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## Essential University Physics: Volume 1 (3rd Edition)
$3.69\times10^6 \ gallons$
We know that there are 600 GW, and they are used in a day, which has 24 hours in it. Thus, we find: $=600\times24=14,400 \ Gwh$ Now, we can convert, using conversions from Appendix C: $14,400 \ Gwh\times \frac{1,000,000 \ kwh}{Gwh}\times \frac{gal}{39 \ kwh}=3.69\times10^6 \ gallons$
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# Quark Matter 2019 - the XXVIIIth International Conference on Ultra-relativistic Nucleus-Nucleus Collisions
3-9 November 2019
Wanda Reign Wuhan Hotel
Asia/Shanghai timezone
## Identified light flavor particle production in "jetty" and "isotropic" pp collisions at √s = 13 TeV with ALICE at the LHC
4 Nov 2019, 17:40
20m
Wanda Han Show Theatre & Wanda Reign Wuhan Hotel
#### Wanda Han Show Theatre & Wanda Reign Wuhan Hotel
Poster Presentation Small systems
### Speaker
Adrian Fereydon Nassirpour (Lund University (SE))
### Description
Identified light flavour particles, such as the $\pi$, K, $\phi$ mesons and the p, $\Lambda$, $\Xi$ baryons, constitute interesting probes to investigate the collective behaviour recently observed in small collision systems. The underlying mechanisms of light flavour production are currently not well understood, and the mechanisms are explained in the framework of different models. pQCD models based on hard scatterings, such as PYTHIA, describe light flavour production via string-breakings and rope hadronization. Other thermal and statistical models, mainly dominated by soft processes, predict a mechanism for the production of light flavour particles based on mass hierarchies in (grand) canonical ensembles.
This analysis is aimed to disentangle and isolate events that are dominated by soft processes ("isotropic") and hard processes ("jetty") by using the transverse spherocity observable. The light flavour production is then studied in both events with jet-like topologies and isotropic topologies, which are assumed to be dominated by hard and soft processes, respectively. This is done in an effort to pin-point the underlying mechanisms of the collective behaviour observed in small systems, such as radial flow and long-range angular correlations.
In this contribution we report about the measurement of transverse momentum spectra of strange and non-strange mesons and baryons in transverse spherocity selected events. The results are obtained by exploiting the data collected with ALICE in pp collisions at a center-of-mass energy, $\sqrt{s}$, of 13 TeV.
### Primary author
Adrian Fereydon Nassirpour (Lund University (SE))
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Problem Solving on Subtraction
Problem solving on subtraction will help us to get the idea on how to solve the basic subtraction statement problems.
1. Eight birds sat on a wire. Three birds flew away. How many were left?
Total number of birds sat on a wire = 8
Number of birds flew away = 3
Therefore, number of birds left = 8 - 3 = 5
2. Sam had 7 dollars. He spent 4 dollars. How many dollars is he left with?
Total amount of money Sam had = $7 He spent =$4
Therefore, amount of money left with him = $7 -$4 = \$3
3. Five boats were tied up. Four of the boats sailed away. How many were left?
Total number of boats tied up = 5
Number of boats sailed away = 4
Therefore, number of boats were left = 5 - 4 = 1
4. Ron had 10 stamps. His father took 2 stamps. How many stamps does Ron have now?
Total number of stamps Ron had = 10
Number of stamps his father took = 2
Therefore, number of stamps he have now = 10 - 2 = 8
5. Diana had 18 toffees. She gave 5 toffees to her friend. How many toffees left with her?
Total number of toffees Diana had = 18
Number of toffees she gave to her friend = 5
Therefore, number of toffees left = 18 - 5 = 13
More examples on statement problem solving on subtraction:
6. Mr. Daniel had 39 goats in a pasture. When he opened the pasture gate, 13 goats went out. How many goats remained in?
Total number of goats in a pasture Mr. Daniel had = 39
Number of goats went out = 13
Therefore, number of goats remained in = 39 - 13 = 26
7. Derek’s father is 47 years old. His mother is 35 years old. What is the difference of their ages?
Age of Derek’s father = 47 years
Age of his mother = 35 years
Therefore, difference of their ages = 47 - 35 = 12 years
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# Chapter 28 - Think and Discuss: 106
Both moon halos and rainbows are produced by light refracting from water (however, it's ice in the first case, liquid water in the second case).
#### Work Step by Step
A moon halo, also known as a moon ring or a $22^{\circ}$ halo, occurs when ice crystals refract moonlight into a circle. This is similar to how water droplets reflect and refract light into rainbows.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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Average Hamming distance between strings after some number of random substitutions in a population of initially identical elements - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T15:46:32Z http://mathoverflow.net/feeds/question/65928 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65928/average-hamming-distance-between-strings-after-some-number-of-random-substitution Average Hamming distance between strings after some number of random substitutions in a population of initially identical elements unknown (yahoo) 2011-05-25T06:31:54Z 2011-05-25T17:15:23Z <p>Let's say I have a set $S$, $(s_1, ..., s_i, ..., s_P) \in S$, of $P$ identical strings over a $k$-letter alphabet, each of length $|s_i| = L$. With uniform random probability across all strings in $S$ (and all string positions in any $s_i$), I randomly substitute one character for another. And I do so $N$ times.</p> <p>For $N >> 1$, all $s_i$ will approximate random sequences. But what is the average Hamming distance between any two strings as a function of $N$?</p> http://mathoverflow.net/questions/65928/average-hamming-distance-between-strings-after-some-number-of-random-substitution/65930#65930 Answer by Robert Israel for Average Hamming distance between strings after some number of random substitutions in a population of initially identical elements Robert Israel 2011-05-25T07:22:12Z 2011-05-25T17:15:23Z <p>At each move, I assume you choose one of the character positions in one of the strings (with equal probabilities for all), and replace the character in that position by a randomly chosen character (with equal probabilities for all - note that this allows the possibility that the new character is the same as the old one). Let $X(n)$ be the event that after $n$ moves, the $i$'th character in string $j_1$ is the same as the $i$'th character in string $j_2$. Now if in move $n$ the position chosen was anything other than the $i$'th character in string $j_1$ or $j_2$, $X(n) = X(n-1)$, while if it was either of those, $X(n)$ has probability $1/k$. Thus ${\rm P}(X(n)) = (1 - \frac{2}{PL}) {\rm P}(X(n-1)) + \frac{2}{PLk}$ with ${\rm P}(X(0)) = 1$. The solution of this recurrence is ${\rm P}(X(n)) = \frac{1}{k} + \frac{k-1}{k} \left( 1-\frac{2}{PL} \right)^n$. The expected Hamming distance between strings $j_1$ and $j_2$ after $n$ moves is $L (1 - {\rm P}(X(n)))$.</p> <p>(added in response to unknown(yahoo)'s further question): if the new character must be different from the existing one at that position, the recurrence becomes ${\rm P}(X(n)) = (1 - \frac{2}{PL}) {\rm P}(X(n-1)) + \frac{2}{PL} \frac{1-P(X(n-1))}{k-1}$, and the solution is ${\rm P}(X(n)) = \frac{1}{k} + \frac{k-1}{k} \left(1 - \frac{2k}{PL(k-1)}\right)^n$. Again the expected Hamming distance after $n$ moves is $L (1 - {\rm P}(X(n)))$.</p>
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# The Difference between String and StringBuilder in C#
AramT
35.6K views
A String is basically an immutable sequence of characters. Each character is a Unicode character in the range U+0000 to U+FFFF.
Immutable means that its state cannot be modified after it is created.
Thus, if you assign a value to a string then reassign. The first value will stay in the memory and a new memory location will be assigned to accept the new value.
Let’s take these string allocations for example:
string string1 = "Tech";
string string2 = "io";
string string3 = "";
string3 = string1 + "." + string2;
This leads to 4 memory addresses to be allocated: Tech, io, [EmptyString] , and the Concatenation between tech , [dot] and io
When using the above method for a specific purpose or when you know the number of strings to be concatenated at compile time will be few, then the wasted memory will be negligible, however, this will generate a big issue when in a loop with lots of iterations, where a string is adding (Concatenating) to itself another string or value, it is just constantly reallocating, in other words, it is copying to a new memory location when the memory manager can’t expand the requested amount in place. At this point, the amount of memory required (or wasted) and the processing time used for the creation of the numerous objects may become a bottleneck in the application.
This case is mostly common when, for instance, dynamically generating a TABLE and its content, or building an XML document. With this type of scenario in mind, Microsoft included the StringBuilder class in .NET.
By the way, you might wonder what is the difference between string and String, they both refer to the string type. string, in its small letter form, is only an alias to the String class in C#.
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# Definition:Electric Field Strength
## Definition
Electric field strength is the measure of the intensity of an electric field.
It is a vector quantity.
### Symbol
The usual symbol used to denote electric field strength is $\mathbf E$.
### Dimension
Electric field strength has the dimension $\mathsf {M L T}^{-3} \mathsf I^{-1}$.
### Units
The SI unit for electric field strength can be given either as:
volt per metre: $\mathrm V / \mathrm m$
newton per coulomb: $\mathrm N / \mathrm C$
## Also known as
Some sources give this as electric field intensity.
## Also see
• Results about electric fields can be found here.
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I do not have a calculator that does matrix stuff that would be great for my classes this semester. I don't want to spend $100+ on a graphing calculator since I use matlab at home strictly now. Im looking to solve simultaneous equations, matrix operations like det,eigen,mult,...etc Any help would be great! PhysOrg.com science news on PhysOrg.com >> Leading 3-D printer firms to merge in$403M deal (Update)>> LA to give every student an iPad; $30M order>> CIA faulted for choosing Amazon over IBM on cloud contract Blog Entries: 3 The TI line (83, 89, etc.) handles matrix math just fine, and some of them go for under 100. Recognitions: Gold Member My Ti-89 is still a classic. The best you can get for the money. I wont bother with anything below Ti-89 ## cheap scientific calculator that does matrix operations I was kind of aiming towards$20-30...
Blog Entries: 3
Quote by bassplayer142 I was kind of aiming towards $20-30... ebay? I haven't seen a graphing calculater under$80 in years, and I don't think non-graphing ones handle matrices.
uhhg, maybe I can borrow one. Thanks everyone.
I just bought a TI-86 on craigslist for \$30. It didn't seem like that was an exceptionally low price or anything, so you shouldn't have a problem finding one in your price range.
See online matrix calculator http://calculator-online.org/s/matrix/
TI-89 is good but price differs for place to place better to get an used one it would be better
casio fx 115 es
There is an excellent software calculator called Mathwizard that does Matrices,algebra,calculus,scientific calculator and plot graph.check it out. you can also find mobile applications that does matrices,algebra,calculus,differential equations , and plot graphs
Similar discussions for: cheap scientific calculator that does matrix operations Thread Forum Replies Precalculus Mathematics Homework 13 Precalculus Mathematics Homework 5 Calculators 2 General Discussion 4 Calculators 12
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Corpus ID: 118245496
# Pressure-induced enhancement of superconductivity and superconducting-superconducting transition in CaC$\_6$
@article{Gauzzi2006PressureinducedEO,
title={Pressure-induced enhancement of superconductivity and superconducting-superconducting transition in CaC\$\\_6\$},
author={Andrea Gauzzi and Shinya Takashima and Nao Takeshita and Chieko Terakura and Hidenori Takagi and Nicolas Emery and Claire H{\'e}rold and Philippe Lagrange and Genevi{\e}ve Loupias},
journal={arXiv: Superconductivity},
year={2006}
}`
• A. Gauzzi, +6 authors G. Loupias
• Published 2006
• Materials Science, Physics
• arXiv: Superconductivity
We measured the electrical resistivity, $\varrho(T)$, of superconducting CaC$\_6$ at ambient and high pressure up to 16 GPa. For $P \leq$8 GPa, we found a large increase of $T\_c$ with pressure from 11.5 up to 15.1 K. At 8 GPa, $T\_c$ drops and levels off at 5 K above 10 GPa. Correspondingly, the residual $\varrho$ increases by $\approx$ 200 times and the $\varrho(T)$ behavior becomes flat. The recovery of the pristine behavior after depressurization is suggestive of a phase transition at 8 GPa… Expand
#### References
SHOWING 1-2 OF 2 REFERENCES
The Electron-Phonon Interaction
• Materials Science
• 1963
Publisher Summary The interaction of the conduction electrons with the lattice vibrations is at the center of the whole theory of the transport properties of solids. It determines the electrical andExpand
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시간 제한 메모리 제한 제출 정답 맞은 사람 정답 비율
3 초 512 MB 19 5 5 26.316%
## 문제
Bessie and Farmer John enjoy goat kart racing. The idea is very similar to Go-Kart racing that others enjoy, except the karts are pulled by goats and the track is made from nearby farmland. The farmland consists of $N$ meadows and $M$ roads, each connecting a pair of meadows.
Bessie wants to make a course from nearby farms. A farm is a subset of two or more meadows within which every meadow can reach every other meadow along a unique sequence of roads.
The nearby farmland may contain multiple farms. Suppose there are $K$ farms. Bessie would like to make a goat kart loop by connecting all $K$ farms by adding $K$ roads of length $X$. Each farm should be visited exactly once and at least one road must be traversed inside each farm.
To make the course interesting for racers, the total length of the track should be at least $Y$. Bessie wants to know the sum, over all such interesting tracks, of the total track lengths. A track is different from another if there are two meadows which are adjacent (after adding the roads between farms) in one track but not the other. Please note that only the roads chosen matter, and not the direction the goat karts will travel along those roads.
## 입력
The first line of input contains $N$, $M$, $X$, and $Y$ where $1 \leq N \leq 1500$, $1 \leq M \leq N-1$, and $0 \leq X, Y \leq 2500$.
Each of the $M$ following lines describe roads. The lines are of the form: $A_i$ $B_i$ $D_i$, meaning that meadows $A_i$ and $B_i$ are connected with a road of integer length $D_i$ ($1 \leq A_i, B_i \leq N$, $0 \leq D_i \leq 2500$). Each meadow is incident to at least one road, and there are no cycles of roads.
In at least 70% of the test cases, it is also guaranteed that $N \leq 1000$ and $Y \leq 1000$.
## 출력
Output a single integer, giving the sum of track lengths over all interesting tracks. As the sum of track lengths can be quite large, print the sum of lengths modulo $10^9+7$.
## 예제 입력 1
5 3 1 12
1 2 3
2 3 4
4 5 6
## 예제 출력 1
54
## 힌트
This example has 6 possible tracks
• 1 --> 2 --> 4 --> 5 --> 1 (length 11)
• 1 --> 2 --> 5 --> 4 --> 1 (length 11)
• 2 --> 3 --> 4 --> 5 --> 2 (length 12)
• 2 --> 3 --> 5 --> 4 --> 2 (length 12)
• 1 --> 2 --> 3 --> 4 --> 5 --> 1 (length 15)
• 1 --> 2 --> 3 --> 5 --> 4 --> 1 (length 15)
The answer is $12+12+15+15=54$, adding up only the tracks where the length is at least $12$.
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Product of phase velocity and group velocity is equal to square of speed of light
Prove that $\rightarrow$
The Product of phase velocity and group velocity is equal to the square of the speed of light i.e. $\left( V_{p}.V_{g}=c^{2} \right)$
Proof → We know that
$V_{p}=\nu \lambda \qquad(1)$
And de Broglie wavelength-
$\lambda =\frac{h }{mv}\qquad(2)$
According to Einstein's mass-energy relation-
$E=mc^{2}$
$h\nu=mc^{2}$
$\nu=\frac{mc^{2}}{h}\qquad(3)$
Now put the value of $\lambda$ and $\nu$ in equation$(1)$
$V_{p}= [\frac{mc^{2}}{h}] [\frac{h}{mv}]$
$V_{p}=\frac{C^{2}}{v}$
Since group velocity is equal to particle velocity i.e. $V_{g}=v$. So above equation can be written as
$V_{p}=\frac{C^{2}}{V_{g}}$
$V_{p}.V_{g}=C^{2}$
Note →
➢ $V_{g}=V_{p}$ for a non-dispersive medium ( in a non-dispersive medium all the waves travel with phase velocity).
➢ $V_{g}< V_{p}$ for normal dispersive medium
➢ $V_{g}> V_{p}$ for anomalous dispersive media.
Dispersive medium → The medium in which the phase velocity varies with wavelength or frequency is called a dispersive medium. In such a medium, waves of different wavelengths travel with different phase velocities.
Non-dispersive medium → The medium in which the phase velocity does not vary with wavelength or frequency is called a Non-dispersive medium.
Dispersive waves → Those waves in the medium for which phase velocity varies with wavelength or frequency are called dispersive waves.
Non-dispersive waves → Those waves in which phase velocity does not vary with wavelength are called non-dispersive waves. So phase velocity independent of wavelength.
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# Balancing REDOX reactions in a basic solution, when no H+ ions are present
I am trying to balance the following redox reaction, in basic solution. From what I have learned, to balance a redox reaction in basic solution you simply balance it as you normally would in acidic solution, then make sure the H+ ions are neutralized by adding OH- ions. In the example below, is my answer balanced in basic solution already? Or do I need to somehow cancel out the H present in H2O?
The bottom equation looks balanced to me. There are the same number of atoms of each element on each side of the equation and charge is conserved. The hydrogen in the $\ce{H2O}$ is balanced by the hydrogen in the $\ce{HAsO2}$ on the other side so it looks fine.
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# Mod-Gaussian convergence and Its applications for models of statistical mechanics
Méliot, Pierre-Loïc; Nikeghbali, Ashkan (2015). Mod-Gaussian convergence and Its applications for models of statistical mechanics. In: Donati-Martin, Catherine; Lejay, Antoine; Rouault, Alain. In Memoriam Marc Yor - Séminaire de Probabilités XLVII. Cham, Switzerland: Springer, 369-425.
## Abstract
In this paper we complete our understanding of the role played by the limiting (or residue) function in the context of mod-Gaussian convergence. The question about the probabilistic interpretation of such functions was initially raised by Marc Yor. After recalling our recent result which interprets the limiting function as a measure of “breaking of symmetry” in the Gaussian approximation in the framework of general central limit theorems type results, we introduce the framework of L$^{1}$-mod-Gaussian convergence in which the residue function is obtained as (up to a normalizing factor) the probability density of some sequences of random variables converging in law after a change of probability measure. In particular we recover some celebrated results due to Ellis and Newman on the convergence in law of dependent random variables arising in statistical mechanics. We complete our results by giving an alternative approach to the Stein method to obtain the rate of convergence in the Ellis-Newman convergence theorem and by proving a new local limit theorem. More generally we illustrate our results with simple models from statistical mechanics.
