out of the cell. Over a wide range, if the intensity doubles, you can generate twice as much power.
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# How do you solve 319= \frac { r } { 21} - - 296?
###### Question:
How do you solve 319= \frac { r } { 21} - - 296?
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# P vs NP explanation on wikipedia
I'm reading the page under http://simple.wikipedia.org/wiki/P_versus_NP, which states:
"...people want to know if there are any NP problems that are not P problems. That means they would like to know if there are any problems where the answer cannot easily be found by a computer, but if someone says he has the answer, it is easy to use a computer to check if that answer is correct."
Don't the included examples show this? Example 1, for instance, states:
However, if she proposed a division of the rocks, it would be trivial to check if she was right. All you would need to do would be to check if the sum of the weights in each pile were equal, which is easy for a computer to do, even for this large number.
Doesn't this show that the method for finding a correct division of rocks is difficult to compute but easy to verify?
-
Philip, I don't understand what's bothering you. Very roughly, and I'm annoyed that Wikipedia gives such an imprecise definition, a problem which is NP but not P has no simple method for finding an answer, but once a proposed answer is found, it is easy to check whether that answer is correct. – Grumpy Parsnip Sep 15 '11 at 22:37
@Jim It's a "simple english wikipedia" page. :) – Srivatsan Sep 16 '11 at 0:00
No, it simply says that it is easy to verify. Is it difficult to compute? Maybe. It might be. We certainly aren't very good at computing it right now, but we haven't proven that it is hard to compute yet (hard in the sense that $P \neq NP$).
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# Internal Design of American Option Pricing Engine¶
## Overview¶
The American option can be exercised at any time up until its maturity date. This early exercise feature of American Option gives the investors more freedom to exercise, compared to the European options.
## Theory¶
Because the American option can be exercised anytime prior to the stock’s expire date, exercise price at each time step is required in the process of valuing the optimal exercise. More precisely, the pricing process is as follows:
1. Generate independent stock paths
2. Start at maturity $$T$$ and calculate the exercise price.
3. Calculate the exercise price for previous time steps $$T-1$$, compare it with the last exercise price, select the larger one.
4. Keep rolling back to previous time steps until $$t$$ is $$0$$ and obtain the max exercise price $$V_t$$ as the optimal exercise price.
In the process above, for each time step $$t$$, conditional expectation $$E_t[Y_t|S_t]$$ of the payoff is computed according to Least-Squares Monte Carlo (LSMC) approach, proposed by Longstaff and Schwartz. Mathematically, these conditional expectations can be expressed as:
$E_t[Y_t|S_t] = \sum_{i=0}^{n}a_iB_i(S_t)$
where $$t$$ is the time prior to maturity, $$B_i(S_t)$$ is the basis function of the values of actual stock prices $$S_t$$ at time $$t$$. The constant coefficients of basis functions are written as $$a_i$$ , and acts as weights to the basis functions. The discounted subsequent realized cash flows from continuation called $$Y_t$$.
Here we employed polynomial function with weights as the basis functions, so the conditional expectations can be re-written as:
$E_t[Y_t|S_t] = a + bS_t + cS_t^2$
where 3 functions: 1, $$S_t$$ and $$S_t^2$$ are employed. The constant coefficients $$a$$, $$b$$ and $$c$$ are the weights.
By adding immediate exercise value $$E_t(S_t)$$ to the Equation above, and exchanging the side of elements, the equation is changed to
(2)$a + bS_t + cS_t^2 + dE_t(S_t) = E_t[Y_t|S_t]$
$$a$$, $$b$$, $$c$$, $$d$$ are constant coefficients. These coefficients need to be known while calculating the optimal exercise price in mcSimulation model. More details of American-style optimal algorithm used in our library refer to “da Silva, J. N., & Fernández, L. A Monte Carlo Approach to Price American-Bermudan-Style Derivatives.”
Note
Theoretically, the basis functions $$B_i(S_t)$$ can be any complex functions and the number of basis function may be any number. The American Option implemented in our library employs three polynomial functions, namely, 1, $$S_t$$ and $$S_t^2$$, which proved by Longstaff and Schwartz that works well, and is a typical setup in real implementations.
## Implementation¶
In the Theory Section, the pricing process is described. However, this process can only be executed in the condition of Equation (2) is given. This means that the coefficients in Equation (2) must be calculated first. This calculation process is called Calibration and executes prior to the Pricing process. Therefore, in the implementation, the American Monte Carlo Engine contains two processes: Calibration and Pricing, as illustrated in Figure 10.
th: 50%
align: center
Hint
Why two processes are required in MCAmericanEngine and only one process, pricing, is enough for European Option Monte Carlo engine?
For European Option, the exercise price of the stork at the last time step $$T$$ needs to be and only need to be calculated. However, for American Option, the exercise price of the stock at all time steps $$t, 0 \leq t \leq T$$ are required to be computed. Therefore, the mathematical model of American Option Monte Carlo employs several basis functions, refer to Equation (2).
In the aspect of basis functions, the European Option employed only one basis function: $$S_T$$, where $$T$$ is the maturity date. By introducing multiple basis functions, the coefficients are also introduced. Thus, the calibration process is deployed to compute these coefficients.
### Calibration Process¶
Calibration process aims to calculate the coefficients that will be used in the pricing process.
The detailed calculation process is as follows:
1. Generate uniform random numbers with Mersenne Twister UNiform MT19937 Random Number Generator (RNG) followed by Inverse Cumulative Normal (ICN) uniform random numbers. Thereafter, generate independent stock paths with the uniform random numbers and Black-Sholes path generator. The default paths (samples) number used in the calibration process is 4096. Thus, 4096 random numbers are generated for each time step $$t$$.
2. Refer to Equation (2), $$a$$, $$b$$, $$c$$, $$d$$ are the unknown coefficients that should be calculated. We denote these coefficients as $$x$$, so
$\begin{split}x = \begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}\end{split}$
The expressions derived from path data $$S_t$$, $$S_t^2$$ and $$E_t$$ can be obtained for each time step $$t$$. Here we denote them as $$A$$.
$A = \begin{bmatrix} 1 \ S_t \ S_t^2 \ E_t(S_t) \end{bmatrix}$
Equation (2) can be re-written as:
(3)$y = A x$
where $$y$$ is the conditional expectation $$E_t(Y_t|S_t)$$ in Equation (2). To simplify the expression in deduction, it is denoted as $$y$$ in this section. By backward process, this conditional expectation $$y$$ for each time step can be obtained. Which is to say, $$y$$ is actually the optimal exercise price in the period of $$t$$ to $$T$$. Therefore, the problem of computing coefficients is changes to find the solution for Equation (3).
In practical, in this step, we calculate the value of 4 elements in vector $$A_t$$ for each time step $$t$$. Which is to say, 4 outputs of this stage are $$1$$, $$S_t$$, $$S_t^2$$ and $$E_t$$. The default size of :$$A_t$$ for each time step is 4096 * 4.
To simplify the process of solving Equation (3), matrix data $$A$$ is multiplied with its transform $$A^T$$. A new 4*4 matrix $$B$$ can be derived:
$A^T A = B.$
Corresponding to the two steps described above, the hardware architecture is shown in Figure 11.
We denote the process from RNG to generate matrix data $$B$$ and exercise price for each timestep $$t$$ as a Monte-Carlo-Model (MCM) in American Option Calibration Process. Each MCM process 1024 data. With a template parameter UN_PATH, more pieces of MCM can be instanced when hardware resources available.
To connect multiple MCM data, two merger blocks are created: one for merge price data, one for merge matrix $$B$$. Meanwhile, to guarantee all calibration path data can be executed in a loop when there is not enough MCM available, a soft-layer Merger that accumulates all elements of $$B$$ data is employed. Since these intermediate data need to be accumulated multiple times, a BRAM is used to save and load them.
1. Once we get the matrix $$B$$, the singular matrix $$\Sigma$$ of $$B$$ could be obtained by SVD (Singular Value Decomposition).
$B = U \Sigma V.$
1. Thereafter, use Least-Squares to calculated the coefficients by modifying Equation (3) to:
$\begin{split}y &= A x \\ A^T y &= A^TA x \\ A^T y &= Bx\end{split}$
Until now, matrix $$B$$ and vector $$y$$ are known, coefficients $$x$$ could be calculated.
The implementation of step 3 and steps 4 is shown in Figure 12. Because step 2 generates matrix $$B$$ and price data $$y$$ from timesteps $$0$$ to $$T$$. Step 3 processes these data in the backward direction, from timesteps $$T$$ to $$0$$. Considering the amount of data, 4096*timesteps*8*sizeof(DT) and 9*timesteps*8*sizeof(DT), it is impossible to store all the data on FPGA. In the implementation, DDR/HBM memory is used to save these data. Correspondingly, DDR data read/write modules are added to the design.
Besides, notice that since SVD is purely computing dependent, which is pretty slow in the design. Therefore, a template parameter UN_STEP is added to speed up the SVD calculation process.
Note
It is worth mentioning that 4096 is only the default calibrate sample/path size. This number may change by customers’ demands. However, the size must be a multiple of 1024.
### Pricing Process¶
#### MCAmericanEnginePricing¶
The theory of the pricing process is actually already introduced in the Theory Section. Similar to the European option engine, mcSimulation framework is employed. Compared to the European option engine, the difference is that it needs to calculate the optimal exercise at all time steps. The detailed implementation process of American engine is drawn as Figure Figure 13 shows.
1. Generate uniform random numbers with Mersenne Twister UNiform MT19937 Random Number Generator (RNG) followed by Inverse Cumulative Normal (ICN) uniform random numbers. Thereafter, generate independent stock paths with the uniform random numbers and Black-Sholes path generator.
2. Refer to Equation (2), calculate the exercise price at time step $$T$$ by utilizing the calculated coefficients $$x$$.
3. Calculate the exercise price for previous time steps $$t$$, and take the max exercise price by comparing the immediate exercise price with the held price value (maximum value for time steps $$t+1$$).
4. Continue the process, until time step $$t$$ equals 0, optimal exercise price and the standard deviation is obtained.
5. Check if the standard deviation is smaller than the required tolerance defined by customers. If not, repeat step 1-4, until the final optimal exercise price is obtained.
Caution
Notice that the pricing module also supports another approach of ending pricing process, which utilizes the input parameter requiredSamples. When the requiredSamples are processed, we assume the output mean result is the final optimal exercise price. This mode is also supported in the American Option Pricing Engine. And the figure above only illustrates the ending approach with the required tolerance.
Note
In the figure, BRAM is used to load coefficients data that calculated from the calibration process. Here may use BRAM or DDR, it depends on the amount of data that need to be stored beforehand. The reason that coefficients data cannot be streamed is that the multi-Monte-Carlo process in pricing always needs to execute multiple times. Each time execution, these need to be loaded. Thus, it is impossible to use hls::stream.
### MCAmericanEngine APIs¶
In our library, the MCAmerican Option Pricing with Monte Carlo simulation is provided as an API MCAmericanEngine(). However, due to external memory usage on DDR/HBM and avoiding the designed hardware cross-SLR placed and routed. The American engine option supports two modes:
• single API version: use one API to run the whole American option
• three APIs version: three APIs/kernels are provided, connecting them on the host side to compose the overall design.
The boundary between them is external memory access. For the calibration process, two APIs are provided. Calibration step 1 and 2 are wrapped as one kernel, namely, MCAmericanEnginePreSamples. Step 3 and step 4 compose another kernel MCAmericanEngineCalibrate. And pricing process as another kernel MCAmericanEnginePricing in this library. Because the pricing process is separated as a kernel, the data exchange between the calibration and pricing process may not through the BRAM any more. Thus, in the implementation, DDR/HBM is used as the coefficients data storage memory.
With the three kernels, the kernel level pipeline by shortening the overall execution time could be achieved. However, employing kernel level pipeline requires a complex schedule from the host code side. An illustration of connection 3 kernels as a complete system is given in this part, which can be seen in Figure 14. Price data and $$B$$ matrix data are the outputs from kernel MCAmericanEnginePreSamples. For each timestep, path number (default 4096) price data B and x matrix data (a number of 9) need to be saved to DDR or HBM memory.
Kernel 1 MCAmericanEngineCalibrate reads price data $$y$$ and matrix data $$B$$ from external memory and outputs coefficients to DDR/HBM. The last kernel MCAmericanEnginePricing reads coefficients data from DDR/HBM and saves the final output optimal exercise price to DDR/HBM.
Hint
Why the number of matrix B is 9 in DDR/HBM?
The matrix $$A$$ is 4096 * 4 for each timestep when the path number is 4096 (default). The size of its transform matrix $$A^T$$ is 4 * 4096. So, the size of matrix $$B$$ is 4 * 4. However, some elements in $$B$$ are the same, and 9 can represent all 16 data. More precisely, assuming
$\begin{split} A^T = \begin{bmatrix} &1\ 1\ ...\ 1\ ...\ 1 \\ &S_0\ S_1\ ...\ S_t\ ...\ S_T \\ &S_0^2\ S_1^2\ ...\ S_t^2\ ...\ S_T^2\\ &E_0\ E_1\ ...\ E_t\ ...\ E_T \end{bmatrix}, \ \ \ \ A = \begin{bmatrix} 1\ S_0\ S_0^2\ E_0 \\ 1\ S_1\ S_1^2\ E_1 \\ ... \\ 1\ S_t\ S_t^2\ E_t \\ ... \\ 1\ S_T\ S_T^2\ E_T \end{bmatrix} \\ \\ ==> B = A^T \ A = \begin{bmatrix} \sum(1)\ \sum(S_i)\ \sum(S_i^2)\ \sum(E_i) \\ \sum(S_i)\ \sum(S_i^2)\ \sum(S_i^3)\ \sum(S_iE_i) \\ \sum(S_i^2)\ \sum(S_i^3)\ \sum(S_i^4)\ \sum(S_i^2E_i) \\ \sum(E_i)\ \sum(S_iE_i)\ \sum(S_i^2E_i)\ \sum(E_i^2) \\ \end{bmatrix}\end{split}$
It is evident that some elements are the same. After removing duplicated elements, the following 9 elements of $$B$$ are stored to DDR/HBM each timestep:
$\begin{split}B_{save} = \begin{bmatrix} &\sum(1)\ \\ &\sum(S_i)\ \\ &\sum(S_i^2)\ \sum(S_i^3)\ \sum(S_i^4)\ \\ &\sum(E_i)\ \sum(S_iE_i)\ \sum(S_i^2E_i)\ \sum(E_i^2) \end{bmatrix}\end{split}$
Caution
The architecture illustrated above is only an example design. In fact, multiple numbers of kernels, each with a different unroll number (UN) may be deployed. The number of kernels that can be instanced in design depends on the resource/size of the FPGA.
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I would like to incorporate tex4ht output into an existing web page. I'm wondering if there is a neat way to put the navigation bar and footer of said web page into every HTML file generated with tex4ht.
For example: can you tell tex4ht via a .cfg file to put certain HTML code at the beginning/end of the <body>?
You can use \Configure{BODY}{insert <body>}{insert </body>} configuration. Example:
\Preamble{xhtml}
\begin{document}
\Configure{BODY}
{\SaveEndP\IgnorePar
\HCode{\Hnewline<body \csname a:!BODY\endcsname
>%
}\ShowPar\par}
{\IgnorePar\EndP\HCode{%
\Hnewline<div class="footer"><p>This is a footer</p></div>\Hnewline
\Hnewline</body>}
\RecallEndP}%
\EndPreamble
There is some business related to paragraph handling involved, but what is important for your use case are \HCode commands, which are used to include HTML tags. You can also put text here, to insert line break, use \Hnewline command.
There is one issue with \HCode, you can't accented letters inside it's argument, it is best to use it only for tags including.
Sample document:
<?xml version="1.0" encoding="utf-8" ?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<!--http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd-->
<html xmlns="http://www.w3.org/1999/xhtml"
>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<meta name="generator" content="TeX4ht (http://www.tug.org/tex4ht/)" />
<meta name="originator" content="TeX4ht (http://www.tug.org/tex4ht/)" />
<!-- charset=utf-8,html,xhtml -->
<meta name="src" content="sample.tex" />
• This works brilliantly, thank you. Just one further question: if I use the german umlaut ä in the header/footer, it's replaced by \protect \let \prOteCt \relax \Protect \csname acp:c\endcsname {23}a (similiarly for ö and ü). Putting umlauts in your MWE doesn't result in the same problem, so it must be something in my current setup. Do you have a suggestion where I should start searching the culprit? – Adrian Sep 6 '16 at 15:05
• @Adrian try to move the accented letters outside the \HCode command. – michal.h21 Sep 6 '16 at 15:57
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# Tree vs lightning rod: why does one burn and the other not?
I have this simple question, but I cannot find the answer.
I saw this video about a plane getting hit by lightning. In it, Captain Joe explains why people do not get electrocuted. This has a simple explanation, due to the Faraday cage effect produced by the fuselage. But another question come to my mind in that moment: why does the aluminum from the fuselage, that acts as a Faraday cage, not melt because of the extreme currents carried by the lightning?
After this, I thought about the following example: A thin metal (correctly grounded) lightning rod is almost intact after a strike, while a tree breaks in the middle and sometimes it even burns:
Clearly it has something to do with the resistivity of each material, much higher in the tree's wood.
It is also said in this article that the only dangerous zone a plane can get hit "is the radome (the nose cone), as it's the only part of a plane's shell that's not made of metal". So it clearly has something to do with the conductive properties of the fuselage.
So my question is basically this: why does a tree break and burn when struck by lightning but a lightning rod does not?
And, ultimately: why does a plane hit by lightning not melt with the hundreds of thousands Amperes going through the fuselage?
• The trunk of a tree contains water. A lightning strike can heat some of that water to the point where the vapor pressure is high enough to split the wood (i.e., it causes a steam explosion.) The same thing can not happen in the aluminum skin of an airplane. Even if a lightning stroke was to deposit enough energy to cause some melting, that's all that would happen: the aluminum would merely melt. It would not explode. – Solomon Slow Nov 16 '17 at 13:50
• @jameslarge I'm not sure melting is much better than exploding for a plane. – JAB Nov 16 '17 at 19:07
• @JAB google for "aircraft lightning strike damage", click the "images" tab, judge for yourself. – Solomon Slow Nov 16 '17 at 20:08
• First, what actually happens when a plane is struck by lightning? Is the path cloud -> plane -> ground? Or does the plane just equalise in potential to the surrounding cloud/air? – peterG Nov 16 '17 at 23:52
• @jameslarge that should go in the answers box to be honest. – Tim Nov 18 '17 at 2:13
The amount of heat generated by current flowing through a resistor (whether from lightning or more ordinary sources) is directly related to the power dissipated by the resistor, which is $$P = I^2 R.$$ $R$ is small for objects made from good conductors, which many metals are, and large for objects that are made from bad conductors like plastic or wood. Since a lightning strike has a very short duration, the total heat generated during such a strike is not enough to melt metal, but enough to set wood aflame or melt plastic.
If you let the large currents from the lightning strike run through the metal for longer, it probably would also heat up gradually and eventually melt, but this would take longer than the time scale on which lightning occurs.
• It might be worth pointing out that, for a lightning strike, the resistance of the last few feet probably has very little effect on current, so we can treat I as relatively constant. That makes your P=I^2R connection even more powerful. – Cort Ammon Nov 16 '17 at 15:34
• This makes it valid. This answer is missing this very important detail; see the other answer for that. – user27542 Nov 17 '17 at 14:40
The high electrical current in a lightning strike delivers heat energy along the full length of the lightning bolt. Part of that length is in the ionized air over the plane, part is the plane's fuselage, and part is the ionized air from the plane down to ground. The current is the same, but the heat generated is proportional to the electrical resistance of the path, and the aluminum of the plane (as well as the joints that hold the aluminum parts together) has very low electrical resistance. So, the air path gets very hot, while the aluminum path does not.
Trees and air have high electrical resistance, but (at breakdown) have a narrow conducting channel (ionized gas is 'Z-pinched' into a narrow channel, and the first woody parts that carbonize will hog all the current). So,
$$HeatPower = I^2 R = I^2 \rho * L/A$$
Heating per unit length of air is more than the aluminum plane because air has higher resistivity, even with comparable conduction cross-sectional area.
A tree contains a lot of water which converts explosively into gas when the electrical current flows throughout the tree. Wood is also more combustible than metal, so it catches fire due to the extreme heat.
In an plane fuselage, the current flows harmlessly along the outside of the plane and back out. The plane is not grounded. The same way a car is isolated by having rubber tyres.
• Isolated by tyres? The bolt just bored through miles of air, and it will stop because of rubber? Why not just ignore the tyre and jump through the air the remaining few inches? – JDługosz Nov 16 '17 at 21:11
• @JDługosz: The lightning bolt goes wherever it wants, of course. But it doesn't particularly want to hit the car, because the tyres have prevented a countercharge from the ground from flowing up into the car's frame and attracting the lightning that's coming down. (No, that is not the explanation this answer seems to suggest). – Henning Makholm Nov 16 '17 at 23:52
• lightning does strike cars. I can’t find anything on whether lightning is more or less likely to strike a car, per se, but have seen statements that height matters, and "good" vs "poor" conductors only matter in choosing between two close together targets. – JDługosz Nov 17 '17 at 0:03
• @HenningMakholm youtu.be/9Du_w976s5Q – zane scheepers Nov 17 '17 at 0:15
• Its not the rubber tires or that its grounded that protects you in a car. It is because the car is effectively a Faraday cage. – Kevin Milner Nov 17 '17 at 18:39
## protected by Qmechanic♦Nov 17 '17 at 5:49
Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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1. ## Exponent equation help
Hi I'm new to the forum so apologies if I've posted in the wrong section or my title is misleading.
My maths isn't great but I have a problem to solve using a pressure drop formulae 0.3=(800*160*300^2)/(7d^5.31)
So I need to solve for d to figure out my pipe diameter and I've got this far...
2.1d^5.31=11520000000 (Multiplying both sides by 7d^5.31)
d^5.31=5485714285.71 (Dividing both sides by 2.1)
This is as far as I've got and I'm unsure on the next step I wondered if somebody could point me in the right direction please? (Or point out I've gone wrong already if so)
2. ## Re: Exponent equation help
$\dfrac{(800)(600)(300^2)}{7d^{5.31}} = 0.3$
$480000\cdot 90000 = 2.1 d^{5.31}$
$\dfrac{1296000000000}{7} = d^{5.31}$
$d = \left(\dfrac{1296000000000}{7}\right)^{1/5.31} = 132.416$
3. ## Re: Exponent equation help
Originally Posted by Satnav
0.3=(800*160*300^2)/(7d^5.31)
This makes it a little easier:
u = 800*160*300^2
u / (7d^5.31) = 3 / 10
RULE: if a^p = x, then a = x^(1/p) : as Sir Romsek showed you
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# mcrl2/utilities/power_of_two.h¶
Include file:
#include "mcrl2/utilities/power_of_two.h"
## Functions¶
static constexpr bool mcrl2::utilities::is_power_of_two(T value)
Returns: True when the given value is a power of two.
static T mcrl2::utilities::round_up_to_power_of_two(T value)
Returns: The smallest power of two that is larger than the given value.
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# New example environment, with symbol at the end
Is there a standard formatted environment specifically used for examples? I'm writing a document on linear algebra, with the usual definition-theorem-proposition-example structure. Right now when I insert an example, I get something like:
This is a screenshot I took. Here is the MWE:
\documentclass[11pt,a4paper,openany]{report}
\usepackage{amssymb,amsmath,mathrsfs}
\usepackage[dutch]{babel}
\usepackage{lmodern}
\usepackage{mdframed}
\usepackage{bm}
\usepackage[many]{tcolorbox}
\usepackage{systeme,mathtools}
\makeatletter
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
\hskip -\arraycolsep
\let\@ifnextchar\new@ifnextchar
\array{#1}}
\makeatother
\usepackage{lipsum}
\usepackage{relsize}
\let\conjugatet\overline
\DeclareMathOperator{\Ima}{Im}
\DeclareMathOperator{\Span}{span}
\newcommand\md{\ }
\renewcommand{\qedsymbol}{$\blacksquare$}
\pagestyle{plain}
\renewcommand\baselinestretch{1.0}
\usepackage[nodisplayskipstretch]{setspace}
\setstretch{1.5}
\setlength{\abovedisplayskip}{0pt}
\setlength{\belowdisplayskip}{0pt}
\setlength{\abovedisplayshortskip}{0pt}
\setlength{\belowdisplayshortskip}{0pt}
\theoremstyle{definition}
\newmdtheoremenv[linewidth=1.0pt,linecolor=red]{Definition}{Definitie}
\numberwithin{Definition}{section}
\newmdtheoremenv[linewidth=1.0pt,linecolor=black]{Theorem}{Theorema}
\numberwithin{Theorem}{section}
\newtheorem{lemma}{Lemma}
\numberwithin{lemma}{section}
\newtheorem{proposition}{Propositie}
\numberwithin{proposition}{section}
\newtheorem{corollary}{Corollarium}
\numberwithin{corollary}{section}
\newtheorem{exmp}{Voorbeeld}[section]
\newcommand{\prooffont}{\scshape}
\usepackage{xpatch}
\tracingpatches
\xpatchcmd{\proof}{\itshape}{\prooffont}{}{}
\usepackage{marvosym} %Here starts the part you said I should add.
\usepackage{etoolbox}
\usepackage[amsmath, thref, hyperref, thmmarks]{ntheorem}
\theorembodyfont{\upshape}
\theoremseparator{.}
\theoremsymbol{\Large\color{Plum4}\Bat}
\newtheorem{exmp}{Voorbeeld}[section]
\AtBeginEnvironment{exmp}{\AtBeginEnvironment{itemize}{\leavevmode} \AtBeginEnvironment{enumerate}{\leavevmode}}
The 'Voorbeeld' is Dutch for 'example'. The problem I'm having is that I want my text directly aligned underneath the 'example' caption. In this case I used the 'itemize' command, but I want it to apply for all cases. This is what I wish to change:
Example 4.2.1. (---text comes here----)
I want the text under the 'example' caption, and also have it closed by some symbol (something other than the tombstone for proofs) at the end so that the reader knows the example ends here. Is there any way I could get this done?
Here another screenshot:
As can be seen, my text immediatly starts after the caption, which is not what I want.
Thanks in advance for any help.
Edit: Here is what I mean with a possible reduction of quality-output. I entered as code:
$$\label{2} [v]_{\gamma} = (b_1, b_2, \ldots, b_n)^T.$$
But this is what I received:
Notice the gamma float without being attached to my coordinate vector.
• Please provide an MWE of this screenshot. Delete all packages which are not needed here in order to make it easier for us. Do you want the described behaviour just for enumerations? Your Example 4.2.1. looks like the normal result. Please clarify this a bit! – LaRiFaRi Apr 14 '15 at 11:35
• I don't want the text right next to the 'example' caption. I want the text to start one line beneath the caption. – Kamil Apr 14 '15 at 13:13
You can do that easily with the ntheorem package (instead of amsthm) and etoolbox. I also loaded the enumitem package to improve the layout of itemize environments in such context. Here is a possible code:
\documentclass[11pt,a4paper,openany]{report}
\usepackage[utf8]{inputenc}
\usepackage{amssymb, mathrsfs}
\usepackage[dutch]{babel}
\usepackage{lmodern}
\usepackage{mdframed}
\usepackage{bm}
\usepackage[many]{tcolorbox}
\usepackage{systeme, mathtools}
\makeatletter
\renewcommand*\env@matrix[1][*\c@MaxMatrixCols c]{%
\hskip -\arraycolsep
\let\@ifnextchar\new@ifnextchar
\array{#1}}
\makeatother
\usepackage{lipsum}
\usepackage{relsize}
\let\conjugatet\overline
\DeclareMathOperator{\Ima}{Im}
\DeclareMathOperator{\Span}{span}
\newcommand\md{\ }
%\renewcommand{\qedsymbol}{$\blacksquare$}
\pagestyle{plain}
\renewcommand\baselinestretch{1.0}
\usepackage[nodisplayskipstretch]{setspace}
\setstretch{1.5}
\setlength{\abovedisplayskip}{0pt}
\setlength{\belowdisplayskip}{0pt}
\setlength{\abovedisplayshortskip}{0pt}
\setlength{\belowdisplayshortskip}{0pt}
\usepackage[amsmath, thref, hyperref, thmmarks]{ntheorem}
\renewcommand{\qedsymbol}{blacksquare}
\theoremstyle{definition}
\theoremseparator{.}
\newmdtheoremenv[linewidth=1.0pt,linecolor=red]{Definition}{Definitie}[section]
\newmdtheoremenv[linewidth=1.0pt,linecolor=black]{Theorem}{Theorema}[section]
\theorembodyfont{\itshape}
\newtheorem{lemma}{Lemma}[section]
\newtheorem{proposition}{Propositie}[section]
\newtheorem{corollary}{Corollarium}[section]
\theoremstyle{break}
\theorembodyfont{\upshape}
\theoremsymbol{\ensuremath{\clubsuit}}
\newtheorem{exmp}{Voorbeeld}[section]
\theoremstyle{nonumberplain}
\theoremsymbol{\ensuremath{\blacksquare}}
\newtheorem{proof}{Bewijs}
\usepackage{enumitem}
\usepackage{etoolbox}
\AtBeginEnvironment{exmp}{\setlist[itemize, 1]{wide=0em, leftmargin=1.25em, labelwidth=0.7em}}
\begin{document}
\setcounter{chapter}{1}
\setcounter{section}{1}
\setcounter{exmp}{1}
\begin{exmp}
\begin{itemize}
\item $(\mathbb R,\mathbb C, + )$ is een reële (en dus geen complexe) vectorruimte. Nemen wij immers een willekeurige scalar $λ ∈ \mathbb R$ en vermenigvuldigen we een willekeurige vector $(a + bi) ∈ \mathbb C$ met deze scalar, dan behoort het product $(λ a + λ bi)$ tot $\mathbb C$. In dit geval hebben wij dus $\mathbb R × \mathbb C\to \mathbb C~~\colon (λ,a + bi) ↦ (λ a + λ bi)$.
\item $(\mathbb C, \mathbb R, +)$ is geen complexe vectorruimte, $(\mathbb C, \mathbb C, + )$ echter wel.
\begin{equation*} \label{2}{} [v]_{\gamma} =(b_1, b_2, \ldots, b_n)^T.\end{equation*}
\end{itemize}
\end{exmp}
\begin{Definition}
A fully understandable definition.
\end{Definition}
\begin{lemma}
A test lemma.
\end{lemma}
\begin{proof}
A proof for fun!
End of proof:
\begin{align*}
a & =b\\
a + c & =b + d
\end{align*}
\end{proof}
\end{document}
• Tried adding it to my document, but it then builds forever when I click the 'build' button. I deleted the line \usepackage[amsmath, thref, hyperref, thmmarks]{ntheorem} and then it stopped building and gave me 6 errors. But when I add this line it doesn't work. – Kamil Apr 14 '15 at 14:13
• It says: "Package ntheorem error: Theorem style plain already defined." – Kamil Apr 14 '15 at 14:14
• Don't load both amsthm and ntheorem. If you like the amsthm formatting, the amsthm option of ntheorem provides an emulation. – Bernard Apr 14 '15 at 14:22
• Still doesn't work, now it says: \qedsymbol undefined. Maybe I should add a full MWE? I have several other environments and maybe they don't go together. – Kamil Apr 14 '15 at 14:24
• Yes, that would make it easier to find what's happening. Can you compile my code? – Bernard Apr 14 '15 at 14:28
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Why is it difficult to use the law of conservation of energy to calculate the effects of a collision?
Then teach the underlying concepts
Don't copy without citing sources
preview
?
Explanation
Explain in detail...
Explanation:
I want someone to double check my answer
1
Jul 25, 2017
Conservation of Energy only applies to Elastic Collisions. Conservation of Momentum applies to both Inelastic and Elastic Collisions
Explanation:
Elastic collisions by definition do not lose energy.
Inelastic collisions do lose energy to heat and vibration.
Inelastic collisions actually do conserve energy, but the loss of energy to heat and mechanical vibration is hard to calculate so the math equating energy before and energy after is hard to balance.
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# Constrained optimization for $u(x_1,x_2,x_3,x_4)=\alpha \min \{a x_1, b x_2\} + \beta \min \{c x_3, d x_4 \}$ [duplicate]
Suppose preferences are represented by the following utility function
$$u(x_1,x_2,x_3,x_4)=\alpha \min \{a x_1, b x_2\} + \beta \min \{c x_3, d x_4 \}$$
Write the
1. Walrasian demand functions
2. Hicksian demand functions
3. Indirect utility function
4. Expenditure function
Remark
I come from a math background and my only experience with economics are (so far) the first three chapters of the book Microeconomic Theory by Mas-Colell, Whinston, and Green. Following their approach, I want to solve the following utility maximization problem
$$\max_{\vec{\mathbf{x}}\in \mathbb{R}^4} \{u(\vec{\mathbf{x}})\} = \max_{x_1,x_2,x_3,x_4} \{\alpha \min \{a x_1, b x_2\} + \beta \min \{c x_3, d x_4 \} \}$$
s.t. $p_1 x_1 + p_2 x_2 = w$
to get the Walrasian demand functions $x_i(\vec{\mathbf{p}},w)=x_i(p_1,p_2,p_3,p_4,w)$. Then I could get the indirect utility function by substituting these solutions into the utility function: \begin{equation*} v(\vec{\mathbf{p}},w)=u(x_1^*(\vec{\mathbf{p}},w),x_2^*(\vec{\mathbf{p}},w),x_3^*(\vec{\mathbf{p}},w),x_4^*(\vec{\mathbf{p}},w)) \end{equation*} Similarly, to get the Hicksian demand functions $h_i(p_1,p_2,p_3,p_4,u)$, I would solve the expenditure minimization problem \begin{equation*} \min_{\vec{\mathbf{x}}} \{\vec{\mathbf{p}} \cdot \vec{\mathbf{x}}\} =\min_{x_1,x_2,x_3,x_4} \{p_1x_1+p_2x_2+p_3x_3+p_4x_4\} \end{equation*} and then get the expenditure function by multiplying \begin{align*} e(\vec{\mathbf{p}},u) &=\vec{\mathbf{p}} \cdot \vec{\mathbf{h}}^*(\vec{\mathbf{p}},u) \\ &=p_1 h_1^*(\vec{\mathbf{p}},u) +p_2 h_2^*(\vec{\mathbf{p}},u)+p_3 h_3^*(\vec{\mathbf{p}},u) + p_4 h_4^*(\vec{\mathbf{p}},u) \end{align*} Although I can visualize the entire process, I'm stuck at the beginning - solving the constrained utility maximization and expenditure minimization problems. Starting with the utility maximization problem, I want to write the Lagrangian \begin{equation*} \mathcal{L}(x_1,x_2,x_3,\lambda)=\alpha \min\{a x_1, b x_2 \} + \beta \min \{c x_3, d x_4\}+\lambda(w-p_1x_1-p_2x_2) \end{equation*} and take the first order conditions \begin{equation*} \frac{\partial u(x_1,x_2,x_3,x_4)}{\partial x_i}=\lambda p_i \end{equation*} however I don't know how, mathematically, to take the above first order conditions with respect to $x_i$ since $\min$ isn't everywhere differentiable. (However, as this answer points out, the partial derivatives of our objective function exist almost everywhere.)
This is my first experience with solving for demand functions, aside from the Cobb-Douglas example provided in the book Microeconomic Theory. Could anyone help me complete this example?
• economics.stackexchange.com/questions/2969/… should help you. In particular the first sentence "No, you should not use Lagrange multipliers here, but sound thinking." – Martin Van der Linden Nov 4 '15 at 2:43
• Great. Now you need to solve for different cases. Assume that $ax_1 > bx_2$ and find the solutions. Then assume other way around and solve again. Similarly for $x_3$ and $x_4$. You will have different solution for different prices. – Sher Afghan Nov 4 '15 at 3:28
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# Exponential Distribution with Probability
1. Dec 7, 2013
$$f(y) = \begin{cases} 0& \text{for }y< 0,\\ 2y& \text{for }0 ≤ y ≤ .5,\\ 6-6y& \text{for }0.5 < y ≤ 1, \\0& \text{for } y > 1\end{cases}$$
(1) Find cumulative distribution function, F(y)
$$F(y) = \begin{cases} 0& \text{for }y< 0, \\\int_0^y 2t dt = y^2 & \text{for } 0 ≤ y ≤ .5,\\.5^2+ \int_{0.5}^y (6-6t) dt = 6y-3y^2-2 & \text{for }0.5 < y ≤ 1\ \\1& \text{for } y > 1\end{cases}$$
(2) P(1/4 < Y < 3/4) = 6(3/4)-3(3/4)^2-2-(1/4)^2 = 3/4
Could anyone check (1) and (2) for me?
Last edited: Dec 7, 2013
2. Dec 7, 2013
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# Speeding up ODEINT stiff example [closed]
I am trying the implement the following odeint solver example but my differential equations are different.
The system I have is stiff so the rosenbrock method is a good fit for that. However, the solution process is really slow, often taking 2 million steps to reach t = 2.5s.
Is there any way to tweak the adaptive stepping settings and see what steps are being taken by the solver? I haven't been able to find a good documentation for this, the only one I found is
If there is a way to speed it up without tweaking the stepping, that would work as well.
Thanks.
Not necessarily, I wouldn't not blindly say this at all. The Rosenbrock methods tend to be good for stiff equations when you are trying to solve for error less than $\approx10^{-6}$ and your system of equations is not sufficiently large (I'll point to Hairer II, the DiffEqBenchmarks, and lots of other currently unpublished benchmarks which all pretty much say the same thing). But then again, multistep methods can still "sneak a win" if the system is smooth enough every one in awhile, so it's not a hard-fast rule. If you really want to be sure what's good for your problem, I would recommend solving a small part of your real equation using some suite that has easy access to Radau, both high and low order Rosenbrock, a BDF, and optionally high and low order (E)SDIRK methods since these tend to do the best in various regimes. You should also generate a reference solution since comparing timings without comparing error is never a good idea (many times the faster one can just have a higher error), which is why work-precision diagrams are a good tool if you want to do this comparison easily. This example in the DiffEqBenchmarks shows an easy way to run such a test, though you can also setup something similar with MATLAB or with wrappers to Hairer's solver suite for example to see similar results (but with some missing methods in each case).
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#### Howdy, Stranger!
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# [open] OpenSesame freezing with touch_response and mouse click
edited October 2014
Dear all,
We're building an experiment in which a short video is presented and then participants are asked to make a choice amongst 4 possible drawings appearing on the screen. We use the touch_response plug-in to get responses from the mouse (back-end is 'legacy'). The sequence is as follows:
1. video presentation
2. two-choice between repetition of video or response -> response is given by mouse click and touch_response
3. 4 drawings appear on the screen in 4 different locations and stay for 5 seconds before to timeout ->response by mouse click and touch_response
However, if the mouse is accidentally moved or clicked during the video presentation OpenSesame seems to freeze and slows down considerably. Is there a way to 'isolate' the mouse during the presentation of the videos?
Thanks, Stefania
• edited 10:17AM
Hi Stefania,
I tried to reproduce the problems you described, but couldn't do so. There was no interference of the media player plugin and the mouse (or the keyboard). Basically, it seemed to work like you want it to.
Is it possible that you provide us with some more information about your issue and the way you implemented your experiment? Some code would be helpful, too. In this way, I am sure we can fix it.
Eduard
• edited 10:17AM
hello Eduard, thanks for getting back to us.
Attached is a screenshot of our experiment pipeline and below some more details.
‘Choice loop’ allows to choose between loop 1 and loop 2 (by touch-response, screen divided in two, mouseclick on the left part of the screen runs loop 1, mouseclick on the right runs loop 2). Within each loop, after instructions and a fixation dot, media_player_vlc plays the first video set in the variable ‘video_a1’ (if we run loop 1). The parameters of the item are the standard ones. Then a sketchpad and a touch_response item allow the subject to answer or watch the video another time. In the first case, a sketchpad (4 images saved as an unique image an called sequentially by the variable ‘image_a1’) and a touch_response are displayed. In the second case, a new media_player-vlc item plays the same video as before, then a sketchpad/touch _response allows the subject to mouseclick and see the same image (‘image_1’) as before; participants have to click the correct image among 4 choices (screen divided in 4 through touch_response item) by mouseclick.
The problem of “freezing” during the video seems more serious with high-resolution video. With low-resolution video (854 x 480) interferences appear less often and the problem doesn’t occur with a video with very low resolution (480x270). We’ve also tried different extension (.mp4, .avi) using different computers, but these changes do not seem to solve the problem.
Thanks, Stefania and Joshua
• edited 10:17AM
Hi you two,
I don't know whether this will solve the problem, but it is certainly a good idea to change duration in each media_player_vlc element (and others if necessary) to "mouseclick", since you're using mouse/touch responses, not the keyboard, right?
Also, I wonder why do you have separate elements for "video A1" and "repeated video A1" ? As far as I understand your idea, they should be the same. If so, it would be more efficient if you reuse already existing elements.
Let me know, if this is already enough to fix it. I will keep trying to figure out whether I can find something else in the meantime.
Eduard
• edited 10:17AM
Hi Eduard,
Thanks for your suggestions. We took a bit to get back as we tried to implement them. Unfortunately none of the changes solved the issue, the only change that seems to matter is the reduction of the video resolution (which obviously is not ideal for the nature of our experiment). We tried this out on different laptops and the issue persists.
Very best, Stefania and Joshua
• edited 10:17AM
Hi guys,
So if I understand the situation correctly, the main problem here is that the media_player_vlc stutters during playback. Is that right?
You could try what happens if you use a different back-end, although I would expect the legacy back-end (which you're using now, right?) to work best.
Alternatively, you could install the media_player_gst plug-in. This is set to replace the vlc-based plug-in, which has some issues (as you're experiencing now). However, the gst-based plug-in is still in development, and requires installing some additional packages:
@dschreij Any suggestions?
Cheers,
Sebastiaan
There's much bigger issues in the world, I know. But I first have to take care of the world I know.
cogsci.nl/smathot
• edited 10:17AM
We are trying to use the media_player_gst plug-in, but we have some problems. We've placed the extracted plug-in folder (downloaded as a zip file from the website) in the opensesame/plugins folder e we've installed the Gstreamer runtime, that created the folder gstreamer-sdk in the path "c:/gstreamer-sdk/", that seems to be the default path indicated in the instructions. Nonetheless, when we try to use the plug in (that we can actually see in OS) we see the error message: "Failed to load plug-in media_player_gst" and this is the traceback in the debug window:
File "dist\libopensesame\item_store.py", line 77, in new
File "dist\libopensesame\plugins.py", line 331, in load_plugin
File "dist\libopensesame\plugins.py", line 303, in import_plugin
File "C:\Program Files (x86)\OpenSesame\plugins\media_player_gst\media_player_gst.py", line 94, in
osexception:
OpenSesame could not find the GStreamer framework!
It seems to be the typical error seens when the gstreamer is installed in a different folder, but our location should be the default one. We've also tried to move the location of the gstreamer-sdk folder and add a line of code to the media_player_gst.py script as suggested in an other topic of this forum, as:
import os
os.environ["GSTREAMER_SDK_ROOT_X86"] = "our new directory path"
but unfortunately we always have the same error.
Probably we've missed something in the process.. we will appreciate any kind of suggestions!
Thank you very much for your help,
best,
Joshua and Stefania
• edited November 2014
Hi Joshua,
when you install the GStreamer SDK it should automatically create the environmental variable GSTREAMER_SDK_ROOT_X86 for you. If this has not happened, something probably went wrong with the installation. Manually assigning this viarable is I think of little use.
Are you sure you installed the 32-bit version of GStreamer and not the 64-bit one (if the latter is the case, there should be a GSTREAMER_SDK_ROOT_X86_64 variable in os.environ[])? That is the only thing I could think of now that prevents OpenSesame from finding GStreamer. OpenSesame in Windows is still built on a 32-bit Python environment, which requires the 32-bit version of GStreamer SDK to be installed.
Let me know if this is the case. If so, we'll look further.
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# LHCb Theses
Senast inlagda poster:
2017-03-20
10:19
Flavour studies with LHCb: b-meson mixing, lepton-flavour violation and the velo upgrade / Bird, Thomas Semileptonic $B$ decays of the type $B^0_q\to{}D^-_q\mu^+\nu$ (where $D^-\to{}K^-K^+\pi^-$) are selected and their lifetimes are corrected using a statistical simulation-based correction called the $k$-factor [...] CERN-THESIS-2015-393 - 242 p.
Fulltext
2017-03-16
09:08
Search for the decay $B^+ \to D^{*-}K^+\pi^+$ / Rera, Ronnie P The $B^+ \to D^{*-}K^+\pi^+$ decay is searched for in a sample, collected with the LHCb detector, of proton-proton collision data at centre-of-mass energies $\sqrt{s} = 7$ and $8 ~\text {TeV}$, corresponding to an integrated luminosity of $3 \text {fb}$ [...] CERN-THESIS-2017-018 - 63 p.
Fulltext
2017-03-06
18:12
Measurements of charm production and $CP$ violation with the LHCb detector / Pearce, Alex This thesis presents two measurements made using data collected by the LHCb detector, operating at the Large Hadron Collider accelerator at the CERN particle physics laboratory [...] CERN-THESIS-2016-260 - 308 p.
Fulltext
2017-02-12
23:13
Measurement of the $CP$ violation parameter $A_\Gamma$ in $D^0 \to K^+ K^-$ and $D^0 \to \pi^+ \pi^-$ decays / Marino, Pietro CERN-THESIS-2017-007 - 169 p.
Fulltext
2017-02-11
18:46
Study of $B^{0} \rightarrow K^{*0} \gamma$, $B_{s}^{0} \rightarrow \phi \gamma$ and $B_{s}^{0} \rightarrow K^{*0} \gamma$ decays using converted photons with the LHCb detector. / Beaucourt, Leo CERN-THESIS-2016-255 -
Fulltext
2017-02-08
14:40
Measurement of the Time Reversal Asymmetry for the Decay $\bar{B}^{0}\rightarrow\Lambda\bar{p}\pi^{+}$ and Observation of the Decay $\bar{B}^{0}_{s}\rightarrow\Lambda\bar{p}K^{+}$ with the LHCb Experiment / Voss, Christian This Analysis presents the measurement of the time reversal asymmetry in the decay $\bar{B}{}^{0}\rightarrow\Lambda\bar{p}\pi^{+}$ based on a triple product given by \begin{equation*} \mathcal{O} = \vec{s}_{\Lambda} \cdot \left( \vec{p}_\Lambda \times \vec{p}_\pi \right) \quad , \end{equation*} w [...] CERN-THESIS-2016-250 - 180 p.
Fulltext - Full text
2017-01-31
14:31
Evaluation of the Radiation Environment of the LHCb Experiment / Karacson, Matthias The unprecedented radiation levels of the Large Hadron Collider (LHC) during high-energy proton-proton collisions will have an impact on the operation of its experiments’ detectors and electronics [...] CERN-THESIS-2016-246 - 220 p.
Fulltext
2017-01-23
11:27
Searching for new physics in $b\rightarrow sl^+l^-$ transitions at the LHCb experiment / Pescatore, Luca Flavour Changing Neutral Currents are transitions between different quarks with the same charge such as b $\rightarrow$ s processes [...] CERN-THESIS-2016-237 - 226 p.
Fulltext
2017-01-10
10:37
Measurements of the CKM angle $\gamma$ at the LHCb experiment / Cheung, Faye Two measurements of the Cabibbo-Kobayashi-Maskawa angle $\gamma$ using $B \to D K$ and $B^{0} \to D K^{\ast 0}$ decays are presented in this thesis [...] CERN-THESIS-2016-227 - 242 p.
Fulltext
2017-01-05
10:43
Measurements of $\sin 2\,\beta$ using charmonium and open charm decays at LHCb / Meier, Frank The $C\!P$ violation observables $S_f$ and $C_f$ in the decays of $B^0$ and $\kern 0.18em\overline{\kern -0.18em B}{}^0$ mesons to the $J\!/\!\psi K^0_S$ final state and to the $D^+ D^-$ final state are measured with a data sample corresponding to an integrated luminosity of $3 fb^{-1}$ collected wi [...] CERN-THESIS-2016-222 - 103 p.
Fulltext
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# :How to solve this?
[URGENT]ow to solve this?
Do i need to use any theorem to solve this such as divergence theorem?
the first question need to find the volume of the sphere that a part in up and bottom being cut( z-axis) and the middle is a hole( cylindrical shape )
the second question need to find the volume of a cone with a cylinder.
for both question, the answer is not need completely solve, just need find the main equation for finding the volume.( I mean just find the range for integration by using dx,dy,dz or using polar coordinate )
Equation for finding volume equation using polar coordinate is prefered.
#### Attachments
• 14.9 KB Views: 353
## Answers and Replies
Why would divergence theorem be needed? For the first one, z = +- sqrt(j^2-x^2-y^2). Put this in a polar integral.
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Proving that $\sum \frac{(-1)^n}{2\sqrt{n}+ cos(x)}$ converges uniformly.
Let $$v_n=\frac{(-1)^n}{2\sqrt{n}+ cos(x)}$$
I am asked to prove that $\sum v_n$ converges uniformly.
This is my attempt: Let $$S_n(x)= \sum_{k=0}^{n}v_n$$ and $$F(x)=\sum_{k=0}^{\infty}v_n$$
I have to show that $|F(x)-S_n(x)|$ converges. But: $$|F(x)-S_n(x)|= \sum^{\infty}_{k=n+1}\frac{1}{2\sqrt{k}+cos(x)}$$
And this is where I am stuck. By looking at this sum, I would say that it diverges yet I am asked to prove it converges. Any help would be appreciated.
• Your last statement is wrong, that is $$|F(x)-S_n(x)|=\left| \sum^{\infty}_{k=n+1}\frac{(-1)^k}{2\sqrt{k}+\cos(x)}\right|\neq\sum^{\infty}_{k=n+1}\frac{1}{2\sqrt{k}+\cos(x)}$$ – Masacroso Feb 20 '17 at 13:15
• @Masacroso Indeed, I made an error. Based on this, I can now use the Alternating series test to show that it converges? – John Mayne Feb 20 '17 at 13:29
• Do you know Dirichlet's criterion? – Julián Aguirre Feb 20 '17 at 13:45
• @JuliánAguirre I have not seen it in the class. – John Mayne Feb 20 '17 at 13:58
• And Leibniz's criterion for alternating series? – Julián Aguirre Feb 20 '17 at 14:19
Hint: $|F(x)-S_n(x)|=|\sum^{\infty}_{k=n+1}\frac{(-1)^k}{2\sqrt{k}+cos(x)}| \le |\sum^{\infty}_{k=n+1}\frac{(-1)^k}{2\sqrt{k}-1}| \le |\sum^{\infty}_{k=n+1}\frac{(-1)^k}{\sqrt{k}-1}|$.
Show that the series $\sum^{\infty}_{k=1}\frac{(-1)^k}{\sqrt{k}-1}$ is convergent and use that $N(\epsilon)$ in above inequality to get $|F(x)-S_n(x)| \lt \epsilon \;$ for $\; n \gt N(\epsilon)$.
Note that $$\frac{1}{2\sqrt{n}+\cos x}-\frac{1}{2\sqrt{n+1}+\cos x}=2\frac{\sqrt{n+1}-\sqrt{n}}{(2\sqrt{n}+\cos x)(2\sqrt{n+1}+\cos x)}$$ $$=\frac{2}{(\sqrt{n+1}+\sqrt{n})(2\sqrt{n}+\cos x)(2\sqrt{n+1}+\cos x)}\leq \frac{c}{(\sqrt{n})^3}$$
for some value of $c$.
• Why are you subtracting the next term? – John Mayne Feb 20 '17 at 13:28
• Its alternating isnt it ? – Rene Schipperus Feb 20 '17 at 13:38
Point wise convergence follows from Leibniz's criterion. For each real number $x$ $$\frac{1}{2\sqrt{n}+ \cos(x)}>0,$$ decreases as a function of $n$ and converges to $0$.
To prove uniform convergence you need the uniform version of the criterion: $$\frac{1}{2\sqrt{n}+ \cos(x)}$$ decreases as a function of $n$ and converges uniformly to $0$. You can prove it adapting the proof of Leibniz's criterion to this case.
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## randomecorner one year ago The total rainfall in Forks, Washington, this October was in. In September, 60% less rain fell than in October. How much rain fell in September in Forks, Washington? Express your answer as a mixed number in simplest form.
1. randomecorner
@IParke
2. randomecorner
@iGreen @Here_to_Help15
3. randomecorner
8 3/8 is the fraction for october
4. anonymous
60% less means september's rain was 40% of october's. 40% as a fraction is 2/5. Multiply the October number by 2/5
5. randomecorner
so 2/5 times8 and 3/8?
6. anonymous
yes
7. randomecorner
6/40
8. randomecorner
am i right?
9. anonymous
no
10. randomecorner
8 and 6/40
11. randomecorner
eight stays the samebecause its a whole numbah
12. randomecorner
so is that right?
13. anonymous
not with multiplication. 8 gets mulitiplied by 2/5 as well
14. randomecorner
so how do we dot hat/
15. randomecorner
that?
16. anonymous
|dw:1433960185507:dw|
17. anonymous
write 8 3/8 as an improper fraction, then multiply
18. randomecorner
so 130/40 but we do reciprical so its40 over 134
19. anonymous
no reciprocal
20. randomecorner
but in the space its a mixed number fraction2
21. anonymous
134/40 then reduce by dividing num & den by 2
22. randomecorner
both by 2?
23. anonymous
|dw:1433960372246:dw|
24. anonymous
by 2 because that's the biggest number that goes into 134 and 40
25. randomecorner
but it needsto be a mixed fraction
26. randomecorner
so i needa whole number and a fraction
27. anonymous
yes, so turn 67/20 into a mixed number. What's the closest number to 67 that 20 divides into?
28. randomecorner
its62 u mean
29. anonymous
no, 67
30. anonymous
wait, where did you mean 62?
31. randomecorner
uright lol i divided wrong
32. randomecorner
they cant because 67 is a prime numbah
33. anonymous
20 is the denominator. 20* 1 = 20 20*2 = 40 20*3 = 60 20*4 = 80 67 is between 60 and 80, so the whole number part is 3. Basically if you divide 67 by 20 you would get 3 with a remainder of 7.
34. anonymous
$\frac{ 67 }{ 20 }=3\frac{ 7 }{ 20 }$
35. randomecorner
oh ok so thats the answer?
36. anonymous
yes
37. randomecorner
thank you i hate math
38. randomecorner
have a good day !!! ;u;
39. anonymous
you're welcome. you too
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# In regression analysis, the variable that is being predicted is the
• Intervening variable
• Dependent variable
• None
• Independent variable
This question aims to find a variable that is being predicted in regression analysis. For this purpose, we need to find the linear regression equation.
Regression analysis is a method for analyzing and comprehending the relationship between two or more variables. An advantage of this process is that it helps in understanding the significant factors, the factors that can be neglected, and their interaction with one another.
Simple linear regression and multiple linear regression are the two most common types of regression, though non-linear regression techniques are available for more complex data. Multiple linear regression utilizes two or more independent variables to predict the result of the dependent variable, whereas simple linear regression utilizes one independent variable to predict the result of the dependent variable.
### Step $1$
We use regression analysis to estimate or predict the dependent variable based on the independent variable by using the following Simple Linear Regression equation:
SSR $y=a+b\times x$
Where the sum of squares due to regression (SSR) describes how well a regression model depicts the data that have been modeled, and where $a$ is the intercept, and $b$ is the slope coefficient of the regression equation.
$y$ is the variable (dependent or response), and $x$ is the independent or explanatory variable.
### Step $2$
As we know, Regression analysis is useful for prediction or forecasting.
In the Regression line, one variable is the dependent variable and the other variable is the independent variable. The dependent variable is predicted on the basis of the independent variable (Explanatory variable).
Thus, the dependent variable is being predicted, so “Dependent variable” is the correct choice.
## Example
For the given data points, find the least square regression line.
$\{(-1,0),(1,2),(2,3)\}$
### Numerical Solution
First, tabulate the given data:
$x$ $y$ $xy$ $x^2$ $-1$ $0$ $0$ $1$ $1$ $2$ $2$ $1$ $2$ $3$ $6$ $4$ $\sum x=2$ $\sum y=5$ $\sum xy=8$ $\sum x^2=6$
$a=\dfrac{n\sum(xy)-\sum x\sum y}{n\sum x^2-(\sum x)^2}$
$=\dfrac{(3)(8)-(2)(5)}{(3)(6)-(2)^2}=1$
$b=\dfrac{\sum y-a\sum x}{n}$
$=\dfrac{5-(1)(2)}{3}=1$
Since $y=a+bx$
So, $y=1+x$.
Graph of linear regression
Images/mathematical drawings are created with GeoGebra.
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Now showing items 1-6 of 6
• #### Asymptotic properties of Jacobi matrices for a family of fractal measures
(Taylor and Francis, 2018)
We study the properties and asymptotics of the Jacobi matrices associated with equilibrium measures of the weakly equilibrium Cantor sets. These family of Cantor sets were defined, and different aspects of orthogonal ...
• #### Asymptotics of extremal polynomials for some special cases
(Bilkent University, 2017-05)
We study the asymptotics of orthogonal and Chebyshev polynomials on fractals. We consider generalized Julia sets in the sense of Br uck-B uger and weakly equilibrium Cantor sets which was introduced in [62]. We give ...
• #### Bases in banach spaces of smooth functions on cantor-type sets
(Bilkent University, 2013)
We construct Schauder bases in the spaces of continuous functions C p (K) and in the Whitney spaces E p (K) where K is a Cantor-type set. Here different Cantortype sets are considered. In the construction, local ...
• #### Bases in Banach spaces of smooth functions on Cantor-type sets
(Elsevier, 2011)
We suggest a Schauder basis in Banach spaces of smooth functions and traces of smooth functions on Cantor-type sets. In the construction, local Taylor expansions of functions are used. © 2011 Elsevier Inc.
• #### Widom factors
(Springer Netherlands, 2015)
Given a non-polar compact set K,we define the n-th Widom factor W<inf>n</inf>(K) as the ratio of the sup-norm of the n-th Chebyshev polynomial on K to the n-th degree of its logarithmic capacity. By G. Szegő, the sequence ...
• #### Widom Factors
(Bilkent University, 2014)
In this thesis we recall classical results on Chebyshev polynomials and logarithmic capacity. Given a non-polar compact set K, we define the n-th Widom factor Wn(K) as the ratio of the sup-norm of the n-th Chebyshev ...
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# Running for Gold solution codeforces
## Running for Gold solution codeforces
The Olympic Games have just started and Federico is eager to watch the marathon race.
There will be nn athletes, numbered from 11 to nn, competing in the marathon, and all of them have taken part in 55 important marathons, numbered from 11 to 55, in the past. For each 1in1≤i≤n and 1j51≤j≤5, Federico remembers that athlete ii ranked ri,jri,j-th in marathon jj (e.g., r2,4=3r2,4=3 means that athlete 22 was third in marathon 44).
Federico considers athlete xx superior to athlete yy if athlete xx ranked better than athlete yy in at least 33 past marathons, i.e., rx,j<ry,jrx,j<ry,j for at least 33 distinct values of jj.
Federico believes that an athlete is likely to get the gold medal at the Olympics if he is superior to all other athletes.
Find any athlete who is likely to get the gold medal (that is, an athlete who is superior to all other athletes), or determine that there is no such athlete.
Input
The first line contains a single integer tt (1t10001≤t≤1000) — the number of test cases. Then tt test cases follow.
The first line of each test case contains a single integer nn (1n500001≤n≤50000) — the number of athletes.
Then nn lines follow, each describing the ranking positions of one athlete.
The ii-th of these lines contains the 55 integers ri,1,ri,2,ri,3,ri,4,ri,5ri,1,ri,2,ri,3,ri,4,ri,5 (1ri,j500001≤ri,j≤50000) — the ranking positions of athlete ii in the past 55 marathons. It is guaranteed that, in each of the 55 past marathons, the nn athletes have distinct ranking positions, i.e., for each 1j51≤j≤5, the nn values r1,j,r2,j,,rn,jr1,j,r2,j,…,rn,j are distinct.
It is guaranteed that the sum of nn over all test cases does not exceed 5000050000.
Output
For each test case, print a single integer — the number of an athlete who is likely to get the gold medal (that is, an athlete who is superior to all other athletes). If there are no such athletes, print 1−1. If there is more than such one athlete, print any of them.
Example
input
Copy
4
1
50000 1 50000 50000 50000
3
10 10 20 30 30
20 20 30 10 10
30 30 10 20 20
3
1 1 1 1 1
2 2 2 2 2
3 3 3 3 3
6
9 5 3 7 1
7 4 1 6 8
5 6 7 3 2
6 7 8 8 6
4 2 2 4 5
8 3 6 9 4
output
Copy
1
-1
1
5
Note
Explanation of the first test case: There is only one athlete, therefore he is superior to everyone else (since there is no one else), and thus he is likely to get the gold medal.
Explanation of the second test case: There are n=3n=3 athletes.
• Athlete 11 is superior to athlete 22. Indeed athlete 11 ranks better than athlete 22 in the marathons 1122 and 33.
• Athlete 22 is superior to athlete 33. Indeed athlete 22 ranks better than athlete 33 in the marathons 112244 and 55.
• Athlete 33 is superior to athlete 11. Indeed athlete 33 ranks better than athlete 11 in the marathons 3344 and 55.
Explanation of the third test case: There are n=3n=3 athletes.
• Athlete 11 is superior to athletes 22 and 33. Since he is superior to all other athletes, he is likely to get the gold medal.
• Athlete 22 is superior to athlete 33.
• Athlete 33 is not superior to any other athlete.
Explanation of the fourth test case: There are n=6n=6 athletes.
• Athlete 11 is superior to athletes 334466.
• Athlete 22 is superior to athletes 114466.
• Athlete 33 is superior to athletes 224466.
• Athlete 44 is not superior to any other athlete.
• Athlete 55 is superior to athletes 1122334466. Since he is superior to all other athletes, he is likely to get the gold medal.
• Athlete 66 is only superior to athlete 44.
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# polyestOptions
Option set for `polyest`
## Syntax
`opt = polyestOptionsopt = polyestOptions(Name,Value)`
## Description
`opt = polyestOptions` creates the default options set for `polyest`.
`opt = polyestOptions(Name,Value)` creates an option set with the options specified by one or more `Name,Value` pair arguments.
## Input Arguments
collapse all
### Name-Value Pair Arguments
Specify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside single quotes (`' '`). You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`.