## Abstract
In this paper we complete our understanding of the role played by the limiting (or residue) function in the context of mod-Gaussian convergence. The question about the probabilistic interpretation of such functions was initially raised by Marc Yor. After recalling our recent result which interprets the limiting function as a measure of “breaking of symmetry” in the Gaussian approximation in the framework of general central limit theorems type results, we introduce the framework of L$^{1}$-mod-Gaussian convergence in which the residue function is obtained as (up to a normalizing factor) the probability density of some sequences of random variables converging in law after a change of probability measure. In particular we recover some celebrated results due to Ellis and Newman on the convergence in law of dependent random variables arising in statistical mechanics. We complete our results by giving an alternative approach to the Stein method to obtain the rate of convergence in the Ellis-Newman convergence theorem and by proving a new local limit theorem. More generally we illustrate our results with simple models from statistical mechanics.
## Statistics
### Citations
Dimensions.ai Metrics
1 citation in Web of Science®
1 citation in Scopus®
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# Conversion of annual interest rate compounded monthly to monthly effective interest rate [closed]
I am given that the annual interest rate is $r=4\%$ and that it is compounded monthly. I have to find the monthly effective interest rate.
If I wanted the annual effective interest rate, I would use the formula $r_e=(1+\frac{.04}{12})^{12}-1=.0407$ to find the yearly effective interest rate.
Then to go from yearly effective interest rate to monthly effective interest rate I would use: $r_e=(1+.0407)^\frac{1}{12}-1=.0033$.
Is this method correct? $.33\%$ does not seem high enough. Is there a more direct conversion? Thank you for your help.
• I am voting to close this question for being too basic as per quant.stackexchange.com/help/on-topic. This topic is covered in the early chapters of most introductory textbooks such as Hull's "Options, Futures and Other Derivatives" or Sundaresan's "Fixed Income Markets and Their Derivatives". – LocalVolatility Jan 28 '18 at 1:38
your result and your reasoning is correct. Just notice that there is a shorter way to the answer: $4/12 = 0.33$. Indeed, if the annual monthly compounded interest rate is $4\%$ than this means you get $4/12\%$ of interest every month!
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# Reference in C++
A reference is a synonym for another variable. The value of the variable changes, if the value of variable or value of reference changes. Here is an example of reference.
#include<bits/stdc++.h>
using namespace std;
int main(){
int n = 50;
int& pn = n;
cout<<"n = "<<n<<" pn = "<<pn<<endl;
n = n+10;
cout<<"n = "<<n<<" pn = "<<pn<<endl;
pn = pn*2;
cout<<"n = "<<n<<" pn = "<<pn<<endl;
return 0;
}
As the value of n changes, the value of rn changes. And as the value of rn changes, the value of n changes.
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시간 제한메모리 제한제출정답맞힌 사람정답 비율
1 초 32 MB88252228.205%
문제
Our heroes, Mirko and Slavko, plant Christmas wheat every year on Saint Lucy’s Day. It is well known that stalks of wheat grow at different speeds so, after a certain time, the wheat becomes quite messy. The guys are determined to solve this problem by playing the following game:
• When it’s Mirko’s turn, he chooses a stalk of wheat with the minimal height and prolongs its height so it’s of the same height as the first stalk longer than it.
• When it’s Slavko’s turn, he chooses a stalk of wheat with the maximal height and cuts it to be of the same height as the first stalk shorter than it.
• The game lasts while there are at least three stalks of different heights and the loser is the player who can’t make his move.
For given heights of all stalks of wheat and the assumption that Mirko is the one starting the game, determine the winner of the game and the height of the shortest and longest stalk when the game is finished.
입력
The first line of input contains the integer N (1 ≤ N ≤ 105), the number of wheat stalks. The second line of input contains N space separated integers that denote the heights of individual wheat stalks. The height of each stalk will be less than or equal to 105.
출력
The first line of output must contain the word “Mirko” if Mirko is the winner of the game, or “Slavko” if Slavko is the winner of the game.
The second line of output must contain the height of the shortest and longest stalk when the game is finished.
예제 입력 1
3
3 3 3
예제 출력 1
Slavko
3 3
예제 입력 2
4
3 1 2 1
예제 출력 2
Slavko
1 2
예제 입력 3
7
2 1 3 3 5 4 1
예제 출력 3
Slavko
2 3
힌트
Clarification of the first example: Mirko can’t make his move because there are no three stalks of different heights. Therefore, Slavko is the winner.
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Peggy’s Ribbon World makes award rosettes. Following is information about the company:
Variable cost per rosette-----------$1.11 Sales price per rosette--------------2.90 Total fixed costs per month--------879.00 1: Suppose Peggy’s would like to generate a profit of$750. Determine...
Mark B
(1) Target units = (Fixed cost profit)/Contribution per unit = (879 750)/(2.9-1.11) Target units =910 units (2) Breakeven units = 879/(2.9-1.11) = 491 units Margin of safety in units = 1070 - 491 = 579 units Margin of safety in dollar = 579 x 2.79 = 1615.41 Percentage of sales = (1070 - 491)/1070 x 100 = 54.11% (3)...
Looking for Something Else? Ask a Similar Question
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# A Why Energy of 4s is higher than 3d?
1. May 12, 2016
### ImShiva
According to Aufbau principle energy of 3d should be greater than 4s but here 4s is having higher energy. Please anyone explain this.
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2. May 12, 2016
### BvU
Hi ImShiva,
The principle you mention isn't about energy; it's about the order in which orbits are filled.
3. May 12, 2016
### ImShiva
But isn't lower energy implying that orbital must be filled first?
4. May 12, 2016
### BvU
Apparently not !
5. May 12, 2016
### ImShiva
How...? Please explain
6. May 12, 2016
### Khashishi
Lower energy states do indeed fill up before higher energy states. But, the Aufbau principle refers to how additional electrons fill up. An additional electron placed into the 3d state will have higher energy than an electron that is excited into the 3d state with the states below empty. That's because there is repulsion between the electrons. For helium, which has two electrons, there isn't too much interaction between an electron in 3d and in the 1s position, so, the energy isn't that high. Consider hydrogen, which has only one electron. Then the 3d state has the same energy as the 3p and 3s states. When you fill up the lower states, then the 3d state energy rises because it has more interaction with the inner electrons than the 3s and 3p states.
7. May 12, 2016
### ImShiva
So I guess (n+l) energy rule won't work here. What's the rule for energy of states then ?
8. May 12, 2016
### blue_leaf77
The (n+l) rule is for the individual energy of electrons in a subshell denoted by the quantum number $nl$ to form a ground state of the atom, The energy $E_{nl}$ governed by that rule is the energy of each subshell in the ground state of an atom and not the total energy of the atom. In your diagram, it is the total energy of the atom (He atom) which is shown, in particular it shows energy levels including its ground state and the excited states of the form $1s\, nl$ where one electron stays in $1s$ subshell and the other in $nl$ subshell. Again, the (n+l) rule is irrelevant here because the diagram is not about the energy of each subshell in a He atom.
The only rule I know regarding the energy diagram of a $1s\, nl$ He atom is that the energy for orthohelium is always lower than for parahelium for the same $nl$.
9. May 13, 2016
### DrDu
In Helium, the only orbital filled is 1s, so the aufbau principle is correct. The other orbitals are virtual orbitals and not filled, so they are irrelevant for the aufbau principle.
10. May 13, 2016
### Khashishi
For atoms with more than one electron, you need to go into lengthy calculations to come up with energy levels. There are various approximations, to varying degrees of accuracy (e.g. Hartree-Fock). But nothing so simple as the hydrogen atom.
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# Scalar Transport of Multiple Species
AcuSolve has the capability to track multiple species in a fluid flow by using the scalar transport equations for each of them individually.
The species transport equation is similar to the generalized governing equations for the fluid flow and it is expressed as: (1)
where
• ${\phi }_{i}$ represents the ith scalar species.
• ${\text{Ψ}}_{i}$ is the diffusion flux vector for species ${\phi }_{i}$ .
• ${\sigma }_{i}$ is the source per unit mass for species ${\phi }_{i}$ .
The diffusivity flux vector is expressed as: (2)
$\text{Ψ}=d\nabla \phi$
where $d$ is the diffusivity for species $\phi$ . The diffusivity can be modeled as constant, ramped against time step or customized using variable property and user functions.
The material properties can be set to be functions of species concentration. This models a ‘miscible’ property relative to mixing of multiple fluids.
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The Neal Kernel and Random Kitchen Sinks
So you read a book on Reproducing Kernel Hilbert Spaces and you’d like to try out this kernel thing. But you’ve got a lot of data and most algorithms will give you an expansion that requires a number of kernel functions linear in the amount of data. Not good if you’ve got millions to billions of instances.
You could try out low rank expansions such as the Nystrom method of Seeger and Williams, 2000, the randomized Sparse Greedy Matrix Approximation of Smola and Schölkopf, 2000 (the Nyström method is a special case where we only randomize by a single term), or the very efficient positive diagonal pivoting trick of Scheinberg and Fine, 2001. Alas, all those methods suffer from a serious problem: at training you need to multiply by the inverse of the reduced covariance matrix, which is $$O(d^2)$$ cost for a d dimensional expansion. An example of an online algorithm that suffers from the same problem is this (NIPS award winning) paper of Csato and Opper, 2002. Assuming that we’d like to have d grow with the sample size this is not a very useful strategy. Instead, we want to find a method which has $$O(d)$$ cost for d attributes yet shares good regularization properties that can be properly analyzed.
Enter Radford Neal’s seminal paper from 1994 on Gaussian Processes (a famous NIPS reject). In it he shows that a Neural Network with an infinite number of nodes and a Gaussian Prior over coefficients converges to a GP. More specifically, we get the kernel
$$k(x,x’) = E_{c}[\phi_c(x) \phi_c(x’)]$$
Here $$\phi_c(x)$$ is a function parametrized by c, e.g. the location of a basis function, the degree of a polynomial, or the direction of a Fourier basis function. There is also a discussion regarding RKHS in a paper by Smola, Schölkof and Müller, 1998 that discusses this phenomenon in regularization networks. These ideas were promptly forgotten by its authors. One exception is the empirical kernel map where one uses a generic design matrix that is generated through the observations directly.
It was not until the paper by Rahimi and Recht, 2008 on random kitchen sinks that this idea regained popularity. In a nutshell the algorithm works as follows: Draw d values $$c_i$$ from the distribution over c. Use the corresponding basis functions in a linear model with quadratic penalty on the expansion coefficients. This method works whenever the basis functions are well bounded. For instance, for the Fourier basis the functions are bounded by 1. The proof of convergence of the explicit function expansion to the kernel is then a simple consequence of Chernoff bounds.
In the random kitchen sinks paper Rahimi and Recht discuss RBF kernels and binary indicator functions. However, this works more generally for any set of well behaved set of basis functions used in generating a random design matrix. A few examples:
• Fourier basis with Gaussian parameters. Take functions of the form $$e^{i w^\top x}$$ where the coefficients $$w$$ are drawn from a Gaussian. This is the random kitchen sinks paper. Obviously you can use hash functions rather than an actual random number generator. This ensures that you don’t need to store all parameters $$w$$.
• Pick random separating hyperplanes. This will effectively give you functions of bounded variation.
• Use the empirical kernel map, i.e. we use some function $$k(x,x’)$$ for which we employ for $$x’$$ a random subset of the data we wish to train on.
1. cna-certification-a reblogged this from smolix
2. smolix posted this
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Volume 282 - 38th International Conference on High Energy Physics (ICHEP2016) - Poster Session
Measurement of the top quark mass from leptonic observables in pp collisions
C. Suarez* on behalf of the CMS collaboration
*corresponding author
Full text: pdf
Pre-published on: February 06, 2017
Published on: April 19, 2017
Abstract
A novel technique for measuring the top quark mass using only leptonic observables is discussed. Top and anti-top quark decays with one electron and one muon and at least one jet in the final state are selected in proton-proton collision data collected by the CMS experiment at a center-of-mass energy of 8 TeV, corresponding to an integrated luminosity of 19.7 fb$^{-1}$. Several variables are studied and the transverse momentum distribution of the charged lepton pair originated from the top quark decay is chosen to extract the top quark mass. The measurement is calibrated using simulated events.
DOI: https://doi.org/10.22323/1.282.0742
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
Open Access
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triangle circle inside it, heights prove exercise
Point $O$ lie inside $ABC$ triangle. Points $A1,B1,C1$ are projections of $O$ on heights led from $A,B,C$ Prove that if $AA1=BB1=CC1$ then $AA1=2r$, where $r$ is radius of circle inscribed in $ABC$ triangle. I 'd be happy with a hint that will lead me to answer
• there can be only one point $O$ such that $AA_1=BB_1=CC_1$
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# nLab Bisognano-Wichmann theorem
****, , ## Surveys, textbooks and lecture notes * __ * __ * , *** , , , * **** * , , , , * * * * * , * , * , * * * * * * * * * * * * * * , * * **** * * * and * * **** * * * * * * * * **** * * * * * **** * Axiomatizations * * * * * * * * * * * * * , * * * * -theorem * * Tools * , * * * * * * * * * * , * * , * Structural phenomena * * * * * * * * * Types of quantum field thories * * * * * * , * * * * * * * * * * * , , * examples * , * * * , * * * * * * * * * * * * * , , , , * , , * * * * * * *** **** (, , ) ## Concepts ****: , , , **** * * * * * * , * * * * * , * **** * * , ****, **** * * * * * * * , , * * * * * * * , * / * , * * * , * * * ** ** * , * * , * * **** * * , * * * * * ** ** * , * * , * * , * , * * * * * * * * **** * * * * * * * * / * * * * * , ## Theorems {#Theorems} ### States and observables * * * * * * * * * * ### Operator algebra * * * , * * * * ### Local QFT * * * * * * () ### Perturbative QFT * *
# Contents
## Idea
In the Haag-Kastler approach to quantum field theory the central object is a local net of operator algebras. The modular theory says that the local algebras have an associated modular group and a modular conjugation (see modular theory). The result of Bisognano-Wichmann that this page is about, describes the relation of these to the Poincare group.
## The Theorem
Assume a Wightman field theory of a scalar neutral field is such that the smeared field operators generate the local algebras, see Wightman axioms.
Then:
1. The modular group of the algebra associated with a wedge and the vacuum vector coincides with the unitary representation of the group of Lorentz boosts which maps the wedge onto itself.
2. The modular conjugation of the wedge W is given by the formula
$J_W = \Theta U(R_W(\pi))$
Here $\Theta$ denotes the PCT-operator of the Wightman field theory and $U(R_W(\pi))$ is the unitary representation of the rotation which leaves the characteristic two-plane of the wedge invariant. The angle of rotation is $\pi$.
3. The theory fulfills wedge duality, that is the commutant of the algebra associated to a wedge is the algebra associated to the causal complement of the wedge.
## References
The original work is:
• Bisognano, J. and Wichmann, E.H.: On the duality condition for a Hermitian scalar field , J. Math. Phys. 16 (1975), 985-1007.
There are a lot of secondary references, one is for example:
• Daniele Guido: Modular Theory for the von Neumann Algebras of Local Quantum Physics (arXiv)
Last revised on June 21, 2010 at 13:55:15. See the history of this page for a list of all contributions to it.
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Two-Round Man-in-the-Middle Security from LPN
David Cash, Eike Kiltz, and Stefano Tessaro
Abstract
Secret-key authentication protocols have recently received a considerable amount of attention, and a long line of research has been devoted to devising efficient protocols with security based on the hardness of the learning-parity with noise (LPN) problem, with the goal of achieving low communication and round complexities, as well as highest possible security guarantees. In this paper, we construct 2-round authentication protocols that are secure against sequential man-in-the-middle (MIM) attacks with tight reductions to LPN, Field-LPN, or other problems. The best prior protocols had either loose reductions and required 3 rounds (Lyubashevsky and Masny, CRYPTO'13) or had a much larger key (Kiltz et al., EUROCRYPT'11 and Dodis et al., EUROCRYPT'12). Our constructions follow from a new generic deterministic and round-preserving transformation enhancing actively-secure protocols of a special form to be sequentially MIM-secure while only adding a limited amount of key material and computation.
Available format(s)
Category
Secret-key cryptography
Publication info
Keywords
Secret-key authenticationMan-in-the-Middle securityLPNField LPN.
Contact author(s)
tessaro @ cs ucsb edu
History
Short URL
https://ia.cr/2015/1220
CC BY
BibTeX
@misc{cryptoeprint:2015/1220,
author = {David Cash and Eike Kiltz and Stefano Tessaro},
title = {Two-Round Man-in-the-Middle Security from LPN},
howpublished = {Cryptology ePrint Archive, Paper 2015/1220},
year = {2015},
note = {\url{https://eprint.iacr.org/2015/1220}},
url = {https://eprint.iacr.org/2015/1220}
}
Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.
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# July, 2018
## A probability puzzle
A nice prompt from @shahlock, some time ago: Math Prompt #apstats #mtbosTwo players A, B. A is 4-0 against B. How would you estimate probability A wins next match? Assume independence — M Shah (@shahlock) November 27, 2016 Stand back, everyone: I'm going to apply Bayes's Theorem. A prior Let's
## Ask Uncle Colin: Stuck on some trig
Dear Uncle Colin, I'm trying to solve $2\cos(3x)-3\sin(3x)=-1$ (for $0\le \theta \lt 90º$) but I keep getting stuck and/or confused! What do you recommend? - Losing Angles, Getting Ridiculous Answers, Nasty Geometric Equation Hi, LAGRANGE, and thank you for your message! There are a couple of ways to approach this:
## Can you find a centre and angle of rotation without any construction?
Some time ago, I had a message from someone who - somewhat oddly - wanted to find a centre of rotation (with an unknown angle) without constructing any bisectors. (Obviously, if it was a right-angle rotation, they could use the set-square trick; if it was a half-turn, the centre of
## Ask Uncle Colin: inverses and all sorts
Dear Uncle Colin, I'm stuck on a trigonometry question: find $\cos\br{\frac{1}{2}\arcsin\br{\frac{15}{17}}}$. Any bright ideas? - Any Rules Calculating Some Inverse Notation? Hi, ARCSIN, and thanks for your message! That's a nasty one! Let's start by thinking of a triangle with an angle of $\arcsin\br{\frac{15}{17}}$ - the opposite side is 15
## Wrong, But Useful: Episode 58
In this episode of Wrong, But Useful, we are joined by freelance mathematician @becky_k_warren, formerly of NRICH Becky likes sharing maths with people who "don't like maths" and the #beingmathematical twitter chat Number of the podcast: 157, which is the middle of a sexy prime triplet. Colin goes all Rees-Mogg
## Why the factor and remainder theorem work
So there I was, merrily teaching the factor and remainder theorems, and my student asked me one of my favourite questions: "I accept that the method works, but why does it?" (I like that kind of question because it makes me think on my feet in class, and that makes
Dear Uncle Colin, I had to find the $n$th term of a quadratic sequence (1, 6, 17, 34, 57). I remember my teacher saying something about a table, but I couldn't figure it out. Can you help? Struggles Expressing Quadratics Using Educator's Notation - Concrete Explanation? Hi, SEQUENCE, and thank
## Two coins, one fair, one biased
When the redoubtable @cuttheknotmath (Alexander Bogomolny) poses the following question: Two Coins: One Fair, one Biased https://t.co/Rz2zR3LRDj #FigureThat #math #probability pic.twitter.com/HHhnyGjhkq — Alexander Bogomolny (@CutTheKnotMath) March 5, 2018 ... you know there must be Something Up. Surely (the naive reader thinks) the one with two heads out of three is
## Ask Uncle Colin: A Coin Toss Conundrum
Dear Uncle Colin, Please can you settle an argument? I say, if you toss a coin three times, the probability of getting all heads is one in four, because the only possibilities are HHH, HHT, HTT and TTT. My friend says it's one in eight, being $\frac{1}{2}\times \frac{1}{2} \times \frac{1}{2}$.
## A calculator puzzle
"Your calculator has broken, leaving you with only the buttons for $\sin$, $\cos$, $\tan$ and their inverses, the equals button and the 0 that starts on the screen. Show that you can still produce any positive rational number." When this showed up on Reddit, I knew I was in for
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# How do you find the equation of a circle that passes through the points (1, 1), (1, 5), and (5, 5)?
Jun 25, 2016
${\left(x - 3\right)}^{2} + {\left(y - 3\right)}^{2} = 8$
#### Explanation:
Note that the centre of the circle is equidistant from all three points, the distance being the radius of the circle.
Any point equidistant from two points must lie on the perpendicular bisector of the line segment joining those two points. That is, on the line through the midpoint of the line segment, perpendicular to the line segment.
The perpendicular bisector of the line segment joining the points $\left(1 , 1\right)$ and $\left(1 , 5\right)$ is $y = 3$
The perpendicular bisector of the line segment joining the points $\left(1 , 5\right)$ and $\left(5 , 5\right)$ is $x = 3$
These intersect at the centre of the circle $\left(3 , 3\right)$
The distance between $\left(1 , 1\right)$ and $\left(3 , 3\right)$ is:
$\sqrt{{\left(3 - 1\right)}^{2} + {\left(3 - 1\right)}^{2}} = \sqrt{{2}^{2} + {2}^{2}} = \sqrt{8}$
So the equation of the circle may be written:
${\left(x - 3\right)}^{2} + {\left(y - 3\right)}^{2} = 8$
graph{((x-3)^2+(y-3)^2-8)((x-3)^2+(y-3)^2-0.02)((x-1)^2+(y-1)^2-0.02)((x-1)^2+(y-5)^2-0.02)((x-5)^2+(y-5)^2-0.02)(y-3)(x-3+0.0001y)=0 [-7.625, 12.375, -2.72, 7.28]}
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# Measurement of the Charm-Mixing Parameter ${y}_{CP}$
11 January 2019
Abstract: A measurement of the charm-mixing parameter ${y}_{CP}$ using ${D}^{0}\to {K}^{+}{K}^{-}$, ${D}^{0}\to {\pi }^{+}{\pi }^{-}$, and ${D}^{0}\to {K}^{-}{\pi }^{+}$ decays is reported. The ${D}^{0}$ mesons are required to originate from semimuonic decays of ${B}^{-}$ and ${\overline{B}}^{0}$ mesons. These decays are partially reconstructed in a data set of proton-proton collisions at center-of-mass energies of 7 and 8 TeV collected with the LHCb experiment and corresponding to an integrated luminosity of $3\text{}\text{}{\mathrm{fb}}^{-1}$. The ${y}_{CP}$ parameter is measured to be $\left(0.57±0.13\left(\mathrm{stat}\right)±0.09\left(\mathrm{syst}\right)\right)%$, in agreement with, and as precise as, the current world-average value.
Published in: Physical Review Letters 122 (2019)
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Speaker : Professor Cao, Peng Scattering Elements in the Banach Algebra 2012-09-19 (Wed) 10:30 - 12:00 Seminar Room 617, Institute of Mathematics (NTU Campus) It is well known that in a Banach algebra, if the set of quasinilpotent elements is a linear subspace, or a semigroup, then this set equals the Jacobson Radicals. In this talk, we will consisder the similar case for the elements with at most countable spectrum, which are called scattered elements. Firstly, we will give the definition of scattered radical, and show that the scattered radical has many properties as the Jacobson Radical. Then we will give some equilavent conditions: (i) $S(A)+S(A)\subset S(A);$ (ii) $S(A)S(A)\subset S(A);$ (iii) $[S(A),A]\subset R_{sc}(A).$ where, $S(A)$ means the set of scattered elements in the Banach algebras $A$ , and $R_{sc}(A)$ means the scattered radical of $A$.
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### Home > CCAA > Chapter 2 Unit 4 > Lesson CC3: 2.1.5 > Problem2-51
2-51.
Simplify the following expressions by combining like terms, if possible. Homework Help ✎
1. $x+x−3+4x^2+2x−x$
Rearrange and put like terms together.
Organize the terms from the greatest to the smallest power.
$4x^{2} + 3x − 3$
1. $8x^2+3x−13x^2+10x^2−25x−x$
Follow the steps in part (a).
$5x^{2} − 23x$
1. $4x+3y$
Are there any like terms to combine?
1. $20+3xy−3+4y^2+10−2y^2$
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# Finding and proving if a relation is reflexive/transitive/symmetric/anti-symmetric
Let R be binary relation on N (natural numbers) defined by xRy if and only if x − 2 ≤ y ≤ x + 2. Is R reflexive? Is R symmetric? Is R antisymmetric? Is R transitive?
I'm not sure if I'm understanding this right or how I can prove it. This is what I currently have:
Reflexive: Yes, because $$x-2≤x≤x+2$$ so $$xRx$$ (How can I prove this for all $$x$$ though?)
Symmetric: Yes, because $$xR(x+2)$$ therefore $$(x+2)R((x+2)-2)$$ (I can tell this is wrong, but basically I'm trying to say since x always relates to x+2, x+2 will always relate to x).
Anti-Symmetry and Transitive just need examples of them not working, so I think these work:
Anti-Symmetric: No, e.g. $$2R4$$ and $$4R2$$, but $$4 ≠ 2$$
Transitive: No, e.g. $$2R4$$ and $$4R6$$, but $$2! R 6$$
Thanks for any help.
• To show symmetric you must show if xRy then yRx – J. W. Tanner Apr 30 at 13:54
• All good except for the proof of symmetry. Check the definition again. – A. Pongrácz Apr 30 at 14:03
Note that $$R$$ is just $$|x-y| \le 2$$, that is the distance between two points are less than or equal to $$2$$.
It is reflexive because, we have $$-2 \le 0 \le 2$$, add $$x$$ to the inequalities and we have the result.
It is symmetric, if $$|x-y| \le 2$$, then $$|y-x|=|x-y|\le 2$$.
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# What is the relation between universality and chaotic behavior in Elementary Cellular Automata?
So i know that rule 110 from Elementary Cellular Automata is universal but i am wondering if we had other chaotic behaviors let's say there exists a system that shows nearly infinite chaotic behaviors {a structure at a time} similar to ECA structures, how can we prove or disprove universality from that system ?
Suppose we have a system with one specific rule to execute and the input can be defined by different number of cells {i.e input = 5 cells} these 5 cells for example can show (2000) chaotic behavior... 4 cells can show nearly (1800), for example to get a structure from the (1800) structures, i can change the 4 cells by changing the rule set of these 4 cells just like in ECA => 'two white cells will result in a black cell', i can define a similar rule set on the 4 cells, 5 cells or more generally on x cells where x < infinite.
That system with same input {4 cells} with different rule set can simulate most of the structures in Elementary Cellular Automata like rule 30, 90 & 60 and others, what can be proved about such system and how universality is related to it ?
• You prove (or disprove) universality by proving that the thing can (or cannot) simulate some other universal system. – David Richerby Oct 13 '15 at 23:01
• Can you define "chaotic behaviors", or what it means for a system to "show nearly infinite chaotic behaviors", in a precise or rigorous way? If not then you certainly can't prove a result of the form you seem to hope exists. Personally, I doubt that any such result exists. There are chaotic systems that are very simple and I doubt are universal. The question seems to be based on a hope that somehow there is some relationship between chaos vs universality, but I'm skeptical about whether that hope is actually valid. – D.W. Oct 13 '15 at 23:09
For instance, here's an example to challenge your intuition that the two concepts are somehow related. Wikipedia lists the dynamic system given by the map $x \mapsto 4x (1-x)$ and $y \mapsto x + y \bmod 1$ as an example of a chaotic system. However I doubt that this system is universal in any meaningful sense.
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# Is such transformed beta a known distribution?
I have a distribution $Y \sim 1/(1-Beta(\alpha,\beta))$. I would like to understand its properties.
I was able to write down its density function, but I was not able to integrate it to get its mean, for example (except through simulation).
The Beta is mirror-symmetric $1-\mathrm{Beta}(\alpha, \beta) =\mathrm{Beta}(\beta, \alpha)$, so you are essentially looking for the distribution of the reciprocal of a Beta random variable.
Also, if $X\sim \mathrm{Beta}(\beta, \alpha)$, then $1/X-1\sim\mathrm{BetaPrime}(\alpha, \beta)$, where $\mathrm{BetaPrime}$ is the Beta prime distribution.
So, I would say that your $Y-1$ is distributed $\mathrm{BetaPrime}(\alpha, \beta)$.
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# Cantor set
The Cantor set $\mathcal{C}$ is a subset of the real numbers that exhibits a number of interesting and counter-intuitive properties. It is among the simplest examples of a fractal. Topologically, it is a closed set, and also a perfect set. Despite containing an uncountable number of elements, it has Lebesgue measure equal to $0$.
The Cantor set can be described recursively as follows: begin with the closed interval $[0,1]$, and then remove the open middle third segment $(1/3,2/3)$, dividing the interval into two intervals of length $\frac{1}{3}$. Then remove the middle third of the two remaining segments, and remove the middle third of the four remaining segments, and so on ad infinitum.
$[asy] int max = 7; real thick = 0.025; void cantor(int n, real y){ if(n == 0) fill((0,y+thick)--(0,y-thick)--(1,y-thick)--(1,y+thick)--cycle,linewidth(3)); if(n != 0) { cantor(n-1,y); for(int i = 0; i <= 3^(n-1); ++i) fill( ( (1.0+3*i)/(3^n) ,y+0.1)--( (1.0+3*i)/(3^n) ,y-0.1)--( (2.0+3*i)/(3^n) ,y-0.1)--( (2.0+3*i)/(3^n) ,y+0.1)--cycle,white); } } for(int i = 0; i < max; ++i) cantor(i,-0.2*i); [/asy]$
Equivalently, we may define $\mathcal{C}$ to be the set of real numbers between $0$ and $1$ with a base three expansion that contains only the digits $0$ and $2$ (including repeating decimals).
Another equivalent representation for $\mathcal{C}$ is: Start with the interval $[0,1]$, then scale it by $\frac{1}{3}$. Then join it with a copy shifted by $\frac{2}{3}$, and repeat ad infinitum.
A distorted version of $\mathcal{C}$ can be found by repeatedly applying the function $f(x)=a(x-\frac{1}{2})^2+1-\frac{a}{4},a>4$, and keeping the values of x for which the values always remain bounded. This constructs $\mathcal{C}$ by repeatedly removing the middle. This works since for $x\notin [0,1]$ the values will always diverge, and the values of $x$ for which $f(x)\in [0,1]$ is the union of intervals $\left[0,\frac{1}{2}-\sqrt{\frac{1}{4}-\frac{1}{a}}\right]\cup \left[\frac{1}{2}+\sqrt{\frac{1}{4}-\frac{1}{a}},1\right]$, which are disjoint when $a>4$. -- EVIN-
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# What to do with questions that ask more than one question?
From time to time, we get questions such as this one that ask two separate questions, whose answers are unlikely to have much in common, and for which people are likely to be interested in answering one question but not both.
What should one do when encountering these?
Should I edit one of the questions out, so that the post only asks one question? (presumably with a pointer to the OP to post a new question, and where the OP can find the text originally posted)
Does the answer differ if the question already has answers to one of the questions? More than one?
Previously I have commented to the poster that they should split the question. I do not recall ever seeing this lead to a resolution of the issue.
• Related older question: Posting multiple questions as one?. Other posts linked there might be of interest, too. I'll point out that there is also a comment template. – Martin Sleziak Aug 18 '17 at 14:37
• OTOH the question I linked is almost 5 years old - I believe since than even close reasons have been revamped. So perhaps it would not hurt to have a new discussion about this...? (I'd say it's partly up to Hurkyl -as the OP - whether they are satisfied with the answers on the old question or whether they think some new points should be addressed.) – Martin Sleziak Aug 18 '17 at 14:40
• When unrelated problems are being combined in a Question, I tend to ask (by Commenting) for the OP to do the editorial work. It has happened that the OP is more interested in the second problem (say) than the first, and it is really up to them to decide. My Comment will typical feature the magic-link [ask] about how to ask. The stick is to vote to close (assuming inaction by the OP), which allows any existing Answers to remain in the original context – hardmath Aug 18 '17 at 17:28
• I surveyed Why we should avoid asking multi-question questions on meta.Islam.SE (but it's relevant here). Also note the "too broad" close reason has changed to cover this: "Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question." – Rebecca J. Stones Aug 27 '17 at 8:26
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The influence of temperature and strain rate on the upper yield point of ingot iron was studied. Torsion tests were conducted using strain rates of 12.5/sec, 0.25/sec, and 0.0001/sec over the temperature range 77 to 525 deg K. The upper yield point showed a rapid increase as the temperature was lowered. An increase in the strain rate also caused an increase in the yield point. An apparent activation energy can be associated with the strain rate and temperature dependence of the yield point. This energy is influenced by stress level, and it appears from the present study that the relationship can be described by an equation of the form
$ΔH=ΔH¯τ¯−ττ¯b.$
If this relationship is substituted for ΔH in a modification of the Boltzmann relation, the following result is obtained:
$logγ˙γ˙1$
$=MΔH¯RT1τ¯−τ1τ¯b$
$1−T1Tτ¯−ττ¯−τ1b.$
This equation describes the experimental data within ± 3000 psi. The results of this investigation compared with tensile test data from other investigators confirm that state of stress is an important factor in determining whether a material will behave in a ductile or brittle fashion.
This content is only available via PDF.
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### Home > CCAA8 > Chapter 9 Unit 3 > Lesson CC3: 9.2.7 > Problem9-154
9-154.
Simplify each expression.
1. $\frac { 12 } { 5 } \div \frac { 7 } { 10 }$
Dividing by a fraction is the same as multiplying by its reciprocal.
$\frac{12}{5}\left(\frac{10}{7}\right)$
Simplifying at the beginning is easier than simplifying after multiplying.
$\frac{12}{1}\left(\frac{2}{7}\right)$
$\frac{24}{7}$
1. $\frac{9}{4}\div\left(-\frac{1}{3}\right)$
See the help for part (a).
$-\frac{27}{4}$
1. $-\frac{3}{5}\div\left(-\frac{1}{6}\right)$
See the help for part (a).
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# The modulus
Find the modulus of the complex number 2 + 5i
Correct result:
m = 5.3852
#### Solution:
$m=\sqrt{{2}^{2}+{5}^{2}}=\sqrt{29}=5.3852$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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The vector u = (3.9, u3) and the length of the vector u is 12. What is is u3?
• Vector
Calculate length of the vector v⃗ = (9.75, 6.75, -6.5, -3.75, 2).
• Vector 7
Given vector OA(12,16) and vector OB(4,1). Find vector AB and vector |A|.
• Linear imaginary equation
Given that ? "this is z star" Find the value of the complex number z.
• Unit vector 2D
Determine coordinates of unit vector to vector AB if A[-6; 8], B[-18; 10].
• Imaginary numbers
Find two imaginary numbers whose sum is a real number. How are the two imaginary numbers related? What is its sum?
• Vector perpendicular
Find the vector a = (2, y, z) so that a⊥ b and a ⊥ c where b = (-1, 4, 2) and c = (3, -3, -1)
• Vector v4
Find the vector v4 perpendicular to vectors v1 = (1, 1, 1, -1), v2 = (1, 1, -1, 1) and v3 = (0, 0, 1, 1)
• Log
Calculate value of expression log |3 +7i +5i2| .
• Add vector
Given that P = (5, 8) and Q = (6, 9), find the component form and magnitude of vector PQ.
• Mappings of complex numbers
Find the images of the following points under mappings: z=3-2j w=2zj+j-1
• Coordinates of square vertices
The ABCD square has the center S [−3, −2] and the vertex A [1, −3]. Find the coordinates of the other vertices of the square.
• Bearing
A plane flew 50 km on a bearing 63°20' and the flew on a bearing 153°20' for 140km. Find the distance between the starting point and the ending point.
• Linear combination of complex
If z1=5+3i and z2=4-2i, write the following in the form a+bi a) 4z1+6z2 b) z1*z2
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# Publications by Neville Harnew
## Observation of the suppressed decay Λb0 → pπ−μ+μ−
Journal of High Energy Physics IOP Publishing 2017 (2017) 029-
The suppressed decay Λb0 → pπ−μ+μ−, excluding the J/ψ and ψ(2S) → μ+μ− resonances, is observed for the first time with a significance of 5.5 standard deviations. The analysis is performed with proton-proton collision data corresponding to an integrated luminosity of 3 fb−1 collected with the LHCb experiment. The Λb0 → pπ−μ+μ− branching fraction is measured relative to the Λb0 → J/ψ(→μ+μ−)pπ− branching fraction giving $\frac{\mathrm{\mathcal{B}}\left({\Lambda}_b^0\to p{\pi}^{-}{\mu}^{+}{\mu}^{-}\right)}{\mathrm{\mathcal{B}}\left({\Lambda}_b^0\to \mathrm{J}/\psi \left(\to {\mu}^{+}{\mu}^{-}\right)p{\pi}^{-}\right)}=0.044\pm 0.012\pm 0.007,$ where the first uncertainty is statistical and the second is systematic. This is the first observation of a b → d transition in a baryonic decay.
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# Beam Displacement Calculator
View All Technical Tools
### Refractive Index of Medium, n:
Beam Displacement, s (mm) : --
Displacement of Secondary Reflection, x (mm): --
$$s = t \cdot \sin{\theta} \left( 1 - \frac{\cos{\theta}}{\sqrt{n^2 - \sin^2{\theta}}} \right)$$
$$x = \frac{t \cdot \sin{\left( 2 \theta \right)}}{\sqrt{n^2 - \sin^2{\theta}}}$$
s Beam Displacement x Displacement of secondary reflection θ Angle of incidence between beam and plate normal
t Thickness of plate n Refractive index of medium
Note: This calculator is valid within the following limits: t > 0mm, 0° < θ < 90°, n > 0
## Description
Determine the displacement of a collimated beam incident on a tilted plane-parallel plate, or window. This displacement is dependent on the angle of incidence, the thickness of the plate, and the refractive index of the plate. This calculator also determines the displacement of the secondary reflection resulting from a small portion of the beam reflecting off the glass-air interface of the plate.