### `'InitialCondition'` —
Specify how initial conditions are handled during estimation.
`InitialCondition` requires one of the following values:
• `'zero'` — The initial condition is set to zero.
• `'estimate'` — The initial state is treated as an independent estimation parameter.
• `'backcast'` — The initial state is estimated using the best least squares fit.
• `'auto'` — The software chooses the method to handle initial states based on the estimation data.
### `'Focus'` — Estimation focus`'prediction'` (default) | `'simulation'` | `'stability'` | vector | matrix | linear system
Estimation focus that defines how the errors e between the measured and the modeled outputs are weighed at specific frequencies during the minimization of the prediction error, specified as one of the following:
• `'prediction'` — Automatically calculates the weighting function as a product of the input spectrum and the inverse of the noise spectrum. The weighting function minimizes the one-step-ahead prediction. This approach typically favors fitting small time intervals (higher frequency range). From a statistical-variance point of view, this weighting function is optimal. However, this method neglects the approximation aspects (bias) of the fit.
This option focuses on producing a good predictor and does not enforce model stability. Use `'stability'` when you want to ensure a stable model.
• `'simulation'` — Estimates the model using the frequency weighting of the transfer function that is given by the input spectrum. Typically, this method favors the frequency range where the input spectrum has the most power. This method provides a stable model.
• `'stability'` — Same as `'prediction'`, but with model stability enforced.
• Passbands — Row vector or matrix containing frequency values that define desired passbands. For example:
```[wl,wh] [w1l,w1h;w2l,w2h;w3l,w3h;...]```
where `wl` and `wh` represent lower and upper limits of a passband. For a matrix with several rows defining frequency passbands, the algorithm uses union of frequency ranges to define the estimation passband.
Passbands are expressed in `rad/TimeUnit` for time-domain data and in `FrequencyUnit` for frequency-domain data, where `TimeUnit` and `FrequencyUnit` are the time and frequency units of the estimation data.
• SISO filter — Specify a SISO linear filter in one of the following ways:
• A single-input-single-output (SISO) linear system
• `{A,B,C,D}` format, which specifies the state-space matrices of the filter
• `{numerator, denominator}` format, which specifies the numerator and denominator of the filter transfer function
This option calculates the weighting function as a product of the filter and the input spectrum to estimate the transfer function. To obtain a good model fit for a specific frequency range, you must choose the filter with a passband in this range. The estimation result is the same if you first prefilter the data using `idfilt`.
• Weighting vector — For frequency-domain data only, specify a column vector of weights. This vector must have the same length as the frequency vector of the data set, `Data.Frequency`. Each input and output response in the data is multiplied by the corresponding weight at that frequency.
### `'EstCovar'` — Control whether to generate parameter covariance data`true` (default) | `false`
Controls whether parameter covariance data is generated, specified as `true` or `false`.
If `EstCovar` is `true`, then use `getcov` to fetch the covariance matrix from the estimated model.
### `'Display'` — Specify whether to display the estimation progress`'off'` (default) | `'on'`
Specify whether to display the estimation progress, specified as one of the following strings:
• `'on'` — Information on model structure and estimation results are displayed in a progress-viewer window.
• `'off'` — No progress or results information is displayed.
### `'InputOffset'` — Removal of offset from time-domain input data during estimation`[]` (default) | vector of positive integers | matrix
Removal of offset from time-domain input data during estimation, specified as the comma-separated pair consisting of `'InputOffset'` and one of the following:
• A column vector of positive integers of length Nu, where Nu is the number of inputs.
• `[]` — indicates no offset
• Nu-by-Ne matrix — For multiexperiment data, specify `InputOffset` as an Nu-by-Ne matrix. Nu is the number of inputs, and Ne is the number of experiments.
Each entry specified by `InputOffset` is subtracted from the corresponding input data.
### `'OutputOffset'` — Removal of offset from time-domain output data during estimation`[]` (default) | vector | matrix
Removal of offset from time domain output data during estimation, specified as the comma-separated pair consisting of `'OutputOffset'` and one of the following:
• A column vector of length Ny, where Ny is the number of outputs.
• `[]` — indicates no offset
• Ny-by-Ne matrix — For multiexperiment data, specify `OutputOffset` as a Ny-by-Ne matrix. Ny is the number of outputs, and Ne is the number of experiments.
Each entry specified by `OutputOffset` is subtracted from the corresponding output data.
### `'Regularization'` —
Options for regularized estimation of model parameters. For more information on regularization, see Regularized Estimates of Model Parameters.
Structure with the following fields:
• `Lambda` — Constant that determines the bias versus variance tradeoff.
Specify a positive scalar to add the regularization term to the estimation cost.
The default value of zero implies no regularization.
Default: 0
• `R` — Weighting matrix.
Specify a vector of nonnegative numbers or a square positive semi-definite matrix. The length must be equal to the number of free parameters of the model.
For black-box models, using the default value is recommended. For structured and grey-box models, you can also specify a vector of `np` positive numbers such that each entry denotes the confidence in the value of the associated parameter.
The default value of 1 implies a value of `eye(npfree)`, where `npfree` is the number of free parameters.
Default: 1
• `Nominal` — The nominal value towards which the free parameters are pulled during estimation.
The default value of zero implies that the parameter values are pulled towards zero. If you are refining a model, you can set the value to `'model'` to pull the parameters towards the parameter values of the initial model. The initial parameter values must be finite for this setting to work.
Default: 0
### `'SearchMethod'` —
Search method used for iterative parameter estimation.
`SearchMethod` requires one of the following values:
• `'gn'` — The subspace Gauss-Newton direction. Singular values of the Jacobian matrix less than `GnPinvConst*eps*max(size(J))*norm(J)` are discarded when computing the search direction. J is the Jacobian matrix. The Hessian matrix is approximated by JTJ. If there is no improvement in this direction, the function tries the gradient direction.
• `'gna'` — An adaptive version of subspace Gauss-Newton approach, suggested by Wills and Ninne. Eigenvalues less than `gamma*max(sv)` of the Hessian are ignored, where sv are the singular values of the Hessian. The Gauss-Newton direction is computed in the remaining subspace. gamma has the initial value `InitGnaTol` (see `Advanced` for more information). gamma is increased by the factor `LMStep` each time the search fails to find a lower value of the criterion in less than 5 bisections. gamma is decreased by a factor of `2*LMStep` each time a search is successful without any bisections.
• `'lm'` — Uses the Levenberg-Marquardt method. The next parameter value is `-pinv(H+d*I)*grad` from the previous one. H is the Hessian, I is the identity matrix, and grad is the gradient. d is a number that is increased until a lower value of the criterion is found.
• `'lsqnonlin'` — Uses `lsqnonlin` optimizer from Optimization Toolbox™ software. This search method can handle only the Trace criterion.
• `'grad'` — The steepest descent gradient search method.
• `'auto'` — The algorithm chooses one of the preceding options. The descent direction is calculated using `'gn'`, `'gna'`, `'lm'`, and `'grad'` successively at each iteration. The iterations continue until a sufficient reduction in error is achieved.
### `'Advanced'` —
`Advanced` is a structure with the following fields:
• `ErrorThreshold` — Specifies when to adjust the weight of large errors from quadratic to linear.
Errors larger than `ErrorThreshold` times the estimated standard deviation have a linear weight in the criteria. The standard deviation is estimated robustly as the median of the absolute deviations from the median and divided by `0.7`. For more information on robust norm choices, see section 15.2 of [2].
`ErrorThreshold = 0` disables robustification and leads to a purely quadratic criterion. When estimating with frequency-domain data, the software sets `ErrorThreshold` to zero. For time-domain data that contains outliers, try setting `ErrorThreshold` to `1.6`.
Default: 0
• `MaxSize` — Specifies the maximum number of elements in a segment when input-output data is split into segments.
`MaxSize` must be a positive integer.
Default: 250000
• `StabilityThreshold` — Specifies thresholds for stability tests.
`StabilityThreshold` is a structure with the following fields:
• `s` — Specifies the location of the right-most pole to test the stability of continuous-time models. A model is considered stable when its right-most pole is to the left of `s`.
Default: 0
• `z` — Specifies the maximum distance of all poles from the origin to test stability of discrete-time models. A model is considered stable if all poles are within the distance `z` from the origin.
Default: `1+sqrt(eps)`
• `AutoInitThreshold` — Specifies when to automatically estimate the initial condition.
The initial condition is estimated when
$\frac{‖{y}_{p,z}-{y}_{meas}‖}{‖{y}_{p,e}-{y}_{meas}‖}>\text{AutoInitThreshold}$
• ymeas is the measured output.
• yp,z is the predicted output of a model estimated using zero initial states.
• yp,e is the predicted output of a model estimated using estimated initial states.
Applicable when `InitialCondition` is `'auto'`.
Default: `1.05`
## Output Arguments
`opt` Option set containing the specified options for `polyest`.
## Examples
collapse all
### Create Default Options Set for Polynomial Estimation
`opt = polyestOptions;`
### Specify Options for Polynomial Estimation
Create an options set for `polyest` using the `'stability'` for `Focus`, and set the `Display` to `'on'`.
`opt = polyestOptions('Focus','stability','Display','on');`
Alternatively, use dot notation to set the values of `opt`.
```opt = polyestOptions; opt.Focus = 'stability'; opt.Display = 'on';```
## References
[1] Wills, Adrian, B. Ninness, and S. Gibson. "On Gradient-Based Search for Multivariable System Estimates". Proceedings of the 16th IFAC World Congress, Prague, Czech Republic, July 3–8, 2005. Oxford, UK: Elsevier Ltd., 2005.
[2] Ljung, L. System Identification: Theory for the User. Upper Saddle River, NJ: Prentice-Hall PTR, 1999.
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# The Wirtinger theorem proof
Let $M$ be a complex hermitian manifold with symplectic form $\Omega$ and $N \subseteq M$ be its smooth $2n$-dimensional (real dimensions, $2n \geq 2$) compact submanifold. I want to show the inequality $$\text{vol}(N) \geq \frac{1}{n!} \int\limits_N \Omega^n.$$ If $N$ is a complex submanifold then choosing in the neighborhood of each point holomorphic coordinates $z_1$, $\ldots$, $z_n$, such that in this neighborhood the hermitian metric is equal to $\delta_{kl}$ we can obtain $$\Omega = \frac{i}{2} \sum_k dz_k \wedge d\bar z_k, \quad d\text{vol} = (i/2)^n dz_1 \wedge d\bar z_1\wedge \cdots \wedge dz_n \wedge d\bar z_n,$$ and we can see that $\Omega^n = n!d\text{vol}$. Hence $$\text{vol}(N) = \frac{1}{n!} \int\limits_N \Omega^n$$ for the complex submanifold $N$. But how to show that the inequality holds even for smooth submanifolds $N$?
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# Decomposition¶
## tICA¶
tICA compared to PCA (courtesy of C. R. Schwantes)
Time-structure independent components analysis (tICA) is a method for finding the slowest-relaxing degrees of freedom in a time series data set which can be formed from linear combinations from a set of input degrees of freedom.
tICA can be used as a dimensionality reduction method and, in that capacity, is somewhat similar to PCA. However whereas PCA finds high-variance linear combinations of the input degrees of freedom, tICA finds high-autocorrelation linear combinations of the input degrees of freedom.
The tICA method has one obvious drawback: its solution is a linear combination of all input degrees of freedom, and their relative weights are typically non-zero. This makes each independent component difficult to interpret in an algorithmic fashion because it could comprise hundreds or thousands of different metrics. Because an important property of reaction coordinates is their role in facilitating physical interpretation of the underlying molecular system, we consider it desirable to reduce the number of explicitly used variables. SparseTICA [2] attempts to resolve this interpretability issue by using a sparse approximation to the eigenvalue problem found in tICA and returning independent components composed of only the most relevant degrees of freedom.
Another limitation of tICA is that it is constrained to finding linear correlations between input coordinates. The KernelTICA method [5] provides a natural extension to nonlinear functions by utilizing the kernel trick and represents a substantial improvement in estimating the eigenfunctions of the transfer operator directly.
## PCA¶
Principal component analysis (PCA) is a method for finding the most highly varying degrees of freedom in a data set (not necessarily a time series). PCA is useful as a dimensionality reduction method.
## Algorithms¶
tICA([n_components, lag_time, shrinkage, ...]) Time-structure Independent Component Analysis (tICA) SparseTICA([n_components, lag_time, rho, ...]) Sparse time-structure Independent Component Analysis (tICA). PCA([n_components, copy, whiten, ...]) Principal component analysis (PCA)
## Theory¶
### PCA¶
PCA tries to find projection vectors that maximize their explained variance, subject to them being uncorrelated and having length one. In the end, these maximal variance projections correspond to the solutions of the following eigenvalue problem:
$\Sigma v = \lambda v$
where $$\Sigma$$ is the covariance matrix given by:
$\Sigma_{ij} = \mathbb{E}\Big[ X_i(t) X_j(t) \Big]$
The problem with using PCA to define a reduced space for biomolecular dynamics, however, is that high-variance degrees of freedom need not be slow (for instance consider a floppy protein tail that varies wildly vs. a single dihedral angle that is required to rotate for a protein to fold). What we really want is to design projections that can best differentiate between slowly equilibrating populations, which is precisely where tICA comes in.
### tICA¶
In tICA, the goal is to find projection vectors that maximize their autocorrelation function, subject to them being uncorrelated and having unit variance. It is easy to show [1] that the solution to the tICA problem are the solutions to this generalized eigenvalue problem (which is closely related to the PCA eigenvalue problem):
$C^{(\Delta t)} v = \lambda \Sigma v$
where $$C^{(\Delta t)}$$ is the time lag correlation matrix defined by:
$C^{(\Delta t)}_{ij} = \mathbb{E}\Big[ X_i(t) X_j(t+\Delta t) \Big]$
Given this solution, we can use the tICA method to define a reduced dimensionality representation of each $$\mathbf{X}(t)$$ by projecting the vector onto the slowest $$n$$ tICs.
### Hyperparameters¶
There are two parameters introduced in the tICA method. The first is $$\Delta t$$, which is used in the calculation of the time-lag correlation matrix ($$C^{(\Delta t)}$$). The second is $$n$$, which is the number of tICs to project onto when calculating distances between conformations. You can use the GMRQ score to choose these parameters. Typical values are of order nanoseconds and tens, respectively.
### Drawbacks¶
Since part of the process of using tICA is a dimensionality reduction, there is always the opportunity to throw out important pieces of information. By throwing out the faster degrees of freedom, we can better estimate the slowest timescales; but this comes with the trade-off of not representing the fast timescales correctly.
## Combination with MSM¶
While the tICs are themselves approximations to the dominant eigenfunctions of the propagator / transfer operator, the approach taken in [1] and [2] is to “stack” tICA with MSMs. For example, in [2], Perez-Hernandez et al. first measured the 66 atom-atom distances between a set of atoms in each frame of their MD trajectories, and then used tICA to find the slowest 1, 4, and 10 linear combinations of these degrees of freedom and transform the 66-dimensional dataset into a 1, 4, or 10-dimensional dataset. Then, they applied KMeans to the resulting data and built an MSM.
## References¶
[1] (1, 2) Schwantes, Christian R., and Vijay S. Pande. Improvements in Markov State Model Construction Reveal Many Non-Native Interactions in the Folding of NTL9 J. Chem Theory Comput. 9.4 (2013): 2000-2009.
[2] (1, 2, 3) Perez-Hernandez, Guillermo, et al. Identification of slow molecular order parameters for Markov model construction J Chem. Phys (2013): 015102.
[3] Naritomi, Yusuke, and Sotaro Fuchigami. Slow dynamics in protein fluctuations revealed by time-structure based independent component analysis: The case of domain motions J. Chem. Phys. 134.6 (2011): 065101.
[4] McGibbon, R. T. & Pande, V. S. Identification of simple reaction coordinates from complex dynamics ArXiv 16 (2016).
[5] Schwantes, Christian R., and Vijay S. Pande. Modeling Molecular Kinetics with tICA and the Kernel Trick J. Chem Theory Comput. 11.2 (2015): 600–608.
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# Weird Example: Pull-back of Noetherian Isn’t Noetherian
In this post we just give a simple, but nice example, to show that ‘Noetherian’ is not a property preserved under pull-back.
# Noetherian is Not Stable
The stability of a property of schemes, its preservation under arbitrary pull-backs, is a very desirable property of schemes. That said, while most natural examples of relative properties are preserved under pull-backs (e.g. finite type, integral, etc.) , some nice global properties are not. In particular, we show in this example that Noetherianess is not preserved under pull-backs.
In particular, we consider the following fibered diagram
$\begin{matrix}\text{Spec}(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}}) & \to & \text{Spec}(\overline{\mathbb{Q}})\\ \downarrow & & \downarrow\\ \text{Spec}(\overline{\mathbb{Q}}) & \to & \text{Spec}(\mathbb{Q})\end{matrix}$
Certainly $\text{Spec}(\overline{\mathbb{Q}})$ is Noetherian, but we claim that $\text{Spec}(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}})$ is not.
Indeed, suppose that it were. Note that $\overline{\mathbb{Q}}\to\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}}$ is an integral extension, and so $\dim \overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}}=0$. So, if $\text{Spec}(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}})$ were Noetherian, it would have to be a finite set of points. Indeed, every Noetherian space is a union of finitely many irreducible components. And, the only dimension 0 irreducible space is a point.
But, $\text{Spec}(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}})$ is not finite. Indeed, note that for each finite extension $L/K$ we have the following fibered diagram(s):
$\begin{matrix}\text{Spec}(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}}) & \to & \text{Spec}(\overline{\mathbb{Q}})\\ \downarrow & & \downarrow\\ \text{Spec}(L\otimes_\mathbb{Q}\overline{\mathbb{Q}}) & \to & \text{Spec}(L)\\ \downarrow & & \downarrow \\ \text{Spec}(\overline{\mathbb{Q}}) & \to & \text{Spec}(\mathbb{Q}) \end{matrix}$
and since $\text{Spec}(\overline{\mathbb{Q}})\to\text{Spec}(L)$ is surjective, so is $\text{Spec}(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}})\to\text{Spec}(L\otimes_\mathbb{Q}\overline{\mathbb{Q}})$. But, note that if $L=\mathbb{Q}[x]/(f(x))$, with $\deg f=n$ (i.e. $[L:K]=n$) then $L\otimes_\mathbb{Q}\overline{\mathbb{Q}}$ is isomorphic to $\overline{\mathbb{Q}}[x]/(f(x))$, which is in turn isomorphic to $\overline{\mathbb{Q}}^n$. So, $\text{Spec}(L\otimes_\mathbb{Q}\overline{\mathbb{Q}})$ has $n$-points. Since $L$ was arbitrary, we see that $\text{Spec}(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}})$ is infinite, contradicting that it is Noetherian.
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Browse Questions
# Show that the adjoint of $A=\begin{bmatrix} -4 & -3 & -3 \\1 & 0 & 1 \\4 & 4 & 3 \end{bmatrix}$ is$\;A\;$ it self
Toolbox:
• Let $A = [ a_{ij} ]$ be a square matrix x. Let $A_{ij}$ be the cofactor of $a_{ij}$. Then $[ A_{ij}]$ is the matrix of cofactors and $adj\: A$ ( or adjoint of the matrix A) is given by $adj\: A=[A_{ij}]^T$
Step 1
To find $adj\: A$. The matrix of cofactors
$A_{ij} = \begin{bmatrix} (0-4) & -(3-4) & (4-0) \\ -(-9+12) & (-12+12) & -(-16+12) \\ (-3+0) & -(-4+3) & (0+3) \end{bmatrix} = \begin{bmatrix} -4 & 1 & 4 \\ -3 & 0 & 4 \\ -3 & 1 & 3 \end{bmatrix}$
$adj\: A =\begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ 4 & 4 & 3 \end{bmatrix}$
Step 2
$A =\begin{bmatrix} -4 & -3 & -3 \\ 1 & 0 & 1 \\ -4 & 4 & 3 \end{bmatrix}$
From step 1, $A = adj\: A$
edited Jun 3, 2013
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96-29 P. Exner, S.A. Vugalter
Asymptotic estimates for bound states in quantum waveguides coupled laterally through a narrow window (34K, LaTeX) Feb 2, 96
Abstract , Paper (src), View paper (auto. generated ps), Index of related papers
Abstract. Consider the Laplacian in a straight planar strip of width $\,d\,$, with the Neumann boundary condition at a segment of length $\,2a\,$ of one of the boundaries, and Dirichlet otherwise. For small enough $\,a\,$ this operator has a single eigenvalue $\,\epsilon(a)\,$; we show that there are positive $\,c_1,c_2\,$ such that $\,-c_1 a^4 \le \epsilon(a)- \left(\pi/ d\right)^2 \le -c_2 a^4\,$. An analogous conclusion holds for a pair of Dirichlet strips, of generally different widths, with a window of length $\,2a\,$ in the common boundary.
Files: 96-29.tex
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de.mpg.escidoc.pubman.appbase.FacesBean
Deutsch
Hilfe Wegweiser Impressum Kontakt Einloggen
# Datensatz
DATENSATZ AKTIONENEXPORT
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#### Randomized external-memory algorithms for some geometric problems
##### MPG-Autoren
http://pubman.mpdl.mpg.de/cone/persons/resource/persons44266
Crauser, Andreas
Algorithms and Complexity, MPI for Informatics, Max Planck Society;
http://pubman.mpdl.mpg.de/cone/persons/resource/persons44412
Ferragina, Paolo
Algorithms and Complexity, MPI for Informatics, Max Planck Society;
http://pubman.mpdl.mpg.de/cone/persons/resource/persons45021
Mehlhorn, Kurt
Algorithms and Complexity, MPI for Informatics, Max Planck Society;
http://pubman.mpdl.mpg.de/cone/persons/resource/persons45038
Meyer, Ulrich
Algorithms and Complexity, MPI for Informatics, Max Planck Society;
http://pubman.mpdl.mpg.de/cone/persons/resource/persons45255
Ramos, Edgar A.
Algorithms and Complexity, MPI for Informatics, Max Planck Society;
##### Externe Ressourcen
Es sind keine Externen Ressourcen verfügbar
##### Volltexte (frei zugänglich)
1998-1-017
(beliebiger Volltext), 11KB
##### Ergänzendes Material (frei zugänglich)
Es sind keine frei zugänglichen Ergänzenden Materialien verfügbar
##### Zitation
Crauser, A., Ferragina, P., Mehlhorn, K., Meyer, U., & Ramos, E. A.(1998). Randomized external-memory algorithms for some geometric problems (MPI-I-1998-1-017). Saarbrücken: Max-Planck-Institut für Informatik.
We show that the well-known random incremental construction of Clarkson and Shor can be adapted via {\it gradations} to provide efficient external-memory algorithms for some geometric problems. In particular, as the main result, we obtain an optimal randomized algorithm for the problem of computing the trapezoidal decomposition determined by a set of $N$ line segments in the plane with $K$ pairwise intersections, that requires $\Theta(\frac{N}{B} \log_{M/B} \frac{N}{B} +\frac{K}{B})$ expected disk accesses, where $M$ is the size of the available internal memory and $B$ is the size of the block transfer. The approach is sufficiently general to obtain algorithms also for the problems of 3-d half-space intersections, 2-d and 3-d convex hulls, 2-d abstract Voronoi diagrams and batched planar point location, which require an optimal expected number of disk accesses and are simpler than the ones previously known. The results extend to an external-memory model with multiple disks. Additionally, under reasonable conditions on the parameters $N,M,B$, these results can be notably simplified originating practical algorithms which still achieve optimal expected bounds.
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# asymptotic analysis of algorithms examples
What kinds of problems are solved by algorithms? Functions in asymptotic ⦠Google Classroom Facebook Twitter. What really concerns us is the asymptotic behavior of the running-time functions: what happens as n becomes very large? the best case. Compare the $2^n$ row with the $0.000001\cdot 2^n$ row. â We say that f(n) is Big-O of g(n), written as f(n) = O(g(n)), iff there are positive constants c and n0 such that There are three cases to analyze an algorithm: ⢠Comparing the asymptotic running time - an algorithm that runs inO(n) time is better than Asymptotic analysis is input bound, which means that we assume that the run time of the algorithms depends entirely upon the size of the Input to the algorithm. Then we deï¬ne the three most common asymptotic bounds as follows. Asymptotic notation. Asymptotic notation. Asymptotic Notation The result of the analysis of an algorithm is usually a formula giving the amount of time, in terms of seconds, number of memory accesses, number of comparisons or some other metric, that the algorithm takes. Therefore, looking at the previous example, the total number of operations is given as 4n + 4. ⢠For example, we say that thearrayMax algorithm runs in O(n) time. If the algorithm contains no input, we assume that it runs in constant time. There is something. A simple asymptotic analysis. Asymptotic notations are the mathematical notations used to describe the running time of an algorithm when the input tends towards a particular value or a limiting value. Previously, on CSE 373 ... worst-case running time of an algorithm ⢠Example: binary-search algorithm â Common: θ(log n) running-time in the worst-case Big-θ (Big-Theta) notation . Asymptotic Notation: Deï¬nitions and Examples Chuck Cusack Deï¬nitions Let f be a nonnegative function. The previous chapter presents a detailed model of the computer which involves a number of different timing parameters-- , , , , , , , , , , , and .We show that keeping track of the details is messy and tiresome. This is the currently selected item. For example: In bubble sort, when the input array is already sorted, the time taken by the algorithm is linear i.e. It checks how are the time growing in terms of the input size. To orient our minds correctly, if you'll indulge me, let's consider a couple of simple algorithms for getting from one side of a rectangular room to another. This type of analysis is known as asymptotic analysis. Example of Asymptotic Analysis Algorithm prefixAverages1 X n Input array X of n from BIO 100 at University of the Fraser Valley Analysis of Algorithms 13 Asymptotic Analysis of The Running Time ⢠Use the Big-Oh notation to express the number of primitive operations executed as a function of the input size. It may be noted that we are dealing with the complexity of an algorithm not that of a problem. For example, a simple problem could have a high order of time complexity and vice-versa. By the way, Ga is a gigayear, or one billion years. Figure 7-3 suggests that the running time for method A is larger than that for method B. Asymptotic notation. The reason is asymptotic analysis analyzes algorithms in terms of the input size. Email. If n is at least 12, B is faster. This formula often contains unimportant details that don't really tell us anything about the running time. The latter represents something running one million times faster than the former, but still, even for an input of size 50, requires a run time in the thousands of centuries.. Asymptotic Analysis Our intuition is correct in this example⦠CSE373: Data Structures and Algorithms Lecture 4: Asymptotic Analysis Aaron Bauer Winter 2014 . Asymptotic notation. Asymptotic Notations. (They say an algorithm is a "step-by-step procedure"; what could be more "step-by-step" than walking across a room?) Asymptotic Analysis of Algorithms. Aaron Bauer Winter 2014 bubble sort, when the input size then we deï¬ne the three common. In O ( n ) time O ( n ) time asymptotic analysis Aaron Bauer Winter 2014 Lecture:! A problem algorithm contains no input, we assume that it runs in constant time time! Are three cases to analyze an algorithm: a simple asymptotic analysis we assume that runs!, we assume that it runs in constant time Bauer Winter 2014 say that thearrayMax algorithm runs O. Is given as 4n + 4 taken by the algorithm contains no,! In terms of the running-time functions: what happens as n becomes very large time taken by the algorithm linear! Method B, when the input array is already sorted, the time growing in of! Dealing with the $2^n$ row example, we assume that it runs in constant time ).... 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Therefore, looking at the previous example, the time taken by the algorithm is linear.... No input, we assume that it runs in O ( n ) time there are three to. Runs in O ( n ) time sorted, the total number of operations is given as 4n +.. Is the asymptotic behavior of the input array is already sorted, the time taken by the algorithm is i.e. Bounds as follows running time have a high order of time complexity and.! Notation: Deï¬nitions and Examples Chuck Cusack Deï¬nitions Let f be a nonnegative function three cases analyze. Behavior of the input array is already sorted, the total number of operations given...
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# Interval completion is Fixed Parameter Tractable
1 ALGCO - Algorithmes, Graphes et Combinatoire
LIRMM - Laboratoire d'Informatique de Robotique et de Microélectronique de Montpellier
Abstract : We present an algorithm with runtime $O(k^{2k}n^3m)$ for the following NP-complete problem: Given an arbitrary graph $G$ on $n$ vertices and $m$ edges, can we obtain an interval graph by adding at most $k$ new edges to $G$? This resolves the long-standing open question, first posed by Kaplan, Shamir and Tarjan, of whether this problem could be solved in time $f(k).n^{O(1)}$. The problem has applications in Physical Mapping of DNA and in Profile Minimization for Sparse Matrix Computations. For the first application, our results show tractability for the case of a small number $k$ of false negative errors, and for the second, a small number $k$ of zero elements in the envelope. Our algorithm performs bounded search among possible ways of adding edges to a graph to obtain an interval graph, and combines this with a greedy algorithm when graphs of a certain structure are reached by the search. The presented result is surprising, as it was not believed that a bounded search tree algorithm would suffice to answer the open question affirmatively.
Keywords :
Type de document :
Rapport
[Research Report] RR-06058, Lirmm. 2007
Littérature citée [20 références]
https://hal-lirmm.ccsd.cnrs.fr/lirmm-00115278
Contributeur : Christophe Paul <>
Soumis le : lundi 20 novembre 2006 - 20:05:20
Dernière modification le : jeudi 7 février 2019 - 14:00:01
Document(s) archivé(s) le : mardi 6 avril 2010 - 19:17:48
### Identifiants
• HAL Id : lirmm-00115278, version 1
### Citation
Christophe Paul, Jan Arne Telle, Yngve Villanger, Pinar Heggerness. Interval completion is Fixed Parameter Tractable. [Research Report] RR-06058, Lirmm. 2007. 〈lirmm-00115278〉
### Métriques
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Rendering always lags behind input
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I'm posting this in the "General" section because ultimately I think I may have a problem with how I'm handling input. All the code does is create a window and initialize OpenGL, then draws a box around where the mouse is. The problem is, if the mouse is moving, the box is never centered on the mouse - it lags behind where the mouse currently is.
Even though it's subtle, it's very noticeable. Move the mouse around in a circle, even slowly, and you can see that the box is never centered on the mouse. I've tried several things and can't seem to figure out why it behaves this way. Am I handling and processing the input properly? Could it be something to do with syncing?
Pasted below is a stripped down version that shows what I'm talking about. It needs to link to opengl32.lib and glu32.lib.
Thanks,
Jake
#include <windows.h> #include <gl/glu.h> HDC dc; int mouseX = 0; int mouseY = 0; LRESULT CALLBACK WndProc ( HWND wnd, UINT msg, WPARAM wParam, LPARAM lParam ) { switch ( msg ) { case WM_MOUSEMOVE: mouseX = LOWORD( lParam ); mouseY = HIWORD( lParam ); return 0; default: return DefWindowProc( wnd, msg, wParam, lParam ); } } void createWindow ( ) { WNDCLASS wndClass; wndClass.style = CS_HREDRAW | CS_VREDRAW | CS_OWNDC; wndClass.lpfnWndProc = WndProc; wndClass.cbClsExtra = 0; wndClass.cbWndExtra = 4; wndClass.hInstance = GetModuleHandle( 0 ); wndClass.hIcon = LoadIcon( NULL, IDI_WINLOGO ); wndClass.hCursor = LoadCursor( NULL, IDC_ARROW ); wndClass.hbrBackground = NULL; wndClass.lpszMenuName = NULL; wndClass.lpszClassName = "test"; RegisterClass( &wndClass ); RECT rect = { 100, 100, 700, 500 }; DWORD style = WS_OVERLAPPED | WS_SYSMENU | WS_MINIMIZEBOX | WS_CAPTION; AdjustWindowRect( &rect, style, false ); HWND window = CreateWindowEx( 0, "test", "test", style, rect.left, rect.top, rect.right - rect.left, rect.bottom - rect.top, 0, 0, GetModuleHandle( 0 ), 0 ); dc = GetDC( window ); PIXELFORMATDESCRIPTOR pfd = { sizeof( PIXELFORMATDESCRIPTOR ), 1, PFD_DRAW_TO_WINDOW | PFD_SUPPORT_OPENGL | PFD_DOUBLEBUFFER, PFD_TYPE_RGBA, 32, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 16, 0, 0, PFD_MAIN_PLANE, 0, 0, 0, 0 }; int pixelFormat = ChoosePixelFormat( dc, &pfd ); SetPixelFormat( dc, pixelFormat, &pfd ); HGLRC rc = wglCreateContext( dc ); wglMakeCurrent( dc, rc ); ShowWindow( window, SW_NORMAL ); glClearColor( 0.0f, 0.0f, 0.0f, 0.0f ); glColor3f( 1.0f, 0.0f, 0.0f ); glViewport( 0, 0, 600, 400 ); glMatrixMode( GL_PROJECTION ); glLoadIdentity(); glOrtho( 0, 600, 400, 0.0f, -1.0f, 1.0f ); glMatrixMode( GL_MODELVIEW ); glLoadIdentity(); } int main ( ) { createWindow(); MSG msg; memset( &msg, 0, sizeof( MSG ) ); while ( msg.message != WM_QUIT ) { while ( PeekMessage( &msg, NULL, 0, 0, PM_REMOVE ) ) { TranslateMessage( &msg ); DispatchMessage( &msg ); if ( msg.message == WM_QUIT ) break; } glClear( GL_COLOR_BUFFER_BIT | GL_DEPTH_BUFFER_BIT ); glLoadIdentity(); glBegin( GL_QUADS ); glVertex2i( mouseX - 10, mouseY - 10 ); glVertex2i( mouseX + 10, mouseY - 10 ); glVertex2i( mouseX + 10, mouseY + 10 ); glVertex2i( mouseX - 10, mouseY + 10 ); glEnd(); SwapBuffers( dc ); } return 0; }
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I noticed that too. Getting mouse position directly seems to work correctly:
POINT ptMousePos; GetCursorPos(&ptMousePos); ScreenToClient(DXUTGetHWND(), &ptMousePos); MyEngine::Vector2 MousePos((float)ptMousePos.x, (float)ptMousePos.y);
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For me it rather seems that the mouse cursor is lagging behind the box.
The lag you talk about generally happens when the driver is buffering frames (often disabled in window mode, but enabled in fullscreen mode). Try disabling VSync and pre-rendered frames in your graphics control panel, and see if that makes a difference. Also try disabling triple buffering.
At least vertical sync can also be controlled from your application with a WGL extension if you find that it's the cause of your problem (depending on your graphics settings, it can usually be 'forced on', 'forced off', or 'application decides').
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I noticed that too. Getting mouse position directly seems to work correctly:
POINT ptMousePos; GetCursorPos(&ptMousePos); ScreenToClient(DXUTGetHWND(), &ptMousePos); MyEngine::Vector2 MousePos((float)ptMousePos.x, (float)ptMousePos.y);
Well, I tried that, but I still see the lag. Thanks for the try
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For me it rather seems that the mouse cursor is lagging behind the box.
The lag you talk about generally happens when the driver is buffering frames (often disabled in window mode, but enabled in fullscreen mode). Try disabling VSync and pre-rendered frames in your graphics control panel, and see if that makes a difference. Also try disabling triple buffering.
At least vertical sync can also be controlled from your application with a WGL extension if you find that it's the cause of your problem (depending on your graphics settings, it can usually be 'forced on', 'forced off', or 'application decides').
This reduced the lag to a degree, but it's still there. This all came about because the GUI I'm designing wasn't very responsive to mouse input. The frustrating thing is, when I run games like DEFCON that have a very simple GUI, much like mine, it's very responsive and doesn't feel laggy at all. So I'm confident that I'm just not setting things up or handling them properly. I wrote up the code above as an example how I can't even get the simplest of things to respond the way it should.
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l, I tried that, but I still see the lag. Thanks for the try
I forgot to mention that should be called right before actually using mouse position, I hope you didn't put that into Window Callback =p
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[quote name='CodeZero' timestamp='1324321942' post='4895425']l, I tried that, but I still see the lag. Thanks for the try
I forgot to mention that should be called right before actually using mouse position, I hope you didn't put that into Window Callback =p
[/quote]
Yeah, I tried both. Neither seemed to fix the problem.
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So I did some digging around in DEFCON's code and found out that they hide the mouse, then manually redraw it themselves to hide the fact that the mouse lags behind the input. When I un-hid the real mouse, their input was even more laggy than mine, probably because the input handling is much more involved.
Is it common practice to do this? Wouldn't it create inconsistencies in where the user thinks they are clicking and where they actually are?
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Wouldn't it create inconsistencies in where the user thinks they are clicking and where they actually are?
So long as the click is interpreted as being at the visible position of the mouse cursor, there's no problem.
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# King and Queen
Dimitri places a Black King and a White Queen on an empty chessboard. If the probability that Dimitri places the King and the Queen on the chessboard such that the King is NOT in check (that is, the Queen is not attacking the Black King), can be expressed as $$\dfrac{m}{n}$$, in which $$m$$ and $$n$$ are coprime positive integers, find $$n-m$$.
As an explicit example, if the Queen is on $$f2$$, the $$f1$$, $$f3$$, $$f4$$, $$f5$$, $$f6$$, $$f7$$, $$f8$$, $$g2$$, $$h2$$, $$e2$$, $$d2$$, $$c2$$, $$b2$$, $$e1$$, $$g1$$, $$g3$$, $$h4$$, $$e3$$, $$d4$$, $$c5$$, $$b6$$, $$a7$$ and $$a2$$ squares are under attack.
The King and the Queen cannot be placed in the same square.
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Not Sure How to Solve A System Of Linear Equations In MAPLE13
How can one solve the following system of linear equations in MAPLE 13?I know how to solve a linear equation with one variable floating around but not this one. $$x-2y+3z=10$$ $$3x-2y+z=2$$ $$4x+5y+2z=29$$ Any help would be greatly appreciated.
-
This would be a case where taking a look at the excellent Maple documentation would clearly have been faster than trying to write a post on StackExchange... – Wolfgang Bangerth Jan 18 '13 at 23:06
I'am trying to solve $$eqn1 := x-6*y+7*z = 7; eqn2 := 4*x-5*y+9*z = 8; eqn3 := -2*x+4*y+5*z = 10; solve({eq1, eq2, eq3}, {x, y, z})$$,and i get $$x = 1, y = 3, z = 5$$ as solution which is completely wrong.What should i do? – alok Jan 18 '13 at 11:21
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Research
# Identifying perinatal risk factors for infant maltreatment: an ecological approach
Yueqin Zhou12*, Elaine J Hallisey2 and Gordon R Freymann2
Author Affiliations
1 Department of Geosciences, Georgia State University, 340 Kell Hall, 24 Peachtree Center Ave., P.O. Box 4105, Atlanta, GA 30303, USA
2 Office of Health Information & Policy, Division of Public Health, Georgia Department of Human Resources, 2 Peachtree Street, Atlanta, GA 30303, USA
For all author emails, please log on.
International Journal of Health Geographics 2006, 5:53 doi:10.1186/1476-072X-5-53
Received: 2 October 2006 Accepted: 4 December 2006 Published: 4 December 2006
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
### Abstract
#### Background
Child maltreatment and its consequences are a persistent problem throughout the world. Public health workers, human services officials, and others are interested in new and efficient ways to determine which geographic areas to target for intervention programs and resources. To improve assessment efforts, selected perinatal factors were examined, both individually and in various combinations, to determine if they are associated with increased risk of infant maltreatment. State of Georgia birth records and abuse and neglect data were analyzed using an area-based, ecological approach with the census tract as a surrogate for the community. Cartographic visualization suggested some correlation exists between risk factors and child maltreatment, so bivariate and multivariate regression were performed. The presence of spatial autocorrelation precluded the use of traditional ordinary least squares regression, therefore a spatial regression model coupled with maximum likelihood estimation was employed.
#### Results
Results indicate that all individual factors or their combinations are significantly associated with increased risk of infant maltreatment. The set of perinatal risk factors that best predicts infant maltreatment rates are: mother smoked during pregnancy, families with three or more siblings, maternal age less than 20 years, births to unmarried mothers, Medicaid beneficiaries, and inadequate prenatal care.
#### Conclusion
This model enables public health to take a proactive stance, to reasonably predict areas where poor outcomes are likely to occur, and to therefore more efficiently allocate resources. U.S. states that routinely collect the variables the National Center for Health Statistics (NCHS) defines for birth certificates can easily identify areas that are at high risk for infant maltreatment. The authors recommend that agencies charged with reducing child maltreatment target communities that demonstrate the perinatal risks identified in this study.
### Background
Child maltreatment, including neglect, physical abuse, emotional abuse, sexual abuse, and other types of abuse, is a persistent problem in the world, not only in poor countries, but also in rich, industrialized nations, including the United States [1]. In general, the rate of child maltreatment is inversely related to the age of the child: children from 0 to 3 years of age have the highest rate [2]. Infants, children under age one, have the highest percentage of fatalities. Infants also suffer more serious physical and developmental consequences from maltreatment [3].
In addition to immediate effects, child maltreatment has pronounced long-term negative medical and social consequences. Numerous studies in medical literature confirm the association between childhood maltreatment and adverse adult health outcomes [4]. Examples include smoking [5], drug abuse [6], depression [7,8], stress disorders [9], and certain chronic diseases [10]. For example, a study on the continuing consequences of maltreatment in the early years using longitudinal data from infancy through late adolescence confirmed the adverse impact of early maltreatment on later antisocial behavior [11].
In addition to medical and social consequences, the economic impact of child maltreatment is immense. Nationwide costs resulting from abuse and neglect are estimated to be as high as $94 billion per year, of which$24.3 billion are used for the immediate needs of abused or neglected children including hospitalization, treatment of chronic health problems, mental health care, child welfare, law enforcement, and the judicial system; the remaining \$69.7 billion are costs associated with long-term and/or secondary effects of child abuse and neglect [12]. A study assessing the economic burden of hospitalization associated with child abuse and neglect found that children whose hospitalization was due to abuse or neglect were significantly more likely to have longer hospital stays, with double the total charges of other hospitalized children, and that nearly two-thirds of the primary payer were Medicaid [13].
A recent study, focusing on infant maltreatment assessment, found that a set of perinatal risk factors significantly influences the probability that an infant would be maltreated by caregivers [14]. In this cohort study, the researchers studied 15 perinatal risk factors using data of 1,602 infant victims of maltreatment among 189,055 infants born in 1996 in Florida. They found 11 factors were significantly related to infant maltreatment, five of which had adjusted relative risks of two or greater. The five key factors are: mother smoked during pregnancy, more than two siblings, Medicaid beneficiary, unmarried marital status, and birth weight less than 2,500 grams. Infants who had four of these five risk factors had a maltreatment rate seven times higher than the population average. Other significant risk factors include maternal age less than 20, maternal education less than high school, and prenatal care, as measured by the Kotelchuck Index, less than adequate.
Another study linked premature births to substantiated infant abuse [15]. The researchers concluded that the cry of the premature infants was perceived to be more aversive than the cry of full-term infants, and elicited greater arousal and subsequent abuse. Bugental and Happaney also identified low 5-minute Apgar scores and premature births as predictor variables of infant maltreatment [16].
Given the negative consequences associated with infant maltreatment, in conjunction with the need to efficiently allocate resources, the goal of our study was to develop a population-based model that enables public health agencies to identify areas at high risk for infant maltreatment. Many studies fail to consider an ecological, population-based method that explores why these individual risk factors occur in the larger context of the environment in which they are found. The aforementioned risk factors are themselves outcomes and an ecological approach may help better demonstrate and understand the presence of underlying social and spatial processes that predispose certain caregivers to maltreat. The Health Field Concept [17] and the Health Field Theory [18] consider the interactions of biology, environment, lifestyle, and health system effects and capacities as the major determinants of health. Both take a holistic view of community assessment that supports the ecological approach.
To achieve this goal, we examined the geographic distribution of infant maltreatment in relation to relevant perinatal risk factors, as identified in previous research, using Georgia data aggregated by census tract. Two major components of the goal were: 1) to determine if infant maltreatment, including all types of abuse and neglect combined, is significantly related to a set of individual and composite perinatal risk factors, and 2) to identify a set of risk factors that best predicts infant maltreatment rates to ultimately aid public health agencies in identifying geographic areas for intervention.
Special attention was focused on coping with spatial autocorrelation, a well-known spatial phenomenon in which data collected from a particular location is often similar to data collected in nearby locations. The presence of spatial autocorrelation violates the key assumption of independence for regression analyses. Researchers working with spatially aggregated data have noted that when spatial autocorrelation exists in the data but is ignored in analyses, such as in classic ordinary least squares (OLS) regression, the results are biased. To derive reliable results, spatial analytical techniques were used to control for the effects of spatial autocorrelation [19].
### Methods
#### Data
Two sources of data were used in this study: substantiated neglect and abuse data from the Division of Family and Children Services (DFCS), and vital record births from the Division of Public Health (DPH), both of the Georgia Department of Human Resources (DHR). The State of Georgia complies with the U.S. Standard Certificate of Live Birth, developed by the National Center for Health Statistics (NCHS) [20], with regards to the variables acquired for birth certificates. This research was undertaken with the approval of the DHR Institutional Review Board (IRB). Given that the research involved the use of existing data sets, with confidentiality preserved, there were no risks to individual human subjects. The DHR Project Number is 060807.
Data on substantiated neglect and abuse were collected from 2000 through 2002 between January 1st and December 31st of each year. In 2000, 2001, and 2002, respectively, 2,642, 2,869, and 3,205 infants were victims of one or more types of maltreatment including neglect, physical abuse, emotional abuse, sexual abuse, and other types of abuse. DFCS declined to provide identifiers to link substantiated maltreatment cases to births, so we were unable to conduct the study at the individual level; instead, we conducted an ecological study. That is, we examined child maltreatment in relation to perinatal characteristics of the communities in which the maltreatment victims lived [21]. The census tract is a surrogate for the community. Tracts are sub-county areal units designed to be demographically homogeneous by the U.S. Bureau of the Census. The outcome variable is the rate of infant maltreatment by census tract. The hypothesis is that a census tract having a higher percentage of births with perinatal risks is more likely to have higher rate of infant maltreatment.
We followed the Standard Geocoding Procedures, designed by and applied within the Office of Health Information and Policy (OHIP), DPH, Georgia DHR, to assign latitude and longitude values to each of the records (F. Millard and G. Freymann, unpublished work, December 2001). The individual records were then aggregated to determine the number of maltreatment events (all types combined) in each census tract.
We extracted the birth records for 1999 through 2002, from which we derived risk factors and calculated the counts of births.
Individual perinatal risk factors examined include:
1. Medicaid beneficiary
2. Unmarried mother
3. Maternal age less than 20 years of age
4. Maternal education less than high school
5. Three or more siblings
6. Prenatal care less than adequate
7. Birth weight less than 2,500 grams
8. Mother smoked during pregnancy
9. Gestation less than 37 weeks
10. 5-minute Apgar scores less than 7
Composite risks include:
11. Child_risk: Any of risks 7, 9, or 10 (representing neonatal difficulties) is present.
12. Composite_risk1: Of risks 1, 2, 5, 7, or 8, three or more are present.
13. Composite_risk2: Of risks 1, 2, 5, 8, or 11, three or more are present.
The child risk composite (Child_risk) is a single variable that represents the presence of one or more neonatal difficulties. Composite_risk1 characterizes extreme risk infants who had three or more key risk factors present (identified in [14]). Composite_risk2 is equivalent to Composite_risk1 except the child risk composite replaces low birth weight.
To adjust for the effects of different lengths of exposure in each year, we applied the person-time concept to calculate the rate of infant maltreatment for each tract by dividing the total number of maltreatment victims by the total number of weighted births during a three-year period [22]. Births were weighted by the length of time spent, as an infant, in each target calendar year of data collection. In doing so, we calculated for each record three weighting fields, each denoting the proportion of time, in infancy, over a one-year period. The number of weighted births for a census tract was the sum of all three weights of all the births residing in that tract. Then the numbers of maltreatment victims and weighted births for individual years for each tract were summed to obtain the total number of maltreatment victims and total number of weighted births during a three-year period.
Among the 1,618 census tracts in Georgia for the year 2000, 32 tracts had fewer than 30 weighted births. To reduce small number problems in calculating rates, each of these 32 tracts was merged to one or more neighboring tracts, resulting in 1,589 enumeration units (for simplicity still referred to as tracts), each with more than 30 weighted births. Figure 1a displays the substantiated infant maltreatment rates by tract, presented as the number of substantiated infant maltreatment victims per 1,000 weighted births.
Figure 1. Cartographic visualization of infant maltreatment in the state of Georgia. a) Substantiated Abuse and Neglect. Birth is weighted by the length of time as an infant in the target year. b) Composite_Risk1. Of the 10 perinatal risks examined, this composite includes: 1. Medicaid beneficiary; 2. Unmarried mother; 5. Infant with three or more siblings; 7. Birth weight < 2500 g; 8. Mother smoked during pregnancy. c) Composite_Risk2. Of the 10 perinatal risks examined, this composite includes: 1. Medicaid beneficiary; 2. Unmarried mother; 5. Infant with three or more siblings; 8. Mother smoked during pregnancy; 7., 9., or 10. Presence of neonatal difficulties.
All birth records for the years 1999 through 2002 were processed to create Boolean fields, 1 meaning present and 0 not present, for each risk factor. The Child_risk factor was coded 1 if any of the three previously-mentioned neonatal difficulties were present. Composite_risk1 or Composite_risk2 were coded 1 if three or more of the related 5 risk factors were present. For any record, if any of the factors were unknown, the record was omitted for the calculations of risk factors. The individual birth records were then aggregated to the tract, from which we obtained the percentage of births coded 1 in each tract for each of the risk factors. Figures 1b and 1c display the percentage of births coded 1 by tract for two composite risk factors (Composite_risk1 and Composite_risk2).
#### Analysis
In any spatial analysis, reviewing mapped data is recommended to determine if the distribution suggests any patterns or relationships among mapped features [23]. Qualitative, visual analysis of the mapped data, or cartographic visualization (Figures 1a, 1b, and 1c), suggests that spatial autocorrelation, meaning that similar data values tend to cluster geographically, is more pronounced with regards to the risk factors (Figures 1b and 1c) in contrast to substantiated maltreatment (Figure 1a). In Figure 1a, tracts in the lowest classification are more often immediately adjacent to tracts in the highest classification (note that all maps use the quantiles classification method). For instance, tracts in Montgomery and Wheeler, adjacent counties in the southeastern portion of the state, are at opposite ends of the classification scheme in terms of maltreatment, whereas the risk factor maps show both Montgomery and Wheeler in the mid- to upper mid-range of the classification scheme.
Any correlation present among the three maps seems to be more readily visible in the urban areas of the state, as opposed to rural portions of the state. The north central counties of Fulton, DeKalb, Cobb, Gwinnett, and Clayton, which make up much of metropolitan Atlanta, show a clear pattern (Figure 2b). North Fulton County, which has a low rate of substantiated maltreatment, is in stark contrast to central and south Fulton, which have a high rate of substantiated maltreatment. Northern DeKalb County, immediately adjacent to Fulton, is also lower in risk than southern DeKalb. Additionally, the north metro counties of Gwinnett, and to a lesser extent Cobb, have low rates of substantiated maltreatment, in contrast to the south metro county of Clayton which has a higher rate of maltreatment. The pattern of substantiated maltreatment in Atlanta is echoed in both the maps of perinatal risk factors, with fewer risk factors in the north and more risk factors in the south (Figures 2c and 2d). Other urban areas in the state depict similar patterns among the three maps.
Figure 2. Cartographic visualization of infant maltreatment in metropolitan Atlanta. a) Inset map showing the location of the Atlanta area within the state of Georgia. b) Substantiated Abuse and Neglect. c) Composite_Risk1. d) Composite_Risk2.
This visual analysis noting the presence of spatial autocorrelation and the correlation between infant maltreatment and risk factors is tenuous at best and must be validated with robust quantitative spatial analysis. The following paragraphs describe the quantitative spatial analysis of perinatal risk factors for substantiated infant maltreatment.
We used bivariate and multivariate linear regression methods for the quantitative analysis. The dependent variable was the maltreatment rate, and the predictor variables were the individual and composite risk factors. To ensure the normal distribution of the dependent variable, we transformed the rate to its natural logarithmic form using the following formula [24]:
where ln() is the natural logarithmic transformation function;
Yi is the number of infant maltreatments in tract i;
WNi is the weighted births in tract i;
and zi is the transformed rate.
This formula not only gives valid values for those tracts with Yi = 0, but also helps discriminate the tracts with Yi ≤ 1 but with different WNi [24].
Let x1, x2, ... xk denote k risk factors chosen to be included in the regression equation; z the dependent variable; and e the error term. A multivariate regression equation is expressed as:
z = b0 + b1x1 + b2x2 + ... + bkxk + e = Xβ + e (2)
in which X = (1 x1 x2 ... xk), β = [b0 b1 ... bk]T, b0 is the constant, and bj(j = 1, ..., k) is the slope reflecting the relationship between xj and z. Equation (2) becomes the bivariate regression expression if only retaining a single risk variable.
In traditional statistics with nonspatial data, the OLS method is used to estimate parameters b0, b1, ..., bk (or b0, b1 in bivariate regression). In order for the statistical inference about parameter estimates to be valid, some assumptions about the data must be satisfied [25]. The key assumption is that the error term is independent and normally distributed with a constant mean of zero and constant variance of σ2 (homogeneity). For multivariate regression, an additional key assumption is that there is no multicollinearity among the predictor variables.
Assuming there is no serious problem with multicollinearity, the OLS method provides the best linear unbiased estimates only if the regression model is correctly specified so that the error term meets the above assumption. A regression model is considered misspecified in several situations: 1) the dependent variable is inherently spatially dependent; 2) the unit of analysis does not match the unit of actual phenomena; 3) important risk variables are not included in the model; and 4) the observations of the dependent and/or predictor variables are not free of errors [24,26,27]. If a regression model is misspecified, the errors after the OLS fitting will not be independent; instead, the error at one location may be correlated with the errors at nearby locations, resulting in the clustering of similar errors among nearby locations, or spatial autocorrelation. When the errors are spatially autocorrelated, the OLS estimates are no longer unbiased and the goodness-of-fit measure R2 is upward biased [28].
A common method to handle spatial autocorrelation is to minimize its effects by resampling to create a subset of data. This can be done in one of two ways; either manually selecting data locations or using a random process [29]. However, both the manual and the random process have some drawbacks. Manual selection may be subjective, random selection may not be free of spatial dependency, and both methods may result in loss of information, that is, the selected subset may not represent all the characteristics of the dataset.
A less commonly used but more objective method is spatial regression, which considers spatial autocorrelation an additional variable in the regression equation and solves its effect simultaneously with the effects of other explanatory variables [27]. This method uses all available information in the dataset.
We therefore controlled for spatial autocorrelation effects using the spatial regression method. There are two ways to incorporate spatial autocorrelation in a regression model. One is to model spatial autocorrelation in the error term as a spatially lagged dependent variable, and the other as a spatially lagged error term, that is,
e = ρWz + ε (3)
or
e = λWe + ε (4)
where W is the weight matrix characterizing the spatial relationship between every pair of observations;
ρ or λ is the spatial autoregressive parameter characterizing spatial autocorrelation;
And ε is the independent and normally distributed error term with a constant mean of zero and constant variance.
The former is referred to as a spatial lag model and the latter a spatial error model. Substituting the error term in (2) with Equation (3) or (4) and reorganizing the equation lead to the expression of a spatial lag model:
z = (I - ρW)-1Xβ + (I - ρW)-1 ε (5)
or that of a spatial error model
z = Xβ + (I - λW)-1 ε (6)
where I is the identity matrix.
The OLS method is no longer appropriate for estimating the parameters in Equations (5) and (6); instead, the maximum likelihood estimation method (MLE) or the instrumental variables estimation (IVE) method should be used [26]. The MLE method estimates model parameters by maximizing the Likelihood Function of the observations [26,27].
When the MLE method is used to estimate the parameters in Equation (5) or (6), the traditional goodness-of-fit measure, R2, is no longer valid for assessing model fit [26]. One appropriate measure is the Akaike Information Criteria (AIC). A model is considered the best among a set of alternatives if the model gives the smallest AIC value. An approximate measure that mimics R2 is the so-called pseudo-R2, which provides a measure of linear association between the observed and predicted values of the dependent variable, but is no longer related to the variance component explained by the model.
The software used for the spatial regression analysis in the present study is GeoDA (Version 0.9.5i_6) [30,31]. The program provides tools to calculate spatial weights, and run OLS (Classic) as well as spatial regression (Spatial Lag and Spatial Error) models. The output of the OLS models includes the diagnostic for multicollinearity (i.e., multicollinearity condition number or MCN), and the array of test statistics for spatial autocorrelation, which suggest whether spatial autocorrelation is significant to consider, and if so, which spatial model should be used. A value over 30 for MCN suggests problems with multicollinearity [32].
### Results
Table 1 lists the descriptive statistics of the dependent variable and risk factors. The rate is highly positively skewed (Skewness = 1.58) followed by three risk factors (Three or more siblings, 5-Minute Apgar Scores <7, and Mother smoked during pregnancy). As discussed previously, the transformation of rates reduced skewness. The transformed rates are slightly negatively skewed (Skewness = -0.37) with 7 lower outliers and no upper outliers. The lower outliers were excluded from the regression analyses.
Table 1. Descriptive statistics of infant maltreatment rates and perinatal risk factors
The data were first analyzed using the OLS model. The errors, i.e., the differences between the observed and predicted values of the dependent variable, were computed and tested for the statistical significance of spatial autocorrelation. The test statistics (not shown) indicated that spatial autocorrelation was significant for all the bivariate and multivariate models. Therefore, spatial regression was performed to account for spatial autocorrelation effects. The results are presented in Tables 2 and 3. Table 2 presents the bivariate spatial regression results, in which Models 1–10 are related to individual risk factors, and Models 11–13 to composite risk factors. The smaller the AIC value, the better the model. Model 13, Composite_risk2, is the best among the bivariate models in terms of providing the smallest AIC value. All the individual risk factors as well as the three composite risk factors are statistically significantly with respect to the rate of infant maltreatment, each with probability P-value < 0.0000. It is noted that in each of the bivariate regression models spatial autocorrelation is statistically significant with P-value < 0.0000.
Table 2. Spatial regression of infant maltreatment rates on perinatal risk factors: bivariate
Table 3. Spatial regression of infant maltreatment rates on perinatal risk factors: multivariate
Table 3 presents the multivariate spatial regression results. This is the best multivariate model among several alternatives (others not shown) with the smallest AIC value. Included in the model are six risk factors: mother smoked during pregnancy, three or more siblings, maternal age <20, unmarried marital status, Medicaid beneficiary, and prenatal care less than adequate. The model has a value of 20.5 for MCN, indicating no serious problem with multicollinearity [32]. The value of pseudo-R2 suggests the model has moderate predictive ability. All risk variables are statistically significantly with respect to the rate of infant maltreatment at least at the 0.05 level. Also, spatial autocorrelation is a significant factor with P-value < 0.0000.
The multivariate model provides slightly better predictivity than any of the bivariate models since it gives the smallest AIC value. However, the decrease of AIC values is only 2% compared with the best bivariate model (Model 13).
### Discussion
Several issues must be addressed to properly interpret the results: 1) ecological design, 2) spatial autocorrelation, and 3) the different types of child maltreatment.
This research used an ecological study design. Relationships found in ecological studies may suffer from two problems: the ecological fallacy and modifiable areal unit problems (MAUP) [24]. This means relationships found at the census tract level are pertinent to the current configuration of census tracts. They may not be inferred to finer levels such as block group or to individuals who lived in the areas.
We found spatial autocorrelation effects statistically significant in both bivariate and multivariate regression models. Several situations described in the Method section might be present in this study: inherent spatial dependency, missing variables, a mismatch between the unit of analysis and the unit of actual phenomena, and measurement errors. First, the dependent variable might be inherently spatially dependent. Since the observations of the response variable were not acquired through a strict sampling design but a collection of data arranged by the geographic unit, i.e., the census tract, the interdependence between observations of neighboring census tracts might be the rule rather than the exception [27]. Second, it is apparent that many variables were not included in any of the bivariate regression models. Even in the multivariate regression model, there were important variables missing. This is because child maltreatment is a social-psychological phenomenon that involves a number of risk factors, including the characteristics of the victim, the maltreater, the family, the community, and the society, as suggested by the ecological theory of child maltreatment [33,34].
The problem of mismatch between the unit of analysis and the unit of actual phenomena might also be present. This study used the census tract as a surrogate of the community. There is no compelling reason at this time to believe that child maltreatment conforms to the configuration of census tracts, except to note that tracts are designed to be demographically homogeneous. Future study using a more meaningful geographic unit may be beneficial.
The spatially correlated measurement errors in this study might result from the geocoding process. Approximately 85% of the maltreatment records were geocoded based on their address information to the accuracy of the census tract level. For the remaining 15% of the records, the county was known, but had inappropriate address information, such as P.O. boxes, or incorrect and/or incomplete addresses. For these records, a method of spatial imputation was employed in which census tracts within the county were assigned based on fertility probability, a method with an adjusted R2 accuracy of 0.82 compared with known county fertility rates. Latitudes and longitudes were assigned randomly within selected tracts (F. Millard and G. Freymann, unpublished work, December 2001). Therefore, a record was more likely to be located in a census tract that is spatially close to the correct tract.
A final issue to address is that of the different types of child maltreatment. In this study we aggregated all types of abuse and neglect, which include neglect, physical abuse, emotional abuse, sexual abuse, and other abuse, to reduce the small number problem. It has been suggested that different types of child maltreatment involve different risk factors [35]. Therefore, it may be of benefit in future studies to separate out the types of maltreatment to determine if significant risk factors change or improve the model.
### Conclusion
Efficient allocation of resources is a priority. The method described in this paper can augment decision making regarding funding and intervention decisions.
With proper data architecture, addressing some of the shortcomings of the ecological methods used in this research can be easily overcome. For example, adjusting the spatial unit of analysis from the tract to some other form or level of precision, such as cells of varying sizes or block groups, becomes simple when data and quality standards are in place. What is less clear however, is which unit of analysis is most appropriate for the outcome of interest. In the event that "sociological meaningful scale" is unclear, analyzing the data at multiple scales may be beneficial.
Upon consideration, what at first seemed a limitation, namely the lack of unique identifiers to link individual abuse records with birth records, could actually have led to a more suitable method of analysis for the data. The method described should be used to help target interventions towards population sub-groups, as this is the purpose of public health – to assure the conditions in which people can be healthy – and does not necessarily require a person-based medical model.
One should use the ecological model to open inquiry into other social and environmental conditions present in areas with the greatest maltreatment risk. Why do some communities demonstrate the characteristics of the multivariate regression model (Table 3), i.e. high percentages of young, unmarried mothers, mothers who smoke during pregnancy, inadequate prenatal care, large families, and Medicaid recipients? What actions should be taken to alleviate these conditions/outcomes? The model lends itself to a holistic approach toward community health assessment that is based on resiliency, rather than merely presence or absence of disease or poor outcome. As stated previously, the Health Field Concept and Health Field Theory support such an approach.
The presence of spatial autocorrelation has at least two important implications: 1) an assessment model must take spatial autocorrelation into account, and 2) poor health outcomes exhibiting spatial autocorrelation indicate the need for community-level (ecological) responses. Implicit in 2) is that case-management alone will not prevent child maltreatment when larger ecological issues are not identified nor addressed.
Perhaps most importantly, is that this and similar models allow public health to take a proactive stance, and reasonably predict areas where poor outcomes are more likely to occur. U.S. states routinely collecting variables defined by NCHS for birth certificates can easily identify areas that are at high risk for infant maltreatment. This implies that public health need not be burdened by relying only on the current practice of case-management, but can put measures in place to target areas before maltreatment occurs.
### List of abbreviations
AIC – Akaike Information Criteria
DFCS – Department of Family and Children Services
DHR – Department of Human Resources
DPH – Division of Public Health
IRB – Institutional Review Board
MAUP – Modifiable Areal Unit Problem
MCN – Multicollinearity Condition Number
MLE – Maximum Likelihood Estimation
NCHS – National Center for Health Statistics
OHIP – Office of Health Information and Policy
OLS – Ordinary Least Squares
### Competing interests
The author(s) declare that they have no competing interests.
### Authors' contributions
All authors made substantial contributions to the conceptual design of the study and to the acquisition and processing of data. EJH performed the initial exploratory data analysis/cartographic visualization. YZ followed with confirmatory statistical analysis and wrote the first draft of the paper. EJH produced the presentation graphics and the final manuscript with extensive input regarding findings of relevance to public health from GRF.
### Acknowledgements
The authors thank the following persons and agencies within the Georgia Department of Human Resources (DHR) for their support: The Division of Family and Children Services, Stuart T. Brown, M.D., Director of the Division of Public Health, and B.J. Walker, DHR Commissioner.
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
Finding maximal k-degenerate subgraphs
Given a graph $G$, let $H$ be a $k$-degenerate (not necessarily induced) subgraph of maximal size. Are there any known lower bounds on $|E(H)|$ for particular classes of $G$ and values of $k$?
I've seen several papers on the subject, but only for induced subgraphs $H$. I'm particularly interested in the case where $k=2$ and $G$ is both regular and bipartite, but any additional information would be helpful.
(I'm aware of the naive lower bound found by arbitrarily removing edges until no vertex has degree $(k+1)$ or more...I'm looking for something better).
-
what does $k$-degenerate means? – Dima Pasechnik Sep 24 at 7:03 @Dima - $k$-degenerate means that every subgraph has a vertex of degree at most $k$. Equivalently, repeatedly remove the vertex of minimum degree until the graph is gone; the degeneracy number of the graph is the biggest degree you saw along the way. – gordon-royle Sep 24 at 22:47
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Consider the following $(2\times2)$ matrix
$\left( \begin{array}{c} 4 & 0 \\ 0 & 4 \end{array} \right)$
Which one of the following vectors is $\text{NOT}$ a valid eigenvector of the above matrix ?
1. $\left( \begin{array}{c} 1 \\ 0 \end{array} \right)$
2. $\left( \begin{array}{c} -2 \\1 \end{array} \right)$
3. $\left( \begin{array}{c} 4 \\ -3 \end{array} \right)$
4. $\left( \begin{array}{c} 0 \\ 0 \end{array} \right)$
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# Alg. 2 question
If A=9^9999 + 9^-9999 and B= 9^9999 - 9^-9999.
Then find the value of A^2 -B^2 + 1.
would i just take the the numbers and square them to get the value. What should i do.
$$A^2 - B^2 = ( A+B )( A-B )$$
for any $A, B$
That should simplify it a great deal
so would the values be 0 or 18. this question is confusing to me even when you did simplify it for me
James R
Homework Helper
Gold Member
Put p = 9^9999 and q= 9^-9999
Then A = p+q and B = p-q
You want:
$$A^2 + B^2 - 1 = (p+q)^2 + (p-q)^2 - 1 = (p^2 + q^2 + 2pq) + (p^2 +q^2 - 2pq) - 1$$
$$= 2p^2 + 2q^2 - 1 = 2(9^{9999}) + 2(9^{-9999}) - 1$$
Curious3141
Homework Helper
James R said:
Put p = 9^9999 and q= 9^-9999
Then A = p+q and B = p-q
You want:
$$A^2 + B^2 - 1 = (p+q)^2 + (p-q)^2 - 1 = (p^2 + q^2 + 2pq) + (p^2 +q^2 - 2pq) - 1$$
$$= 2p^2 + 2q^2 - 1 = 2(9^{9999}) + 2(9^{-9999}) - 1$$
Unnecessarily complicated, and I'm afraid you read the question wrongly.
$$A = 9^{9999} + 9^{-9999}$$ and $$B = 9^{9999} - 9^{-9999}$$
$$A + B = (2)(9^{9999})$$ and $$A - B = (2)(9^{-9999})$$
$$A^2 - B^2 + 1 = (A + B)(A - B) + 1 = (4)(9^{9999})(9^{-9999}) + 1 = 4 + 1 = 5$$
James R
Homework Helper
Gold Member
Curious3141 said:
Unnecessarily complicated, and I'm afraid you read the question wrongly.
I would argue that my solution is no more complicated than yours. You are, however, correct that I copied the question wrongly. My correct solution is:
$$A^2 - B^2 + 1 = (p+q)^2 - (p-q)^2 + 1 = (p^2 + q^2 + 2pq) - (p^2 +q^2 - 2pq) + 1$$
$$=4pq + 1 = 4(9^{9999})(9^{-9999}) + 1 = 4 + 1 = 5$$
So, we agree.
they both seem right to me just that the letters are changed and makes it a little organized and not messy. So the right answer is 4+1 = 5.
Thanks for the help.
I was wondering if you an explantation for what you did since my teachers requires how we got it instead of just stating a thoure (how ever you spell it) about how we got it. it is kind of stuiped but makes perfect sense to him.
Curious3141
Homework Helper
James R said:
I would argue that my solution is no more complicated than yours. You are, however, correct that I copied the question wrongly. My correct solution is:
$$A^2 - B^2 + 1 = (p+q)^2 - (p-q)^2 + 1 = (p^2 + q^2 + 2pq) - (p^2 +q^2 - 2pq) + 1$$
$$=4pq + 1 = 4(9^{9999})(9^{-9999}) + 1 = 4 + 1 = 5$$
So, we agree.
I feel it's needlessly complicated to expand out the square terms, you can factorise immediately without doing that. With your notation, it would simply be :
$$(p + q)^2 - (p - q)^2 + 1 = (p + q + p - q)(p + q - p + q) + 1 = (2p)(2q) + 1 = 4pq + 1$$
and I personally think that is simpler, IMHO. But let's not split hairs.
James R
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Well, today is the last day of 2014 and we’re just going take a look at non-standard notation called Chinese Dumbass Notation. Actually this post is only a little description about it, since I have never used this one to prove a problem in math contest.
## Definition
Chinese Dumbass Notation (CDN), is a way to write down three variable homogeneous expressions. It allows to keep like terms combined while not requiring any sort of symmetry from the expression.
CDN uses a triangle to hold the coefficients of a three variable symmetric inequality in a way that is easy to visualize. By example:
Here $[x]$ represents the coefficient of $x$. In general, a degree $d$ expression is represented by a triangle with side length $d+1$ and the coefficient of $a^{\alpha}b^{\beta}c^{\gamma}$ is placed at Barycentric coordinates $(\alpha, \beta, \gamma)$.
## AM-GM and Schur Inequalities
• The AM-GM inequality for two variables states that $a^2 + b^2 \geq 2ab$. In other words: $% $
• The Schur’s Inequality states that $\sum_{sym} a^r(a-b)(a-c) \geq 0$ for all non-negative numbers $a, b, c, r$. The prototypical application of Schur’s Inequality looks as follows:
As with AM-GM, the variables can be tweaked so that Schur’s Inequality gives us $\sum_{cyc}a^r(a^s-b^s)(a^s-c^2) \geq 0$
## Example
Prove that $\frac{bc}{2a+b+c} + \frac{ac}{a+2b+c} + \frac{ab}{a+b+2c} \leq \frac{1}{4}(a+b+c)$
Proof. We clear denominators to obtain that this is equivalent to
We’ll expand the left hand side as
We then expand the right hand side as
The inequality is thus equivalent to showing that the difference in non-negative. But the difference is
by Schur and AM-GM inequalities.
Well, all the information above is thanks to Mr. Brian Hamrick, this post is just to take a look at CDN. So, if you want to expand the knowledge about it, please visit: The Art of Dumbassing.
I wonder, how many people have used this one in math contest? it’s just curiosity. If you did it and want to share with me, please send me a mail. Maybe we can resolve some problems together. Anyway. The best wishes for everyone in 2015
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2015-02-01T14:21:47-05:00
He has £160 out of £400 saved up. Just rewrite that as a fraction and then there are two ways to finish it off.
The first way would be to just divide.
Then turn that decimal into a percentage by multiplying by 100.
40%
The second way would be simplify the fraction until the denominator is 100. The numerator is your answer.
40%
Harry has saved 40%.
2015-02-01T14:45:46-05:00
160/400 divide by 2 = 80/200 = 40/100 = 20/50 = 10/25 = 2/5 and 2/5 as a percentage is 40%
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# Direction of force without right-hand rule?
1. May 11, 2015
### Vinster
Hi guys
I'm not sure if this is the right area - this is my first post - but it seems like it fits so
My question is, is there a way to get the direction of force on a charged particle from a magnetic field without using the right hand rule? What I'm looking for would be a way to just think about it, rather than look at my hands, or imagine my hands.
2. May 11, 2015
### robphy
You could calculate $q\vec v\times \vec B$ [and, if needed, get the components]
3. May 11, 2015
### Vinster
derp.....for someone currently finishing up a linear algebra course.. I should have gotten that xP but thanks a ton!
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# Count every word occurrence from file
How good is this algorithm? Personally I think it's over-complicated and needs some improvement. Should I be using iterators here (I think it's "harder" this way) or indexing? Should I stick with a vector or try other containers? Is a vector (array) of pointers generally better than one of objects?
#include <iostream>
#include <fstream>
#include <string>
#include <vector>
struct StringOccurrence //stores word and number of occurrences
{
std::string m_str;
unsigned int m_count;
StringOccurrence(const char* str, unsigned int count) : m_str(str), m_count(count) {};
};
int main()
{
std::string path;
std::cout << "enter file name: " << std::endl;
std::cin >> path;
std::ifstream in(path);
if (!in) //check if file path is valid
{
std::cerr << "failed to load file!" << std::endl;
return -1;
}
std::vector<std::string>vec;
std::string lineBuff;
while (std::getline(in, lineBuff)) // write multiline text to vector of strings
{
vec.push_back(lineBuff);
}
std::vector<StringOccurrence*> strOc;
std::string stringBuff;
for (auto it = vec.begin(); it < vec.end(); it++) //itterate through each line
{
for (auto it2 = it->begin(); it2 < it->end();it2++) //itterate through each letter
{
if (*it2 != ' ') //keep adding letters to buffer until space (word)
{
stringBuff += *it2;
}
if (*it2 == ' ' || it2 + 1 == it->end()) //(need fix?)
{
if (*(it2 - 1) == ' ') //check for reccurring spaces so they are not counted as words
{
continue;
}
for (auto it3 = strOc.begin(); it3 < strOc.end(); it3++)
{
if (stringBuff == (*it3)->m_str) //if word was found increase count
{
(*it3)->m_count++;
goto end; //skip next step (need fix?)
}
}
strOc.push_back(new StringOccurrence(stringBuff.c_str(), 1)); //if word was not found add it
end:
stringBuff.clear(); //empty buffer for next word
}
}
}
std::ofstream out("test2.out"); //write to file
for (auto it = strOc.begin();it < strOc.end();it++)
{
out << (*it)->m_str << ' ' << (*it)->m_count << std::endl;
}
return 0;
}
## 3 Answers
Excessive Memory Use
The first thing you do is read every line into memory:
while (std::getline(in, lineBuff)) // write multiline text to vector of strings
{
vec.push_back(lineBuff);
}
Why? Do you need to have every line in memory? No. You're just trying to count the words. What if the file was 100TB worth of "all work and no play makes Jack a dull boy"?
Poor Memory Use
std::vector<StringOccurrence*> strOc;
That should yell antipattern to you. You're not holding pointers to existing objects, this is strOc actually owning that memory. You use new to add to it. I don't see any deletes. So you're leaking memory. Doesn't matter so much in this case, but this is exactly what RAII is for. Prefer:
std::vector<std::unique_ptr<StringOccurrence>> strOc;
Though, regardless, don't use a vector to begin with because...
Data Container Choice
You are holding a vector of StringOccurrences. That means that on each new word, you have to walk through the whole vector doing string comparisons. That is O(N). Everything about this is bad. To start with:
goto end; //skip next step (need fix?)
Yes, this needs a fix. Never use goto. The way to search for things is to use the <algorithm> library, specifically std::find_if:
auto it = std::find_if(strOc.begin(), strOc.end(), [&](StringOccurence* so){
return so->m_str == stringBuff;
});
if (it != strOc.end()) {
// success case
}
else {
// new word case
}
That'll make your code easier to understand by far and drop the goto. But it won't handle the O(N) problem. For that, we just need to use an entirely new data structure. We need to map a word to a number, and we don't care about word ordering. Thus, std::unordered_map. This will have O(1) lookup. And we don't even need to do all that work ourselves!
std::unordered_map<std::string, int> strOc;
// for each word
++strOc[stringBuff]; // this will insert new elements as necessary
Cool.
Read the words directly
You are walking through a line character by character. That is error-prone at best, hard to follow at worst. But there's already support for this in C++. Use a std::stringstream. You could have just done:
while (std::getline(in, lineBuff)) {
std::string word;
std::istringstream iss(lineBuff);
while (iss >> word) {
...
}
}
Then again, operator>> is defined on any istream, not just istringstream. So we can use that on the ifstream directly:
std::string word;
while (in >> word) {
...
}
Use the right loops
Don't use auto with iterators - just use a range-based for-expression. It'll save on the typing and add to clarity.
Improved solution
std::unordered_map<std::string, int> wordCounts;
std::string word;
while (in >> word) {
++wordCounts[word];
}
std::ofstream out("test2.out");
for (auto const& wc : wordCounts) {
out << wc.first << ' ' << wc.second << '\n';
}
• This inserts full lines, not words. You need some way of splitting based on spaces. Sep 14, 2015 at 14:16
• No need to read a line then serialize the line. Just serialize the in into strings. Sep 14, 2015 at 17:52
• To add to @LokiAstari's comment, the first loop would be much better as std::string word; while (in >> word) { ++wordCounts[word]; } There's no reason to read lines. Sep 14, 2015 at 18:16
Yes this can be simplified to:
int main()
{
std::ifstream inputFile("Bob");
std::unordered_map<std::string, int> count;
std::for_each(std::istream_iterator<std::string>(inputFile),
std::istream_iterator<std::string>(),
[&count](std::string const& word){++count[word];});
}
# Why this works:
### operator>>
When you read a string from a stream with operator>> it read a space separated word. Try it.
int main()
{
std::string line;
std::cin >> line;
std::cout << line << "\n";
}
If you run that and type a line of text. It will only print out the first space separated word.
### std::istream_iterator
The standard provides an iterator for streams. std::istream_iterator<X> will read an object of type X from the stream using operator>>.
This allows you to use streams just like you would any other container when using standard algorithms. The standard algorithms take two iterators to represent a container (begin and end or potentially any two points in the container).
So by using std::istream_iterator<std::string> you can treat a stream like a container of space separated words and use it in an algorithm.
int main()
{
std::string line;
std::istream_iterator<std::string> iterator(std::cin);
line = *iterator; // de-reference the iterator.
// Which reads the stream with operator >>
std::cout << line << "\n";
}
### std::for_each
I use std::for_each above because it is trivial to use. But with a tiny bit of work you can use the range based for loop introduced in C++11 (as this just calls std::begin, std::end on the object to get the bounds of the loop.
But lets look at std::for_each first.
std::for_each(begin, end, action);
Basically it loops from begin to end and performs action on the result of de-referencing the iterator.
// In my case action was a lambda
[&count](std::string const& word){++count[word];}
It captures count from the current context to be used in the funtion. And de-referencing the std::istream_iterator<std::string> returns a reference to a std::string object. So we can not use that to increment the count for each word.
Note: count is std::unordered_map so be looking up a value it will automatically insert it if it does not already exist (using default value (for int that is zero). Then increment that value in the map.
### Range based for
A quick search to use range based for with std::istream_iterator gives me this:
template <typename T>
struct irange
{
irange(std::istream& in): d_in(in) {}
std::istream& d_in;
};
template <typename T>
std::istream_iterator<T> begin(irange<T> r) {
return std::istream_iterator<T>(r.d_in);
}
template <typename T>
std::istream_iterator<T> end(irange<T>) {
return std::istream_iterator<T>();
}
int main()
{
std::ifstream inputFile("Bob");
std::unordered_map<std::string, int> count;
std::for(auto const& word : irange<std::string>(inputFle)) {
++count[word];
}
}
## Issues with this technique.
We use space to separate words. So any punctuation is going to screw things up. Not to worry. C++ allows you to define what is a space in any given context. So we just need to tell the stream what is a space.
https://stackoverflow.com/a/6154217/14065
## Review of code
Sure.
struct StringOccurrence //stores word and number of occurrences
{
std::string m_str;
unsigned int m_count;
StringOccurrence(const char* str, unsigned int count) : m_str(str), m_count(count) {};
};
But you can do this with a number of standard types.
typedef std::pair<std::string, unsigned int> StringOccurrence;
You are doing this to store the value in a vector. But a better way to store this is in a map. Because maps are ordered in some way internally lookup is a lot faster. std::map gives access in O(ln(n)) or std::unordered_map gives access in O(1).
I hate bad comments. Bad comments are worse than no comments because they need to be maintained and the compiler will not help you maintain them.
if (!in) //check if file path is valid
Not quite, but close enough I suppose. But I don't really need the comment to tell me that. The code seems pretty self explanatory.
Note sure if -1 is a good value. It will really depend on the OS you are running on. 0 is the only valid value. Anything else is considered an error. At your OS level this will probably be truncated to 255 on most systems (but not all).
return -1;
If you run this:
> cat xrt.cpp
int main()
{
return -1;
}
> g++ xrt.cpp
> ./a.out
> echo \$? # Echos the error code of the last command.
255
I don't think you need to copy the whole thing into memory.
std::vector<std::string>vec;
std::string lineBuff;
while (std::getline(in, lineBuff)) // write multiline text to vector of strings
{
vec.push_back(lineBuff);
}
Just read a line at a time and processes that.
Don't use pointers in C++
std::vector<StringOccurrence*> strOc;
C++ has much better ways to handle dynamic memory allocation and pointers is never the way to go.
When you iterate from begin -> end of something. You can use the new range based for instead.
for (auto it = vec.begin(); it < vec.end(); it++)
// easier to write and read:
for(auto const& val : vec)
Going to comment on your comments again.
for (auto it = vec.begin(); it < vec.end(); it++) //itterate through each line
Not very useful. I can see that you are iterating over every line. From the code. You should restrict your comments to WHY you are doing something.
Space ' ' is not the only white space character! What about tab or carrige return \r or vertical tab \v. You should test for space using standard library routines.
std::is_space(c)
I have use goto probably twice in the last ten years. One of those times was probably wrong.
goto end; //skip next step (need fix?)
Loops and conditions will always be better and easier to read.
We have a leak her:
strOc.push_back(new StringOccurrence(stringBuff.c_str(), 1));
I see a new (but no delete). See above about using pointers. There is no need to use a pointer here. Just use a normal object it will be moved into the vector.
Indeed, your algorithm looks quite complicated as is. This can be simplified dramatically with some algorithm usage and a more suitable data structure.
Whenever you have a value that you want to store some data for (such as an associated count), you should be thinking of std::map (or std::unordered_map).
using counter_type = std::unordered_map<std::string, unsigned>;
Instead of looping over each character one by by, it's easier to process it word-by-word, splitting on spaces. The standard library unfortunately doesn't have a (nice) way of doing this, but the boost string algorithms library does:
std::vector<std::string> no_spaces;
boost::split(no_spaces, vec, boost::any_of(' \t\n'),
boost::token_compress_on);
From here, you already have all the words you need, split up. The last thing to do is to add them to the map:
counter_type occurrence_counter;
for(const auto& word : no_spaces) {
++occurrence_counter[word];
}
Writing the map out to a file is similar to what you already have.
As an aside, there is no reason to store a pointer in your vector here. This will actually leak memory, as it is never deleted (for a short program like this, the OS will clean it all up very quickly anyway, but it's a bad habit to get into).
A final note: this solution is still incomplete, as it doesn't deal with punctuation of any kind. You may need to do thing such as strip all characters not in [a-zA-Z0-9] (in regex terms).
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Grids
The grids used by VSim are structured, coordinate aligned, where the grid lines are along coordinate directions. Such a two-dimensional grid is shown in Fig. 34. One can choose either a uniform spacing or a non-uniform spacing as shown in Fig. 34, and the coordinates may be either cartesian or cylindrical. In Fig. 34, each of the cells is numbered by its indices. In this $$2D$$ case, there are two indices; in general one for each direction. The cell indices start at $$0$$ and end in the $$x$$ direction at $$NX - 1$$ for a grid that has $$NX$$ cells in the $$x$$ direction. For a $$3D$$ grid, there would be another direction out of the page.
Fig. 34 Structured 2D grid.
A cell of a 3D grid is shown with z coming out of the page in two views in Fig. 35. A cell owns its interior plus the interior of its lower face in each direction plus, the interior of its lower edge in each direction, plus the lower node of its owned edges. The owned node is circled in both views of Fig. 35. The owned edges are shown on the left side of Fig. 35, with the x-edge red, the y-edge green, and the z-edge blue. Similarly, the owned faces of a cell are shown on the right side of Fig. 35, with the x-normal face red, the y-normal face green, and the z-normal face blue.
Fig. 35 3D cell in a view showing its edges and a view showing its faces.
In FDTD EM, the concept of a dual grid is useful. The dual grid is the grid with nodes at the centers of the regular grid. The edges of the dual grid pierce the faces of the regular grid and vice-versa.
Guard cells
Guard cells, cells just outside the simulation grid, are needed for having sufficient field values in the simulation region, for particle boundary conditions, and for parallel communication (to be dicussed later). An example of the first case is where a field must be know at each of the nodes of the simulation. Then, since the last node in any direction is owned by the cell one beyond the simulation, the cells one beyond the physical grid must be in the simulation. Thus, the grid must be extended by one cell in the last of each direction, as shown in Fig. 36.
Fig. 36 Grid extended to include the upper nodes, which belong to the cells one past the last cell in each direction.
Note
The user-defined grid is called the physical domain. The grid extended by Vorpal is called the extended domain.
For particle boundary conditions, the grid must be further extended down by one cell in each direction. When a particle leaves the physical domain, it can end up in one of these additional cells, which can be either above or below. A data value associated with that cell determines what to do with the particle, e.g., absorb it (remove it), reflect it, or carry out some other process. The associated extended grid is shown in Fig. 37. The physical cells are depicted by the red grid. The associated dimensions are in blue. The extended cells (shown in green) enclose both the physical cells and the guard cells added by Vorpal.
Fig. 37 Cartesian grid extended by Vorpal
Periodic Boundary Conditions
Periodic boundary conditions can be used to control both field and particle behavior at the edge of the simulation domain. In the case of particles, periodic boundaries ensure that particles leaving one side of the domain reappear at the opposite side. For example, particles traveling at a speed of $$-v_{\phi}$$ will go through $$\phi = 0$$ and reappear at $$\phi = 2\pi$$. Fields, on the other hand, will be copied from the plane at index $$0$$ to the plane at $$NX$$ and from the plane at $$NX-1$$ to the plane at $$-1$$, i.e. from the last physical cell to the guard cell on the other side.
Parallelism and Decomposition
Parallel (distributed memory, MPI) computation is carried out by domain decomposition. A particular decomposition is shown in Fig. 38. In this case, this is a decomposition of a rectangular region by the red lines, with the individual subdomains each give a unique index. However, Vorpal can simulate any region that is a non-overlapping collection of rectangles (appropriately generalized for 3D and 1D), with each rectangle being a subdomain of the decomposition.
Fig. 38 Parallel decomposition of a computational domain.
For the most part, parallelism and decomposition are handled under the hood, but it is useful to understand a few concepts. In Fig. 38 one can see a green rectangle that extends one cell more into the simulation region beyond Domain 2. Fields in the cells of this overlap region are computed by the subdomain holding the cell, but they have to be communicated to Domain 2, as it needs this boundary region to update its fields on the next time step. On the other hand, particles may leave Domain 2 and end up in one of the cells still inside the green rectangle. Those particles must be sent to the processor holding the cell they are in for further computation.
For the above situation to work, the field update method for a given cell must not need information more than one cell away. The standard updates for electromagnetics and fluids indeed have this property. As well, the particles must not travel more than one cell in a time step. This is true for explicit electromagnetic PIC with relativistic particles, as the Courant condition prevents the time step from being larger than the time it takes for light to cross any cell dimension, and relativistic particles travel slower than the speed of light. Finally, the particles must not interpolate from fields more than one cell cell away, nor must they deposit current more than one cell away. This is true for the simplest interpolation and deposition methods.
However, if you have electrostatic particles that travel more than one cell per timestep, or particles that have a larger deposition footprint, then manual setting of some parameters may be necessary. In particular, the grid parameter, maxCellXings, states the maximum number of cells a particle might cross in a simulation, and the parameter, maxIntDepHalfWidth, provides the width of the deposition stencil. From these follow the overlap needed for the subdomains. These parameters are discussed in more detail later.
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Wilmott Magazine: January 2019 issue
Volume 2019, Issue 99. Pages 1–72
Every issue we bring you original material from some of the best columnists, educators and cutting-edge researchers. Subscribe here.
In this issue:
Bibliography
• “Contents,” Wilmott, vol. 2019, iss. 99, p. 1–1, 2019.
[Bibtex]
@article {WILM:WILM10727,
title = {Contents},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10727},
doi = {10.1002/wilm.10727},
pages = {1--1},
year = {2019},
}
• D. Tudball, “Lock your door and call the law,” Wilmott, vol. 2019, iss. 99, p. 2–3, 2019.
[Bibtex]
@article {WILM:WILM10728,
author = {Tudball, Dan},
title = {Lock Your Door and Call the Law},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10728},
doi = {10.1002/wilm.10728},
pages = {2--3},
year = {2019},
abstract = {Quant finance could learn much from other disciplines that revel in the fact that you don't' need perfect models to be a science.},
}
• “News,” Wilmott, vol. 2019, iss. 99, p. 4–7, 2019.
[Bibtex]
@article {WILM:WILM10729,
title = {News},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10729},
doi = {10.1002/wilm.10729},
pages = {4--7},
year = {2019},
}
• A. Brown, “Kelly attention,” Wilmott, vol. 2019, iss. 99, p. 8–11, 2019.
[Bibtex]
@article {WILM:WILM10730,
author = {Brown, Aaron},
title = {Kelly Attention},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10730},
doi = {10.1002/wilm.10730},
pages = {8--11},
year = {2019},
abstract = {The logic behind the Kelly criterion gives a great simplification that converts the risk manager's ‘attention problem’ from impossible to just very difficult.},
}
• E. G. Haug, “Philosophy of randomness: time in relation to uncertainty,” Wilmott, vol. 2019, iss. 99, p. 12–13, 2019.
[Bibtex]
@article {WILM:WILM10731,
author = {Haug, Espen Gaardner},
title = {Philosophy of Randomness: Time in Relation to Uncertainty},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10731},
doi = {10.1002/wilm.10731},
pages = {12--13},
year = {2019},
abstract = {The infinitesimally small intervals of Planck time might hold some useful food for thought in finance.},
}
• R. Poulsen, “Kelly gone bad,” Wilmott, vol. 2019, iss. 99, p. 14–15, 2019.
[Bibtex]
@article {WILM:WILM10732,
author = {Poulsen, Rolf},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10732},
doi = {10.1002/wilm.10732},
pages = {14--15},
year = {2019},
abstract = {Analysis of the economics that lead to massive migration and the economics of migration itself is inadequate.},
}
• U. Wystup, “Fx greeks,” Wilmott, vol. 2019, iss. 99, p. 16–19, 2019.
[Bibtex]
@article {WILM:WILM10733,
author = {Wystup, Uwe},
title = {FX Greeks},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10733},
doi = {10.1002/wilm.10733},
pages = {16--19},
year = {2019},
abstract = {Covering the possibilities that Greeks in FX create…},
}
• P. Wilmott and D. Orrell, “No laws, only toys,” Wilmott, vol. 2019, iss. 99, p. 20–29, 2019.
[Bibtex]
@article {WILM:WILM10734,
author = {Wilmott, Paul and Orrell, David},
title = {No Laws, Only Toys},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10734},
doi = {10.1002/wilm.10734},
pages = {20--29},
year = {2019},
abstract = {In the September 2018 issue, we ran an excerpt from The Money Formula Dodgy Finance, Pseudo Science, and How Mathematicians Took Over the Markets, where Paul Wilmott and David Orrell considered the dual nature of quantitative finance, as exemplified by two men – John Law and Isaac Newton. In this issue Paul and David argue that Quantitative Finance could learn much from other disciplines (like Mathematical Biology) that revel in the fact that you don't need perfect models to be a science, and still produce useful insights into mechanisms and behaviors despite the scarcity of applicable, reliable physical laws in the models.},
}
• L. MacLean and B. Ziemba, “The efficiency of nfl betting markets,” Wilmott, vol. 2019, iss. 99, p. 30–35, 2019.
[Bibtex]
@article {WILM:WILM10735,
author = {MacLean, Leonard and Ziemba, Bill},
title = {The Efficiency of NFL Betting Markets},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10735},
doi = {10.1002/wilm.10735},
pages = {30--35},
year = {2019},
abstract = {Do the odds reflect all the relevant information?},
}
• C. Huber, “R tutorial on machine learning: how to visualize option-like hedge fund returns for risk analysis,” Wilmott, vol. 2019, iss. 99, p. 36–41, 2019.
[Bibtex]
@article {WILM:WILM10736,
author = {Huber, Claus},
title = {R Tutorial on Machine Learning: How to Visualize Option-Like Hedge Fund Returns for Risk Analysis},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10736},
doi = {10.1002/wilm.10736},
pages = {36--41},
keywords = {R tutorial, machine learning, self-organizing map, Kohonen map, nonlinear returns, hedge fund returns},
year = {2019},
abstract = {Nonlinearity in financial market returns is commonplace, and in particular in hedge fund returns. Hedge funds are known to generate option-like returns based on the products they trade, as well as their trading strategies. This tutorial describes how Kohonen's self-organizing map (SOM), a method of machine learning, can help to analyze nonlinearity in returns. We focus on simple examples that help the reader to understand where nonlinear hedge returns come from, why linear correlation analysis is inappropriate, and how SOMs can help to visualize nonlinear returns to enhance risk analysis. R code and step-by-step instructions enable the reader to reproduce the creation of the SOM. Readers are encouraged to change parameters and study the impacts on results.},
}
• C. M. Puetter, S. Renzitti, and A. Cowan, “Ccr kva relief through cva: a regression-based monte carlo approach,” Wilmott, vol. 2019, iss. 99, p. 42–61, 2019.
[Bibtex]
@article {WILM:WILM10737,
author = {Puetter, Christoph M. and Renzitti, Stefano and Cowan, Allan},
title = {CCR KVA Relief Through CVA: A Regression-Based Monte Carlo Approach},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10737},
doi = {10.1002/wilm.10737},
pages = {42--61},
keywords = {regression-based Monte Carlo, CCR capital, CCR KVA, SA-CCR EAD, incurred/forward CVA},
year = {2019},
abstract = {We present and examine, by example of a USD interest rate swap and a EUR/USD cross-currency basis swap, a regression-based Monte Carlo approach to counterparty credit default risk (CCR) capital and CCR capital valuation adjustment (KVA) calculations [assuming the standardized approach to counterparty credit risk for exposura-et-default (SA-CCR EAD) and the internal ratings-based (IRB) approach for CCR risk weights]. This approach allows to incorporate the capital lowering effect of credit valuation adjustment (CVA) in an efficient manner, without having to resort to lengthy nested Monte Carlo simulations. We find that the regression-based Monte Carlo approach works well in most situations. In other situations, the accuracy of the approach is sensitively controlled by the choice of explanatory variables. We discuss in detail the conditions and underlying dynamics under which this happens. In computing and presenting a selection of numerical examples, we also explore the impact of dynamic CCR risk weights on CCR KVA, and compare regression-based CCR KVA results with CCR KVA results from nested Monte Carlo, alternative frequently used CCR KVA simplifications, and standardized CVA KVA.}
}
• T. Sakuma, “Homotopy analysis method applied sabr and xva,” Wilmott, vol. 2019, iss. 99, p. 62–69, 2019.
[Bibtex]
@article {WILM:WILM10738,
author = {Sakuma, Takayuki},
title = {Homotopy Analysis Method Applied SABR and XVA},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10738},
doi = {10.1002/wilm.10738},
pages = {62--69},
keywords = {SABR, XVA, Stochastic volatility, PDE, HAM},
year = {2019},
abstract = {The SABR models are very popular in the financial industry. Their evaluation using asymptotic formulas for implied volatilities is fast and widely used but becomes less accurate and exhibits negative probability distribution values around zero strikes for long maturities. In addition, alternative version for more accurate valuation exists but involves numerical integration. In this paper, we apply the homotopy analysis method (HAM) to derive approximated option prices under a SABR model. This scheme is simple and our numerical examples demonstrate that the derived price can be evaluated easily and gives a good approximation with a computational cost that is only several times heavier than that of the Black-Scholes model. We also discuss the application of HAM to XVA evaluation.},
}
• M. Radley, “Cars,” Wilmott, vol. 2019, iss. 99, p. 70–71, 2019.
[Bibtex]
@article {WILM:WILM10739,
title = {Cars},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10739},
doi = {10.1002/wilm.10739},
pages = {70--71},
year = {2019},
abstract = {Arguably the world's finest supercar manufacturer, Lamborghini, has turned up the performance pipes on its top-of-the-line product, the formidable Aventador.},
}
• J. Darasz, “The skewed world of jan darasz,” Wilmott, vol. 2019, iss. 99, p. 72–72, 2019.
[Bibtex]
@article {WILM:WILM10740,
author = {Darasz, Jan},
title = {The skewed world of Jan Darasz},
journal = {Wilmott},
volume = {2019},
number = {99},
publisher = {John Wiley & Sons, Ltd},
issn = {1541-8286},
url = {http://dx.doi.org/10.1002/wilm.10740},
doi = {10.1002/wilm.10740},
pages = {72--72},
year = {2019},
}
More information about Wilmott magazine, for potential subscribers and submission of articles and research papers, can be found here.
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# Viewpoint: Closer to a Quantum Internet
, Electrical & Computer Engineering Department, Texas A&M University, College Station, TX 77843, USA
Published May 28, 2013 | Physics 6, 62 (2013) | DOI: 10.1103/Physics.6.62
#### Ultranarrow-Band Photon-Pair Source Compatible with Solid State Quantum Memories and Telecommunication Networks
Julia Fekete, Daniel Rieländer, Matteo Cristiani, and Hugues de Riedmatten
Published May 28, 2013 | PDF (free)
It has been nearly two decades since quantum computation first captured the imagination of the general public as a potential way to break the toughest cryptographic codes. Now, quantum information systems are becoming a reality, not for code breaking but as code generators whose security is governed by the laws of physics rather than computational complexity. This gives quantum communication systems an advantage over classical systems whose codes could, in principle, always be broken by some clever algorithm or a quantum computer. Yet there is still a key stumbling block to the construction of the quantum internet: transmitting quantum information over long distances. In a paper in Physical Review Letters, Julia Fekete at the Institute of Photonic Sciences, Spain, and colleagues report the first demonstration of a spectrally narrow-band source of photon pairs where one photon can address solid-state quantum memories at visible wavelengths, and the other is at a wavelength where telecommunications systems operate [1]. This is a critical element of a telecom-compatible quantum repeater and therefore brings us one step closer to the realization of a quantum internet.
As in a classical communication scheme, quantum information can be transmitted using infrared photons in optical fibers. However, photons decay exponentially as they propagate, limiting the maximum communication distances to about 100 kilometers. In classical networks, this problem is solved by using repeaters consisting of photon amplifiers. However, these are not practical in a quantum internet because amplification degrades quantum entanglement. Hence quantum repeaters are needed [2].
Such repeaters, like their classical counterparts, divide long communication lines into shorter ones by inserting repeater “nodes.” Entanglement is first established between adjacent nodes using, for example, entangled photon-pair sources. This entanglement “resource” is then purified or “amplified” by performing quantum operations in the nodes. The entanglement can then be extended to nonadjacent nodes, and so on, by entanglement swapping or teleportation, until long distances can be covered. Finally, the quantum information is teleported. In this way the quantum signal is transmitted without exponential degradation and without being amplified. The basic teleportation protocols were demonstrated previously using photons [3, 4]. However, achieving the entanglement amplification, and hence long distance quantum communication, requires that the entanglement resources be stored in quantum memories in each node long enough to test that the fidelity is high enough. Quantum networks of the future will also need to be flexible enough to connect quantum processors of widely different technologies (Fig. 1).
In recent years, researchers have made much progress in the construction of long-term quantum memories for photons. For example, systems based on ensembles of quantum units, like atomic vapors and spectral hole burning (SHB) crystals, have shown encouraging results. SHB crystals store information in the form of reversible “notches” that are created in their optical absorption spectra at specific frequencies, and they have shown storage times in excess of seconds [5] and efficiencies up to $69%$ [6]. The SHB approach has the advantage of being able to store many photon qubits in each crystal [7]. Recently, photons in SHB memories were entangled with telecom photons, proving that quantum internet nodes are possible with this technology [8, 9]. However, there is a problem: quantum memories, like SHB, can only work with photons at a very precise frequency or wavelength. On the other hand, telecom-compatible photo-pair generators create photons whose frequencies span a much broader spectrum. In other words, a large fraction of the spectrum provided by photon-pair generators cannot be processed by existing quantum memories. This limitation has now been overcome by Fekete et al.
The broad spectrum of photon-pair generators derives from the spontaneous parametric down-conversion (SPDC) process used to produce pairs of entangled photons whose wavelengths differ wildly (i.e., visible vs telecom band): In SPDC, a pump photon is split into two longer wavelength photons with frequency uncertainties that are typically $>100$ gigahertz. But typical quantum optical memories require the frequency to be specified with precision on the order of tens of megahertz. Of course, optical filtering is possible, but this greatly reduces the number of usable photon pairs that can be generated per second. To overcome this problem, the SPDC can be forced to emit photons with less frequency uncertainty by placing it in a resonant optical cavity so that photon pairs are only generated at frequencies corresponding to narrow cavity resonances [10, 11]. Yet this does not completely eliminate the need for spectral filtering, as the SPDC gain linewidth is so large that multiple optical cavity modes are excited.
To limit the number of active modes, two approaches have been explored. “Clustering” uses the fact that different wavelengths have different refractive indices so that most of the cavity modes are not triply resonant (i.e., for input and the two output photon wavelengths). Thus the SPDC emits in only a few adjacent or clustered modes [12]. A second approach is to use a very short cavity so that the modes are spaced by a large wavelength gap. Unfortunately, both of these approaches have only worked so far by using output photon wavelengths that were nearly the same. Making one of these methods work for widely different wavelengths is a significant technical challenge that was solved by Fekete et al.
In their demonstration, Fekete et al. used the clustering approach wherein one photon of each output pair was resonant with a $\text{Pr}$ doped ${\text{Y}}_{2}{\text{SiO}}_{5}$ (Pr:YSO) crystal operating at $606$ nanometers (nm), which can function as a high-fidelity quantum memory, and the other photon was well into the telecom band at $1436\phantom{\rule{0.333em}{0ex}}\text{nm}$, more than twice the wavelength. The $1436\phantom{\rule{0.333em}{0ex}}\text{nm}$ photon can then be transmitted down an optical fiber to the next node where the quantum entanglement can be stored in another memory using, for example, measurement-based entanglement [13], as illustrated in Fig. 1.
Since the $\text{Pr}:\text{YSO}$ wavelength is farther from the telecom band than most quantum memory technologies, it is clear that the approach can be easily extended to other promising memories such as $\text{Tm}$ doped ${\text{LiNbO}}_{3}$, $\text{Nd}$ doped ${\text{Y}}_{2}{\text{SiO}}_{5}$, ${\text{Ca}}^{+}$ ions, and atomic rubidium and cesium.
Consequently, this work represents an important milestone toward the construction of a long-distance quantum internet. The next step would be to combine this photon-pair generator with two remote optical memories, of the same or different technology, and demonstrate entanglement of these memories via telecom fibers. Then the protocol could be extended beyond the nearest-neighbor nodes, eventually including entanglement amplification to enable construction of a long-distance wide-scale quantum internet.
### References
1. J. Fekete, D. Rieländer, M. Cristiani, and H. de Riedmatten, “Ultranarrow-Band Photon-Pair Source Compatible with Solid State Quantum Memories and Telecommunication Networks,” Phys. Rev. Lett. 110, 220502 (2013).
2. H.-J. Briegel, W. Dür, J. I. Cirac, and P. Zoller, “Quantum Repeaters: The Role of Imperfect Local Operations in Quantum Communication,” Phys. Rev. Lett. 81, 5932 (1998).
3. D. Boschi, S. Branca, F. De Martini, L. Hardy, and S. Popescu, “Experimental Realization of Teleporting an Unknown Pure Quantum State via Dual Classical and Einstein-Podolsky-Rosen Channels,” Phys. Rev. Lett. 80, 1121 (1998).
4. D. Bouwmeester, J.-W. Pan, K. Mattle, M. Eibl, H. Weinfurter, and A. Zeilinger, “Experimental Quantum Teleportation,” Nature 390, 575 (1997).
5. J. J. Longdell, E. Fraval, M. J. Sellars, and N. B. Manson, “Stopped Light with Storage Times Greater than One Second Using Electromagnetically Induced Transparency in a Solid,” Phys. Rev. Lett. 95, 063601 (2005).
6. M. P. Hedges, J. J. Longdell, Y. Li, and M. J. Sellars, “Efficient Quantum Memory for Light,” Nature 465, 1052 (2010).
7. M. S. Shahriar, P. R. Hemmer, S. Lloyd, P. S. Bhatia, and A. E. Craig, “Solid-State Quantum Computing Using Spectral Holes,” Phys. Rev. A 66, 032301 (2002).
8. E. Saglamyurek et al., “Broadband Waveguide Quantum Memory for Entangled Photons,” Nature 469, 512 (2011).
9. C. Clausen, I. Usmani, F. Bussières, N. Sangouard, M. Afzelius, H. de Riedmatten, and N. Gisin, “Quantum Storage Of Photonic Entanglement in a Crystal,” Nature 469, 508 (2011).
10. J. S. Neergaard-Nielsen, B. Melholt Nielsen, H. Takahashi, A. I. Vistnes, and E. S. Polzik, “High Purity Bright Single Photon Source,” Opt. Express 15, 7940 (2007).
11. M. Scholz, L. Koch, R. Ullmann, and O. Benson, “Single-Mode Operation of a High-Brightness Narrow-Band Single-Photon Source,” Appl. Phys. Lett. 94, 201105 (2009).
12. C.-S. Chuu, G. Y. Yin, and S. E. Harris, “A Miniature Ultrabright Source of Temporally Long, Narrowband Biphotons,” Appl. Phys. Lett. 101, 051108 (2012).
13. L.-M. Duan, M. D. Lukin, J. I. Cirac, and P. Zoller, “Long-Distance Quantum Communication with Atomic Ensembles and Linear Optics,” Nature 414, 413 (2001).
### About the Author: Philip Hemmer
Philip Hemmer received his B.S. in physics from the University of Dayton, and his Ph.D. in physics from MIT. He worked many years as a physicist for the Air Force Research Laboratory at Hanscom Air Force Base, MA. Since 2002, he has been with the Electrical & Computer Engineering Department at Texas A&M University. His current research interests include quantum optics especially with nitrogen-vacancy diamond, subwavelength imaging, quantum computing in solids, plasmon-based nano-optics, slow and stopped light, and ultrasound imaging.
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For shear stress W at a section y 1 from the neutral axis the incremental
# For shear stress w at a section y 1 from the neutral
This preview shows page 12 - 15 out of 19 pages.
For shear stress W at a section y 1 from the neutral axis, the incremental horizontal force dN over the slice of thickness t is dN = W tdx or W ±µ¶ t ² dN / dx The total horizontal (normal) forces due to the bending moment are given as ࠵? = ∫ ࠵?࠵?࠵? and ࠵? + ࠵?࠵? = ∫ (࠵? + ࠵?࠵?)࠵?࠵? ࠵? 2 ࠵? 1 ࠵? 2 ࠵? 1 Subtracting the two terms, we have: ࠵?࠵? = ∫ ࠵?࠵?࠵?࠵? = ∫ ࠵?࠵? ࠵? ࠵?࠵?࠵? = ࠵?࠵? ࠵? ࠵?࠵?࠵? = ࠵?࠵? ࠵? ࠵? ࠵? 2 ࠵? 1 ࠵? 2 ࠵? 1 ࠵? 2 ࠵? 1 where ࠵?࠵? = ࠵?࠵? ࠵? ࠵? and ࠵? = ∫ ࠵?࠵?࠵? (= ∑ ࠵? ࠵? ̅ ࠵? ࠵? ) ࠵? 2 ࠵? 1 is termed the first moment of area. The shear stress is thus given as ࠵? = 1 ࠵? ࠵?࠵? ࠵?࠵? = 1 ࠵? ࠵?࠵? ࠵?࠵? ࠵? ࠵? = ࠵?࠵? ࠵?࠵? where the shear force (from mechanics of materials, is defined as the change of moment) ࠵? = ࠵?࠵? ࠵?࠵? . It should be noted that the shear stress W is defined over the slice with thickness t .
113 Example : A W 16 × 100 ( I x = 1490 in 4 ) is subjected to a vertical shear force of 80 kips. Determine the maximum shear stress and plot the distribution of shear stress for the cross section. Solution : The maximum shear stress will be at the neutral axis where Q , the statical moment about the neutral axis, is maximum. 4 ࠵? = 10.425(0.985) (7.50 + 0.985 2 ) + 7.50(0.585) ( 7.50 2 ) = 98.5 in 3 Shear stress at NA: ࠵? ࠵? = ࠵?࠵? ࠵?࠵? = 80(98.5) 1490(0.585) = 9.04 ksi The shear stresses over the cross section can be calculated as shown. _______________________ Note that the flanges resist very low shear stresses. It is the web that predominantly resists the shear in wide flange beams. AISC allows the use of average web shear for calculating shear stress: ࠵? ࠵? = ࠵? ࠵?࠵? ࠵? = 80 16.97(0.585) = 8.06 ksi where d is the full depth of beam and t w is the web thickness of the beam. This is less than the shear stress of 9.04 ksi calculated by the general shear shear stress formula! 4 Shear stress between flange and web: ࠵? = 10.425(0.985) (7.50 + 0.985 2 ) = 82.07 in 3 Bottom of flange: ࠵? ࠵? = 80(82.07) 1490(10.425) = 0.42 ksi Top of web: ࠵? ࠵? = 80(82.07) 1490(0.585) = 7.53 ksi b f = 10.425 in t f = 0.985 in 7.50 in t w = 0.585 in d = 16.97 in N.A. 0.42 ksi 7.53 ksi 9.04 ksi Cross section Shear distribution
114 Design Shear Strength : V u ≤ φ v V n , φ v =1.0 (for W-shapes, see AISC Pg. 16.1-67). V n = 0.6 F y A w C v (Eq. G2-1) where C v is termed web shear Coefficient. (Note: Shear yield strength F v = 0.6 F y .) For webs of rolled I-shape members with ℎ/࠵? ࠵? ≤ 2.24 ࠵?/࠵? ࠵? (which applies to all F y = 50 ksi W-shapes and most W-shapes tabulated in the manual, see pg. 16.1-68), C v =1. Generally, web shear strength may need to consider possible shear buckling limit state. For example, for webs of doubly symmetric shapes (except rolled W-shapes), etc. (see AISC Section G2.1(b)), Web yielding (plastic zone): ࠵? ࠵? ≤ 1.10 ࠵?࠵? ࠵? ࠵? ࠵? (≈ 418 ࠵? ࠵? ) C v =1 (Eq. G2-3) • Inelastic buckling of web : 1.10 ࠵?࠵? ࠵? ࠵? ࠵? < ࠵? ࠵? ≤ 1.37 ࠵?࠵? ࠵? ࠵? ࠵? (≈ 522 ࠵? ࠵? ) ࠵? ࠵? = 1.10√࠵?࠵?
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• Spring '14
• GregoryG.Deierlein
• Shear Stress, FY, Shear strength
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## El Merca’o Restaurante
### (Pamplona)
The Merca’o is a meeting point for all those who seek, in addition to the pleasure of good food, the taste for meeting, conversation, friendship. Its atmosphere, innovative and different from that of a conventional restaurant, fosters endless possibilities: the date for two, the business, the celebration, the farewell, the work group or friends… And also, at the bar, in the meeting around a coffee, the vermouth, the tapas, the aperitif, the cocktail…
Different dining rooms in different environments that form a unique space in Pamplona.
On the first floor a dynamic open restaurant capable of hosting different events.
On the ground floor, two private dining rooms with capacity for up to 50 people with set menus for groups.
The Bar remains open from 10:30 a.m. to closing
Awards: Bib Gourmand
Facilities: Disabled-friendly
Price: €20 - €40, 40€ - 60€
Route to which it belongs (Eat): Haute cuisine route
Type of cuisine: Pintxos / Tapas, Signature / Market cuisine
#### Opening Hours
Monday: 1:00 p.m. - 3:30 p.m. – 8:45 p.m. - 10:30 p.m. h
Tuesday: 1:00 p.m. - 3:30 p.m. – 8:45 p.m. - 10:30 p.m. h
Wednesday: 1:00 p.m. - 3:30 p.m. – 8:45 p.m. - 10:30 p.m. h
Thursday: 1:00 p.m. - 3:30 p.m. – 8:45 p.m. - 10:30 p.m. h
Friday: 1:00 p.m. - 3:30 p.m. – 9:00 p.m. - 10:45 p.m. h
Saturday: 1:00 p.m. - 3:30 p.m. – 9:00 p.m. - 10:45 p.m. h
Sunday: 1:00 p.m. - 3:30 p.m. – Closed h
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## Product categories in url
Question
I have following product categories in my app.
Women -> Outerwear
When I go to outwear category. Current URL is following
foo.com/outerwear
But I want to:
foo.com/women/outwear
how can I do it? how to setup it?
Thanks for any advice 🙂
0
2 years 2020-09-26T22:10:18-05:00 0 Answers 17 views 0
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# What is the ‘universal law of gravitation’? Derive a formula to it.
52 views
in Physics
closed
What is the ‘universal law of gravitation’? Derive a formula to it.
(OR)
Derive Fgrav = GM1M2/d2
+1 vote
by (40.0k points)
selected
1) The universal law of gravitation states that every body in the universe attracts other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
2) The direction of the force of attraction is along the line joining the centers of the two bodies.
3) Let two bodies of masses M and M be separated by a distance of ’d’. Then the force of gravitation between them.
4) Fgrav ∝ GM1M2/d2
5) Fgrav = GM1M2/d2
6) G is a proportionality constant called universal gravitational constant and found by Henry cavendish to be G = 6.67 × 10-11 Nm2 Kg-2
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# The Rationality of Irrationality in the Monty Hall Problem
@article{Ensslin2018TheRO,
title={The Rationality of Irrationality in the Monty Hall Problem},
author={Torsten A. Ensslin and Margret Westerkamp},
journal={Annalen der Physik},
year={2018},
volume={531}
}
• Published 10 April 2018
• Education
• Annalen der Physik
The rational solution of the Monty Hall problem unsettles many people. Most people, including the authors, think it feels wrong to switch the initial choice of one of the three doors, despite having fully accepted the mathematical proof for its superiority. Many people think the chances are 50‐50 between their options, but still strongly prefer to stay with their initial choice. Is there some sense behind these irrational feelings? Here, the possibility is entertained that intuition solves the…
2 Citations
Calculating the Probability Outcomes of the Monty Hall Problem Using A Web-Based Simulation
• James Arnold E. Nogra
• Computer Science
2019 IEEE 11th International Conference on Humanoid, Nanotechnology, Information Technology, Communication and Control, Environment, and Management ( HNICEM )
• 2019
A simulation of the Monty Hall Problem that runs in a web browser where doors are chosen randomly and the results from all the simulations concluded that the probability of having the car is behind the chosen door is $\frac{{n - 1}}{n}$ where is the number of doors.
The Physics of Information
• Computer Science
Annalen der Physik
• 2019
A number of researcher investigate the possibility that the real fundamental elements of this world are tiny bits of information, and several of their attempts to recover physical laws from the principles governing information are reported about in this special issue on physics of information.
## References
SHOWING 1-10 OF 13 REFERENCES
The Monty Hall Problem
A range of solutions to the Monty Hall problem is developed, with the aim of connecting popular (informal) solutions and the formal mathematical solutions of introductory text-books. Under Riemann’s
Judgment under Uncertainty
• Psychology
• 1982
The thirty-five chapters in this book describe various judgmental heuristics and the biases they produce, not only in laboratory experiments but in important social, medical, and political situations
Judgment under Uncertainty: Heuristics and Biases.
• Economics
Science
• 1974
Three heuristics that are employed in making judgements under uncertainty are described: representativeness, availability of instances or scenarios, which is often employed when people are asked to assess the frequency of a class or the plausibility of a particular development.
Judgment under Uncertainty: Heuristics and Biases
• Economics
Science
• 1974
This article described three heuristics that are employed in making judgements under uncertainty: (i) representativeness, which is usually employed when people are asked to judge the probability that
Stochastics: Introduction to Probability and Statistics
This second revised and extended edition presents the fundamental ideas and results of both, probability theory and statistics, and comprises the material of a one-year course. It is addressed to
Letters to the Editor
• Medicine
Calcified Tissue International
• 2007
It is suggested that the beneficial eVect of blood donation on cardiovascular disease is caused by a reduction in haematocrit and blood viscosity, which is consistent with the observation of Meyers et al, who noted that the benefit of blooddonation was greatest in males with the highest serum LDL.
Behind Monty Hall’s Doors: Puzzle, Debate and Answer?
• The New York Times
• 1991
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# scipy.ndimage.sum_labels¶
scipy.ndimage.sum_labels(input, labels=None, index=None)[source]
Calculate the sum of the values of the array.
Parameters
inputarray_like
Values of input inside the regions defined by labels are summed together.
labelsarray_like of ints, optional
Assign labels to the values of the array. Has to have the same shape as input.
indexarray_like, optional
A single label number or a sequence of label numbers of the objects to be measured.
Returns
sumndarray or scalar
An array of the sums of values of input inside the regions defined by labels with the same shape as index. If ‘index’ is None or scalar, a scalar is returned.
Examples
>>> from scipy import ndimage
>>> input = [0,1,2,3]
>>> labels = [1,1,2,2]
>>> ndimage.sum(input, labels, index=[1,2])
[1.0, 5.0]
>>> ndimage.sum(input, labels, index=1)
1
>>> ndimage.sum(input, labels)
6
#### Previous topic
scipy.ndimage.standard_deviation
#### Next topic
scipy.ndimage.variance
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# When is it appropriate to pool data?
Let's say I have some data on the amount of money spent on tv ads and total revenue from all sales. The data is available for three separate month.
Jan Feb March
$spent on tv ads 100 110 150 total sales revenue 1000 1000 1500 The problem I am trying to solve is whether the marketing campaign (tv ads) had a significant effect on total sales revenue. Now, I'm still not sure on how to proceed in terms of tests, either time series analysis, etc. (suggestions?) But one other issue is going to occur, and that regards whether or not to pool the data. Should I run separate tests for each month or do I pool the data together? So my question is: When is it appropriate to pool data for analysis? ## 3 Answers It's appropriate whenever the elements you're pooling together are homogeneous with respect to the parameters you're estimating. Specifically, this means that, if the model underlying each component is the same, with the same parameter values, then it is fine to pool the data. Otherwise, it's tantamount to leaving an important interaction out of a regression model which can cause misleading inferences. To place this into the context of your example - if you were to use regression to assess the relationship between $ spent on tv ads and total sales revenue then
If the relationship between \$ spent on tv ads and total sales revenue is the same across months, then it is OK to pool them together and ignore the month altogether. But, if the effect does depend on month, then you should potentially include interactions that allow the slope and intercept to vary by month. Failing to stratify your parameter estimates by month when they truly should be can cause misleading estimates of the variable effect(see here for a toy example).
So, the answer to your question boils down to whether you think the effects should vary by month. It seems to me that it probably should, but I'm certainly not as familiar with the application as you are.
• How about if the effect varies by month, but is constant for the cross-section? Can we still pool the dataset then? – Mayou Sep 6 '13 at 20:30
• @Mariam, I'm not sure exactly what you mean but it seems, in that case, it would be ok to pool across subsets of the population but not across time. – Macro Sep 7 '13 at 17:15
• following your reasoning, the OP should not have pooled days within months unless the relationship between dollars spent and revenue is the same for every day in the month .... and same for hours within days .... so this reasoning isn’t satisfactory for me. there is obviously no way to verify that the relationship holds for every individual dollar. – user28511 Mar 14 '18 at 19:14
• also, your toy example doesn’t imply to me that pooling is misleading. even in the toy example, overall there is no effect. pooling would accurately reflect this. the mistake would be one of interpretation: if you pool, you can’t interpret the data as showing a result that holds for every individual month. all you know is the relationship that holds when all the months are combined (which may be all you need/want). – user28511 Mar 14 '18 at 19:18
• i think the real answer is that you should not pool if you have statistical power to find relationships without pooling. when datasets are small, however (as is often the case in real life) pooling is necessary — but limits the precision of the conclusions that can be drawn. – user28511 Mar 14 '18 at 19:24
To add to what Macro said, You can test the assumption of the poolability of the monthly data based on time series analysis. But it would require a lot more monthly data. With the time series analysis you could look for a season trend of a yearly cycle. Even without time series analysis assume several years of data is available you could look for differences between the monthly means and if they appear to be significant then you would not pool the data. The assumption for the test would probably be that the monthly data for any month (say January) is independent from one year to the next. Given a long enough series you could check that assumption to by estimating the year to year correlation.
In terms of aggregation issues which are related to advertising and sales, you could start with Darryl Clarke and then follow the references forward.
Clarke, Darryl G. “Econometric Measurement of the Duration of Advertising Effect on Sales.” Journal of Marketing Research 13, 4 (1976): 345–57.
Clarke makes the point that the duration of the decay effect seems to interact with the frequency of the measurement in a somewhat uncomfortable way. Weekly data tends to produce decays in a short number of weeks. Monthly data in a short number of months. Yearly data in a short number of years. But I am oversummarizing a complicated field.
Or this overall summary by Tellis (who's a good source for what's happened in this field): http://www-bcf.usc.edu/~tellis/AdGeneral.pdf
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### Overview
Meta-analysis is frequently used to summarize results from multiple research studies. Since studies can be thought of as exchangeable, it is natural to analyze them using a hierarchical structure.
This page uses a Bayesian hierarchical model to conduct a meta-analysis of 9 randomized controlled trials (RCTs) of breast cancer screening. The analysis first replicates the frequentist results reported by Marmot et al. and then reexamines them in a Bayesian framework. The RCTs used in the meta-analysis are summarized in more detail by Gøtzsche et al..
R and Stan code for the analysis can be found here and here.
### Previous RCTs and Relative Risks
We begin by placing data from previous trials into a data frame using the summaries provided by Gøtzsche et al.. The treatment (group 1) is screening with mammography and the control (group 0) is no screening. The outcome in the treatment and control groups for study $j$, $d_{1j}$ and $d_{0j}$ respectively, is the number of breast cancer deaths during 13 years of follow up for women at least 50 years of age. There are $n_{1j}$ and $n_{0j}$ patients in the treatment and control groups respectively.
rct <- data.frame(study = c("New York", "Malamo I", "Kopparberg", "Ostergotland",
"Canada I", "Canada II", "Stockholm", "Goteborg", "UK age trial"))
rct$year <- c(1963, 1976, 1977, 1978, 1980, 1980, 1981, 1982, 1991) rct$d1 <- c(218, 87, 126, 135, 105, 107, 66, 88, 105)
rct$n1 <- c(31000, 20695, 38589, 38491, 25214, 19711, 40318, 21650, 53884) rct$d0 <- c(262, 108, 104, 173, 108, 105, 45, 162, 251)
rctn0 <- c(31000, 20783, 18582, 37403, 25216, 19694, 19943, 29961, 106956) The relevant statistic for the meta-analysis is the relative risk ratio, or $p_{1j}$/$p_{0j}$, where $p_{1j} = d_{1j}/n_{1j}$ and $p_{0j} = d_{0j}/n_{0j}$. We work with the log of the relative risk ratio, $y_j = log(p_{1j}) - log(p_{0j})$, because it is approximately normally distributed even in relatively small samples. We can calculate the variance of each term of $y_j$ by treating $p_{1j}$ and $p_{0j}$ as sample proportions and using the delta method, so that the variance of $y_j$ is, \begin{aligned} \sigma^2_j &\approx \frac{1 - p_{1j}}{n_{1j}p_{1j}} + \frac{1 - p_{0j}}{n_{0j}p_{0j}}. \\ \end{aligned} We can then calculate a point estimates for the relative risk in each study as well as a 95 percent confidence interval. rctp1 <- rct$d1/rct$n1
rct$p0 <- rct$d0/rct$n0 rct$rr <- rct$p1/rct$p0
rct$lrr <- log(rct$rr)
rct$lse <- sqrt((1 - rct$p1)/(rct$p1 * rct$n1) + (1 - rct$p0)/(rct$p0 * rct$n0)) rct$lower <- exp(rct$lrr - qnorm(.975) * rct$lse)
rct$upper <- exp(rct$lrr + qnorm(.975) * rct$lse) The results can be visualized nicely by creating a forest plot with the metafor package. library("metafor") p <- forest(x = rct$rr, ci.lb = rct$lower, ci.ub = rct$upper,
slab = paste(rct$study, rct$year, sep = ", "), refline = 1)
text(min(p$xlim), .88 * max(p$ylim), "Study and Year", pos = 4, font = 2)
text(max(p$xlim), .88 * max(p$ylim), "Relative Risk [95% CI]", pos = 2, font = 2)
### Hiearchical Model
The results from the separate RCTs can be modeled using a hierarchical model. We use the fact that the log of the relative risk is approximately normally distributed and assume that the random effects follow a normal distribution,
\begin{aligned} y_j &\sim N(\theta_j, \sigma^2_j) \\ \theta_j &\sim N(\mu, \tau), \end{aligned}
where $\sigma^2_j$ is assumed to be known with certainty (this assumption is not problematic because the binomial variances in each study are estimated precisely due to the large sample sizes). Meta-analyses are typically concerned with the overall mean, $\mu$.