The exit beam will be parallel to the input beam and magnification remains constant when imaging through a plane-parallel plate.
## Example
Question: What is the displacement of a beam and its secondary reflection when the beam passes through a 10mm thick N-BK7 plate at an angle of incidence of 15 degrees?
Answer: The beam displacement and secondary reflection displacement can be determined by:
$$s = \left( 10 \text{mm} \right) \sin{\left( 15° \right)} \left( 1 - \frac{\cos{\left( 15° \right)}}{\sqrt{1.517^2 - \sin^2{\left(15° \right)}}} \right) = 0.916 \text{mm}$$
$$x = \frac{\left( 10 \text{mm} \right) \sin{\left( 2 \cdot 15° \right)}}{\sqrt{1.517^2 - \sin^2{\left( 15° \right)}}} = 3.499 \text{mm}$$
The displacement of the beam is 0.916mm and the displacement of the beam’s secondary reflection is 3.499mm.
## Related Resources
Application Note
Application Note
View All Now
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# What is the specific heat of a material if a 6.0 g sample absorbs 50 J when it is heated from 30°C to 50°C?
Nov 9, 2015
$C s$= 0.4167$\frac{J}{g C}$
#### Explanation:
Use the equation $q = m {C}_{s} \Delta T$
You are given the $q$ in the form of J. You are given the mass in grams and also the initial and final temperature. $\Delta$T = ${T}_{f}$-${T}_{i}$.
If you plug in the known variables the equation will look like:
$50 J$=$6 g$$\left({C}_{s}\right)$$20 C$
At this point you just solve for ${C}_{s}$ using some algebra.
$50 J$/$\left(6 g\right)$$\left(20 C\right)$
It is really helpful to keep units in the equation when solving. Let the units lead the way in solving thermochemistry problems.
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Attention! You have been redirected to a symmetry.
Av(132, 213, 231)
Permutation examples
length 2: 12, 21
length 3: 123, 312, 321
length 4: 1234, 4123, 4312, 4321
length 5: 12345, 51234, 54123, 54312, 54321
Generating function
$A(x) = \frac{x^{2} - x + 1}{x^{2} - 2 x + 1}$
Coefficients
1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ...
System of equations
$\operatorname{F_{2}}{\left (x \right )} = \operatorname{F_{0}}{\left (x \right )} + \operatorname{F_{1}}{\left (x \right )}$
$\operatorname{F_{0}}{\left (x \right )} = 1$
$\operatorname{F_{1}}{\left (x \right )} = \frac{x}{\left(x - 1\right)^{2}}$
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• Share
# Design of Machine for GATE ME Quiz 2 – Attempt Now!
7 months ago .
Save
GATE exam is round the corner. If you are appearing for this year then here is Design Machine GATE Quiz 2 to help you prepare for your upcoming GATE 2019 exam. GATE ME syllabus covers several subjects, each one is equally important. One of the most important subjects among all subjects in GATE ME is Design of Machine. The subject is vast, but with proper practice you can ace it.
This quiz contains important questions which match the pattern of the GATE exam. Check your preparation level in every chapter of Design of Machine for GATE ME by taking the quiz and comparing your ranks. Learn about Design for static and dynamic loading; failure theories; fatigue strengthshafts, gears, rolling and sliding contact bearings, brakes and clutches, springs and more!
Design of Machine Elements for GATE ME Quiz 2
Que. 1
A band brake having band-width of 80 mm, drum diameter of 250 mm, coefficient of friction of 0.25 and angle of wrap of 270 degrees is required to exert a friction torque of 1000 N-m. The maximum tension (in kN) developed in the band is
1.
1.88
2.
3.56
3.
6.76
4.
11.56
Que. 2
A lightly loaded full journal bearing has a journal of 50 mm, bush bore of 50.05 mm and bush length of 20 mm. if rotational speed of journal is 1200 rpm and average viscosity of liquid lubricant is 0.03 Pa s, the power loss (in W) will be
1.
37
2.
74
3.
118
4.
237
Que. 3
If $$\frac{x}{y} = 4$$, then stress concentration due to hole given in the figure will be
Que. 4
Two identical ball bearings P and Q are operating at loads 30kN and 45kN respectively. The ratio of the life of bearing P to the life of bearing Q is
1.
81/16
2.
27/8
3.
9/4
4.
3/2
Que. 5
A spur gear has a module of 3 mm, number of teeth 16, a face width of 36 mm and a pressure angle of 20°. It is transmitting a power of 3 kW at 20 rev/s. Taking a velocity factor of 1.5, and Modified Lewis form factor of 0.3, the stress in the gear tooth is about
1.
32 MPa
2.
46 MPa
3.
58 MPa
4.
70 MPa
As we all know, practice is the key to success. Therefore, boost your preparation by starting your practice now.
Furthermore, chat with your fellow aspirants and our experts to get your doubts cleared on Testbook Discuss:
7 months ago
7 months ago
7 months ago
7 months ago
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# Fractional part integration
$\large \int_1^\infty \dfrac{\{ x \}} {x^5} \, \mathrm{d}x$
If the value of the integral above is equal to $\dfrac1A - \dfrac{\pi^B}C$ for positive integers $A$, $B$ and $C$, find $A+B+C$.
Bonus: Find the general form of $\displaystyle \int_1^\infty \dfrac{\{x\}}{x^n} \, \mathrm{d}x$.
Clarification: $\{x\}$ denotes the fractional part function.
×
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My Math Forum rearranging
New Users Post up here and introduce yourself!
February 26th, 2018, 07:51 AM #1 Newbie Joined: Feb 2018 From: Wales Posts: 1 Thanks: 0 rearranging Which one of the following products cannot be rearranged to give a correct product? 26x3=78 16x3=48 Last edited by greg1313; February 26th, 2018 at 10:58 AM.
February 26th, 2018, 10:21 AM #2 Global Moderator Joined: Dec 2006 Posts: 19,178 Thanks: 1646 The first can be rearranged as 38 × 2 = 76. Thanks from MathLover2005
February 26th, 2018, 12:23 PM #3 Global Moderator Joined: May 2007 Posts: 6,540 Thanks: 591 The second is also $4\times 12\ or\ 2\times 24$. What's the point?
February 26th, 2018, 12:52 PM #4 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,762 Thanks: 860 or 6+4-3+1 = 8 and 6*4*1/3 = 8 (if all 5 original digits must be used)
Thread Tools Display Modes Linear Mode
Contact - Home - Forums - Cryptocurrency Forum - Top
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# Greenland Telescope (GLT) Project: "A Direct Confirmation of Black Hole with Submillimeter VLBI"
The GLT project is deploying a new submillimeter (submm) VLBI station in Greenland. Our primary scientific goal is to image a shadow of the supermassive black hole (SMBH) of six billion solar masses in M87 at the center of the Virgo cluster of galaxies. The expected SMBH shadow size of 40-50 $\mu$as requires superbly high angular resolution, suggesting that the submm VLBI would be the only way to obtain the shadow image. The Summit station in Greenland enables us to establish baselines longer than 9,000 km with ALMA in Chile and SMA in Hawaii as well as providing a unique $u$–$v$ coverage for imaging M87. Our VLBI network will achieve a superior angular resolution of about 20 $\mu$as at 350 GHz, corresponding to $\sim2.5$ times of the Schwarzschild radius of the supermassive black hole in M87. We have been monitoring the atmospheric opacity at 230 GHz since August. 2011; we have confirmed the value on site during the winter season is comparable to the ALMA site thanks to high altitude of 3,200 m and low temperature of $-50\degr$C. We will report current status and future plan of the GLT project towards our expected first light on 2015–2016.
Date added: Tue, 8 Oct 13
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# Classify each of the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper traingular, a lower triangular matrix. [9 2 -3] - Mathematics and Statistics
Sum
Classify each of the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper traingular, a lower triangular matrix.
[9 sqrt(2) -3]
#### Solution
Let A = [9 sqrt(2) -3]
As matrix A has only one row.
∴ A is a row matrix.
Concept: Determinant of a Matrix
Is there an error in this question or solution?
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# Coincidence point
In mathematics, a coincidence point (or simply coincidence) of two mappings in their domain having the same image point under both mappings.
Formally, given two mappings
$f,g \colon X \rightarrow Y$
we say that a point x in X is a coincidence point of f and g if f(x) = g(x).
Coincidence theory (the study of coincidence points) is, in most settings, a generalization of fixed point theory, the study of points x with f(x) = x. Fixed point theory is the special case obtained from the above by letting X = Y and taking g to be the identity mapping.
Just as fixed point theory has its fixed-point theorems, there are theorems that guarantee the existence of coincidence points for pairs of mappings. Notable among them, in the setting of manifolds, is the Lefschetz coincidence theorem, which is typically known only in its special case formulation for fixed points.
Coincidence points, like fixed points, are today studied using many tools from mathematical analysis and topology. An equaliser is a generalization of the coincidence set.
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# Problem with tensor equation in Cartan formalism
1. Sep 8, 2014
### kurre
I have some problems verifying a particular equation in the Cartan formalism of GR. Basically I want to prove the following equation in three dimensions
$$\epsilon_{abc} e^a\wedge R^{bc}=\sqrt{|g|}Rd^3 x$$
where R is the Ricci scalar and $R^{bc}$ is the Ricci curvature
Attempt at a solution:
$$\epsilon_{abc} e^a\wedge R^{bc}=\epsilon_{abc} e_\mu^ae_\alpha^be_\beta^c R^{\alpha\beta}_{\nu\rho} dx^\mu\wedge dx^\nu\wedge dx^\rho$$
Now the idea is that the number of dimensions and the Levi-Civita tensor and the antisymmetry of the three-form forces the set {alpha,beta}={nu,rho}. This will give the expression
\begin{align}\epsilon_{abc} e^a\wedge R^{bc}&=\epsilon_{abc} e_0^ae_1^be_2^c R^{12}_{12} dx^0\wedge dx^1\wedge dx^2+\\&\epsilon_{abc} e_0^ae_1^be_2^c R^{12}_{21} dx^0\wedge dx^2\wedge dx^1+\\&\epsilon_{abc} e_0^ae_2^be_1^c R^{21}_{21} dx^0\wedge dx^2\wedge dx^1+\\&\epsilon_{abc} e_0^ae_2^be_1^c R^{21}_{12} dx^0\wedge dx^1\wedge dx^2+({\rm cyclic\,permutations})\end{align}
The problem now is that the Ricci scalar is $$R^{12}_{12}+R^{21}_{21}+({\rm cyclic\,permutations})$$, so when counting the number of terms I obtain $$2\sqrt{|g|}Rd^3 x$$ which is wrong by a factor of 2. Can anyone see where I made a mistake?
(Maybe this post belongs in the Homework section since the question is definitely homework-like, but since it is NOT homework and the rules/guidelines do not say anything about "homework-like questions", I wasnt sure...)
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1. ## Linear Transformations
Verify that the mapping T : R4 -> R2 defined by
T (x1; x2; x3; x4) = (x3 - x2; 5x1 + 3x4)
is a linear transformation.
Find a basis for Im (T ) and Ker (T ).
2. Showing linearity:
Choose 2 vectors in $\displaystyle v_1,v_2\in \mathbb{R}^4$.
$\displaystyle v_1 = (a_1,a_2,a_3,a_4)$
$\displaystyle v_2 = (b_1,b_2,b_3,b_4)$
Choose $\displaystyle \lambda_1,\lambda_2\in \mathbb{R}$
We want to show that $\displaystyle T(\lambda_1v_1+\lambda_2v_2)= \lambda_1T(v_1)+\lambda_2T(v_2)$:
Can you do that? It's just writing out.
Observe that $\displaystyle v\in$ ker$\displaystyle (T) \Leftrightarrow T(v)= (x_3-x_2,5x_1+3x_4) = (0, 0)$. Hence v satisfies $\displaystyle x_3= x_2, x_4 = -\frac{5}{3}x_1$.
Find two independant vectors in $\displaystyle v_1,v_2\in \mathbb{R}^4$ that satisfy these relations and you have a basis for ker(T).
Find 2 more independant vectors: $\displaystyle v_3,v_4\in \mathbb{R}^4$ and you have a basis for Im(T)
(Dim(Im(T))+ Dim(ker(T)) = 4).
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Conference paper Open Access
# Age influence in gender stereotypes related to Internet use in young people: a case study
Verdugo-Castro, S.; García-Holgado, A.; Sánchez-Gómez, M. C.
### JSON-LD (schema.org) Export
{
"inLanguage": {
"alternateName": "eng",
"@type": "Language",
"name": "English"
},
"description": "<p>The existence of gender stereotypes in relation to the use of the Internet led to the need to carry out the present study, which approaches young people perception of the use of the Internet and technologies. Based on knowledge of the existence of gender stereotypes in relation to the use of the Internet, the objective of the study is to detect whether the age of young people, with some previous experience on the Internet, marks differences in relation to the use they make of the Internet; in other words, whether the use made of the world network and the existing stereotypes may or may not have some type of dependent relationship with the age of the subjects. The study was applied in a group of the University of Salamanca (Spain) during the academic year 2018/2019, after the previous realization of activities during four months of sensitization, under the European project WYRED (netWorked Youth Research for Empowerment in the Digital). The final sample was composed by 48 subjects. For this purpose, a questionnaire was applied with 40 final items, which collected different activities that can be carried out on the Internet in order to find out what trend of use they presented in relation to the Internet, in comparison to their age. The two groups consisted of 26 people aged 20 or younger, and 22 people aged 20 or older. By means of a descriptive analysis, and the application of normality tests and non-parametric tests, no dependent relationships were detected between the use of the Internet after carrying out the survey and the age group in which the subject belonged. For the future it would be of special interest to be able to repeat the study comparing the opinion and use of young people on the Internet </p>",
"creator": [
{
"affiliation": "University of Salamanca",
"@id": "https://orcid.org/0000-0002-9357-1747",
"@type": "Person",
"name": "Verdugo-Castro, S."
},
{
"affiliation": "University of Salamanca",
"@id": "https://orcid.org/0000-0001-9663-1103",
"@type": "Person",
},
{
"affiliation": "University of Salamanca",
"@id": "https://orcid.org/0000-0003-4726-7143",
"@type": "Person",
"name": "S\u00e1nchez-G\u00f3mez, M. C."
}
],
"headline": "Age influence in gender stereotypes related to Internet use in young people: a case study",
"datePublished": "2020-01-14",
"url": "https://zenodo.org/record/3608168",
"version": "1.0",
"keywords": [
"Gender stereotypes, Internet, gender roles, inclusive digital society"
],
"@context": "https://schema.org/",
"identifier": "https://doi.org/10.1145/3362789.3362846",
"@id": "https://doi.org/10.1145/3362789.3362846",
"@type": "ScholarlyArticle",
"name": "Age influence in gender stereotypes related to Internet use in young people: a case study"
}
28
23
views
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# lmomlap: L-moments of the Laplace Distribution In lmomco: L-Moments, Censored L-Moments, Trimmed L-Moments, L-Comoments, and Many Distributions
## Description
This function estimates the L-moments of the Laplace distribution given the parameters (ξ and α) from parlap. The L-moments in terms of the parameters are λ_1 = ξ, λ_2 = 3α/4, τ_3 = 0, τ_4 = 17/22, τ_5 = 0, and τ_6 = 31/360.
For r odd and r ≥ 3, λ_r = 0, and for r even and r ≥ 4, the L-moments using the hypergeometric function {}_2F_1() are
λ_r = \frac{2α}{r(r-1)}[1 - {}_2F_1(-r, r-1, 1, 1/2)]\mbox{,}
where {}_2F_1(a, b, c, z) is defined as
{}_2F_1(a, b, c, z) = ∑_{n=0}^∞ \frac{(a)_n(b)_n}{(c)_n}\frac{z^n}{n!}\mbox{,}
where (x)_n is the rising Pochhammer symbol, which is defined by
(x)_n = 1 \mbox{\ for\ } n = 0\mbox{, and}
(x)_n = x(x+1)\cdots(x+n-1) \mbox{\ for\ } n > 0\mbox{.}
## Usage
1 lmomlap(para)
## Arguments
para The parameters of the distribution.
## Value
An R list is returned.
lambdas Vector of the L-moments. First element is λ_1, second element is λ_2, and so on. ratios Vector of the L-moment ratios. Second element is τ, third element is τ_3 and so on. trim Level of symmetrical trimming used in the computation, which is 0. leftrim Level of left-tail trimming used in the computation, which is NULL. rightrim Level of right-tail trimming used in the computation, which is NULL. source An attribute identifying the computational source of the L-moments: “lmomlap”.
W.H. Asquith
## References
Hosking, J.R.M., 1986, The theory of probability weighted moments: IBM Research Report RC12210, T.J. Watson Research Center, Yorktown Heights, New York.
parlap, cdflap, pdflap, qualap
1 2 3 lmr <- lmoms(c(123,34,4,654,37,78)) lmr lmomlap(parlap(lmr))
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# Understanding proper distance in Schwarzschild solution.
1. Dec 21, 2011
### peter46464
I'm trying to understand the Schwarzschild solution concept of proper distance. Given the proper distance equation
$$d\sigma=\frac{dr}{\left(1-\frac{R_{S}}{r}\right)^{1/2}}$$
how would I calculate the coordinate distance. For example - assuming the distance from the Earth to the Sun is 150,000,000km, is it a valid question to ask what the coordinate distance is, and how would I calculate it?
I know $R_{S}$ is about 3km.
Many thanks
2. Dec 21, 2011
### A.T.
If you use Schwarzschild coordinates, then the radial coordinate distance is simply the difference in the radial coordinate "r" between two points: |rA-rB|
To get the radial proper distance you have to integrate your formula between rA and rB.
3. Dec 21, 2011
### peter46464
Thanks. For me, integrating that looks hard. Can I make any simplifying approximations?
4. Dec 21, 2011
### Staff: Mentor
That general form of integral actually appears in standard tables of integrals, so you can find an exact antiderivative for it (you may have to modify the form of the integrand somewhat to fit it into a standard form). Try here for one such table:
http://integral-table.com/
Or, particularly if r is very large compared to R_s (which it certainly is in your case), you can expand the binomial in the denominator in a power series, as here:
http://hyperphysics.phy-astr.gsu.edu/hbase/alg3.html
You should only need the first couple of terms to see how things will go for the case of R_s / r very small.
5. Dec 21, 2011
### Passionflower
It is very easy to solve it with a math program such as Mable, Mathematica or Matlab. But effectively R = rho as the Sun's mass is not large enough to be considered a strong field. You have to go many numbers behind the decimal point to find a discrepancy.
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## A community for students. Sign up today
Here's the question you clicked on:
## JordanDeja 2 years ago 11 : x = 5 to 1 , what is 'x' ?
• This Question is Closed
1. ghazi
$\frac{ 11 }{ x }=\frac{ 5 }{ 1 }$ use cross multiplication $x=\frac{ 11*1 }{ 5 }$
2. JordanDeja
so then x would = 2.2 ?