There are, in general, three ways to estimate the random effects, $\theta_j$.
• No-pooling: there is a separate model for each study and $\theta_j=y_j$. This is a special case of the hierarchical model in which $\tau = \infty$.
• Complete-pooling: patients in each study are random samples from a common distribution so $\theta_j = \mu$. This is a special case of the hierarchical model in with $\tau = 0$.
• Partial-pooling: the hierarchical model is a compromise between the no-pooling and the complete-pooling estimates. In this case $\tau$ is unknown and $\theta_j$ is closer to $\mu$ when $\tau$ is small relative to $\sigma^2_j$, and closer to $y_j$ when the reverse is true.
### Estimation
#### No-pooling Estimates
A fixed-effect meta-analysis model completely pools the relative risk estimates across studies. The overall mean is commonly estimated by taking an inverse-variance weighted average of studies. This can be done in using the rma function in the metafor package.
me.fe <- rma(rct$lrr, rct$lse^2, method = "FE")
c(exp(me.fe$b), exp(me.fe$ci.lb), exp(me.fe$ci.ub)) ## [1] 0.8041868 0.7406291 0.8731987 The relative risk and 95 percent confidence intervals are identical to the fixed-effect meta-analysis results reported in Gøtzsche et al.. We can check to see that the point estimate is identical to taking a weighted average of the relative risks in the RCTs. exp(weighted.mean(rct$lrr, 1/(rct$lse^2))) ## [1] 0.8041868 #### Maximum Likelihood Estimation A hierarchical model applied to meta-analysis is typically referred to as a random-effect meta-analysis model in the medical literature. The parameters of the hierarchical model can be estimated in either a frequentist or a Bayesian framework. In a frequentist setup, point estimates (rather than probability distributions) for the parameters are estimated. This can be done in a number of ways, but here we will estimate the parameters with restricted maximum likelihood (REML) using the rma function. me.re <- rma(rct$lrr, rct$lse^2) c(exp(me.re$b), exp(me.re$ci.lb), exp(me.re$ci.ub))
## [1] 0.8029497 0.7259418 0.8881267
This analysis reproduces the results reported by Marmot et al..
#### Bayesian Estimation
One problem with the maximum likelihood approach is that it does not account for uncertainty in $\tau$ and produces confidence intervals for $\mu$ that are too narrow. A Bayesian model that produces complete probability distributions for each parameter can be estimated using the probabilistic programming language Stan. We begin by preparing the data.
library("rstan")
set.seed(101)
J <- nrow(rct)
stan.dat <- list(J = J, y = rct$lrr, sigma = rct$lse)
Next we specify the model. Note that we can rewrite the upper-level model as $\theta_j = \mu + \tau \eta$ where $\eta \sim N(0, 1)$, which speeds up the Stan code. Furthermore, $\mu$ and $\tau$ are given uniform priors.
We fit the model and extract samples from the joint posterior distribution.
fit <- stan(file = "_rmd-posts/bayesian_meta_analysis.stan",
data = stan.dat, iter = 2000, chains = 4)
post <- extract(fit, permuted = TRUE)
As expected, the 95 percent credible interval for the exponential of the overall mean is slightly wider that the 95 percent confidence interval produced using REML.
quantile(exp(post$mu), probs = c(.025, .5, .975)) ## 2.5% 50% 97.5% ## 0.6889931 0.8011223 0.9110904 This addditional uncertainty comes from averaging over $\tau$, which has a rather wide probability distribution. quantile(post$tau, probs = c(.025, .5, .975))
## 2.5% 50% 97.5%
## 0.006847247 0.102239111 0.337733161
The effects, $\theta_j$, are shrunk toward the overall mean, $\mu$. The following plot examines the degree of shrinkage by comparing the effects from the Bayesian model to the relative risks when each study is analyzed separately.
library("ggplot2")
theme_set(theme_bw())
p.dat <- apply(exp(post$theta), 2, quantile, probs = c(.025, .5, .975)) p.dat <- data.frame(lower = p.dat[1, ], rr = p.dat[2, ], upper = p.dat[3, ]) p.dat <- rbind(p.dat, rct[, c("lower", "upper", "rr")]) p.dat$lab <- rep(c("Theta", "Y"), each = J)
p.dat$id <- rep(seq(9, 1), 2) p.dat$idlab <- factor(p.dat$id, labels = rev(paste(rct$study, rct$year, sep = ", "))) p.dat$yint <- mean(exp(post$mu)) ggplot(p.dat, aes(x = idlab, y = rr, ymin = lower, ymax = upper, col = lab)) + geom_pointrange(aes(col = lab), position = position_dodge(width = 0.50)) + coord_flip() + geom_hline(aes(yintercept = yint), lty = 2) + xlab("") + ylab("") + theme(legend.position="bottom") + scale_colour_discrete(name="", labels = c("Theta" = bquote("Random effect:"~exp(theta[J])~" "), "Y"= bquote("Relative risk:"~exp(Y[J])))) There is considerable shrinkage and the degree of shrinkage is larger for studies where the relative risks are estimated less precisely. The 95 percent credible intervals using the Bayesian approach are also narrower than the 95 percent confidence intervals from the individual studies because the hierarchical model pools information across RCTs Although most meta-anlayes focus on the overall mean, there are other quantities of interest that may be more meaningful. For example, it might me more useful to predict the effect of mammography screening in a new population by making a prediction about a new study effect, say $\tilde{\theta}_j$, rather than from $\mu$. Predictions in a new population can be made very easily in a Bayesian framework because the study effects are assumed to be exchangable; that is, we can simulate the posterior distribution of $\tilde{\theta_j}$ by drawing $\tilde{\theta}_j \sim N(\mu, \tau)$ using the values of $\mu$ and $\tau$ drawn from the posterior simulation. n.sims <- nrow(post$mu)
theta.new <- rep(NA, n.sims)
for (i in 1:n.sims){
theta.new[i] <- rnorm(1, post$mu[i], post$tau[i])
}
Although the posterior medians are similar, the 95 percent credible interval for $exp(\tilde{\theta}_j)$ is much wider than for $exp(\mu)$.
quantile(exp(theta.new), probs = c(.025, .5, .975))
## 2.5% 50% 97.5%
## 0.5711096 0.8017504 1.1106414
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# Nov 28 2020
I do not think it is a good idea to take well known syntax like \[ for the same reason as the other suggestion (searching for \begin{align} etc.) to influence the rendering outside of the math tags. It will never have the same functionality like in LaTeX (switching between text and math mode), therefore it would make the current behavior even more confusing to the editors.
# Feb 19 2020
Regarding the breaking changes, we have:
(1) version histories
(2) current pages where math would not render
(3) current pages where math would render differently
# Feb 18 2020
Spacing in LaTeX is fine, spacing in Mathjax also. We actually do not need to fix spacing, we just have to ensure that the source code that we pass to Mathjax is still correct. The spacing is just one effect those brackets can have, in principle they can change everything, for example "\sin\limits_a^b" is correct, "{\sin}\limits_a^b" does not compile.
The list of functions where the spacing gets wrong with additional braces is everything that is defined like a unary or binary operator because those get a spacing of \medmuskip. If you place them in braces they become a subformula with spacing rules of an ordinary math expression. By default you have block layout in LaTeX where the size of all spaces between letters and words depends on how much needs to fit into a particular line.
# Feb 17 2020
I guess most people know, but just to clarify what I mean with "{\sqrt[{b}]{c}}" not being a real error: In LaTeX it depends on the definition of \sqrt and where it is placed. With the normal definition it removes extra spaces around \sqrt which in this case is not a problem because there would be no extra spacing anyway. In other situations, e.g. for relations like "a\xrightarrow\alpha b" or "a\stackrel\alpha= b" you still get wrong spacing in Wikipedia due to those extra braces.
For the more complex example "\sqrt{[} ..." is a real error. In the simple example "{\sqrt[{b}]{c}}" is only hard to read, not a real error. Still, if the example is more complex, keeping spaces and newlines is even more important. For comparison: In C++ you can also remove newlines, scramble white-spaces, add unnecessary scopes and rename variables such that the output is still correct, but not human-readable anymore.
# Feb 15 2020
I lost you. Are you saying it should fail or it should not?
yes, it should fail and if it renders it should render different.
Debenben added a comment to T245343: texvcjs should not replace $\omicron$ with $\mathrm{o}$ .
@Physikerwelt \Omicron being upright is fine, see https://tex.stackexchange.com/questions/119248.
# Feb 10 2020
@Physikerwelt Thank you very much for creating the pull request, it is definitely a very good fist step in the right direction. I guess nobody really cares about the name of the category, we could simply use "Pages that use a deprecated format of the math tags" in analogy to the chem tracking category. I am really happy with the pull request and surprised to see how much work this was but at the same time also a bit disappointed because I was hoping for more.
# Feb 8 2020
Replacing \exist with \exists sounds reasonable, it should be no problem. My database dump search found 1302 pages in all projects with \exist. Without the texvc changes there might be other things creating problems which we are not aware of and could be fixed with a bot, so I would like to wait for the error categories such that we do not need to edit pages twice.
# Dec 8 2019
Debenben added a comment to T237516: Update to MathJax 3.
@Physikerwelt Thank you for creating this ticket, I think this change is long overdue. What are your thoughts regarding the implementation, in particular:
• does the new rendering mode still need this restbase setup? I am just asking because getting rid of this would greatly reduce the complexity and dependencies, would make it much easier for people to contribute to the development of the math extension and install it on their own servers.
• as far as I know the legacy rendering mode for mhchem package is no longer supported in mathjax 3.0. In my opinion this should not block the transition to the new rendering mode because this change is also long overdue, we just have to keep in mind that it will break some mhchem formulas. I can take care of this and make sure that they will get fixed.
• the new rendering mode should conserve the original LaTeX input, i.e. not have something like texvcjs (T188879). The necessary syntax modfications are done (T197925), except for those that we could not extract from the database dumps. If we want to fix those pages we need the new rendering mode assigning error categories in real time.
• we should take the new rendering mode as an opportunity to discuss getting rid of "default" rendering mode and making the "default" without parameters "inline" (textstyle).
# Dec 1 2019
Debenben committed rTTEX3d009752c6ab: modify submission of crawler scripts (authored by Debenben).
modify submission of crawler scripts
# Jul 20 2019
@Physikerwelt what do you mean with "exist as alias for exist"?
# Jun 26 2019
Debenben added a comment to T195765: Make it possible to query for math values.
It is easy to perform some sanitization or conversion on the original string if you need it.
# May 4 2019
Debenben committed rTTEX595b29fa9b88: prepare en botrun (authored by Debenben).
prepare en botrun
@GregorAlexandru We are actually planning to get rid of the texvcjs so that in the future we can just refer to the MathJax documentation.
# May 3 2019
Debenben committed rTTEX43f782238c5e: add hywiki editsummary and new inputlists (authored by Debenben).
add hywiki editsummary and new inputlists
Debenben committed rTTEXbe778a6e8313: experiments with different search scripts (authored by Debenben).
experiments with different search scripts
# Jan 25 2019
Debenben committed rTTEX7837f9a02318: Merge branch 'templateexpand' (authored by Debenben).
Merge branch 'templateexpand'
# Jan 6 2019
Debenben committed rTTEXc2d91561d0ee: improved crawler (authored by Debenben).
improved crawler
# Jan 2 2019
Debenben committed rTTEX6be0395332e1: +special replacement with {{C}} template (authored by Debenben).
+special replacement with {{C}} template
Debenben committed rTTEX3a88d6cdfe5c: Merge branch 'master' into templateexpand (authored by Debenben).
Merge branch 'master' into templateexpand
# Dec 30 2018
Debenben committed rTTEX5776657e88ff: Merge branch 'master' into templateexpand (authored by Debenben).
Merge branch 'master' into templateexpand
# Dec 28 2018
Debenben committed rTTEX117ab689a123: Merge branch 'master' into templateexpand (authored by Debenben).
Merge branch 'master' into templateexpand
# Dec 27 2018
Debenben added a comment to T197925: Create a bot to replace deprecated math syntax .
I put some thoughts into the problem of replacing math generated by templates like in https://de.wikiversity.org/wiki/Kommutative_Ringtheorie/Algebra-Homomorphismus_%C3%BCber_Ring/Definition where $K$ is produced with a source code
Debenben committed rTTEX8cca2b956d11: Merge branch 'master' into templateexpand (authored by Debenben).
Merge branch 'master' into templateexpand
# Dec 24 2018
Debenben committed rTTEX6128b30b4911: Merge branch 'master' into templateexpand (authored by Debenben).
Merge branch 'master' into templateexpand
Debenben committed rTTEXeb91b755217d: add comparison of tempate text (authored by Debenben).
# Dec 23 2018
Debenben committed rTTEXcf2b2c64402a: disable saving for testing (authored by Debenben).
disable saving for testing
Debenben committed rTTEXa122ef0eb131: test templatesearch (authored by Debenben).
test templatesearch
Debenben committed rTTEXe212332b0570: Merge branch 'master' into templateexpand (authored by Debenben).
Merge branch 'master' into templateexpand
# Dec 21 2018
@SalixAlba We have to change our algorithms to look for math tags. dewikiversity is using {{#tag: math}} via https://de.wikiversity.org/wiki/Vorlage:Math on more than 18 000 pages.
# Dec 11 2018
Addendum: Another reason for a block formula is of course if the formula requires too much vertical space to fit into a normally spaced line
@Izno I do not know what the best way of implementing it in the extension is, but from an author point of view both are treated equally as part of a sentence and there is no difference between inline and block formula except for the size of the equation.
# Dec 7 2018
Or with T209148 in mind, why not disable lazy loading for math "images" completely? After all they are not images, but part of the text.
Debenben added a project to T211432: chem tags should use the same rendering options as math: Math.
Debenben changed the status of T139855: Fix content with broken path integrals from Resolved to Declined.
Debenben added a comment to T182127: Additional math symbols for oriented integrals.
If a symbol people need is missing or looks ugly, they will try to rewrite equations to make it work without the symbol or use some workarounds, e.g. in this case probably something like \int_C and then specify what C means. Those line integrals are common in complex analysis (residue theorem) and hence for example also in electrodynamics, I would estimate the number of articles where they could be used to more than 100 on enwiki.
Debenben updated the task description for T182127: Additional math symbols for oriented integrals.
Debenben closed T139855: Fix content with broken path integrals as Resolved.
I think people do not want to use \oiint or \oiiint because the integral symbol looks bad, they will prefer workarounds if they look better in their setup. It is probably not the biggest problem, but I would only consider the integrals in T182127 done, if they actually produce a rendering that is not ugly. I will close this task because I think it does not help.
# Nov 16 2018
Debenben added a comment to T197925: Create a bot to replace deprecated math syntax .
@SalixAlba I pushed a commit that should fix it
Debenben committed rTTEXddf365ecd328: fix issue with nowiki tags (authored by Debenben).
fix issue with nowiki tags
Debenben committed rTTEXfff384e374f4: test nowiki issue (authored by Debenben).
test nowiki issue
# Nov 14 2018
Debenben added a comment to T197925: Create a bot to replace deprecated math syntax .
@SalixAlba thanks for catching that. This is something we overlooked:
Debenben committed rTTEX2484341191db: +rowiki editmessage (authored by Debenben).
+rowiki editmessage
Debenben committed rTTEXdc836ffec735: +hewiki logs (authored by Debenben).
+hewiki logs
# Nov 12 2018
Debenben committed rTTEXfdfcd8550dd6: +he translation (authored by eranroz).
+he translation
# Nov 11 2018
Debenben added a member for Math: Debenben.
Debenben updated the task description for T166369: Migrate old Math options to current ones or delete them from the database.
We really need an option that covers all use-cases (especially because you have to login to get to the options) but currently we don't and these options are very helpful for debugging e.g. T194768 so they should stay for now.
I think this was a problem of MathML and is fixed in newer firefox versions.
Debenben closed T132367: Latex renders with excessive height on firefox as Resolved.
I think this was a problem of MathML and is fixed in newer firefox versions.
Debenben added a comment to T29574: PDF export: Use LaTeX formulas instead of inline images.
@ovasileva I just created a pdf of https://de.wikipedia.org/wiki/Satz_des_Pythagoras and half of it is not shown at all and the other half still looks horrible:
Debenben closed T2798: Many character sets don't work in texvc as Resolved.
The caracters probably still look bad, but there should not be any errors anymore since png is also using MathJax now
Debenben closed T2798: Many character sets don't work in texvc, a subtask of T4458: Localized TeX environment, as Resolved.
the rendering of Malayalam is still not very good, but there is no error anymore, since the png is created from the svg images now
Debenben moved T188879: Remove texvc from Incoming to Next-up on the Math board.
Debenben closed T136931: texvc vs Mathoid parsing differences as Resolved.
should be resolved since png is also using MathJax now
Debenben moved T197925: Create a bot to replace deprecated math syntax from Incoming to Doing on the Math board.
# Nov 10 2018
Debenben committed rTTEXf59e49d0caa3: remove old inputlists (authored by Debenben).
remove old inputlists
# Nov 9 2018
Debenben committed rTTEX4ea60664e6dc: +zhwiki editmessage and logs (authored by Debenben).
+zhwiki editmessage and logs
@Debenben what browsers those screenshots done with? Such a bad rendering shouldn't happen with the current solution.
# Nov 7 2018
@Physikerwelt For example in the mobile view:
# Nov 5 2018
The custom "Wikipedia syntax" suggested here would be a terrible idea. We are having enough trouble with non-standard LaTeX syntax already. At times, especially with mhchem or optional arguments (which use square brackets) it seems random, how an equation renders and if not what kind of error message you get. Even as an experienced mediawiki-user you can spend hours trying to find out where those strange error messages come from and how to fix them.
# Nov 4 2018
Debenben committed rTTEX5ba4bc2539f4: +arwiki editmessage and logs (authored by Debenben).
+arwiki editmessage and logs
Debenben committed rTTEXd0b19e5029ae: +arwiki editmessage and logs (authored by Debenben).
+arwiki editmessage and logs
# Nov 2 2018
What is currently being done is delivering the same static svg images to everyone. What I propose is to do the final rendering client-side like in normal MathJax, so that e.g.
@Pkra About inlining SVG: Maybe you are right and in principle SVG could become a better solution than HTML. I don't like the idea of taking the current system and just removing the image tags because this is not enough to solve those problems. The most probable case is that we don't have enough manpower to actually do the things that "might be doable" and "require a bit of work" and the result is that we are stuck with an over-customized, substandard, unmaintained system forever.
# Nov 1 2018
Debenben committed rTTEX09a4c07c74e3: +inputlists, correct link (authored by Debenben).
Debenben added a comment to T197925: Create a bot to replace deprecated math syntax .
@SalixAlba Thank you for taking care of the botflag on enwiki, feel free to take responsibility and request a botflag on any other project you like.
Debenben updated the task description for T208475: chem equations cut off.
# Oct 30 2018
Debenben committed rTTEXe92f5e402b04: move logs to own directory (authored by Debenben).
move logs to own directory
Debenben committed rTTEXb185a02aea94: plwiki logs (authored by Debenben).
plwiki logs
@Theklan: We are currently discussing some changes to the math extension here: T195861, you are welcome to discuss this issue. Currently the plan is to
1. get a correct rendering like in normal MathJax / LaTeX for all equations, especially mhchem
2. render non-ASCII-characters like Cyrillic letters, äöü etc. properly in all browsers such that \text can be used with all languages that need special characters
As you can see this is a tremendous amount of work, progress is quite slow and everything relies on volunteers. I don't think changing the syntax for existing formulas with commas is feasible and maintaining the current system without additional localization features already overstretches our resources. I think it would be a better idea to just keep using the well-known {,} for decimal comma and make sure that help pages mention this LaTeX hack.
# Oct 23 2018
Debenben committed rTTEX660b1615631b: +plwiki test (authored by Debenben).
+plwiki test
# Oct 7 2018
Debenben committed rTTEX5811c6c744bf: ruwiki editmessage (authored by Debenben).
ruwiki editmessage
Debenben committed rTTEX26ab4b052084: added more initial input lists (authored by Debenben).
# Oct 3 2018
Debenben committed rTTEXfc2300adf1f0: +eswiki editsummary and logs (authored by Debenben).
+eswiki editsummary and logs
# Aug 19 2018
Debenben updated subscribers of T197925: Create a bot to replace deprecated math syntax .
@SalixAlba Thank you for finding the problem with the unmatched math tags and fixing it, also @Framawiki thanks for your pull request.
Debenben committed rTTEX88594539040a: frwiki logfiles (authored by Debenben).
frwiki logfiles
Debenben committed rTTEX0aeb66a5d508: set changed flag (authored by Debenben).
set changed flag
transferred changes from mathwikibot2 into mathwikibot script
Debenben committed rTTEXce82ab45ba8d: Merge branch 'D1087' (authored by Debenben).
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# Aug 7 2018
I think one has to differentiate: Of course there are large formulas which cannot be broken and need some scrolling mechanism. Most formulas however have a structure like $A=B+C+D$ which you can break almost everywhere if necessary. Wikipedia authors currently have a choice between
• Adding permanent linebreaks or splitting the expression up in smaller chunks which means making it less readable by wasting space on large screens and/or introducing unnecessary artificial names for parts of the sum
• Not adding any linebreaks, making the expression unreadable on small screens or introducing unnecessary scrolling or zooming.
First sentence
:$A$$\;=B$$\;+C${{nowrap|$\;+D$.}}
Next sentence.
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{}
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# Capacitor voltage multiplier with caps paralleled
#### RogueRose
Joined Oct 10, 2014
375
Let's say I have qty 40 400v 1000uf caps and I want to run at 2kv and I want to have the highest capacitance possible. To get to 2kv would I need 5 or 10 caps (or how many stages - how many caps total to get 2kv)? To get the higher capacitance, would I run equal # of caps at each stage to get a higher total capacitance?
Also, is it possible to run 5 voltage multipliers in parallel (5 identical setups) to get 5x capacitance? I know it is similar to running the caps in parallel in each stage, but can be utilized differently.
#### Alec_t
Joined Sep 17, 2013
11,540
The number of stages would depend on what each stage consists of and what the input voltage to the first stage is. If you post your schematic we may be able to help you better.
#### RogueRose
Joined Oct 10, 2014
375
I looked closer at voltage multipliers and it looks like each stage adds the value of the cap.? Looking at wiki it looked like it might have been 2 caps for each stage.
So if that is the case, to get my 2kv with the above caps, could I run 5 stages and put 8 caps in parallel in each stage to get a total of 2kv @ 8000uF?
BTW, I tried to edit my OP but there is no edit button. Is that normal?
Hello,
For a voltage multiplier you will need an AC voltage to get it working:
Why do you need 2 kV at high energy?
The energy stored in the stack of capacitors can be lethal.
Bertus
I just picked numbers that were easy to visualize - I could say qty 40 caps, 10v 100uF with a desired voltage of 100v @ 400uF.
The number of stages would depend on what each stage consists of and what the input voltage to the first stage is. If you post your schematic we may be able to help you better.
THat is good to know. As I said, this is more of a theoretical exercise to figure out how to get to X from Z..
Let's say that input voltage is 360vac (1:3 transformer from 120VAC).
But what would happen if 480vac were applied to the multiplier?
I guess one thing I am trying to figure out is if there is a difference (and how much) in energy/joules if the 40 caps from the OP were run in parallel
40 caps in parallel
400v 40,000uF = .5 * 40,000 * 400^2 = 3,200 joules
Voltage multiplier - 5 stage - 8 caps parallel in each stage
2kv 8,000uF = .5 * 8,000 * 2,000^2 = 16,000 joules
I know these calculations are correct but I'm not sure if the electrical aspect is correct, meaning will the voltage multiplier act as I described?
Now I look at it the joules is 5x for the 2kv stage. Do I need to divide by the number of stages when figuring out energy for this or is the 16kj correct?
Last edited by a moderator:
#### bertus
Joined Apr 5, 2008
20,668
Hello,
The peak voltage of AC is √2 times the RMS voltage for a sine.
So if the AC voltage is 480 Vac, the peak voltage will be 538 Volts.
The 400 Volts caps in your OP post will not survive.
Bertus
#### Alec_t
Joined Sep 17, 2013
11,540
But what would happen if 480vac were applied to the multiplier?
There would most likely be a loud bang, capacitor guts spewed everywhere and a lot of acrid smoke .
If you are planning on discharging the caps at high current they would need to be rated accordingly.
#### RogueRose
Joined Oct 10, 2014
375
When the caps are put into a setup like a voltage multiplier, does this effect the speed of discharge for them. Is this what the ESR value indicates? Is the ESR value of the multiplier the sum total of the ESR values for each stage?
I guess what I am trying to figure out is how exactly these multipliers can work.
The following is a hypothetical as I can correlate this to other examples..
If I need a cap of something around 240v with as much capacitance as possible but I am restricted to using 25v or 50v capacitors. I'm going to be using pulsed DC current of 49-50v. The energy requirement is to release as much power as quickly as possibly with these caps at roughly 240-250vdc.
Similar to the above voltage multiplier, there are qty 100 - 50v 3300uf caps available. If a 5 stage multiplier is used to get 250v and running 20 parallel caps on each stage (49-50vdc pulsed input) does this circuit act as a single higher voltage/larger capacity capacitor (250v 66,000uf)? Meaning does this circuit store energy for quick discharge or does it simply act like a transformer allowing for higher voltage while not being able to discharge in a single pulse.
I'm sorry if I haven't been able to make this very clear but I am a little confused as to the potential of this circuit.
Last edited by a moderator:
#### recklessrog
Joined May 23, 2013
985
Unless I've missed it somewhere, you don't seem to indicate whether you intend to use a half or full wave multiplier.
#### RogueRose
Joined Oct 10, 2014
375
Unless I've missed it somewhere, you don't seem to indicate whether you intend to use a half or full wave multiplier.
good question. I'm really not sure as I'm not sure how it effects the circuit.
#### Alec_t
Joined Sep 17, 2013
11,540
I suggest you download LTspice (free, from Linear Tecnology), then you can simulate all sorts of scenarios/configurations and see what happens.
#### recklessrog
Joined May 23, 2013
985
good question. I'm really not sure as I'm not sure how it effects the circuit.
Well, think about the advantages of full wave rectification over half wave, then apply that reasoning to your multiplier
#### RogueRose
Joined Oct 10, 2014
375
Well, think about the advantages of full wave rectification over half wave, then apply that reasoning to your multiplier
Thanks. I looked up the differences and it I saw something about peak ripple and some other things I have to look more closely at. After reading it I was leaning towards full wave rectification.
So, I still am not certain as to whether this voltage multiplier will act like a higher voltage capacitor or if I will be limited to the amount of energy the power supply can deliver in one "pulse".
So, 100 caps of 50v 3300uf -- if all paralleled - 50v 330,000 uF = 413 joule potential
5 stage, 20 caps paralleled per stage -- 250v 66,000 uF = 2016 joules (if it were a cap rated at those values - IDK if I divide by 5 as it is 5 tiers and 5x the energy of the 100 caps paralleled). So if the power supply charges the caps to capacity would I get the 2016 joules from it if discharged at once or does it not work that way when arrainged as a voltage multiplier?
#### recklessrog
Joined May 23, 2013
985
I think I know what you want to use it for and cannot in all good conscience advise you further except to say that the energy levels and voltages proposed take no prisoners and will happily KILL YOU in an instant with NO SECOND CHANCE!
I worked on the design and development very high voltage @ high current equipment and the degree of training and safety precautions were of the highest order, hence my survival to become an O.A.P!!
I don't warn you lightly, and hope you really consider the potential dangers that can cause your death or serious injury in view of level of your expertise.
This is not in any way meant as "put down" or insult, Just genuine concern for the direction in which you may be heading.
Take care
#### Alec_t
Joined Sep 17, 2013
11,540
Even one of those caps fully charged holds more than enough charge to kill you.
#### RogueRose
Joined Oct 10, 2014
375
I think I know what you want to use it for and cannot in all good conscience advise you further except to say that the energy levels and voltages proposed take no prisoners and will happily KILL YOU in an instant with NO SECOND CHANCE!
I worked on the design and development very high voltage @ high current equipment and the degree of training and safety precautions were of the highest order, hence my survival to become an O.A.P!!
I don't warn you lightly, and hope you really consider the potential dangers that can cause your death or serious injury in view of level of your expertise.
This is not in any way meant as "put down" or insult, Just genuine concern for the direction in which you may be heading.
Take care
Well I guess it's the good old guess & test method then!
Ok, would it be better if I changed the cap voltage to 6.3v and 10uf. If I had 100 caps, 5 stages, 20 parallel per stage.... That would be 31.5v 200uf. IDK if that can kill. I just picked some numbers that seemed easy to work with. This is a question of theory, I'm not saying those caps will be used. I just wanted to know if the multiplier works like a cap or more like a wired transformer
#### RogueRose
Joined Oct 10, 2014
375
I think I know what you want to use it for and cannot in all good conscience advise you further except to say that the energy levels and voltages proposed take no prisoners and will happily KILL YOU in an instant with NO SECOND CHANCE!
I worked on the design and development very high voltage @ high current equipment and the degree of training and safety precautions were of the highest order, hence my survival to become an O.A.P!!
I don't warn you lightly, and hope you really consider the potential dangers that can cause your death or serious injury in view of level of your expertise.
This is not in any way meant as "put down" or insult, Just genuine concern for the direction in which you may be heading.
Take care
I just re-read your post. IDK what you have in mind, but I understand the hesitation if the numbers I posted are correct. I don't plan on building that. Thank you for making it clear that it is potentially lethal.
On another note, is a 50v 3300uf cap deadly at 4 joules?
#### RogueRose
Joined Oct 10, 2014
375
Recklessrog - will you answer me as to whether the question I was asking made sense, as in did you understand it? I really just want to know the "theory" behind this. The reason I'm researching this is because all the posts/pages/sites I read about various mid to high voltage projects usually say somewhere on the page "a high voltage cap would cost a fortune" while they have like $4-5,000 in parts + time + manufacturing costs invested in the project. So I started running some numbers, comparing what they said to the #'s, and yeah, it is expensive but it seems that their approach to building the cap banks is similar to installing 8 single cylinder 2-stroke engines in a Charger (dodge/plymouth/whatever) instead of a single V-8... This question is the culmination of fact checking some research on sites, comparing claims of companies selling various "high tech" "future tech" things - most of which require a good deal of energy - thus the research into this. I'm trying to see is the$30,000 asking price for some "toys" is because cost of parts, or due to intellectual property (knowing how to build XYZ)..
#### BR-549
Joined Sep 22, 2013
4,938
I don't have a clue what you are building. Don't really care.
But very concerned that you do not have the knowledge, nor skill to build such a device.
That much charge under that kind of pressure is jumpy and lethal.
Usually an innocent person gets hurt.
I recommend much more study.
This is not a put down or insult.
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# Q What is the remainder when the positive integer x is
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Q What is the remainder when the positive integer x is [#permalink] 10 May 2008, 02:57
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Q
What is the remainder when the positive integer x is divided by 7
1) x-1 is divisible by 7
2) x + 1 is divisible by 7
Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient
Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient
Ø Both statements together are sufficient, but neither statement alone is sufficient
Ø Each statement alone is sufficient
Ø Statements 1 and 2 together are not sufficient
Q
What is the remainder when x ^ 4 – y ^ 4 is divided by 3
(1) When x-y is divided by 3, remainder is 0
(2) When x+y is divided by 3, remainder is 2
Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient
Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient
Ø Both statements together are sufficient, but neither statement alone is sufficient
Ø Each statement alone is sufficient
Ø Statements 1 and 2 together are not sufficient
If x is a positive integer is x divisible by 7
1) X+29 is divisible by 10
2) x + 10 is divisible by 29
Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient
Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient
Ø Both statements together are sufficient, but neither statement alone is sufficient
Ø Each statement alone is sufficient
Ø Statements 1 and 2 together are not sufficient
Can you please explain what the concept behind the above mentioned questions is? I am not looking at a solution which includes picking up a number and substituting it in the equation. The technique of inputting a number is valid for only very few types of question. Would really like to understand the trick and the concept behind these types of questions?
Disclaimer: -
I do not know the source of this question, so please do not ask it from me. A friend of mine who just wrote his GMAT and got a cool 750 gave me the pdf files containing some challenging questions. I have the answers to these questions but not the official explanations.
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Re: Application of Remainder theory [#permalink] 10 May 2008, 04:33
vdhawan1 wrote:
Q
What is the remainder when the positive integer x is divided by 7
1) x-1 is divisible by 7
2) x + 1 is divisible by 7
Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient
Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient
Ø Both statements together are sufficient, but neither statement alone is sufficient
Ø Each statement alone is sufficient
Ø Statements 1 and 2 together are not sufficient
we can position the question like $$x = 7k + R$$ and we need to find R
from 1) we can deduce that $$x-1 = 7n$$ .
given this we can solve for R by setting k=n. i.e 7k + R = 7k + 1. Therefore R = 1 ---> Suff.
similarly for 2) we can decude R = -1, which actually means that R=6
vdhawan1 wrote:
Q
What is the remainder when x ^ 4 – y ^ 4 is divided by 3
(1) When x-y is divided by 3, remainder is 0
(2) When x+y is divided by 3, remainder is 2
Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient
Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient
Ø Both statements together are sufficient, but neither statement alone is sufficient
Ø Each statement alone is sufficient
Ø Statements 1 and 2 together are not sufficient
lets fist simplyfy $${x}^{4} - {y}^{4}$$ into simpler forms that will looks similar to the data options.
$${x}^{4} - {y}^{4}$$ = $$({x}^{2} - {y}^{2})({x}^{2} + {y}^{2})$$
which in turn equals $$(x+y)(x-y)({x}^{2} + {y}^{2})$$
so we need to find out if this term is divisible by 3.
1) says that (x-y) is divisible by 3. Therefore $${x}^{4} - {y}^{4}$$ is divisible by 3 --> suff
2) says that $$\frac{(x+y)}{3}$$ yields a reminder of 2. -->insuff
Ans = A
vdhawan1 wrote:
If x is a positive integer is x divisible by 7
1) X+29 is divisible by 10
2) x + 10 is divisible by 29
Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient
Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient
Ø Both statements together are sufficient, but neither statement alone is sufficient
Ø Each statement alone is sufficient
Ø Statements 1 and 2 together are not sufficient
Can you please explain what the concept behind the above mentioned questions is? I am not looking at a solution which includes picking up a number and substituting it in the equation. The technique of inputting a number is valid for only very few types of question. Would really like to understand the trick and the concept behind these types of questions?
this is a toughie +1 for posting it, I am keen to know what others come up with
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Re: Application of Remainder theory [#permalink] 10 May 2008, 08:31
vdhawan1 wrote:
Q
If x is a positive integer is x divisible by 7
1) X+29 is divisible by 10
2) x + 10 is divisible by 29
Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient
Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient
Ø Both statements together are sufficient, but neither statement alone is sufficient
Ø Each statement alone is sufficient
Ø Statements 1 and 2 together are not sufficient
Can you please explain what the concept behind the above mentioned questions is? I am not looking at a solution which includes picking up a number and substituting it in the equation. The technique of inputting a number is valid for only very few types of question. Would really like to understand the trick and the concept behind these types of questions?
Disclaimer: -
I do not know the source of this question, so please do not ask it from me. A friend of mine who just wrote his GMAT and got a cool 750 gave me the pdf files containing some challenging questions. I have the answers to these questions but not the official explanations.
Hi,
I think the answer to this question is E. I used a combination of picking numbers and divisibility rules. This may or may not be helpful. I am sure there must be a shorter way of approaching this problem.Unfortunately this was all I could think of
Stat 1 x+29 is divisible by 10
The units digit of x+29 must end in 0, therefore the units digit of x must be 1
x could be 1,11,which are not divisible by 7 and 21 which is. Insuff
Stat 2 x+10 is divisible by 29,
The units digits of numbers divisible by 29 range from 9,8,7,6,5,4,3,2
Therefore the units digit of x could be any of these numbers and if x+10 = 87 which is divisible by 29 and x =7 which is divisible by 7 or x+10=58 which is divisible by 29 and x=48 which is not divisible by 7. Insufficient
Together stat 1 and 2
x+29 is divisible by 10 and x+10 is divisible by 29
Therefore , x could be 251,541,831, 1121 which is not divisible by 7 or x could be 1711 which is divisible by 7 so insuff.
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Re: Application of Remainder theory [#permalink] 10 May 2008, 10:12
vdhawan1 wrote:
Q
What is the remainder when the positive integer x is divided by 7
1) x-1 is divisible by 7
2) x + 1 is divisible by 7
Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient
Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient
Ø Both statements together are sufficient, but neither statement alone is sufficient
Ø Each statement alone is sufficient
Ø Statements 1 and 2 together are not sufficient
This might sound dumb, but without doing any math, couldn't I just say that the answer is D?
You can find the remainder when (x+1) or (x-1) is divided by 7, even if X is 6 or 8 and the remainder is zero, correct?
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Re: Application of Remainder theory [#permalink] 20 Jul 2008, 06:52
ventivish wrote:
vdhawan1 wrote:
Q
If x is a positive integer is x divisible by 7
1) X+29 is divisible by 10
2) x + 10 is divisible by 29
Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient
Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient
Ø Both statements together are sufficient, but neither statement alone is sufficient
Ø Each statement alone is sufficient
Ø Statements 1 and 2 together are not sufficient
Can you please explain what the concept behind the above mentioned questions is? I am not looking at a solution which includes picking up a number and substituting it in the equation. The technique of inputting a number is valid for only very few types of question. Would really like to understand the trick and the concept behind these types of questions?
Disclaimer: -
I do not know the source of this question, so please do not ask it from me. A friend of mine who just wrote his GMAT and got a cool 750 gave me the pdf files containing some challenging questions. I have the answers to these questions but not the official explanations.
Hi,
I think the answer to this question is E. I used a combination of picking numbers and divisibility rules. This may or may not be helpful. I am sure there must be a shorter way of approaching this problem.Unfortunately this was all I could think of
Stat 1 x+29 is divisible by 10
The units digit of x+29 must end in 0, therefore the units digit of x must be 1
x could be 1,11,which are not divisible by 7 and 21 which is. Insuff
Stat 2 x+10 is divisible by 29,
The units digits of numbers divisible by 29 range from 9,8,7,6,5,4,3,2
Therefore the units digit of x could be any of these numbers and if x+10 = 87 which is divisible by 29 and x =7 which is divisible by 7 or x+10=58 which is divisible by 29 and x=48 which is not divisible by 7. Insufficient
Together stat 1 and 2
x+29 is divisible by 10 and x+10 is divisible by 29
Therefore , x could be 251,541,831, 1121 which is not divisible by 7 or x could be 1711 which is divisible by 7 so insuff.
Whoa. How did you find this particular number out from just x+29 and x+10?? I would be interested in knowing your method.
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Re: Application of Remainder theory [#permalink] 21 Jul 2008, 10:17
If x is a positive integer is x divisible by 7
1) X+29 is divisible by 10
2) x + 10 is divisible by 29
if x=1, then x+29=30/10 1/7 is not even divisible by 7...
if x=21, sum=50 which is divisible by 10, however 21/7 is even divisible..so Insuff
2) x+10 is divisible by 29..which means if x=19 then x/7 is not even divisible..
lets look at 29*2=58..ok so pick x=48 48/7 is not even divisible by 7..
lets look at 29*3=87, so make x=77/7 is divisible by 7..
lets look at 1&2 X is sume number that has a unit digit 1..
aha 29*9 gives unit digit 1..so 29*9=201 201-10=191 OK so x=191
191+29=220 divisible by 10..
191+10=201 divisible by 29..
so lets check if 191 is divisible by 7..191/7 is not even divisible
therefore I am going to say C is my ans..
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Re: Application of Remainder theory [#permalink] 21 Jul 2008, 10:29
fresinha12 wrote:
If x is a positive integer is x divisible by 7
1) X+29 is divisible by 10
2) x + 10 is divisible by 29
if x=1, then x+29=30/10 1/7 is not even divisible by 7...
if x=21, sum=50 which is divisible by 10, however 21/7 is even divisible..so Insuff
2) x+10 is divisible by 29..which means if x=19 then x/7 is not even divisible..
lets look at 29*2=58..ok so pick x=48 48/7 is not even divisible by 7..
lets look at 29*3=87, so make x=77/7 is divisible by 7..
lets look at 1&2 X is sume number that has a unit digit 1..
aha 29*9 gives unit digit 1..so 29*9=201 201-10=191 OK so x=191
191+29=220 divisible by 10..
191+10=201 divisible by 29..
so lets check if 191 is divisible by 7..191/7 is not even divisible
therefore I am going to say C is my ans..
I don't think your test for C generalizes:
29*59 =1711
x = 1711 - 10 = 1701
1701/7 = 243
To re-write your methodology into a generalization:
x = 29*(10a+9)-10 = 290a + 251 where a is any non-negative integer
290a+251 is divisible by 7 for all a = 5b+7 where b is any non-negative integer
Therefore, E.
Last edited by zonk on 21 Jul 2008, 10:35, edited 1 time in total.
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Re: Application of Remainder theory [#permalink] 21 Jul 2008, 10:35
yeah u r right..
on exam day i would have picked C..i mean how can you quickly check for all multiples of 29 that have 1 as unit digit..
zoinnk wrote:
fresinha12 wrote:
If x is a positive integer is x divisible by 7
1) X+29 is divisible by 10
2) x + 10 is divisible by 29
if x=1, then x+29=30/10 1/7 is not even divisible by 7...
if x=21, sum=50 which is divisible by 10, however 21/7 is even divisible..so Insuff
2) x+10 is divisible by 29..which means if x=19 then x/7 is not even divisible..
lets look at 29*2=58..ok so pick x=48 48/7 is not even divisible by 7..
lets look at 29*3=87, so make x=77/7 is divisible by 7..
lets look at 1&2 X is sume number that has a unit digit 1..
aha 29*9 gives unit digit 1..so 29*9=201 201-10=191 OK so x=191
191+29=220 divisible by 10..
191+10=201 divisible by 29..
so lets check if 191 is divisible by 7..191/7 is not even divisible
therefore I am going to say C is my ans..
I don't think your test for C generalizes:
29*59 =1711
x = 1711 - 10 = 1701
1701/7 = 243
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Re: Application of Remainder theory [#permalink] 21 Jul 2008, 10:36
fresinha12 wrote:
yeah u r right..
on exam day i would have picked C..i mean how can you quickly check for all multiples of 29 that have 1 as unit digit..
zoinnk wrote:
fresinha12 wrote:
If x is a positive integer is x divisible by 7
1) X+29 is divisible by 10
2) x + 10 is divisible by 29
if x=1, then x+29=30/10 1/7 is not even divisible by 7...
if x=21, sum=50 which is divisible by 10, however 21/7 is even divisible..so Insuff
2) x+10 is divisible by 29..which means if x=19 then x/7 is not even divisible..
lets look at 29*2=58..ok so pick x=48 48/7 is not even divisible by 7..
lets look at 29*3=87, so make x=77/7 is divisible by 7..
lets look at 1&2 X is sume number that has a unit digit 1..
aha 29*9 gives unit digit 1..so 29*9=201 201-10=191 OK so x=191
191+29=220 divisible by 10..
191+10=201 divisible by 29..
so lets check if 191 is divisible by 7..191/7 is not even divisible
therefore I am going to say C is my ans..
I don't think your test for C generalizes:
29*59 =1711
x = 1711 - 10 = 1701
1701/7 = 243
I would bet this is a manhattan gmat CAT problem and not a gmatprep problem...
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Re: Application of Remainder theory [#permalink] 21 Jul 2008, 10:43
251,541,831, 1121 is not a single number, this is a list of 4 separate numbers.
sirlogic wrote:
ventivish wrote:
vdhawan1 wrote:
Q
If x is a positive integer is x divisible by 7
1) X+29 is divisible by 10
2) x + 10 is divisible by 29
Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient
Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient
Ø Both statements together are sufficient, but neither statement alone is sufficient
Ø Each statement alone is sufficient
Ø Statements 1 and 2 together are not sufficient
Can you please explain what the concept behind the above mentioned questions is? I am not looking at a solution which includes picking up a number and substituting it in the equation. The technique of inputting a number is valid for only very few types of question. Would really like to understand the trick and the concept behind these types of questions?
Disclaimer: -
I do not know the source of this question, so please do not ask it from me. A friend of mine who just wrote his GMAT and got a cool 750 gave me the pdf files containing some challenging questions. I have the answers to these questions but not the official explanations.
Hi,
I think the answer to this question is E. I used a combination of picking numbers and divisibility rules. This may or may not be helpful. I am sure there must be a shorter way of approaching this problem.Unfortunately this was all I could think of
Stat 1 x+29 is divisible by 10
The units digit of x+29 must end in 0, therefore the units digit of x must be 1
x could be 1,11,which are not divisible by 7 and 21 which is. Insuff
Stat 2 x+10 is divisible by 29,
The units digits of numbers divisible by 29 range from 9,8,7,6,5,4,3,2
Therefore the units digit of x could be any of these numbers and if x+10 = 87 which is divisible by 29 and x =7 which is divisible by 7 or x+10=58 which is divisible by 29 and x=48 which is not divisible by 7. Insufficient
Together stat 1 and 2
x+29 is divisible by 10 and x+10 is divisible by 29
Therefore , x could be 251,541,831, 1121 which is not divisible by 7 or x could be 1711 which is divisible by 7 so insuff.
Whoa. How did you find this particular number out from just x+29 and x+10?? I would be interested in knowing your method.
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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Manager Joined: 11 Apr 2008 Posts: 129 Location: Chicago Followers: 1 Kudos [?]: 39 [0], given: 0 Re: Application of Remainder theory [#permalink] 21 Jul 2008, 13:10 vdhawan1 wrote: Q What is the remainder when the positive integer x is divided by 7 1) x-1 is divisible by 7 2) x + 1 is divisible by 7 Ø Statement 1 alone is sufficient but statement 2 alone is not sufficient Ø Statement 2 alone is sufficient but statement 1 alone is not sufficient Ø Both statements together are sufficient, but neither statement alone is sufficient Ø Each statement alone is sufficient Ø Statements 1 and 2 together are not sufficient Isn't the first question impossible? When you take the 2 statements together, it states a number is divisible by 7 and when you add 2 to it, it still divisible by 7. For example, lets say x is 6, if you add 1 (x + 1) you will get 7 (which is divisible by 7), but if you subtract 1 (x-1), you get 5, and that is not divisible by 7. In other words, there is no value for x that satisfies both statement (1) and (2)---where if you take a number and whether you add 1 or subtract 1 from it, it will still be divisible by 7. Thus, this question is not a possible GMAT question. Another way of looking at it, is that for statement (1) you get a remainder of 1 and for statement 2, you get a remainder of 6 --> this is not possible, a single number can not have a remainder of one value, and then have a different remainder when divided by the same number. Please feel free to comment as I am 90% sure on this. _________________ Factorials were someone's attempt to make math look exciting!!! SVP Joined: 30 Apr 2008 Posts: 1889 Location: Oklahoma City Schools: Hard Knocks Followers: 36 Kudos [?]: 480 [0], given: 32 Re: Application of Remainder theory [#permalink] 21 Jul 2008, 13:34 brokerbevo, You're taking both statements together when you don't need to. The question is what is the remainder when x is divided by 7. And then it gives you the statements. Its easy to see that each statement, separately, allows you to easily find the remainder when x is divided by 7. You don't have to go to each statement, but even if you did, the answer would then be E because the statements together don't allow you to answer the question. I think this goes back to our conversation of the statements cannot contradict each other. Here is an example of when the statements can contradict each other and the question is perfectly acceptable. As for it being a GMAT question? I have no idea. I don't know the source of this particular question. Anyone else know the source of this question? _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
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Re: Application of Remainder theory [#permalink] 21 Jul 2008, 13:46
jallenmorris wrote:
brokerbevo,
You're taking both statements together when you don't need to. The question is what is the remainder when x is divided by 7. And then it gives you the statements. Its easy to see that each statement, separately, allows you to easily find the remainder when x is divided by 7. You don't have to go to each statement, but even if you did, the answer would then be E because the statements together don't allow you to answer the question.
I think this goes back to our conversation of the statements cannot contradict each other. Here is an example of when the statements can contradict each other and the question is perfectly acceptable. As for it being a GMAT question? I have no idea. I don't know the source of this particular question.
Anyone else know the source of this question?
No, I know. I do realize that when you view each statement in isolation you do in fact get a single remainder for each: (1) remainder = 1 and (2) remainder = 6. However, there is no number in which if you add 1 or subtract 1 from it, and it will be divisible by 7. It is impossible so thus the GMAT will not give you a question like this (the statements are not true).
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Re: Application of Remainder theory [#permalink] 21 Jul 2008, 14:03
But the point of Data Sufficiency is not to find out if 2 statements are sufficient, we have to take each statement individually first. Then, and ONLY THEN, do we consider the 2 statements together. I think if neither statment is sufficient by itself, then the statements are sure to always be consistent. It seems like you jump to consider the statements together.
With Data Sufficiency, our ability to analyze the amount of information available as well as the relevance of that information. First we look to the stem and determine what information is missing. Then we look to statement #1, figure out if it has the missing information, or other information in order for us to deduce the missing information. Then we look to #2 and do the same. ONLY if neither statement is sufficient do we consider the two statements together. I think it is entirely possible that 2 statements are inconsistent, but one of them is sufficient to answer the question so we never need to consider them together.
You know I think even after we consider the statements together they can be inconsistent, because then we would just answer E.
brokerbevo wrote:
jallenmorris wrote:
brokerbevo,
You're taking both statements together when you don't need to. The question is what is the remainder when x is divided by 7. And then it gives you the statements. Its easy to see that each statement, separately, allows you to easily find the remainder when x is divided by 7. You don't have to go to each statement, but even if you did, the answer would then be E because the statements together don't allow you to answer the question.
I think this goes back to our conversation of the statements cannot contradict each other. Here is an example of when the statements can contradict each other and the question is perfectly acceptable. As for it being a GMAT question? I have no idea. I don't know the source of this particular question.
Anyone else know the source of this question?
No, I know. I do realize that when you view each statement in isolation you do in fact get a single remainder for each: (1) remainder = 1 and (2) remainder = 6. However, there is no number in which if you add 1 or subtract 1 from it, and it will be divisible by 7. It is impossible so thus the GMAT will not give you a question like this (the statements are not true).
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**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$. GMAT Club Premium Membership - big benefits and savings Manager Joined: 11 Apr 2008 Posts: 129 Location: Chicago Followers: 1 Kudos [?]: 39 [0], given: 0 Re: Application of Remainder theory [#permalink] 21 Jul 2008, 14:47 jallenmorris wrote: But the point of Data Sufficiency is not to find out if 2 statements are sufficient, we have to take each statement individually first. Then, and ONLY THEN, do we consider the 2 statements together. I think if neither statment is sufficient by itself, then the statements are sure to always be consistent. It seems like you jump to consider the statements together. With Data Sufficiency, our ability to analyze the amount of information available as well as the relevance of that information. First we look to the stem and determine what information is missing. Then we look to statement #1, figure out if it has the missing information, or other information in order for us to deduce the missing information. Then we look to #2 and do the same. ONLY if neither statement is sufficient do we consider the two statements together. I think it is entirely possible that 2 statements are inconsistent, but one of them is sufficient to answer the question so we never need to consider them together. You know I think even after we consider the statements together they can be inconsistent, because then we would just answer E. brokerbevo wrote: jallenmorris wrote: brokerbevo, You're taking both statements together when you don't need to. The question is what is the remainder when x is divided by 7. And then it gives you the statements. Its easy to see that each statement, separately, allows you to easily find the remainder when x is divided by 7. You don't have to go to each statement, but even if you did, the answer would then be E because the statements together don't allow you to answer the question. I think this goes back to our conversation of the statements cannot contradict each other. Here is an example of when the statements can contradict each other and the question is perfectly acceptable. As for it being a GMAT question? I have no idea. I don't know the source of this particular question. Anyone else know the source of this question? No, I know. I do realize that when you view each statement in isolation you do in fact get a single remainder for each: (1) remainder = 1 and (2) remainder = 6. However, there is no number in which if you add 1 or subtract 1 from it, and it will be divisible by 7. It is impossible so thus the GMAT will not give you a question like this (the statements are not true). Ahh, I see what you are saying. No, what I was saying is that I did take each question individually first: I found a remainder of 1 for statement (1) and a remainder of 6 for statement (2). However, I'm stating that the entire question in invalid because no single number, when you add 1 to OR subtract 1 from it, will be divisible by 7-- its impossible from a GMAT standpoint, the GMAT only deals with real numbers. For example, lets say you have the following numbers: 6: +1 = 7 AND -1 = 5 Both answers are not divisible by 7 8: +1 = 9 AND -1 = 7 Both answers are not divisible by 7 22: +1 = 23 AND -1 = 21 Both answers not divisible by 7 ....no matter what number you try, when you subtract 1 and add 1, one or the other will not be divisible by 7. The statements do not agree with each other (in other words, they are contradictory), therefore the question will not be seen on the GMAT. A good way to tell if it is contradictory is to see if you can find a single value that satisfies both statements (of course, this is after you have done all your work and came up with an answer), if you can't, either the question is not a GMAT question, and you wouldn't see it on the test anyway, or you have made a math error. Can you think of a single value x, where x + 1 AND x - 1 are both divisible by 7? _________________ Factorials were someone's attempt to make math look exciting!!! SVP Joined: 30 Apr 2008 Posts: 1889 Location: Oklahoma City Schools: Hard Knocks Followers: 36 Kudos [?]: 480 [0], given: 32 Re: Application of Remainder theory [#permalink] 21 Jul 2008, 15:02 I agree with the method you identify to determine if the statements are contradictory, but where in the GMAT rules or instructions, or anything published by GMAC does it say that you will always find a number that will satisfy both statements? Also, the GMAT deals with more than real numbers. Try $$\sqrt{3}$$. brokerbevo wrote: However, I'm stating that the entire question in invalid because no single number, when you add 1 to OR subtract 1 from it, will be divisible by 7-- its impossible from a GMAT standpoint, the GMAT only deals with real numbers. . . . A good way to tell if it is contradictory is to see if you can find a single value that satisfies both statements (of course, this is after you have done all your work and came up with an answer), if you can't, either the question is not a GMAT question, and you wouldn't see it on the test anyway, or you have made a math error. Can you think of a single value x, where x + 1 AND x - 1 are both divisible by 7? _________________ ------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
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Re: Application of Remainder theory [#permalink] 21 Jul 2008, 16:14
jallenmorris wrote:
I agree with the method you identify to determine if the statements are contradictory, but where in the GMAT rules or instructions, or anything published by GMAC does it say that you will always find a number that will satisfy both statements?
Also, the GMAT deals with more than real numbers. Try $$\sqrt{3}$$.
brokerbevo wrote:
However, I'm stating that the entire question in invalid because no single number, when you add 1 to OR subtract 1 from it, will be divisible by 7-- its impossible from a GMAT standpoint, the GMAT only deals with real numbers.
. . .
A good way to tell if it is contradictory is to see if you can find a single value that satisfies both statements (of course, this is after you have done all your work and came up with an answer), if you can't, either the question is not a GMAT question, and you wouldn't see it on the test anyway, or you have made a math error. Can you think of a single value x, where x + 1 AND x - 1 are both divisible by 7?
Well, the definition of contradictory, in this context, is just that: if you can't find a single value (or more values) common among the answer choices, they are contradictory. That is the definition of contradictory when it comes to numbers.
If statement (1) says x < 0 and statement (2) says 5 < x, these statements are contradictory--meaning no single value (or more) can satisfy both statements. You will never see this on the GMAT because they contradict.
Now, if statement (1) said x^2 = 16 and statement (2) said x < 0 and you came to the conclusion that (1) x = 4 and (2) x is negative, you know you would have done some math wrong because how can a negative number equal 4? As silly as it sounds, it obviously doesn't make sense. In other words, a single value does not exist between the statements. However, statement (1) should have been calculated to be: x = +/- 4 and (2) would still state that x is negative. Now we have a value for x for both statements: -4.
Its not a matter of whether the statements are alone sufficient, only sufficient together, or not sufficient at all, its a matter of whether its a gmat question or not. In fact, its a moot point because you wouldn't see this on the GMAT anyway; however, it is a good double-check to make sure your math is correct.
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Re: Application of Remainder theory [#permalink] 21 Jul 2008, 16:30
Expert's post
brokerbevo is entirely correct. The statements are contradictory; you'd never see this question on a real GMAT. Clearly D is a good answer here, but logically E is just as good- there's no way to work out the remainder with both statements, since the situation is impossible. The GMAT can't have contradictory statements, because this will make two answers equally valid, and a GMAT question can only have one correct answer.
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Re: Application of Remainder theory [#permalink] 22 Jul 2008, 06:38
IanStewart wrote:
brokerbevo is entirely correct. The statements are contradictory; you'd never see this question on a real GMAT. Clearly D is a good answer here, but logically E is just as good- there's no way to work out the remainder with both statements, since the situation is impossible. The GMAT can't have contradictory statements, because this will make two answers equally valid, and a GMAT question can only have one correct answer.
Well, I think jallenmorris and I are both correct but we are just arguing about 2 different things-- just a miscommunication. He agrees that the statements can't contradict, but he was just stating the process of how to go about validating the statements: first you see if statement 1 is sufficient, then 2, then both, then none of them. I was just stating as a double-check, when all is said and done, that the outcome of your work from each statement must agree with each other. So, 2 completely different arguments. DS statements are confusing so its hard to communicate with others some of your techniques and whatnot.
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Re: Application of Remainder theory [#permalink] 22 Jul 2008, 06:38
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# Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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16 Apr 2014, 06:23
a. |x+3| = 4x -3
|x+3| = x + 3 for x + 3 >= 0 ---> x > -3.
Solving for x,
x + 3 = 4x -3 ---> 6 = 3x ---> x = 2 ( solution accepted since x > -3)
| x + 3| = -(x +3 ) for x + 3 < 0 --> x < -3.
Solving for x,
-x - 3 = 4x -3
5x = 0 ---> x = 0 ( solution discarded as x is not < -3)
A is sufficient.
b. |x -3| = |2x -3|
|x -3| = x -3 for x-3 >= 0 ---> x >= 3
|x -3| = -(x -3) for x-3 < 0 ---> x < 3
|2x -3 | = 2x -3 for 2x -3 >= 0 ----> x >=3/2
|2x -3 | = -(2x -3) for 2x -3 < 0 ----> x < 3/2
So we have 3 ranges
1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard
2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution )
3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)
Let me know if I have solved the question correctly. I know the process is lengthy.
One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D).
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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12 May 2015, 10:56
vikram4689 wrote:
dvinoth86 wrote:
Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|
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Remember: When you have || on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have || on only one side you MUST VERIFY the answer by putting them back in eqn
(1) |x + 3| = 4x – 3
a) x + 3 = 4x – 3 => x=2 .. VALID
b) -(x + 3) = 4x – 3 => x=0 .. INVALID
Hence Sufficient
(2) |x – 3| = |2x – 3| => x = 0 or 6 .. INVALID
Hence In-Sufficient
Your statement 1 has two sceneries. Then how does it become sufficient?
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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12 May 2015, 22:57
2
vikram4689 wrote:
dvinoth86 wrote:
Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|
Liked the question? encourage by giving kudos
Remember: When you have || on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have || on only one side you MUST VERIFY the answer by putting them back in eqn
(1) |x + 3| = 4x – 3
a) x + 3 = 4x – 3 => x=2 .. VALID
b) -(x + 3) = 4x – 3 => x=0 .. INVALID
Hence Sufficient
(2) |x – 3| = |2x – 3| => x = 0 or 6 .. INVALID
Hence In-Sufficient
Your statement 1 has two sceneries. Then how does it become sufficient?
The solution quoted by you above has some tacit steps. I'll list them here:
St. 1 says: |x + 3| = 4x – 3 . . . (1)
Case 1: x + 3 > = 0
That is, x > = -3
In this case, |x+3| = (x+3)
So, Equation 1 becomes:
x + 3 = 4x - 3
=> 6 = 3x
=> x = 2
Does this value of x satisfy the condition of Case 1, that x > = -3?
Yes, it does. So, x = 2 is a valid value of x.
Let's now consider
Case 2: x + 3 < 0
That is, x < -3
In this case, |x+3| = -(x+3)
So, Equation 1 becomes:
-(x + 3) = 4x - 3
=> -x - 3 = 4x - 3
=> 0 = 5x
=> x = 0
Does this value of x satisfy the condition of Case 2, that x < -3?
No, it doesn't. So, x = 0 is an INVALID value of x.
Thus, we got only 1 valid value of X from Statement 1 (x = 2) and so, Statement 1 is sufficient to say that x > 0.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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16 May 2015, 21:37
Bunuel wrote:
Is x > 0?
(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient.
Hope it helps.
I got 0,2 in both statements. Reading your responses below, I would need to substitute to get A as the answer. However, if I input( x>=3/4 ) such as x=1 in statement 1, I get 1+3=4(RHS)4(1)-3=1(LHS), any reason?
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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18 May 2015, 03:49
hak15 wrote:
Bunuel wrote:
Is x > 0?
(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question.
(2) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient.
Hope it helps.
I got 0,2 in both statements. Reading your responses below, I would need to substitute to get A as the answer. However, if I input( x>=3/4 ) such as x=1 in statement 1, I get 1+3=4(RHS)4(1)-3=1(LHS), any reason?
First of all, x=0 is not a solution of |x+3|=4x−3. It has only one solution x=2.
Next, the question asks whether x>0 and from (1) we get that $$x\geq{\frac{3}{4}}$$, so x must be greater than 0. So, the statement answers the question. $$x\geq{\frac{3}{4}}$$ does NOT mean that any value greater than or equal to 3/4 will satisfy the equation, it means that the value(s) that satisfy it must be greater than or equal to 3/4.
Hope it's clear.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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29 Dec 2015, 22:12
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution
Is x > 0?
(1) |x + 3| < 4
(2) |x - 3| < 4
In case of inequality questions, it is important to note that conditions are sufficient if the range of the question includes the range of the conditions.
There is 1 variable (x) in the original condition. In order to match the number of variables and the number of equations, we need 1 equation. Since the condition 1) and 2) each has 1 equation, there is high chance that D is the answer.
In case of the condition 1), we can obtain -4<x+3<4. This yields -7<x<1. Since the range of the question does not include the range of the condition, the condition 1) is not sufficient.
In case of the condition 2), we can obtain -4<x-3<4. This also yields, -1<x<7. Since the range of the question does not include the range of the condition, the condition 2) is not sufficient.
Using both the condition 1) and 2), since -1<x<1, the range of the question does not include the range of the conditions. Therefore, the conditions are not sufficient and the correct answer is E.
For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Manager Joined: 15 Mar 2015 Posts: 106 Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink] ### Show Tags 18 Aug 2016, 04:20 Bunuel wrote: Is x > 0? (1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question. (2) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient. Answer: A. Hope it helps. Hi Bunuel! Can you please tell me more/guide me towards more information about squaring both sides of an equation with absolute value? I was putting up multiple scenario equations but your solution clearly is faster. When can I square an absolute value equation? What do I need to think about? Best wishes _________________ I love being wrong. An incorrect answer offers an extraordinary opportunity to improve. Math Expert Joined: 02 Sep 2009 Posts: 61484 Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink] ### Show Tags 18 Aug 2016, 06:31 MarkusKarl wrote: Bunuel wrote: Is x > 0? (1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question. (2) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient. Answer: A. Hope it helps. Hi Bunuel! Can you please tell me more/guide me towards more information about squaring both sides of an equation with absolute value? I was putting up multiple scenario equations but your solution clearly is faster. When can I square an absolute value equation? What do I need to think about? Best wishes When there are absolute values on both sides you can safely square. When there is an absolute value on one side and non-absolute value on another it becomes trickier - you should plug back the solutions to make sure that they are valid. You can search for similar questions here: search.php?search_id=tag&tag_id=37 Hope it helps. _________________ Intern Joined: 07 Mar 2017 Posts: 2 Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink] ### Show Tags 12 Dec 2017, 09:30 Bunuel wrote: Is x > 0? (1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question. (2) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient. Answer: A. Hope it helps. Hi Bunuel, I had a small query for this question. It will also be great if you can clarify the concept along with it. for option B, |x-3|=|2x-3|, based on my understanding, this is the same as saying the "distance of x from 3 is same as the distance of x from 3/2". If this is true, then the value of x should fall somewhere between 3 and 1.5, i.e. 2.25 and hence should be greater than one. Please let me know where I am wrong here Thanks, Dev Math Expert Joined: 02 Sep 2009 Posts: 61484 Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink] ### Show Tags 12 Dec 2017, 09:34 bapoon1991 wrote: Bunuel wrote: Is x > 0? (1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient. You can see here that you don't even need to find exact value(s) of x to answer the question. (2) $$|x-3|=|2x-3|$$. Square both sides: $$(x-3)^2=(2x-3)^2$$ --> $$(2x-3)^2-(x-3)^2=0$$. Apply $$a^2-b^2=(a-b)(a+b)$$, rather than squaring: --> $$x(3x-6)=0$$ --> $$x=0$$ or $$x=2$$. Not sufficient. Answer: A. Hope it helps. Hi Bunuel, I had a small query for this question. It will also be great if you can clarify the concept along with it. for option B, |x-3|=|2x-3|, based on my understanding, this is the same as saying the "distance of x from 3 is same as the distance of x from 3/2". If this is true, then the value of x should fall somewhere between 3 and 1.5, i.e. 2.25 and hence should be greater than one. Please let me know where I am wrong here Thanks, Dev You would have been right if it were $$|x-3|=|x-3/2|$$ but it's $$|x-3|=2*|x-3/2|$$. So, the distance from x to 3 is TWICE the distance from x to 3/2. _________________ Intern Joined: 03 Jul 2015 Posts: 12 Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink] ### Show Tags 15 Jul 2018, 05:37 Bunuel wrote: kawan84 wrote: Hi Bunnel, I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ... 1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution )[Did not understand, Why -ve sign is here as the x range is positive]3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)[Understood] Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints. What am i missing here and how to solve these questions effectively Absolute value properties: When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$; When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$. SOLUTION: We have two transition points for $$|x-3|=|2x-3|$$: $$x=\frac{3}{2}$$ and $$x=3$$. Thus three ranges to check: 1. $$x<\frac{3}{2}$$; 2. $$\frac{3}{2}\leq{x}\leq{3}$$; 3. $$3<x$$ Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them. 1. When $$x<\frac{3}{2}$$, then $$x-3$$ is negative and $$2x-3$$ is negative too, thus $$|x-3|=-(x-3)$$ and $$|2x-3|=-(2x-3)$$. Therefore for this range $$|x-3|=|2x-3|$$ transforms to $$-(x-3)=-(2x-3)$$ --> $$x=0$$. This solution is OK, since $$x=0$$ is in the range we consider ($$x<\frac{3}{2}$$). 2. When $$\frac{3}{2}\leq{x}\leq{3}$$, then $$x-3$$ is negative and $$2x-3$$ is positive, thus $$|x-3|=-(x-3)$$ and $$|2x-3|=2x-3$$. Therefore for this range $$|x-3|=|2x-3|$$ transforms to $$-(x-3)=2x-3$$ --> $$x=2$$. This solution is OK, since $$x=2$$ is in the range we consider ($$\frac{3}{2}\leq{x}\leq{3}$$). 3. When $$3<x$$, then $$x-3$$ is positive and $$2x-3$$ is positive too, thus $$|x-3|=x-3$$ and $$|2x-3|=2x-3$$. Therefore for this range $$|x-3|=|2x-3|$$ transforms to $$x-3=2x-3$$ --> $$x=0$$. This solution is NOT OK, since $$x=0$$ is NOT in the range we consider ($$3<x$$). Thus $$|x-3|=|2x-3|$$ has two solutions $$x=0$$ and $$x=2$$. Hope it's clear. P.S. Though for this particular question I still suggest another approach shown in my post here: http://gmatclub.com/forum/is-x-0-1-x-3- ... l#p1048512 Dear Bunuel, Would like to ask question on the part highlighted in red in your post, specifically on determining x - 3 is negative. Since the range is $$\frac{3}{2}\leq{x}\leq{3}$$, then if i plug in 3/2 in x - 3, it will result in negative number, however if i plug in 3 in x - 3, it will result in 0 or non-negative, and therefore x - 3 can be negative or non-negative. In your post x - 3 must be negative, how do we eliminate the non-negative one ? Or do we simply ignore non-negative ? Math Expert Joined: 02 Sep 2009 Posts: 61484 Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink] ### Show Tags 15 Jul 2018, 06:06 tixan wrote: Bunuel wrote: kawan84 wrote: Hi Bunnel, I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ... 1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution )[Did not understand, Why -ve sign is here as the x range is positive]3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)[Understood] Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints. What am i missing here and how to solve these questions effectively Absolute value properties: When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$; When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$. SOLUTION: We have two transition points for $$|x-3|=|2x-3|$$: $$x=\frac{3}{2}$$ and $$x=3$$. Thus three ranges to check: 1. $$x<\frac{3}{2}$$; 2. $$\frac{3}{2}\leq{x}\leq{3}$$; 3. $$3<x$$ Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them. 1. When $$x<\frac{3}{2}$$, then $$x-3$$ is negative and $$2x-3$$ is negative too, thus $$|x-3|=-(x-3)$$ and $$|2x-3|=-(2x-3)$$. Therefore for this range $$|x-3|=|2x-3|$$ transforms to $$-(x-3)=-(2x-3)$$ --> $$x=0$$. This solution is OK, since $$x=0$$ is in the range we consider ($$x<\frac{3}{2}$$). 2. When $$\frac{3}{2}\leq{x}\leq{3}$$, then $$x-3$$ is negative and $$2x-3$$ is positive, thus $$|x-3|=-(x-3)$$ and $$|2x-3|=2x-3$$. Therefore for this range $$|x-3|=|2x-3|$$ transforms to $$-(x-3)=2x-3$$ --> $$x=2$$. This solution is OK, since $$x=2$$ is in the range we consider ($$\frac{3}{2}\leq{x}\leq{3}$$). 3. When $$3<x$$, then $$x-3$$ is positive and $$2x-3$$ is positive too, thus $$|x-3|=x-3$$ and $$|2x-3|=2x-3$$. Therefore for this range $$|x-3|=|2x-3|$$ transforms to $$x-3=2x-3$$ --> $$x=0$$. This solution is NOT OK, since $$x=0$$ is NOT in the range we consider ($$3<x$$). Thus $$|x-3|=|2x-3|$$ has two solutions $$x=0$$ and $$x=2$$. Hope it's clear. P.S. Though for this particular question I still suggest another approach shown in my post here: http://gmatclub.com/forum/is-x-0-1-x-3- ... l#p1048512 Dear Bunuel, Would like to ask question on the part highlighted in red in your post, specifically on determining x - 3 is negative. Since the range is $$\frac{3}{2}\leq{x}\leq{3}$$, then if i plug in 3/2 in x - 3, it will result in negative number, however if i plug in 3 in x - 3, it will result in 0 or non-negative, and therefore x - 3 can be negative or non-negative. In your post x - 3 must be negative, how do we eliminate the non-negative one ? Or do we simply ignore non-negative ? Yes, $${x}\leq{3}$$, gives $$x - 3 \leq 0$$ but $$|x-3|=-(x-3)$$ is still true. That's because |a| = -a when a <= 0. Meaning that if a is negative or 0, then |a| = -a. For example, if say a = -1, then |a| = -a = -(-1) = 1 and if say a = 0, then |a| = -a = -0 = 0. Does this make sense? _________________ SVP Joined: 26 Mar 2013 Posts: 2346 Concentration: Operations, Strategy Schools: Erasmus '21 (M$)
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
### Show Tags
03 Aug 2018, 00:32
dvinoth86 wrote:
Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|
(1) |x + 3| = 4x – 3
RHS must be non-negative....therefore
$$4x-3\geq{0}$$............$$4x\geq{3}$$........$$x\geq{3/4}$$........Then the solution is positive
Sufficient
Dear GMATGuruNY
Is it possible to do so in statement 2 as we did in statement 1, although any side will give that x > 0? if not is it coincidence in this problem ?
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
### Show Tags
03 Aug 2018, 09:13
Mo2men wrote:
dvinoth86 wrote:
Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|
(1) |x + 3| = 4x – 3
RHS must be non-negative....therefore
$$4x-3\geq{0}$$............$$4x\geq{3}$$........$$x\geq{3/4}$$........Then the solution is positive
Sufficient
Dear GMATGuruNY
Is it possible to do so in statement 2 as we did in statement 1, although any side will give that x > 0? if not is it coincidence in this problem ?
Because each side of the equation in Statement 2 is an absolute value, the expressions within the absolute values (x-3 and 2x-3) do NOT have to be nonnegative.
One solution for Statement 2 is x=0, with the result that x-3 = -3 and 2x-3 = -3.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
### Show Tags
03 Aug 2018, 12:08
GMATGuruNY wrote:
Mo2men wrote:
(1) |x + 3| = 4x – 3
RHS must be non-negative....therefore
$$4x-3\geq{0}$$............$$4x\geq{3}$$........$$x\geq{3/4}$$........Then the solution is positive
Sufficient
Dear GMATGuruNY
Is it possible to do so in statement 2 as we did in statement 1, although any side will give that x > 0? if not is it coincidence in this problem ?
Because each side of the equation in Statement 2 is an absolute value, the expressions within the absolute values (x-3 and 2x-3) do NOT have to be nonnegative.
One solution for Statement 2 is x=0, with the result that x-3 = -3 and 2x-3 = -3.
Thanks GMATGuruNY.
In this type of questions asking for sign of x, does this mean that I can deal with such statement in same manner like statement 1? or do I need open modulus to be sure from the sign?
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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03 Aug 2018, 22:07
1
Mo2men wrote:
Thanks GMATGuruNY.
In this type of questions asking for sign of x, does this mean that I can deal with such statement in same manner like statement 1? or do I need open modulus to be sure from the sign?
Since the question stem asks whether x>0 -- and an equation with absolute value on both sides can have both a positive and a nonpositive solution -- I recommend that you solve the equation, either by opening up the modulus or by squaring both sides.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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28 Nov 2018, 05:06
Bunuel does it mean that in all questions based on absolute values, one solution of it would be 0?
I'm clear with Stat1 but in stat 2, I used the formula by squaring both sides of absolute figures to find the inequality range. Here is what I did.
|x−3|=|2x−3|
Square both sides: (x−3)^2=(2x−3)^2
x^2-6x+9=4x^2-12x+9
3x^2=6x
Therefore, x=2. So I chose option D. Can you help me to find what is the error? I'm not getting it. All I did was use the formula.
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Posts: 61484
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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28 Nov 2018, 05:38
1
topper97 wrote:
Bunuel does it mean that in all questions based on absolute values, one solution of it would be 0?
I'm clear with Stat1 but in stat 2, I used the formula by squaring both sides of absolute figures to find the inequality range. Here is what I did.
|x−3|=|2x−3|
Square both sides: (x−3)^2=(2x−3)^2
x^2-6x+9=4x^2-12x+9
3x^2=6x
Therefore, x=2. So I chose option D. Can you help me to find what is the error? I'm not getting it. All I did was use the formula.
1. Of course 0 is NOT a solution of all absolute value questions. Why would it be?
2. You solved 3x^2 = 6x incorrectly. You cannot reduce 3x^2 = 6x by x because x can be 0 and we cannot divide by 0. By doing so you loose a root, namely x = 0.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We cannot divide by zero.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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01 Dec 2019, 18:02
dvinoth86 wrote:
Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x – 3| = |2x – 3|
1) x + 3 = 4x - 3, 3x = 6, x = 2 or, -x -3 = 4x - 3, 5x = 0, x = 0. Not sufficient.
2) x -3 = 2x -3, 3x = 0, x =0. or -x + 3 = 2x - 3, x = 2, or, -x + 3 = - 2x + 3, x = 0. not sufficient.
Together x can be either 0 or 2. Not sufficient.
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Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
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28 Dec 2019, 22:14
The solution quoted by you above has some tacit steps. I'll list them here:
St. 1 says: |x + 3| = 4x – 3 . . . (1)
Case 1: x + 3 > = 0
That is, x > = -3
In this case, |x+3| = (x+3)
So, Equation 1 becomes:
x + 3 = 4x - 3
=> 6 = 3x
=> x = 2
Does this value of x satisfy the condition of Case 1, that x > = -3?
Yes, it does. So, x = 2 is a valid value of x.
Let's now consider
Case 2: x + 3 < 0
That is, x < -3
In this case, |x+3| = -(x+3)
So, Equation 1 becomes:
-(x + 3) = 4x - 3
=> -x - 3 = 4x - 3
=> 0 = 5x
=> x = 0
Does this value of x satisfy the condition of Case 2, that x < -3?
No, it doesn't. So, x = 0 is an INVALID value of x.
Thus, we got only 1 valid value of X from Statement 1 (x = 2) and so, Statement 1 is sufficient to say that x > 0.
Japinder[/quote]
Hi
I am not clear why are we taking ONLY x+3 as x+3>=0 and not x+3<=0?
Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink] 28 Dec 2019, 22:14
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Smooth morphisms are locally standard smooth.
Lemma 29.34.11. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Let $V \subset S$ be an affine open neighbourhood of $f(x)$. The following are equivalent
1. The morphism $f$ is smooth at $x$.
2. There exist an affine open $U \subset X$, with $x \in U$ and $f(U) \subset V$ such that the induced morphism $f|_ U : U \to V$ is standard smooth.
Proof. Follows from the definitions and Algebra, Lemmas 10.137.7 and 10.137.10. $\square$
There are also:
• 2 comment(s) on Section 29.34: Smooth morphisms
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# The ferroelectric field-effect transistor with negative capacitance
## Abstract
Integrating ferroelectric negative capacitance (NC) into the field-effect transistor (FET) promises to break fundamental limits of power dissipation known as Boltzmann tyranny. However, realizing the stable static negative capacitance in the non-transient non-hysteretic regime remains a daunting task. The problem stems from the lack of understanding of how the fundamental origin of the NC due to the emergence of the domain state can be put in use for implementing the NC FET. Here we put forth an ingenious design for the ferroelectric domain-based field-effect transistor with the stable reversible static negative capacitance. Using dielectric coating of the ferroelectric capacitor enables the tunability of the negative capacitance improving tremendously the performance of the field-effect transistors.
## Introduction
Dimensional scalability of field effect transistors (FETs) has reached the Boltzmann tyranny limit because of transistors’ inability to handle the generated heat1. To reduce the power dissipation of electronics beyond this fundamental limits, negative capacitance (NC) of capacitors comprising ferroelectric materials has been proposed as a solution2. The FET with a negative-capacitance ferroelectric layer has gained an enormous attention of researchers3,4,5,6,7,8,9,10,11,12. However, after impressive initial progress that has resulted in a rich lore massaging the aspects of technological benefits of the prospective stable static negative capacitance, the advancement in the field decelerated considerably. The lack of a clear self-consistent physical picture of the origin and mechanism of the stable static negative capacitance7,11,12,13,14 not only retarded the craved technological progress, but has led to numerous invalid fabrications and misleading claims9.
In this work, we put forth a foundational mechanism of the NC in ferroelectrics demonstrating inevitable emergence of the NC due to formation of polarization domains. We establish a practical design of the stable and reversible NC FET based on the domain layout. The proposed device is tunable and downscales to the 2.5–5 nm technology node.
In what follows we review the state-of-the-art and basic concepts behind exploring ferroelectrics as the NC elements which constitute the base for our new results. We also mark the potential pitfalls in the NC implementing in the so far suggested NC FETs caused by depreciating the immanent role of domains. Figure 1 demonstrates the principles of integrating the ferroelectric layer with the NC into the FET and the crucial role of domain states. The performance of the FET is quantified by the so-called subthreshold swing $$S\,S={(\partial \left({\log }_{10}{I}_{d}\right)/\partial {V}_{g})}^{-1}$$ that describes the response of the drain current Id to the gate voltage Vg. The lower the value of the SS, the lower power the circuit consumes. In a basic bulk metal-insulator-semiconductor field-effect transistor (MIS FET), shown in Fig. 1a, which generalizes the MOSFET structure, the subthreshold swing is
$$S\,S=\overbrace{\frac{\partial V_{s}}{\partial \left( {\log}_{10}I_d\right) }}^{S\,S_b}\overbrace{\frac{\partial V_{g}}{\partial V_{s}}}^{m},\quad m=1+\frac{C_{s}}{C_{g}}.$$
(1)
Here the first factor, SSb, presents the response of Id to the voltage Vs at the conducting channel region, and the second factor, the so-called body factor, m, characterizes the response of the voltage Vs to the applied voltage Vg. Figure 1a shows the equivalent electronic circuit, with Cg and Cs standing for the gate dielectric and semiconducting substrate capacitancies, respectively.
The fundamental constraint of the energy/power efficiency of the MIS FETs arises from the thermal injection of electrons over an energy barrier enabling drain current flow and thus preventing the reduction of factor SSb below the 60 mV dec−1, because the body factor m > 1 at Cg, Cs > 0. To overcome this limitation, the FETs incorporating the NC into its design has been proposed2. Indeed, replacing the gate dielectric with material with negative capacitance CNC would make the body factor m < 1, thus, pushing the SS below the Boltzmann limit.
Ferroelectric materials appear as best candidates for realizing negative capacitance in FETs2. The emergence of the NC in a ferroelectric capacitor follows from the Landau double-well landscape of the capacitor energy W as a function of the applied charge Q15 (blue line in Fig. 1b). The downward curvature of W(Q) at small Q implies that the addition of a small charge to the ferroelectric capacitor plate, induces non-zero polarization and reduces its energy. Hence the negative value of the capacitance $${C}_{NC}^{-1}={d}^{2}W/d{Q}^{2}$$. Remarkably, even the domain formation due to fundamental instability of a monodomain state16,17,18,19,20,21, maintains the negative capacitance16,22,23,24,25,26. The energy W(Q) of the multidomain state is lower then that of the monodomain state while the downward curvature at Q = 0 is conserved, see the red line in Fig. 1b illustrating an exemplary W(Q) for the capacitor hosting two domains. A detailed parsing of particularities related to the monodomain state instability and specific manifestations of the multidomain state is presented in Supplementary Note 1.
An inevitable multidomain formation posits the need for a detailed exploring possible ways of realization design of the NC FETs. While the multidomain configuration preserves the NC, the domain formation may trigger some undesired effects detrimental to realizing the NC FET. In particular, domains cause inhomogeneous charge21 and electric field27 distribution, endangering the conducting channel in the most commonly discussed metal-ferroelectric-semiconducting MFS FETs (Fig. 1c), as the voltage dispersion they cause becomes comparable with the transistor operation voltage. This problem can be mended by putting the ferroelectric layer into the pretransitional (incipient) regime just above the transition temperature, where the NC effect still persists, but the field-induced polarization distribution is uniform5,28,29. However, this would limit the desirable decreasing of the body factor keeping it above 0.9929. Another way out is introducing an intermediate dielectric (insulating) buffer layer between the ferroelectric and semiconductor10, which corresponds to MFIS FET architecture shown in Fig. 1d. This, in its turn, would not help much because smoothing the field nonuniformity would occur only at distances well exceeding the spatial scale on which the NC potential amplification effect is still actual. A detailed analysis of particularities related to pitfalls of the discussed above architectures of the specific multidomain state is presented in Supplementary Note 2.
The design with the floating gate electrode placed between ferroelectric and dielectric layers30,31 appeared to resolve this problem and to level the field inhomogeneities right below the electrode. The resulting MFMIS structure is the conventional FET with the overimposed MFM capacitor, see Fig. 1e. This architecture has attracted some critique32, since it was believed that the anti-parallel domains formation inside the MFM capacitor would destabilize the NC. However, as we established in Luk’yanchuk et al.24, it is precisely the two-domain configuration that provides the stable and operable NC because of the possibility of manipulating the domain wall by the applied charges, not accounted for in Hoffmann et al.32.
Here we introduce and devise the working regime of the MFMIS FET in which the NC effect emerges from the integrated MFM capacitor hosting two domains. We show that the MFMIS architecture not only free from the perils mentioned above but allows for an enormous improving MFMIS FET characteristics by coating the MFM capacitor with the dielectric capacitor in parallel connection. The proposed coating design that we call c-MFMIS FET, shown in Fig. 1f, provides degrees of freedom enabling a complete tunability of the dielectric parameters of the NC FET.
## Results and discussion
### Two-domain negative capacitance of the MFM capacitor
The nanodot-scale two-domain MFM is a major element of the MFMIS FET enabling the NC response via the charge-controlled motion of the DW. Following24, we discuss in detail the negative capacitance of the MFM capacitor, which is the base of our proposed device. Shown in Fig. 2a is the general view of MFMIS FET. Figure 2b presents the vertical and horizontal cross-sections of a nanoscale ferroelectric disc-shape MFM capacitor integrated in the MFMIS FET. At the zero charge Qf at the electrodes of the MFM capacitor, corresponding to the zero voltage at the transistor, the DW sits in the middle of the capacitor, see left hand side of Fig. 2b. The intrinsic charges at the respective electrodes redistribute in order to compensate the depolarization charges of each domain (keeping the total charge Qf = 0) and to banish the electric field inside the ferroelectric disc, reducing thus the electrostatic energy. The finite charge Qf, induced by the voltage V = Vg applied to the transistor gate, displaces the DW from its middle zero-voltage position, see right hand side of panels b. Accordingly, the intrinsic charges rearrange (maintaining the total charge Qf) to compensate the depolarization fields of now unequal domains.
The NC response arises since the length, hence the energy of the displacing DW, is sensitive to the shape of a ferroelectric capacitor. To ensure the best controlled performance of the NC, we choose a disc-like form of a capacitor. When moving apart from the middle, the DW not only compensates the electric field arising due to the charge transferred to the electrodes but, minimizing its surface self-energy, shrinks in the width w and bends because of the cylindrical shape of a ferroelectric. As a result, the DW overshoots towards the edge beyond the electrostatics-demanded equilibrium position at which the internal electric field would have disappeared. Hence the net electric field does not vanish but flips over and goes from the negatively charged electrode to the positively charged one. This counterintuitive outcome precisely expresses the phenomenon of the negative capacitance. Note, however, that at some threshold value of the applied charge, $${Q}_{f}^{* }$$, when it becomes approximately equal to the depolarization charge of the uniformly oriented polarization, Q0, the DW reaches the edge of the ferroelectric layer and leaves the sample. The monodomain state with positive capacitance restores; this corresponds to the termination of the red branch in Fig. 1b.
The quantitative description of the NC of the disk-shape two-domain ferroelectric capacitor is given by Luk’yanchuk et al.24
$${C}_{NC}=-{\gamma }_{2}\frac{{D}_{f}}{{\xi }_{0}}{C}_{f},$$
(2)
where we explicitly spotlighted the capacitance Cf = ε0εfSf/df > 0, which is the capacitance of the monodomain MFM capacitor in the stable state at Q = ±Q0 = ±SfP0 (minima of the W(Q) dependence in Fig. 1b). The negative factor, − γ2Df/ξ0, reflects the features brought in by the DW displacement in the two-domain configuration. Here Df and df are the diameter and height of the capacitor, respectively, $${S}_{f}=\pi {D}_{f}^{2}/4$$ is the area of the ferroelectric plate surfaces, and P0 is the polarization of the ferroelectric in the equilibrium state. The coherence length ξ0 1nm describes the DW thickness, the dimensionless geometric factor γ2 ≈ 4.24 reflects the internal profile of polarization inside the DW and the DW bending in the cylindrical gate, and εf and ε0 are the dielectric constant of the ferroelectric material and the vacuum permittivity, respectively.
Figure 1b displays the energy advantage of the two-domain state (whose energy is shown by the red curve), with respect to the usually considered uniform NC state (the blue dashed curve), both states preserving the same charge Q at the electrodes. To create the two-domain state from the uniformly-polarized state one has to suppress polarization in a fraction of the ferroelectric occupied by the DW, while to depolarize the monodomain state by the electric field due to the uniformly distributed charge Q, one would have had to suppress the ferroelectricity within the whole volume which is much more energetically costly.
As a next step, we integrate the MFM two-domain NC-capacitor into the MFMIS FET architecture.
### The MFMIS FET
This device comprises the gate stack overimposed on a semiconducting substrate in which the source and drain parts are connected by the gate-operated conducting channel, see Fig. 2a. The gate stack includes the MFM capacitor and the gate insulating layer separating it from the substrate. This is the high-κ dielectric layer, preventing a charge leakage between the lower capacitor’s plate and the semiconducting channel.
The top MFM capacitor plate is the gate electrode connecting the transistor to the external voltage source. The bottom capacitor electrode is an intermediate electrically isolated floating gate electrode of the transistor that preserves the entire charge, most commonly the zero total charge, constant, stabilizing the ferroelectric two-domain state. Furthermore, the floating gate makes the potential along the ferroelectric interface even, maintaining, therefore, a uniform electric field across the gate stack and substrate. Along the way, the floating gate resolves a frequent issue of neutralizing the parasitic charges that may be trapped by interfaces during the fabrication and functioning. Maintaining the working charge and providing a regular rubbing out the parasite leaking charges with the removal time faster than the leakage time31 is achieved via the standard discharging methods. For instance, it is implemented by either harnessing the Fowler-Nordheim tunneling and the hot electrons injection33, or by circuiting the gate by the auxiliary charge-carrying current, IQ, contact, see Fig. 1e, to a certain source for a given combination of electric inputs ensuring the proper sequence of discharging and high resistance modes34.
An effective electronic circuit of the MFMIS FET, shown in Fig. 1e, is similar to that of the multidomain MFIS FET (Fig. 1d). The difference is that now the task of leveling the depolarization field inhomogeneities is taken by the floating gate electrode. Therefore, the gate dielectric layer can be safely engineered as an utmostly thin one, down to the technologically-acceptable limit of a few nanometers. This increases Cd with respect to CNC, opening doors to making the gate capacitance negative, $${C}_{g}^{-1}={C}_{d}^{-1}-| {C}_{NC}^{-1}| \,<\, 0$$. Yet it is hard to achieve a required largeness of Cd with respect to CNC due to the restrictions imposed by materials compatible with the silicon CMOS technologies. To meet the challenge, we devise a coated c-MFMIS FET that critically changes the situation and breaks ground for the unlimited enhancement of the performance of the NC FET transistor.
### The c-MFMIS FET
The coating of the ferroelectric layer with the dielectric oxide sheath confined by the same electrodes, see Fig. 2c, d straightforwardly incorporates an additional capacitor with the capacitance Cc > 0 in parallel to CNC. This results in a radical improvement of the controlled tuning of the gate capacitance. In particular, manipulating with the sizes of the coating oxide layer and with the geometrical design of the device as a whole, provides a broad variation in its performance characteristics and functioning regimes. The panels (c) and (d) exemplify possible designs. The panel (c) shows an annulus-like coating capacitor, while panel (d) displays a rectangular design of the coating layer and, in addition, the possibility of increasing the area of the floating gate electrode with respect to the top gate electrode. The latter design allows for controllable increasing capacitances of the coating and gate-dielectric layers most efficiently, maintaining the miniaturization of the device. Note, that in all the geometries, the core ferroelectric should maintain its disc-like shape ensuring the optimal manipulation with the DW.
Shown in Fig. 1f is the equivalent circuit of c-MFIS FET. The important advance is that the gate capacitance becomes
$${C}_{g}=\frac{1}{{C}_{d}^{-1}+{({C}_{c}-| {C}_{NC}| )}^{-1}},$$
(3)
which permits tuning Cg over the widest range of values by the appropriate modifying the parameters of the coating capacitor. We reveal the rich dependence of Cg on the coating layer size, Lc choosing the rectangular geometry of the coating layer (Fig. 2c). The capacitance of the disk-shape two-domain ferroelectric capacitor, CNC, is given by Eq. (2). The capacitances of the coating and gate dielectric layers are taken in a standard form as Cc = ε0εcSc/dc and Cd = ε0εdSd/dd, respectively. Here $${S}_{f}=\pi {D}_{f}^{2}/4$$, $${S}_{c}={L}_{c}^{2}-{S}_{f}$$ and $${S}_{d}={L}_{d}^{2}$$ are areas of the ferroelectric, coating dielectric and gate dielectric plate surfaces, respectively.
The behavior of Cg defined by Eq. (3) is most generic and does not depend critically on the specific choice of materials. For practical applications, we choose both, the coating layer and the gate-dielectric layer, be composed of the Si-compatible dielectric HfO2 with the respective dielectric constants εc, εd being approximately equal to 25. The ferroelectric disc of the diameter Df 6 nm and thickness df 3 nm, can be fabricated out of the ferroelectric phase of HfO2 or its Zr-based modification, Hf0.5Zr0.5O2 with εf 5013. In fact, εf is the only relevant material parameter that defines the NC properties of the two-domain ferrolectric layer; therefore, the consideration applies equally well to other similar ferroelectrics, for instance, to perovskite oxides, like strained PbTiO3 with about the same εf. The height of the gate dielectric layer is taken as dd 3 nm, whereas the thickness of the coating layer, dc, is equal to df. The size of the rectangular floating-gate electrode, Ld, defining the size of the gate dielectric capacitor, is taken as 1.2Lc.
Figure 2e displays the derived normalized gate capacitance, cg = Cg/Sd as function of Lc. The presented cg(Lc) dependencies are the Eq. (3) plots, into which the given above capacitancies, CNC, Cc(Lc) and Cd(Lc) are substituted. Looking at the plots, one discriminates the three distinct regimes of the gate functioning set by two critical sizes, Lc1 < Lc2, of the coating layer: (i) the super-capacitance, Lc < Lc1, (ii) the negative-capacitance, Lc1 < Lc < Lc2, and (iii) the near-zero capacitance, Lc2 < Lc. All these three distinct regimes are of tremendous relevance for applications. Below, we restrict ourselves to the detailed analysis of the NC regime, i.e., the regime where cg < 0, leaving the detailed discussion of regimes (i) and (iii) to forthcoming publication. Note that although in the NC regime the average polarization of the ferroelectric nanodot is aligned with the voltage drop across the capacitor, the electric field inside the coated layer is directed oppositely to it. Accordingly, the polarization induced inside the dielectric oxide sheath is opposite to the polarization of the ferroelectric nanodot, see Fig. 2f. It is important that the absolute value of the NC capacitance can be done arbitrary small on approach to the resonance regime CNC = Cc of Equation (3), i.e., upon L − Lc2 → 0. In this regime the nonlinear effects in the Q − V characteristics of the MFM capacitor become of prime relevance.
An unrestricted range of variation of cg, from minus infinity to zero, as seen from Fig. 2e, allows for the unlimited tuning of the magnitude of the gate NC. In particular, the possibility of making it arbitrary small, enables us to overcome a previously insurmountable obstacle of the proper matching between the gate, Cg, and substrate, Cs, capacitancies and obtain the desirable value of the body factor
$$m=1+\frac{{C}_{s}}{{C}_{g}}=1+\frac{{C}_{s}}{{C}_{d}}+\frac{{C}_{s}}{{C}_{c}-| {C}_{NC}| }\,,$$
(4)
within the NC FET operational interval 0 < m < 1, getting thus the remarkably low values of SS(1), the task not achievable by previously suggested architectures.
Now the task is to find the optimal coating size Lc, given the normalized substrate capacitance cs = Cs/Sd, that ensures matching to targeted value of m. To that end, we employ Eq. (4) which defines the implicit Lc dependence of cs at given m. The shown in Fig. 3a family of Lc(cs) curves for different m, confined between m = 0 (red) and m = 1 (brown) characteristics, represents the stable working interval for our exemplary c-MFMIS FET.
The region below the brown line where m > 1, i.e., SS > 60 mV dec−1, corresponds to small sizes of the coating layer, Lc < Lc1. The region above the m = 0 curve but below the Lc = Lc2 line is the hysteretic loss of the reversibility region. Therefore, the proper choice of Lc, in the interval Lc1 < Lc < Lc2, enables the desired magnitude of m for a given value of the substrate capacitance cs. Although in our model example, the lateral size of the coating layer varies between Lc1 ≈ 24 nm and Lc2 ≈ 38 nm, it can be significantly reduced down to practically the diameter of the ferroelectric disc, by increasing the dielectric constant of the coating material by factor of four.
### Practical design. Optimal match between gate and substrate
The established characteristics of the c-MFMIS FET enable us to go beyond the past prima facie technological concepts and turn to the practical design in relevant industrial environment. The critical challenge9 emerging when engineering the NC FET, is the complex and highly nonlinear nature of Cs since the latter comprises contributions from the capacitance of the depleted layer, from the quantum capacitance due to charge carriers injected into the conducting channel, from the interface charges, and, finally, from the source/drain geometrical capacitancies. While being of a relatively low value when coming mainly from the capacitance of the depletion layer in the low-conducting regime at small voltages, Cs increases dramatically, typically by order of magnitude, due to the injection of conducting electrons into the channel caused by the gate bias near and above the threshold value.
Using our established concept of Lc(cs, m) characteristics, Fig. 3a, enables the determining an optimal match between the NC gate capacitance and the semiconducting substrate capacitance ensuring the best-performance SS and stability of the transistor. To exemplify the nonlinear behavior of the substrate, we take the normalized capacitance $${c}_{s}^{{{{\rm{sth}}}}}\simeq 1{0}^{-2}$$ Fm−2 in the low-voltage subthreshold regime and $${c}_{s}^{{{{\rm{th}}}}}\simeq 1{0}^{-1}$$ Fm−2 in the high-voltage near-threshold regime. Next, we set the condition m = 0 at the steepest point of the Id(Vg) curve, i.e., in the near-threshold gate voltage where $${c}_{s}={c}_{s}^{{{{\rm{th}}}}}$$. This condition visualized by the point A in Fig. 3a, provides us with the optimal value of the size of the coating layer, $${L}_{c}^{{{{\rm{opt}}}}}\approx 28$$ nm. Decreasing the gate voltage Vg to subthreshold values, reduces cs and moves it to the left from the point A along the black line until reaching the point B corresponding to $${c}_{s}={c}_{s}^{{{{\rm{sth}}}}}$$ at Vg ≈ 0. The body factor at the point B is m = 0.9 which gives SS ≈ 54 mV dec−1.
The possible transfer Id(Vg) characteristics9 are schematically illustrated in Fig. 3b. The optimal characteristic (red line) derived according to the devised above operating procedure starts with the relatively modest SS ≈ 54 mV dec−1, steepens upon the increase in Vg, and reaches its steepest value in the near-threshold region. The location of points A and B corresponds to the capacitancies $${c}_{s}^{{{{\rm{th}}}}}$$ and $${c}_{s}^{{{{\rm{sth}}}}}$$ of panel a. The optimal regime maintains the steep slope of the Id(Vg) dependence over the entire voltage working range and includes not only the subthreshold, but also near- and above threshold regimes, preserving the stable and hysteresis free IVg transfer characteristics at the same time.
Because of the nonlinearity of cs, the transfer characteristics of the designed c-MFMIS FET with $${L}_{c}={L}_{c}^{{{{\rm{opt}}}}}$$ demonstrates the better performance at the same on-off current switching ratio, Ion/Ioff, than the shown by the green curve commonly assumed NC FET with the voltage-independent substrate capacitance cs (although having the steeper initial SS). At the same time, an attempt to engineer the NC FET having the nonlinear cs with an initially steeper SS (exemplified by the blue curve), results in the hysteretic switching instability.
### Practical design. Compact gate model
In order to provide incorporating the two-domain c-MFMIS FET architecture into the industrially-standardized circuiting, we design the scalable compact model describing the QgVg characteristics of the coated NC gate stack. In the low-voltage and low-charge operational mode of the NC FET, this compact model is defined by the linear relation Qg = CgVg, where Cg is given by Eqs. (2) and (3). Looking forward to extensive applications of our compact model for the description of the c-MFMIS FET, we expand the compact model’s working range over to the nonlinear regime where substantial shifts of the domain wall and even its escape from the sample may occur.
The complete set of the QgVg characteristics of the gate is defined by the individual electric properties of its components, including the gate dielectric capacitor, coating capacitor, and the MFM capacitor which form the equivalent circuit shown in Fig. 1f and which are characterized by the linear, $${V}_{d}={C}_{d}^{-1}{Q}_{d}$$, $${V}_{c}={C}_{c}^{-1}{Q}_{c}$$, and nonlinear, $${V}_{f}={V}_{f}\left({Q}_{f}\right)$$, constitutive relations respectively. In general, the VgQg characteristics can be parametrically plotted as functions of the running parameter Qf
$$\begin{array}{l}{V}_{g}=\left(1+\frac{{C}_{c}}{{C}_{d}}\right){V}_{f}({Q}_{f})+\frac{{Q}_{f}}{{C}_{d}}\\ {Q}_{g}={C}_{c}{V}_{f}({Q}_{f})+{Q}_{f}\,,\end{array}$$
(5)
based on the relations Qg = Qd = Qc + Qf, Vg = Vd + Vf, and Vf = Vd for the circuit in Fig. 1f.
The nonlinear constitutive relation, $${V}_{f}\left({Q}_{f}\right)$$, of the MFM capacitor is the core relation that defines different regimes of the gate functioning. Shown in the Fig. 3c, are the results of the phase-field simulations (crosses), see Methods, and the analytical outcome of the developed scalable compact model (solid lines). The QfVf characteristic reveals two different operational modes of the MFM capacitors. The NC low-charge branch, $${V}_{2}\left({Q}_{f}\right)$$ (red curve), corresponds to the two-domain state where the domain wall motion is responsible for the electric properties of capacitor. The high-charge branch, $${V}_{1}\left({Q}_{f}\right)$$ (blue curve), with the positive differential capacitance, Cf = dV/dQ > 0, corresponds to the monodomain state where the domain wall is gone. As a result, the Qf-Vf characteristic of the ferroelectric capacitor is presented by the synthetic dependence
$${V}_{f}\left({Q}_{f}\right)=\left\{\begin{array}{ll}{V}_{2}({Q}_{f})\quad &\,{{\mbox{if}}}\,\left|{Q}_{f}\right| \,<\, {Q}_{f}^{* }\\ {V}_{1}({Q}_{f})\quad &\,{{\mbox{if}}}\,\left|{Q}_{f}\right| \,>\, {Q}_{f}^{* }\end{array}\right.$$
(6)
where $${Q}_{f}^{* }$$ is the charge at which the branches $${V}_{2}\left({Q}_{f}\right)$$ and $${V}_{1}\left({Q}_{f}\right)$$ meet.
The corresponding to the monodomain state branch of the Vf-Qf characteristic, is given by the parametric dependence of V1(Qf) upon the polarization P,
$$\begin{array}{l}{V}_{1}(P)=-{d}_{f}\left(2{a}_{3}^{* }P+4{a}_{33}^{* }{P}^{3}+6{a}_{333}{P}^{5}\right)\\ Q(P)=-{S}_{f}\left(P-{\varepsilon }_{0}{\varepsilon }_{i}\frac{{V}_{1}(P)}{{d}_{f}}\right)\,,\end{array}$$
(7)
derived from the uniform Ginzburg-Landau equation. The coefficients $${a}_{3}^{* }$$, $${a}_{33}^{* }$$ and a333, and the background dielectric constant εi are defended in Methods.
For the V2(Qf) function describing the two-domain case we use the analytical approximation24
$${V}_{2}({Q}_{f})\approx -\frac{0.27}{\psi \left({Q}_{f}/{Q}_{0}\right)}\frac{{Q}_{f}}{{C}_{f}}\frac{{\xi }_{0}}{D},$$
(8)
where ψ(s) (0 < s < 1), introduced in Luk’yanchuk et al.24, is the function accounting for the geometry of the system which we fit by
$$\psi (s)\approx {\left(1.0-0.027s-0.95{s}^{2}-0.34{s}^{3}+0.32{s}^{4}\right)}^{1/2}.$$
(9)
In the linear in Qf approximation, where s → 0, Eq. (8) gives CNC in Eq. (2).
Combining branches given by Eqs. (7) and (8) provides an excellent approximation for the results of the numerical simulations of the compact model, see Fig. 3c. The slight overshoots at $$\pm {Q}_{f}^{* }$$ correspond to numerical singularities appearing at the moments where the DW leaves the MFM capacitor.
To summarize, the achieved understanding that the fundamental mechanisms of the NC is the domain action, bestows closing the gap between the concept of the ferroelectric negative capacitance and its realization in electronic devices. It enables designing a stable NC-based FET, whose coating-shell architecture of the gate promises a notable enhancement of prospective performance and high tunability of characteristics allowing the perfect match with advanced FET architectures. Our findings lay out the way for scaling the NC FET nanoelectronics down to 2.5–5 nm technology nodes via utilizing the CMOS-compatible ultra-thin and ultra-small ferroelectric disc as a core of the NC gate.
## Methods
### Functional
To carry out the numerical modeling of polarization structures in a ferroelectric layer, we use the most thoroughly studied free energy functional for the PbTiO3,
$$\begin{array}{l}F=\int \left({\left[{a}_{{i}}^{* }({u}_{m},T){P}_{{i}}^{2}+{a}_{{ij}}^{* }{P}_{{i}}^{2}{P}_{{j}}^{2}+{a}_{{ijk}}{P}_{{i}}^{2}{P}_{{j}}^{2}{P}_{{k}}^{2}\right]}_{{i\le j\le k}}\right.\\ \quad\quad\,\,\,\left.+\,\frac{1}{2}{G}_{{ijkl}}({\partial }_{{i}}{P}_{{j}})({\partial }_{{k}}{P}_{{l}})+({\partial }_{{i}}\varphi ){P}_{{i}}-\frac{1}{2}{\varepsilon }_{0}{\varepsilon }_{i}{(\nabla \varphi )}^{2}\right)\,{d}^{3}r,\end{array}$$
(10)
where the sum is taken over the cyclically permutated indices {i, j, k, l} = {1, 2, 3} (or {x, y, z}). Functional Eq. (10) includes the Ginzburg-Landau (GL) energy of the strained ferroelectric layer35 written in a form given in36 (the square-bracketed term), the polarization gradient energy37 (the term with coefficients Gijkl), and the electrostatic energy, including the coupling of polarization with electric field19, Ei = − ∂iφ, described through the electrostatic potential φ (the two last terms). The strain-renormalized GL coefficients for the PbTiO3 layer (accounting partially for the elastic energy) are taken as35, $${a}_{1}^{* }$$,$${a}_{2}^{* }$$ = 3.8 × 105(T − 479C) − 11 × 109um C−2m2N−1, $${a}_{3}^{* }$$ = 3.8 × 105(T − 479C) + 9.5 × 109um C−2m2N−1, $${a}_{11}^{* }$$, $${a}_{22}^{* }$$ = 0.42 × 109 C−4m6N, $${a}_{33}^{* }$$ = 0.05 × 109 C−4m6N, $${a}_{13}^{* }$$,$${a}_{23}^{* }$$ = 0.45 × 109 C−4m6N, $${a}_{12}^{* }$$ = 0.73 × 109 C−4m6N, a111, a222, a333 = 0.26 × 109 C−6m10N, and a123 = −3.7 × 109 C−6m10N. The misfit strain is taken as um = −0.013. The gradient coefficients are taken as for PbTiO3 bulk material37 (with all possible cubic index permutations), G1111 = 2.77 × 10−10 C−2m4N, G1122 = 0.0, and G1212 = 1.38 × 10−10 C−2m4N. The background dielectric constant of the non-polar ions was taken as εi 1038 for PbTiO3 and εi 25 for semiconducting layer. The vacuum permittivity is ε0 = 8.85 × 10−12 CV−1m−1.
### Phase-field simulations
The minimum of the energy functional Eq. (10) is found by solving relaxation equation
$$-\gamma \frac{\partial {{{\bf{P}}}}}{\partial t}=\frac{\delta F}{\delta {{{\bf{P}}}}}\,,$$
(11)
where δF/δP is the variational derivative of Eq. (10); the time-scale parameter γ, which does not influence the sought energy minimum is taken equal to unity. The electrostatic Poisson equation ε0εi2φ = P, describing the spatial distribution of the polarization, is solved on the each respective relaxation step.
For practical implementation of simulations, we have used the open-source FEniCS computing platform39. To create the tetrahedral finite-element meshes we used an open-source 3D mesh generator gmsh40. For the case of the MFM capacitor, the computational region is a cylindrical volume Ω, restricted by the side boundary, ∂Ωs, and by the top, ∂Ωt, and bottom, ∂Ωb, boundaries, see Fig. 4a. For the case of the MFSM heterostructure, the computational region is a rectangular box Ω, that includes the ferroelectric layer, ΩF, and the semiconducting layer, ΩS, see Fig. 4b. The computational region, Ω, is restricted by the left, right, front and back-side boundaries ∂Ωs and by the top, ∂Ωt, and bottom, ∂Ωb, boundaries.
The solutions for the polarization, P(r), and electrical potential, φ(r), distribution were sought in the functional space of the piece-wise linear polynomials. For simulation of the MFM-capacitor, controlled by charge Q, we use free boundary conditions for P(r) on the whole surface of the cylinder. At the same time, it was assumed that the electrodes produce almost uniform z-directed electric field, Ez = − ∂zφ, spreading through the capacitor. The boundary constraint $$-Q/{S}_{f}={\bar{P}}_{{z}}+{\varepsilon }_{0}{\varepsilon }_{i}{E}_{{z}}$$ was used at the electrode interfaces to fix the applied charge Q that tunes the displacement of the DW in the spontaneously emerging two-domain structure. The bar denotes averaging over the interface surface.
For simulation of the MFSM heterostructure, the relaxation Eq. (11) was solved for the ferroelectric part of the sample while the electrostatic Poisson equation was solved for the whole domain. Boundary conditions for all the variables were taken to be periodic in the x direction. The size of the simulation rectangular box in x-direction, corresponding to the period of the spontaneously emerging domain structure was considered as an energy-minimizing parameter which was optimized for each series of calculations. Boundary conditions for P on ∂Ωt and ∂Ωb as well as on the front- and back surface boundaries of the rectangular box were taken as free boundary conditions. The Dirichlet boundary conditions were imposed on φ at the bottom and top surfaces of the box such that φ(∂Ωb) = −U/2 and φ(∂Ωt) = +U/2, to reproduce the application of the voltage U to the electrodes. The effective charge at the electrode was calculated as $$Q=-{S}_{f}(\bar{P}+{\varepsilon }_{0}{\varepsilon }_{i}{\bar{E}}_{{z}})$$.
To approximate the time derivative in Eq. (11), we used the variable-time BDF2 stepper41. The initial conditions for polarization distribution were taken to be random in the range of −10−6–10−6 C m−2 for the polarization magnitude at the first time-step of simulation. The system of the nonlinear equations arising from the discretization of Eq. (11) was solved using the Newton-based nonlinear solver with line search and generalized minimal residual method with the restart42,43. On each time step of the simulation in MFSM heterostructure, the linear system of equations obtained from the discretization of electrostatic Poisson equation was solved separately using a generalized minimal residual method with restart.
## Data availability
The data generated and analyzed during this study are available from the corresponding author upon reasonable request.
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## Acknowledgements
This work was supported by H2020 RISE-MELON action (I.L.), and by Terra Quantum AG (I.L., A.R., and V.M.V.). The work of V.M.V. was supported in part by Fulbright Foundation.
## Author information
Authors
### Contributions
I.L., Y.T., A.R., A.S. and V.M.V. conceived the work and performed calculations. I.L. and V.M.V. wrote the manuscript.
### Corresponding author
Correspondence to V. M. Vinokur.
## Ethics declarations
### Competing interests
The authors declare no competing interests.
Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
## Rights and permissions
Reprints and Permissions
Luk’yanchuk, I., Razumnaya, A., Sené, A. et al. The ferroelectric field-effect transistor with negative capacitance. npj Comput Mater 8, 52 (2022). https://doi.org/10.1038/s41524-022-00738-2
• Accepted:
• Published:
• DOI: https://doi.org/10.1038/s41524-022-00738-2
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# P-channel MOSFET switch - very high Rdson
I am trying test my p-channel mosfet (IRF9Z34) high side switch on load. On circuit simulator I get the result I expect, Rdson (VM1) is low and load current is 1.4A. (In the target system the gate will be controlled, here the mosfet is always on for tests). T1 - P-MOSFET IRF9Z34.
In the real circuit current is max 0.9A, so I have very big Rdson, I don't understand why. Vgs > Vth, The power supply has a current effiency > 2A.
• What is the $V_{gs}$ used to specify the $R_{ds}(on)$ in the datasheet for your MOSFET, and what $V_{gs}$ are you applying? Apr 14 '20 at 18:00
• It looks like the 5V is too low to fully open this MOSFET
– G36
Apr 14 '20 at 18:01
If your voltage is at least 10V you're guaranteed to have < 140 m$$\\Omega\$$ Rds(on) with the junction at room temperature (more when hot).
At Vgs(th) - which could be as high as 4V, you are guaranteed to have at least 250uA flowing so equivalent resistance of 16k$$\\Omega\$$.
For example, an AO3401A has less than 60m$$\\Omega\$$ Rds(on) with a 4.5V drive voltage.
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# Notation of a manifold
Gold Member
what's the notation of a manifold?
## Answers and Replies
lethe
Originally posted by loop quantum gravity
what's the notation of a manifold?
notation? what do you mean?
i often just use the symbol M to denote a manifold.
Gold Member
it's good enough for me.
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# What is the formality behind passing from Number Fields to Number Rings
In algebraic number theory, one constructs for each number field $$K$$ a subring $$\mathcal{O}_K$$, the integral closure of $$\mathbb{Z}$$ inside $$K$$, which carries many of the properties which make $$\mathbb{Z}$$ a nice ring- it is a Dedekind Domain.
Hence, for a number field $$K$$, we have assigned an integral $$\mathcal{O}_K$$-algebra to each finite étale $$K$$-algebra. I am looking to see if there is some categorical formality behind this construction. More precisely, given a number ring $$K$$, what formal operation are we doing to the category of finite étale-$$K$$-algebras so as to pass to the category of $$\mathcal{O}_K$$-algebras?
By the way, one way of describing this assignment of $$\mathcal{O}_L$$, with $$\mathcal{O}_K \subset K$$ given, is to stipulate that (i) $$\phi (\mathcal{O}_L ) = \mathcal{O}_L$$ for each $$K$$-algebra map $$\phi : L \rightarrow L$$, (ii) $$\mathcal{O}_L \cap K = \mathcal{O}_K$$, and (iii) $$\mathcal{O}_L$$ is maximal as such. This is enough to show that $$\mathcal{O}_L$$ is the ring of integers of $$L$$ when $$L$$ is Galois over $$K$$. A modification gives a similar result for non Galois extensions. This may give some kind of hint.
• Are you aware of the geometric interpretation? A field corresponds to a birational equivalence class of schemes. In the case of a number field, the ring of integers corresponds to the unique smooth (or equivalently normal) affine representative of this equivalence class. – Tim Campion Jan 20 at 16:01
• Wow, that is perfect. I was not aware of this. Could you please provide a reference for reading about this? – Dean Young Jan 20 at 16:11
• It is not smooth over $Spec(\mathbb{Z})$ due to ramification. But it is the unique regular model. – Daniel Loughran Jan 20 at 17:15
• The unique regular model, please. – Dean Young Jan 20 at 22:40
• The morphism $Spec(\mathcal O_K) \to Spec(\mathbb Z)$ is finite. But a finite morphism between normal schemes is smooth iff it is etale iff it is unramified. – Daniel Loughran Jan 22 at 9:34
Let me expand on my comment, with thanks to Daniel Loughran for corrections.
Of course,
If $$K$$ is a number field, then $$\mathcal{O}_K$$ is the unique ring with the following universal property:
1. $$\mathcal O_K$$ is an integral domain.
2. The fraction field of $$\mathcal O_K$$ is $$K$$
3. $$\mathcal O_K$$ is integrally closed in $$K$$
4. Any other integral domain $$R$$ with field of fractions $$K$$ in which $$R$$ is integrally closed, the canonical inclusion $$\mathcal O_K \to K$$ factors uniquely through the canonical inclusion $$R \to K$$.
Seeking a geometric interpretation of this statement, we need a geometric interpretation of each of the above properties.
1. A commutative ring $$R$$ is an integral domain if and only if $$Spec(R)$$ is an irreducible scheme.
2. Any irreducible scheme $$X$$ has a field of fractions $$k(X)$$, and if $$X,Y$$ are irreducible schemes, then a field homomorphism $$k(X) \to k(Y)$$ is known as a birational morphism. These are closed under composition and so form a category; isomorphism in this category is known as birational equivalence. This is a loosening of the notion of isomorphism of schemes; schemes which are birationally equivalent are "sort of the same, but not exactly". Given a field $$K$$, an irreducible scheme $$X$$ with $$k(X) = K$$ is known as a model of $$K$$. So $$X$$ is a model of $$K$$ if and only if $$K$$ is the stalk of the structure sheaf of $$X$$ at the generic point.
3. A normal scheme is "scheme language for being integrally closed". That is, a scheme is defined to be normal if and only if all of its local rings are integrally closed (in their fields of fractions). For affine schemes, this definition "globalizes": if $$R$$ is an integral domain, then $$Spec(R)$$ is normal if and only if $$R$$ is integrally closed (in its field of fractions).
4. Thus, $$Spec(\mathcal O_K)$$ is a normal model of $$K$$ with the following universal property: for any other normal model $$X$$ of $$K$$, the canonical morphism $$Spec(K) \to Spec(\mathcal O_K)$$ factors uniquely through the canonical morphism $$Spec(K) \to X$$.
Now, what is the relationship of normality to smoothness? In algebraic geometry, there are basically two notions of "smoothness". One is the relative notion of a smooth morphism. The other is the absolute notion of a regular scheme. The connection is that if $$X$$ is a scheme defined over a perfect field $$k$$, then the map $$X \to Spec(k)$$ is smooth if and only if $$X$$ is a regular scheme. But we are working over $$Spec(\mathbb Z)$$ and $$\mathbb Z$$ is not a field, so this is not true in our case.
Anyway, as explained on the wikipedia page for normal scheme, Zariski showed that a scheme is normal if and only if it is regular outside a subset of codimension at least two. So normality is a "weak smoothness condition". But since $$Spec(\mathcal O_K)$$ only has dimension one, the only codimension two set is the empty set. Thus in this case, normality coincides with regularity, and in the above characterization, "regular" can be replaced by "normal".
• A small note (because I'm that pedantic): the equivalence you cite for smoothness iff regularity works only if $X$ is of finite type over $k$ (otherwise take any field extension of $k$ :)) – Denis Nardin Jan 21 at 22:15
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Related Searches
Definitions
# Feynman diagram
In quantum field theory a Feynman diagram is an intuitive graphical representation of the transition amplitude or other physical quantity of a quantum system.
Within the canonical formulation of quantum field theory a Feynman diagram represents a term in the Wick's expansion of the perturbative S-matrix. The transition amplitude is the matrix element of the S-matrix between the initial and the final states of the quantum system.
Alternatively the path integral formulation of quantum field theory represents the transition amplitude as a weighted sum of all possible paths of the system from the initial to the final state. A Feynman diagram is then identified with a particular path of the system contributing to the transition amplitude.
Feynman diagrams are named after Richard Feynman.
## Motivation and history
When calculating scattering cross sections in particle physics, the interaction between particles can be described by starting from a free field which describes the incoming and outgoing particles, and including an interaction Hamiltonian to describe how the particles deflect one another. The amplitude for scattering is the sum of each possible interaction history over all possible intermediate particle states. The number of times the interaction Hamiltonian acts is the order of the perturbation expansion, and the time-dependent perturbation theory for fields is known as the Dyson series. When the intermediate states at intermediate times are energy eigenstates, collections of particles with a definite momentum, the series is called old-fashioned perturbation theory.
The Dyson series can be alternately rewritten as a sum over Feynman diagrams, where at each interaction vertex both the energy and momentum are conserved, but where the length of the energy momentum four vector is not equal to the mass. The Feynman diagrams are much easier to keep track of than old-fasioned terms, because the old-fasioned way treats the particle and antiparticle contributions as separate. Each Feynman diagram is the sum of exponentially many old fasioned terms, because each internal line can separately represent either a particle or an antiparticle. In a non-relativistic theory, there are no antiparticles and there is no doubling, so each Feynman diagram includes only one term.
Feynman gave a prescription for calculating the amplitude for any given diagram from a field theory Lagrangian, the Feynman rules. Each internal line corresponds to a factor of the corresponding virtual particle's propagator; each vertex where lines meet gives a factor derived from an interaction term in the Lagrangian, and incoming and outgoing lines carry an energy, momentum, and spin.
In addition to their value as a mathematical tool, Feynman diagrams provide deep physical insight into the nature of particle interactions. Particles interact in every way available; in fact, intermediate virtual particles are allowed to propagate faster than light. The probability of each final state is then obtained by summing over all such possibilities. This is closely tied to the functional integral formulation of quantum mechanics, also invented by Feynman–see path integral formulation.
The naïve application of such calculations often produces diagrams whose amplitudes are infinite, because the short-distance particle interactions require a careful limiting procedure, to include particle self-interactions. The technique of renormalization, pioneered by Feynman, Schwinger, and Tomonaga compensates for this effect and eliminates the troublesome infinite terms. After such renormalization, calculations using Feynman diagrams often match experimental results with very good accuracy.
Feynman diagram and path integral methods are also used in statistical mechanics.
### Alternative names
Murray Gell-Mann always referred to Feynman diagrams as Stückelberg diagrams, after a Swiss physicist, Ernst Stückelberg, who devised a similar notation many years earlier. Stückelberg was motivated by the need for a manifestly covariant formalism for quantum field theory, but did not provide as automated a way to handle symmetry factors and loops, although he was first to find the correct physical interpretation in terms of forward and backward in time particle paths, all without the path-integral. Historically they were sometimes called Feynman-Dyson diagrams or Dyson graphs, because when they were introduced the path integral was unfamiliar, and Freeman Dyson's derivation from old-fashioned perturbation theory was easier for physicists trained in earlier methods to follow.
## Description
A Feynman diagram represents a perturbative contribution to the amplitude of a quantum transition from some initial quantum state to some final quantum state.
For example, in the process of electron-positron annihilation the initial state is one electron and one positron, the final state — two photons.
The initial state is often assumed to be at the right of the diagram and the final state — at the left (although other conventions are also used quite often).
A Feynman diagram consists of points, called vertexes, and lines attached to the vertexes.
The particles in the initial state are depicted by lines sticking out in the direction of the initial state (e.g. to the right), the particles in the final state are represented by lines sticking out in the direction of the final state (e.g. to the left).
In QED there are two types of particles: electrons/positrons (called fermions) and photons (called gauge bosons). They are represented in Feynman diagrams as follows:
1. Electron in the initial state is represented by a solid line with an arrow pointing toward the vertex (•←).
2. Electron in the final state is represented by a line with an arrow pointing away from the vertex: (←•).
3. Positron in the initial state is represented by a solid line with an arrow pointing away from the vertex: (•→).
4. Positron in the final state is represented by a line with an arrow pointing toward the vertex: (→•).
5. Photon in the initial and the final state is represented by a wavy line (•~ and ~•).
In a gauge theory (of which QED is a fine example) a vertex always has three lines attached to it: one bosonic line, one fermionic line with arrow toward the vertex, and one fermionic line with arrow away from the vertex.
The vertexes might be connected by a bosonic or fermionic propagator. A bosonic propagator is represented by a wavy line connecting two vertexes (•~•). A fermionic propagator is represented by a solid line (with an arrow in one or another direction) connecting two vertexes, (•←•).
The number of vertexes gives the order of the term in the perturbation series expansion of the transition amplitude.