3. ghazi
exactly
4. JordanDeja
thanks :D
5. ghazi
:D
6. JordanDeja
I have a few more :o
7. ghazi
keep posting, in a new thread i guess this is what OS says lol
#### Ask your own question
Sign Up
Find more explanations on OpenStudy
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# System for producing shortest form of natural numbers
Wondering what the way of producing the shortest length string form of the natural numbers $$\mathbb{N}$$. I am thinking about this in terms of a binary representation (base 2 representation), but I think I could figure that out starting from a standard base 10 representation. It seems like this has something to do with polynomial equations, so I started with that. (Not totally sure I'm doing it right)....
$$1 = 0^1$$
$$2 = 2^1$$
$$3 = 3^1$$
$$4 = 2^2 = 4^1$$
$$10000 = 10^4 = 100^2 = 10000^1$$
$$1000000 = 10^6$$
At larger numbers it becomes apparent that you can represent them with smaller "strings" than the original number. So 1000000 is 7 digits, but $$10^6$$ is only 3. If the number was large like $$10^{50}$$ then it would be $$10000000000000000000000000000000000000000000000000000...$$, so there is a big gain.
Some other thoughts...
For even larger numbers you can perhaps start nesting the exponents, so $$10^{50^{50}}$$, and gain an even larger advantage.
Basically I would like to know if there are any equations or systems for figuring out the ideal "smallest" representation of the number as a polynomial equation, where the number could be represented as any base. So $$1000000$$ is better represented as the more compact $$10^6$$ since it's only 3 digits instead of 7. But I don't see any equations here on how to figure this out (yet).
• If there is such an algorithm then it implies there is a way to quickly factor the product of two large primes which is NP-hard. Therefore this problem is at least NP hard. – cr001 Feb 12 at 8:49
• Didn't you ask it about a week ago, and someone rightfully stated that you still need at least $\log n$ bits to represent numbers from $1$ to $n$? Which leads to conclusion, that some of the numbers use more bits to be represented, than they do now in any system, that succeeds in saving bits for representing $10^{50^{30}}$. – dEmigOd Feb 12 at 10:33
• Google "Kolmogerov-complexity". There are infinite many numbers that cannot be expressed "way shorter" than by simply printing it. But there is no algorithm to determine the complexity in general because this would imply that the halting problem is solveable which is not the case. – Peter Feb 12 at 13:35
• If you choose , lets say, a $10^9$-digit number randomly, the probability that it can be significantly compressed (lets say, at $90$% of the original length) is almost $0$. – Peter Feb 12 at 13:39
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/* Copyright (c) 2002, Joerg Wunsch
Redistribution and use in source and binary forms, with or without
modification, are permitted provided that the following conditions are met:
* Redistributions of source code must retain the above copyright
notice, this list of conditions and the following disclaimer.
* Redistributions in binary form must reproduce the above copyright
notice, this list of conditions and the following disclaimer in
the documentation and/or other materials provided with the
distribution.
* Neither the name of the copyright holders nor the names of
contributors may be used to endorse or promote products derived
from this software without specific prior written permission.
THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS"
AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE
LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR
CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF
SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS
INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN
CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE)
ARISING IN ANY WAY OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE
POSSIBILITY OF SUCH DAMAGE.
*/
/* $Id$ */
#include <stdio.h>
#include "sectionname.h"
#include "stdio_private.h"
ATTRIBUTE_CLIB_SECTION
size_t
fread(void *ptr, size_t size, size_t nmemb, FILE *stream)
{
size_t i, j;
uint8_t *cp;
int c;
if ((stream->flags & __SRD) == 0)
return 0;
for (i = 0, cp = (uint8_t *)ptr; i < nmemb; i++)
for (j = 0; j < size; j++) {
c = getc(stream);
if (c == EOF)
return i;
*cp++ = (uint8_t)c;
}
return i;
}
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Using Macros in MiniLatex
User defined macros in math mode are totally legit in MiniLatex. One way to insert them is like this:
That is, enclose the definitions in double dollar signs. Do this in the body of the text. You can now use these macros in the usual way, e.g. $\bra a | b \ket$.
If you are using www.knode.io , there is another way. Make a plain text document titled, say “TeX Macros.” The actual title is irrelevant. Put the macro definitions in this document. Take not of the document ID number (it is displayed in the footer). Let’s suppose that the ID is 453. Then, in the keywords field of the document that is to use the macros, put the text “texmacros:453”, as in the figure below. That’s all there is to it!
You can try this out using the Demo App.
Bibliographies
Today I posted changes to MiniLatex that allow users to create bibliographies in the customary way using \cite in the body of the document and \bibitem for entries in the bibliography. Whenever the user presses the full-render button, the citations are resolved in the text as live links in the HTML rendered on-screen. Since the syntax used is standard LaTeX, running an exported document through pdflatex also resolves the bibliography.
Here is a full example:
\begin{tableofcontents}
\bibitem[A64]{abraham64} Abraham, Hank. Lives of Famous Hackers.
Caveman Press 1964, pp 1001.
\bibitem[U79]{dave} Ungar, Dave. Principles of Fermentation,
Better Bread Publishing, 1979, pp. 32.
\end{tableofcontents}
In the body of the article, one would refer to the Ungar article as
as noted in \cite{dave}, fermentation causes ...
When rendered, the bibliography would look like the below:
[A64] Abraham, Hank. Lives of Famous Hackers.
Caveman Press 1964, pp 1001.
[U79] Ungar, Dave. Principles of Fermentation,
Better Bread Publishing, 1979, pp. 32.
The citation would be an active link as in the text “as noted in [U79], fermentation …”
Please see the MiniLatex Demo App for an example of the bibliography in action.
Building MiniLatex
The goal of the MiniLatex project is to put a defined subset of LaTeX in the browser. This means being able to “live edit” LaTeX in a web app and to immediately see the result rendered as web page — well, if not immediately, at least very, very quickly! Another goal is to never, ever have to rewrite text. Documents written for the web in MiniLatex should be exportable at the click of a button to a file on the author’s computer that can be typeset using pdlatex or some other standard LaTeX tool.
Both of these goals are attainable. Here is a proof of concept:
MiniLatex Demo
Progress
This blog will report on progress on the development of MiniLatex and provide a forum for comments and discussion. Some articles will be technical, even quite technical, while others are intended for the general reader who may use MiniLatex as a writing tool. We are still in the research and development phase, so your feedback is both welcome and essential.
Articles
Here are some articles of interest.
And here are examples of documents written in MiniLatex
The examples above were written using the web app www.knode.io. It hosts a MiniLatex rendering engine written in Elm.
On the Horizon
• MiniLatex, as a defined subset of LaTeX, is an entity independent of whatever engine is used to transform MiniLatex source text into HTML. At the moment, there is just one engine, implemented in Elm. I plan to write a Haskell engine at some point.
• Once Elm 0.19 is released, MiniLatex will run much, much faster.
• When MathJax 3.0 is released, it will be integrated into Elm. This will result in an estimated 10-fold increase in processing speed for the MathJax part of the processing pipeline.
Good company in a journey makes the way seem shorter. — Izaak Walton
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# Language of Turing machines that never visit some given state
Can someone help me to determine and prove if the following language is decidable or not?
I tried to think on some reductions but I can't figure it out...
$$A=\{\langle M\rangle|\text{M is TM and there is a state q in M that M never visits}\}$$
my try:
I built mapping reduce $E_{TM}\le_{m} B$:
$<M>\rightarrow <M>$
so I need to prove: $$<M>\in E_{TM}\iff <M>\in B$$
First direction:
$<M>\in E_{TM}\rightarrow L(M)=\emptyset \rightarrow M \text{ never visits } q_{accept} \rightarrow <M>\in B$
The problem is in the second direction. How can I promiss that if the language is empty $M$ visits all the states?
Thanks!
• Hint: $q$ could be the accepting state. Translate the problem and see if you can apply Rice's Theorem. – A.Schulz May 7 '16 at 18:52
• Or a direct reduction from the halting problem. – David Richerby May 7 '16 at 19:43
• @A.Schulz are you trying to say that I can see the language as $E_{TM}=\{\text{$M$is$TM$and$L(M) =\emptyset$}\}$? If so, I don't know how to write it correctly... It is confusing me. – user9 May 7 '16 at 20:09
• @user9: I wanted to say, if you can deicide $A$ then you can decide $E_\text{TM}$. (Not quite sure how $q$ is quantified in your question.) – A.Schulz May 8 '16 at 8:17
• @A.Schulz I edited my question, Can you help me in the second direction? – user9 May 8 '16 at 17:50
I will assume that you're using $E_{TM}$ to mean the language $HALT=\{\langle M\rangle|\text{$M$does not halt}\}$ and $B=A=\{\langle M\rangle|\text{$M$is$TM$and there is a state$q$in$M$that$M$never visits}\}$. Also, you should state what your mapping is because if it is just the identity (which it seems to be here) then this won't work. This is because we can have a turing machine that halts right away but also happens to have an extra state that it never reaches, so it is not the case that $B\subset E_{TM}$.
Instead, it may help to think about this more informally. If have the ability to tell whether or not a turing machine visits all of its states, how might we use this superpower to decide the halting problem? The simplest, most direct approach is to see if we can answer the question of whether $M$ visits a specific state $q\in Q$ using knowledge of $B$. Then we can reduce to the halting problem by simply asking if a given TM $M$ visits any of its accepting states.
To do this, let's see what "subsets" of $M$ are still in $B$. By this I mean that we will choose a subset $P\subset Q$ and then on any transition that goes from a state in $P$ to one in $Q\setminus P$ we will instead reroute it to a halting state. Let's call the new machine with the new accepting state along with $P$ as its states as $N_P$.
Now, we start off by looking asking if $N_P\in B$ for $P$ being the set containing just the starting state. If this TM is in $B$ then we know $M$ never visits any state other than its initial state. If it is not in $B$ then we continue by adding a new state to $P$ and asking if the new TM is in $B$, continuing until we get to a TM which is in $B$. The $P$ we end on will have to be the set of states the TM visits, and all other states are unvisited. Thus, we can tell whether a $M$ visits a certain state $Q$ by simply checking if $q$ is in the final version of $P$.
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Definition:Complex Number/Imaginary Unit
(Redirected from Definition:Imaginary Unit)
Definition
Let $\C = \set {a + b i: a, b \in \R}$ be the set of complex numbers.
The entity $i := 0 + 1 i$ is known as the imaginary unit.
Also denoted as
Using the formal definition of complex numbers it is the ordered pair $\tuple {0, 1}$.
The non-italicized $\mathrm i$ can also be seen.
Historical Note
The symbol $i$ that is in widespread use for the imaginary unit was due to Leonhard Paul Euler's influence.
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## Differential and Integral Equations
### Variational models for prestrained plates with Monge-Ampère constraint
#### Abstract
We derive a new variational model in the description of prestrained elastic thin films. The model consists of minimizing a biharmonic energy of the out-of plane displacements $v\in W^{2,2}(\Omega, \mathbb{R})$, satisfying the Monge-Ampèere constraint: $$\det\nabla^2v = f .$$ Here, $f=-\mbox{curl}^T\mbox{curl} (S_g)_{2\times 2}$ is the linearized Gauss curvature of the incompatibility (prestrain) family of Riemannian metrics $G^h= \mbox{Id}_3 + 2 h^\gamma S_g+ h.o.t.$, imposed on the referential configurations of the thin films with midplate $\Omega$ and small thickness $h$. We further discuss multiplicity properties of the minimizers of this model in some special cases.
#### Article information
Source
Differential Integral Equations, Volume 28, Number 9/10 (2015), 861-898.
Dates
First available in Project Euclid: 23 June 2015
https://projecteuclid.org/euclid.die/1435064543
Mathematical Reviews number (MathSciNet)
MR3360723
Zentralblatt MATH identifier
1363.74063
Subjects
Primary: 74K20: Plates 74B20: Nonlinear elasticity
#### Citation
Lewicka, Marta; Ochoa, Pablo; Pakzad, Mohammad Reza. Variational models for prestrained plates with Monge-Ampère constraint. Differential Integral Equations 28 (2015), no. 9/10, 861--898. https://projecteuclid.org/euclid.die/1435064543
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# Why does ImportString[“1c”, “Table”] eat the letter c?
I have some whitespace-separated matrix data that I read with Import[..., "Table"]. The data contained mixed strings and numbers (the strings are for row and column names).
I noticed this weird behaviour:
ImportString["123c", "Table"]
(* ==> {{123}} *)
Mathematica ate the letter c!! Why?
It doesn't eat any other letters:
ImportString["123a", "Table"]
(* {{"123a"}} *)
ImportString["123e", "Table"]
(* {{"123e"}} *)
What is the explanation and what is a good workaround?
Update:
It seems that this happens even if the labels is quoted in the file:
ImportString["\"24c\"", "CSV"]
(* ==> {{24}} *)
• It's either a bug or it's interpreting 123c as some non-string datatype, I just can't figure out what. – Szabolcs Jul 18 '14 at 16:50
• This is because it is SMILES :). Try this StringFormat["123c"] !Mathematica graphics – Nasser Jul 18 '14 at 17:01
• Also notice what it says in help if the format specification is not given: "attempts to determine the format of the string from its contents." the keyword is attempt. I do not think this is a good way to make functions. This is all fuzzy type programming. What does "Attempt" actually mean? How do I grade this attempt? A grade? B grade? 85% attempt? A function should be clear. It should tell exactly what is the input and what is the output. – Nasser Jul 18 '14 at 17:17
• @Nasser The reason why I consider this very bad is that I didn't even notice the incorrect import until I started doing some consistency checks on the data. Only after that did I discover that this lead to incorrect results (I had both labels "24" and "24c", and they needed to be distinguishable). Most people wouldn't suspect that importing a simple CSV or similar file will change their data. It's completely unexpected, and thus difficult to discover, and will easily lead to ruined work. "CurrencyTokens" -> None should be the default. – Szabolcs Jul 18 '14 at 18:40
Mathematica is interpreting c as a currency marker. This is controlled by the "CurrencyTokens" import option for "Table".
The default setting for "CurrencyTokens" is
{{"\$ ", "£", "¥", "€"}, {"c ", "¢", "p ", "F "}}
so this also happens with the letters p or F.
The workaround is
ImportString["123c", "Table", "CurrencyTokens" -> None]
(* {{"123c"}} *)
Notice: The same applies to the "CSV", "TSV" and "List" import formats.
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## FANDOM
6 Pages
$\Sigma(13)>f_{\omega}(2046)$ (Wythagoras 2016)
## Historical table of boundsEdit
Bound Discoverer Year and month
$f_{\omega}(2046)$ Wythagoras September 2016
## Analysis of the current record holderEdit
### Ackermannian growth using 8 statesEdit
0 1 1 r 0
0 _ 1 r 1
1 _ _ r halt
1 1 _ r 2
2 _ _ l 6
2 1 1 l 7
3 1 1 l 3
3 _ _ r 4
4 1 1 r 5
5 1 _ r 0
6 _ 1 l 3
7 1 1 l 7
7 _ _ l 8
8 1 1 l 7
8 _ 1 r 9
9 _ 1 r 0
Note that the following half-states can be paired: (4+6), (5+9).
This machine executes a function $f(a_1,a_2,a_3,\cdots,a_n)$ of $n$ variables, defined as
• $f(a_1,a_2,a_3,\cdots,a_n)=f(a_1+3,a_2-1,a_3,\cdots,a_n)$ if $a_2>1$.
• $f(a_1,\underbrace{1,1,\cdots,1,1}_{k},1,a_k,\cdots,a_n)=f(3,\underbrace{1,1,\cdots,1,1}_{k},a_1+1,a_k-1,\cdots,a_n)$ if $a_k>1$.
• $f(a_1,\underbrace{1,1,\cdots,1,1}_{k})=a_1+k+1$.
We have $f(3,a_2)=f(3a_2,1)$, $f(3,a_2,2)=f(3a_2,1,2)=f(3,3a_2+1,1)>f(3,3a_2,1)=f(9a_2,1,1)$ and in general $f(3,a_2,a_3)>f(3^{a_3}a_2,1,1)$. $f(3,1,a_3)>f(3^{a_3},1,1)$. This gives $f(3,1,1,a)>f(3\uparrow\uparrow a,1,1,1)$, and in general, by induction $f(3,\underbrace{1,1,\cdots,1,1}_{b},a)>f(3\uparrow^b a,\underbrace{1,1,\cdots,1,1}_{b+1})>3\uparrow^b a$
### Input using 5 statesEdit
There exists a 5 state TM with the following output[1]:
11 (011)2047 1
It halts with the head on the second one. Instead of halting, we let it go to state 0 of the Ackermannian growth machine. This amounts to simulating $f(2,\underbrace{2,2,\cdots,2,2}_{2046},3) > f(3,\underbrace{1,1,\cdots,1,1}_{2046},3) > 3 \uparrow^{2046} 3 > f_{\omega}(2046)$
## ReferencesEdit
1. H. Marxen, 5-state TM #3 from MaBu-List
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Minimal enclosing parallelepiped in 3D
1 REMAP - Regularity and massive parallel computing
Inria Grenoble - Rhône-Alpes, LIP - Laboratoire de l'Informatique du Parallélisme
Abstract : We investigate the problem of finding a minimal volume parallelepiped enclosing a given set of n three-dimensional points. We give two mathematical properties of these parallelepipeds, from which we derive two algorithms of theoretical complexity O(n^6). Experiments show that in practice our quickest algorithm runs in O(n^2) (at least for $n \leq 10^5$n 10^5). We also present our application in structural biology.
Keywords :
Document type :
Reports
Domain :
Complete list of metadata
https://hal.inria.fr/inria-00071901
Contributor : Rapport De Recherche Inria Connect in order to contact the contributor
Submitted on : Tuesday, May 23, 2006 - 7:12:29 PM
Last modification on : Wednesday, March 2, 2022 - 1:28:06 PM
Identifiers
• HAL Id : inria-00071901, version 1
Citation
Frédéric Vivien, Nicolas Wicker. Minimal enclosing parallelepiped in 3D. [Research Report] RR-4685, LIP RR-2002-49, INRIA, LIP. 2002. ⟨inria-00071901⟩
Record views
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## anonymous 5 years ago can anyone help me get started on this? dh/dt= 0.016(2.5-h) Solve the differential equation to find h in terms of t.
1. anonymous
dh/(2.5-h)=0.016 dt now integrating, -log(2.5-h)=0.016 t+c
2. anonymous
Is this enough, ersp?
3. anonymous
i think so my brains kinda died doing this work XD thanks :)
4. anonymous
ok :)
5. anonymous
actually i dont get where the minus log comes from when integrating
6. anonymous
Comes from the -h...
7. anonymous
$\int\limits_{}{}\frac{dh}{2.5-h}=-\int\limits_{}{}\frac{(2.5-h)'dh}{2.5-h}=-\log (2.5-h)+c$
8. anonymous
I got bumped. Is this explanation okay?