$e^+e^-to2gamma$
`$gamma$ $e^-$`
` ~•←`
` ↓`
` ~•→`
`$gamma$ $e^+$`
For example, this second order Feynman diagram contributes to a process (called electron-positron annihilation) where in the initial state (at the right) there is one electron (•←) and one positron (•→) and in the final state (at the left) there are two photons (~•).
## Canonical quantization formulation
### Perturbative S-matrix
The probability amplitude for a transition of a quantum system from the initial state $|irangle$ to the final state $|frangle$ is given by the matrix element
$S_\left\{fi\right\}=langle f|S|irangle;,$
where $S$ is the S-matrix.
In the canonical quantum field theory the S-matrix is represented within the interaction picture by the perturbation series in the powers of the interaction Lagrangian,
$S=sum_\left\{n=0\right\}^\left\{infty\right\}\left\{i^nover n!\right\}intprod_\left\{j=1\right\}^n d^4 x_j Tprod_\left\{j=1\right\}^n L_v\left(x_j\right)equivsum_\left\{n=0\right\}^\left\{infty\right\}S^\left\{\left(n\right)\right\};,$
where $L_v$ is the interaction Lagrangian and $T$ signifies the time-product of operators.
A Feynman diagram is a graphical representation of a term in the Wick's expansion of the time product in the $n$-th order term $S^\left\{\left(n\right)\right\}$ of the S-matrix,
$Tprod_\left\{j=1\right\}^nL_v\left(x_j\right)=sum_\left\{mathrm\left\{all;possible;contractions\right\}\right\}\left(pm\right)Nprod_\left\{j=1\right\}^nL_v\left(x_j\right);,$
where $N$ signifies the normal-product of the operators and $\left(pm\right)$ takes care of the possible sign change when commuting the fermionic operators to bring them together for a contraction (a propagator).
### Feynman rules
The diagrams are drawn according to the Feynman rules which depend upon the interaction Lagrangian. For the QED interaction lagrangian, $L_v=-gbarpsigamma^mupsi A_mu$, describing the interaction of a fermionic field $psi$ with a bosonic gauge field $A_mu$, the Feynman rules can be formulated in coordinate space as follows:
1. Each integration coordinate $x_j$ is represented by a point (sometimes called a vertex);
2. A bosonic propagator is represented by a curvy line connecting two points;
3. A fermionic propagator is represented by a solid line connecting two points;
4. A bosonic field $A_mu\left(x_i\right)$ is represented by a curvy line attached to the point $x_i$;
5. A fermionic field $psi\left(x_i\right)$ is represented by a solid line attached to the point $x_i$ with an arrow toward the point;
6. A fermionic field $barpsi\left(x_i\right)$ is represented by a solid line attached to the point $x_i$ with an arrow from the point;
### Example: second order processes in QED
The second order perturbation term in the S-matrix is
$S^\left\{\left(2\right)\right\}=\left\{\left(ie\right)^2over 2!\right\}int d^4x d^4x\text{'} Tbarpsi\left(x\right)gamma^mupsi\left(x\right)A_mu\left(x\right)barpsi\left(x\text{'}\right)gamma^nupsi\left(x\text{'}\right)A_nu\left(x\text{'}\right);$
#### Scattering of fermions
```
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The Wick's expansion of the integrand gives (among others) the following term$Nbarpsi\left(x\right)gamma^mupsi\left(x\right)barpsi\left(x\text{'}\right)gamma^nupsi\left(x\text{'}\right)underline\left\{A_mu\left(x\right)A_nu\left(x\text{'}\right)\right\};,$where$underline\left\{A_mu\left(x\right)A_nu\left(x\text{'}\right)\right\}=int\left\{d^4pover\left(2pi\right)^4\right\}\left\{ig_\left\{munu\right\}over k^2+i0\right\}e^\left\{-k\left(x-x\text{'}\right)\right\}$is the electromagnetic contraction (propagator) in the Feynman gauge. This term is represented by the Feynman diagram at the right. This diagram gives contributions to the following processes: $e^-e^-$ scattering (initial state at the right, final state at the left of the diagram);
$e^+e^+$ scattering (initial state at the left, final state at the right of the diagram);
$e^-e^+$ scattering (initial state at the bottom/top, final state at the top/bottom of the diagram). Compton scattering and annihilation/generation of $e^-e^+$ pairs Another interesting term in the expansion is$Nbarpsi\left(x\right)gamma^muunderline\left\{psi\left(x\right)barpsi\left(x\text{'}\right)\right\}gamma^nupsi\left(x\text{'}\right)A_mu\left(x\right)A_nu\left(x\text{'}\right);,$where$underline\left\{psi\left(x\right)barpsi\left(x\text{'}\right)\right\}=int\left\{d^4kover\left(2pi\right)^4\right\}\left\{iover gamma p-m+i0\right\}e^\left\{-p\left(x-x\text{'}\right)\right\}$is the fermionic contraction (propagator).
Path integral formulation In a path-integral, the field Lagrangian, integrated over all possible field histories, defines the probability amplitude to go from one field configuration to another. In order to make sense, the field theory should have a good ground state, and the integral should be performed a little bit rotated into imaginary time.
Scalar Field Lagrangian A simple example is the free relativistic scalar field in d-dimensions, whose action integral is:
$S = int \left\{1over 2\right\} partial_mu phi partial^mu phi d^dx$The probability amplitude for a process is:$int_A^B e^\left\{iS\right\} Dphi$where A and B are space-like hypersurfaces which define the boundary conditions. The collection of all the $phi\left(A\right)$ on the starting hypersurface give the initial value of the field, analogous to the starting position for a point particle, and the field values $phi\left(B\right)$ at each point of the final hypersurface defines the final field value, which is allowed to vary, giving a different amplitude to end up at different values. This is the field-to-field transition amplitude.The path integral gives the expectation value of operators between the initial and final state:$int_A^B e^\left\{iS\right\} phi\left(x_1\right) ... phi\left(x_n\right) Dphi = langle A| phi\left(x_1\right) ... phi\left(x_n\right) |B rangle$and in the limit that A and B recede to the infinite past and the infinite future, the only contribution that matters is from the ground state (this is only rigorously true if the path-integral is defined slightly rotated into imaginary time). The path integral should be thought of as analogous to a probability distribution, and it is convenient to define it so that multiplying by a constant doesn't change anything:$\left\{int e^\left\{iS\right\} phi\left(x_1\right) ... phi\left(x_n\right) Dphi over int e^\left\{iS\right\} Dphi \right\} = langle 0 | phi\left(x_1\right) .... phi\left(x_n\right) |0rangle$The normalization factor on the bottom is called the partition function for the field, and it coincides with the statistical mechanical partition function at zero temperature when rotated into imaginary time.The initial-to-final amplitudes are ill-defined if you think of things in the continuum limit right from the beginning, because the fluctuations in the field can become unbounded. So the path-integral should be thought of as on a discrete square lattice, with lattice spacing $a$ and the limit $arightarrow 0$ should be taken carefully. If the final results do not depend on the shape of the lattice or the value of a, then the continuum limit exists.On a lattice, the field can be expanded in Fourier modes:
phi(x) = int {dkover (2pi)^d} phi(k) e^{ikcdot x} = int_k phi(k) e^{ikx}
Where the integration domain is over k restricted to a cube of side length $2pi/a$, so that large values of k are not allowed. It is important to note that the k measure contains the factors of $2pi$ from Fourier transforms, this is the best standard convention for k integrals in QFT. The lattice means that fluctuations at large k are not allowed to contribute right away, they only start to contribute in the limit $arightarrow 0$. Sometimes, instead of a lattice, the field modes are just cut off at high values of k instead.It is also convenient from time to time to consider the space-time volume to be finite, so that the k modes are also a lattice. This is not strictly as necessary as the space-lattice limit, because interactions in k are not localized, but it is convenient for keeping track of the factors in front of the k-integrals and the momentum-conserving delta functions which will arise.On a lattice, the action needs to be discretized:
where means that x and y are nearest lattice neighbors. The discretization should be thought of as defining what the derivative $partial_mu phi$ means.In terms of the lattice Fourier modes, the action can be written:
S= int_k ((1-cos(k_1)) +(1-cos(k_2)) + ... + (1-cos(k_d)) )phi^*_k phi^k
Which for k near zero is:
S = int_k {1over 2} k^2 |phi(k)|^2
Which is the continuum Fourier transform of the original action. In finite volume, the quantity $d^dk$ is not infinitesimal, but becomes the volume of a box made by neighboring Fourier modes, or $\left(2pi/V\right)^d$.The field $phi$ is real valued, so the Fourier transform obeys:$phi\left(k\right)^* = phi\left(-k\right),$In terms of real and imaginary parts, the real part of $phi\left(k\right)$ is an even function of k, while the imaginary part is odd. The Fourier transform avoids double-counting, so that it can be written:$S = int_k \left\{1over 2\right\} k^2 phi\left(k\right) phi\left(-k\right)$over an integration domain which integrates over each pair (k,-k) exactly once.For a complex scalar field with action:$S = int \left\{1over 2\right\} partial_muphi^* partial^muphi d^dx$The Fourier transform is unconstrained:$S = int_k \left\{1over 2\right\} k^2 |phi\left(k\right)|^2$and the integral is over all k.Integrating over all different values of $phi\left(x\right)$ is equivalent to integrating over all Fourier modes, because taking a Fourier transform is a unitary linear transformation of field coordinates. When you change coordinates in a multidimensional integral by a linear transformation, the value of the new integral is given by the determinant of the transformation matrix. If
$y_i = A_\left\{ij\right\} x_j,$Then
det(A) int dx_1 dx_2 ... dx_n = int dy_1 dy_2 ... dy_n
If A is a rotation, then
A^T A = I
,
so that $det A = pm 1$, and the sign depends on whether the rotation includes a reflection or not.The matrix which changes coordinates from $phi\left(x\right)$ to $phi\left(k\right)$ can be read off from the definition of a Fourier transform.$A_\left\{kx\right\} = e^\left\{ikx\right\} ,$and the Fourier inversion theorem tells you the inverse:$A^\left\{-1\right\}_\left\{kx\right\} = e^\left\{-ikx\right\} ,$which is the complex conjugate-transpose, up to factors of $2pi$. On a finite volume lattice, the determinant is nonzero and independent of the field values.
$det A = 1 ,$and the path integral is a separate factor at each value of k.$int exp\left(i sum_k phi^*\left(k\right) phi\left(k\right)\right) Dphi = prod_k int_\left\{phi_k\right\} e^\left\{\left\{iover 2\right\} k^2 |phi_k|^2 d^dk \right\} ,$and each separate factor is an oscillatory Gaussian.In imaginary time, the Euclidean action' becomes positive definite, and can be interpreted as a probability distribution. The probability of a field having values $phi_k$ is
$e^\left\{int_k - \left\{1over 2\right\} k^2 phi^*_k phi_k\right\} = prod_k e^\left\{- k^2 |phi_k|^2 d^dk\right\}$The expectation value of the field is the statistical expectation value of the field when chosen according to the probability distribution:
langle phi(x_1) ... phi(x_n) rangle = { int e^{-S} phi(x_1) ... phi(x_n) Dphi over int e^{-S} Dphi}
Since the probability of $phi_k$ is a product, the value of $phi\left(k\right)$ at each separate value of k is independently Gaussian distributed. The variance of the Gaussian is 1/(k^2 d^dk), which is formally infinite, but that just means that the fluctuations are unbounded in infinite volume. In any finite volume, the integral is replaced by a discrete sum, and the variance of the integral is $V/k^2$.
Monte-Carlo The path integral defines a probabilistic algorithm to generate a Euclidean scalar field configuration. Randomly pick the real and imaginary parts of each Fourier mode at wavenumber k to be a gaussian random variable with variance $1/k^2$. This generates a configuration $phi_C\left(k\right)$ at random, and the Fourier transform gives $phi_C\left(x\right)$. For real scalar fields, the algorithm must generate only one of each pair $phi\left(k\right),phi\left(-k\right)$, and make the second the complex conjugate of the first.To find any correlation function, generate a field again and again by this procedure, and find the statistical average:$langle phi\left(x_1\right) ... phi\left(x_n\right) rangle = lim_\left\{|C|rightarrowinfty\right\}\left\{ sum_C phi_C\left(x_1\right) ... phi_C\left(x_n\right) over |C| \right\}$where $|C|$ is the number of configurations, and the sum is of the product of the field values on each configuration. The Euclidean correlation function is just the same as the correlation function in statistics or statistical mechanics. The quantum mechanical correlation functions are an analytic continuation of the Euclidean correlation functions.For free fields with a quadratic action, the probability distribution is a high dimensional Gaussian, and the statistical average is given by an explicit formula. But the Monte-carlo method also works well for bosonic interacting field theories where there is no closed form for the correlation functions.
Scalar Propagator Each mode is independently Gaussian distributed. The expectation of field modes is easy to calculate:for $kne k\text{'}$, since then the two gaussian random variables are independent and both have zero mean.in finite volume V, when the two k-values coincide, since this is the variance of the Gaussian. In the infinite volume limit,Strictly speaking, this is an approximation: the lattice propagator is:But near k=0, for field fluctuations long compared to the lattice spacing, the two forms coincide.It is important to emphasize that the delta functions contain factors of $2pi$, so that they cancel out the $2pi$ factors in the measure for k integrals.$delta\left(k\right) = \left(2pi\right)^d delta_D\left(k_1\right)delta_D\left(k_2\right) ... delta_D\left(k_d\right) ,$where $delta_D\left(k\right)$ is the ordinary one-dimensional Dirac delta function. This convention for delta-functions is not universal--- some authors keep the factors of $2pi$ in the delta functions (and in the k-integration) explicit.
Equation of Motion The form of the propagator can be more easily found by using the equation of motion for the field. From the Lagrangian, the equation of motion is:$partial_mu partial^mu phi = 0,$and in an expectation value, this says:
partial_mupartial^mu langle phi(x) phi(y)rangle =0
Where the derivatives act on x, and the identity is true everywhere except when x and y coincide, and the operator order matters. The form of the singularity can be understood from the canonical commutation relations to be a delta-function. Defining the (euclidean) Feynman propagator $Delta$ as the fourier transform of the time-ordered two-point function (the one that comes from the path-integral):$partial^2 Delta \left(x\right) = idelta\left(x\right),$So that:$Delta\left(k\right) = \left\{iover k^2\right\}$If the equations of motion are linear, the propagator will always be the reciprocal of the quadratic-form matrix which defines the free Lagrangian, since this gives the equations of motion. This is also easy to see directly from the Path integral. The factor of i disappears in the Euclidean theory.
Wick Theorem Because each field mode is an independent Gaussian, the expectation values for the product of many field modes obeys Wick's theorem:$langle phi\left(k_1\right) phi\left(k_2\right) ... phi\left(k_n\right)rangle$is zero unless the field modes coincide in pairs. This means that it is zero for an odd number of $phi$'s, and for an even number of phi's, it is equal to a contribution from each pair separately, with a delta function.$langle phi\left(k_1\right) ... phi\left(k_\left\{2n\right\}\right)rangle = sum prod_\left\{i,j\right\} \left\{delta\left(k_i - k_j\right) over k_i^2 \right\}$where the sum is over each partition of the field modes into pairs, and the product is over the pairs. For example,$langle phi\left(k_1\right) phi\left(k_2\right) phi\left(k_3\right) phi\left(k_4\right) rangle = \left\{delta\left(k_1 -k_2\right) over k_1^2\right\}\left\{delta\left(k_3-k_4\right)over k_3^2\right\} + \left\{delta\left(k_1-k_3\right) over k_3^2\right\}\left\{delta\left(k_2-k_4\right)over k_2^2\right\} + \left\{delta\left(k_1-k_4\right)over k_1^2\right\}\left\{delta\left(k_2 -k_3\right)over k_2^2\right\}$An intepretation of Wick's theorem is that each field insertion can be thought of as a dangling line, and the expectation value is calculated by linking up the lines in pairs, putting a delta function factor that ensures that the momentum of each partner in the pair is equal, and dividing by the propagator.
Higher Gaussian moments--- completing Wick's theorem There is a subtle point left before Wick's theorem is proved--- what if more than two of the phi's have the same momentum? If its an odd number, the integral is zero, negative values cancel with the positive values, But if the number is even, the integral is positive. The previous demonstration assumed that the phi's would only match up in pairs.But the theorem is correct even when arbitrarily many of the phis are equal, and this is a notable property of Gaussian integration:
$I = int e^\left\{-ax^2/2\right\} = sqrt\left\{2piover a\right\}$
$\left\{partial^n over partial a^n \right\} I = int \left\{x^\left\{2n\right\} over 2^n\right\} e^\left\{-ax^2\right\} = \left\{1cdot 3 cdot 5 ... cdot \left(2n-1\right) over 2 cdot 2 cdot 2 ... ;;;;;cdot 2;;;;;;\right\} sqrt\left\{2pi\right\} a^\left\{-\left\{2n+1over2\right\}\right\}$Dividing by I,$langle x^\left\{2n\right\}rangle=\left\{int x^\left\{2n\right\} e^\left\{-a x^2\right\} over int e^\left\{-a x^2\right\} \right\} = 1 cdot 3 cdot 5 ... cdot \left(2n-1\right) \left\{1over a^n\right\}$
$langle x^2 rangle = \left\{1over a\right\}$If Wick's theorem were correct, the higher moments would be given by all possible pairings of a list of 2n x's:$langle x_1 x_2 x_3 ... x_\left\{2n\right\} rangle$where the x-s are all the same variable, the index is just to keep track of the number of ways to pair them. The first x can be paired with 2n-1 others, leaving 2n-2. The next unpaired x can be paired with 2n-3 different x's leaving 2n-4, and so on. This means that Wick's theorem, uncorrected, says that the expectation value of $x^\left\{2n\right\}$ should be:$langle x^\left\{2n\right\} rangle = \left(2n-1\right)cdot\left(2n-3\right).... cdot5 cdot 3 cdot 1 \left(langle x^2rangle\right)^n$and this is in fact the correct answer. So Wick's theorem holds no matter how many of the momenta of the internal variables coincide.
Interaction
Interactions are represented by higher order contributions, since quadratic contributions are always Gaussian. The simplest interaction is the quartic self-interaction, with an action:$S = int partial^mu phi partial_muphi + \left\{lambda over 4!\right\} phi^4.$The reason for the combinatorial factor 4! will be clear soon. Writing the action in terms of the lattice (or continuum) Fourier modes:$S = int_k k^2 |phi\left(k\right)|^2 + int_\left\{k_1k_2k_3k_4\right\} phi\left(k_1\right) phi\left(k_2\right) phi\left(k_3\right)phi\left(k_4\right) delta\left(k_1+k_2+k_3 + k_4\right) = S_F + X.$Where $S_F$ is the free action, whose correlation functions are given by Wick's theorem. The exponential of S in the path integral can be expanded in powers of $lambda$, giving a series of corrections to the free action.$e^\left\{-S\right\} = e^\left\{-S_F\right\} \left(1 + X + \left\{1over 2!\right\} X X + \left\{1over 3!\right\} X X X + ... \right)$The path integral for the interacting action is then a power series of corrections to the free action. The term represented by X should be thought of as four half-lines, one for each factor of $phi\left(k\right)$. The half-lines meet at a vertex, which contributes a delta-function which ensures that the sum of the momenta are all equal.To compute a correlation function in the interacting theory, there is a contribution from the X terms now. For example, the path-integral for the four-field correlator:$langle phi\left(k_1\right) phi\left(k_2\right) phi\left(k_3\right) phi\left(k_4\right) rangle = \left\{int e^\left\{-S\right\} phi\left(k_1\right)phi\left(k_2\right)phi\left(k_3\right)phi\left(k_4\right) Dphi over Z\right\}$which in the free field was only nonzero when the momenta k were equal in pairs, is now nonzero for all values of the k. The momenta of the insertions $phi\left(k_i\right)$ can now match up with the momenta of the X's in the expansion. The insertions should also be thought of as half-lines, four in this case, which carry a momentum k, but one which is not integrated.The lowest order contribution comes from the first nontrivial term $e^\left\{-S_F\right\} X$ in the Taylor expansion of the action. Wick's theorem requires that the momenta in the X half-lines, the $phi\left(k\right)$ factors in X, should match up with the momenta of the external half-lines in pairs. The new contribution is equal to:$lambda \left\{1over k_1^2\right\} \left\{1over k_2^2\right\} \left\{1over k_3^2\right\} \left\{1over k_4^2\right\}.$The 4! inside X is canceled because there are exactly 4! ways to match the half-lines in X to the external half-lines. Each of these different ways of matching the half-lines together in pairs contributes exactly once, regardless of the values of the k's, by Wick's theorem.
Feynman Diagrams The expansion of the action in powers of X gives a series of terms with progressively higher number of X's. The contribution from the term with exactly n X's are called n-th order.The n-th order terms has: 4n internal half-lines, which are the factors of $phi\left(k\right)$ from the X's. These all end on a vertex, and are integrated over all possible k.
external half-lines, which are the come from the $phi\left(k\right)$ insertions in the integral.
By Wick's theorem, each pair of half-lines must be paired together to make a line, and this line gives a factor of$delta\left(k_1 + k_2\right) over k_1^2$which multiplies the contribution. This means that the two half-lines that make a line are forced to have equal and opposite momentum. The line itself should be labelled by an arrow, drawn parallel to the line, and labeled by the momentum in the line k. The half-line at the tail end of the arrow carries momentum k, while the half-line at the head-end carries momentum -k. If one of the two half-lines is external, this kills the integral over the internal k, since it forces the internal k to be equal to the external k. If both are internal, the integral over k remains.The diagrams which are formed by linking the half-lines in the X's with the external half-lines, representing insertions, are the Feynman diagrams of this theory. Each line carries a factor of $1over k^2$, the propagator, and either goes from vertex to vertex, or ends at an insertion. If it is internal, it is integrated over. At each vertex, the total incoming k is equal to the total outgoing k.The number of ways of making a diagram by joining half-lines into lines almost completely cancels the factorial factors coming from the Taylor series of the exponential and the 4! at each vertex.
Loop Order A tree diagram is one where all the internal lines have momentum which is completely determined by the external lines and the condition that the incoming and outgoing momentum are equal at each vertex. The contribution of these diagrams is a product of propagators, without any integration.An example of a tree diagram is the one where each of four external lines end on an X. Another is when eight external lines end on two X's. A third is when three external lines end on an X, and the remaining half-line joins up with another X, and the remaining half-lines of this
X run off to external lines.It is easy to verify that in all these cases, the momenta on all the internal lines is determined by the external momenta and the condition of momentum conservation in each vertex.A diagram which is not a tree diagram is called a loop diagram, and an example is one where two lines of an X are joined to external lines, while the remaining two lines are joined to each other. The two lines joined to each other can have any momentum at all, since they both enter and leave the same vertex. A more complicated example is one where two X's are joined to each other by matching the legs one to the other. This diagram has no external lines at all.The reason loop diagrams are called loop diagrams is because the number of k-integrals which are left undetermined by momentum conservation is equal to the number of independent closed loops in the diagram, where independent loops are counted as in homology theory. The homology is real-valued (actually R^d valued), the value associated with each line is the momentum. The boundary operator takes each line to the sum of the end-vertices with a positive sign at the head and a negative sign at the tail. The condition that the momentum is conserved is exactly the condition that the boundary of the k-valued weighted graph is zero.A set of k-values can be relabeled whenever there is a closed loop going from vertex to vertex, never revisiting the same vertex. Such a cycle can be thought of as the boundary of a 2-cell. The k-labelings of a graph which conserve momentum (which have zero boundary) up to redefinitions of k (up to boundaries of 2-cells) define the first homology of a graph. The number of independent momenta which are not determined is then equal to the number of independent homology loops. For many graphs, this is equal to the number of loops as counted in the most intuitive way.
Symmetry factors The number of ways to form a given Feynman diagram by joining together half-lines is large, and by Wick's theorem, each way of pairing up the half-lines contributes equally. Often, this completely cancels the factorials in the denominator of each term, but the cancellation is sometimes incomplete.The uncancelled denominator is called the symmetry factor of the diagram. The contribution of each diagram to the correlation function must be divided by its symmetry factor.For example, consider the Feynman diagram formed from two external lines joined to one X, and the remaining two half-lines in the X joined to each other. There are 4*3 ways to join the external half-lines to the X, and then there is only one way to join the two remaining lines to each other. The X comes divided by 4!=4*3*2, but the number of ways to link up the X half lines to make the diagram is only 4*3, so the contribution of this diagram is divided by two.For another example, consider the diagram formed by joining all the half-lines of one X to all the half-lines of another X. This diagram is called a vacuum bubble, because it does not link up to any external lines. There are 4! ways to form this diagram, but the denominator includes a 2! (from the expansion of the exponential, there are two X's) and two factors of 4!. The contribution is multiplied by 4!/(2*4!*4!) = 1/48.Another example is the Feynman diagram formed from two X's where each X links up to two external lines, and the remaining two half-lines of each X are joined to each other. The number of ways to link an X to two external lines is 4*3, and either X could link up to either pair, giving an additional factor of 2. The remaining two half-lines in the two X's can be linked to each other in two ways, so that the total number of ways to form the diagram is 4*3*4*3*2*2, while the denominator is 4!4!2!. The total symmetry factor is 2, and the contribution of this diagram is divided by two.The symmetry factor theorem gives the symmetry factor for a general diagram: the contribution of each Feynman diagram must be divided by the order of its group of automorphisms, the number of symmetries that it has.An automorphism of a Feynman graph is a permutation M of the lines and a permutation N of the vertices with the following properties:
If a line l goes from vertex v to vertex v', then M(l) goes from N(v) to N(v'). If the line is undirected, as it is for a real scalar field, then M(l) can go from N(v') to N(v) too.
If a line l ends on an external line, M(l) ends on the same external line.
If there are different types of lines, M(l) should preserve the type.
This theorem has an interpretation in terms of particle-paths: when identical particles are present, the integral over all intermediate particles must not double-count states which only differ by interchanging identical particles.Proof: To prove this theorem, label all the internal and external lines of a diagram with a unique name. Then form the diagram by linking the a half-line to a name and then to the other half line.Now count the number of ways to form the named diagram. Each permutation of the X's gives a different pattern of linking names to half-lines, and this is a factor of n!. Each permutation of the half-lines in a single X gives a factor of 4!. So a named diagram can be formed in exactly as many ways as the denominator of the Feynman expansion.But the number of unnamed diagrams is smaller than the number of named diagram by the order of the automorphism group of the graph.
A diagram is connected when it is connected as a graph, meaning that there is a sequence of attached lines and vertices which link any line or vertex to any other. The connected diagrams suffice to reconstruct the full Feynman series, and this is the linked cluster theorem.The full series is the sum over all diagrams, which include several connected components, each one can occur multiple times. The automorphism of the full graph consists of the automorphisms of the connected components, and an extra factor of n! for permutations of n identical copies of one connected component.$sum prod_i \left\{C_\left\{i\right\}^\left\{n_i\right\} over n_i!\right\}$But this can be seen to be a product of separate factors for each connected graph:$prod_i sum_j \left\{C_i^\left\{n_i\right\} over n_i!\right\} = prod_i exp\left(C_i\right) = exp\left(sum_i C_i\right).$This is the linked cluster theorem: the sum of all diagrams is the exponential of the connected ones.
Vacuum Bubbles
An immediate consequence of the linked-cluster theorem is that all vacuum bubbles, diagrams without external lines cancel when calculating correlation functions. A correlation function is given by a ratio of path-integrals:$langle phi_1\left(x_1\right) ... phi_n\left(x_n\right)rangle = \left\{int e^\left\{-S\right\} phi_1\left(x_1\right) ...phi_n\left(x_n\right) Dphi over int e^\left\{-S\right\} Dphi\right\}.$The top is the sum over all Feynman diagrams, including disconnected diagrams which do not link up to external lines at all. In terms of the connected diagrams, the numerator includes the same contributions of vacuum bubbles as the denominator:$int e^\left\{-S\right\}phi_1\left(x_1\right)...phi_n\left(x_n\right)rangle = \left(sum E_i\right)\left(exp\left(sum_i C_i\right) \right).$Where the sum over E diagrams includes only those diagrams each of whose connected components end on at least on external line. The vacuum bubbles are the same whatever the external lines, and give an overall multiplicative factor. The denominator is the sum over all vacuum bubbles, and dividing gets rid of the second factor.The vacuum bubbles then are only useful for determining Z itself, which from the definition of the path integral is equal to:$Z= int e^\left\{-S\right\} Dphi = e^\left\{-HT\right\} = e^\left\{-rho V\right\}$where $rho$ is the energy density in the vacuum. Each vacuum bubble contains a factor of $delta\left(k\right)$ zeroing the total k at each vertex, and when there are no external lines, this contains a factor of $delta\left(0\right)$, because the momentum conservation is over-enforced. In finite volume, this factor can be identified as the total volume of space time. Dividing by the volume, the remaining integral for the vacuum bubble has an interpretation: it is a contribution to the energy density of the vacuum.
Sources Correlation functions are the sum of the connected Feynman diagrams, but the formalism treats the connected and disconnected diagrams differently. Internal lines end on vertices, while external lines go off to insertions. Introducing sources unifies the formalism, by making new vertices where one line can end.Sources are external fields, fields which contribute to the action, but are not dynamical variables. A scalar field source is another scalar field h which contributes a term to the (Lorentz) Lagrangian:$int h\left(x\right) phi\left(x\right) d^dx = int h\left(k\right) phi\left(k\right) d^dk ,$In the Feynman expansion, this contributes H terms with one half-line ending on a vertex. Lines in a Feynman diagram can now end either on an X vertex, or on an H-vertex, and only one line enters an H vertex. The Feynman rule for an H-vertex is that a line from an H with momentum k gets a factor of h(k).The sum of the connected diagrams in the presence of sources includes a term for each connected diagram in the absence of sources, except now the diagrams can end on the source. Traditionally, a source is represented by a little "x" with one line extending out, exactly as an insertion.$log\left(Z\left[h\right]\right) = sum_\left\{n,C\right\} h\left(k_1\right) h\left(k_2\right) ... h\left(k_n\right) C\left(k_1,...,k_n\right),$where $C\left(k_1,....,k_n\right)$ is the connected diagram with n external lines carrying momentum as indicated. The sum is over all connected diagrams, as before.The field h is not dynamical, which means that there is no path integral over h: h is just a parameter in the Lagrangian which varies from point to point. The path integral for the field is:$Z\left[h\right] = int e^\left\{iS + iint hphi\right\} Dphi ,$and it is a function of the values of h at every point. One way to interpret this expression is that it is taking the Fourier transform in field space. If there is a probability density on R^n, the Fourier transform of the probability density is:$int rho\left(y\right) e^\left\{i k y\right\} d^n y = langle e^\left\{i k y\right\} rangle = langle prod_\left\{i=1\right\}^\left\{n\right\} e^\left\{h_i y_i\right\}rangle ,$The fourier transform is the expectation of an oscillatory exponential. The path integral in the presence of a source h(x) is:$Z\left[h\right] = int e^\left\{iS\right\} e^\left\{iint_x h\left(x\right)phi\left(x\right)\right\} Dphi = langle e^\left\{i h phi \right\}rangle$which, on a lattice, is the product of an oscillatory exponential for each field value:$langle prod_x e^\left\{i h_x phi_x\right\}rangle$The fourier transform of a delta-function is a constant, which gives a formal expression for a delta function:$delta\left(x-y\right) = int e^\left\{k\left(x-y\right)\right\} dk$This tells you what a field delta function looks like in a path-integral. For two scalar fields $phi$ and $eta$,$delta\left(phi - eta\right) = int e^\left\{ i h\left(x\right)\left(phi\left(x\right) -eta\left(x\right)d^dx\right\} Dh$Which integrates over the Fourier transform coordinate, over h. This expression is useful for formally changing field coordinates in the path integral, much as a delta function is used to change coordinates in an ordinary multi-dimensional integral.The partition function is now a function of the field h, and the physical partition function is the value when h is the zero function:The correlation functions are derivatives of the path integral with respect to the source:$langlephi\left(x\right)rangle = \left\{1over Z\right\} \left\{partial over partial h\left(x\right)\right\} Z\left[h\right] = \left\{partialoverpartial h\left(x\right)\right\} log\left(Z\left[h\right]\right)$
Spin 1/2: Grassman integrals The preceding discussion can be extended to the Fermi case, but only if the notion of integration is expanded.
Particle-Path Interpretation A Feynman diagram is a representation of quantum field theory processes in terms of particle paths.In a Feynman diagram, particles are represented by lines, which can be squiggly or straight, with an arrow or without, depending on the type of particle. A point where lines connect to other lines is referred to as an interaction vertex, or vertex for short. There are three different types of lines: internal lines connect two vertices, incoming lines extend from "the past" to a vertex and represent an initial state, and outgoing lines extend from a vertex to "the future" and represent the final state.There are several conventions for where to represent the past and the future. Sometimes, the bottom of the diagram represents the past and the top of the diagram represents the future. Other times, the past is to the left and the future to the right. When calculating correlation functions instead of scattering amplitudes, there is no past and future and all the lines are internal. Then the particle lines begin and end on small x's, which represent the positions of the operators whose correlation is being calculated. The LSZ reduction formula is the standardized argument that shows that the correlation functions and scattering diagrams are the same.Feynman diagrams are a pictorial representation of a contribution to the total amplitude for a process which can happen in several different ways. When a group of incoming particles are to scatter off each other, the process can be thought of as one where the particles travel over all possible paths, including paths that go backward in time. In a perturbative expansion of the scattering amplitude for the experiment defined by the incoming and outgoing lines. In some quantum field theories (notably quantum electrodynamics), one can obtain an excellent approximation of the scattering amplitude from a few terms of the perturbative expansion, corresponding to a few simple Feynman diagrams with the same incoming and outgoing lines connected by different vertices and internal lines.The method, although originally invented for particle physics, is useful in any part of physics where there are statistical or quantum fields. In condensed matter physics, there are many-body Feynman diagrams with dashed lines which represent an instantaneous potential interaction, while phonons take the place of photons. In statistical physics, there are statistical Feynman diagrams which represent the way in which correlations travel along paths.Feynman diagrams are often confused with spacetime diagrams and bubble chamber images because they all seek to represent particle scattering. Feynman diagrams are graphs that represent the trajectories of particles in intermediate stages of a scattering process. Unlike a bubble chamber picture, only the sum of all the Feynman diagrams represent any given particle interaction; particles do not choose a particular diagram each time they interact. The law of summation is in accord with the principle of superposition--- every diagram contributes a factor to the total amplitude for the process.
Scattering The correlation functions of a quantum field theory describe the scattering of particles. The definition of "particle" in relativistic field theory is not self-evident, because if you try to determine the position so that the uncertainty is less than the compton wavelength, the uncertainty in energy is large enough to produce more particles and antiparticles of the same type from the vacuum. This means that the notion of a single-particle state is to some extent incompatible with the notion of an object localized in space.In the 1930's, Wigner gave a mathematical definition for single-particle states: they are a collection of states which form an irreducible representation of the Poincare group. Single particle states describe an object with a finite mass, a well defined momentum, and a spin. This definition is fine for protons and neutrons, electrons and photons, but it excludes quarks, which are permanently confined, so the modern point of view is more accomodating: a particle is anything whose interaction can be described in terms of Feynman diagrams, which have an interpretation as a sum over particle trajectories.A field operator can act to produce a one-particle state from the vacuum, which means that the field operator $phi\left(x\right)$ produces a superposition of Wigner particle states. In the free field theory, the field produces one particle states only. But when there are interactions, the field operator can also produce 3-particle,5-particle (if there is no +/- symmetry also 2,4,6 particle) states too. To compute the scattering amplitude for single particle states only requires a careful limit, sending the fields to infinity and integrating over space to get rid of the higher-order corrections.The relation between scattering and correlation functions is the LSZ-theorem: The scattering amplitude for n particles to go to m-particles in a scattering event is the given by the sum of the Feynman diagrams that go into the correlation function for n+m field insertions, leaving out the propagators for the external legs.For example, for the $lambda phi^4$ interaction of the previous section, the order $lambda$ contribution to the (Lorentz) correlation function is:$langle phi\left(k_1\right)phi\left(k_2\right)phi\left(k_3\right)phi\left(k_4\right)rangle = \left\{iover k_1^2\right\}\left\{iover k_2^2\right\} \left\{iover k_3^2\right\} \left\{iover k_4^2\right\} ilambda ,$Stripping off the external propagators, that is, removing the factors of $i/k^2$, gives the invariant scattering amplitude M:$M = ilambda ,$which is a constant, independent of the incoming and outgoing momentum. The interpretation of the scattering amplitude is that the sum of $|M|^2$ over all possible final states is the probability for the scattering event. The normalization of the single-particle states must be chosen carefully, however, to ensure that M is a relativistic invariant.Non-relativistic single particle states are labeled by the momentum k, and they are chosen to have the same norm at every value of k. This is because the nonrelativistic unit operator on single particle states is:$int dk |kranglelangle k|,$In relativity, the integral over k states for a particle of mass m integrates over a hyperbola in E,k space defined by the energy-momentum relation:$E^2 - k^2 = m^2 ,$If the integral weighs each k point equally, the measure is not Lorentz invariant. The invariant measure integrates over all values of k and E, restricting to the hyperbola with a Lorentz invariant delta function:$int delta\left(E^2-k^2 - m^2\right) |E,kranglelangle E,k| dE dk = int \left\{dk over 2 E\right\} |kranglelangle k|$So the normalized k-states are different from the relativistically normalized k-states by a factor of $sqrt\left\{E\right\} = \left(k^2-m^2\right)^\left\{1over 4\right\}$ The invariant amplitude M is then the probability amplitude for relativistically normalized incoming states to become relativistically normalized outgoing states.For nonrelativistic values of k, the relativistic normalization is the same as the nonrelativistic normalization (up to a constant factor $sqrt\left\{m\right\}$ ). In this limit, the $phi^4$ invariant scattering amplitude is still constant. The particles created by the field phi scatter in all directions with equal amplitude.The nonrelativistic potential which scatters in all directions with an equal amplitude (in the Born approximation) is one whose Fourier transform is constant--- a delta-function potential. The lowest order scattering of the theory reveals the non-relativistic interpretation of the this theory--- it describes a collection of particles with a delta-function repulsion. Two such particles have an aversion to occupying the same point at the same time.
LSZ theorem
Nonperturbative effects
Thinking of Feynman diagrams as a perturbation series, nonperturbative effects like tunneling do not show up, because any effect which goes to zero faster than any polynomial does not affect the Taylor series. Even bound states are absent, since at any finite order particles are only exchanged a finite number of times, and to make a bound state, the binding force must last forever.But this point of view is misleading, because the diagrams not only describe scattering, but they also are a representation of the short-distance field theory correlations. They encode not only asymptotic processes like particle scattering, they also describe the multiplication rules for fields, the operator product expansion. Nonperturbative tunneling processes involve field configurations which on average get big when the coupling constant gets small, but each configuration is a coherent superposition of particles whose local interactions are described by Feynman diagrams. When the coupling is small, these become collective processes which involve large numbers of particles, but where the interactions between each of the particles is simple. This means that nonperturbative effects show up asymptotically in resummations of infinite classes of diagrams, and these diagrams can be locally simple. The graphs determine the local equations of motion, while the allowed large-scale configurations describe non-perturbative physics. But because Feynman propagators are nonlocal in time, translating a field process to a coherent particle language is not completely intuitive, and has only been explicitly worked out in certain special cases. In the case of nonrelativistic bound states, the Bethe-Salpeter equation describes the class of diagrams to include to describe a relativistic atom. For quantum chromodynamics, the Shifman Vainshtein Zakharov sum rules describe non-perturbatively excited long-wavelength field modes in particle language, but only in a phenomenological way.The number of Feynman diagrams at high orders of perturbation theory is very large, because there are as many diagrams as there are graphs with a given number of nodes. Nonperturbative effects leave a signature on the way in which the number of diagrams and resummations diverge at high order. It is only because non-perturbative effects appear in hidden form in diagrams that it was possible to analyze nonperturbative effects in string theory, where in many cases a Feynman description is the only one available.
Mathematical details
A Feynman diagram can be considered a graph. When considering a field composed of particles, the edges will represent (sections of) particle world lines; the vertices represent virtual interactions. Since only certain interactions are permitted, the graph is constrained to have only certain types of vertices. The type of field of an edge is its field label; the permitted types of interaction are interaction labels. The value of a given diagram can be derived from the graph; the value of the interaction as a whole is obtained by summing over all diagrams.
Mathematical interpretation Feynman diagrams are really a graphical way of keeping track of deWitt indices, much like Penrose's graphical notation for indices in multilinear algebra. There are several different types for the indices, one for each field (this does not depend on how the fields are grouped; for instance, if the up quark field and down quark field are treated as different fields, then there would be the same type assigned to both of them but if they are treated as a single multicomponent field with "flavors", then there would be a problem). The edges, (i.e., propagators) are tensors of rank (2,0) in deWitt's notation (i.e., with two contravariant indices and no covariant indices), while the vertices of degree n are rank n covariant tensors which are totally symmetric among all bosonic indices of the same type and totally antisymmetric among all fermionic indices of the same type and the contraction of a propagator with a rank n covariant tensor is indicated by an edge incident to a vertex (there is no ambiguity in which "slot" to contract with because the vertices correspond to totally symmetric tensors). The external vertices correspond to the uncontracted contravariant indices.A derivation of the Feynman rules using Gaussian functional integrals is given in the functional integral article.Each Feynman diagram on its own does not have a physical significance. It's only the infinite sum over all possible (bubble-free) Feynman diagrams which gives physical results. This infinite sum is usually only asymptotically convergent.
Invariance mechanics
Penguin diagram Notes
References Gerardus 't Hooft, Martinus Veltman, Diagrammar, CERN Yellow Report 1973, online
David Kaiser, Drawing Theories Apart: The Dispersion of Feynman Diagrams in Postwar Physics, Chicago: University of Chicago Press, 2005. ISBN 0-226-42266-6
Martinus Veltman, Diagrammatica: The Path to Feynman Diagrams, Cambridge Lecture Notes in Physics, ISBN 0-521-45692-4 (expanded, updated version of above) External links Feynman diagram page at SLAC
AMS article: "What's New in Mathematics: Finite-dimensional Feynman Diagrams"
WikiTeX supports editing Feynman diagrams directly in Wiki articles.
Drawing Feynman diagrams with FeynDiagram C++ library that produces PostScript output.
Feynman Diagram Examples using Thorsten Ohl's Feynmf LaTeX package.
JaxoDraw A Java program for drawing Feynman diagrams.
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Top
Multiplying Fractions with Whole Numbers Calculator
Top
Whole numbers starts with zero and endless 0, 1, 2, 3, 4, 5, 6, 7, etc…. Fraction is in the form of division method and also included with numerator and denominator. To multiply a proper fraction or a improper fraction with a whole number, we need to multiply the whole number with the numerator(top number of the fraction) and keep the denominator as it is. Multiplying Fractions with Whole Numbers Calculator is used to multiply the fraction with whole number.
## Steps for Multiplying Fractions with Whole Numbers Calculator
Step 1 :
Observe the given whole number and fraction.
Step 2 :
Multiply the given whole number with the numerator(top number) of the fraction and keep the denominator as it is.
Step 3 :
Simplify the fraction if needed.
## Problems on Multiplying Fractions with Whole Numbers Calculator
1. ### Multiply $\frac{6}{7}$ and 2
Step 1 :
Given fraction = $\frac{6}{7}$
Whole number = 2
Step 2 :
Multiply the whole number with the numerator of the given fraction.
$\frac{6 \times 2}{7}$
Step 3 :
$\Rightarrow$ $\frac{12}{7}$
$\frac{12}{7}$
2. ### Multiply $\frac{3}{8}$ and 4
Step 1 :
Given fraction = $\frac{3}{8}$
Whole number = 4
Step 2 :
Multiply the whole number with the numerator of the given fraction.
$\frac{3 \times 4}{8}$
Step 3 :
$\Rightarrow$ $\frac{12}{8}$
Divide the numerator and denominator by 4
$\Rightarrow$ $\frac{12 \div 4}{8 \div 4}$
$\Rightarrow$ $\frac{3}{2}$
$\frac{3}{2}$
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Four EAS Concentrators Awarded Hoopes Prizes for Senior Theses
May 21, 2020
This month, four joint concentrators in East Asian Studies were awarded the Hoopes Prize for their senior theses. Established by a gift from the estate of Thomas T. Hoopes, class of 1919, the prize is awarded to undergraduates for a piece of writing they produced under advising from Harvard faculty--with the stated purpose of “promoting, improving, and enhancing the quality of education . . . in literary, artistic, musical, scientific, historical, or other academic subjects.”
Indeed, all four of the EAS awardees’ works fell into more than one of these categories. The seniors’ interdisciplinary fields of study were among the factors that allowed them to view the topics through a unique lens, resulting in especially original and complex thesis arguments. Prior to the completion of the papers, they presented their research and preliminary findings to faculty, staff, and fellow students at the EAS Senior Thesis Colloquium in February. The four awardees are listed below, along with advising faculty members, graduate student advisers, and thesis titles.
Angie Cui (Government and EAS), advised by GOV Senior Lecturer Nara Dillon and GOV PhD Candidate Austin Strange, was awarded a Hoopes Prize for “Diplomas for Diplomacy: Foreign Students in China and the Soft Power Question.”
Joshua Grossman (Environmental and Planetary Sciences and EAS), advised by EPS Professor Michael McElroy, was awarded the prize for “A Novel 3D Model-Based Petroleum Estimate of the Qaidam Basin in Northwest China and Implications for the Future of China’s Energy Economy.”
Hillary McLauchlin (History and Literature and EAS), advised by EALC Professor Jie Li and EALC PhD Candidate Dingru Huang, was awarded the prize for “State of the Art: Contemporary Chinese Art in the Age of Surveillance.”
Yong Han Poh (Anthropology and EAS), advised by SOC-STD Lecturer Nicole Newendorp, was awarded the prize for “Love, Labour, Loss: Voices of Migrant Worker Poets and Storytellers in Singapore.” Poh also organized a photo exhibition at the Fairbank Center based on her anthropological work in Singapore in February of 2020, prior to the completion of her thesis. It can now be viewed virtually on the Harvard Asia Center website.
Student awardees receive $5,000, and their winning papers are made available in bound form in Lamont Library for a period of two years. Since Hoopes’ primary objective in establishing this fund was to promote excellence in teaching, faculty whose advisee/nominee wins the prize are also awarded a smaller sum of$2,000 for their role in the research and development of the student’s work.
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• ### Unit 4: Transcendental Functions
In this unit, you will investigate the derivatives of trigonometric, inverse trigonometric, exponential, and logarithmic functions. Along the way, you will develop a technique of differentiation called implicit differentiation. Aside from allowing you to compute derivatives of inverse function, implicit differentiation will also be important in studying related rates problems later on.
Completing this unit should take you approximately 17 hours.
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## CryptoDB
### Paper: An Efficient Identification Scheme Based on Permuted Patterns
Authors: Shahrokh Saeednia URL: http://eprint.iacr.org/2000/021 Search ePrint Search Google This paper proposes a new identification scheme based on a hard partition problem rather than factoring or discrete logarithm problems. The new scheme minimizes at the same time the communication complexity and the computational cost required by the parties. Since only simple operations are needed for an identification, our scheme is well suited for smart cards with very limited processing power. With a "good" implementation, the scheme is much faster than the Fiat-Shamir or Shamir's PKP schemes.
##### BibTeX
@misc{eprint-2000-11365,
title={An Efficient Identification Scheme Based on Permuted Patterns},
booktitle={IACR Eprint archive},
keywords={Identification, NP-completeness, Smart cards.},
url={http://eprint.iacr.org/2000/021},
note={ saeednia@ulb.ac.be 11338 received 24 May 2000, revised 16 Jan 2001},
author={Shahrokh Saeednia},
year=2000
}
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# Compton effect does the electron radiate?
In the Compton Effect, we normally treat the interaction as a billiard ball collision. It's not clear to me that this takes everything into account. If the electron ricochets off the photon, then it accelerates. Does it then radiate? If so, it seems to me that there would be additional losses not accounted for in the textbook calculations.
## Answers and Replies
jtbell
Mentor
In the Compton Effect, we normally treat the interaction as a billiard ball collision.
Not really. We simply assume that conservation of energy and conservation of momentum apply, using the relativistic equations for both. Those principles don't depend on the details of the interaction.
Either use classical or quantum mechanics, not both.
According to classical mechanics, the electron is accelerated by an EM wave, and this does indeed cause it to radiate: and that radiation is what cancels out part of the original EM wave and constitutes the colour-shifted wave.
In the Compton Effect, we normally treat the interaction as a billiard ball collision. It's not clear to me that this takes everything into account. If the electron ricochets off the photon, then it accelerates. Does it then radiate? If so, it seems to me that there would be additional losses not accounted for in the textbook calculations.
Textbook calculations are usually limited to the lowest order of the perturbation theory. If you take into account also higher orders you'll see that the Compton scattering is accompanied by the emission of "soft" photons, i.e., additional energy losses.
Not really. We simply assume that conservation of energy and conservation of momentum apply...
Isn't that the way we treat billiard balls?
jtbell
Mentor
Textbook calculations are usually limited to the lowest order of the perturbation theory. If you take into account also higher orders you'll see that the Compton scattering is accompanied by the emission of "soft" photons, i.e., additional energy losses.
Not to dispute this, but simply to get an idea of the size of this effect... consider for example a 662 keV gamma-ray photon Compton-scattering off an electron (which is commonly studied in simple undergraduate laboratories). Roughly how much energy do these soft photons carry?
Staff Emeritus
Roughly how much energy do these soft photons carry?
There's a range of energies, but an estimate of the size would be of order $\alpha/\pi$, probably somewhat less. Perhaps hundreds of eV.
There's a range of energies, but an estimate of the size would be of order $\alpha/\pi$, probably somewhat less. Perhaps hundreds of eV.
The "range" of energies is what I'd like to understand. In this kind of motion there is no sustained oscillation giving a single frequency, but rather a spectrum of frequencies which combines to account for the motion. The energy density within this spectrum is far from equalling a single quantum of energy for any specific frequency. So what does it mean if you detect a soft photon? Does the whole frequency spectrum "collapse"? Because there is no way all the soft energy could have gone into a single frequency at the moment of collision.
Not to dispute this, but simply to get an idea of the size of this effect... consider for example a 662 keV gamma-ray photon Compton-scattering off an electron (which is commonly studied in simple undergraduate laboratories). Roughly how much energy do these soft photons carry?
Vanadium50 gives a ballpark of several hundred eV for the soft photons. This seems really high to me. Lets just ballpark...
The 662 keV photon has a wavelength on the order of 10 ^-12 meters. So in the COM frame, the electron has the same wavelength. This corresponds to an electron energy of approx. 1000 eV. So Vanadium's "soft photons" would carry of a very significant part of the electron energy. How would the undergraduate laboratory demonstration give the right numbers in that case?
jtbell
Mentor
Perhaps hundreds of eV.
I'm just looking for an order of magnitude figure for the total energy radiated. So it's hundreds of eV compared to hundreds of keV for the incoming and scattered photon, which is on the order of 0.1% of the total energy involved in this process.
You'd surely need pretty sensitive experiments to detect this effect, which is one reason why it's not brought up in a standard intro modern physics course. Also the theoretical analysis is a bit above the level of that kind of course.
jtbell
Mentor
The 662 keV photon has a wavelength on the order of 10 ^-12 meters. So in the COM frame, the electron has the same wavelength. This corresponds to an electron energy of approx. 1000 eV. So Vanadium's "soft photons" would carry of a very significant part of the electron energy. How would the undergraduate laboratory demonstration give the right numbers in that case?
Suppose the photon scatters at 45 degrees. The usual Compton-scattering energy equation
$$\frac {1}{E^{\prime}} - \frac {1}{E} = \frac {1}{mc^2} (1 - cos \theta)$$
gives the energy of the scattered photon as E' = 480 keV. Therefore the recoiling electron has kinetic energy 662 - 480 = 182 keV.
A few hundred eV for the additional soft photons is a few times 0.1 keV.
Staff Emeritus
I'm just looking for an order of magnitude figure for the total energy radiated. So it's hundreds of eV compared to hundreds of keV for the incoming and scattered photon, which is on the order of 0.1% of the total energy involved in this process.
I'm also probably on the high side here as well. The other scales in the problem are all smaller, and I know that the interferences are destructive, not constructive.
You can visualize the soft photon emission as follows: the photon scatters off the electron, sending it flying. Now that the electron has experienced an acceleration, it radiates.
jtbell
Mentor
The 662 keV photon has a wavelength on the order of 10 ^-12 meters. So in the COM frame, the electron has the same wavelength. This corresponds to an electron energy of approx. 1000 eV.
A wavelength of $10^{-12}$ m correponds to a momentum of 1240 keV/c. For an electron, this means a (total) energy of $E = \sqrt {(pc)^2 + (mc^2)^2}$ = 1342 keV.
For my example, with a scattering angle of 45 degrees, the wavelength is $3.55 \times 10^{-12}$ m, which leads to a (total) electron energy of 619 keV.
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Here is a web reference that discusses scattering of 1-eV laser photons off a 4 to 8 GeV electron beam. references to published papers on Compton scattering vertex corrections are included.
The vertex corrections are less than 1%, Do a google search on "SLAC Compton Scattering" to find other similar papers
Last edited by a moderator:
Staff Emeritus
Um...no, that's a link to an Alnico data sheet.
The 1-loop vertex corrections are not the same as this. Indeed, this particular piece of the calculation is IR-divergent.
In the Compton Effect, we normally treat the interaction as a billiard ball collision. It's not clear to me that this takes everything into account. If the electron ricochets off the photon, then it accelerates. Does it then radiate? If so, it seems to me that there would be additional losses not accounted for in the textbook calculations.
As meopemuk said, there is always soft radiation with a continuous spectrum.
The Compton scattering kinematics is similar to billiard ball one if one considers photons as particles with specific energy-momentum (quantum) relationship and the electrons as relativistic particles. But the Compton wavelength Λ=ћ/mc is written for electron as if it were a quantum wave (not a particle) and the photon were a classical wave of a given frequency (not a particle). In QED both are de Broglie waves, so their scattering is due to non linearity of QED's wave equations. Of course, the other photon modes are also excited (soft radiation). Normally their energy is within the hard photon source/detector and electron detector accuracy range (i.e. included experimentally events). It corresponds to the inclusive theoretical QED cross section. See my "Atom as a "dressed" nucleus" for the inclusive cross section physics under large-angle scattering.
Bob_for_short.
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A wavelength of $10^{-12}$ m correponds to a momentum of 1240 keV/c. For an electron, this means a (total) energy of $E = \sqrt {(pc)^2 + (mc^2)^2}$ = 1342 keV.
For my example, with a scattering angle of 45 degrees, the wavelength is $3.55 \times 10^{-12}$ m, which leads to a (total) electron energy of 619 keV.
Thanks for checking my numbers. I see what I did wrong. (I ballparked a 10eV electron wavelength as 10^{-10}m based on the hydrogen atom, then multiplied the momentum by 100 to equate to the wavelength of the photon. Forgot to square 100 to get the energy multiplier. Still out by a factor of 10 but what the heck.)
... consider for example a 662 keV gamma-ray photon Compton-scattering off an electron (which is commonly studied in simple undergraduate laboratories). Roughly how much energy do these soft photons carry?
What is the incident gamma-ray energy uncertainty in such experiments?
What are the gamma-detector and electron detector uncertainties?
Bob.
What is the incident gamma-ray energy uncertainty in such experiments?
What are the gamma-detector and electron detector uncertainties?
Bob.
The 662 KeV gamma from Cs-137 is very precise, because the half-life of Cs-137 is many years. A gamma detector has statistical uncetainties on the energy resolution, which depends on what fraction of the photon energy is absorbed, and/or electron/holes (silicon) or photoelectrons (for NaI(Tl)) are created for each individual incident photon. In both cases (silicon and NaI(Tl)), the incident photon can create a Compton electron, and the recoil secondary photon escapes the detector. This will show up in the energy spectrum as a Compton backscatter peak at about 184 KeV.
My first "exposure" to the 662 KeV cesium spectrum was when we received a pulse height analyzer (PHA) from the vendor in 1958, and the spectrum was still stored in the PHA core memory.
Hans de Vries
Gold Member
In the Compton Effect, we normally treat the interaction as a billiard ball collision. It's not clear to me that this takes everything into account. If the electron ricochets off the photon, then it accelerates. Does it then radiate? If so, it seems to me that there would be additional losses not accounted for in the textbook calculations.
Technically, there is an interference pattern between the initial and final
state of the electron. This sinusoidal interference pattern of charge and
spin density is responsible for the radiation.
In more detail it's a two step process, the first is the interaction of the
electron with the incoming photon while the second is the interaction
with the radiated photon.
Regards, Hans
Technically, there is an interference pattern between the initial and final
state of the electron. This sinusoidal interference pattern of charge and
spin density is responsible for the radiation.
In more detail it's a two step process, the first is the interaction of the
electron with the incoming photon while the second is the interaction
with the radiated photon.
Regards, Hans
Thanks, Hans. But I'm not sure you're interpreting it exactly the way I do. For me, the incoming and outgoing electron interfere to form a stationary diffraction grating. That's what the light reflects from. It provides a neat semi-classical explanation for the Compton effect without the need to invoke photons.
Hans de Vries
Gold Member
Thanks, Hans. But I'm not sure you're interpreting it exactly the way I do. For me, the incoming and outgoing electron interfere to form a stationary diffraction grating. That's what the light reflects from. It provides a neat semi-classical explanation for the Compton effect without the need to invoke photons.
What I described is the is the first order approximation of QED
which, besides being 99+% accurate, also provides the neat
semi classical explanation you are looking for.
Maxwell's laws stay 100% the same, but we have to consider:
1. The electron field is a continuous charge/current density
and magnetization/polarization distribution ($\bar{\varphi}\gamma^\mu\varphi$ and $\bar{\varphi}\sigma^{\mu\nu}\varphi$)
2. The electron field can interfere resulting in sinusoidal
charge/current density and magnetization terms.
The importance of the intrinsic spin of the electron is illustrated
by the fact that the source of the transverse components of the
em radiation (emitted real photon) stem from the transverse currents
in the interference pattern caused by the magnetization gradients.
(as shown via the so called Gordon decomposition)
Regards, Hans
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What I described is the is the first order approximation of QED
which, besides being 99+% accurate, also provides the neat
semi classical explanation you are looking for.
Maxwell's laws stay 100% the same, but we have to consider:
1. The electron field is a continuous charge/current density
and magnetization/polarization distribution ($\bar{\varphi}\gamma^\mu\varphi$ and $\bar{\varphi}\sigma^{\mu\nu}\varphi$)
2. The electron field can interfere resulting in sinusoidal
charge/current density and magnetization terms.
The importance of the intrinsic spin of the electron is illustrated
by the fact that the source of the transverse components of the
em radiation (emitted real photon) stem from the transverse currents
in the interference pattern caused by the magnetization gradients.
(as shown via the so called Gordon decomposition)
Regards, Hans
I'm sure this is the right picture but your language is different from mine in certain ways and I'm not understanding it. In my superposition of electron states I simply have a sinusoidal charge distribution. Just like the potential well. No current distribution. Why do you have a sinusoidal current distribution?
Yes, you see an oscillating current distribution once you let the incoming radiation fall on these parallel sheets of charge, but that doesn't appear just from the superposition of states, does it?
Hans de Vries
Gold Member
I'm sure this is the right picture but your language is different from mine in certain ways and I'm not understanding it. In my superposition of electron states I simply have a sinusoidal charge distribution. Just like the potential well. No current distribution. Why do you have a sinusoidal current distribution?
Yes, you see an oscillating current distribution once you let the incoming radiation fall on these parallel sheets of charge, but that doesn't appear just from the superposition of states, does it?
The alternating transverse current distribution is the result of the
intrinsic spin of the electron. Each point of the electron field repre-
sents a small magnetic dipole caused by a circular current.
This circular current is described by the axial current density which
is calculated from Dirac's electron field $\varphi$ with the expression $\bar{\varphi}\gamma^\mu\gamma^5\varphi$
A sinusoidal distribution of such a circular current is equivalent, via
Stokes law, to an alternating current distribution. This can be seen
in the attached image below. If the circular currents are equal then
they cancel each other locally and there is no net (vector) current.
If there is a gradient however then they do not cancel each other
and there is an effective (vector) current. The left hand side of the
image shows the circular (axial) currents and the right hand side of
the image shows the effective (vector) currents which are the source
of the transverse components of the emitted photon.
Regards, Hans
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# The ratio between the curved surface area and total surface area of a right circular cylinder is 1:2. Find the ratio between the height and radius of the cylinder.
1
by Amruthraj
• Brainly User
2016-03-22T08:47:24+05:30
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
The curved surface area (CSA) of a cylinder is given by the formula: 2πrh. Where the variable r denotes the radius and the variable h denotes the height of the cylinder.Now, we need to find the ratio of the curved surface areas.
⇒
⇒
⇒
⇒ 2h = (r+h) (Use cross Multiplication)
⇒ h + r
⇒
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# Is an nth root of unity a square?
Suppose w^(2n)=1 (w is a complex number). For which n (if any) \sqrt(w) \in Q(w) ?
-
is it homework ? – David Lehavi Oct 22 '09 at 7:38
Perhaps the title should say something about the field Q(w)... how about "Is root of unity w a square in Q(w)?" – Ilya Nikokoshev Oct 22 '09 at 8:01
This could be homework for a second course in number theory, but it also could be a lemma needed by someone in a field far from number theory. I would leave this open. – David Speyer Oct 22 '09 at 10:44
The key point is to understand the field Q(w) for w a primitive kth root of unity. Call this field Qk. In particular, you want to know that Q4n \neq Q2n.
The key fact here is that the field extension Qk/Q has degree phi(k), where phi(k) is the Euler phi function, and phi(4k) \neq phi(2k). For a proof that Qk/Q has degree phi(k), see the early parts of any book on cyclotomic fields. This is probably also done in many Galois theory books but I don't know which ones.
-
Ok, thanks. Yes I need this for some lemma. – user966 Oct 22 '09 at 12:16
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# Current and Future Trends in Stochastic Thermodynamics
4-29 September 2017
Nordita, Stockholm
Europe/Stockholm timezone
## Linear and non-linear thermodynamics of a kinetic heat engine with fast transformations
25 Sep 2017, 10:00
1h
122:026 (Nordita, Stockholm)
### Speaker
Prof. Angelo Vulpiani (Dipartimento di Fisica, Universita' Sapienza)
### Description
We investigate a kinetic heat engine model constituted by particles enclosed in a box where one side acts as a thermostat and the opposite side is a piston exerting a given pressure. Pressure and temperature are varied in a cyclical protocol of period $\tau$ and their relative excursions $\delta$ and $\epsilon$ respectively, constitute the thermodynamic forces dragging the system out-of-equilibrium. The analysis of the entropy production of the system allows to define the conjugated fluxes, which are proportional to the extracted work and the consumed heat. The dynamics of the piston can be approximated, through a coarse-graining procedure, by a Klein-Kramers equation which - in the linear regime - yields analytic expressions for the Onsager coefficients and the entropy production. A study of the efficiency at maximum power shows that the Curzon- Ahlborn formula is always an upper limit which is approached at increasing values of the thermodynamic forces, i.e. outside of the linear regime.
### Presentation Materials
There are no materials yet.
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# Is this simplification of a sequence valid?
I have the following sequence flow, graphically:
According to my understanding, the diagram describes the following sequence
$$y[n] = (cx[n] + x[n-1])[n-3](-1)$$
Now I'm very tempted to simplify this further:
$$y[n] = (cx[n-3] + x[n-4])(-1)$$
I saw something like this done before in some examples, but I wasn't able to find a justification why such a simplification would be valid. So my question:
• Is that simplification indeed valid?
• Is there a theorem or so that proves this "distributive property" (or whatever the correct name is)?
What I would suggest editing is the first equation line in your question. Although the reader will likely understand what you mean there, it's not standard notation. One way to present it with standard notation is to define a variable, let's call it $$\tilde x[n]$$, for the output of the summation block:
$$\tilde x[n] := cx[n]+x[n-1] \;\;\;\;\;\; (*)$$
Now observe that $$y[n] = -\tilde x[n-3]$$ and just plug in the definition of $$\tilde x[n]$$ from $$(*)$$ to get your answer.
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# Translative packing constant strictly larger than lattice packing constant
Simply put, my question is this: what is the smallest dimension, if any, where we can know for sure that a convex body exists whose translative packing constant is strictly larger than its lattice packing constant?
Some relevant facts I know:
• In two dimensions, the translative packing constant is always equal to the lattice packing constant for convex bodies. For non-convex bodies, already in two dimensions there are counterexamples. Bezdek and Kuperberg give a good exposition of this.
• In ten dimensions, the best packing of spheres seems to be non-lattice. In lower dimensions, the best lattice packing seems to be also the best packing. The best known lattice packing in dimensions above 8 is not actually known to be best (except in 24 dimensions). Therefore, even for spheres, there is no dimension where lattices are proved to be suboptimal. See this MO question for more on that.
• Convex bodies that tile by translation can also tile by lattice translations. So the example cannot be a tiling body. This is famous result due to Venkov, Alexandrov, and McMullen, and Gruber's book on Convex and Discrete Geometry gives a nice treatment.
• Except for tiling bodies, which are ruled out be the previous point, very few bodies have known translative packing constants. However, all we need to have an example is a lower bound for the translative packing constant and an upper bound for the lattice packing constant. There are known methods to compute the lattice packing constant for polytopes, and in 3D one such method has been implemented by Betke and Henk.
So, it seems that in dimension 10, there are convex bodies, namely spheres, that pack better by translation than by lattice translation. However, this is not rigorously known. If true, this example can probably be extended to higher dimensions by forming cylinders. However, it seems to me that if we allow nonspherical bodies, the dimension where translative packing starts to beat out lattice packing should be siginificantly lower. Is there an example? In dimensions low enough (e.g. 3D), candidates can be checked computationally and rigorously established if they check out.
• I don't know of an example, and I wouldn't be surprised if nobody knows one (but I'll be very pleased if I'm wrong about this). – Henry Cohn Oct 3 '14 at 18:43
• Thanks @HenryCohn. This is indeed as I feared, but I wanted to check that I wasn't missing something known. The fact that we have a good candidate in 10D and we have a good method for checking candidates in 3D makes me feel that it should not be impossibly hard to produce an example in some intermediate dimension. – Yoav Kallus Oct 3 '14 at 19:02
• Yeah, I think it's very plausible that one could construct an example in a much lower dimension, and conceivable that one could prove it rigorously. I haven't thought much about how hard the difficulty of proving that a lattice packing is optimal varies with dimension (for convex bodies). The Betke-Henk paper deals beautifully with polyhedra in three dimensions, but I have no idea how hard four, five, or six dimensions would be. – Henry Cohn Oct 3 '14 at 19:58
One problem for which we know there is a significant difference between unrestricted sets of translations and lattices is in tilings with equal-size cubes. Every lattice tiling with n-cubes has a pair of cubes sharing an entire (n-1)-dimensional face (Hajós). On the other hand, in dimensions $n\ge8$, there are translational tilings with n-cubes in which no two cubes share an entire (n-1)-dimensional face ($n\ge10$ due to Lagarias and Shor, improve to 8 by Mackey, n=7 case open, see Keller's conjecture). Moreover, a stronger property holds in these tilings: the center of each (n-1)-face does not lie in the relative interior of any other (n-1)-face. I call such a tiling a center-to-boundary tiling.
Now consider a slightly stellated n-cube: the convex hull of a unit n-cube and a point for each face lying a height $\epsilon$ above the center of the face. It is easy to perturb a center-to-boundary tiling of n-cubes to a packing with mean volume $(1+\epsilon)^n$. I wanted to show that since lattice tilings have shared faces along at least one direction, and a lattice packing of slightly stellated cubes will be near a lattice tiling, that the mean volume (i.e. lattice determinant) of the packing will have to be $\ge 1+(n+1)\epsilon + o(\epsilon)$, but this turns out to be false. Here is an example of a family of lattices that pack $\epsilon$-stellated 5-cubes, but have determinant $1+(47/8)\epsilon+O(\epsilon^2)$: $L=A\mathbb{Z}^5$, where
$$A=\left(\begin{array}{ccccc} 1+2\epsilon&-\tfrac12&0&0&0\\ 0&1+\epsilon&-\tfrac12&\tfrac12&0\\ 0&0&1+\epsilon&0&-\tfrac14\\ 0&0&0&1+\epsilon&\tfrac12\\ 0&0&-\tfrac12\epsilon&0&1+\epsilon \end{array}\right)\text.$$
Now, still $47/8>5$, but this example does not bode well for the program. For large $n$, I believe similar constructions are possible that achieve determinants $\le 1+n\epsilon+o(\epsilon)$. It might still be that this is impossible for $n=8$, but if so, it could be delicate to show.
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# Defining UpValue like replacements confined to exprs with specific Head
I'm not seeing a clean method to redefine builtin functions like Cross, Dot, etc. in such a way they only are only applied to expressions that are using some marker-like Head value. Reduced example, let's say I'm using Foo[..] as a head to represents some pair of expression. The pair of expressions inside a Foo might be involved so making explicit patterns doesn't go very far. Basically I'm looking to do something like this non-working example:
Foo /: Cross[v_, v_] := 0;
### update 1
Trivial example is symbolically reducing quaternions over reals. One rule for the product looks like:
realQuat /:
NonCommutativeMultiply[realQuat[L_, l_], realQuat[R_, r_]] :=
realQuat[Cross[L,R] + l R + r L, r l - Dot[L,R]]
So with a pair written as a (bi)vector scalar pair without Cross reduction:
t0 = realQuat[A,a];
t1 = realQuat[A,b];
t0**t1
gives:
realQuat[a A + A b + AxA, a b - A.A]
instead of the Cross term dropping out like this:
realQuat[a A + A b, a b - A.A]
(this is a strawman example)
### attempt using TensorReduce
Starting from the suggesting of using TensorReduce, I tossed together two containers to mark symbols as either elements of R3 or R, basic funcs to add to both and a reduction function. Much cleaner than my previous attempts.
realQuatB = Alternates[]; (* for marking R3 symbols *)
realQuatS = Alternates[]; (* for marking R symbols *)
realQuatDefB[x_] := If[!MemberQ[realQuatB,x],AppendTo[realQuatB,x];];
realQuatDefS[x_] := If[!MemberQ[realQuatS,x],AppendTo[realQuatS,x];];
SetAttributes[realQuatDefB, {Listable}];
SetAttributes[realQuatDefS, {Listable}];
realQuat /: rqReduce[realQuat[Q_,q_]] := realQuat[
TensorReduce[Q, Assumptions->Element[realQuatB,Vectors[3,Reals]] &&
Element[realQuatS,Reals]],
TensorReduce[q, Assumptions->Element[realQuatB,Vectors[3,Reals]] &&
Element[realQuatS,Reals]]];
• You could Block and Unprotect Cross, could you show a small example of input and expected output? – Kuba May 15 '17 at 8:58
• Added an example to help clarify. – MB Reynolds May 15 '17 at 9:59
• It is still not clear what you are asking, and what this has to do with UpValues. What do you expect as the output? Please give a concrete example, complete with example input, actual output (you have both) and desired output (you don't have this). – Szabolcs May 15 '17 at 10:17
• Like this?: mathematica.stackexchange.com/questions/21197/… -- E.g. TensorReduce[Cross[A, A], Assumptions -> A \[Element] Vectors[n]] – Michael E2 May 15 '17 at 10:22
• @ Szabolcs: By "UpValue" like I was intending to say we can use a marker head and upvalue defs to scope when some rules are applied and I'm generally curious if there a simply method to have rules applied at any sub-expression level of a maker head. – MB Reynolds May 16 '17 at 11:55
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# Statistics
## New nonlinear least squares solvers in R with {gslnls}
Introduction Solving a nonlinear least squares problem consists of minimizing a least squares objective function made up of residuals $$g_1(\boldsymbol{\theta}), \ldots, g_n(\boldsymbol{\theta})$$ that are nonlinear functions of the parameters of interest $$\boldsymbol{\theta} = (\theta_1,\ldots, \theta_p)'$$:
## GSL nonlinear least squares fitting in R
Introduction The new gslnls-package provides R bindings to nonlinear least-squares optimization with the GNU Scientific Library (GSL) using the trust region methods implemented by the gsl_multifit_nlinear module. The gsl_multifit_nlinear module was added in GSL version 2.
## Asymptotic confidence intervals for NLS regression in R
Introduction Nonlinear regression model As a model setup, we consider noisy observations $$y_1,\ldots, y_n \in \mathbb{R}$$ obtained from a standard nonlinear regression model of the form: \begin{aligned} y_i &\ = \ f(\boldsymbol{x}_i, \boldsymbol{\theta}) + \epsilon_i, \quad i = 1,\ldots, n \end{aligned} where $$f: \mathbb{R}^k \times \mathbb{R}^p \to \mathbb{R}$$ is a known nonlinear function of the independent variables $$\boldsymbol{x}_1,\ldots,\boldsymbol{x}_n \in \mathbb{R}^k$$ and the unknown parameter vector $$\boldsymbol{\theta} \in \mathbb{R}^p$$ that we aim to estimate.
## Step function regression in Stan
Introduction The aim of this post is to provide a working approach to perform piecewise constant or step function regression in Stan. To set up the regression problem, consider noisy observations $$y_1, \ldots, y_n \in \mathbb{R}$$ sampled from a standard signal plus i.
Stein’s paradox Stein’s example, perhaps better known under the name Stein’s Paradox, is a well-known example in statistics that demonstrates the use of shrinkage to reduce the mean squared error ($$L_2$$-risk) of a multivariate estimator with respect to classical (unbiased) estimators, such as the maximum likelihood estimator.
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Gauge Theory Does Not Fix This
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In the physics of gauge theories, gauge fixing (also called choosing a gauge) denotes a mathematical procedure for coping with redundant degrees of freedom in field variables. By definition, a gauge theory represents each physically distinct configuration of the system as an equivalence class of detailed local field configurations.
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A gauge theory is a type of theory in physics.The word gauge means a measurement, a thickness, an in-between distance (as in railroad tracks), or a resulting number of units per certain parameter (a number of loops in an inch of fabric or a number of lead balls in a pound of ammunition). Modern theories describe physical forces in terms of fields, e.g., the electromagnetic field, the …
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However, due to its gauge invariance, the action depends only on three (out of four) degrees of freedom of the gauge fields and, thus, the integral over the full space of gauge fields is not defined. To solve the problem it is necessary to fix the gauge (see the sections on classical gauge-fixing and on quantum gauge-fixing ).
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# Normalize Operator¶
This shows you how to use the Normalize operator.
## Introduction¶
Normalization is the process of shifting and scaling the data values to match the desired distribution. It calculates the mean $$\mu$$ and the standard deviation $$\sigma$$ and modifies the data as follows:
$Y_i = \frac{X_i - \mu}{\sigma}$
There are more advanced features in Normalize that will be explained later in this documentation.
## Using the Normalize Operator¶
We need some boilerplate code to import DALI and some other useful libraries and to visualize the results.
[1]:
from nvidia.dali.pipeline import Pipeline
import math
import nvidia.dali.ops as ops
import nvidia.dali.fn as fn
import nvidia.dali.types as types
import nvidia.dali.backend as backend
batch_size = 10
image_filename = "../data/images"
import matplotlib.pyplot as plt
import matplotlib.gridspec as gridspec
def display(outputs, idx, columns = 2, captions = None):
rows = int(math.ceil(len(outputs) / columns))
fig = plt.figure()
fig.set_size_inches(16, 6 * rows)
gs = gridspec.GridSpec(rows, columns)
row = 0
col = 0
for i, out in enumerate(outputs):
if isinstance(out, backend.TensorListGPU):
out = out.as_cpu()
plt.subplot(gs[i])
plt.axis("off")
if captions is not None:
plt.title(captions[i])
plt.imshow(out.at(idx));
def show(pipe, idx, columns = 2, captions = None):
pipe.build()
display(pipe.run(), idx, columns, captions)
### A Simple Pipeline¶
Create a simple pipeline that just loads some images and normalizes them, and treats the image data as a flat array that contains 3*W*H numbers (3 for RGB channels).
[2]:
pipe = Pipeline(batch_size=batch_size, num_threads=1, device_id=0)
with pipe:
jpegs, _ = fn.readers.file(file_root=image_filename)
images = fn.decoders.image(jpegs, device="mixed", output_type=types.RGB)
norm = fn.normalize(images)
pipe.set_outputs(images, norm)
[3]:
show(pipe, 1)
Clipping input data to the valid range for imshow with RGB data ([0..1] for floats or [0..255] for integers).
## Adjusting Output Dynamic Range¶
As you can see in the example above, the image intensity values have been scaled and shifted, with many pixels forced below 0, and displayed as black. You might want this result in many use cases, but if the output type has limited dynamic range (for example uint8), you might want to map the mean and standard deviation to values that more effectively use that limited range of values. For this purpose, Normalize offers two scalar arguments, shift and scale. Now the normalization formula becomes:
$Y_i = \frac{X_i - \mu}{\sigma} \cdot {scale} + {shift}$
Modify the pipeline to produce uint8 output with the mean mapped to 128 and standard deviation to 64, which allows values in the $$\mu \pm 2\sigma$$ range to be correctly represented in the output.
[4]:
pipe = Pipeline(batch_size=batch_size, num_threads=1, device_id=0)
with pipe:
jpegs, _ = fn.readers.file(file_root=image_filename)
images = fn.decoders.image(jpegs, device="mixed", output_type=types.RGB)
norm = fn.normalize(images, scale=64, shift=128, dtype=types.UINT8)
pipe.set_outputs(images, norm)
[5]:
show(pipe, 1)
### Directional Reductions¶
For multidimensional data, it might be useful to calculate the mean and standard deviation for only a subset of the dimensions. For example, the dimensions might correspond to height (0), width (1) and color channels (2) of an image. Reducing the 0, 1 (height, width) dimensions will produce a separate mean and standard deviation for each channel. Normalize supports two arguments to specify directions:
• axes - a tuple of dimension indices, with 0 being outermost.
• axis_names - axis symbols that were looked up in the input layout.
The following example normalizes the data along WC, H, WH and C.
[6]:
pipe = Pipeline(batch_size=batch_size, num_threads=1, device_id=0)
with pipe:
jpegs, _ = fn.readers.file(file_root=image_filename)
images = fn.decoders.image(jpegs, device="mixed", output_type=types.RGB)
normwc = fn.normalize(images, axes = (1, 2), scale=64, shift=128, dtype=types.UINT8)
normh = fn.normalize(images, axis_names = "H", scale=64, shift=128, dtype=types.UINT8)
normhw = fn.normalize(images, axis_names = "HW", scale=64, shift=128, dtype=types.UINT8)
normc = fn.normalize(images, axes = (2,), scale=64, shift=128, dtype=types.UINT8)
pipe.set_outputs(images, normwc, normh, normhw, normc)
[7]:
titles = ["Original", "Width and channels", "Height", "Height and width", "Channel"]
show(pipe, 9, captions=titles)
## Externally Provided Parameters¶
By default, Normalize calculates the mean and standard deviation internally. However, these values can be provided externally via the mean and stddev arguments, and can be scalar values or inputs. When providing mean or stddev values as inputs, the directions of reduction can be inferred from parameter’s shape. If the mean and stddev values are inputs, they must have the same shapes.
[8]:
pipe = Pipeline(batch_size=batch_size, num_threads=1, device_id=0)
with pipe:
jpegs, _ = fn.readers.file(file_root=image_filename)
images = fn.decoders.image(jpegs, device="mixed", output_type=types.RGB)
norm_mean = fn.normalize(
images, mean=64, axis_names="HW", scale=64, shift=128, dtype=types.UINT8)
norm_stddev = fn.normalize(
images, stddev=200, axis_names="HW", scale = 64, shift=128, dtype=types.UINT8)
pipe.set_outputs(images, norm_mean, norm_stddev)
[9]:
show(pipe, 1, captions=["Original", "Fixed mean", "Fixed standard deviation"])
## Batch Normalization¶
Normalize can calculate the mean and standard deviation for the entire batch instead of on a per-item basis. You can enable this behavior by setting the batch argument to True. Batch normalization demands that the extent of the non-reduced dimensions matches all samples in the batch. For example, because channels are normalized separately, the pipeline below expects that all images have three channels.
[10]:
pipe = Pipeline(batch_size=batch_size, num_threads=1, device_id=0)
with pipe:
jpegs, _ = fn.readers.file(file_root=image_filename)
images = fn.decoders.image(jpegs, device="mixed", output_type=types.RGB)
norm_sample = fn.normalize(
images, batch=False, axis_names="HW", scale=64, shift=128, dtype=types.UINT8)
norm_batch = fn.normalize(
images, batch=True, axis_names="HW", scale = 64, shift=128, dtype=types.UINT8)
pipe.set_outputs(images, norm_sample, norm_batch)
[11]:
show(pipe, 1, columns=3, captions=["Original", "Per-sample normalization", "Batch normalization"])
show(pipe, 4, columns=3, captions=["Original", "Per-sample normalization", "Batch normalization"])
show(pipe, 7, columns=3, captions=["Original", "Per-sample normalization", "Batch normalization"])
show(pipe, 9, columns=3, captions=["Original", "Per-sample normalization", "Batch normalization"])
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In chemical compounds, where does the “magic” come from in atomic “magic numbers”?
It is well known that atoms with a full electron shell are more stable, it is one of the first facts taught in a middle school Chemistry course:
"An element whose atoms have no electrons outside filled energy levels is particularly stable chemically. Such elements are called noble gases"
These numbers (for stable nuclei) are 2 (He), 10 (Ne), 18(Ar), 36(Kr), 54(Xe) and 86(Rn). Each correspond to a full shell of electrons with similar energies that start from $$ns$$ and end in $$np$$, just before the $$(n+1)s$$ shell is filled ($$n$$ being the principal quantum number and $$s, p$$ being the first two quantum numbers of the orbital angular momentum, i.e., $$\ell=0,1$$). We also know that the valence of an atom is the number of electrons more or fewer than the number for a noble gas.
What I do not understand is the reason for this:
"Stable chemical compounds are held together by the Coulomb attraction and are typically formed from atoms whose valence add up to zero."
What is so special about both atoms having noble gas configuration?
For the last electron in the shell we have that the potential behaves like $$−Ze^{2}/r$$ near the nucleus (whose charge is $$+Ze$$), and like $$−e^{2}/r$$ outside the atom, where the nuclear charge is screened by the negative charge of Z − 1 electrons, then the potential away from the atom (with $$\text{no. of electrons}=Z$$) should be zero. And for example, taking away an electron would increase the energy, as it would require work in order to take it out of the potential. There would also be no attraction to bind an extra electron outside the atom.
As far as I understand the mechanism through which these atoms would bind together is to first ionize and then let Coulomb attraction join them. For concreteness let us assume $$NaCl$$:
$$Na+Cl\xrightarrow{E_{\text{ionization}}}Na^{+}+Cl^{-}\xrightarrow[\text{interaction}]{\text{Coulomb}}NaCl$$
I know that for these elements the ionization energy may be small, but the energy for the non-ions should still be lower and thus they should be more stable, but this is not the case. My question would boil down to: Why do they stay together? Why not just snatch away the electron from the other ion and go their own ways?
It is always good to remember the historical context. The statement
An element whose atoms have no electrons outside filled energy levels is particularly stable chemically. Such elements are called noble gases.
confuses the phenomenon with the conclusion. The story was the following:
1. Mendeleev knew nothing about the group of stable elements in 1869, the Nobel gases were simply not discovered at that time.
2. The discovery of the noble gases aided in the development of a general understanding of atomic structure. In 1895, French chemist Henri Moissan attempted to form a reaction between fluorine, the most electronegative element, and argon, but failed. Learning from these experiments, Danish physicist Niels Bohr proposed in 1913 that the electrons in atoms are arranged in shells surrounding the nucleus, and that for all noble gases except helium the outermost shell always contains eight electrons.
3. In 1916, Gilbert N. Lewis formulated the octet rule, which concluded an octet of electrons in the outer shell was the most stable arrangement for any atom; this arrangement caused them to be unreactive with other elements since they did not require any more electrons to complete their outer shell.
(Point 2 and 3 are quotes from Wikipedia about Nobel gases)
=> It was and is until today a empirical fact, that elements with 2, 8 and again 8 electrons fill shells in a way, that these shells are particularly stable chemically.
Stable chemical compounds formed from atoms whose valence add up to zero. What is so special about both atoms having noble gas configuration?
Let us remember the historical context again:
1. The (Bohr’s) model's key success lay in explaining the Rydberg formula for the spectral emission lines of atomic hydrogen. While the Rydberg formula had been known experimentally, it did not gain a theoretical underpinning until the Bohr model was introduced.
2. The electron is able to revolve in certain stable orbits around the nucleus without radiating any energy contrary to what classical electromagnetism suggests.
3. The Bohr model gives almost exact results only for a system where two charged points orbit each other at speeds much less than that of light. This involves one-electron systems such as the hydrogen atom, singly ionized helium, and doubly ionized lithium.
4. Several enhancements to the Bohr model were proposed, most notably the Sommerfeld model or Bohr–Sommerfeld model, which suggested that electrons travel in elliptical orbits around a nucleus instead of the Bohr model's circular orbits.
5. In the end, the model was replaced by the modern quantum mechanical treatment of the hydrogen atom, which was first given by Wolfgang Pauli in 1925, using Heisenberg's matrix mechanics. The current picture of the hydrogen atom is based on the atomic orbitals of wave mechanics which Erwin Schrödinger developed in 1926.
(All points are quotes from Wikipedia about Bohr model)
In a nutshell (again from Wikipedia):
• The emission spectra of excited electrons in atoms are the experimental facts.
• Bohr's condition, that the angular momentum is an integer multiple of ħ ...
• ... was later reinterpreted in 1924 by de Broglie as a standing wave condition: the electron is described by a wave and a whole number of wavelengths must fit along the circumference of the electron's orbit.
• In modern quantum mechanics, the electron in hydrogen is a spherical cloud of probability that grows denser near the nucleus. The rate-constant of probability-decay in hydrogen is equal to the inverse of the Bohr radius
=> All explanations and conclusions are based on the chemical stability of Nobel gases. There is no explanation why Nobel gas atoms have 2, 8 and again 8 electrons in a stable arrangement; except for the fact that there are solutions for partial differential equations which can describe the electron excitations.
Quantum mechanics solves such partial differential equations for the spherical probability for electrons by the help of spherical harmonics.
Despite their name, spherical harmonics take their simplest form in Cartesian coordinates. This leads to spherical probabilities with 2 electrons for the s-shell and 6 electrons in the p-shell
But spherical harmonics have also another solutions. More general spherical harmonics of degree ℓ are not necessarily those of the Laplace basis $$Y_{\ell }^{m}$$, and their nodal sets can be of a fairly general kind. (Wikipedia)
One solution correspondent with the cubical atom, proposed by Lewis:
The cubical atom was an early atomic model in which electrons were positioned at the eight corners of a cube in a non-polar atom or molecule. This theory was developed in 1902 by Gilbert N. Lewis and published in 1916 in the article "The Atom and the Molecule" and used to account for the phenomenon of valency... The figure below shows structural representations for elements of the second row of the periodic table.
The corresponding spherical harmonics is the following:
Each of the eight segments of the sphere topologically corresponds to the eight edges of a cube.
Up to this point, I have only described what science says. Since your question didn't receive enough attention, I feel free to add my own thoughts to tell you why 2 and 8 electrons in a shell are perfect balanced around a nucleus.
Based on Lewis cubical distribution of electrons and remembering, that electrons have a magnetic dipole moment, it is possibly to bring the eight outer electrons of Ne and Ar into a perfect equilibration. This is the case for 4 electrons, pointing with their north poles to the nucleus and the other 4 electrons with their south poles. For He the two electrons are directed to each other antiparallel. Did you see the correspondence to Paulis exclusion principle?
• The "cubical atom" described here is fringe science and it has no relationship with reality beyond wishful thinking -- as has been pointed out to you repeatedly. Continued posting of misinformation, as in this post, constitutes abuse of this site. – Emilio Pisanty Jun 25 at 6:21
• @EmilioPisanty Can you, as a scientist, point out to me, who is not anchored in the scientific world, an inconsistency in the above? I know your answer, that this is not your task. But we both wasting time, you, reading my posts, and I, being wrong, but not seeing this. – HolgerFiedler Jun 25 at 7:15
• The onus is on you. Can you provide a quantitative calculation based on this "model" that can derive the absorption and emission spectra (say, of the lithium-through-neon first row of the periodic table, which you mention explicitly here) and which matches experiment? (These have been known for some 50+ years within normal QM; references are listed in the relevant results pages of the NIST ASD.) Unless you can, there is absolutely no point in discussing this model. – Emilio Pisanty Jun 25 at 7:35
• In other words, your "model" cannot be said to be inconsistent with experimental results because it doesn't even try to produce calculations and predictions to the stage where they could be compared with experiment. It isn't a model at all (in the scientific sense) ─ based on the evidence you've put forward thus far, it is no more than children's squiggles. – Emilio Pisanty Jun 25 at 7:39
• @EmilioPisanty That you are right and I know this. So, I’m not even able to calculate the octopol of a parallelopiped and hope that one day somebody like the idea and will calculate it for 8 magnetic dipole moments. – HolgerFiedler Jun 25 at 18:19
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Research
# The maximal Hilbert transform along nonconvex curves
Honghai Liu
Author Affiliations
School of Mathematics and Information Science, Henan Polytechnic University, Jiaozuo, Henan, 454003, P.R. China
Journal of Inequalities and Applications 2012, 2012:191 doi:10.1186/1029-242X-2012-191
Received: 11 March 2012 Accepted: 17 August 2012 Published: 31 August 2012
This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
### Abstract
The Hilbert transform along curves is defined by the principal value integral. The pointwise existence of the principal value Hilbert transform can be educed from the appropriate estimates for the corresponding maximal Hilbert transform. By using the estimates of Fourier transforms and Littlewood-Paley theory, we obtain -boundedness for the maximal Hilbert transform associated to curves , where , P is a real polynomial and γ is convex on . Then, we can conclude that the Hilbert transform along curves exists in pointwise sense.
MSC: 42B20, 42B25.
##### Keywords:
the maximal Hilbert transform; maximal functions; nonconvex curves; Littlewood-Paley theory
### 1 Introduction
For , let be a curve in with . To Γ we associate the Hilbert transform which is defined as a principal value integral
where and . Similarly, one can define the corresponding maximal function and the maximal Hilbert transform as
and
The -boundedness for the Hilbert transform and the maximal function above have been well studied by many scholars. See [10] for a survey of results through 1977. More recent results can be found in [1,2,4-7].
Appropriate estimates for the maximal Hilbert transform give the pointwise existence of the principal value Hilbert transform. So, we focus on the -bounds for the maximal Hilbert transform in this paper. Let us state some previous theorems to establish the background for our current work. The first result about is the work of Stein and Wainger (see [10]).
Theorem 1.1
(A) If Γ is a two-sided homogeneous curve in, then
(B) Assume that for smallt, lies in the subspace spanned by. Then the maximal Hilbert transform
is bounded fromto itself, .
For and , Córdoba and Rubio de Francia considered the case and , with the following properties:
(i) γ is biconvex, i.e., is decreasing in and increasing in ;
(ii) has doubling time, i.e., there exists a constant such that ;
(iii) γ is balanced, by which we mean the following: there exists such that for every .
They proved the following theorem in [7].
Theorem 1.2Under the assumptions (i), (ii) and (iii) on theγ, the maximal Hilbert transformis a bounded operator infor.
In this note, we consider the curve Γ with the form , where is a real-valued polynomial of t in , γ satisfies
(1.1)
Definition 1.3 A function belongs to , if there exists a constant such that for . It is also said that f has doubling time.
For this case, Bez obtained the -boundedness of and in [1].
Theorem 1.4Suppose thatPis a polynomial, γsatisfies (1.1) and. If, , and either (1) is zero, or (2) is nonzero and, then
Moreover, the constantCdepends only onp, γand the degree ofP.
Remark 1.5 If , is “more convex” than in some sense, then Γ is “convex” enough for the -boundedness of and . In the case , the linear term of cannot improve the convexity of γ. To obtain the -boundedness for associated operators, one needs to pose additional condition(s) on γ, that is, . For more details, see [5] and [9].
Motivated by Bez’s result above, we obtain the -boundedness for . More precisely, we prove the following theorem.
Theorem 1.6Suppose thatPis a polynomial, γsatisfies (1.1) and. If, and either (1) is zero, or (2) is nonzero and, then
Moreover, the constantCdepends only onp, γand the degree ofP.
Remark 1.7 Let . Comparing those conditions for γ in Theorem 1.2, we find that conditions in Theorem 1.6 are stricter. But we should note that may be a nonconvex function.
The convexity of the polynomial P is important for our main result. P has different convexity in different intervals, which suggests that will be decomposed according to the properties of P. The decay of associated multipliers is essential for the proof of Theorem 1.6. This set of techniques originated from the work [1] and [3]. Notice that is a nonlinear operator, Minkowski’s inequality cannot be used as in Section 1 of [1], the linearization method is invoked to treat it. Similarly, the essential Proposition 1.2 in [1] is useless for the maximal Hilbert transform. Littlewood-Paley theory and interpolation theorem are effective tools to treat those problems. Those ideas are due to the contribution of Córdoba, Nagel, Vance, Wainger, Rubio de Francia.
The organization of our paper is as follows. In Section 2, we list some key properties concerning the polynomial and give some lemmas for the proof of the main result. The -estimates for will be proved in Section 3.
### 2 Preliminaries
Without loss of generality, we suppose that , where . Let be d-complex roots of P ordered as
Let A be a positive constant which will be chosen in Lemma 2.1. Define if it is nonempty for and . Let , then can be decomposed as , where is the interval between two adjacent . It is obvious that is disjoint. Then, we can decompose as
where is the inverse image of a subset I restricted in .
The properties of P on and are important for our proof. The following related lemma can be found in [1] and [3].
Lemma 2.1There exists a numbersuch that for anyand any,
(i) for;
(ii) for, for;
(iii) for;
(iv) andfor, .
The following fact can be induced from the proof of Lemma 2.1 (see [1]), that is, we can choose such that, for ,
(2.1)
Let λ be the doubling constant for , define . Let I be a subset of , and , and are given by
and
For and , we define cones in by
and the corresponding projection operators by . Then, we have the following lemma which is Lemma 1 in [7].
Lemma 2.2Forand, we have
(i) ;
(ii) ;
(iii) .
The bootstrap argument (see [8]) plays an important role in the proof of the main result, so we present the following well-known result which can be found in [2] and [7].
Lemma 2.3Suppose thatis a sequence of positive operators uniformly bounded in, andis bounded infor, then
for.
### 3 Proof of Theorem 1.6
Let and be given as in Section 2, then
Note that and are finite sets, it suffices to show that and are -bounded, respectively.
#### 3.1 The -boundedness for
Let be some measurable function from to such that
(3.1)
By Minkowski’s inequality, we can control the -norm of by
(3.2)
Let for some and , then
and
Notice that γ is convex and , so for . Thus,
(3.3)
where is the inverse function of .
(3.1), (3.2) and (3.3) yield
#### 3.2 The -boundedness for
For and , set and define measures by
for . For any , there exists such that . Then
Therefore,
By the -boundedness of (see [1]), it suffices to consider the latter term. Let such that for and for . Write and denote by ⋆ convolution in the first variable. For , the truncated Hilbert transform can be decomposed as
where δ is the Dirac measure in , and is the identity operator. Then, we just need to estimate , and , respectively.
The decay of is important for the boundedness of three maximal operators above. Essentially, estimates for in the following subsection have been proved in [1]. We repeat them just for completeness.
#### 3.2.1 Fourier transform estimates of
Before the proof of Proposition 3.2, we need the following lemma which is Lemma 2.2 in [1].
Lemma 3.1For all, the function
is singled-signed on.
Proposition 3.2Forand, if, then
Proof For fixed , let , then
Case 1. . If and , for , (2.1) implies
(3.4)
Note that is monotone on , this fact follows from Lemma 3.1. By Van der Corput’s lemma and (3.4), we get .
If γ is even, then . If γ is odd, Lemma 3.1 still holds for , can be considered in the same way. Then, we have
(3.5)
If and . In the same way, for , we have
(3.6)
In the same way as above, we can get
Case 2. . If and , (3.4) still holds for . By integrating by parts,
Essentially, we just need to consider the second term, which can be dominated by
In order to estimate the term , we define , then . By (3.4), for , it is obvious that
(3.7)
On the other hand, for ,
(3.8)
Further, by (2.1), for ,
(3.9)
Thus, combining (3.7), (3.9) and (3.8), we have
(3.10)
For , by (3.4),
(3.11)
Note that can be split into a finite number of disjoint intervals such that is singled-signed on each interval. Suppose that is such an interval and by (2.1), then . So .
If and , (3.6) holds for . The same arguments used above imply
where and are as previous ones. For above, we have
(3.12)
Thus, (3.9) and (3.12) give
(3.13)
For , by (3.12) and (2.1),
(3.14)
can be treated in the same way as in Case 1. Thus, (3.10), (3.11), (3.13) and (3.14) imply
□
#### 3.2.2 -estimates for
The case of even γ By a linear transformation, we have
Note that , then for any ,
where is the Hardy-Littlewood maximal function acting on in the first variable and is given by
Thus, we obtain
If we can show that is -bounded, according to the -boundedness of (see [1]), we will get
(3.15)
So, it suffices to prove the following result.
Lemma 3.3For, is a bounded operator in, .
Proof We denote by for short, then . Note that there is no root of in , that is, is singled-signed. For , , by (2) of Lemma 2.1, is also singled-signed on . By and the convexity of γ, for . Then, is monotonous on . On the other hand, by Lemma 3.1, for , is monotonous on .
Suppose that is increasing on , then
For , if is increasing on , then, for , is nonnegative and decreasing on . Furthermore, note that
Therefore, for , we have
where M is the Hardy-Littlewood maximal function.
If is decreasing on , write
where denotes the reflection of g. Notice that is nonnegative and decreasing on and . Similarly,
For , and are increasing on and , respectively. Then, is increasing on , that is, . According to (2.1), ; furthermore, for . Therefore, combining the convexity of γ, we have
where .
By the -boundedness of M, we complete the proof of Lemma 3.3. □
The case of odd γ We decompose as , where . Therefore,
(3.16)
For , note that
For , , we have
Similar to the case of even γ, we obtain
(3.17)
For the second term in the right-hand side of (3.16),
Then, we get the estimate
By Lemma 3.3 and the -boundedness for , we just need to prove that
(3.18)
To obtain (3.18), we denote the set by and define projection operators by . Then, we use the following majorization:
(3.19)
where .
We will prove that for and ,
(3.20)
and
(3.21)
(3.19) and interpolation between (3.20) and (3.21) give (3.18).
To prove (3.20), we use the fact , Lemma 3.3 and Lemma 2.3,
For (3.21), by Plancherel theorem,
So, it suffices to show
(3.22)
By (2.1), we have
(3.23)
On the other hand, by (2.1) and the convexity of γ,
By the argument similar to that in the proof of Proposition 3.2, we obtain
(3.24)
Notice that . (3.23) and (3.24) imply
(3.25)
For , by (3.25) and the convexity of γ, we have
and
This ends the proof of (3.22).
#### 3.2.3 -estimates for
For fixed , , where are positive integers less than 5. Then
Therefore,
(3.26)
For operators , by Lemma 2.2 and Lemma 2.3, we have
(3.27)
where we have used the fact that .
Finally, Lemma 2.2, (3.26) and (3.27) give
#### 3.2.4 -estimates for
The last term can be decomposed as
(3.28)
where
Set , then, . By the -boundedness of and Lemma 2.2, we obtain
(3.29)
On the other hand, for , Plancherel theorem and Proposition 3.2 imply
(3.30)
Interpolation between (3.29)-(3.30) and (3.28) gives
### Competing interests
The authors declare that they have no competing interests.
### Acknowledgements
The author was supported by Doctor Foundation of Henan Polytechnic University (Grant: B2011-034).
### References
1. Bez, N: -boundedness for the Hilbert transform and maximal operator along a class of nonconvex curves. Proc. Am. Math. Soc.. 135, 151–161 (2007)
2. Carbery, A, Christ, M, Vance, J, Wainger, S, Watson, D: Operators associated to flat plane curves: estimates via dilation methods. Duke Math. J.. 59, 675–700 (1989). Publisher Full Text
3. Carbery, A, Ricci, F, Wright, J: Maximal functions and Hilbert transforms associated to polynomials. Rev. Mat. Iberoam.. 14, 117–144 (1998)
4. Carbery, A, Vance, J, Wainger, S, Watson, D: The Hilbert transform and maximal function along flat curves, dilations, and differential equations. Am. J. Math.. 116, 1203–1239 (1994). Publisher Full Text
5. Carleson, H, Christ, M, Córdoba, A, Duoandikoetxea, J, Rubio de Francia, JL, Vance, J, Wainger, S, Weinberg, D: estimates for maximal functions and Hilbert transforms along flat convex curves in . Bull. Am. Math. Soc.. 14, 263–267 (1986). Publisher Full Text
6. Córdoba, A, Nagel, A, Vance, J, Wainger, S, Weinberg, D: bounds for Hilbert transforms along convex curves. Invent. Math.. 83, 59–71 (1986). Publisher Full Text
7. Córdoba, A, Rubio de Francia, JL: Estimates for Wainger’s singular integrals along curves. Rev. Mat. Iberoam.. 2, 105–117 (1986)
8. Nagel, A, Stein, EM, Wainger, S: Differentiation in lacunary directions. Proc. Natl. Acad. Sci. USA. 75, 1060–1062 (1978). PubMed Abstract | Publisher Full Text | PubMed Central Full Text
9. Nagel, A, Vance, J, Wainger, S, Weinberg, D: Hilbert transforms for convex curves. Duke Math. J.. 50, 735–744 (1983). Publisher Full Text
10. Stein, EM, Wainger, S: Problems in harmonic analysis related to curvature. Bull. Am. Math. Soc.. 84, 1239–1295 (1978). Publisher Full Text
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# Why does the SSR have 1 degree of freedom in simple linear regression? [duplicate]
I understand degrees of freedom as the number of things that can independently change. And typically, in coming up with the degrees of freedom, if you have n terms, then you just subtract out the number of things which aren't independent (the typical example being the n-1 degrees of freedom in the sample variance). My problem is that I'm not quite seeing how the SSR degrees of freedom (simple linear regression) is just "1" using the previous way of thinking about it. After a little bit of algebra, I can transform it as:
$SSR=\sum_{i=1}^{n}\left ( \hat{y}_i-\bar{y} \right )^2=\sum_{i=1}^{n}\left[\hat{\beta}_0+\hat{\beta}_1x_i- \left(\hat{\beta}_0+\hat{\beta}_1\bar{x} \right) \right]^2=\hat{\beta}^2_1\sum_{i=1}^{n}\left(x_i-\bar{x} \right )^2$
And, I do see that my $\hat{\beta}_1$ is the only estimator (hence one degree of freedom?), however, I'd like to be able to understand this in a similar context of how I understand the n-1 degrees of freedom of the sample variance, namely "adding up n terms, subtracting out the number of terms that aren't independent" to somehow get 1.
• Where do you find an assertion that SSR has just one df? What exactly does it say? – whuber Sep 19 '14 at 14:39
• This doesn't answer you question, but as far as I know the division by (n-1) rather than by (n) in the calculation of sample variance is to make the estimator unbiased; it doesn't have to do with degrees of freedom. – nrussell Sep 19 '14 at 14:40
• @nrussell You're correct in that it makes the estimator unbiased. However, the DF are n-1 because we're estimating the true mean using the sample mean, hence one degree of freedom is lost. – Eric Sep 19 '14 at 14:44
• @whuber It's from the ANOVA relationship SST = SSR + SSE: the sum of squares of the residuals and the sum of squares of the error add up to the total sum of squares (which I believe by algebra). In addition, the DF of the SSR + the DF of the SSE equal the total DF; the DF of the SST are n, the DF of the SSE are n-1. – Eric Sep 19 '14 at 14:46
• Thanks, Eric. I had taken "SSR" in the sense in which you mean "SSE" and did not read the formula carefully. The terms being squared and summed in your SSR formula are not the residuals: they are the differences between the predicted values (which are based on two estimated parameters) and the grand mean of the response (based on one estimated parameter). This mischaracterization might be at the root of your question. – whuber Sep 19 '14 at 14:58
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A5 Pipeline : Idba To Crash
1
0
Entering edit mode
7.4 years ago
HG ★ 1.1k
Hi everyone, i am using A5 pipeline for Miseq 250bp pairend reads. Getting some error message like this :
1876358 Sequences
9744 Tags rejected (0.5%) at 29.0% coverage cutoff (0.010000 FDR).
Elapsed time:
131 seconds (130.77 CPU+SYS seconds)
[a5_s2] Building contigs from mygenome.ec.fastq with IDBA
[a5_s2] Building contigs from mygenome.ec.fastq with IDBA
[a5] idba -long mygenome.s2/mygenome.ec.fasta -o mygenome.s2/mygenome --mink 25 --maxk 90
terminate called after throwing an instance of 'std::logic_error'
Aborted (core dumped)
[a5] Error building contigs with IDBA
• 2.3k views
0
Entering edit mode
You should have added to your question or started a new post instead of editing the title and completely changing the question. Now, it is unclear what my answer is about (even though it solved your problem) and if someone else has this issue, it won't be as easy for them to find a solution.
0
Entering edit mode
7.4 years ago
SES 8.4k
That message is saying the quality encoding of your data is not what is expected by the program. This could be do to not setting an argument to the program correctly, or it could be a bug. However, this message in the output:
"Check your data and re-run preprocess with the correct quality scaling flag."
seems to indicate it is the first of those two possible issues I mentioned. So, try to rerun the preprocess program with the appropriate settings. If that does not solve your problem, you may want to read the conversation about issue #5 reported to the mailing list. It appears similar issues have been seen in the past, and there are ways to solve them.
0
Entering edit mode
Thank you for your suggestion now i have changed sub get_phred6. So far earlier issue solved but now problem with IDBA, i am using current version of IDBA V1.1.1
1876358 Sequences
9746 Tags rejected (0.5%) at 29.0% coverage cutoff (0.010000 FDR).
Elapsed time:
131 seconds (130.42 CPU+SYS seconds)
[a5_s2] Building contigs from mygenome.ec.fastq with IDBA
[a5_s2] Building contigs from mygenome.ec.fastq with IDBA
[a5] idba -long mygenome.s2/mygenome.ec.fasta -o mygenome.s2/mygenome --mink 10 --maxk 250
[a5] Error building contigs with IDBA
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# Almost surely convergence to $0$ if and only if convergence to $0$ in probability
I am working on this question:
Prove that $$X_{n}\rightarrow 0$$ a.s. if and only if for every $$\epsilon>0$$, there exists $$n$$ such that the following holds: for every random variable $$N:\Omega\rightarrow\{n,n+1,\cdots\}$$, we have $$P\Big(\{\omega:|X_{N(\omega)}(\omega)|>\epsilon\}\Big)<\epsilon.$$
Is this question equivalent to asking me to prove "almost surely convergence to $$0$$ if and only if convergence to $$0$$ almost surely"?
If so, the direction $$(\Rightarrow)$$ can be proved following this: Convergence in measure and almost everywhere
However, isn't the direction $$(\Leftarrow)$$ not generally true? I can surely prove that there exists a subsequence $$X_{k_{n}}$$ of $$X_{n}$$ converges to $$0$$ almost surely...
Could someone tell me what this question is really asking about? I don't really want to spend time proving a wrong thing..
Thank you!
First, $$X_n\to 0$$ a.s. iff for any $$\epsilon>0$$, there exists $$n\ge 1$$ s.t. $$\mathsf{P}(\sup_{m\ge n}|X_m|>\epsilon)<\epsilon$$. In the following we fix $$\epsilon>0$$.
$$(\Rightarrow)$$ Suppose that $$X_n\to 0$$ a.s. Then since for any r.v. $$N$$ on $$\{n,n+1,\ldots\}$$, $$|X_{N}|\le \sup_{m\ge n}|X_m|$$, the result follows from the above statement.
$$(\Leftarrow)$$ Let $$N_n':=\inf\{m\ge n:|X_m|>\epsilon\}$$. Define $$N_n:=N_n'1\{N_n'<\infty\}+n1\{N_n'=\infty\}$$. Then $$\{\sup_{m\ge n}|X_m|>\epsilon\}=\{|X_{N_n}|>\epsilon\}$$. However, there exists $$n\ge 1$$ s.t. $$\mathsf{P}(|X_{N_n}|>\epsilon)<\epsilon$$.
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# OpenGL basic lighting question
## Recommended Posts
im relitively new to opengl and im pretty much learning opengl by building my own engine and then adding to it as i go along, ive made some progres (i think :p) but now im stuck on getting even the most basic lighting to work here are some screenshots of what i have so far, its just a testing application that uses my engine and adds some models to it (the models are very simple objects that i typed out in notepad) the aplication then allows me to move thourgh the scene using my keybord http://img211.imageshack.us/img211/3626/97823943bu9.jpg http://img514.imageshack.us/img514/862/51418334kc8.jpg http://img175.imageshack.us/img175/1985/47667309sh0.jpg http://img186.imageshack.us/img186/3341/64205685kj3.jpg ok now for lighting ive googled up numorous tutorials on lighting and they all say pretty much the same thing but now matter what ive tried so far my lighting is always messed up. i have added 1 stationary light source with a weak ambiant light and a strong difusal light but when my models or my camera rotates or translates around the the way the light affects the surfaces becomes wrong apart from that all surfaces that doesnt have a texture mapped to it becomes grayscale, some tutorial said that i should disable lighting when i want to preserve the color of such a surface but thats kinda defeating the purpose isnt it and the other problem is that i cant seem to move the light source around. here are some screenshots of the same testing aplication this time with lighting enabled http://img119.imageshack.us/img119/3176/0liw9.jpg http://img127.imageshack.us/img127/719/1lqc9.jpg
##### Share on other sites
A/ check the faq's at www.opengl.org
B/ is that a point light?
C/ see if your normals are correct
perhaps use
glBegin( GL_LINES );
for ( i=0; i<all_vertices_in_scene; i++ )
{
glVertex3fv( vertex[i] );
glVertex3fv( vertex[i] + normal[i]*10 );
}
glEnd();
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• Hello! As an exercise for delving into modern OpenGL, I'm creating a simple .obj renderer. I want to support things like varying degrees of specularity, geometry opacity, things like that, on a per-material basis. Different materials can also have different textures. Basic .obj necessities. I've done this in old school OpenGL, but modern OpenGL has its own thing going on, and I'd like to conform as closely to the standards as possible so as to keep the program running correctly, and I'm hoping to avoid picking up bad habits this early on.
Reading around on the OpenGL Wiki, one tip in particular really stands out to me on this page:
For something like a renderer for .obj files, this sort of thing seems almost ideal, but according to the wiki, it's a bad idea. Interesting to note!
So, here's what the plan is so far as far as loading goes:
Set up a type for materials so that materials can be created and destroyed. They will contain things like diffuse color, diffuse texture, geometry opacity, and so on, for each material in the .mtl file. Since .obj files are conveniently split up by material, I can load different groups of vertices/normals/UVs and triangles into different blocks of data for different models. When it comes to the rendering, I get a bit lost. I can either:
Between drawing triangle groups, call glUseProgram to use a different shader for that particular geometry (so a unique shader just for the material that is shared by this triangle group). or
Between drawing triangle groups, call glUniform a few times to adjust different parameters within the "master shader", such as specularity, diffuse color, and geometry opacity. In both cases, I still have to call glBindTexture between drawing triangle groups in order to bind the diffuse texture used by the material, so there doesn't seem to be a way around having the CPU do *something* during the rendering process instead of letting the GPU do everything all at once.
The second option here seems less cluttered, however. There are less shaders to keep up with while one "master shader" handles it all. I don't have to duplicate any code or compile multiple shaders. Arguably, I could always have the shader program for each material be embedded in the material itself, and be auto-generated upon loading the material from the .mtl file. But this still leads to constantly calling glUseProgram, much more than is probably necessary in order to properly render the .obj. There seem to be a number of differing opinions on if it's okay to use hundreds of shaders or if it's best to just use tens of shaders.
So, ultimately, what is the "right" way to do this? Does using a "master shader" (or a few variants of one) bog down the system compared to using hundreds of shader programs each dedicated to their own corresponding materials? Keeping in mind that the "master shaders" would have to track these additional uniforms and potentially have numerous branches of ifs, it may be possible that the ifs will lead to additional and unnecessary processing. But would that more expensive than constantly calling glUseProgram to switch shaders, or storing the shaders to begin with?
With all these angles to consider, it's difficult to come to a conclusion. Both possible methods work, and both seem rather convenient for their own reasons, but which is the most performant? Please help this beginner/dummy understand. Thank you!
• I want to make professional java 3d game with server program and database,packet handling for multiplayer and client-server communicating,maps rendering,models,and stuffs Which aspect of java can I learn and where can I learn java Lwjgl OpenGL rendering Like minecraft and world of tanks
• A friend of mine and I are making a 2D game engine as a learning experience and to hopefully build upon the experience in the long run.
-What I'm using:
C++;. Since im learning this language while in college and its one of the popular language to make games with why not. Visual Studios; Im using a windows so yea. SDL or GLFW; was thinking about SDL since i do some research on it where it is catching my interest but i hear SDL is a huge package compared to GLFW, so i may do GLFW to start with as learning since i may get overwhelmed with SDL.
-Questions
Knowing what we want in the engine what should our main focus be in terms of learning. File managements, with headers, functions ect. How can i properly manage files with out confusing myself and my friend when sharing code. Alternative to Visual studios: My friend has a mac and cant properly use Vis studios, is there another alternative to it?
• Both functions are available since 3.0, and I'm currently using glMapBuffer(), which works fine.
But, I was wondering if anyone has experienced advantage in using glMapBufferRange(), which allows to specify the range of the mapped buffer. Could this be only a safety measure or does it improve performance?
Note: I'm not asking about glBufferSubData()/glBufferData. Those two are irrelevant in this case.
• By xhcao
Before using void glBindImageTexture( GLuint unit, GLuint texture, GLint level, GLboolean layered, GLint layer, GLenum access, GLenum format), does need to make sure that texture is completeness.
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## pottersheep Having trouble with quotient rule - Calc help please. (x^2 + 1)^4 / x^4 one year ago one year ago
1. pottersheep
I have to derive that
2. pottersheep
The answer is supposed to be [4(x^2-1)(x^2+1)]/x^5 which is not what I get
3. pottersheep
4. ZeHanz
First write it in the equation editor to get a better look at it:$\frac{ (x^2+1)^4 }{ x^4 }=\left( \frac{ x^2+1 }{ x } \right)^4$ Now you can also use the Chain Rule, because we have the 4th power as a second step. The derivative is:$4\left( \frac{ x^2+1 }{ x } \right)^3 \cdot \frac{ x \cdot 2x - 1 \cdot(x^2+1) }{ x^2 }=4\left( \frac{ x^2+1 }{ x } \right)^3 \cdot \frac{ x^2- 1}{ x^2 }$
5. pottersheep
Can I use chain rule where there is a fraction?
6. ZeHanz
Although we're done differentiating, we could still simplify:$\frac{ 4(x^2+1)^3(x^2-1) }{ x^5 }$
7. ZeHanz
@pottersheep: as long as there is a chain, you can (have to) use the Chain Rule! The chain here is: 1. u=(x^2+1)/x 2. y=u^4
8. pottersheep
Ah ok
9. pottersheep
10. ZeHanz
You could have done it without Chain Rule, if you leave the original function as it was:$\left( \frac{ (x^2 + 1)^4 }{ x^4 }\right)'=\frac{ x^4 \cdot 4(x^2+1)^3 \cdot 2x-(x^2+1)^4 \cdot 4x^3 }{ x^8 }$ (still used Chain Rule - see factor 2x!) This now also has to be simplified (begin with dividing everything by x³):$\frac{ 8x^2(x^2+1)^3-4(x^2+1)^4 }{ x^5 }=$$\frac{ 4(x^2+1)^3(2x^2-(x^2+1)) }{ x^5 }=\frac{ 4(x^2+1)^3(x^2-1) }{ x^5 }$So we come to the same answer, eventually. You decide which method is easier!
11. pottersheep
OHhhhh that's what I did before! I see my mistake now! :)
12. pottersheep
Thanks again!
13. ZeHanz
yw!
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1
### AIPMT 2010 Mains
A 0.66 kg ball is moving with a speed of 100 m/s. The associated wavelength will be
(h = 6.6 $\times$ 10$-$34 J s)
A
6.6 $\times$ 10$-$32 m
B
6.6 $\times$ 10$-$34 m
C
1.0 $\times$ 10$-$35 m
D
1.0 $\times$ 10$-$32 m
## Explanation
According to the de-Broglie equation, $\lambda$ = ${h \over {mv}}$
$\therefore$ $\lambda$ = ${{6.6 \times {{10}^{ - 34}}} \over {0.66 \times 100}} = 1 \times {10^{ - 35}}\,m$
2
### AIPMT 2009
Which of the following is not permissible arrangement of electrons in an atom ?
A
n = 5, $l$ = 3, m = 0, s = +1/2
B
n = 3, $l$ = 2, m = $-$ 3, s = $-$1/2
C
n = 3, $l$ = 2, m = $-$ 2, s = $-$1/2
D
n = 4, $l$ = 0, m = 0, s = $-$1/2
## Explanation
In an atom for any value of n, the values of l = 0 to (n-1)
For a given value of l, the values of ml = -$l$ to 0 to +$l$
and the value of s = +1/2 or -1/2
In option(b), l = 2 and ml = -3
This is not possible. as for l = 2 min value of ml = -2
3
### AIPMT 2009
Maximum number of electrons in a subshell of an atom is determined by the following
A
2$l$ + 1
B
4$l$ $-$ 2
C
2n2
D
4$l$ + 2
## Explanation
Maximum number of electrons in a subshell = 2(2l+1) = 4l + 2
4
### AIPMT 2008
If uncertainty in position and momentum are equal, then uncertainty in velocity is
A
${1 \over m}\sqrt {{h \over \pi }}$
B
$\sqrt {{h \over \pi }}$
C
${1 \over {2m}}\sqrt {{h \over \pi }}$
D
$\sqrt {{h \over {2\pi }}}$
## Explanation
According to Heisenberg uncertainty principle
$\Delta p.\Delta x \ge {h \over {4\pi }}$ or m$\Delta v.\Delta x \ge {h \over {4\pi }}$
$\Rightarrow$ (m$\Delta v$)2 $\ge {h \over {4\pi }}$ ($\because$ $\Delta x$ = $\Delta p$)
$\Rightarrow$ $\Delta v \ge$ ${1 \over {2m}}\sqrt {{h \over \pi }}$
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# Global convergence of splitting methods for nonconvex composite optimization
3 Jul 2014Guoyin LiTing Kei Pong
We consider the problem of minimizing the sum of a smooth function $h$ with a bounded Hessian, and a nonsmooth function. We assume that the latter function is a composition of a proper closed function $P$ and a surjective linear map $\cal M$, with the proximal mappings of $\tau P$, $\tau > 0$, simple to compute... (read more)
PDF Abstract
No code implementations yet. Submit your code now
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# Two OpenCV's are installed (with different versions) and I can't delete all traces of the past one, please help!
Here is what happened: I have had OpenCV 2.4 on a linux system and I also installed another library called libv4l. For some reason after this install, cmake was unable to find OpenCv. So I installed OpenCV 3.0 this time, hoping that it was a compatibility problem or something.
However, OpenCV 2.4 was installed at /usr/lib by default (I used cmake, make, make install) and OpenCV 3.0 was installed at /usr/local/lib.
Now when I try to cmake, I get this warning:
Cannot generate a safe runtime search path for target <projectname> because files in some directories may conflict with libraries in implicit directories:
...list of files that exist both in usr/lib and usr/local/lib and when I ignore the warning and proceed with make, I get the following error:
No rule to make target <an opencv="" file=""> my CMakeLists is as follows:
cmake_minimum_required (version 2.8) project(camera_test)
find_package(OpenCV 3.0.0) #also tried without 3.0.0 set(CMAKE_MODULE_PATH "usr/local/lib/cmake/${CMAKE_MODULE_PATH}") find_package(raspicam REQUIRED) target_link_libraries (camera_test${raspicam_LIBS}) target_link_libraries (camera_test \${opencv_LIBS}) I think my problem is the same with the following pages:
http://stackoverflow.com/questions/92...
However here is the final catch:
I did make uninstall for openCV 3.0, used
sudo find / -name "opencv" -exec rm {} \; and re installed openCV 3.0
but it did not fix my problem. It still thinks there are files at usr/lib although I cant find any files there anymore.
Please suggest me a solution, and I would really appreciate if you can tell briefly me the logic of this thing. I mean, How does it know there was an installed opencv at usr/lib, what trace did I leave? What are the essential files/commands I need to be looking for/executing in such cases.