9. anonymous
i think so cheers
10. anonymous
so what method is this?
11. anonymous
Method? To solve the differential equation?
12. anonymous
The d.e. is separable.
13. anonymous
Or, you can just take the reciprocal of both sides and integrate directly, but it's the same thing.
14. anonymous
thanks makes a bit more sense to me now, my mind is really failing me today!
15. anonymous
np ;)
16. anonymous
shouldnt both sides be logged?
17. anonymous
No. It goes like this:$\frac{dh}{dt}=0.016(2.5-h) \rightarrow \frac{dh}{2.5-h}=0.016dt \rightarrow \int\limits_{}{}\frac{dh}{2.5-h}=\int\limits_{}{}0.016dt$$\rightarrow -\log (2.5-h) = 0.016t + c \rightarrow \log (2.5-h)=-0.016t+c_1 \rightarrow 2.5-h=c_2e^{-0.016t}$$\rightarrow h=2.5-ce^{-0.016t}$
18. anonymous
I added subscripts as I went along, but we just lump constants into constants, so normally, the subscripts are left off...which is what I did in the end.
19. anonymous
ah that makes sense thanks very much, this work ive been throwing my head in to a wall for the last few days!
20. anonymous
lol, it happens :p
21. anonymous
Thanks for fanning me, ersp :D
22. anonymous
after that much help it seems logical! XD
23. anonymous
ah it doesnt seem to fit for the next part of the question, at max h is 2.5, how long does it take to get to this point, so i have to rearrange it to make t the subject but the values dont seem to fit at all!
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# Have they solved this exercise correct?(banach space, function space).
Please look at this exercise: It is the last question I have a problem with
Here is the solution:
They say that $\|I(1)\|=\sup|g(t)|$. But isn't $\|I(1)\|=\int_0^1g(t)dt$? If so, what is the correct answer?
• Do you realize that it's difficult to read the images? – user21820 May 29 '14 at 7:38
• @user21820 I understand, I have no problem reading it, I really don't understand why it may be viewed differently for you. Here are the links directly to the images:, can you see them clear here: ? image 1: s27.postimg.org/y8d4c57wz/image.png , image 2: s1.postimg.org/am6ii8anz/image.png – user119615 May 29 '14 at 7:41
• @user119615: The image is too small to read even in the link. Can you somehow made it bigger? – user99914 May 29 '14 at 8:10
• @John Does it work if you open them as .jpg? Try these links: s27.postimg.org/y8d4c57wz/image.jpg and s1.postimg.org/am6ii8anz/image.jpg , if that works maybe it is a problem with the .png images – user119615 May 29 '14 at 8:22
You are right. Since $g(t) \geq 0$, it follows that $I(1)$ is a non-decreasing function on $[0,1]$, and hence, $||I(1)|| = (I(1))(1) =\int_0^1 g(t) dt$. In fact, one can use the estimate $$\int_{0}^1 |f(t)g(t)|\, dt \leq \int_0^1 \left(\sup|f(t)|\right)|g(t)|\, dt = ||f||\int_0^1 |g(t)|\, dt$$ to show that one can take $M = \int_0^1 |g(t)|\, dt$ for all $g \in C$, which is better than $M = \sup |g(t)|$.
• Thanks, and it is also the fact that $\int_0^1g(t)dt$, is the least of all possible M's you can chose, making that the operator-norm? – user119615 May 29 '14 at 8:37
• Yes. For $g$ positive, we have $\int_0^1 g(t)\, dt = ||I(1)|| \leq ||I||\cdot||1|| = ||I||$. On the other hand, the inequality in my answer shows that for all $g \in C$, we have $||I|| \leq \int_0^1 |g(t)|\,dt$. – ivanpenev May 29 '14 at 9:45
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- Current Issue - Ahead of Print Articles - All Issues - Search - Open Access - Information for Authors - Downloads - Guideline - Regulations ㆍPaper Submission ㆍPaper Reviewing ㆍPublication and Distribution - Code of Ethics - For Authors ㆍOnlilne Submission ㆍMy Manuscript - For Reviewers - For Editors
Boehmians on the torus Bull. Korean Math. Soc. 2006 Vol. 43, No. 4, 831-839 Printed December 1, 2006 Dennis Nemzer California State University Abstract : By relaxing the requirements for a sequence of functions to be a delta sequence, a space of Boehmians on the torus $\beta(T^d)$ is constructed and studied. The space $\beta(T^d)$ contains the space of distributions as well as the space of hyperfunctions on the torus. The Fourier transform is a continuous mapping from $\beta(T^d)$ onto a subspace of Schwartz distributions. The range of the Fourier transform is characterized. A necessary and sufficient condition for a sequence of Boehmians to converge is that the corresponding sequence of Fourier transforms converges in $\mathcal{D}'(\RR^d)$. Keywords : Boehmian, Fourier transform, distribution MSC numbers : 44A40, 46F12, 42B05 Downloads: Full-text PDF
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# ASP.NET Core 2.2 and Angular template with monorepo
The template’s default structure has everything in one place, like this:
/
bin/
obj/
ClientApp/
myproject.csproj
Startup.cs
etc.
My structure has many libraries and angular apps – i.e. a monorepo – so it must be more organized:
/
libs
client
client2
server
bin/
obj/
myproject.csproj
Startup.cs
lib1
lib2
I edited various references to reflect this structure, most importantly Startup.cs:
services.AddSpaStaticFiles(configuration => {
configuration.RootPath = "../../../../client/dist";
});
// and
app.UseSpa(spa => {
spa.Options.SourcePath = "../../../../client";
});
But when running I get: InvalidOperationException: Failed to start 'npm'.
When I run the server and client separately, they work… so the problem is with how the “spa services” is configured. I tried both ../../../../client (from bin directory) and ../client (from server project’s base directory).
How do I reconfigure the project structure? (Is there a working sample repo somewhere?)
Source: AngularJS
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AGING WORKFORCE What changes in employment relationships are likely to occur as the population ages
# AGING WORKFORCE What changes in employment relationships are likely to occur as the population ages
S
Asked by 2 years ago
0 points
What changes in employment relationships are likely to occur as the population ages
Do you think increasing age diversity will create new challenges for managers? What types of challenges do you expect will be most profound?
What types of policies might lead to charges of age discrimination, and how can they be changed to eliminate these problems?
Discuss some other challenges organizations often encounter while trying to build a diverse workforce.
Describe some of the actions an organization should take to overcome diversity challenges, explaining why those actionswould be helpful. Provide examples to support your conclusions.
What are some of the important benefits of having a diverse workf
America Aging Work Force
As the U.S. population is becoming gradually diverse, so its'...
AGING WORKFORCE
solarc
### 1 Answer
S
Answered by 2 years ago
0 points
#### Oh Snap! This Answer is Locked
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### Your Answer
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Views: 20
Asked: 2 years ago
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# Notebook Entry
uh, trying to use IPopt in Python.
# Ipopt installation
I tried to install from the Ubuntu repos, but had trouble linking to it or running the example:
$sudo aptitude install coinor-libipopt1 coinor-libipopt-dev coinor-libipopt-doc So then I removed it: $ sudo aptitude remove coinor-libipopt1 coinor-libipopt-dev coinor-libipopt-doc
The Ipopt documentation gives very nice instructions for installing it from source, which basically goes like this:
$cd ~/src$ svn co https://projects.coin-or.org/svn/Ipopt/stable/3.11 CoinIpopt
$cd CoinIpopt/ThirdParty/Blas$ ./get.Blas
$cd ../Lapack$ ./get.Lapack
$cd ../ASL$ ./get.ASL
$cd ../Mumps$ ./get.Mumps
$cd ../Metis$ ./get.Metis
This downloads the BLAS and LAPACK reference implementations, but I think the compilation process looks in the system for the system installed implementations and uses them if you have them.
It is a good idea to get the HSL code (but not necessarily required because Mumps can be used) and drop it in the ThirdParty/HSL directory. You have to get the free academic license 2011 code and sign a form to use it. It also takes a day to get the download link by email. (Note that this can be linked as a shared lib after compiling ipopt, so recompiling is required, but it may be easier to just recompile).
First change into the build directory:
$cd ../../build And run configure: $ ./configure
These are some potential options I may want to set in the future:
--prefix /usr/local # for system install
--with-blas="-L$HOME/lib -lmyblas" # link to better blas implementations Now compile, test, and install: $ make -j5
$make test$ make install # sudo if system install
This ended up putting everything in ~/src/CoinIpopt/ like ~/src/CoinIpopt/include, ~/src/CoinIpopt/lib, etc. So I did something wrong, as I thought it should have ended up in ~src/CoinIpopt/build/.
# cyipopt
I'm hoping to use Ipopt through a Python wrapper. Seems like there are two existing standalone wrappers pyipopt and cyipopt. I like cython and cyipopt was newer, so I gave it a shot.
I first created a conda Python 2.7 environment:
$conda create -n cyipopt numpy scipy cython matplotlib sphinx$ source activate cyipopt
Then got the source code from bitbucket:
$cd ~/src$ hg clone https://bitbucket.org/amitibo/cyipopt
$cd cyipopt I installed on Ubuntu 14.04 so I had to do some pruning in the setup.py file. Basically just remove some odd stuff from main_unix() so it looks like this: def main_unix(): setup( name=PACKAGE_NAME, version=VERSION, description=DESCRIPTION, author=AUTHOR, author_email=EMAIL, url=URL, packages=[PACKAGE_NAME], cmdclass={'build_ext': build_ext}, ext_modules=[ Extension( PACKAGE_NAME + '.' + 'cyipopt', ['src/cyipopt.pyx'], **pkgconfig('ipopt') ) ], ) Since I didn't install IPopt system wide I needed to export these two environment variables to get things to work: $ export PKG_CONFIG_PATH=$PKG_CONFIG_PATH:~/src/CoinIpopt/lib/pkgconfig$ export LD_LIBRARY_PATH=$LD_LIBRARY_PATH:~/src/CoinIpopt/lib And finally: $ python setup.py install
$python test/examplehs071.py$ python test/lasso.py
And it worked.
# Other Things
Here are a bunch of other notes about things I found today:
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Volume 7, Issue 2
Generalized and Unified Families of Interpolating Subdivision Schemes
Numer. Math. Theor. Meth. Appl., 7 (2014), pp. 193-213.
Published online: 2014-07
Cited by
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• Abstract
We present generalized and unified families of $(2n)$-point and $(2n-1)$-point $p$-ary interpolating subdivision schemes originated from Lagrange polynomial for any integers $n ≥ 2$ and $p ≥ 3$. Almost all existing even-point and odd-point interpolating schemes of lower and higher arity belong to this family of schemes. We also present tensor product version of generalized and unified families of schemes. Moreover, error bounds between limit curves and control polygons of schemes are also calculated. It has been observed that error bounds decrease when complexity of the scheme decrease and vice versa. Furthermore, error bounds decrease with the increase of arity of the schemes. We also observe that in general the continuity of interpolating scheme do not increase by increasing complexity and arity of the scheme.
65D17, 65D07, 65D05
• BibTex
• RIS
• TXT
@Article{NMTMA-7-193, author = {}, title = {Generalized and Unified Families of Interpolating Subdivision Schemes}, journal = {Numerical Mathematics: Theory, Methods and Applications}, year = {2014}, volume = {7}, number = {2}, pages = {193--213}, abstract = {
We present generalized and unified families of $(2n)$-point and $(2n-1)$-point $p$-ary interpolating subdivision schemes originated from Lagrange polynomial for any integers $n ≥ 2$ and $p ≥ 3$. Almost all existing even-point and odd-point interpolating schemes of lower and higher arity belong to this family of schemes. We also present tensor product version of generalized and unified families of schemes. Moreover, error bounds between limit curves and control polygons of schemes are also calculated. It has been observed that error bounds decrease when complexity of the scheme decrease and vice versa. Furthermore, error bounds decrease with the increase of arity of the schemes. We also observe that in general the continuity of interpolating scheme do not increase by increasing complexity and arity of the scheme.
}, issn = {2079-7338}, doi = {https://doi.org/10.4208/nmtma.2014.1313nm}, url = {http://global-sci.org/intro/article_detail/nmtma/5871.html} }
TY - JOUR T1 - Generalized and Unified Families of Interpolating Subdivision Schemes JO - Numerical Mathematics: Theory, Methods and Applications VL - 2 SP - 193 EP - 213 PY - 2014 DA - 2014/07 SN - 7 DO - http://doi.org/10.4208/nmtma.2014.1313nm UR - https://global-sci.org/intro/article_detail/nmtma/5871.html KW - Interpolating subdivision scheme, even-ary schemes, odd-ary schemes, Lagrange polynomial, parameters, error bounds, tensor product. AB -
We present generalized and unified families of $(2n)$-point and $(2n-1)$-point $p$-ary interpolating subdivision schemes originated from Lagrange polynomial for any integers $n ≥ 2$ and $p ≥ 3$. Almost all existing even-point and odd-point interpolating schemes of lower and higher arity belong to this family of schemes. We also present tensor product version of generalized and unified families of schemes. Moreover, error bounds between limit curves and control polygons of schemes are also calculated. It has been observed that error bounds decrease when complexity of the scheme decrease and vice versa. Furthermore, error bounds decrease with the increase of arity of the schemes. We also observe that in general the continuity of interpolating scheme do not increase by increasing complexity and arity of the scheme.
Ghulam Mustafa, Pakeeza Ashraf & Jiansong Deng. (2020). Generalized and Unified Families of Interpolating Subdivision Schemes. Numerical Mathematics: Theory, Methods and Applications. 7 (2). 193-213. doi:10.4208/nmtma.2014.1313nm
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Plasmaphysik Seminar
# Collective Laser Ion Acceleration by the Use of Mass Limited Near Critical Plasmas at the PHELIX Laser Facility
## by Peter Hilz (HI Jena)
Tuesday, April 17, 2018 from to (Europe/Berlin)
at GSI Darmstadt ( KBW Hörsaal )
Description Laser plasma based acceleration mechanisms are investigated by a rapidly growing scientific community. This talk will present experimental results on proton acceleration obtained at the Phelix laser by the use of mass limited targets. Hereby the Phelix laser was used to irradiate single levitated spheres with sub focus diameter (1µm). The particles were positioned and stabilized into the interaction region by the use of a Paul Trap. Prior to the main pulse the target already expands to near/subcritical densities. The acceleration dynamics in such a system lead to mono energetic proton beams with a mean kinetic energy of 25-30 MeV and a few MeV bandwidth. The experimental results were reproduced in a 3D PIC simulation. The simulation further allowed to identify the underlying processes and indicates possibilities to optimize this process in the near future.
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# Quotients of group actions on varieties
Let $Y$ be an affine algebraic variety over $\mathbb{C}$ and let $X$ be its closed subvariety. Let $G$ be a reductive group acting on $Y$ and let $H$ be a reductive subgroup of $G$ preserving $X$ such that the induced map $\phi: X//H \to Y//G$ is $1$-$1$ and a finite map. Question: Is $\phi$ a closed immersion?
Rmk. If the image of $\phi$ is normal then one can prove that the answer is affirmative.
-
Let's assume that X is normal, if that helps. – Adam S Sikora Jun 21 '12 at 19:13
Here is a counterexample. Let $Y = \mathbb A^2$, and let $G$ be a cyclic group of order $2$; a generator acts via $(x, y) \mapsto (-x, y)$. We take as $H$ the trivial subgroup, and as $X$ the curve $x^3= y$.
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697views
A cut has to be made 12m deep, inclined at an angle of 35$^{\circ}$ to the horizontal.
The possible circular failure surface has a radius equal to 20.2m, and is passing through the toe of cut slope and through the point 4m away on the top ground from the edge of cut. The C.G of the failure mass is at a distance of 9.4m from the center of the failure circle, The properties of soil are C=30kN/m$^{2}$, $\phi =15°$ and $\gamma =20kN/m^{2}$. Find the factor of safety that would be available on failure surface by friction circle method.
Being watched by a moderator
I'll actively watch this post and tag someone who might know the answer.
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# How do you solve y = -2x^2 + 5x - 1?
May 11, 2018
The roots are:
$\left(\frac{5 + \sqrt{17}}{4} , 0\right)$ and $\left(\frac{5 - \sqrt{17}}{4} , 0\right)$
The approximate roots are:
$\left(0.2192 , 0\right)$ and $\left(2.281 , 0\right)$
#### Explanation:
Solve:
$y = - 2 {x}^{2} + 5 x - 1$ is a quadratic equation in standard form:
$y = a {x}^{2} + b x + c$,
where:
$a = - 2$, $b = 5$, $c = - 1$
Substitute $0$ for $y$.
$0 = - 2 {x}^{2} + 5 x - 1$
Solve using the quadratic equation.
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 \cdot a}$
Plug in the known value.
$x = \frac{- 5 \pm \sqrt{{5}^{2} - 4 \cdot - 2 \cdot - 1}}{2 \cdot - 2}$
Simplify.
$x = \frac{- 5 \pm \sqrt{25 - 8}}{- 4}$
$x = \frac{- 5 \pm \sqrt{17}}{- 4}$
Simplify.
$x = \frac{5 \pm \sqrt{17}}{4}$
$x = \frac{5 + \sqrt{17}}{4} ,$ $\frac{5 - \sqrt{17}}{4}$
The roots are:
$\left(\frac{5 + \sqrt{17}}{4} , 0\right)$ and $\left(\frac{5 - \sqrt{17}}{4} , 0\right)$
The approximate roots are:
$\left(0.2192 , 0\right)$ and $\left(2.281 , 0\right)$
graph{y=-2x^2+5x-1 [-10, 10, -5, 5]}
May 11, 2018
$x = - 0.22 \text{ and } x = 2.28$ for $y = 0$
#### Explanation:
When we say "solution" instead of the entire curve, we usually mean the place(s) were the function is zero. That is, the "roots" of the equation are at:
$- 2 {x}^{2} + 5 x - 1 = 0$
Now you can solve this be number of ways - factoring, quadratic formula, and graphing.
This one may be solved most quickly with the quadratic formula.
x =( −b ± sqrt(b^2−4ac))/(2a)
in this case, $a = - 2 , b = 5 , c = - 1$
x =( −5 ± sqrt(5^2−4(-2)(-1)))/(2(-2))
x = (−5 ± sqrt(17))/(−4)
x = 5/4 ± sqrt(17)/4 ; x = 5/4 ± 1.03
$x = - 0.22 \text{ and } x = 2.28$
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NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
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Let $$z\left(t\right)=x\left(t\right)\ast y\left(t\right)$$, where "*" denotes convolution. Let C be a positive real-val...
GATE EE 2017 Set 1
Consider the system with following input-output relation $$y\left[n\right]=\left(1+\left(-1\right)^n\right)x\left[n\righ... GATE EE 2017 Set 1 The value of$$\int_{-\infty}^{+\infty}e^{-t}\partial\left(2t-2\right)dt$$. where$$\partial\left(t\right)$$is the Dira... GATE EE 2016 Set 1 A moving average function is given by$$y\left(t\right)=\frac1T\int_{t-T}^tu\left(\tau\right)d\tau$$. If the input u is ... GATE EE 2015 Set 1 For a periodic signal the$$v\left(t\right)=30\sin100t\;+\;10\cos300t\;+\;6\sin\left(500t\;+\;\frac{\mathrm\pi}4\right)...
GATE EE 2013
A point Z has been plotted in the complex plane, as shown in figure below. The plot of the complex number $$y=\frac1z... GATE EE 2011 The convolution of the function$$f_1\left(t\right)=e^{-2t}\;u\left(t\right)$$and$$f_2\left(t\right)=e^t\;u\left(t\rig...
GATE EE 1995
The value of the integral $$\int_{-5}^{+6}e^{-2t}\delta\left(t-1\right)dt$$ is equal to ________.
GATE EE 1994
## Marks 2
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The function shown in the figure can be represented as ...
GATE EE 2014 Set 1
If u(t), r(t) denote the unit step and unit ramp functions respectively and u(t)*r(t) their convolution, then the functi...
GATE EE 2007
### Joint Entrance Examination
JEE Main JEE Advanced WB JEE
### Graduate Aptitude Test in Engineering
GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN
NEET
Class 12
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• Gurobi Staff
Hi,
I understand that you want to set a feasible MIP solution using a callback.
Concerning
But can we force this feasible solution to be considered as the actual optimal solution at that node (instead of the LP solution), so that ObjBound could be improved accordingly ?
The branch-and-bound algorithm solves LP relaxations at the nodes, so I don't understand what you mean by forcing the feasible MIP solution to be considered as the actual optimal solution of a node.
Concerning the Objective Bound improvement, if the solution you provide is indeed better as the best known feasible solution at that stage, then the $$\texttt{Incumbent}$$ value will be updated accordingly. The $$\texttt{BestBd}$$ value is related to the LP-relaxations, and cannot be changed by providing feasible MIP solutions.
I hope this answer helps.
Elisabeth
Hi Elisabeth, thank you and indeed it was probably not clear, sorry. I try to explain below, but it is OK if there is no answer for now: it is more a wish for a feature, than a question.
Just like the Gurobi MILP solver allows for lazy constraints, I was asking if it could manage "lazy solutions" when the MILP model provided to the solver is just a relaxation of the problem we want to solve at end.
In my situation (but it may also arise in bilevel prog for example), I minimize a MINLP by solving its MILP relaxation with the Gurobi LP B&C. The original nonlinear constraints and the incumbent update are managed "by hand" within a callback at MIPSOL nodes.
A MINLP solution that is eventually obtained "by hand" at a MIPSOL node may differ (in the value of the fractional variables) from the solution of the LP relaxation at that node and its cost may also be higher. My wish would be to tell the solver that the MINLP solution is not only a feasible solution but also the actual optimal solution at that node, so that its higher value will be taken into account also in the LB (BestBd) update.
My workaround is to store the MINLP solution and eventually discard the node from the search (with a lazy cut) in order to push up the LB: the search will still be complete but I suspect that it confuses the pseudocost computation a lot. Thank you again, Sophie.
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We’re delighted to announce the introduction of LaTeX to Edublogs.
The plugin, released by Patrick Chia, allows you to simply post really really complex equations, like this one:
$latex i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right>$
Just turn it on in your ‘Plugins’ area and enjoy.
Provide some details for your blog
No stress, you can always change this later on.
.edublogs.org
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# zbMATH — the first resource for mathematics
##### Examples
Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.
##### Operators
a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
Controllability of linear stochastic systems in Hilbert spaces. (English) Zbl 1031.93032
Linear stochastic control systems defined in infinite-dimensional Hilbert spaces are considered. Three types of stochastic controllability are studied: approximate, complete, and $S$-controllability. The relationships between different types of controllability are explained using methods of stochastic differential equations and the theory of stochastic processes. Moreover, several conditions for these types of controllability are formulated and proved. As an application of the theoretical results, the controllability of the wave equation with a distributed control and disturbances described by a Wiener process is studied. The relationships to deterministic controllability are also discussed.
Finally, it should be stressed that similar stochastic controllability problems have been considered in the papers [J. Klamka and L. Socha, IEEE Trans. Autom. Control. 22, 880-881 (1997; Zbl 0363.93048)] and [N. I. Mahmudov and A. Denker, Int. J. Control 73, 144-151 (2000; Zbl 1031.93033)].
##### MSC:
93B05 Controllability 93E03 General theory of stochastic systems 93C25 Control systems in abstract spaces
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# How do you graph x – 2y = 6 by plotting points?
Jun 7, 2016
Choose two values for $y$ and compute the related $x$ values
#### Explanation:
First of all, you need to note that you have the equation of a line. In fact, a line is the only function you can plot knowing only two of its points (you simply need to connect them!)
So, we need two points which surely lie on the line.
To do so, we express the $x$ values as dependent on the $y$ ones:
$x - 2 y = 6 \setminus \implies x = 2 y + 6$
So let's choose, for example, $y = 0$ and $y = 1$. We deduce that, if $y = 0$, then $x = 2 \cdot 0 + 6 = 6$. So, a first point is $\left(6 , 0\right)$.
Similarly, if $y = 1$, then $x = 1 \cdot 2 + 6 = 8$, and there is our second point.
Now, simply draw the points $\left(6 , 0\right)$ and $\left(8 , 1\right)$, connect them, and there you go!
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1. Feb 4, 2013
### Artusartos
1. The problem statement, all variables and given/known data
Let X1, X2, …, Xn be a random sample from a Poisson(λ) distribution. Let $\bar{X}$ be their sample mean and $S^2$ their sample variance.
a) Show that $\frac{\sqrt{n}[\bar{X}-\lambda]}{\sqrt{\bar{X}}}$ and $\frac{\sqrt{n}[\bar{X}-\lambda]}{S}$ both have a standard normal limiting distribution.
b) Find the limiting distribution of $\sqrt{n}[\bar{X}-\lambda]^2$
c) Find the limiting distribution of $\sqrt{n}[\bar{X}^2-\lambda^2]$
2. Relevant equations
3. The attempt at a solution
a) For $\frac{\sqrt{n}[\bar{X}-\lambda]}{S}$, we know that $\frac{\sqrt{n}[\bar{X}-\lambda]}{S}$ = $(\frac{\sqrt{n}[\bar{X}-\lambda]}{\sigma})(\frac{\sigma}{S})$. Since
$\frac{\sqrt{n}[\bar{X}-\lambda]}{\sigma}$ appraches N(0,1) in distribution by CLT, and since $(\frac{\sigma}{S})$ appraches 1 in probability, the whole thing approaches N(0,1).
For
$\frac{\sqrt{n}[\bar{X}-\lambda]}{\sqrt{\bar{X}}}$, we get the same result...since the mean is equal to the variance in a poisson distribution.
c) From a theorem in my textbook, I know that if $\sqrt{n}(X_n - \theta) \rightarrow N(0,\sigma^2)$ and if there is a differentiable function g(x) at theta where the derivative at theta is not zero...then $\sqrt{n}(g(X_n)-g(\theta)) \rightarrow N(0,\sigma^2(g'(\theta))^2)$.
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### Home > APCALC > Chapter 8 > Lesson 8.3.3 > Problem8-124
8-124.
You have designed a model of a square-based pyramid in honor of your calculus teacher. The height will be $10$ inches. The equation of one side of the pyramid is: $f(x) = -0.5x + 10$.
1. Sketch a diagram of your pyramid. A complete diagram includes the function, the $x$‑ and $y$‑axes, and a typical slice labeled with the appropriate dimensions.
Pyramids have pointy tips. If the height is $10$ inches, the base would start at $y = 0$.
2. Set up and evaluate an integral that calculates the exact volume of your pyramid.
Notice that the cross-sections are perpendicular to the $y$-axis, so set up and evaluate an integral in terms of $y$.
$V=\int_{y=a}^{y=b}\text{area }dy$
Use the eTool below to help solve the problem.
Click the link at right for the full version of the eTool: Calc 8-124 HW eTool
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# Figure 12
Top panels: The combined differential production cross section of charged-particle anti-$k_{\rm T}$ $R=0.4$ b jets measured in \pp (left) and \pPb (right) collisions at $\snn=5.02$\,TeV. The data are compared with a NLO pQCD prediction by the POWHEG dijet tune with PYTHIA~8 fragmentation~. Systematic and statistical uncertainties are shown as boxes and error bars, respectively. The additional common normalization uncertainty due to luminosity, $\sigma_{\mathcal{L}}^{\rm Sys}$, is quoted separately. Bottom panels: Ratio of the theory calculations to the data.
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Imandra comes with several different directives for loading files or including external modules.
It can be a bit daunting initially to undertstand which directive to use in the appropriate context, let's look at each in detail:
## use¶
Syntax: #use "filename.iml";; or [@@@use "filename.iml"]
The behavior of use is that of reading the content of "filename.iml" and evaluating it into the current environment, as if each definition had been directly entered into the toplevel.
The use directive is meant to used in interactive sessions rather than in library/production code, and is not recursive.
There exist a functional equivalent of the directive, called System.use that can be used to (recursively) load files at runtime, this is often useful to implement the common pattern of having a top.iml file responsible for loading the entire model one is working on, or more generally for programmatic loading of imandra files.
If filename.iml is a relative path, it is resolved to an absolute path by searching for it in Imandra's load path.
The current directory is always implicitly part of the load path, and additional directories can be added dynamically via the directory directive, or at startup using the -dir argument.
## mod_use¶
Syntax: #mod_use "filename.iml";; or [@@@mod_use "filename.iml"]
mod_use behaves exactly like use, with the only exception that the contents of the file will be evaluated in the toplevel as if they were wrapped in a module called Filename
The functional equivalent of the directive is System.mod_use
## use_gist¶
Syntax: #use_gist "my-user/gist-hash";; or [@@@use_gist "my-user/gist-hash"]
use_gist behaves exactly like use, except instead of loading a local path it will download the gist content and load that in the current environment.
## import¶
Syntax: #import "filename.iml";; or [@@@import "filename.iml"]
The behavior of import is similar to that of mod_use, with a number of significant differences:
• toplevel directives are disabled in the file being imported
• import is idempotent, as it caches on first import
• Imandra resolves the import at compile time, so the imported module is available when typechecking the remainder of the file
As such, import can be considered a version of mod_use to be used in IML libraries as opposed to during interactive development.
Like use and mod_use, import uses Imandra's load path to search for files, when importing an IML library, the current directory will be temporary set to that of the library itself.
## require¶
Syntax: #require "some-lib";; or [@@@require "some-lib"] or as an argument to imandra: -require somelib
The behavior of require is to search for some-lib in the list of installed libraries (using topfind) and load it in the current environment.
If some-lib comes packaged with a some-lib.iml file, it will be automatically loaded by Imandra using #import after the library has been required.
## require_use¶
Syntax: #require_use "some-lib";; or [@@@require_use "some-lib"] or as an argument to imandra: -require-use somelib
Similar to #require "some-lib";; except it loads the file some-lib.iml with #use instead of #import, so no module is created, and it's not idempotent.
## require_mod_use¶
Syntax: #require_mod_use "some-lib";; or [@@@require_mod_use "some-lib"] or as an argument to imandra: -require-mod-use somelib
Similar to #require "some-lib";; except it loads the file some-lib.iml with #mod_use instead of #import, so it's not idempotent.
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The pdf $f(x)$ of an exponential distribution with parameter $\lambda$ is
$$f(x) = \lambda \exp(-\lambda x)\quad x\geq 0$$
The cdf $F(x)$ of such a distribution is
$$F(x) = \int^x_0 f(x') dx'$$
1. Must the pdfs of two different exponential distributions intersect?
2. In what cases do $F(x)$ and $f(x)$ intersect?
3. Can the cdfs for two exponential distributions intersect?
4. Can the pdfs for three different exponential distibutions intersect at the same point?
You might want to experiment with specific choices of $\lambda$ before constructing more general arguments.
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# The Unimodality of the Crank on Overpartitions
Research paper by Wenston J. T. Zang, Helen W. J. Zhang
Indexed on: 25 Nov '18Published on: 25 Nov '18Published in: arXiv - Mathematics - Combinatorics
#### Abstract
Let $N(m,n)$ denote the number of partitions of $n$ with rank $m$, and let $M(m,n)$ denote the number of partitions of $n$ with crank $m$. Chan and Mao proved that for any nonnegative integers $m$ and $n$, $N(m,n)\geq N(m+2,n)$ and for any nonnegative integers $m$ and $n$ such that $n\geq12$, $n\neq m+2$, $N(m,n)\geq N(m,n-1)$. Recently, Ji and Zang showed that for $n\geq 44$ and $1\leq m\leq n-1$, $M(m-1,n)\geq M(m,n)$ and for $n\geq 14$ and $0\leq m\leq n-2$, $M(m,n)\geq M(m,n-1)$. In this paper, we analogue the result of Ji and Zang to overpartitions. Note that Bringmann, Lovejoy and Osburn introduced two type of cranks on overpartitions, namely the first residue crank and the second residue crank. Consequently, for the first residue crank $\overline{M}(m,n)$, we show that $\overline{M}(m-1,n)\geq \overline{M}(m,n)$ for $m\geq 1$ and $n\geq 3$ and $\overline{M}(m,n)\geq \overline{M}(m,n+1)$ for $m\geq 0$ and $n\geq 1$. For the second residue crank $\overline{M2}(m,n)$, we show that $\overline{M2}(m-1,n)\geq \overline{M2}(m,n)$ for $m\geq 1$ and $n\geq 0$ and $\overline{M2}(m,n)\geq \overline{M2}(m,n+1)$ for $m\geq 0$ and $n\geq 1$. Moreover, let $M_k(m,n)$ denote the number of $k$-colored partitions of $n$ with $k$-crank $m$, which was defined by Fu and Tang. They conjectured that when $k\geq 2$, $M_k(m-1,n)\geq M_k(m,n)$ except for $k=2$ and $n=1$. With the aid of the inequality $\overline{M}(m-1,n)\geq \overline{M}(m,n)$ for $m\geq 1$ and $n\geq 3$, we confirm this conjecture.
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## Wednesday, July 30, 2008 ... //
### LHC: FAQ
Update: Physics World spreads a rumor about the first proton beams to be injected on August 9th into parts of the tunnel and on September 2nd-3rd to the whole ring.
Original text from July 28th: Martin Coles at The (Montreal) Gazette wrote a couple of excellent articles about the Large Hadron Collider. Click at the following:
Frequently asked questions (illuminating, a lot of understandable numbers!)
Deep thinking (from Galileo to the LHC, a story)
In their realm, outside Canada (Montrealers at CERN)
For a photographer and a journalist, these sensible articles are extraordinary and deserve an explanation. The main explanation is probably that Martin Coles has studied math and philosophy in Cambridge, Old England, so he is not quite just a photographer or a journalist. ;-)
Incidentally, all eight sectors of the LHC except for "78" are already below the temperature of 5 K, usually close to the final 1.9 K. Only the "78" sector is at 15 K in average (update, July 30th: close to 5 K, too). It should be cooled down soon.
I don't really like rap but if you do, play the video above. Ms Alpinekat explains the basic purpose of the four LHC experiments very well.
Hat tip: Tommaso Dorigo.
#### snail feedback (0) :
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# Mean Absolute Scaled Error [duplicate]
Right now, I am analyzing the prediction quality of a dynamic model that has variables with different units (e.g. $x_{1,t}$ is in meters, $x_{2,t}$ is in kilograms etc.). I have discovered a great tool called Mean Absolute Scaled Error: $$\frac{1}{n}\sum_{t=1}^n\left( \frac{\left| \hat x_{t,i} - x_{t,i} \right|}{\frac{1}{n-1}\sum_{\tau=2}^n \left| x_{\tau,i}-x_{\tau-1,i}\right|} \right)$$ where $i$ is index of variable.
The MASE is denoted as scale-free. I would be curious how to say that the error is large or small. My experience is that $MASE=3$ is not bad, but I would appreciate some more rigorous answer or reference.
• Is this time series data? – forecaster Jan 5 '15 at 22:09
• As far as I understand, the magnitude of MASE will differ from case to case not only due to differences in the quality of forecasting methods but also due to the special features of the problem. MASE=3 does not sound impressive, though... Something closer to 1 would be more convincing. (I assume you understand the interpretation of MASE=1.) You may check this question out for another perspective. – Richard Hardy Jan 6 '15 at 9:41
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# How to embed a song in a pdf?
How can I create a document that has a song playing in the back with (La)TeX? Is it even possible?
My hunch is that if you can embed video, there should be a way to embed music... but what is the best way to do it? I guess having a black box playing a song is possible, but something tells me there should be a beTeX way to do this.
The media9 package has support for embedding audio only, as well as video.
An example from the package documentation:
\documentclass{article}
\usepackage{media9}
\begin{document}
\includemedia[
flashvars={
source=http://www.openbsd.org/songs/song49.mp3
&autoPlay=true
},
transparent
]{\color{blue}\fbox{Listen to OpenBSD 4.9 release song}}{APlayer.swf}
\end{document}
The UI for this is a simple button to download the song and play it.
Note: The PDF output will only work in Adobe Reader 9, but no version greater than 9.4.1 on Linux, as per the media9 documentation.
• Nice. Do you know if evince supports it? Mine doesn't. – Sigur Apr 23 '14 at 1:08
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# The Institutional Life of Algorithmic Risk Assessment
Alicia Solow-Niederman, YooJung Choi, and Guy Van den Broeck.
In Berkeley Technology Law Journal (BTLJ), 2019
### Abstract
As states nationwide turn to risk assessment algorithms as tools for criminal justice reform, scholars and civil society actors alike are increasingly warning that this technological turn comes with complications. Research to date tends to focus on fairness, accountability, and transparency within algorithmic tools. Although attention to whether these instruments are fair or biased is normatively essential, this Article contends that this inquiry cannot be the whole conversation. Looking at issues such as fairness or bias in a tool in isolation elides vital bigger-picture considerations about the institutions and political systems within which tools are developed and deployed. Using California’s Money Bail Reform Act of 2017 (SB 10) as an example, this Article analyzes how risk assessment statutes create frameworks within which policymakers and technical actors are constrained and empowered when it comes to the design and implementation of a particular instrument. Specifically, it focuses on the tension between, on one hand, a top-down, global understanding of fairness, accuracy, and lack of bias, and, on the other, a tool that is well-tailored to local considerations. It explores three potential technical and associated policy consequences of SB 10’s framework: proxies, Simpson’s paradox, and thresholding. And it calls for greater attention to the design of risk assessment statutes and their allocation of global and local authority.