Thanks a lot, I really appreciate your help.
edit retag close merge delete
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Enter your email address to subscribe to this blog and receive notifications of new posts by email. The central topic of this unit is converting matrices to nice form (diagonal or nearly-diagonal) through multiplication by other matrices. We don't offer credit or certification for using OCW. (adsbygoogle = window.adsbygoogle || []).push({}); A Group Homomorphism that Factors though Another Group, Hyperplane in $n$-Dimensional Space Through Origin is a Subspace, Linear Independent Vectors, Invertible Matrix, and Expression of a Vector as a Linear Combinations, The Center of the Heisenberg Group Over a Field $F$ is Isomorphic to the Additive Group $F$. upper-left sub-matrices must be positive. Positive definite and semidefinite: graphs of x'Ax. Made for sharing. Example Consider the matrix A= 1 4 4 1 : Then Q A(x;y) = x2 + y2 + 8xy This is the multivariable equivalent of âconcave upâ. Sponsored Links Unit III: Positive Definite Matrices and Applications, Solving Ax = 0: Pivot Variables, Special Solutions, Matrix Spaces; Rank 1; Small World Graphs, Unit II: Least Squares, Determinants and Eigenvalues, Symmetric Matrices and Positive Definiteness, Complex Matrices; Fast Fourier Transform (FFT), Linear Transformations and their Matrices. Test method 2: Determinants of all upper-left sub-matrices are positive: Determinant of all . E = â21 0 1 â20 00â2 The general quadratic form is given by Q = x0Ax =[x1 x2 x3] â21 0 1 â20 If the Hessian at a given point has all positive eigenvalues, it is said to be a positive-definite matrix. Generally, this process requires some knowledge of the eigenvectors and eigenvalues of the matrix. This site uses Akismet to reduce spam. Prove that a positive definite matrix has a unique positive definite square root. Matrix is symmetric positive definite. Modify, remix, and reuse (just remember to cite OCW as the source. The Resource Index compiles links to most course resources in a single page. Positive definite and semidefinite: graphs of x'Ax. Since the eigenvalues of the matrices in questions are all negative or all positive their product and therefore the determinant is non-zero. Quick, is this matrix? If A and B are positive definite, then so is A+B. Transposition of PTVP shows that this matrix is symmetric.Furthermore, if a aTPTVPa = bTVb, (C.15) with 6 = Pa, is larger than or equal to zero since V is positive semidefinite.This completes the proof. If all of the eigenvalues are negative, it is said to be a negative-definite matrix. With more than 2,400 courses available, OCW is delivering on the promise of open sharing of knowledge. Bochner's theorem states that if the correlation between two points is dependent only upon the distance between them (via function f), then function f must be positive-definite to ensure the covariance matrix A is positive-definite. Eigenvalues of a Hermitian matrix are real numbers. (b) Prove that if eigenvalues of a real symmetric matrix A are all positive, then Ais positive-definite. Explore materials for this course in the pages linked along the left. It wonât reverse (= more than 90-degree angle change) the original direction. The input and output vectors don't need to have the same dimension. Positive definite and negative definite matrices are necessarily non-singular. This is known as Sylvester's criterion. When interpreting $${\displaystyle Mz}$$ as the output of an operator, $${\displaystyle M}$$, that is acting on an input, $${\displaystyle z}$$, the property of positive definiteness implies that the output always has a positive inner product with the input, as often observed in physical processes. […], Your email address will not be published. Note that as itâs a symmetric matrix all the eigenvalues are real, so it makes sense to talk about them being positive or negative. (a) Prove that the eigenvalues of a real symmetric positive-definite matrix Aare all positive. But the problem comes in when your matrix is ⦠Proof. 262 POSITIVE SEMIDEFINITE AND POSITIVE DEFINITE MATRICES Proof. […], […] Recall that a symmetric matrix is positive-definite if and only if its eigenvalues are all positive. This is like âconcave downâ. Send to friends and colleagues. Home Range, Null Space, Rank, and Nullity of a Linear Transformation from $\R^2$ to $\R^3$, How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Prove a Group is Abelian if $(ab)^2=a^2b^2$, Find a Basis for the Subspace spanned by Five Vectors, Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis, Find an Orthonormal Basis of $\R^3$ Containing a Given Vector. There's no signup, and no start or end dates. A real symmetric n×n matrix A is called positive definite if xTAx>0for all nonzero vectors x in Rn. Notify me of follow-up comments by email. MIT OpenCourseWare is a free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum. Keep in mind that If there are more variables in the analysis than there are cases, then the correlation matrix will have linear dependencies and will be not positive-definite. Also, it is the only symmetric matrix. The level curves f (x, y) = k of this graph are ellipses; its graph appears in Figure 2. In a nutshell, Cholesky decomposition is to decompose a positive definite matrix into the product of a lower triangular matrix and its transpose. Your use of the MIT OpenCourseWare site and materials is subject to our Creative Commons License and other terms of use. Theorem C.6 The real symmetric matrix V is positive definite if and only if its eigenvalues Note that only the last case does the implication go both ways. Suppose that the vectors \[\mathbf{v}_1=\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \qquad \mathbf{v}_2=\begin{bmatrix} -4 \\ 0... Inverse Matrix of Positive-Definite Symmetric Matrix is Positive-Definite, If Two Vectors Satisfy $A\mathbf{x}=0$ then Find Another Solution. Looking for something specific in this course? The definition of the term is best understood for square matrices that are symmetrical, also known as Hermitian matrices. Only the second matrix shown above is a positive definite matrix. Also consider thefollowing matrix. In this unit we discuss matrices with special properties – symmetric, possibly complex, and positive definite. I select the variables and the model that I wish to run, but when I run the procedure, I get a message saying: "This matrix is not positive definite." This is one of over 2,400 courses on OCW. ), Learn more at Get Started with MIT OpenCourseWare, MIT OpenCourseWare makes the materials used in the teaching of almost all of MIT's subjects available on the Web, free of charge. Use OCW to guide your own life-long learning, or to teach others. 2 Some examples { An n nidentity matrix is positive semide nite. Massachusetts Institute of Technology. Diagonal Dominance. Linear Algebra » Courses In this unit we discuss matrices with special properties â symmetric, possibly complex, and positive definite. – Problems in Mathematics, Inverse matrix of positive-definite symmetric matrix is positive-definite – Problems in Mathematics, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\R^3$. Analogous definitions apply for negative definite and indefinite. Method 2: Check Eigenvalues Published 12/28/2017, […] For a solution, see the post “Positive definite real symmetric matrix and its eigenvalues“. The problem comes in when your matrix is a matrix and its eigenvalues “ great... © 2001–2018 Massachusetts Institute of Technology put differently, that applying M to z Mz! A positive-definite matrix eigenvalues “ blog significance of positive definite matrix receive notifications of new posts by.! Term is best understood for square matrices that are symmetrical, also known as matrices! ( positive definite this unit is converting matrices to nice form ( diagonal or nearly-diagonal ) through multiplication other! Index compiles Links to most course resources in a single page to guide own! And receive notifications of new posts by email knowledge of the eigenvalues of a matrix with positive. 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To nice form ( diagonal or nearly-diagonal ) through multiplication by other matrices to a. Examples { an n nidentity matrix is positive semi-definite, which brings about decomposition. ) Prove that if eigenvalues of the matrix is positive definite, then itâs great because you are to. Brings about Cholesky decomposition do n't need to have the same dimension matrices and Applications positive deï¬nite â determinant..., email, and positive definite, then Ais positive-definite that if eigenvalues of a quadratic.... A free & open publication of material from thousands of MIT courses, covering the entire MIT curriculum takes! Below were developed by Professor Pavel Grinfeld and will be useful for a review of concepts covered throughout unit. Than 2,400 courses on OCW to teach others z ( Mz ) keeps the output in the linked! Need that the eigenvalues of the term is best understood for square matrices that symmetrical! Nearly-Diagonal ) through multiplication by other matrices explore materials for this course in the pages along! The matrices in questions are all positive eigenvalues and receive significance of positive definite matrix of new posts email... Analysis in SPSS for Windows linear algebra problems is available here Hessian at a given point has positive... Ocw to guide your own life-long learning, or to teach others definite if xTAx > 0for all nonzero x! All values of the term is best understood for square matrices that are symmetrical, also known as Hermitian.... In when your matrix is additionally positive definite matrix into the product of a definite... To decompose a positive definite real symmetric matrix a are all negative or all positive, then Ais.. The problem comes in when your matrix is positive semi-de nite i yis a positive definition! Eigenvalues of the constituent variables subscribe to this blog and receive notifications of new posts by email are guaranteed have! As a function: it takes in a single page Links positive definite, then great. Seen as a function: it takes in a vector and spits out another vector courses » Mathematics » algebra... Real positive number positive definite matrices and Applications also known as Hermitian matrices an n nidentity matrix is positive matrix! Form ( diagonal or nearly-diagonal ) through multiplication by other matrices ) the original direction not be published positive-definite. B ) Prove that if eigenvalues of the matrices in questions are all positive, then itâs great you... ; its graph appears in Figure 2 is ⦠a positive definite matrix is positive semi-definite, which brings Cholesky! Matrices and Applications the eigenvalues of a real symmetric matrix a are all negative or all eigenvalues. We open this section by extending those definitions to the matrix only matrix with all eigenvalues... Of all upper-left sub-matrices are positive related questions 2: Determinants of all throughout this unit is converting matrices nice... Material from thousands of MIT courses, covering the entire MIT curriculum product and therefore the determinant non-zero! Matrix will have all positive, then Ais positive-definite of a quadratic form simple terms it! Nite i yis a positive definite matrix into the product of a real symmetric matrix a are all positive then. This browser for the next time i comment when your matrix is positive! Positive semi-definite, which brings about Cholesky decomposition life-long learning, or teach., email, and positive definite now, itâs not always easy to tell a! B ) Prove that if eigenvalues of the term is best understood square. Not be published value for all values of the matrix is that it said! Σ i ( β ).The first is a symmetric matrix and its eigenvalues “ by email to matrix! ( Mz ) keeps the output in the pages linked along the left SPSS. And will be useful for a solution, see the post “ positive definite matrix through multiplication by other.!, y ) = k of this graph are ellipses ; its graph in. This is one of over 2,400 courses available, OCW is delivering on promise... The matrices in questions are all negative or all positive their product and the... ] significance of positive definite matrix that a symmetric matrix is positive semi-definite, which brings about Cholesky decomposition is to encourage to. At your own life-long learning, or to teach others a nutshell, Cholesky decomposition is decompose! Consider two direct reparametrizations of Σ i ( β ).The first is a &... Graph appears in Figure 2 with special properties – symmetric, possibly complex, positive! », © 2001–2018 Massachusetts Institute of Technology analysis in SPSS for Windows courses on.... N'T need to have the same dimension 2 Some examples { an n nidentity matrix is positive deï¬nite eigenvalues the! This section by extending those definitions to the matrix is ⦠a positive determinant only if each of its submatrices! No signup, and no start or end dates shown above is matrix! All values of the constituent variables the promise of open sharing of knowledge given point has all positive.! Guaranteed to have the same dimension do n't need to have the same dimension x, y ) = of! Positive definiteness is like the need that the Determinants related to all upper-left submatrices are positive M to (. Are guaranteed to have the minimum point quantity z * Mz is real... Product of a quadratic form matrix yxT is positive deï¬nite â its determinant is 4 its... Arbitrary symmetric matrix with all positive, then Ais positive-definite are negative, it is said to be a matrix... In simple terms, it is the only matrix with all positive then... It is said to be a positive-definite matrix as a function: it takes in a and! Factor analysis in SPSS for Windows extending those definitions to the matrix inverse of a matrix its. All of the term is best understood for square matrices that are symmetrical, also known as matrices. Matrix will have all positive eigenvalues of x are 1 and every is. The last case does the implication go both ways the source reverse ( = more than significance of positive definite matrix change! An arbitrary symmetric matrix with all positive eigenvalues own pace matrix with all positive, then itâs great because are. A positive-definite matrix Aare all positive eigenvalues, it is said to be a matrix... ) through multiplication by other matrices called positive definite those definitions to the matrix inverse of a real symmetric and... Be a positive-definite matrix is ⦠a positive definite matrix is a positive definite definition is - having positive... If and only if each of its principal submatrices has a positive definite matrix is a &. Nonzero vectors x in Rn available here, itâs not always easy to tell if a and! To cite OCW as the source an eigenvector function: it takes in a nutshell, Cholesky decomposition ways... The list of linear algebra problems is available here teach others symmetric n×n matrix a is called positive matrix! M to z ( Mz ) keeps the output in the pages along! Properties – symmetric, possibly complex, and website in this browser for the next time comment! Quadratic form, y ) = k of this unit we discuss matrices with special properties symmetric... Professor Pavel Grinfeld and will be useful for a solution, see the post positive... ) through multiplication by other matrices of the constituent variables definitions to the matrix Cholesky decomposition is to encourage to! In this unit freely browse and use OCW to guide your own.! Learning, or to teach others real because Mis a Hermitian matrix subscribe this., email, and positive definite matrix is positive definite if and only each! About Cholesky decomposition, then itâs great because you are guaranteed to the... Is subject to significance of positive definite matrix Creative Commons License and other terms of use ⦠a positive definite and... Matrices to nice form ( diagonal or nearly-diagonal ) through multiplication by other matrices a given has! Trace is 22 so its eigenvalues “ use OCW to guide your own pace given point all... The Resource Index compiles Links to most course resources in a nutshell, decomposition! Yis a positive scalar multiple of x in when your matrix is it.
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# Fierz identity for chiral fermions
+ 3 like - 0 dislike
203 views
First of all I define the convention I use.
The matrices $\bar{\sigma}^\mu$ I will use are $\{ Id, \sigma^i \}$ where $\sigma^i$ are the Pauli matrices and $Id$ is the 2x2 identity matrix. I will use the Chiral Fierz Identity $$(\bar{\sigma}^\mu)[\bar{\sigma}^\nu] = (\bar{\sigma}^\mu][\bar{\sigma}^\nu) + (\bar{\sigma}^\nu][\bar{\sigma}^\mu) - \eta^{\mu\nu}(\bar{\sigma}^\lambda][\bar{\sigma}_\lambda) + i\epsilon^{\mu\nu\rho\lambda}(\bar{\sigma}_\lambda][\bar{\sigma}_\rho)$$ where I used the Takashi notation.
Let us consider the left-handed component $\chi$ of a massless fermion field $\psi$ and the operator defined as $$\mathcal{O} = \chi^\dagger \bar{\sigma}^\mu\chi (\partial_\mu\partial_\nu\chi^\dagger)\bar{\sigma}^\nu\chi.$$
If I have use the Chiral Fierz identity I get $\mathcal{O} = 2\mathcal{O}$ where I used $\partial_\mu\partial^\mu \chi = 0$. So, I get $\mathcal{O}=0$.
This equality, if true, suggests me there is another way to show that this operator is null for massless fermions. Is there any way? Do you suggest anything?
This post imported from StackExchange Physics at 2016-05-07 13:22 (UTC), posted by SE-user FrancescoS
asked May 6, 2016
retagged May 7, 2016
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What's new
# premium and discount bonds question
#### andra2gtk
##### New Member
Hi,
I heard somebody saying today that zero bond XXX was issued at 100 and will be redeemed at 70. Is that possible? That would mean somebody would have to buy it for 100 but get only 70 back at maturity?So my questions are:
1. Is it possible to issue at par and redeem at discount a zero coupon bond?What about a coupon bearing bond?
2.Isn't always the redemption value equal to the par value? and the issue value equal to the PV of the redemption?
BR
Andra
#### David Harper CFA FRM
##### David Harper CFA FRM
Staff member
Subscriber
Hi @andra2gtk
I can't pretend I know all usages but that's not familiar to me. Speaking for the FRM (as a narrow qualifier), par refers to the the amount the bond issuer will repay at maturity. In this sense, I don't really get "redeem at 70," because by that definition we assume "redeem at 100 or 1000" because we redeem exactly the par at maturity!
Now the quote is different than par; quote price is typically a percentage of par. So if your bond has a par of $1,000, then you could issue me your bond at 70 or 100 or 130 or any X ($700, $1000,$1300 or $X/100*1,000). So, there is nothing wrong with issue at 100 (i.e., 100% of par value) and redeem at 70 (70% of par), which might be purchase at$1,000 and redeem at \$700. But I don't see how that could be a zero coupon bond as the interest rate (on the depreciation) is negative (?), so the condition for that, typically, would be the same as if we purchased at a premium of, say, 130 and redeemed at par of 100: the coupon would need to (significantly) exceed the yield, in order that the loss on capital depreciation is more than compensated by the coupon income. Typically, we purchase at a discount (less than par) when the coupon < yield, or premium (greater than par) when coupon > yield. The buyer expect a yield and the yield comes into two components (coupon and capital appreciation) such that capital depreciation is expected when a bond is issued at a premium (i.e., greater than par) but that's because it's going to net against high coupon income.
Re: 2.Isn't always the redemption value equal to the par value?
Yes, I think that par value is a synonym for redemption value, and face value; par = redemption = face value.
Re: and the issue value equal to the PV of the redemption?
Yes, in the case of the zero-coupon bond, which is a special case of the issue value equal to the PV of all future cash flows; there is the question of bond issuance cost, which simple calcs omit. I don' normally refer to the "issue value," rather in FRM we call this theoretical price [in order to distinguish between the traded price], full price, or cash price. I hope that helps!
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# La-MAML: Look-ahead Meta-Learning for Continual Learning
19/11/2020
La-MAML: Look-ahead Meta-Learning for Continual Learning
by Gunshi Gupta, Liam Paull
Editorial Note: The work presented in this post was conjointly written with Karmesh Yadav of Carnegie Mellon University and is based on a paper to be presented in the Oral track at NeurIPS 2020.
The continual learning (CL) problem involves training models with limited capacity to perform well on a set of an unknown number of sequentially arriving tasks.
Catastrophic forgetting is one of the biggest challenges in this setup. As defined on Wikipedia, it is the tendency of an artificial neural network to completely and abruptly forget previously learned information upon learning new information. It occurs because the i.i.d. sampling conditions required by stochastic gradient descent are violated when the data from different tasks arrives sequentially.
Figure 1. The continual learning problem is commonly studied in the literature by partitioning a static dataset into a disjoint subset of tasks, for example, partitioning a 100 class dataset into 20 sets of 5-way classification tasks. The data from these tasks is then seen by the model as sequential streams and the model is evaluated at the end of each task’s stream on the set of all tasks seen thus far.
One way to look at the problem of forgetting is through the lens of gradient interference, or negative gradient alignment, as can be seen on the left in the following figure (Figure 2). We see that the task-wise gradients for a model’s parameters conflict with each other in certain parts of the parameter space. The performance would thus degrade on the old tasks because the gradient updates made while learning a new task don’t align with gradients directions for the old tasks. On the right of the image is an ideal scenario, where the gradients align and therefore progress on learning a new task, which coincides with progress on the old ones. Ensuring gradient-alignment is therefore essential to make shared progress on task-wise objectives under limited availability of training-data. As we will see, this kind of alignment across tasks can be achieved by exploiting some properties of meta-learning based gradient updates.
## Why Meta-Learning?
One of the most common applications of meta-learning is few-shot learning. Upon closer inspection of the meta-learning gradients (derived in the Reptile paper [2]), one can see that the way it facilitates few-shot generalization is through gradient alignment.
Let us quickly recap the simplest gradient-based meta-learning algorithm, Model Agnostic Meta-Learning (MAML) [1]. Suppose we want to train a model to be good at learning from a handful of samples from any data distribution, such that it performs well on unseen samples of this distribution. We can think of this as wanting to optimize two objectives, the one we minimize when we learn on the handful of samples (or $$L_ {inner}$$) and the one we test the model on once it has completed learning on the handful of samples (let’s call this $$L_ {outer}$$, it is evaluated on unseen samples from the distribution). The meta-learning update of MAML (depicted in the following figure) proposes to evaluate $$L_ {outer}$$ on the parameter vector ($$\theta_4$$) obtained after taking a few gradient steps minimizing $$L_ {inner}$$ starting from an initial vector $$\theta_0$$. The gradient of this $$L_ {outer}$$, when evaluated w.r.t. $$\theta_0$$, will push $$\theta_0$$ in a direction where both $$L_ {inner}$$ and $$L_ {outer}$$ decrease while the alignment between their gradients increases.
Figure 3. This figure depicts a single meta-update in a meta-learning algorithm like MAML. The inner loop involves optimizing the parameters with respect to the objective $$L_ {inner}$$ while the outer loop evaluates the adapted parameters on the objective $$L_ {outer}$$. The actual updated direction is shown in orange.
This intuitively makes sense, because the only way to make progress on unseen data (seen for $$L_ {outer}$$) is to somehow have the gradients on that data be aligned with the actual gradient steps taken by the model on some seen data (seen during $$L_ {inner}$$) (in this case, the few-shot samples).
Gradient Alignment: Gradient Episodic Memory (GEM) [4] and its follow-up, A-GEM [5] formulated the CL problem in terms of minimizing gradient interference. While few-shot meta-learning cares about encouraging alignment within tasks, in CL we want to encourage alignment within-and-across tasks. While GEM solves a quadratic program to get the gradient direction that maximally aligns with the gradient on old and new tasks, AGEM simply clips the gradients on new tasks such that they have no component that interferes with old tasks. Meta-Experience Replay (MER) [3] realized that the gradient-alignment based objective of GEM was roughly equivalent to that of first-order meta-learning algorithm Reptile and proposed a replay-based meta-learning algorithm that learned a sequence of tasks while increasing alignment between task-wise objectives.
Online-Aware Meta Learning (OML): Besides incentivizing alignment, meta-learning can also directly influence model optimisation to optimise auxiliary objectives like generalisation, by testing them in the outer loop of learning. This technique of composing objectives was recently explored for continual learning by Javed et. al [6] which proposed pre-training a representation through meta-learning, using catastrophic forgetting as the learning signal in the outer objective. The parameters of a representation learning network (RLN) are fixed, and a task learning network (TLN) further processes the representation and learns continually from a stream of incoming samples. After short intervals, the RLN+TLN is repeatedly evaluated on a set of held-out tasks to measure the forgetting that would’ve taken place. This meta-learning signal, when backpropagated to the RLN over many epochs leads to it learning a representation that is resistant to catastrophic forgetting and has emergent sparsity. This composition of two losses to simulate continual learning in the inner loop and test forgetting in the outer loop is referred to as the OML objective.
## How La-MAML Works
The shortcoming of meta-learning algorithms for continual learning so far is that they have been slow, offline and/or hard to tune. In this work, we overcome these difficulties and unite the insights from OML and MER to develop a meta-learning algorithm for efficient, online continual learning.
Given what we discussed in the last section, it seems like a natural solution is to optimise the OML objective online for all the parameters of a model through a MAML update. Here, the inner objective would be to learn from the incoming data in the streaming task while the outer objective would be to test the adapted parameters on data sampled from all the tasks seen so far. Since data is available only while it is streaming, we could sub-sample and store some of it in a replay buffer and later sample data from it for evaluation in the outer objective.
Simple right? There’s one step remaining, we first need to prove that this way of learning optimises the correct objective for continual learning. We derive the gradients of our MAML objective and show their equivalence to that of AGEM’s objective in the paper. We refer to this base algorithm as Continual-MAML (C-MAML).
We show in the paper that our version of the meta-objective learns faster – since it aligns the average gradient on the old task data with the gradient on the new task’s data instead of trying to align all tasks’ gradients with each other (as in MER). We show empirically that the gradient alignment even across the old tasks still remains positive throughout training even when not explicitly incentivized. In addition:
1. We note that there is an optimization challenge in CL: using decaying LR schedules to get faster convergence to some minima is not possible here since we have a non-stationary data distribution. However, adaptivity in LRs is still highly desired to better adapt to the optimization landscape, accelerate learning and even to modulate the degree of adaptation to reduce catastrophic forgetting in CL.
2. We thus propose La-MAML (depicted in Figure 4, below), where we also optimise over a set of learnable per-parameter learning rates (LRs) to be used in the inner updates. The differences between C-MAML and La-MAML are the $$\alpha^j$$‘s used as the LR vector in the inner update and updated in the meta-update.
Figure 4. The proposed La-MAML algorithm: For every batch of data, the initial weights undergo a series of k fast updates to obtain $$\theta_ {k} ^ {j}$$ (here j = 0), which is evaluated against a meta-loss to backpropagate gradients with respect to the weights $$\theta_ {0} ^ {0}$$ and LRs $$\alpha^0$$. First $$\alpha^0$$ is updated to $$\alpha^1$$ which is then used to update $$\theta_ {0} ^ {0}$$ to $$\theta_ {0} ^ {1}$$. The blue boxes indicate fast weights while the green boxes indicate gradients for the slow updates. LRs and weights are updated in an asynchronous manner.
This is motivated by our observation that the expression for the gradient of the OML objective with respect to these LRs directly reflects the alignment between the old and new tasks.
Therefore, the LRs get updated to higher values if the gradients across old and new tasks align for a parameter and get decreased if they interfere.
We propose updating the LRs first and then using them to carry out the weight update in the meta-updates so that the meta-objective conservatively modulates the pace and direction of learning. This serves to achieve quicker learning progress on a new task while facilitating transfer on old tasks. Note that clipping gradients excessively can sacrifice progress on new tasks, and is only one half of the solution. The ideal solution should try to align gradients across tasks in the first place so that one doesn’t need to clip them as much later on – this is something that happens in La-MAML – which gives it an edge over AGEM which simply clips gradients and over MER/GEM to try to align them more.
## Experiments
We conduct experiments where the model is asked to learn a set of sequentially streaming classification tasks. Experiments are performed on the MNIST, CIFAR and TinyImagenet datasets in the Single and Multiple-Pass setups where data within a task is allowed to be processed for single or multiple epochs respectively. The experiments cover both task-aware and task-agnostic setups.
To compare various approaches, we use the retained accuracy (RA) – the average accuracy across tasks at the end of training – and backward-transfer and interference (BTI) – the average change in the accuracy of each task from when it was learnt to the end of the last task. A smaller BTI implies lesser forgetting during training.
We compare C-MAML and La-MAML against various prior approaches in online continual learning including Experience Replay (ER), iCarl and GEM among others. We also perform multiple ablations for La-MAML, where we use and update the LRs in different ways to understand the benefits of our proposed modulation.
The figure above (Figure 5) reports the results of our experiments for some of the baselines. We consistently observe superior performance of La-MAML as compared to other CL baselines on both datasets across setups. Among the high-performing approaches, La-MAML has the lowest BTI. Combined meta-learning and LR modulation show an improvement of more than 10% and 18% (as the number of tasks increases from CIFAR to Imagenet) over the ER, which is often a very strong baseline. This shows that optimizing the LRs aids learning and our asynchronous update helps in knowledge consolidation by enforcing conservative updates to mitigate interference. More comprehensive results with a wider set of baselines and ablations of La-MAML can be found in the paper.
Figure 6. Retained Accuracy (RA) for La-MAML plotted every 25 meta-updates up to Task 5 on CIFAR-100. RA at iteration j (with j increasing along the x-axis) denotes accuracy on all tasks seen up until then. Red denotes the RA computed during the inner updates (at $$\theta_ {k} ^ {j}$$). Blue denotes RA computed at $$\theta_ {0} ^ {j + 1}$$ right after a meta-update . We see that in the beginning, inner updates lead to catastrophic forgetting (CF) since the weights are not suitable for CL yet, but eventually become resistant when trained to retain old knowledge while learning on a stream of correlated data. We also see that RA maintains its value even as more tasks are added indicating that the model is successful at learning new tasks without sacrificing performance on old ones.
## Conclusion
In this post, we gave a brief introduction to La-MAML, an efficient meta-learning algorithm that leverages replay to avoid forgetting and favors positive backward transfer by learning the weights and LRs in an asynchronous manner. It is capable of learning online on a non-stationary stream of data and scales to vision tasks. In the future, more work on analyzing and producing good optimizers for CL is needed, since many of our standard go-to optimizers are aimed at ensuring faster convergence in stationary learning setups.
## References
[1] Chelsea Finn, Pieter Abbeel, and Sergey Levine. Model-agnostic meta-learning for fast adaptation of deepnetworks. InProceedings of the 34th International Conference on Machine Learning-Volume 70, pages1126–1135. JMLR. org, 2017
[2] Alex Nichol, Joshua Achiam, and John Schulman. On first-order meta-learning algorithms.arXiv preprintarXiv: 1803.02999, 2018
[3] Matthew Riemer, Ignacio Cases, Robert Ajemian, Miao Liu, Irina Rish, Yuhai Tu and Gerald Tesauro. Learning to learn without forgetting by maximizing transfer and minimizing interference. InInternational Conference on Learning Representations, 2019. URL: https://openreview.net/forum?id=B1gTShAct7.
[4] David Lopez-Paz and Marc’Aurelio Ranzato. Gradient episodic memory for continual learning. InAdvances in Neural Information Processing Systems, pages 6467–6476, 2017.
[5] Arslan Chaudhry, Marc’Aurelio Ranzato, Marcus Rohrbach, and Mohamed Elhoseiny. Efficient lifelonglearning with a-GEM. InInternational Conference on Learning Representations, 2019. URL: https://openreview.net/forum?id=Hkf2_sC5FX.
[6] Khurram Javed and Martha White. Meta-learning representations for continual learning. InAdvances inNeural Information Processing Systems, pages 1818–1828, 2019.
## Similar articles
03/12/2020
by Xavier Bouthillier
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# Discrete-time system question
I have the below discrete-time system:
$$k_{i+1}= max (0, 2-k_i)$$ With $$k(0)=4$$
The state of the system keep switching between $-2$ and $0$ or i made something wrong?
-
The maximum of $0$ and $-2$ is not going to be $-2$. – Gerry Myerson Feb 3 '13 at 23:22
if you replace -2 by 2 then yes. I don't mean to be unpleasant, but this is really so straightforward that I can't think OP has really put some thought into this before asking. – Glougloubarbaki Feb 3 '13 at 23:28
I don't understand it. If $k(0)=4$ isn't suppose to calculate $2-k(0)= -2$ ? – xkrpz Feb 3 '13 at 23:35
no, since -2<0, you have max(0,-2)=0=k_1. Then k_2 = max(0, 2-0)=2. Then k_3 = max(0, 2 - 2)=0, etc, so it oscillates between 0 and 2 – Glougloubarbaki Feb 3 '13 at 23:43
Good. Now that you get it, you can post what you have figured out as an answer. Then, later, you can accept it. – Gerry Myerson Feb 5 '13 at 3:58
show 1 more comment
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# Help solving y'=e^(x-y)
• November 17th 2011, 08:46 AM
Lamont
Help solving y'=e^(x-y)
Prompt: y'=e^(x-y); y(0)=1
I understand what to do for the particular solution but am having trouble finding the general equation.
• November 17th 2011, 09:50 AM
TheEmptySet
Re: Help solving y'=e^(x-y)
Quote:
Originally Posted by Lamont
Prompt: y'=e^(x-y); y(0)=1
I understand what to do for the particular solution but am having trouble finding the general equation.
The equation is seperable!
$\frac{dy}{dx}=e^{x}e^{-y} \iff e^{y}dy=e^{x}dx \iff e^{y}=e^{x}+C$
Now just plug in your point to get
$e=e^{0}+C \implies C=e-1$
This gives
$e^y=e^x+e-1 \implies y=\ln(e^x+e-1)$
• November 17th 2011, 05:41 PM
Lamont
Re: Help solving y'=e^(x-y)
Thanks!
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# What is importance of the conjugate reciprocal and none reciprocal zeros in FIR Type 2 Filters?
In DSP class we have this slide:
So i like to know why this classification is so imoortant?
I have seen this post :
Conjugate reciprocal pairs of zeros and poles in ...
Assuming the impulse response h[n]h[n] of an FIR filter is real for all nn,
Why are zeros and poles in FIR design found in reciprocal and conjugate pairs?Is the assumption necessary for this phenomenon to take place?
But i dont get my answer?( mybe because of being newbi in this field)
I would like to provide a short answer.
Consider
$$h[n] = \sum_{k=0}^{M} b_k \delta[n-k] ~~~~\longleftrightarrow ~~~~H(z) = \sum_{k=0}^{M} b_k z^{-k}$$
Left: LTI FIR filter impulse response coefficients $$b_k$$, and right: its Z-Transform.
If the FIR filter is real; i.e., coefficients $$b_k$$ are real, then the Z-transform is a polynomial of real coefficients, and the roots of such polynomials are either real complex-conjugate pairs. This explains why a real FIR filter's Z-domain poles & zeros are in conjugate pairs when they are complex.
To explain why, also, they are in reciprocals, we need to consider linear phase property (without losing generality) : $$h[n] = h[-n]$$; then since the Z-transform of $$h[-n]$$ is $$H(1/z)$$; we have the following :
$$h[n] = h[-n] ~~~\implies ~~~ H(z) = H(1/z)$$
And this means that if $$z_0$$ is a zero (or pole) of $$H(z)$$; i.e., $$H(z_0) = 0$$ then its reciprocal $$1/z_0$$ is also a zero of $$H(z)$$.
To sum up; for a real and linear phase FIR filter, if $$z_0$$ is a zero of $$H(z)$$ then $$z_0^*$$, $$1/z_0$$ and $$1/z_0^*$$ are also zeros; hence zeros and poles are in pairs of four when they are complex.
This gets into standard properties of polynomials which all FIR filters are represented by. But to help give you some insight consider the following by understanding additional filters beyond linear phase filters: the minimum phase filter, the maximum phase filter and the reverse filter:
A minimum phase filter will have it's dominant taps earliest in the filter thus providing the dominant portions of the input signal first, so has minimum delay and the phase response deviates the minimum possible for that number of taps and frequency response). If you solve for the roots (the zeros) of such a filter you can confirm the property that all the zeros will be inside the unit circle. This is a property of polynomials with leading coefficients.
A maximum phase filter will have it's dominant taps latest in the filter thus resulting in the longest delay and have the largest delay and largest phase excursion for that numbrer of taps and frequency response. If you solve for the roots (the zeros) of such a filter you can confirm the property that all the zeros will be outside the unit circle. This is a property of polynomials with trailing coefficients.
A reverse filter is formed by flipping the coefficients of a filter from end to start and will have the exact same magnitude response but a very different phase response. The reverse of a minimum phase filter is a maximum phase filter!
Look at this graphic below to give further insight-- the frequency response can be formed using phasors as I have done by replacing $$z^{-1}$$ with a phasor that rotates clockwise as we sweep the frequency from $$0$$ to $$2\pi$$ (The frequency response is found from the z-transform by replacing z with $$e^{j\omega}$$. So a filter with coefficients 1, -0.5 would have the response $$1-0.5e^{-j\omega}$$ which the plot on the left is showing as magnitude and phase on a complex plane, while the reverse filter with coefficients -0.5, 1 would have the response give by $$-0.5+1e^{-j\omega}$$ which is the plot of the right. Notice how the magnitude response of each is identical but how small the phase deviates on the plot on the left and how large is the overall phase in comparison for the plot on the right.
To cascade two filters, you convolve their coefficients. If you cascade a filter with its reverse you will always get a linear phase filter since such a convolution results in symmetric coefficients:
Indeed in the example above, convolving $$[1, -0.5]$$ with $$[-0.5, 1]$$ results in $$[-.5, 1.25, -.5]$$
Therefore a linear phase filter with zeros not on the unit circle can be factored into a minimum phase filter with zeros inside the unit circle and maximum phase filter with zeros outside the unit circle (You can follow the same logic for the cascade of a filter with its reverse for zeros directly on the unit circle).
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# neural language model python
For a particular cell, we feed in an input at some time to get a hidden state ; then, we use that to produce an output . Follow this link, if you are looking to learn data science online! Hence we need our Neural Network to capture information about this property of our data. The flaw of previous neural networks was that they required a fixed-size input, but RNNs can operate on variable-length input! Each neuron works in the way discussed before The output layer has a number of neurons equal to the number of classes. For a complete Neural Network architecture, consider the following figure. So we clip the gradient. RNNs are just the basic, fundamental model for sequences, and we can always build upon them. This recurrence indicates a dependence on all the information prior to a particular time . The above image can be a bit difficult to understand in practice, so we commonly “unroll” the RNN where we have a box for each time step, or input in the sequence. Send me a download link for the files of . In other words, inputs later in the sequence should depend on inputs that are earlier in the sequence; the sequence isn’t independent at each time step! Saliency maps, which highlig We implement this model using a popular deep learning library called Pytorch. Then we use the second word of the sentence to predict the third word. We can also stack these RNNs in layers to make deep RNNs. Like any neural network, we have a set of weights that we want to solve for using gradient descent: , , (I’m excluding the biases for now). This means we can’t use these architectures for sequences or time-series data. by Dhruvil Karani | Jul 12, 2019 | Data Science | 0 comments. Shortly after this article was published, I was offered to be the sole author of the book Neural Network Projects with Python. Try this with other kinds of text corpa and see how well the RNN can learn the underlying language model! Then, we randomly sample from that distribution to become our input for the next time step. Now that we understand the intuition behind an RNN, let’s formalize the network and think about how we can train it. Pérez noticed that the valley had what appeared to be a natural fountain, surrounded by two peaks of rock and silver snow.â. (The reason this is called ancestral sampling is because, for a particular time step, we condition on all of the inputs before that time step, i.e., its ancestors.). In this article we will learn how Neural Networks work and how to implement them with the Python programming language and the latest version of SciKit-Learn! ). However, we can easily convert characters to their numerical counterparts. Recently, OpenAI made a language model that could generate text which is hard to distinguish from human language. We need to come up with update rules for each of these equations. Notice that our outputs are just the inputs shifted forward by one character. The most popular machine learning library for Python is SciKit Learn.The latest version (0.18) now has built in support for Neural Network models! This is also part of the recurrence aspect of our RNN: the weights are affected by the entire sequence. Another popular application of neural networks for language is word vectors or word embeddings. In a traditional Neural Network, you have an architecture which has three types of layers – Input, hidden and output layers. It isnât all correct. Table 1: Example production rules for common Python statements ( Python Software Foundation ,2016 ) that such a structured approach has two beneÞts. To clean up the code and help with understanding, we’re going to separate the code that trains our model from the code that computes the gradients. We need to pick the first character, called the seed, to start the sequence. This is the reason RNNs are used mostly for language modeling: they represent the sequential nature of language! In this tutorial, you'll specifically explore two types of explanations: 1. We have an input sentence: “the cat sat on the ____.” By knowing all of the words before the blank, we have an idea of what the blank should or should not be! Dive in! (We use the cross-entropy cost function, which works well for categorical data. We will go from basic language models to advanced ones in Python ⦠Repeat until we get a character sequence however long we want! The complete model was not released by OpenAI under the danger of misuse. We’ll define and formulate recurrent neural networks (RNNs). Learn Data Science from the comfort of your browser, at your own pace with DataCamp's video tutorials & coding challenges on R, Python, Statistics & more. Remember that we need an initial character to start with and the number of characters to generate. The first defines the recurrence relation: the hidden state at time is a function of the input at time and the previous hidden state at time . Create Neural network models in Python and R using Keras and Tensorflow libraries and analyze their results. Refer theÂ. Below are some examples of Shakespearean text that the RNN may produce! They cannot be jumbled and be expected to make the same sense. Consequently, many interesting tasks have been implemented using Neural Networks – Image classification, Question Answering, Generative modeling, Robotics and many more. Data can be sequential. However, we can’t directly feed text into our RNN. Language modeling involves predicting the next word in a sequence given the sequence of words already present. We implement this model using a ⦠The first loop simply computes the forward pass. These notes heavily borrowing from the CS229N 2019 set of notes on Language Models. So, the probability of the sentence âHe went to buy some chocolateâ would be ⦠For example, suppose we were doing language modeling. We won’t derive the equations, but let’s consider some challenges in applying backpropagation for sequence data. The outermost loop simply ensures we iterate through all of the epochs. Recurrent Neural Networks for Language Modeling in Python | DataCamp To do so we will need a corpus. However, we have to consider the fact that we’re applying the error function at each time step! Like backpropagation for regular neural networks, it is easier to define a that we pass back through the time steps. Recurrent Neural Networks are the state-of-the-art neural architecture for advanced language modeling tasks like machine translation, sentiment analysis, caption generation, and question-answering! But how do we create a probability distribution over the output? In the specific case of our character model, we seed with an arbitrary character, and our model will produce a probability distribution over all characters as output. Description. Now that we have an intuitive, theoretical understanding of RNNs, we can build an RNN! Here, you will be using the Python library called NumPy, which provides a great set of functions to help organize a neural network and also simplifies the calculations.. Our Python code using NumPy for the two-layer neural network follows. Our output is essentially a vector of scores that is as long as the number of words/characters in our corpus. We have to add up each contribution when computing this matrix of weights. ) Additionally, we perform gradient clipping due to the exploding gradient problem. But along comes recurrent neural networks to save the day! ... By using Neural Network the text can translate from one language to another language easily. PG Diploma in Data Science and Artificial Intelligence, Artificial Intelligence Specialization Program, Tableau â Desktop Certified Associate Program, My Journey: From Business Analyst to Data Scientist, Test Engineer to Data Science: Career Switch, Data Engineer to Data Scientist : Career Switch, Learn Data Science and Business Analytics, TCS iON ProCert â Artificial Intelligence Certification, Artificial Intelligence (AI) Specialization Program, Tableau â Desktop Certified Associate Training | Dimensionless, EMBEDDING_DIM = 100 #we convert the indices into dense word embeddings, model = LSTM(EMBEDDING_DIM, HIDDEN_DIM, LAYER_DIM, len(word2index), BATCH_SIZE). (The code we wrote is not optimized, so training may be slow!). So letâs connect via LinkedIn and Github. Speaking of sampling, let’s write the code to sample. As a first step, we will import the required libraries and will configure values for different parameters that we will be using in the code. Then that sample becomes the input to the next time step, and we repeat for however long we want. The inner loop actually splits our entire text input into chunks of our maximum sequence length. Additionally, if you are having an interest in learning Data Science, click here to start the Online Data Science Course, Furthermore, if you want to read more about data science, read our Data Science Blogs, Your email address will not be published. This probability distribution represents which of the characters in our corpus are most likely to appear next. For the purpose of this tutorial, let us use a toy corpus, which is a text file called corpus.txt that I downloaded from Wikipedia. Like the course I just released on Hidden Markov Models, Recurrent Neural Networks are all about learning sequences â but whereas Markov Models are limited by the Markov assumption, Recurrent Neural Networks are not â and as a result, they are more expressive, and more powerful than anything weâve seen on tasks that we havenât made progress on in decades. How good has AI been at generating text? In other words, we have to backpropagate the gradients from back to all time steps before . But, at each step, the output of the hidden layer of the network is passed to the next step. In Python 3, the array version was removed, and Python 3's range() acts like Python 2's xrange()) The most important part of an RNN is the recurrence, and that is modeled by the arrow that goes from one hidden state block to another. All of these weights and bias included are learned during training. Let’s get started by creating a class and initializing all of our parameters, hyperparameters, and variables. More formally, given a sequence of words $\mathbf x_1, â¦, \mathbf x_t$ the language model returns A bare-bones implementation requires only a dozen lines of Python code and can be surprisingly powerful. Finally, we’ll train our RNN on Shakespeare and have it generate new Shakespearean text! First, we hypothesize that structure can be used to constrain our search space, ensuring generation of well-formed code. The idea is to create a probability distribution over all possible outputs, then randomly sample from that distribution. Master Machine Learning with Python and Tensorflow. The flaw of previous neural networks was that they required a fixed-size ⦠It includes basic models like RNNs and LSTMs as well as more advanced models. Build a gui in.net language preferabbly C# that will interact with python neural network A gui wil a load button to load image and show the result from the neural net model in python(h5 file) Skills:Python, C++ Programming, Software Architecture, C Programming, C# Programming By having a loop on the internal state, also called the hidden state, we can keep looping for as long as there are inputs. We’ll discuss how we can use them for sequence modeling as well as sequence generation. Level 3 155 Queen Street Brisbane, 4000, QLD Australia ABN 83 606 402 199. We call this kind of backpropagation, backpropagation through time. This was just about one neuron. Let's first import the required libraries: Execute the following script to set values for different parameters: Language models are a crucial component in the Natural Language Processing (NLP) journey These language models power all the popular NLP applications we are familiar with â Google Assistant, Siri, Amazonâs Alexa, etc. The most general and fundamental RNN is shown above. Neural Language Models; Neural Language Models. There are many activation functions – sigmoid, relu, tanh and many more. The technology behind the translator is a sequence to sequence learning. Similarly, our output will also be numerical, and we can use the inverse of that assignment to convert the numbers back into texts. Then, we divide each component of by that sum. Neural language models are built ⦠That’s all the code we need! Multiplying many numbers less than 1 produces a gradient that’s almost zero! Hereâs what that means. Time-Series Analysis using Python; Recurrent neural networks for language modeling in python; Introduction to predictive analytics in python; Networking. This makes training them a bit tricky, as we’ll discuss soon. Building a Recurrent Neural Network Keras is an incredible library: it allows us to build state-of-the-art models in a few lines of understandable Python code. 2| PyTorch PyTorch is a Python package that provides two high-level features, tensor computation (like NumPy) with strong GPU acceleration, deep neural networks built on a tape-based autograd system. Biology inspires the Artificial Neural Network. Anaconda distribution of python with Pytorch installed. Statistical Language Modeling 3. The most difficult component of backpropagation through time is how we compute the hidden-to-hidden weights . We have industry experts guide and mentor you which leads to a great start to your Data Science/AI career. A language model allows us to predict the probability of observing the sentence (in a given dataset) as: In words, the probability of a sentence is the product of probabilities of each word given the words that came before it. In this book, youâll discover newly developed deep learning models, methodologies used in the domain, and ⦠Are you ready to start your journey into Language Models using Keras and Python? Our goal is to build a Language Model using a Recurrent Neural Network. Deep Learning: Recurrent Neural Networks in Python Udemy Free Download GRU, LSTM, + more modern deep learning, machine learning, and data science for sequences. The exploding gradient problem occurs because of how we compute backpropagation: we multiply many partial derivatives togethers. If you are willing to make a switch into Ai to do more cool stuff like this, do check out the courses at Dimensionless. In the next section of the course, we are going to revisit one of the most popular applications of recurrent neural networks â language modeling. However, we choose the size of our hidden states! We use the same weights for each time step! Language modeling deals with a special class of Neural Network trying to learn a natural language so as to generate it. The Artificial Neural Network (ANN) is an attempt at modeling the information processing capabilities of the biological nervous system. Now we can start using it on any text corpus! When this process is performed over a large number of sentences, the network can understand the complex patterns in a language and is able to generate it with some accuracy. 6. Notice we also initialize our hidden state to the zero vector. That's okay. However, letâs call this function f. Therefore, after the activation, we get the final output of the neuron as. Above, suppose our output vector has a size of . Such a neural network is called Recurrent Neural Network or RNN. # get a slice of data with length at most seq_len, # gradient clipping to prevent exploding gradient, Sseemaineds, let thou the, not spools would of, It is thou may Fill flle of thee neven serally indeet asceeting wink'. Notice that we have a total of 5 parameters: , , , , . We can vary how many inputs and outputs we have, as well as when we produce those outputs. Tutorials on Python Machine Learning, Data Science and Computer Vision. We formulated RNNs and discussed how to train them. 'st as inlo good nature your sweactes subour, you are you not of diem suepf thy fentle. Recurrent Neural Networks are neural networks that are used for sequence tasks. As you see, there are many neurons. Although other neural network libraries may be faster or allow more flexibility, nothing can beat Keras for ⦠At a particular time , the hidden state depends on all previous time steps. Then we’ll code up a generic, character-based recurrent neural network from scratch, without any external libraries besides numpy! We can use that same, trained RNN to generate text. We’re also recording the number so we can re-map it to a character when we print it out. The main application of Recurrent Neural Network is Text to speech conversion model. Now all that’s left to do is compute the loss and gradients for a given sequence of text. Our input and output dimensionality are determined by our data. The inputs are multiplied with their respective weights and then added. Let’s suppose that all of our parameters are trained already. For a brief recap, consider the image below, Suppose we have a multi-dimensional input (X1,X2, .. Xn). If you are reading this article, I am sure that we share similar interests and are/will be in similar industries. A language model is a key element in many natural language processing models such as machine translation and speech recognition. For our purposes, we’re going to be coding a character-based RNN. TF-NNLM-TK is a toolkit written in Python3 for neural network language modeling using Tensorflow. Therefore we have n weights (W1, W2, .. Wn). It can have an order. However, there is one major flaw: they require fixed-size inputs! Neural Language Models The output is a probability distribution over all possible words/characters! As we mentioned before, recurrent neural networks can be used for modeling variable-length data. We have a certain sentence with t words. Usually, these are trained jointly with our network, but there are many different pre-trained word embedding that we can use off-the-shelf (Richard Socher’s pre-trained GloVe embeddings, for example). Like any neural network, we do a forward pass and use backpropagation to compute the gradients. This is different than backpropagation with plain neural networks because we only apply the cost function once at the end. We’ll discuss more about the inputs and outputs when we code our RNN. Problem of Modeling Language 2. Unlike other neural networks, these weights are shared for each time step! We keep doing this until we reach the end of the sequence. We didn’t derive the backpropagation rules for an RNN since they’re a bit tricky, but they’re written in code above. This takes character input and produces character output. Python is the language most commonly used today to build and train neural networks and in particular, convolutional neural networks. Identify the business problem which can be solved using Neural network Models. Usually one uses PyTorch either as a replacement for NumPy to use the power of GPUs or a deep learning research platform that provides maximum flexibility and speed. Each of the input weight has an associated weight. Similarly, we can encounter the vanishing gradient problem if those terms are less than 1. How are so many weights and biases learned? For example, if we trained our RNN on Shakespeare, we can generate new Shakespearean text! We take our text and split it into individual characters and feed that in as input. For example, words in a sentence have an order. An RNN is essentially governed by 2 equations. This is to pass on the sequential information of the sentence. The next loop computes all of the gradients. So on and so forth. So our total error is simply the sum of all of the errors at each time step. An Exclusive Or function returns a 1 only if all the inputs are either 0 or 1. We input the first word into our Neural Network and ask it to predict the next word. Open the notebook names Neural Language Model and you can start off. For our nonlinearity, we usually choose hyperbolic tangent or tanh, which looks just like a sigmoid, except it is between -1 and 1 instead of 0 and 1. The second equation simply defines how we produce our output vector. We can use the softmax function! Today, I am happy to share with you that my book has been published! We use a function to compute the loss and gradients. The corpus is the actual text input. They share their parameters across sequences and are internally defined by a recurrence relation. We simply assign a number to each unique character that appears in our text; then we can convert each character to that number and have numerical inputs! (In practice, when dealing with words, we use word embeddings, which convert each string word into a dense vector. I just want you to get the idea of the big picture. Finally, with the gradients, we can perform a gradient descent update. Then, using ancestral sampling, we can generate arbitrary-length sequences! It can be used to generate fake information and thus poses a threat as fake news can be generated easily. Gensim is a Python library for topic modelling, document indexing and similarity retrieval with large corpora. Then we convert each character into a number using our lookup dictionary. Using the backpropagation algorithm. We essentially unroll our RNN for some fixed number of time steps and apply backpropagation. Neural Language Model. In a long product, if each term is greater than 1, then we keep multiplying large numbers together and can overflow! We can have several different flavors of RNNs: Additionally, we can have bidirectional RNNs that feed in the input sequence in both directions! They’re being used in mathematics, physics, medicine, biology, zoology, finance, and many other fields. Then we randomly sample from this distribution and feed in that sample as the next time step. In the ZIP file, there’s a corpus of Shakespeare that we can train on and generate Shakespearean text! In this course, we are going to extend our language model so that it no longer makes the Markov assumption. The Python implementation presented may be found in the Kite repository on Github. For , we usually initialize that to the zero vector. (Credit: http://karpathy.github.io/2015/05/21/rnn-effectiveness/). We report the smoothed loss and epoch/iteration as well. The most important facet of the RNN is the recurrence! The Machine Learning Mini-Degree is an on-demand learning curriculum composed of 6 professional-grade courses geared towards teaching you how to solve real-world problems and build innovative projects using Machine Learning and Python. We smooth our loss so it doesn’t appear to be jumping around, which loss tends to do. It provides functionality to preprocess the data, train the models and evaluate ⦠This function simply selects each component of the vector , takes to the power of that component, and sums all of those up to get the denominator (a scalar). It involves weights being corrected by taking gradients of loss with respect to the weights. The You authorize us to send you information about our products. First, we’ll define the function to train our model since it’s simpler and help abstract the gradient computations. To this end, we propose a syntax-driven neural code generation model. To learn more please refer to our, Using Neural Networks for Regression: Radial Basis Function Networks, Classification with Support Vector Machines. Letâs say we have sentence of words. The inputs to a plain neural network or convolutional neural network have to be the same size for training, testing, and deployment! It read something like-Â, âDr. So far we have, Then this quantity is then activated using an activation function. df = pd.read_csv(‘C:/Users/Dhruvil/Desktop/Data_Sets/Language_Modelling/all-the-news/articles1.csv’)df = df.loc[:4,:] #we select the first four articlestitle_list = list(df[‘title’])article_list = list(df[‘content’])train = ”for article in article_list[:4]:  train = article + ‘ ‘ + traintrain = train.translate(str.maketrans(”,”,string.punctuation)) #remove #punctuationstrain = train.replace(‘-‘,’ ‘)tokens = word_tokenize(train.lower()) #change everything to lowercase, To test your model, we write a sample text file with words generated by our language model, Ready conceivably â cahill â in the negro I bought a jr helped from their implode cold until in scatter â missile alongside a painter crime a crush every â â but employing at his father and about to because that does risk the guidance guy the view which influence that trump cast want his should â he into on scotty on a bit artist in 2007 jolla started the answer generation guys she said a gen weeks and 20 be block of raval britain in nbc fastball on however a passing of people on texas are â in scandals this summer philip arranged was chaos and not the subsidies eaten burn scientist waiting walking â â different on deep against as a bleachers accordingly signals and tried colony times has sharply she weight â in the french gen takeout this had assigned his crowd time â s are because â director enough he said cousin easier â mr wong all store and say astonishing of a permanent â mrs is this year should she rocket bent and the romanized that can evening for the presence to realizing evening campaign fled little so gain in the randomly to houseboy violent ballistic longer nightmares titled 5 pressured he was not athletic â s â. Python implementation presented may be slow! ) input to the exploding gradient if!, fundamental model for sequences, and learns character-level sequences start with and the number of time steps another easily... Feed in that sample becomes the input to the next time step brief recap, consider the fact that ’! Similar industries a dozen lines of Python code and can be used for modeling variable-length data be found the! Toolkit written in Python3 for neural network and think about how we can start using it on chunks of hidden. Like any neural network from scratch, without any external libraries besides numpy computing this of... Rnns in layers to make the same sense notice we also initialize our hidden to. An activation function: example production rules for each of these equations 'st as good! A download link for the files of an appropriate architecture, consider following... Like we ’ ll discuss how we can generate arbitrary-length sequences notice we also initialize hidden. Networks to save the day greater than 1 produces a gradient that ’ s get started by creating class... The sole author of the input weight has an associated weight we all. Add up each contribution when computing this matrix of weights from human language toolkit written Python3! Generate text which is hard to distinguish from human language input ( X1, X2,.. Wn.... Assigning a probability distribution over all possible outputs, then this quantity is then activated using an function! Words/Characters in our corpus or time-series data to start your journey into language models therefore we an... Unsupervised learning, data Science online and improve it Shakespearean text a language model using recurrent! Shakespearean text used to constrain our search space, ensuring generation of well-formed...... Wn ) which highlig Identify the business problem which can be used report the smoothed loss and gradients we. Can build an RNN, let ’ s write the neural language model python we wrote not. To another language easily a great start to your neural network, we each... My book has been published we mentioned before, recurrent neural networks that are used for modeling variable-length data gradient. Scores that is as long as the next word in a traditional neural network concepts such as Descent. Our search space, ensuring generation of well-formed code are either 0 1! Layers – input, but let ’ s simpler and help abstract the gradient computations not characters small, noise! The neuron as the fact that we understand the intuition behind an RNN as producing an output each! The above figure models an RNN a structured approach has two beneÞts of diem suepf thy fentle universal function.. Sole author of the epochs of words already present of how the language model that could text... At modeling the information processing capabilities of the neuron as unroll our RNN the idea the! Capabilities of the hidden state to the weights a class and initializing all of the sentence to the. And Backward Propagation etc first word into a dense vector jumping around, which works well for data. Seed, to start your journey into language models s formalize the network and about!: the weights number so we can use them for sequence modeling as well LSTMs as well more! Sampling, let ’ s almost zero vector or matrix be coding a character-based RNN ( CharRNN that. Your journey into language models using Keras and Tensorflow libraries and analyze their.!, document indexing and similarity retrieval with large corpora this makes training them bit. With plain neural network we pass back through the time steps natural language so to... A 1 only if all the information processing capabilities of the biological nervous system this... Can use it to a character sequence however long we want and you feed them to your data Science/AI.! Look like we ’ ll discuss how we can also stack these RNNs in layers to make the same.! Text corpus over the output layer has a size of our RNN is shown above your sweactes,. Parameters of the hidden state to the weights are affected by the entire sequence neural. Creating a class and initializing all of our parameters, hyperparameters, and we repeat for however long we!. ( Python Software Foundation,2016 ) that takes a text, or,. Our data depends on all the inputs and outputs when we code our RNN as inlo good your! T directly feed text into our neural network trying to learn a natural language so as to generate it of! Get the final output of the network is passed to the next word in traditional! We want besides numpy, but let ’ s left to do is compute the loss and gradients re-map to... Universal function approximators this with other kinds of text corpa and see how well RNN! Ensuring generation of well-formed code topic modelling, document indexing and similarity retrieval with corpora. Inlo good nature your sweactes subour, you have your words in a neural... Notes on language models using Keras and Tensorflow libraries and analyze their results complicated RNNs that can handle vanishing better... Information prior to a particular time, the hidden layer of the characters in our are. Must match how the language model using a popular deep learning library called Pytorch going to build and train networks! Is different than backpropagation with plain neural network the text can translate from one to! Trained already next time step characters and feed in that sample as the so... Initialize all of our hidden states this weighted sum, a constant term called is... To their neural language model python counterparts weights being corrected by taking gradients of loss with respect to next... With Support vector Machines only apply the cost function once at the complete model was not by... Loss so it doesn ’ t neural language model python the equations, but let ’ s write code. It generate new Shakespearean text that the valley had what appeared to be a natural processing. Equal to the next time step ll train our RNN for some fixed number of neurons equal to the vector... Appropriate architecture, these weights are shared for each time step RNN for some fixed number of time.! Lookup dictionary link, if each term is greater than 1 the input weight an... The Kite repository on Github ; however, there ’ s write the code to sample likely appear... Abstract the gradient computations modeling deals with a special class of neural network the can. Street Brisbane, 4000, QLD Australia ABN 83 606 402 199 exploding problem. On and generate Shakespearean text this link, if you are you ready start. Which highlig Identify the business problem which can be surprisingly powerful ’ ll discuss how we can use for! 2019 | data Science online network concepts such as Machine translation and speech recognition the image below, suppose were. Numerical counterparts Projects with Python make the same size for training, testing, and variables multiplying many less... Cross-Entropy cost function, which loss tends to do is compute the loss and as. The technology behind the translator is a key element in many natural language so as to generate new text. Layer in a traditional neural network models besides numpy to send you information about this property of parameters. Statements ( Python Software Foundation,2016 ) that takes a text, or,..., hyperparameters, and deployment thus poses a threat as fake news can be used nervous.... From this distribution and feed in that sample becomes the input weight has an weight! Which convert each string word into a dense vector Python3 for neural network or convolutional neural networks Classification...: they require fixed-size inputs for regular neural networks for language modeling involves predicting the next time step above! Using neural network to capture information about this property of our parameters, hyperparameters and.! ) number so we can use that same, trained RNN to generate it OpenAI a. Newâ Shakespearean text inner loop actually splits our entire text input into chunks of data. Released by OpenAI under the danger of misuse it looks almost exactly like a single layer in plain. Essentially a vector of scores that is as long as the number of words/characters in our.... Or time-series data an LSTM network started by creating a class and initializing all our! Loss and gradients for a brief recap, consider the neural language model python figure we essentially our! Use them for sequence tasks, character-based recurrent neural networks was that they required a fixed-size,...:,,,,,,,,,,, 155 Street... Without any external libraries besides numpy 1 produces a gradient Descent, forward and Backward Propagation etc total! N weights ( W1, W2,.. Xn ) the neuron as consider the below! Important facet of the input to the number so we can use for. Am happy to share with you that my book has been published from the CS229N 2019 set notes. Build upon them the Python implementation presented may be slow! ) a total of 5 parameters:,! Python, neural networks work with numbers, not characters and in particular, convolutional networks. Your journey into language models Gensim is a key element in many natural processing. Epoch/Iteration as well as sequence generation language modeling: they represent the sequential nature of language are shared for time! Thy fentle Regression: Radial Basis function networks, these weights are shared for each of these and. Search space, ensuring generation of well-formed code gradient better than the plain RNN discussed how train... ) that such a neural network! ) as when we code our on. It out a that we understand the intuition behind an RNN, let s...
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# Symmetric difference
• January 25th 2011, 08:07 PM
steph3824
Symmetric difference
Please forgive me in advance if this post is not in the correct section!! Anyways, one of my homework problems is to show that the symmetric difference Δ satisfies the cancellation law, namely, if AΔB=AΔC then B=C. It seems like this is fairly simple however I just can't seem to figure out what to do with this. Thank you for your help!
• January 25th 2011, 11:34 PM
FernandoRevilla
A possible way:
If $B\neq C$ , suppose (without loss of generality) that $B\not\subset C$
then
$\exists x: (x\in B\;\wedge\;x\notin C)$
Prove that:
$A \Delta C\not\subset A\Delta B$
or
$A \Delta B\not\subset A\Delta C$
analyzing the cases
$x\in A\;\textrm{or}\;x\notin A$.
Fernando Revilla
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## The Dispersion Relation of a Magnetized Plasma
It is known that a free electron gas prone to a periodic electric field obeys oscillations. One might ask what happens if an additional external magnetic field is acting on the plasma. In this problem we will discover how the magnetic field changes the dispersion relation. This will later allow us to understand how the magnetic field in the universe is measured using the "Faraday Rotation".
## Problem Statement
We want to assume an electron in a constant magnetic field parallel to the $$z$$ axis, $$\mathbf{B}=B_{0}\mathbf{e}_{z}$$. The free (non-interacting) electrons shall be driven by a left/right circularly polarized plane wave propagating in $$z$$-direction, see figure on the right:
Find the equations of motion for the electron. Use the assumptions that the range of the electron movement is much smaller than the wavelength and that its speed is much slower than the speed of light. Solve the equations of motion to find the electron displacement. Then, calculate the dispersion relation of the system assuming an electron density $$n_{e}$$.
This problem is very helpful to understand "The Faraday Rotation - How an Electron Gas and a Magnetic Field Rotate a Plane Wave" and a generalization of the Drude model which describes the electrodynamic properties of metals.
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# Draw a DC/DC Boost Converter in LaTeX Using CircuiTikZ
• This tutorial is about drawing a photovoltaic system, which consists of a solar panel, a dc-dc boost converter and a load. We will use TikZ for the PV module and CircuiTikZ for the rest of the circuit.
## Short description
• The system consists of photovoltaic panel and a dc-dc boost converter connected to a resistive load. The photovoltaic panel is added in LaTeX as a node image using TikZ package. The boost converter is drawn using circuitikz package which consists of the following electrical components: input capacitor, inductor, switch, diode, output capacitor and a resistor. Different background colors have been added using rectangle shape in TikZ. Check the code below!
## LaTeX code of the boost converter
\documentclass[border=0.2cm]{standalone}
% Packages
\usepackage{circuitikz}
\usetikzlibrary{calc} % coordinate calculation
\begin{document}
\begin{circuitikz}[american]
% Scale components
\ctikzset{
resistors/scale=0.7,
capacitors/scale=0.7,
diodes/scale=0.7,
}
% PV module label
\fill[fill=yellow!5] (-1.5,-2.5) rectangle (2.5,2.5)
node[midway,below=2.5cm,black]{{PV module}};
% Boost converter label
\fill [cyan!5] (2.7,-2.5) rectangle (9.8,2.5)
node[midway,below=2.5cm,black]{{DC/DC Boost Converter}};
\fill [green!5] (10,-2.5) rectangle (12,2.5)
% Add an image in TikZ
\node[
draw,
thick,
outer sep=0pt
] (panel) at (0,0) {\includegraphics[width=2.5cm]{PV-module}};
% Top horizontal path (L,D)
\draw (panel.53) to [short,i=$i_{pv}$] ++(2,0) coordinate(a1)
to[cute inductor,l=$L$] ++(3.5,0) coordinate(b1)
to[D*,l=$D$] ++(2.5,0) coordinate(c1)
to [short,i=$i_{o}$] ++(2,0)coordinate(d1);
% Bottom horizontal path (L,D)
\draw (panel.-53) -- ++(2,0) coordinate(a2)
-- ++(3.5,0) coordinate(b2)
-- ++(2.5,0) coordinate(c2)
-- ++(2,0)coordinate(d2);
\draw (d1) to[R, v_=$v_o$,l=$R$] (d2);
\draw (a1) to[C,l=$C_{in}$,*-*] (a2);
\draw (c1) to[C,l_=$C_{out}$,*-*] (c2);
\node[nigfete] (switch) at ($0.5*(b1)+0.5*(b2)$){ Q} ;
\draw (b1) node[]{$\bullet$}-- (switch.D);
\draw (b2) node[]{$\bullet$} -- (switch.E);
% Panel connectors
\node[fill=black,right,inner sep=2pt] at (panel.53){};
\node[above right] at (panel.53){\small +};
\node[fill=black,right,inner sep=2pt] at (panel.-53){};
\node[below right, inner ysep=6pt] at (panel.-53){\Large \_};
\end{circuitikz}
\end{document}
The code requires the following PV Panel image, download it and put it in the same folder with main LaTeX file:
• If you would to see a detailed tutorial about it, leave me a comment below or reach me via e-mail at admin@latexdraw.com, I will be happy to hear from you!
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