### Citation
@article{Solow-NiedermanBTLJ19,
author = {Solow-Niederman, Alicia and Choi, YooJung and Van den Broeck, Guy},
title = {The Institutional Life of Algorithmic Risk Assessment},
journal = {Berkeley Technology Law Journal},
month = {aug},
year = {2019},
}
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## Persistent nonhyperbolic transitive diffeomorphisms.(English)Zbl 0852.58066
The paper constructs a large class of $$C^1$$-persistent, nonhyperbolic, transitive diffeomorphisms of a closed manifold of dimension $$m \geq 3$$ that carries a $$C^\infty$$ transitive Anosov vector field. Some of them are isotopic to the identity. Others are partially hyperbolic diffeomorphisms with a central space of dimension greater than one. An essential element of the proofs is the use of so-called center-stable blenders, which are kinds of higher-dimensional horseshoes.
### MSC:
37D99 Dynamical systems with hyperbolic behavior
Full Text:
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Minisymposium: Topologie appliquée et numérique
Mercredi, 19 juin
Salle Du Jardin
15h00
Application of topology to granular materials
Kramar, Miroslav
Rutgers University
The state of granular media can be represented by a persistence diagram. This representation provides an interesting insight into the physical properties of the granular media as demonstrated on a system undergoing compression. Time evolution of the system can be seen as a curve in the space of persistence diagrams. Different notions of distance in this space provide a useful tool for understanding the dynamic. In particular the compressed systems (viewed as a discrete dynamical system) exhibit a few different regimes where dynamics changes from fast to slow. Dependence of the system on its previous state is strongly affected by the sampling rate. We conclude the talk by addressing the problem of determining the 'appropriate' sampling rate.
15h30
Multiparameter persistent homology for shape comparison: from continuous to discrete
Kaczynski, Tomasz
Université de Sherbrooke
The theory of multiparameter persistent homology was initially developed in the discrete setting of filtered simplicial complexes. Stability of persistence was proved for topological spaces filtered by continuous vector-valued functions, that is, for continuous data. This talk aims to provide a bridge between the continuous setting, where stability properties hold, and the discrete setting, where actual computations are carried out. The existence of this bridge is not obvious due to the phenomenon of structural gap between the two settings, called topological aliasing, which appears in the multiparameter case when using the standard piecewise linear interpolation of the discrete model. We solve the problem by introducing an adapted axis-wise linear interpolation and develop a stability preserving method for comparing rank invariants of vector functions obtained from discrete data. These advances support the choice of multiparameter persistent homology as a tool for shape comparison in computer vision. This is a joint work with M. Ethier, P. Frosini, and C. Landi.
16h00
The Topology of Noise
Bobrowski, Omer
Duke University
We study the topology of random Cech complexes, generated by $n$ iid points in a Euclidean space, and a radius $r$. In particular, we are interested in the limiting behavior of the Betti numbers of such complexes, as $n\to\infty$ and $r\to 0$. We consider different cases in which samples are generated by either pure noise, a probability distribution on a close manifold, or a combination of both. The study of the Betti numbers can be done directly, or via a Morse-theoretic approach, by counting critical points of distance functions. The results show that the Betti numbers of a random complex exhibit a different limiting behavior, depending on both the support of the underlying distribution and the rate in which $r \to 0$. In this talk, we will present the known results to date for each of these limiting phenomena. In addition, we will discuss how these results could be applied to the problem of recovering the topology of a hidden manifold from a finite set of samples.
16h30
Letting Loops Loose
Mileyko, Yuriy
University of Illinois at Urbana-Champaign
Problems in applied topology often require computation of an optimal (in some sense) representative among shapes satisfying particular topological constraints. In many cases this is a very challenging, if not infeasible task. The aim of this talk is to show that in some cases the needed optimality is a typical property, meaning that a representative chosen uniformly at random will be close to optimal with an overwhelming probability. In particular, we consider the space of closed polygonal chains which represent a fixed free homotopy class in a (multi-)punctured plane and have a large number of small edges. By employing tools from large deviation theory we show that the uniform measure on this space is sharply concentrated around representatives whose length is arbitrarily close to the length infimum. As a consequence, sampling from the stationary distribution of a Markov process (using, say, the Metropolis-Hastings algorithm) gives us a simple way to approximate the minimal length representative in a free homotopy class. This result has an important application in multi-agent systems, as it allows the agents to move towards the optimal configuration without any control.
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# Controlling overflow and loss of precision during floating point multiplication
I have a large number of floating point numbers (~10,000 numbers) , each having 6 digits after decimal. Now, the multiplication of all these numbers would yield about 60,000 digits. But the double range is for 15 digits only. How can I get my product with minimum loss in precision?
I thought of storing the numbers as 6 digit long long integers and storing their exponents elsewhere. But this appears cumbersome and may not yield correct result. Is there an alternate easier way to do this?
• As it is your question is not precise enough. Are your numbers exact values from the point of view of your problem, or are they only 6 digit approximations to begin with (e.g. results from physical measurements). Most likely, it is the latter, but you should make it clear in the question, either by saying so or giving the context of the problem. In that case, there are precise rules that define the meaningful number of digits in the result (which is not likely to be 60.000), and to give a proper estimate on the computational uncertainty on that result. – babou Jul 12 '15 at 9:48
Multiplication of floating point numbers is considered uncritical with respect to accuracy. If your input is only accurate to 6 digits, there is no point in computing the output to 60,000 digits. The expected relative error after 10,000 multiplications is $\sqrt{10,000}\epsilon=100\epsilon$ with $\epsilon<10^{-14}$ for double precision. This is more than enough precision for your case.
Overflow and underflow on the other hand can indeed happen during multiplication. In practice, I would prefer to work with numbers in a range where I know that this won't happen. However, this may not always be possible. One can use frexp to split floating-point numbers into significant and exponent, multiply the significants separately and add the exponents together, and use ldexp for putting the result back together. However, doing this naively by applying frexp for every number and multiplying all 10,000 significants together at once would be a bad idea, because it would be slow and nearly guarantee underflow. But if done intelligently (appropriate grouping), it will yield the correct result, and the speed difference will be almost unnoticeable.
You can use a multi-precision software floating point package such as MPFR. MPFR will use however many bits for the mantissa as you ask it to. That said, I'm not sure why you need 60,000 digits of precision.
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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
A simple plotting feature we need to be able to do with R is make a 2 y-axis plot. First let’s grab some data using the built-in beaver1 and beaver2 datasets within R. Go ahead and take a look at the data by typing it into R as I have below.
# Get the beaver datasets
beaver1
beaver2
We’re going to plot the temperatures within both of these datasets, which we can see (after punching into R) is the third column.
First let’s check the length of these datasets and make sure they’re the same.
# Get the length of column 3
length(beaver1[,3])
length(beaver2[,3])
[1] 114
[2] 100
Since beaver1 is longer, we’ll only plot rows 1 through 100 of the temperature data, so that it is the same length as beaver2.
# Plot the data
plot(beaver1[1:100, 3], type ="l", ylab = "beaver1 temperature")
Cool, your plot should look like this.
Now, let’s add that second dataset on the right y-axis. So, we have to have to create a plot on top of this plot using the command par(new = TRUE).
# Add the second y-axis
plot(beaver1[1:100, 3], type ="l", ylab = "beaver1 temperature")
par(new = TRUE)
plot(beaver2[,3], type = "l")
Woah, this plot is ugly! We have 2 y-axis labels plotting, 2 y-axis values plotting, and 2 x-axis values and labels plotting. Let’s turn those off using the commands xaxt = “n” and yaxt = “n”.
# updated plot
plot(beaver1[1:100, 3], type ="l", ylab = "beaver1 temperature")
par(new = TRUE)
plot(beaver2[,3], type = "l", xaxt = "n", yaxt = "n",
ylab = "", xlab = "")
Okay, it’s still pretty ugly, so let’s clean it up. Let’s make the margins bigger on the right side of the plot, add a y2 axis label, add a title, change the color of the lines and adjust the x-axis label. Don’t forget the legend! Here’s the code:
# final plot
par(mar = c(5, 5, 3, 5))
plot(beaver1[1:100, 3], type ="l", ylab = "beaver1 temperature",
main = "Beaver Temperature Plot", xlab = "Time",
col = "blue")
par(new = TRUE)
plot(beaver2[,3], type = "l", xaxt = "n", yaxt = "n",
ylab = "", xlab = "", col = "red", lty = 2)
axis(side = 4)
mtext("beaver2 temperature", side = 4, line = 3)
legend("topleft", c("beaver1", "beaver2"),
col = c("blue", "red"), lty = c(1, 2))
Woo! Looks good. That’s all for now.
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• entries
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38978
# Programming Cause and Effect.
913 views
So, it is sunny outside, and that makes you happy, so you go outside to play. But then, it starts raining which makes it rainy, and that causes you to be sad. So, you decide to go to sleep.
I sat down today and thought of a way to program cause and effect. I came up with this notation:(initCondtion, initState, initBehavior) + Action = (deltaState, deltaCondition,deltaBehavior) = Reaction
The initial condition is that it is sunny outside.
The initial state is that you are happy.
This causes an initial behavior-- you go outside to play.
But then, something happens (an action).
It starts to rain. Now it is rainy outside.
Now you are sad.
This causes you to go back inside and go to sleep.
The action triggers a change in state and a change in condition.
This change in state and condition produces a reaction-- You go inside to go to sleep.
PseudoExample:--Statesnormal = truehappy = falsesad = false --Conditionsclear = truerainy= falsesunny = false --Behaviors --Initial Behaviorfunction GoOutToPlay() --Change in Behaviorfunction GoToSleep() --Actionfunction Rain()start to rainend --Reactionfunction ChangeState()sad = true function ChangeCondition()rainy = true function ChangeBehavior()if sad && rainyGoToSleep()
I wonder if this is solid enough to implement. hmm.
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# Linear Algebra Calculators
## Cholesky Factorization
This calculator uses Wedderburn rank reduction to find the Cholesky factorization of a symmetric positive definite matrix $A$. The process constructs the matrix $L$ in stages.
At each stage you'll have an equation $A=LL^T+B$ where you start with $L$ nonexistent and with $B=A$. The next column of $L$ is chosen from $B$. (The $L$ column is scaled.) Then $L$ and $B=A-LL^T$ are updated. Eventually $B=0$ and $A=LL^T$.
At this point $L$ is lower triangular. If the original matrix $A$ wasn't positive definite, the algorithm will fail at some point when confronted with the square root of a negative quantity.
Either choose a size and press this button to get a randomly generated matrix, or enter your matrix in the box below. (Look at the example to see the format.)
Matrix $A$:
Update $L$ and $B$.
The reset button leaves the $A$ matrix alone, but reinitializes $L$ and $B$.
Back to calculator page or home page.
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# The vectors @, @,€ € R2 depicted below are unit vectors The vectors 6,d,f € R2 depicted below have length
###### Question:
The vectors @, @,€ € R2 depicted below are unit vectors The vectors 6,d,f € R2 depicted below have length 2: Adjacent vectors form an angle of "/8 with each other (e.g-, the angle between b and € is w/8). b
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2015 CAP Congress / Congrès de l'ACP 2015
Jun 13 – 19, 2015
University of Alberta
America/Edmonton timezone
Welcome to the 2015 CAP Congress! / Bienvenue au congrès de l'ACP 2015!
QCD Sum Rule Analysis of Heavy-light Hybrids for $J^{P}=1^{-}$
Jun 18, 2015, 9:15 AM
15m
CCIS L1-140 (University of Alberta)
CCIS L1-140
University of Alberta
Oral (Student, In Competition) / Orale (Étudiant(e), inscrit à la compétition) Theoretical Physics / Physique théorique (DTP-DPT)
Speaker
Mr Jason Ho (University of Saskatchewan)
Description
Quantum chromodynamics (QCD) predicts many bound states that have not yet been conclusively identified, and as more charmonium-like XYZ states are being discovered, interest is increasing in matching these theoretical bound states with experimental observation. Among these states are hybrid mesons: bound states of a quark, an antiquark, and a gluon. With upcoming experiments such as GlueX, and PANDA, experimental data within the expected mass ranges of hybrids will be abundant in the next decade, and theoretical predictions are needed to help identify them. We present preliminary aspects of a QCD sum rule analysis of a heavy-light (open-flavour) $J^{P}=1^{-}$ hybrid system, including non-perturbative condensate contributions up to six dimensions.
Primary author
Mr Jason Ho (University of Saskatchewan)
Co-authors
Derek Harnett (University of the Fraser Valley) Tom Steele (University of Saskatchewan)
Presentation materials
There are no materials yet.
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Home/Class 12/Physics/
A wheel with $$10$$ metallic spokes each $$0.5\ \text{m}$$ long is rotated with a speed of $$120\; \text{rev/min}$$ in a plane normal to the horizontal component of earth’s magnetic field $$H_E$$ at a place. If $$H_E=0.4\;G$$ at the place, what is the induced emf between the axle and the rim of the wheel? Note that $$1\;G=10^{-4}\;T$$ .
Speed
00:00
09:38
## QuestionPhysicsClass 12
A wheel with $$10$$ metallic spokes each $$0.5\ \text{m}$$ long is rotated with a speed of $$120\; \text{rev/min}$$ in a plane normal to the horizontal component of earth’s magnetic field $$H_E$$ at a place. If $$H_E=0.4\;G$$ at the place, what is the induced emf between the axle and the rim of the wheel? Note that $$1\;G=10^{-4}\;T$$ .
Given that,
Angular speed of Wheel $$\left(\omega \right)=120\;\frac{\mathrm{rev}}{\mathrm{min}}=120\times \frac{\pi } 3=4\pi \;\mathrm{rad}/\mathrm{sec}$$
Radius of the wheel $$\left(R\right)=0.5\;\text{m}$$
Horizontal magnetic field $$(H)_E\mid B)=0.4\;G$$
Let us consider a small elemental length $$^\prime \mathit{dl}^\prime$$ of the metallic spoke at a distance $$l$$ from the axle $$($$ centre of the wheel to considered as the origin $$)$$ moving at a linear speed $$v$$ in a magnetic field $$B$$ perpendicular to the length and speed.
The emf induced across the elemental length $$\mathit{dl}$$
$${\therefore}$$ $$\mathit{d\varepsilon }=\mathit{vBdl}$$
Since the wheel is rotating at an angular speed $$\omega$$ , we have
$${\therefore}v=\mathit{\omega l}$$
$$\Rightarrow \mathit{d\varepsilon }=\mathit{\omega lBdl}$$
The total emf induced across the metallic spoke is given by:
$${\therefore}\varepsilon =\int _0^R\mathit{d\varepsilon }=\int _0^R\mathit{\omega lBdl}$$
$$=\mathit{\omega B}\int _0^R\mathit{ldl}$$
$$=\frac 1 2\mathit{\omega B}R^2$$
$$=\frac 1 2\times 4\pi \times 0.4\times 10^{-4}\times (0.5)^2$$
$$\Rightarrow \varepsilon =6.28\times 10^{-5}V$$
Since all the metallic spokes are parallel as one end is connected to common axle and other end is connected to common rim. So emf across all the metallic spokes will be the same.
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$\partial \bar{\partial}$ on a complex manifold - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T02:08:08Z http://mathoverflow.net/feeds/question/87968 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87968/partial-bar-partial-on-a-complex-manifold $\partial \bar{\partial}$ on a complex manifold william 2012-02-09T06:42:38Z 2012-02-11T06:13:58Z <p>hallo,</p> <p>i have the following question: let $M$ be a complex $n-$dimensional manifold and $R \subset M$ be a totally real, compact, $n-$dimensional (real) manifold. let $\alpha$ be a smooth nonnegative $(n,n)-$form on $M$. does there exists a smooth plurisubharmonic function $f : U \rightarrow \mathbb{R}$, where $U$ is a open neigbourhood of $R$ in $M$ such that the equation $(\partial \bar{\partial} f)^{n} = \alpha$ is satiesfied. So i am interested in smooth solution in a neigbourhood of the real manifold of the inhomogenous monge-ampere equation. does there exists such a solution ? if yes, can you give me some reference and if no what can be done to ave such a solution ? hope for answers. thanks in advance.</p> <p>marco </p>
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# Stone-Weierstrass theorem proof in Rudin
I'm reviewing Rudin's Principles of analysis now.
He first proves Weierstrass approximation and uses it to generalize to Stone-Weierstrass theorem.
The problem is, this version is very restrictive.
What he proves here is that:
Let $X$ be a compact metric space.
Let $\mathscr{A}$ be a self-adjoint algebra of $C(X,\mathbb{C})$.
Let $C(X,\mathbb{C})$ be equipped with the sup norm.
If $\mathscr{A}$ separates points in $X$ and vanishes at no point of $X$, then $\mathscr{A}$ is dense in $C(X,\mathbb{C})$.
However, the general version in wikipedia states that this is true when $X$ is just compact Hausdorff.
I tried to remove metric hypotheses in Rudin's argument, but since his argument is based on second-countability of a compact metric space, I couldn't.
In general, compact Hausdorff is NOT second countable. So I guess that Rudin's argument cannot generalize it further.
Am I thinking right?
If I'm thinkihg right, then i have the following question.
Textsbooks in my hand right now are Munkres-Topology,Rudin-PMA,RCA, but there is no chapter for this general version of Stone-Weierstrass Theorem in those texts.
So, is there any text introducing this theorem in full strength? And is the proof quite elementary so that an undergraduate (just like me) can understand the proof?
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Canadian Mathematical Society www.cms.math.ca
location: Publications → journals → CMB
Abstract view
# A Classification of Three-dimensional Real Hypersurfaces in Non-flat Complex Space Forms in Terms of Their generalized Tanaka-Webster Lie Derivative
Published:2016-08-19
Printed: Dec 2016
• George Kaimakamis,
Faculty of Mathematics and Engineering Sciences, Hellenic Military Academy, Vari, Attiki, Greece
• Konstantina Panagiotidou,
Faculty of Mathematics and Engineering Sciences, Hellenic Military Academy, Vari, Attiki, Greece
• Juan de Dios Pérez,
Departmento de Geometria y Topologia, Universidad de Granada, 18071, Granada Spain
Format: LaTeX MathJax PDF
## Abstract
On a real hypersurface $M$ in a non-flat complex space form there exist the Levi-Civita and the k-th generalized Tanaka-Webster connections. The aim of the present paper is to study three dimensional real hypersurfaces in non-flat complex space forms, whose Lie derivative of the structure Jacobi operator with respect to the Levi-Civita connections coincides with the Lie derivative of it with respect to the k-th generalized Tanaka-Webster connection. The Lie derivatives are considered in direction of the structure vector field and in directions of any vecro field orthogonal to the structure vector field.
Keywords: $k$-th generalized Tanaka-Webster connection, non-flat complex space form, real hypersurface, Lie derivative, structure Jacobi operator
MSC Classifications: 53C15 - General geometric structures on manifolds (almost complex, almost product structures, etc.) 53B25 - Local submanifolds [See also 53C40]
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© Canadian Mathematical Society, 2018 : https://cms.math.ca/
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